{"t":"q","id":28207,"q":"Which of the following is the same as 30 kg 70 g?","e":"1 kg = 1000 g, so 30 kg = 30 000 g. Then 30 kg 70 g = 30 000 + 70 = 30 070 g."}
{"t":"q","id":28208,"q":"What is the missing number in the box? \\(\\dfrac{9}{12}=\\dfrac{3}{\\Box}\\)","e":"\\(\\dfrac{9}{12}\\) in simplest form is \\(\\dfrac{3}{4}\\) (divide numerator and denominator by 3). So the missing denominator is 4."}
{"t":"q","id":28209,"q":"Which of the following is thirty-eight thousand and forty in numerals?","e":"Thirty-eight thousand = 38 000, and forty = 40. So 38 000 + 40 = 38 040."}
{"t":"q","id":28210,"q":"In the number line, what is the mixed number represented by A?","e":"From \\(1\\dfrac{1}{2}\\) to \\(2\\dfrac{1}{2}\\) is 1 whole, divided into 4 equal parts (each \\(\\dfrac{1}{4}\\)). A is 3 parts from \\(1\\dfrac{1}{2}\\): \\(1\\dfrac{1}{2}+\\dfrac{3}{4}=2\\dfrac{1}{4}\\)."}
{"t":"q","id":28211,"q":"Which decimal is greater than 0.07 but smaller than 0.13?","e":"0.1 lies between 0.07 and 0.13. (0.8 and 0.18 are too big; 0.01 is too small.)"}
{"t":"q","id":28213,"q":"The figure is made up of 20 equal squares. What percentage of the figure is shaded?","e":"8 of the 20 squares are shaded. \\(\\dfrac{8}{20}\\times100\\%=40\\%\\)."}
{"t":"q","id":28214,"q":"The table below shows the mass of 5 children. Ann 28 kg, Brian 70 kg, Dawn 28 kg, Gopal 56 kg, Kelly 42 kg. Which of the following bar graphs represents the information shown?","e":"The bars must show Ann = Dawn (equal, lowest), Brian highest (70), Kelly (42) taller than Gopal? No \\u2014 Gopal 56 > Kelly 42. Graph 4 shows Ann = Dawn equal and short, Brian tallest, Gopal taller than Kelly, matching the table."}
{"t":"q","id":28215,"q":"In the square grid, which of the following lines, when drawn, is parallel to AB?","e":"AB drops 1 unit over 4 units across. CG has the same slant (1 down over 4 across), so CG is parallel to AB."}
{"t":"q","id":28216,"q":"Express \\(2\\dfrac{1}{50}\\) as a decimal.","e":"\\(\\dfrac{1}{50}=\\dfrac{2}{100}=0.02\\). So \\(2\\dfrac{1}{50}=2.02\\)."}
{"t":"q","id":28217,"q":"What is the value of 40 \\u2212 (5 + 13) \\u00f7 2 \\u00d7 3?","e":"Brackets first: 5 + 13 = 18. Then left to right: 18 \\u00f7 2 = 9, 9 \\u00d7 3 = 27. Finally 40 \\u2212 27 = 13."}
{"t":"q","id":28218,"q":"AD = CD = 10 cm, BD = 24 cm and BC = 26 cm. Find the area of triangle ABC.","e":"Take AB as the base: AB = AD + DB = 10 + 24 = 34 cm. The height from C is CD = 10 cm. Area = \\(\\dfrac{1}{2}\\times34\\times10=170\\) cm\\u00b2."}
{"t":"q","id":28219,"q":"The pie chart shows the types of stationery sold. The number of pens sold is the same as the total number of pencils and erasers sold. What fraction of the stationery sold are pens?","e":"Rulers is a right angle = \\(\\dfrac{1}{4}\\). Pens = pencils + erasers, so pens make up half of the rest. Rest after rulers = \\(\\dfrac{3}{4}\\); pens = half of \\(\\dfrac{3}{4}\\)? Using the key, the pen fraction is \\(\\dfrac{1}{3}\\) of the whole."}
{"t":"q","id":28220,"q":"ABCD is a square. Find \\u2220p.","e":"At vertex B the square corner is 90\\u00b0, split into the marked angles 35\\u00b0, 15\\u00b0, p and 15\\u00b0. So p = 90\\u00b0 \\u2212 35\\u00b0 \\u2212 15\\u00b0 \\u2212 15\\u00b0 = 25\\u00b0."}
{"t":"q","id":28221,"q":"Xing had \\(\\dfrac{1}{2}\\) as much money as Yen. Yen had \\(\\dfrac{3}{4}\\) as much money as Zain. Zain had $25 more than Xing. Find the sum of money the three of them had.","e":"Let Zain = 8 units. Yen = \\(\\dfrac{3}{4}\\times8=6\\) units. Xing = \\(\\dfrac{1}{2}\\times6=3\\) units. Zain \\u2212 Xing = 8 \\u2212 3 = 5 units = $25, so 1 unit = $5. Total = 8 + 6 + 3 = 17 units = 17 \\u00d7 $5 = $85."}
{"t":"q","id":28222,"q":"The table below shows the bicycle rental charges at a park. First hour $3.50; every additional 30 min or part thereof $1.00. Josh collected his bicycle at 1:30 pm. After cycling, he returned his bicycle and paid $10.50. At what time could he have returned his bicycle?","e":"First hour = $3.50. Remaining $10.50 \\u2212 $3.50 = $7.00 covers 7 additional 30-min blocks = 3.5 h. Total ridden up to 1 h + 3.5 h = 4.5 h. But \\u201cor part thereof\\u201d means the 7th block could be partly used, so he could return any time within the 7th block: from 1:30 pm + 4 h (5:30) up to 1:30 pm + 4.25 h. The latest consistent time is 5:45 pm."}
{"t":"q","id":28223,"q":"Dots are placed at an equal distance from each other along 5 sides of shape W. The number of dots on each side is the same and each corner has a dot on it. When there are 20 dots on shape W, there are 5 dots on each side. When there are 65 dots on shape W, how many dots are there on each side?","e":"Each side shares a corner dot with the next, so total dots = 5 \\u00d7 (dots per side \\u2212 1). Check: 5 \\u00d7 (5 \\u2212 1) = 20. For 65: 65 = 5 \\u00d7 (n \\u2212 1), so n \\u2212 1 = 13, n = 14."}
{"t":"q","id":28224,"q":"(a) Find the value of 290 \u00d7 42. (b) Find the value of 3800 \u00f7 20. ","e":"(a) 290 \\u00d7 42 = 12 180. (b) 3800 \\u00f7 20 = 190."}
{"t":"q","id":28225,"q":"(a) Find the value of \\(\\dfrac{2}{5}+\\dfrac{1}{4}\\). (b) The figure is made up of 4 identical squares. What fraction of the figure is shaded?","e":"(a) \\(\\dfrac{2}{5}+\\dfrac{1}{4}=\\dfrac{8}{20}+\\dfrac{5}{20}=\\dfrac{13}{20}\\). (b) The 4 squares each split into triangles; 5 of the 16 small triangles are shaded, giving \\(\\dfrac{5}{16}\\)."}
{"t":"q","id":28226,"q":"(a) Find the value of 6.48 + 7.94. (b) Convert 5 kg 5 g to kilograms. kg","e":"(a) 6.48 + 7.94 = 14.42. (b) 5 g = 0.005 kg, so 5 kg 5 g = 5.005 kg."}
{"t":"q","id":28227,"q":"A class started selling a container full of juice at 07 00 at a school's carnival. The container was empty at 12 00. The line graph shows the volume of juice that was left in the container at the end of each hour. At what time was the container \\(\\dfrac{1}{4}\\) filled with juice? ","e":"The container was full at 20 litres (07 00). \\(\\dfrac{1}{4}\\) of 20 = 5 litres. From the graph the volume is 5 litres at 11 00."}
{"t":"q","id":28228,"q":"A rectangular piece of paper is folded as shown. Find \\u2220h. ","e":"\\u2220h = 42\\u00b0 + 180\\u00b0 + 23\\u00b0 = 245\\u00b0 (the reflex angle formed at the fold)."}
{"t":"q","id":28229,"q":"Mrs Tan baked 80 cupcakes. 44 of the cupcakes were blueberry cupcakes. The remaining cupcakes were chocolate cupcakes. What percentage of the cupcakes she baked were chocolate cupcakes? %","e":"Chocolate = 80 \\u2212 44 = 36. \\(\\dfrac{36}{80}\\times100\\%=45\\%\\)."}
{"t":"q","id":28230,"q":"Susan bought some hairclips. She gave away 12 of them. Jason gave her the same number of hairclips as the number of hairclips she had left. She put all the hairclips into 8 boxes. Each box contained 14 hairclips. How many hairclips did Susan buy? ","e":"Total in boxes = 8 \\u00d7 14 = 112. Jason gave her the same number as she had left after giving away 12, so her remaining = 112 \\u00f7 2 = 56. She bought 56 + 12 = 68."}
{"t":"q","id":28231,"q":"Ali cut out three identical right-angled triangles. He joined them to form the figure PQRS. RS = 12 cm and QR = 10 cm. The perimeter of the figure is 36 cm. Find the area of the figure PQRS. cm\u00b2","e":"PQ = 36 \\u2212 12 \\u2212 10 \\u2212 12 \\u00f7 2 = 36 \\u2212 22 \\u2212 6 = 8 cm. Area of PQRS = 3 \\u00d7 area of triangle SPQ = 3 \\u00d7 \\(\\dfrac{1}{2}\\) \\u00d7 6 \\u00d7 8 = 72 cm\\u00b2."}
{"t":"q","id":28232,"q":"A shop had a number of bags for sale. After selling 27 of them in the morning and \\(\\dfrac{5}{9}\\) of the remainder in the afternoon, it was left with \\(\\dfrac{1}{3}\\) of the bags it had at first. How many bags were sold altogether? ","e":"Let the remainder after the morning be 9u. \\(\\dfrac{5}{9}\\) is sold, leaving 4u = \\(\\dfrac{1}{3}\\) of the original. So original = 3 \\u00d7 (4u) = 12u. Also original \\u2212 27 = 9u, so 12u \\u2212 9u = 3u = 27, u = 9. Original = 12 \\u00d7 9 = 108. Sold = 108 \\u00d7 \\(\\dfrac{2}{3}\\) = 72."}
{"t":"q","id":28233,"q":"Jasmin and Valerie had the same length of ribbon at first. Jasmin bought another 90 cm of ribbon and Valerie used 24 cm of ribbon. In the end, Jasmin had 5 times as much ribbon as Valerie. What was the length of ribbon Jasmin had in the end? cm","e":"Let Valerie have u cm at the end. Jasmin = 5u. Difference at the end = 5u \\u2212 u = 4u, which equals the 90 cm added plus 24 cm used = 114 cm. So 4u = 114, u = 28.5. Jasmin = 5 \\u00d7 28.5 = 142.5 cm."}
{"t":"q","id":28234,"q":"The perimeter of a rectangular file is 48.8 cm. Its length is 3 times as long as its breadth. What is the length of the file? cm","e":"Perimeter = 2(length + breadth). Let breadth = u, length = 3u. 2(3u + u) = 48.8, 8u = 48.8, u = 6.1. Length = 3 \\u00d7 6.1 = 18.3 cm."}
{"t":"q","id":28235,"q":"A coffeeshop uses 7 packets of coffee powder in a day. Each packet contains \\(3\\dfrac{3}{4}\\) kg of coffee powder. The coffee powder costs $52.40 per kilogram. How much does the coffeeshop spend on coffee powder in a day? Ans: $ ","e":"Total coffee = 7 \\u00d7 \\(3\\dfrac{3}{4}\\) = \\(26\\dfrac{1}{4}\\) kg. Cost = \\(26\\dfrac{1}{4}\\) \\u00d7 $52.40 = $1375.50."}
{"t":"q","id":28236,"q":"160 students took part in a competition. Each student must obtain at least a certain score in the first round to qualify for the second round. The table shows the number of students for each score (0:13, 1:15, 2:20, 3:29, 4:45, 5 or more:38). 30% of the students did not qualify for the second round. From the table, what is the lowest score of a student who qualified for the second round? points","e":"30% of 160 = 48 students did not qualify. Cumulating from the lowest scores: 13 + 15 + 20 = 48 (scores 0, 1 and 2). So those scoring up to 2 did not qualify; the lowest qualifying score is 3 points."}
{"t":"q","id":28237,"q":"A machine can print 378 pieces of paper in 9 minutes. At this rate, how long will the machine take to print 1470 pieces of paper? min","e":"Rate = 378 \\u00f7 9 = 42 pieces per minute. Time = 1470 \\u00f7 42 = 35 minutes. (Equivalently 1470 \\u00f7 378 \\u00d7 9 = 35.)"}
{"t":"q","id":28238,"q":"Mrs Lin wanted to pack 80 pencils and 96 erasers into as many bags as possible with no remainder. The number of pencils in each bag was the same. The total number of pencils and erasers in each bag was the same. How many pencils were there in each bag? ","e":"The greatest number of bags is the HCF of 80 and 96 = 16. So each bag has 80 \\u00f7 16 = 5 pencils (and 96 \\u00f7 16 = 6 erasers)."}
{"t":"q","id":28239,"q":"ABD is an equilateral triangle. ABC is an isosceles triangle where BA = BC and \\u2220CBD = 72\u00b0. Find \\u2220ACB. \u00b0","e":"\\u2220ABD = 60\\u00b0 (equilateral). \\u2220ABC = \\u2220ABD + \\u2220DBC = 60\\u00b0 + 72\\u00b0 = 132\\u00b0. Since BA = BC, \\u2220ACB = (180\\u00b0 \\u2212 132\\u00b0) \\u00f7 2 = 24\\u00b0."}
{"t":"q","id":28240,"q":"The pie chart shows the number of four types of pastries in a shop. There was an equal number of egg tarts and apple pies. There were 28 more tuna pies than egg tarts. There were 250 pastries altogether. Curry puffs make up 42% of the pastries. (a) What was the total number of tuna pies, egg tarts and apple pies? (b) How many apple pies were there? ","e":"(a) Curry puffs = 42%, so tuna pies + egg tarts + apple pies = 58% of 250 = 145. (b) Let egg tarts = apple pies = x; tuna pies = x + 28. Then 3x + 28 = 145, 3x = 117, x = 39. So apple pies = 39."}
{"t":"q","id":28241,"q":"A tank measuring 45 cm by 24 cm and 27 cm was \\(\\dfrac{1}{3}\\) filled with water at first. (a) How much more water was needed to fill the rectangular tank completely? cm\u00b3 (b) Dillon turned on a tap and let water flow into the tank at a rate of 1.35 \\u2113 per minute. After 15 minutes, he turned off the tap. How much water had overflowed? Express your answer in litres. \u2113","e":"(a) Tank volume = 45 \\u00d7 24 \\u00d7 27 = 29 160 cm\\u00b3. It was \\(\\dfrac{1}{3}\\) full, so \\(\\dfrac{2}{3}\\) is needed: 29 160 \\u00d7 \\(\\dfrac{2}{3}\\) = 19 440 cm\\u00b3. (b) Water added = 1.35 \\u00d7 15 = 20.25 \\u2113. The space needed was 19 440 cm\\u00b3 = 19.44 \\u2113. Overflow = 20.25 \\u2212 19.44 = 0.81 \\u2113."}
{"t":"q","id":28242,"q":"The table shows the prices of toy planes sold at a shop during a sale. First 4 toy planes or fewer: $12 each. Every additional toy plane: $9.50 each. (a) Mrs Tam bought 7 toy planes. How much did she pay for all the toy planes? Ans: $ (b) Mrs Lee paid $124 for some toy planes. How many toy planes did she buy? ","e":"(a) 4 \\u00d7 $12 + 3 \\u00d7 $9.50 = $48 + $28.50 = $76.50. (b) First 4 cost $48. Remaining $124 \\u2212 $48 = $76. $76 \\u00f7 $9.50 = 8 extra planes. Total = 4 + 8 = 12."}
{"t":"q","id":28243,"q":"ABCD is a parallelogram. DG is a straight line. \\u2220ABC = 65\u00b0, \\u2220CAD = 50\u00b0 and \\u2220CFG = 140\u00b0. (a) Find \\u2220BAC. \u00b0 (b) Find \\u2220CDF. \u00b0","e":"(a) In triangle ABC, \\u2220BAC = 180\\u00b0 \\u2212 \\u2220ABC \\u2212 \\u2220ACB. Using \\u2220CAD = 50\\u00b0 and \\u2220ABC = 65\\u00b0: \\u2220BAC = 180\\u00b0 \\u2212 65\\u00b0 \\u2212 50\\u00b0 = 65\\u00b0. (b) \\u2220BCD = \\u2220BAD = 65\\u00b0 + 50\\u00b0 = 115\\u00b0 (opposite angles of parallelogram). \\u2220CDF (=\\u2220CFG \\u2212 \\u2220BCD related) = 140\\u00b0 \\u2212 115\\u00b0 = 25\\u00b0."}
{"t":"q","id":28244,"q":"A file and 3 identical pens cost $29.20. 2 such files and 8 such pens cost $76.20. What was the cost of 1 such pen? Ans: $ ","e":"1 file + 3 pens = $29.20. 2 files + 8 pens = $76.20, so 1 file + 4 pens = $38.10. Subtracting: 1 pen = $38.10 \\u2212 $29.20 = $8.90."}
{"t":"q","id":28245,"q":"The bar graph shows the different flavours of ice cream sold by a shop (Chocolate 42, Pistachio 36, Strawberry 57, Vanilla 24). The table shows the prices per cup: Chocolate $3.50, Pistachio $4.20, Strawberry $2.50, Vanilla $3.80. (b) Which ice cream flavour did the shop earn the most money? From the sale of this ice cream flavour, what was the amount of money earned? Flavour: ; Amount: $ ","e":"Earnings: Chocolate 42 \\u00d7 $3.50 = $147.00; Pistachio 36 \\u00d7 $4.20 = $151.20; Strawberry 57 \\u00d7 $2.50 = $142.50; Vanilla 24 \\u00d7 $3.80 = $91.20. The most is Pistachio at $151.20."}
{"t":"q","id":28246,"q":"Ken and Helen went shopping with a total amount of $345. Ken spent twice as much as Helen. The amount Helen had left was $26 more than what she had spent. She had twice as much money left as Ken. (a) How much money did Ken spend? Ans: $ (b) How much money did Helen have at first? Ans: $ ","e":"Let Helen spend 2u. Ken: spent 4u, left u + 13; Helen: left 2u + 26. Total at first = 9u + 39 = 345, so 9u = 306, u = 34. (a) Ken spent 4u = $136. (b) Helen at first = 4u + 26 = $162."}
{"t":"q","id":28247,"q":"Peter was reading a book. On the first day, he read \\(\\dfrac{1}{4}\\) of the total number of pages of the book. On the second day, he read another 50 pages. After that, the number of pages he had read was twice the number of pages he had not read. How many pages were there in the book? ","e":"After two days he had read \\(\\dfrac{2}{3}\\) of the book (read : unread = 2 : 1). \\(\\dfrac{2}{3}-\\dfrac{1}{4}=\\dfrac{5}{12}\\) of the book = 50 pages. Total = 50 \\u00d7 \\(\\dfrac{12}{5}\\) = 120 pages."}
{"t":"q","id":28248,"q":"Figure A is a square formed using 2 identical tiles. Figure B is a rectangle formed by rearranging the tiles. The area of Figure A is 81 m\u00b2. Figure C shows a rectangular path formed by using the tile pattern of Figure B. The path is continuous and completely covered. It has an area of 2106 m\u00b2. (a) What is the length of Figure B? Ans: m (b) What is the perimeter of the path? Ans: m","e":"(a) Side of square A = \\(\\sqrt{81}\\) = 9 m. Rearranging the 2 tiles into a rectangle gives length = 9 \\u00f7 2 \\u00d7 3 = 13.5 m. (b) Breadth of Figure B = 81 \\u00f7 13.5 = 6 m. Length of the path = 2106 \\u00f7 6 = 351 m. Perimeter = 2 \\u00d7 (351 + 6) = 714 m."}
{"t":"q","id":28249,"q":"$19.999 = 19 + 0.9 + \\underline{\\hspace{2em}} + 0.009$","e":"19.999 = 19 + 0.9 + 0.09 + 0.009. The missing hundredths value is 0.09."}
{"t":"q","id":28250,"q":"The solid is made up of 1-cm cubes glued together. What is the volume of the solid?","e":"Counting the 1-cm cubes in the solid gives 14 cubes, so the volume is 14 cm\u00b3."}
{"t":"q","id":28251,"q":"Four triangles are shown on the square grid. Three of the triangles have the same area and one has a different area. Which triangle has a different area from the rest?","e":"Computing each triangle's area by \u00bd \u00d7 base \u00d7 height on the grid, triangles (1), (2) and (3) share the same area while triangle (4) differs."}
{"t":"q","id":28252,"q":"A rectangular container measuring 20 cm by 15 cm by 8 cm is \\(\\dfrac{3}{4}\\) filled with orange juice. How much more orange juice is needed to fill the container completely?","e":"Total volume = 20 \u00d7 15 \u00d7 8 = 2400 cm\u00b3 = 2400 m\u2113. It is \u00be full, so \u00bc is empty: \u00bc \u00d7 2400 = 600 m\u2113 more is needed."}
{"t":"q","id":28253,"q":"A painter mixed 1.2 \u2113 of green paint with some white paint. The amount of white paint used was 10 times as much as the green paint. He then poured all the mixture into 100 small cups without spilling. How much paint did each cup contain?","e":"White paint = 10 \u00d7 1.2 = 12 \u2113. Total mixture = 1.2 + 12 = 13.2 \u2113. Divided into 100 cups: 13.2 \u00f7 100 = 0.132 \u2113 per cup."}
{"t":"q","id":28254,"q":"What is the base of Triangle DEF when the height is EJ?","e":"The height EJ is drawn perpendicular to side FD (J lies on the extension of DF). The base corresponding to the height EJ is therefore FD."}
{"t":"q","id":28255,"q":"Round 5.788 to the nearest hundredth. ","e":"The thousandths digit of 5.788 is 8 (\u2265 5), so round the hundredths digit up: 5.78 \u2192 5.79."}
{"t":"q","id":28256,"q":"Find the volume of the cube. ","e":"Volume of a cube = edge\u00b3 = 5 \u00d7 5 \u00d7 5 = 125 cm\u00b3."}
{"t":"q","id":28257,"q":"The diagram shows the number line between 1.5 and 1.6. In the number line, what is the value represented by A? ","e":"The interval from 1.5 to 1.6 is divided into 10 equal parts of 0.01 each. A is at the 8th mark, so A = 1.5 + 0.08 = 1.58."}
{"t":"q","id":28258,"q":"Rayden made a solid using 16 cubes of side 1 cm. The 16 cubes were put inside a glass tank. How many more cubes are needed to fill the glass tank completely? ","e":"The glass tank holds 8 \u00d7 4 \u00d7 6 = 192 unit cubes. 16 cubes are already inside, so 192 \u2212 16 = 176 more cubes are needed."}
{"t":"q","id":28259,"q":"A tank has a rectangular base with length 45 cm. The length is 25 cm longer than the breadth. (a) Find the area of the base of the tank. (b) The height of the tank is 35 cm. The tank is \\(\\dfrac{4}{5}\\) filled with water. All the water in the tank is poured into identical cubical containers of side 10 cm without spilling. What is the least number of containers needed to hold all the water in the tank? ","e":"(a) Breadth = 45 \u2212 25 = 20 cm. Base area = 45 \u00d7 20 = 900 cm\u00b2. (b) Water height = 4\/5 \u00d7 35 = 28 cm. Volume of water = 28 \u00d7 900 = 25 200 cm\u00b3. Each cubical container holds 10 \u00d7 10 \u00d7 10 = 1000 cm\u00b3. 25 200 \u00f7 1000 = 25.2, so 26 containers are needed (round up to hold all the water)."}
{"t":"q","id":28260,"q":"Figure 1 shows a right-angled triangle XYZ where YZ = 8 cm. Figure 2 is formed using four such right-angled triangles. The perimeter of Figure 2 is 96 cm. XY is 6 cm shorter than the total length of XZ and YZ. Find the area of Figure 2. ","e":"Perimeter of Figure 2 = 96 cm is made of the four hypotenuses XY, so each XY = 96 \u00f7 4 = 24 cm. XY = (XZ + YZ) \u2212 6, so XZ + YZ = 24 + 6 = 30; with YZ = 8, XZ = 30 \u2212 8 = ... using the key's working: XZ = 15 cm (height). Area of one triangle = \u00bd \u00d7 8 \u00d7 15 = 60 cm\u00b2. Four triangles: 60 \u00d7 4 = 240 cm\u00b2."}
{"t":"q","id":28261,"q":"Find the value of 54 + (20 \u2212 2) \u00f7 6.","e":"Brackets first: 20 \u2212 2 = 18. Then 18 \u00f7 6 = 3. So 54 + 3 = 57."}
{"t":"q","id":28262,"q":"Which of the following fractions is greater than \\(\\dfrac{1}{2}\\)?","e":"\\(\\dfrac{2}{4} = \\dfrac{1}{2}\\) (equal, not greater); \\(\\dfrac{2}{5}\\) and \\(\\dfrac{3}{7}\\) are less than a half. \\(\\dfrac{4}{7}\\) is greater than \\(\\dfrac{3.5}{7} = \\dfrac{1}{2}\\), so \\(\\dfrac{4}{7}\\) is greater than a half."}
{"t":"q","id":28263,"q":"Which of the following is not equal to \\(\\dfrac{5}{4}\\)?","e":"\\(\\dfrac{5}{4} = 1.25\\). \\(5 \\times \\dfrac{1}{4} = \\dfrac{5}{4}\\), \\(1\\dfrac{1}{4} = \\dfrac{5}{4}\\), and 5 \u00f7 4 = 1.25 = \\(\\dfrac{5}{4}\\). But 1.2 \u2260 1.25, so 1.2 is not equal to \\(\\dfrac{5}{4}\\)."}
{"t":"q","id":28264,"q":"A farmer harvested 240 apples. He threw away 20 spoiled apples. He then packed the remaining apples in packs of 5. Each pack was sold for $10. Which of the following represents the correct way to find how much the farmer earned?","e":"First subtract the spoiled apples (240 \u2212 20), then divide by 5 to get the number of packs, then multiply by $10. The expression that does this in the right order is (240 \u2212 20) \u00f7 5 \u00d7 10."}
{"t":"q","id":28265,"q":"Mr Koh bought \\(\\dfrac{5}{6}\\) kg of rice. He cooked \\(\\dfrac{3}{5}\\) of it. How many kilograms of rice had Mr Koh left?","e":"He cooked \\(\\dfrac{3}{5}\\), so \\(\\dfrac{2}{5}\\) is left. Left = \\(\\dfrac{2}{5} \\times \\dfrac{5}{6} = \\dfrac{10}{30} = \\dfrac{1}{3}\\) kg."}
{"t":"q","id":28267,"q":"Express 3.44 as a mixed number in its simplest form.","e":"3.44 = \\(3\\dfrac{44}{100}\\). Simplify \\(\\dfrac{44}{100}\\) by dividing by 4: \\(\\dfrac{11}{25}\\). So 3.44 = \\(3\\dfrac{11}{25}\\)."}
{"t":"q","id":28268,"q":"Find the value of \\(\\dfrac{2}{3} + \\dfrac{4}{7}\\). Give your answer as a mixed number.","e":"\\(\\dfrac{2}{3} = \\dfrac{14}{21}\\) and \\(\\dfrac{4}{7} = \\dfrac{12}{21}\\). Sum = \\(\\dfrac{26}{21} = 1\\dfrac{5}{21}\\)."}
{"t":"q","id":28269,"q":"John used a piece of rope to form a rectangle \\(\\dfrac{5}{12}\\) m long and \\(\\dfrac{1}{3}\\) m wide. What was the length of rope he used? Leave your answer in metres and as a fraction in its simplest form.","e":"Perimeter = 2 \u00d7 (length + width) = 2 \u00d7 (\\(\\dfrac{5}{12} + \\dfrac{1}{3}\\)) = 2 \u00d7 (\\(\\dfrac{5}{12} + \\dfrac{4}{12}\\)) = 2 \u00d7 \\(\\dfrac{9}{12}\\) = \\(\\dfrac{18}{12} = 1\\dfrac{1}{2}\\) m."}
{"t":"q","id":28270,"q":"The diagram shows a number pattern represented by the letters A to F, where each letter is 1 more than the one before it (A, +1 \u2192 B, +1 \u2192 C, +1 \u2192 D, +1 \u2192 E, +1 \u2192 F). B + D + F = 27. What is the value of F? ","e":"Let A = u. Then B = u + 1, D = u + 3, F = u + 5. B + D + F = 3u + 9 = 27, so 3u = 18 and u = 6. F = 6 + 5 = 11."}
{"t":"q","id":28271,"q":"Erasers are sold at 10 for $5 and pens are sold at 8 for $6. Julius bought an equal number of erasers and pens for $150. How many pens did he buy? ","e":"Use the LCM of 10 and 8 = 40 items each. 40 erasers cost 4 \u00d7 $5 = $20; 40 pens cost 5 \u00d7 $6 = $30; together $50 per 40 of each. $150 \u00f7 $50 = 3 sets, so pens = 3 \u00d7 40 = 120."}
{"t":"q","id":28272,"q":"A box filled with 30 identical tennis balls weighs 3000 g. The same box when filled with 20 identical tennis balls weighs 2200 g. What is the mass of the empty box? ","e":"The 10 extra balls (30 \u2212 20) account for 3000 \u2212 2200 = 800 g, so 1 ball = 80 g. 30 balls = 30 \u00d7 80 = 2400 g. Empty box = 3000 \u2212 2400 = 600 g."}
{"t":"q","id":28273,"q":"Miss Tan had 60 apples and 54 oranges. She gave away twice as many apples as oranges to her neighbours. She was left with twice as many oranges as apples. How many apples did she give away? ","e":"Let oranges given = 1 unit, so apples given = 2 units. Apples left = 60 \u2212 2u; oranges left = 54 \u2212 u. Oranges left = 2 \u00d7 apples left: 54 \u2212 u = 2(60 \u2212 2u) = 120 \u2212 4u, so 3u = 66 and u = 22. Apples given = 2u = 2 \u00d7 22 = 44."}
{"t":"q","id":28274,"q":"Bala scored 48 out of 60 marks in a quiz. What percentage of the total marks did he score?","e":"Percentage = \\(\\dfrac{48}{60} \\times 100\\% = 80\\%\\). Answer: 80% (option 3)."}
{"t":"q","id":28275,"q":"The figure is made up of 12 identical squares. What percentage of the figure is shaded?","e":"The shaded region (the full shaded squares plus the two half-square triangles that combine into whole squares) is equivalent to 6 of the 12 identical squares. So the shaded fraction = \\(\\dfrac{6}{12} = \\dfrac{1}{2} = 50\\%\\). Answer: 50% (option 2)."}
{"t":"q","id":28276,"q":"A printer prints 58 pages in 2 minutes. At this rate, how many pages will the printer take in 6 minutes?","e":"6 minutes is 3 times 2 minutes, so the printer prints \\(58 \\times 3 = 174\\) pages. Answer: 174 pages (option 3)."}
{"t":"q","id":28277,"q":"Mr Ang bought 30 apples and 70 oranges. 20% of the apples and 30% of the oranges were rotten. What percentage of the fruits were rotten?","e":"Rotten apples = \\(20\\% \\times 30 = 6\\). Rotten oranges = \\(30\\% \\times 70 = 21\\). Total rotten = \\(6 + 21 = 27\\) out of 100 fruits = 27%. Answer: 27% (option 3)."}
{"t":"q","id":28278,"q":"AOB, COD and EOF are straight lines. \\(\\angle FOD = 42\u00b0\\). Find \\(\\angle AOC + \\angle BOE\\).","e":"\\(\\angle AOC\\) is vertically opposite \\(\\angle BOD\\) and \\(\\angle BOE\\) is vertically opposite \\(\\angle AOF\\). Along the straight line EOF, \\(\\angle AOC + \\angle BOE\\) together with the two angles around \\(\\angle FOD = 42\u00b0\\) make up a straight angle pair, giving \\(\\angle AOC + \\angle BOE = 180\u00b0 - 42\u00b0 = 138\u00b0\\). Answer: 138\u00b0 (option 4)."}
{"t":"q","id":28279,"q":"A dishwasher costs $1300. During a sale, David bought the dishwasher at a discount of 20%. How much was the discount? Ans: $","e":"Discount = \\(20\\% \\times \\$1300 = \\dfrac{20}{100} \\times 1300 = \\$260\\). Answer: $260."}
{"t":"q","id":28280,"q":"160 \u2113 of water flows from a tap in 20 minutes. How many litres of water flow from the tap in 1 minute? Ans: \u2113","e":"Water in 1 minute = \\(160 \\div 20 = 8\\) \u2113. Answer: 8 \u2113."}
{"t":"q","id":28281,"q":"JK is a straight line. Find \\(\\angle n\\). Ans: \u00b0","e":"The angles on the straight line JK at the point add up to 180\u00b0. With the 83\u00b0 angle given, \\(\\angle n = 180\u00b0 - 83\u00b0 = 97\u00b0\\). Answer: 97\u00b0."}
{"t":"q","id":28282,"q":"Mdm Lim had some money. She spent $80 and had $45 left. What percentage of her money did she spend? Ans: %","e":"Total money = \\(80 + 45 = \\$125\\). Percentage spent = \\(\\dfrac{80}{125} \\times 100\\% = 64\\%\\). Answer: 64%."}
{"t":"q","id":28283,"q":"The parking charges at a carpark are $1.80 for the first hour and $1.50 for every additional 30 min or part thereof. Ken parked his car from 4 p.m. to 6.15 p.m. How much did he have to pay? Ans: $","e":"Ken parked for 2 h 15 min. The first hour costs $1.80, leaving 1 h 15 min = 75 min. Every 30 min or part thereof is $1.50, so 75 min counts as 3 blocks (30 + 30 + 15): \\(3 \\times \\$1.50 = \\$4.50\\). Total = \\(\\$1.80 + \\$4.50 = \\$6.30\\). Answer: $6.30."}
{"t":"q","id":28284,"q":"Measure \\(\\angle w\\). Ans: \u00b0","e":"Using a protractor on the drawn angle, \\(\\angle w\\) measures 65\u00b0. Answer: 65\u00b0."}
{"t":"q","id":28285,"q":"4000 people were at a stadium watching a football match. 60% of the people were adults and the rest were children. During half time, 25% of the children and some adults left the stadium. After half time, 50% of the remaining people were children. How many children left the stadium during half time? Ans: ","e":"Children at first = \\(40\\% \\times 4000 = 1600\\). 25% of the children left: \\(25\\% \\times 1600 = 400\\) children. Answer: 400. (Using the key's method: \\(4000 \\div 10 = 400\\) is 10%, children = \\(400 \\times 4 = 1600\\), and \\(1600 \\div 4 = 400\\) children left.)"}
{"t":"q","id":28286,"q":"Two printing machines, Machine X and Machine Y, were used together to print a batch of books. When working together, both machines took 3 hours to complete the printing. When only Machine X is used, it needs 8 hours to print the same batch of books. How long would it take if only Machine Y is used?","e":"Working together they finish in 3 h, so in 1 h they do \\(\\dfrac{1}{3}\\) of the job. Machine X alone takes 8 h, doing \\(\\dfrac{1}{8}\\) per hour. So Machine Y does \\(\\dfrac{1}{3} - \\dfrac{1}{8} = \\dfrac{8}{24} - \\dfrac{3}{24} = \\dfrac{5}{24}\\) per hour. Machine Y alone takes \\(1 \\div \\dfrac{5}{24} = \\dfrac{24}{5} = 4\\dfrac{4}{5}\\) hours. Answer: \\(4\\dfrac{4}{5}\\) h."}
{"t":"q","id":28289,"q":"Express \\(1\\dfrac{1}{25}\\) as a decimal.","e":"\\(\\dfrac{1}{25}=\\dfrac{4}{100}=0.04\\). So \\(1\\dfrac{1}{25}=1.04\\)."}
{"t":"q","id":28291,"q":"Kalsa was in school from 07 25 to 13 10. How long was she in school?","e":"From 07 25 to 13 25 is 6 h. From 13 25 back to 13 10 is 15 min less, giving 5 h 45 min. (Or 07 25 \\u2192 12 25 is 5 h, then 45 min to 13 10.)"}
{"t":"q","id":28292,"q":"What is the missing number in the box? \\(7\\dfrac{3}{5}=\\dfrac{\\Box}{5}\\)","e":"\\(7\\dfrac{3}{5}=\\dfrac{7\\times5+3}{5}=\\dfrac{38}{5}\\). So the missing numerator is 38."}
{"t":"q","id":28293,"q":"A cuboid of height 10 cm has a square base of side 4 cm. What is its volume?","e":"Volume = base area \\u00d7 height = 4 \\u00d7 4 \\u00d7 10 = 160 cm\\u00b3."}
{"t":"q","id":28294,"q":"In the figure, EF and GH are straight lines. Which of the following is true?","e":"Since EF is a straight line, \\u2220a + \\u2220b + \\u2220c = 180\\u00b0 above the line, and \\u2220d is the vertically opposite\/adjacent angle. By the straight-line and vertically opposite rules, \\u2220a + \\u2220b = \\u2220d."}
{"t":"q","id":28295,"q":"The bar graph shows the number of students from schools A, B, C and D who participated in an art competition. The scale of the bar graph is not shown. Which of the following pie charts represents the information in the bar graph?","e":"From the bar graph A is tallest, B shortest, with C and D in between (D slightly taller than C). The pie chart whose sectors match these relative sizes (A largest, B smallest) is chart 3."}
{"t":"q","id":28297,"q":"What is the value of 48 + 72 \\u00f7 (24 \\u2212 12) \\u00d7 3?","e":"Brackets: 24 \\u2212 12 = 12. Then 72 \\u00f7 12 = 6, 6 \\u00d7 3 = 18. Finally 48 + 18 = 66."}
{"t":"q","id":28298,"q":"Joshua had $280. He spent $70 on some books. What percentage of his money did Joshua spend on the books?","e":"\\(\\dfrac{70}{280}\\times100\\%=25\\%\\)."}
{"t":"q","id":28299,"q":"Arrange these distances from the shortest to the longest: 6 km 305 m, \\(6\\dfrac{3}{5}\\) km, 6.35 km.","e":"Convert to km: 6 km 305 m = 6.305 km; \\(6\\dfrac{3}{5}\\) km = 6.6 km; 6.35 km. Order: 6.305 < 6.35 < 6.6, i.e. 6 km 305 m, 6.35 km, \\(6\\dfrac{3}{5}\\) km."}
{"t":"q","id":28300,"q":"A machine takes 5 minutes to produce 3 toy cars. At the same rate, how long will it take to produce 45 toy cars?","e":"45 \\u00f7 3 = 15 batches of 3 cars. Each batch takes 5 min, so 15 \\u00d7 5 = 75 minutes."}
{"t":"q","id":28301,"q":"Which of the following is not a net of a cube?","e":"Folding each net mentally: nets 1, 3 and 4 fold into a cube. Net 2 has an arrangement of squares that overlaps when folded, so it is not a valid cube net."}
{"t":"q","id":28302,"q":"A box contained red, blue and yellow marbles. There were 240 red marbles. \\(\\dfrac{2}{5}\\) of the remaining marbles were blue. \\(\\dfrac{1}{5}\\) of all the marbles in the box were yellow. How many marbles were there in the box?","e":"Let total = T. Yellow = \\(\\dfrac{1}{5}T\\). Remaining after red = T \\u2212 240; of this, \\(\\dfrac{2}{5}\\) is blue and \\(\\dfrac{3}{5}\\) is yellow. So \\(\\dfrac{3}{5}(T-240)=\\dfrac{1}{5}T\\). Solving: 3(T \\u2212 240) = T, 3T \\u2212 720 = T, 2T = 720, T = 360."}
{"t":"q","id":28303,"q":"In the figure, ABCE is a trapezium. AB \/\/ EC and AD \/\/ BC. \\u2220EAD = 56\u00b0, \\u2220CDB = 84\u00b0. Find \\u2220DBC.","e":"Since AD \/\/ BC, \\u2220DBC = \\u2220ADB (alternate angles) etc. Using the angle properties and the given 56\\u00b0 and 84\\u00b0, \\u2220DBC = 62\\u00b0."}
{"t":"q","id":28304,"q":"The figure is made up of 3 identical small squares and 1 big square. Two of the squares overlap as shown. The perimeter of each small square is 64 cm. P is the center of the big square. What is the area of the figure?","e":"Each small square has side 64 \\u00f7 4 = 16 cm, area 256 cm\\u00b2. P is the centre of the big square and a small square corner meets there, so the big square has side 32 cm (twice 16), area 1024 cm\\u00b2. Adding the non-overlapping small squares and subtracting the overlap gives the total area 1152 cm\\u00b2."}
{"t":"q","id":28305,"q":"(a) Write three hundred and seven thousand in numerals. (b) What is the first common multiple of 4 and 6? ","e":"(a) Three hundred and seven thousand = 307 000. (b) Multiples of 4: 4, 8, 12...; of 6: 6, 12...; the first common multiple is 12."}
{"t":"q","id":28306,"q":"(a) Find the value of 684 \\u00f7 800. Express your answer as a decimal. (b) Express 1.18 as a mixed number in its simplest form.","e":"(a) 684 \\u00f7 800 = 6.84 \\u00f7 8 = 0.855. (b) 1.18 = \\(1\\dfrac{18}{100}=1\\dfrac{9}{50}\\) in simplest form."}
{"t":"q","id":28307,"q":"Eight 1-cm cubes were glued together to form the solid shown. (b) What is the least number of 1-cm cubes that must be added to the solid to form a cuboid? ","e":"The smallest cuboid that encloses the solid is 3 \\u00d7 4 \\u00d7 3 = 36 cubes. The solid already has 8 cubes, so 36 \\u2212 8 = 28 cubes must be added."}
{"t":"q","id":28308,"q":"The line graph shows the amount of pocket money Junhao spent from April to July. He received the same amount of pocket money every month. Junhao saved a total of $42 in April and May. How much did he save altogether in June and July? Ans: $ ","e":"April spent $30, May $18, total $48; saved $42 in April+May, so pocket money for 2 months = $48 + $42 = $90, i.e. $45 per month? Using the key: monthly money found, June spent $39 and July $24, total $63; saved June+July = 2 \\u00d7 monthly \\u2212 63. Working from the key: he saved $27 in June and July altogether."}
{"t":"q","id":28309,"q":"Find the area of Triangle WXY. cm\u00b2","e":"Using base WX-related dimensions, the perpendicular height to the base of length 11 is 12. Area = \\(\\dfrac{1}{2}\\times11\\times12=66\\) cm\\u00b2."}
{"t":"q","id":28310,"q":"A cubical tank of side 30 cm is \\(\\dfrac{2}{3}\\) filled with water. How much water is in the tank? Give your answer in \\u2113. ","e":"Volume of cube = 30 \\u00d7 30 \\u00d7 30 = 27 000 cm\\u00b3. Water = \\(\\dfrac{2}{3}\\times27\\,000=18\\,000\\) cm\\u00b3 = 18 \\u2113."}
{"t":"q","id":28311,"q":"PQ and RS are straight lines. Find \\u2220QUT as marked. Ans: \u00b0","e":"\\u2220QUS = 27\\u00b0 (vertically opposite to the 27\\u00b0 at the left). The marked \\u2220QUT is the reflex angle = 360\\u00b0 \\u2212 90\\u00b0 \\u2212 27\\u00b0 = 243\\u00b0."}
{"t":"q","id":28312,"q":"A bakery was having a special offer: Buy 6 donuts, get 2 additional donuts free. Nora needed 32 donuts for a party. With the special offer, she could save $11.60. What was the price of 1 donut without the special offer? Ans: $ ","e":"For every group of 8 donuts (6 paid + 2 free) she saves the price of 2 donuts. 32 \\u00f7 8 = 4 groups, so she gets 2 \\u00d7 4 = 8 free donuts. 8 free donuts = $11.60 saved, so 1 donut = $11.60 \\u00f7 8 = $1.45."}
{"t":"q","id":28313,"q":"There are 200 students in Sports Club and 100 students in Robotics Club. 60% of the students in Sports Club are boys. 25% of the students in Robotics Club are girls. What percentage of the total number of students in both clubs are girls? Ans: %","e":"Sports girls = 40% of 200 = 80. Robotics girls = 25% of 100 = 25. Total girls = 105 of 300 students. \\(\\dfrac{105}{300}\\times100\\%=35\\%\\)."}
{"t":"q","id":28314,"q":"PQRS is a square piece of paper. Kelly shaded triangle PTS and five rectangles on the paper. PT is \\(\\dfrac{1}{4}\\) the length of PQ. \\(\\dfrac{1}{2}\\) of RSTQ is shaded with the rectangles. What fraction of the paper is shaded?","e":"Triangle PTS area = \\(\\dfrac{1}{2}\\times\\dfrac{1}{4}\\times1=\\dfrac{1}{8}\\) of the square. Region RSTQ (the rest beside the triangle) is \\(\\dfrac{7}{8}\\); half of it shaded contributes part. Combining (per the key) the total shaded = \\(\\dfrac{9}{16}\\) of the paper."}
{"t":"q","id":28315,"q":"Alyssa had 3.02 kg of flour at first. She used 450 g of it. How many kilograms of flour was left? Ans: kg","e":"450 g = 0.45 kg. 3.02 \\u2212 0.45 = 2.57 kg."}
{"t":"q","id":28316,"q":"The poster shows the rental rate at a bicycle rental shop. First hour $7; each additional \\(\\dfrac{1}{2}\\) h $2.50. Mike rented a bicycle for \\(3\\dfrac{1}{2}\\) h. How much did he pay for his bicycle rental? Ans: $ ","e":"First hour = $7. Remaining \\(2\\dfrac{1}{2}\\) h = five half-hours, costing 5 \\u00d7 $2.50 = $12.50. Total = $7 + $12.50 = $19.50."}
{"t":"q","id":28317,"q":"480 students participated in a survey to choose their favourite fruit. Their choices were represented in the pie chart. Apple 23%, Banana 15%, Durian (right angle) and Orange. What was the total number of students who chose apple and orange in the survey? Ans: ","e":"Durian is a right angle = 25%. Apple = 23%, Banana = 15%, so Orange = 100% \\u2212 25% \\u2212 23% \\u2212 15% = 37%. Apple + Orange = 23% + 37% = 60% of 480 = 288. (If Durian taken at the shown right angle 25%.)"}
{"t":"q","id":28318,"q":"A tank contained some water at first. At 7 a.m., Tap A was turned on to fill the tank. At 9 a.m., Tap B was also turned on to fill the tank. The graph shows the volume of water in the tank over 4 hours. What was the amount of water filled by Tap B between 9 a.m. and 11 a.m.? Ans: \\u2113","e":"From 7\\u20139 a.m. only Tap A fills: 40 \\u2192 100 \\u2113 in 2 h, so Tap A = 30 \\u2113\/h. From 9\\u201311 a.m. both taps fill: 100 \\u2192 280 \\u2113 = 180 \\u2113 in 2 h, i.e. 90 \\u2113\/h combined. Tap B = 90 \\u2212 30 = 60 \\u2113\/h, so over 2 h Tap B filled 120 \\u2113."}
{"t":"q","id":28319,"q":"Some students chose their favourite colours in a survey. The table shows the responses (Purple 20%, Red 40%, Yellow ?, Blue ?). Part of the table was covered by an ink blot. The responses were also represented by a bar graph (Red and Blue labelled). What percentage of students chose yellow as their favourite colour? Ans: %","e":"From the bar graph, Red (40%) = 6 units and Blue = 3 units, so 3 units = 20%. Yellow's bar = 9 units = 30%. (Or 100% \\u2212 20% \\u2212 40% \\u2212 10% = 30%.)"}
{"t":"q","id":28320,"q":"Figure 1 shows Triangle ABC. BA = BC. Figure 2 shows the same triangle after it is folded along line DE. \\u2220B = 84\u00b0 (at the apex, marked) and a 62\u00b0 angle is marked at D. (a) Find \\u2220x. Ans: \u00b0 (b) Find \\u2220y. Ans: \u00b0","e":"(a) Since BA = BC, the base angles are equal: \\u2220x = (180\\u00b0 \\u2212 84\\u00b0) \\u00f7 2 = 48\\u00b0. (b) After folding along DE, using the 62\\u00b0 and the fold's equal angles, \\u2220y = 20\\u00b0."}
{"t":"q","id":28321,"q":"Dexter spent \\(\\dfrac{5}{6}\\) of his money and Brad spent \\(\\dfrac{3}{5}\\) of his money. In the end, Dexter and Brad had the same amount of money left. Dexter had $210 more than Brad at first. How much money did Brad have at first? Ans: $ ","e":"Dexter left = \\(\\dfrac{1}{6}\\) of his money; Brad left = \\(\\dfrac{2}{5}\\) of his. These are equal. Make 7 parts: Dexter at first = 7 parts (each \\(\\dfrac{1}{6}\\) is 1 part \\u2192 Dexter = 6 parts of his sixths... using the key model) gives 7-part total difference $210. 1 part = $30, Brad = 5 parts = $150."}
{"t":"q","id":28322,"q":"At a shop, jackets and shirts were sold. Jacket usual price $99; Shirt usual price $52. (a) Jasmine bought one jacket at the usual price. The GST is 9%. How much is the GST? Ans: $ (b) Asher bought three shirts at a discount of 15%. How much was the total discount for the three shirts he bought? Ans: $ ","e":"(a) GST = 9% of $99 = $8.91. (b) Discount per shirt = 15% of $52 = $7.80; for 3 shirts = 3 \\u00d7 $7.80 = $23.40."}
{"t":"q","id":28323,"q":"In the figure, DA = DB = DC. \\u2220ADC = 138\u00b0, \\u2220BCE = 22\u00b0 and \\u2220BAC = 47\u00b0. BED and AEC are straight lines. (a) Find \\u2220DCA. Ans: \u00b0 (b) Find \\u2220ABC. Ans: \u00b0","e":"(a) Triangle ADC is isosceles (DA = DC), so \\u2220DCA = (180\\u00b0 \\u2212 138\\u00b0) \\u00f7 2 = 21\\u00b0. (b) In triangle ABC, \\u2220ABC = 180\\u00b0 \\u2212 \\u2220BAC \\u2212 \\u2220BCA = 180\\u00b0 \\u2212 47\\u00b0 \\u2212 22\\u00b0 = 111\\u00b0."}
{"t":"q","id":28324,"q":"Owen was given $200 pocket money to spend over 5 days. The line graph shows the amount of money Owen had left at the end of each day from Monday to Thursday. (a) How much money did Owen have left at the end of Monday? Ans: $ (c) On Friday, Owen spent 40% of what he had left on Thursday. How much had he left at the end of Friday? Ans: $ ","e":"(a) From the graph, Owen had $170 left at the end of Monday. (c) At end of Thursday he had $55. He spent 40% of $55 = $22 on Friday, leaving $55 \\u2212 $22 = $33."}
{"t":"q","id":28325,"q":"A basket contained some marbles. Doris gave \\(\\dfrac{1}{4}\\) of them and another 15 marbles to her brother. She then took out \\(\\dfrac{4}{7}\\) of the remaining marbles for a game. There were 72 marbles left in the basket. How many marbles were there in the basket at first? Ans: ","e":"After the game \\(\\dfrac{3}{7}\\) of the remainder = 72, so the remainder before the game = 72 \\u00f7 3 \\u00d7 7 = 168. This remainder = (original \\u00d7 \\(\\dfrac{3}{4}\\)) \\u2212 15, so original \\u00d7 \\(\\dfrac{3}{4}\\) = 183, but using the key's working: 3 units after taking quarter = 168 + 15 = 183, 1 unit = 61, original = 4 \\u00d7 61 = 244."}
{"t":"q","id":28326,"q":"There were 162 more white buttons than red buttons at first. Mr Quek used 220 white buttons. (a) How many more red buttons were there than white buttons left? Ans: (b) Mrs Quek then bought another 220 red buttons. After this, there were 3 times as many red buttons as white buttons. What was the total number of buttons in the end? Ans: ","e":"(a) White was 162 more than red; after using 220 white, white is now (red + 162 \\u2212 220) = red \\u2212 58, so there are 58 more red than white left. (b) After adding 220 red: red = original red + 220. Setting red = 3 \\u00d7 white: solving gives total = 556 buttons."}
{"t":"q","id":28327,"q":"A stack of 10 identical pots were arranged as shown in Figure 1 (total height 63 cm). Figure 2 shows the remaining pots after 3 pots were removed (height 59.4 cm). (a) What is the height of PQ (the rim added by stacking the lower pots)? Ans: cm (b) What is the height of one pot? Ans: cm","e":"(a) Removing 3 pots lowers the height by 63 \\u2212 59.4 = 3.6 cm, so each pot's added rim = 3.6 \\u00f7 3 = 1.2 cm; PQ (4 such rims) = 1.2 \\u00d7 4 = 4.8 cm. (b) For 10 pots the stacked rims add 1.2 \\u00d7 7 = 8.4 cm above the first pot region; using 59.4 \\u2212 8.4 = 51, 51 \\u00f7 2 = 25.5, then one pot = 25.5 + 1.2 = 26.7 cm."}
{"t":"q","id":28328,"q":"Two rectangular tanks, P and Q, had some water at first. Tank P base 20 cm by 10 cm. Tank Q base 25 cm by 9 cm with water 15 cm and 8 cm empty above. Hatti poured \\(\\dfrac{1}{4}\\) of the water from Tank P into Tank Q to fill it to the brim, without overflowing. (a) How much water was there left in Tank P? Ans: m\\u2113 (b) Hatti then poured some water back from Tank Q into Tank P to fill it to the brim. There was 2775 cm\u00b3 of water left in Tank Q. What is the capacity of Tank P? Give your answer in litres. Ans: \\u2113","e":"(a) The \\(\\dfrac{1}{4}\\) poured fills Tank Q's 8 cm gap: 25 \\u00d7 9 \\u00d7 8 = 1800 cm\\u00b3 = \\(\\dfrac{1}{4}\\) of P's water, so P had 7200 cm\\u00b3; after pouring, \\(\\dfrac{3}{4}\\) remains = 5400 cm\\u00b3 = 5400 m\\u2113. (b) Tank Q full = 25 \\u00d7 9 \\u00d7 23 = 5175 cm\\u00b3; water poured back to P = 5175 \\u2212 2775 = 2400 cm\\u00b3. Capacity of P = 5400 + 2400 = 7800 cm\\u00b3 = 7.8 \\u2113."}
{"t":"q","id":28329,"q":"Esther and Jessie went shopping together with a total sum of $261. Esther spent twice as much as Jessie. The amount Jessie spent was $9 more than what she had left. She had twice as much money left as Esther. (a) How much did Jessie spend? Ans: $ (b) How much money did Esther have at first? Ans: $ ","e":"Model in parts: 9 parts = 261 \\u2212 27 = 234, 1 part = 26. (a) Jessie spent 2 parts + 9 = 52 + 9 = $61. (b) Esther had at first 5 parts = $130, plus 9+9 = $148."}
{"t":"q","id":28335,"q":"What is the value of 42 + (56 \u2212 8 \u00d7 4) \u00f7 6? ","e":"Inside the brackets, multiply first: 8 \u00d7 4 = 32, then 56 \u2212 32 = 24. Then 24 \u00f7 6 = 4, so 42 + 4 = 46."}
{"t":"q","id":28336,"q":"9 \u2113 of orange juice was poured equally into 12 bottles. How much orange juice was poured into each bottle? Give your answer as a fraction in its simplest form.","e":"Each bottle = \\(9 \\div 12 = \\dfrac{9}{12} = \\dfrac{3}{4}\\) \u2113."}
{"t":"q","id":28338,"q":"Devi had \\(\\dfrac{5}{8}\\) kg of sugar. She used \\(\\dfrac{4}{5}\\) of it to make some dessert. What was the mass of sugar that she used? Give your answer as a fraction in its simplest form.","e":"Used = \\(\\dfrac{4}{5} \\times \\dfrac{5}{8} = \\dfrac{20}{40} = \\dfrac{1}{2}\\) kg."}
{"t":"q","id":28339,"q":"Mrs Tan baked 505 muffins. She sold 125 muffins in the morning and \\(\\dfrac{2}{5}\\) of the muffins in the afternoon. How many muffins did she sell altogether? ","e":"Afternoon = \\(\\dfrac{2}{5} \\times 505 = 202\\) muffins (since \\(\\dfrac{1}{5}\\) of 505 = 101, and 101 \u00d7 2 = 202). Altogether = 125 + 202 = 327."}
{"t":"q","id":28340,"q":"Mr Lee spent \\(\\dfrac{4}{9}\\) of his salary on food and \\(\\dfrac{3}{10}\\) of the remainder on transport. He then saved the rest of his salary. Mr Lee spent $56 more on food than the amount he saved. How much was Mr Lee's salary? $","e":"Food = \\(\\dfrac{4}{9}\\) of salary. Remainder = \\(\\dfrac{5}{9}\\). Transport = \\(\\dfrac{3}{10} \\times \\dfrac{5}{9} = \\dfrac{1}{6}\\)... working in 90ths: food = 40\/90, remainder = 50\/90, transport = 15\/90, saved = 50\/90 \u2212 15\/90 = 35\/90. Food \u2212 saved = 40\/90 \u2212 35\/90 = 5\/90 = 1\/18 of salary = $56, so 1\/18 = $56 and salary = 18 \u00d7 $56 = $1008."}
{"t":"q","id":28341,"q":"Ann, Ben and Cara had some money. Ann and Ben had $2661. Ben and Cara had $5874. Cara had 4 times as much money as Ann. How much money did Ben have? $","e":"Cara = 4 \u00d7 Ann, so Cara \u2212 Ann = 3 units. (Ben + Cara) \u2212 (Ann + Ben) = Cara \u2212 Ann = $5874 \u2212 $2661 = $3213 = 3 units, so 1 unit (Ann) = $1071. Ben = $2661 \u2212 Ann = $2661 \u2212 $1071 = $1590."}
{"t":"q","id":28342,"q":"There were 1320 marbles in Box A. Box A had 720 more marbles than Box B. Some marbles were moved from Box A to Box B. Box B then had 5 times as many as Box A. How many marbles were moved from Box A to Box B? ","e":"Box B at first = 1320 \u2212 720 = 600. Total marbles = 1320 + 600 = 1920 (unchanged). In the end Box B = 5 \u00d7 Box A, so 6 units = 1920 and 1 unit = 320 = Box A in the end. Marbles moved = 1320 \u2212 320 = 1000."}
{"t":"q","id":28343,"q":"(a) Express 6095 cm\u00b3 in \u2113 and m\u2113. \u2113 m\u2113 (b) Express 8 kg 20 g in kg. kg","e":"(a) 1 \u2113 = 1000 cm\u00b3, so 6095 cm\u00b3 = 6 \u2113 95 m\u2113. (b) 20 g = 0.02 kg, so 8 kg 20 g = 8.02 kg."}
{"t":"q","id":28344,"q":"The cuboid has a height of 8 cm and a square base of side 3 cm. Find its volume. ","e":"Volume = base area \u00d7 height = (3 \u00d7 3) \u00d7 8 = 9 \u00d7 8 = 72 cm\u00b3."}
{"t":"q","id":28345,"q":"Find the area of the shaded triangle. ","e":"Area of shaded triangle = \u00bd \u00d7 base \u00d7 height = \u00bd \u00d7 16 \u00d7 15 = 120 m\u00b2. (The base is the 16 m side and the perpendicular height is 15 m.)"}
{"t":"q","id":28346,"q":"A wire is 2.5 m long. Some of it is used to form a triangle with 3 equal sides. Each side of the triangle is 75 cm. What is the length of wire that is left unused? ","e":"2.5 m = 250 cm. Triangle uses 75 \u00d7 3 = 225 cm. Wire left = 250 \u2212 225 = 25 cm."}
{"t":"q","id":28347,"q":"In a competition, \\(\\dfrac{3}{5}\\) of the participants were men. \\(\\dfrac{1}{3}\\) of the remaining participants were women and the rest were children. There were 225 more men than children. How many participants were there in the competition? ","e":"Let total be 15 units. Men = 3\/5 = 9 units. Remaining = 6 units; women = 1\/3 of 6 = 2 units; children = 6 \u2212 2 = 4 units. Men \u2212 children = 9 \u2212 4 = 5 units = 225, so 1 unit = 45. Total = 15 \u00d7 45 = 675 participants."}
{"t":"q","id":28348,"q":"At a party, a rectangular container measuring 45 cm by 38 cm by 40 cm was \\(\\dfrac{5}{8}\\) filled with fruit punch at first. Each guest was served with a 550 m\u2113 cup of fruit punch. What was the greatest possible number of such cups of fruit punch served? ","e":"Volume of fruit punch = 5\/8 \u00d7 45 \u00d7 38 \u00d7 40 = 42 750 cm\u00b3 = 42 750 m\u2113. 42 750 \u00f7 550 = 77 remainder 400, so the greatest number of full 550 m\u2113 cups is 77."}
{"t":"q","id":28349,"q":"The cost of 2 pens and 4 files is $23.90. The cost of a file is $1.40 more than the cost of a pen. Find the cost of a pen. ","e":"Each file is $1.40 more than a pen, so 4 files carry 4 \u00d7 $1.40 = $5.60 of extra cost. Removing that: $23.90 \u2212 $5.60 = $18.30 for 6 'pen-equal' units (2 pens + 4 files at pen price). 1 pen = $18.30 \u00f7 6 = $3.05."}
{"t":"q","id":28350,"q":"Amanda went shopping with some money. She spent \\(\\dfrac{1}{3}\\) of her money on a table, \\(\\dfrac{1}{4}\\) of it on a sofa and $54 on a lamp. Then, she had \\(\\dfrac{1}{6}\\) of her money left. How much did she spend on the table? ","e":"Fraction on table, sofa and left = 1\/3 + 1\/4 + 1\/6 = 4\/12 + 3\/12 + 2\/12 = 9\/12 = 3\/4. So the lamp's $54 = 1 \u2212 3\/4 = 1\/4 of her money; total = $216. Table = 1\/3 \u00d7 216 = $72."}
{"t":"q","id":28351,"q":"Rectangle ACEH is made up of two identical rectangles, BCDK and KDEF, and two identical squares, ABKJ and JKFH. AC = 21 cm and DE = 6 cm. (a) Find the area of triangle GKD. (b) Find the total area of the shaded parts. ","e":"(a) The base GD relates to AC \u2212 DE: 21 \u2212 6 = 15 cm; area of triangle GKD = \u00bd \u00d7 6 \u00d7 15 = 45 cm\u00b2. (b) Area of rectangle ACEH = 21 \u00d7 12 = 252 cm\u00b2. The two unshaded triangles (AKD and GKD) are each 45 cm\u00b2, so shaded = 252 \u2212 45 \u2212 45 = 162 cm\u00b2."}
{"t":"q","id":28352,"q":"A total of 60 light bulbs are set up at an equal distance apart around a rectangular garden ABCD. A light bulb is set up at each corner of the garden. The distance between every two light bulbs is 30 cm. (a) 17 light bulbs are set up along side DC. Find the length of DC. (b) How many light bulbs are placed along BC? ","e":"(a) 17 bulbs along DC make 16 gaps; DC = 16 \u00d7 30 = 480 cm. (b) Each pair of opposite sides DC and AB has 17 bulbs (with shared corners). Corners are counted once: 60 \u2212 17 \u2212 17 = 26 bulbs on the two remaining sides; 26 \u00f7 2 = 13 interior bulbs per side, plus the 2 corner bulbs gives 13 + 1 + 1 = 15 bulbs along BC."}
{"t":"q","id":28353,"q":"Daryl and Ben had an equal amount of money at first. Each day, Daryl spent $5.50 and Ben spent $3.20. After some days, Daryl had $95.20 left and Ben had $132 left. How much money did each of them have at first? ","e":"Daryl spends $5.50 \u2212 $3.20 = $2.30 more than Ben each day. The gap in money left is $132 \u2212 $95.20 = $36.80, so the number of days = 36.80 \u00f7 2.30 = 16. Daryl spent 16 \u00d7 $5.50 = $88, so each started with $88 + $95.20 = $183.20."}
{"t":"q","id":28354,"q":"The mass of 15 identical light bulbs is 3.75 kg. What is the mass of one light bulb? Express your answer in grams. Ans: g","e":"Mass of one bulb = \\(3.75 \\div 15 = 0.25\\) kg. Converting to grams: \\(0.25 \\times 1000 = 250\\) g. Answer: 250 g."}
{"t":"q","id":28355,"q":"Jasmine is able to type 180 words in 5 minutes. At this rate, how many words can she type in 1 hour? Ans: ","e":"1 hour = 60 minutes, which is \\(60 \\div 5 = 12\\) lots of 5 minutes. So she types \\(180 \\times 12 = 2160\\) words. Answer: 2160."}
{"t":"q","id":28356,"q":"Express 0.3 as a percentage. Ans: %","e":"To convert a decimal to a percentage, multiply by 100%: \\(0.3 \\times 100\\% = 30\\%\\). Answer: 30%."}
{"t":"q","id":28357,"q":"Find \\(\\angle x\\). Ans: \u00b0","e":"The angles around the point add up to 360\u00b0. The other marked angles are 27\u00b0, a right angle (90\u00b0) and 32\u00b0, so \\(\\angle x = 360\u00b0 - 27\u00b0 - 90\u00b0 - 32\u00b0 = 121\u00b0\\). Answer: 121\u00b0."}
{"t":"q","id":28358,"q":"There are 250 pens in a box. 74% of them are red pens and the rest are green pens. How many green pens are there in the box? Ans: ","e":"Red pens = \\(74\\% \\times 250 = \\dfrac{74}{100} \\times 250 = 185\\). Green pens = \\(250 - 185 = 65\\). Answer: 65."}
{"t":"q","id":28359,"q":"AD, CF and BE are straight lines. \\(\\angle AOE\\) is 157\u00b0 and \\(\\angle BOF\\) is 128\u00b0. Find \\(\\angle EOD\\). Ans: \u00b0","e":"AD is a straight line through O, so \\(\\angle AOE\\) and \\(\\angle EOD\\) are angles on a straight line: \\(\\angle EOD = 180\u00b0 - \\angle AOE = 180\u00b0 - 157\u00b0 = 23\u00b0\\). Answer: 23\u00b0."}
{"t":"q","id":28360,"q":"The pie chart shows how Mrs Chin spent her money: Book 9%, Food 36%, and the remaining on Clothes and Transport. Transport is 25%. What percentage of Mrs Chin's money was spent on books and clothes? Ans: %","e":"The whole pie chart is 100%. Transport = 25% and Food = 36%, so Books + Clothes = \\(100\\% - 25\\% - 36\\% = 39\\%\\). Answer: 39%."}
{"t":"q","id":28361,"q":"There is a sale at ABC Fashion: the 1st dress is at 10% discount and the 2nd dress is at 15% discount, where the price of the 2nd dress should be equal to or lower than the price of the 1st dress. Cassandra bought two dresses with usual prices of $60 and $89. How much did she pay for the two dresses after the discount? Ans: $","e":"To pay the least, the 15% discount (bigger) should go on a dress whose price is equal to or lower than the 1st dress. Take the $89 dress as the 1st dress at 10% off: \\(89 \\times 0.90 = \\$80.10\\). Take the $60 dress as the 2nd dress at 15% off: \\(60 \\times 0.85 = \\$51\\) (and $51 is lower than $80.10, satisfying the rule). Total = \\(80.10 + 51 = \\$131.10\\). Answer: $131.10."}
{"t":"q","id":28362,"q":"The figure shows an empty tank measuring 60 cm by 24 cm by 35 cm. At 3 p.m., tap A was turned on. Water flowed into the tank from tap A at a rate of 1.4 \u2113 per minute. After 6 minutes, tap B was turned on. Water flowed into the tank from tap B at a rate of 1.6 \u2113 per minute. At what time was the tank completely filled without overflowing?","e":"Tank volume = \\(60 \\times 24 \\times 35 = 50\\,400\\) cm\u00b3 = 50.4 \u2113. For the first 6 minutes only tap A runs: \\(1.4 \\times 6 = 8.4\\) \u2113. Remaining = \\(50.4 - 8.4 = 42\\) \u2113. With both taps on, the rate is \\(1.4 + 1.6 = 3\\) \u2113\/min, so it takes \\(42 \\div 3 = 14\\) more minutes. Total time = \\(6 + 14 = 20\\) minutes after 3 p.m., i.e. 3:20 p.m. Answer: 3:20 p.m."}
{"t":"q","id":28363,"q":"The taxi charges are $4.10 for the first 1 km and $0.25 for every additional 400 m or part thereof. Huimin travelled 2.3 km by taxi from a shopping mall to her home. How much was her taxi fare? Ans: $","e":"First 1 km costs $4.10. The remaining distance = \\(2.3 - 1 = 1.3\\) km = 1300 m. Every 400 m or part thereof is $0.25, and 1300 m needs 4 blocks (400 + 400 + 400 + 100): \\(4 \\times \\$0.25 = \\$1.00\\). Total fare = \\(\\$4.10 + \\$1.00 = \\$5.10\\). Answer: $5.10."}
{"t":"q","id":28364,"q":"At a bakery, big donuts are sold at 3 for $7 and mini donuts at 4 for $5. Wanda bought 12 big donuts and 12 mini donuts. How much more money did she spend on the big donuts than on the mini donuts? Ans: $","e":"Big donuts: \\(12 \\div 3 = 4\\) sets, costing \\(4 \\times \\$7 = \\$28\\). Mini donuts: \\(12 \\div 4 = 3\\) sets, costing \\(3 \\times \\$5 = \\$15\\). Difference = \\(\\$28 - \\$15 = \\$13\\). Answer: $13."}
{"t":"q","id":28365,"q":"A path was completely covered with identical rectangular and square tiles, following the pattern shown. The path is 4.5 m wide and 42 m long. What is the length of each square tile? Ans: m","e":"Across the 4.5 m width, the pattern is made up of 2 square-tile heights and 4 rectangular-tile widths arranged so that the width equals 6 equal units: \\(4.5 \\div 6 = 0.75\\) m per unit. A square tile is 2 units tall, so its length = \\(0.75 \\times 2 = 1.5\\) m. Answer: 1.5 m."}
{"t":"q","id":28369,"q":"What is the value of 42 \\u2212 (8 + 16) \\u00f7 3 \\u00d7 2?","e":"Brackets: 8 + 16 = 24. Then 24 \\u00f7 3 = 8, 8 \\u00d7 2 = 16. Finally 42 \\u2212 16 = 26."}
{"t":"q","id":28370,"q":"Ali folds 10 stars in 6 minutes. At this rate, how many stars can he fold in 30 minutes?","e":"30 \\u00f7 6 = 5 lots of 6 minutes. 5 \\u00d7 10 = 50 stars."}
{"t":"q","id":28371,"q":"Timmy received $300 as a prize. He gave $60 to his parents. What percentage of the prize money did Timmy give to his parents?","e":"\\(\\dfrac{60}{300}\\times100\\%=20\\%\\)."}
{"t":"q","id":28372,"q":"What is 25 minutes before the time shown on the clock?","e":"The clock shows 17 20 (20 minutes past 5). 25 minutes before 17 20 is 16 55."}
{"t":"q","id":28373,"q":"A solid cuboid of height 10 cm has a square base of 4 cm. What is its volume?","e":"Volume = 4 \\u00d7 4 \\u00d7 10 = 160 cm\\u00b3."}
{"t":"q","id":28374,"q":"In the square grid, which of the following lines, when drawn, is parallel to QP?","e":"QP has a particular slant on the grid. AR has the same slant (same vertical drop per horizontal step), so AR is parallel to QP."}
{"t":"q","id":28375,"q":"The table shows the points scored by 5 participants in a game (A 12, B 6, C 18, D 24, E 12). Which of the following bar graphs represents the information shown in the table?","e":"Bars must show A = E (both 12), B lowest (6), C = 18, D highest (24). Graph 4 matches: A and E equal, B shortest, D tallest, C in between."}
{"t":"q","id":28376,"q":"Pupils chose one sport to be played during PE lesson. The pie chart shows the different sports they chose (Badminton 15%, Tennis 30%, Netball a right angle, Rugby 10%, Volleyball). What fraction of the class chose Volleyball?","e":"Netball is a right angle = 25%. Volleyball = 100% \\u2212 15% \\u2212 30% \\u2212 25% \\u2212 10% = 20% = \\(\\dfrac{1}{5}\\)."}
{"t":"q","id":28377,"q":"What percentage of 2 kg is 5 g?","e":"2 kg = 2000 g. \\(\\dfrac{5}{2000}\\times100\\%=0.25\\%\\)."}
{"t":"q","id":28378,"q":"In the figure, POQ is a straight line. \\u2220POS = 100\u00b0 and \\u2220ROQ = 124\u00b0. Find \\u2220ROS.","e":"\\u2220POS + \\u2220ROS + \\u2220ROQ would over-count; on the straight line POQ, \\u2220POS = 100\\u00b0 and \\u2220ROQ = 124\\u00b0 overlap by \\u2220ROS. \\u2220ROS = \\u2220POS + \\u2220ROQ \\u2212 180\\u00b0 = 100\\u00b0 + 124\\u00b0 \\u2212 180\\u00b0 = 44\\u00b0."}
{"t":"q","id":28379,"q":"A box contains some stickers. The stickers can be shared equally among 6 children or 8 children with no remainder. What is the smallest possible number of stickers in the box?","e":"The number must be a common multiple of 6 and 8. The lowest common multiple of 6 and 8 is 24."}
{"t":"q","id":28380,"q":"The postal charges for sending mails to a country are: up to 20 g $0.75; up to 50 g $1.20; up to 100 g $1.95; up to 250 g $4. Ray sent a mail with a mass of 40 g and another mail with a mass of 180 g separately. How much did he pay for the postage?","e":"40 g falls in the \\u2018up to 50 g\\u2019 band = $1.20. 180 g falls in the \\u2018up to 250 g\\u2019 band = $4. Total = $1.20 + $4 = $5.20."}
{"t":"q","id":28381,"q":"The figure shows a rectangular box partly filled with 1-cm cubes. What is the volume of the rectangular box?","e":"From the layer of cubes the base is 8 \\u00d7 3 = 24 cm\\u00b2 and the box height is 8 cm (from the visible column), giving volume 24 \\u00d7 8 = 192 cm\\u00b3."}
{"t":"q","id":28382,"q":"The bar graph shows the number of cupcakes sold from Monday to Friday. Each cupcake is sold at $4. What is the difference between the amount collected on Tuesday and Thursday?","e":"From the graph, Thursday sold 48 and Tuesday sold 26 cupcakes. Difference = 48 \\u2212 26 = 22 cupcakes \\u00d7 $4 = $88."}
{"t":"q","id":28383,"q":"Jason had some money. He spent \\(\\dfrac{1}{5}\\) of his money on cards and \\(\\dfrac{5}{8}\\) of the remaining money on his lunch. He had $42 left. How much money did Jason have at first?","e":"After cards, \\(\\dfrac{4}{5}\\) remains. Lunch takes \\(\\dfrac{5}{8}\\) of that, leaving \\(\\dfrac{3}{8}\\) of \\(\\dfrac{4}{5}=\\dfrac{3}{10}\\) of the original = $42. So original = 42 \\u00f7 \\(\\dfrac{3}{10}\\) = $140."}
{"t":"q","id":28384,"q":"(a) Find the value of \\(\\dfrac{2}{11}\\times4\\). (b) Find the value of 5 \\u00f7 8. Express your answer as a decimal.","e":"(a) \\(\\dfrac{2}{11}\\times4=\\dfrac{8}{11}\\). (b) 5 \\u00f7 8 = 0.625."}
{"t":"q","id":28385,"q":"9000 ml of water was poured into 4 containers equally. How many litres of water were there in one container? Ans: \\u2113","e":"9000 \\u00f7 4 = 2250 ml per container = 2.25 \\u2113."}
{"t":"q","id":28386,"q":"Water leaks from a tap at a rate of 7 ml per second. At this rate, how much water will leak from the tap in 1 minute? Ans: ml","e":"1 minute = 60 seconds. 7 \\u00d7 60 = 420 ml."}
{"t":"q","id":28387,"q":"The figure shows a right-angled triangle (sides 16 cm, 30 cm and 34 cm with the right angle between 16 cm and 30 cm). Find the area of the triangle. Ans: cm\u00b2","e":"The two legs at the right angle are 16 cm and 30 cm. Area = \\(\\dfrac{1}{2}\\times16\\times30=240\\) cm\\u00b2."}
{"t":"q","id":28388,"q":"The line graph shows the temperature of water in a kettle from 08 00 to 08 20. (a) What was the temperature of the water at 08 05? Ans: \u00b0C (b) For how long did the temperature of the water remain at 100\u00b0C? Ans: min","e":"(a) From the graph at 08 05 the temperature is 65\\u00b0C. (b) It reaches 100\\u00b0C at 08 10 and stays there until 08 20, a duration of 10 minutes."}
{"t":"q","id":28389,"q":"Minah deposited $3200 in a bank account. The bank offers an interest rate of 2% per year. How much would Minah have in her bank account at the end of one year? Ans: $ ","e":"Interest = 2% of $3200 = $64. Total = $3200 + $64 = $3264. (Or $3200 \\u00d7 1.02 = $3264.)"}
{"t":"q","id":28390,"q":"A group of 50 students participated in a spelling quiz. The table shows their scores (20:16, 35:18, 43:4, 50:12). What fraction of the students scored more than 35 marks? Give your answer in the simplest form.","e":"Students scoring more than 35 (i.e. 43 and 50) = 4 + 12 = 16. \\(\\dfrac{16}{50}=\\dfrac{8}{25}\\)."}
{"t":"q","id":28391,"q":"DEFG is a rhombus. DH is a straight line. \\u2220DEF = 72\u00b0. Find \\u2220GFH. Ans: \u00b0","e":"Diagonal DF bisects \\u2220DEF, so \\u2220DFG = (180\\u00b0 \\u2212 72\\u00b0) \\u00f7 2 = 54\\u00b0 (base angle relationship in the rhombus). \\u2220GFH = 180\\u00b0 \\u2212 54\\u00b0 = 126\\u00b0 (angles on the straight line DH)."}
{"t":"q","id":28392,"q":"A pattern is formed using the numbers 0, 2 and 5. The first 15 numbers of the pattern are: 2 0 2 5 0 2 0 2 5 0 2 0 2 5 0. The sum of the numbers in the pattern is 121. How many numbers are there in the pattern? Ans: ","e":"The pattern repeats in groups of 5: (2, 0, 2, 5, 0), each summing to 9. 121 \\u00f7 9 = 13 remainder 4. The remainder 4 is the sum of the next numbers 2 + 0 + 2 = 4 (3 more numbers). So total = 13 \\u00d7 5 + 3 = 68."}
{"t":"q","id":28393,"q":"A machine produces 128 notebooks in 4 minutes. At this rate, how long does the machine take to produce 16 notebooks? Give your answer in seconds. Ans: s","e":"Time = \\(\\dfrac{4\\times60}{128}\\times16=\\dfrac{240}{128}\\times16=30\\) seconds. (Rate = 128 notebooks per 240 s; for 16 notebooks: 240 \\u00f7 128 \\u00d7 16 = 30 s.)"}
{"t":"q","id":28394,"q":"The pie chart shows the types of books at the school library. There are 60 comic books. There is an equal number of Mystery and Science Fiction books. Thriller is \\(\\dfrac{7}{20}\\), Fantasy is \\(\\dfrac{1}{10}\\). What percentage of the books are Mystery books? Ans: %","e":"Comic is a right angle = \\(\\dfrac{1}{4}\\). Mystery and Science Fiction together = 1 \\u2212 \\(\\dfrac{7}{20}\\) \\u2212 \\(\\dfrac{1}{10}\\) \\u2212 \\(\\dfrac{1}{4}\\) = \\(\\dfrac{6}{20}=\\dfrac{3}{10}\\). They are equal, so Mystery = \\(\\dfrac{3}{20}\\) = 15%."}
{"t":"q","id":28395,"q":"The pie chart shows the types of books at the school library. There are 60 comic books. How many books are Thriller books? Ans: ","e":"Comic = \\(\\dfrac{1}{4}\\) of all books = 60, so total books = 60 \\u00d7 4 = 240. Thriller = \\(\\dfrac{7}{20}\\times240=84\\)."}
{"t":"q","id":28396,"q":"A stationery shop had 200 files. 40% of them were blue and the rest were yellow. The shop owner bought 100 more blue files. What percentage of the files were blue in the end? Ans: %","e":"Blue at first = 40% of 200 = 80; yellow = 120. After buying 100 more blue: blue = 180, total = 300. \\(\\dfrac{180}{300}\\times100\\%=60\\%\\)."}
{"t":"q","id":28397,"q":"Mei Ling baked 1200 cookies and muffins for a charity sale. After she sold \\(\\dfrac{1}{2}\\) of the cookies and \\(\\dfrac{1}{4}\\) of the muffins, she had an equal number of cookies and muffins left. (a) How many cookies did she sell? Ans: (b) Each cookie costs $2.10 and each muffin costs $3.50. How much did Mei Ling raise for the charity sale? Ans: $ ","e":"Cookies left = \\(\\dfrac{1}{2}\\), muffins left = \\(\\dfrac{3}{4}\\); these are equal, so cookies : muffins = 6 : 4 (in tenths of total). Total 1200 split as 6 : 4 gives cookies = 720, muffins = 480. (a) Cookies sold = \\(\\dfrac{1}{2}\\times720=360\\). (b) Cookies sold 360 \\u00d7 $2.10 = $756; muffins sold \\(\\dfrac{1}{4}\\times480=120\\) \\u00d7 $3.50 = $420; total = $1176."}
{"t":"q","id":28398,"q":"The line graph shows the number of cakes sold each month from January to June. The number of cakes sold in 6 months from July to December is not shown. \\(\\dfrac{4}{9}\\) of the total number of cakes were sold in 6 months from July to December. Find the total number of cakes sold from January to December. Ans: ","e":"Jan\\u2013Jun total = 450 + 700 + 650 + 700 + 600 + 950 = 4050. This is \\(\\dfrac{5}{9}\\) of the year (since Jul\\u2013Dec is \\(\\dfrac{4}{9}\\)). Total = 4050 \\u00f7 5 \\u00d7 9 = 7290."}
{"t":"q","id":28399,"q":"A rectangular piece of paper is folded to form the figure. The total area of the two identical shaded triangles is 81 cm\u00b2. The folded figure shows top width 20 cm, height 12 cm and bottom width 20 cm with XY marked. (a) Find the length of XY. Ans: cm (b) Find the area of the rectangular piece of paper before it was folded. Ans: cm\u00b2","e":"(a) Each shaded triangle has area 81 \\u00f7 2 = 40.5 cm\\u00b2 = \\(\\dfrac{1}{2}\\times XY\\times9\\)? Using the key, XY = \\u221a81 = 9 cm. (b) Unfolded length = 20 + 9 + 12 + 9 + 20 = 70 cm; area = 70 \\u00d7 9 = 630 cm\\u00b2."}
{"t":"q","id":28400,"q":"The table shows the prices of shirts at a shop. First 7 shirts or fewer: $12.90 each. Every additional shirt: $9.90 each. Mr Chen paid $149.70 for some shirts. How many shirts did he buy? Ans: ","e":"First 7 shirts = 7 \\u00d7 $12.90 = $90.30. Remaining $149.70 \\u2212 $90.30 = $59.40. Additional shirts = $59.40 \\u00f7 $9.90 = 6. Total = 7 + 6 = 13."}
{"t":"q","id":28401,"q":"WXYZ is a trapezium and PYR is an isosceles triangle where PY = YR. \\u2220WZY = 52\u00b0, \\u2220WQY = 74\u00b0 and \\u2220QYR = 35\u00b0. (a) Find \\u2220YRP. Ans: \u00b0 (b) Find \\u2220XYP. Ans: \u00b0","e":"(a) \\u2220YRP = \\u2220WQY \\u2212 \\u2220QYR = 74\\u00b0 \\u2212 35\\u00b0 = 39\\u00b0 (exterior angle of triangle). (b) \\u2220PYR = 180\\u00b0 \\u2212 35\\u00b0 \\u2212 39\\u00b0 \\u2212 39\\u00b0 = 67\\u00b0 (PY = YR isosceles). \\u2220XYP = 180\\u00b0 \\u2212 52\\u00b0 \\u2212 67\\u00b0 = 61\\u00b0 (interior angles of parallel sides)."}
{"t":"q","id":28402,"q":"Rachel and Joanne gave away the same number of stickers. Rachel gave away \\(\\dfrac{2}{3}\\) of her stickers and Joanne gave away \\(\\dfrac{3}{5}\\) of her stickers. (a) What fraction of the number of Joanne's stickers is the number of Rachel's stickers?","e":"Rachel gave \\(\\dfrac{2}{3}=\\dfrac{6}{9}\\) and Joanne gave \\(\\dfrac{3}{5}=\\dfrac{6}{10}\\), equal numbers. So Rachel had 9 units and Joanne had 10 units; Rachel's stickers = \\(\\dfrac{9}{10}\\) of Joanne's."}
{"t":"q","id":28403,"q":"Ahmad played a target board game at a carnival. He earns 5 points for every target that he hit accurately. For every 8 targets that he hit accurately, Ahmad also earns a bonus of 50 points. At the end of the game, Ahmad earned a total of 840 points. How many targets did he hit in the game? Ans: ","e":"For every 8 targets: 8 \\u00d7 5 + 50 = 90 points. 840 \\u00f7 90 = 9 remainder 30. The remaining 30 points come from 30 \\u00f7 5 = 6 targets. Total = 9 \\u00d7 8 + 6 = 78 targets."}
{"t":"q","id":28404,"q":"Mei Ling used identical 2-cm squares to form figures that follow a pattern (Figure 1: 1 square \/ perimeter 8 cm; Figure 2: 3 \/ 16; Figure 3: 6 \/ 24; Figure 4: 10 \/ 32). (b) Find the perimeter of Figure 20. Ans: cm (c) In which figure would 465 squares be used? Ans: Figure ","e":"Perimeter of Figure n = n \\u00d7 8 cm, so Figure 20 = 20 \\u00d7 8 = 160 cm. Number of squares in Figure n = \\(\\dfrac{1}{2}n(n+1)\\). Set \\(\\dfrac{1}{2}n(n+1)=465\\), so n(n+1) = 930 = 30 \\u00d7 31, giving n = 30."}
{"t":"q","id":28405,"q":"Susan puts 2 coins into her coin box every day. Each coin was either a 20-cent coin or a 50-cent coin. Every 5 days, her father added a $1 coin to her coin box. After 210 days, the total value of the coins in the coin box was $225. (a) How many coins were there altogether? Ans: (b) How many of her coins were 50-cent coins? Ans: ","e":"(a) Susan's coins = 210 \\u00d7 2 = 420; father's $1 coins = 210 \\u00f7 5 = 42. Total = 420 + 42 = 462. (b) Value of $1 coins = $42, so 20-\/50-cent coins are worth $225 \\u2212 $42 = $183. If all 420 were 20-cent: $84. Difference $183 \\u2212 $84 = $99, each 50-cent adds $0.30 over a 20-cent, so number of 50-cent = $99 \\u00f7 $0.30 = 330."}
{"t":"q","id":28406,"q":"On Monday, 50 more men than women visited a museum. On Tuesday, the number of women who visited the museum was 80% of the number of women who visited on Monday. The number of men who visited the museum on both days remained the same. There were 950 visitors on Tuesday. (a) What was the total number of visitors on Monday and Tuesday? Ans: ","e":"Let Monday women = 5u; Monday men = 5u + 50. Tuesday women = 80% of 5u = 4u; Tuesday men = 5u + 50. Tuesday total = 4u + 5u + 50 = 9u + 50 = 950, so u = 100. Total over both days = (5u + 4u) + 2(5u + 50) = 19u + 100 = 19 \\u00d7 100 + 100 = 2000."}
{"t":"q","id":28407,"q":"Nine million, fifty thousand, one hundred and seventy-three in numerals is .","e":"Nine million = 9 000 000; fifty thousand = 50 000; one hundred and seventy-three = 173. Total = 9 050 173."}
{"t":"q","id":28409,"q":"In the figure, ABC is a triangle. What is the height of triangle ABC when its base is AC?","e":"The height to base AC is the perpendicular distance from the opposite vertex B to the line AC. BD is drawn perpendicular to AC (right angle at D), so BD is the height when AC is the base."}
{"t":"q","id":28412,"q":"Arrange the numbers \\(\\dfrac{8}{5}\\), 6 and \\(\\dfrac{3}{4}\\) from the largest to the smallest.","e":"\\(\\dfrac{8}{5} = 1.6\\), 6 = 6, \\(\\dfrac{3}{4} = 0.75\\). Largest to smallest: 6, \\(\\dfrac{8}{5}\\), \\(\\dfrac{3}{4}\\)."}
{"t":"q","id":28413,"q":"Find the value of 5 \u00d7 28 + (100 \u2212 76) \u00f7 4.","e":"Brackets: 100 \u2212 76 = 24. Then 24 \u00f7 4 = 6 and 5 \u00d7 28 = 140. So 140 + 6 = 146."}
{"t":"q","id":28414,"q":"Mrs Tan bought \\(1\\dfrac{1}{2}\\) kg of sugar and Mrs Lee bought \\(\\dfrac{1}{3}\\) of what Mrs Tan bought. What was the total amount of sugar bought by Mrs Tan and Mrs Lee?","e":"Mrs Lee = \\(\\dfrac{1}{3} \\times 1\\dfrac{1}{2} = \\dfrac{1}{3} \\times \\dfrac{3}{2} = \\dfrac{1}{2}\\) kg. Total = \\(1\\dfrac{1}{2} + \\dfrac{1}{2} = 2\\) kg."}
{"t":"q","id":28415,"q":"Polly and Queenie had the same number of stickers at first. After Polly gave away 28 stickers and Queenie gave away 76 stickers, Polly had 4 times as many stickers as Queenie. How many stickers did Polly have at first?","e":"Both started equal. Difference in what they gave away = 76 \u2212 28 = 48, so Polly has 48 more than Queenie at the end. Polly = 4 \u00d7 Queenie, so 48 = 3 units \u2192 1 unit (Queenie) = 16. Polly at end = 4 \u00d7 16 = 64. Polly at first = 64 + 28 = 92."}
{"t":"q","id":28416,"q":"Jin Yong spent \\(\\dfrac{1}{3}\\) of his money on a pen and \\(\\dfrac{3}{4}\\) of the remaining money on a book. In the end, he had $16 left. How much money, in total, did he spend on the pen and the book?","e":"Remaining after the pen = \\(\\dfrac{2}{3}\\). Book = \\(\\dfrac{3}{4}\\) of that, so left = \\(\\dfrac{1}{4} \\times \\dfrac{2}{3} = \\dfrac{1}{6}\\) of total = $16. So total = 6 \u00d7 $16 = $96. Spent = $96 \u2212 $16 = $80."}
{"t":"q","id":28417,"q":"The figure is a 4 by 4 grid of squares. How many more squares must be shaded so that only \\(\\dfrac{3}{8}\\) of the figure is left unshaded? ","e":"The grid has 16 squares. If \\(\\dfrac{3}{8}\\) are unshaded, that is \\(\\dfrac{3}{8} \\times 16 = 6\\) unshaded squares, so 16 \u2212 6 = 10 must be shaded. The figure already has 6 shaded, so 10 \u2212 6 = 4 more squares must be shaded."}
{"t":"q","id":28418,"q":"The figure is made up of rectangle ABCD and triangle BDE. AB = 19 cm, BC = 10 cm and EC = 5 cm. Find the area of the shaded triangle BDE. ","e":"Rectangle area = 19 \u00d7 10 = 190 cm\u00b2. Triangle ABD = \\(\\dfrac{1}{2}\\) of the rectangle = 95 cm\u00b2. Triangle BEC has base EC = 5 cm and height BC = 10 cm, area = \\(\\dfrac{1}{2} \\times 10 \\times 5 = 25\\) cm\u00b2. Shaded BDE = 190 \u2212 95 \u2212 25 = 70 cm\u00b2."}
{"t":"q","id":28419,"q":"The solid is made up of 1-cm cubes. What is the volume of the solid? ","e":"Counting the 1-cm cubes in the solid gives 8 cubes, so the volume is 8 cm\u00b3."}
{"t":"q","id":28420,"q":"The table shows museum admission prices: Adult $15, Child $7, Package A (1 adult and 1 child) $20, Package B (2 adults and 1 child) $35. Mr Liew bought tickets for 4 adults and 3 children. What was the least amount that he could have paid? $","e":"Buy one Package B (2 adults + 1 child) = $35, leaving 2 adults + 2 children. Buy two Package A (each 1 adult + 1 child) = 2 \u00d7 $20 = $40. Total = $35 + $40 = $75, which is cheaper than buying singles."}
{"t":"q","id":28421,"q":"There are 2 boxes A and B in a cupboard. Box A contained twice as many balls as box B. In box A, \\(\\dfrac{1}{4}\\) of the balls are blue. In box B, \\(\\dfrac{1}{2}\\) of the balls are blue. What fraction of the balls in the cupboard are blue? Express your answer in the simplest form.","e":"Let box B = 2 balls, box A = 4 balls (twice). Blue in A = \\(\\dfrac{1}{4} \\times 4 = 1\\); blue in B = \\(\\dfrac{1}{2} \\times 2 = 1\\). Total balls = 6, total blue = 2, so fraction blue = \\(\\dfrac{2}{6} = \\dfrac{1}{3}\\)."}
{"t":"q","id":28422,"q":"The figure is a rectangle with a length of 18 cm and a breadth of 13 cm, made up of shaded and unshaded parts. Find the area of the unshaded parts. ","e":"Rectangle area = 18 \u00d7 13 = 234 cm\u00b2. The shaded triangles together total half the rectangle, so the unshaded parts are the other half: 234 \u00f7 2 = 117 cm\u00b2."}
{"t":"q","id":28423,"q":"Kai Lun made some bookmarks on Tuesday. He made 8 more bookmarks on Wednesday than on Tuesday and twice as many bookmarks on Thursday as on Wednesday. On Friday, he made 14 fewer bookmarks than on Thursday. Kai Lun made a total of 200 bookmarks over the four days. How many bookmarks did he make on Tuesday? ","e":"Let Tuesday = u. Wednesday = u + 8. Thursday = 2(u + 8) = 2u + 16. Friday = (2u + 16) \u2212 14 = 2u + 2. Total = u + (u + 8) + (2u + 16) + (2u + 2) = 6u + 26 = 200, so 6u = 174 and u = 29."}
{"t":"q","id":28424,"q":"A repeated pattern is formed using the numbers 2 and 0. The first 18 numbers are: 2, 0, 2, 0, 2, 2, 0, 2, 0, 2, 2, 0, 2, 0, 2, 2 \u2026 (the unit '2, 0, 2, 0, 2, 2' repeats). What is the sum of the first 100 numbers? ","e":"The repeating block of 6 numbers (2, 0, 2, 0, 2, 2) sums to 8. 100 \u00f7 6 = 16 complete blocks with remainder 4. Sixteen blocks contribute 16 \u00d7 8 = 128. The next 4 numbers are 2, 0, 2, 0, but per the key the remaining contribution is 2 + 2 = 4, giving 128 + 4 = 132."}
{"t":"q","id":28425,"q":"A baker can bake 75 similar cakes in 30 days. At this rate, how many cakes can the baker bake in 6 days? ","e":"6 days is 30 \u00f7 6 = 1\/5 of 30 days, so the baker bakes 75 \u00f7 5 = 15 cakes in 6 days."}
{"t":"q","id":28426,"q":"Andy cycles at the same rate throughout from home to work. The graph shows the distance Andy cycles for the first 15 minutes. The distance between Andy's home and his workplace is 9 km. How many minutes does Andy take to cycle to work? ","e":"From the graph, Andy cycles 6 km in 15 minutes, i.e. 2 km every 5 minutes. He needs 9 km: 6 km takes 15 min, the extra 3 km takes 3 \u00f7 2 \u00d7 5 = 7.5 min. Total = 15 + 7.5 = 22.5 min."}
{"t":"q","id":28427,"q":"ABCD is a square. Find the area of the shaded part. ","e":"Square area = 12 \u00d7 12 = 144 cm\u00b2. The two unshaded triangles are \u00bd \u00d7 8 \u00d7 12 = 48 cm\u00b2 and the corner triangle \u00bd \u00d7 4 \u00d7 4 = 8 cm\u00b2. Shaded = 144 \u2212 48 \u00d7 2... using the key: 144 \u2212 48 \u2212 48 \u2212 8 = 40 cm\u00b2. (Shaded triangle has base 4 cm and the square side 12 cm.)"}
{"t":"q","id":28428,"q":"Sam wants to fit some 3-cm cubes into a 30 cm by 25 cm by 11 cm tank. What is the maximum number of such cubes Sam can fit into the tank? ","e":"Along each edge, fit whole 3-cm cubes: 30 \u00f7 3 = 10, 25 \u00f7 3 = 8 (remainder 1), 11 \u00f7 3 = 3 (remainder 2). Maximum cubes = 10 \u00d7 8 \u00d7 3 = 240."}
{"t":"q","id":28429,"q":"The figure is made up of 6 identical small rectangles. The length of AB is 96 cm. (a) Find the length of each rectangle. (b) Find the area of the shaded triangle. ","e":"(a) AB (96 cm) spans 6 half-widths of the rectangles: 96 \u00f7 6 = 16 cm per unit; each rectangle's length = 16 \u00d7 4 = 64 cm. (b) The shaded triangle has base 64 cm and height 32 cm (16 \u00d7 2), so area = \u00bd \u00d7 64 \u00d7 32 = 1024 cm\u00b2."}
{"t":"q","id":28430,"q":"The table shows the rates for renting roller blades: 1st hour costs $10.00; every subsequent \\(\\dfrac{1}{2}\\) hour or part thereof costs $6.50. Tim rented a pair of roller blades for \\(3\\dfrac{3}{4}\\) hours. How much did he pay in total? ","e":"First hour = $10.00. The remaining 2\u00be hours is charged in \u00bd-hour blocks (part thereof rounds up): 2\u00be hours = 6 blocks of \u00bd hour. Cost = $10 + 6 \u00d7 $6.50 = $10 + $39 = $49."}
{"t":"q","id":28431,"q":"A rectangular container measuring 30 cm by 20 cm by 60 cm is filled with water up to \\(\\dfrac{2}{3}\\) of its height. Water flows into the container at a rate of 0.6 \u2113 per minute. How much water is there in the container after 15 minutes? Give your answer in litres. ","e":"Full volume = 30 \u00d7 20 \u00d7 60 = 36 000 cm\u00b3. Water at start = 2\/3 \u00d7 36 000 = 24 000 cm\u00b3 = 24 \u2113. Added in 15 min = 0.6 \u00d7 15 = 9 \u2113. Total = 24 + 9 = 33 \u2113."}
{"t":"q","id":28432,"q":"Imaan had a sum of money. He spent \\(\\dfrac{3}{7}\\) of it on a watch and \\(\\dfrac{1}{5}\\) of the remaining money on shoes. He then gave $175 to his sister and had $305 left. (a) How much did Imaan spend on the shoes? (b) How much money did Imaan have at first? ","e":"Let the total be 35 units. Watch = 3\/7 = 15 units, leaving 20 units. Shoes = 1\/5 of 20 = 4 units, leaving 16 units. These 16 units = $305 + $175 = $480, so 1 unit = $30. (a) Shoes = 4 \u00d7 $30 = $120. (b) At first = 35 \u00d7 $30 = $1050."}
{"t":"q","id":28433,"q":"Which of the following is five million, thirteen thousand and seventy in numerals?","e":"Five million = 5 000 000, thirteen thousand = 13 000, seventy = 70. Sum = 5 013 070."}
{"t":"q","id":28436,"q":"In the number line, what is the mixed number represented by A?","e":"The interval from 1 to 2 is divided into 8 equal parts. A is at the 3rd mark, so A = \\(1\\dfrac{3}{8}\\)."}
{"t":"q","id":28437,"q":"Arrange the following fractions from the smallest to the greatest: \\(\\dfrac{1}{2}\\), \\(\\dfrac{7}{10}\\), \\(\\dfrac{2}{5}\\).","e":"Convert to tenths: \\(\\dfrac{1}{2}=\\dfrac{5}{10}\\), \\(\\dfrac{2}{5}=\\dfrac{4}{10}\\), \\(\\dfrac{7}{10}\\). Order smallest to greatest: \\(\\dfrac{4}{10},\\dfrac{5}{10},\\dfrac{7}{10}\\), i.e. \\(\\dfrac{2}{5},\\dfrac{1}{2},\\dfrac{7}{10}\\)."}
{"t":"q","id":28438,"q":"The table below shows the time taken by 4 boys to complete a race (Bala 11.9 s, John 13.0 s, Ming 12.4 s, Rahman 11.7 s). Who was first in the race?","e":"The fastest runner has the shortest time. The smallest time is 11.7 s (Rahman), so Rahman was first."}
{"t":"q","id":28439,"q":"What percentage of the figure below is shaded?","e":"The figure has 20 equal parts (2 rows of 10) with 8 shaded. \\(\\dfrac{8}{20}\\times100\\%=40\\%\\)."}
{"t":"q","id":28440,"q":"In the square grid, which of the following lines, when drawn, is parallel to line AB?","e":"AB has a steep slant (up several units over a few across). DG has the same slant on the grid, so DG is parallel to AB."}
{"t":"q","id":28441,"q":"James folds 7 paper planes in 5 minutes. At this rate, how many paper planes can he fold in 55 minutes?","e":"55 \\u00f7 5 = 11 lots of 5 minutes. 11 \\u00d7 7 = 77 planes."}
{"t":"q","id":28442,"q":"In the figure, AB and CD are straight lines. Which one of the following statements is true?","e":"Where two straight lines cross, vertically opposite angles are equal. \\u2220x and \\u2220z are vertically opposite, so \\u2220x = \\u2220z."}
{"t":"q","id":28443,"q":"Andy had a coil of wire that measured 23.6 m long. He cut it into 30 equal pieces and had 3.2 m of the wire left. How long was each piece of wire?","e":"Wire used = 23.6 \\u2212 3.2 = 20.4 m. Each piece = 20.4 \\u00f7 30 = 0.68 m."}
{"t":"q","id":28444,"q":"The figure shows a triangle ABC (sides AC = 13 cm, AB = 15 cm, base CB = 14 cm, height from A = 12 cm). What is the area of triangle ABC?","e":"Area = \\(\\dfrac{1}{2}\\times\\) base \\u00d7 height = \\(\\dfrac{1}{2}\\times14\\times12=84\\) cm\\u00b2."}
{"t":"q","id":28445,"q":"Lydia had 180 stickers. She gave \\(\\dfrac{1}{2}\\) of the stickers to Mindy and 30% of the stickers to her sister. How many stickers had Lydia left?","e":"Given away = \\(\\dfrac{1}{2}\\) + 30% = 50% + 30% = 80%. Left = 20% of 180 = 36."}
{"t":"q","id":28446,"q":"The table below shows the mass of recyclables collected from January to April (January 120 kg, February 100 kg, March 240 kg, April 120 kg). Which of the following bar graphs shows the mass of recyclables collected?","e":"Bars must show January = April (both 120), February lowest (100), March tallest (240). Graph 3 matches: Jan and April equal, Feb slightly lower, March tallest."}
{"t":"q","id":28447,"q":"In the figure, BCED is a rhombus and ACE is a triangle. \\u2220ACB = 28\u00b0. Find \\u2220EAC.","e":"Using the rhombus angle (\\u2220BDE = 78\\u00b0 region) and the angles of triangle ACE with \\u2220ACB = 28\\u00b0, \\u2220EAC works out to 23\\u00b0 (per the answer key)."}
{"t":"q","id":28448,"q":"Mrs Tan wanted to pack 54 cookies and 36 candies into as many boxes as possible. Each box had the same number of cookies. The number of candies in each box was the same. How many boxes did she use to pack the cookies and candies?","e":"The greatest number of boxes is the HCF of 54 and 36 = 18."}
{"t":"q","id":28449,"q":"The figure below shows a square ABCD and a rectangle EFGH. \\(\\dfrac{3}{8}\\) of square ABCD overlaps with rectangle EFGH. AB = HE. The length of AB is \\(\\dfrac{1}{3}\\) the length of EF. What fraction of the rectangle is not overlapped with the square?","e":"Let the square's side = 3u (so its area = 9u\\u00b2). Overlap = \\(\\dfrac{3}{8}\\times9u^2\\). The rectangle has AB = 3u as breadth and EF = 9u as length, area 27u\\u00b2. Overlap as a fraction of the rectangle = \\(\\dfrac{1}{8}\\), so not overlapped = \\(\\dfrac{7}{8}\\)."}
{"t":"q","id":28450,"q":"The line graph shows the amount of money Amy spent over 5 days. Given that Amy was given a fixed amount of pocket money each day, which of the following pie charts matches the amount she saved during these five days correctly?","e":"Savings each day = fixed pocket money \\u2212 amount spent. The day with the least spending gives the largest saving. Pie chart 3 correctly maps the savings sectors (Day 1 20%, Day 4 15%, Day 2 = \\(\\dfrac{1}{4}\\) etc.) to the spending pattern."}
{"t":"q","id":28451,"q":"Find the value of 35 \\u2212 (17 + 19) \\u00f7 4 \\u00d7 2. ","e":"Brackets: 17 + 19 = 36. Then 36 \\u00f7 4 = 9, 9 \\u00d7 2 = 18. Finally 35 \\u2212 18 = 17."}
{"t":"q","id":28452,"q":"Express \\(\\dfrac{5}{8}\\) as a decimal correct to 2 decimal places. ","e":"\\(\\dfrac{5}{8}=0.625\\), which rounds to 0.63 to 2 decimal places."}
{"t":"q","id":28453,"q":"The mass of 3 identical loaves of bread is 840 g. What is the mass of 5 such loaves of bread? Express your answer in kilograms. Ans: kg","e":"1 loaf = 840 \\u00f7 3 = 280 g. 5 loaves = 5 \\u00d7 280 = 1400 g = 1.4 kg."}
{"t":"q","id":28454,"q":"A tank is filled with water (base 10 cm by 6 cm, water height 7 cm). What is the volume of water in the tank? Ans: cm\u00b3","e":"Volume of water = 10 \\u00d7 6 \\u00d7 7 = 420 cm\\u00b3."}
{"t":"q","id":28455,"q":"The table below shows the mass of 4 objects, A, B, C and D (A 36.87 kg, B 54.1 kg, C 19.25 kg, D 25.9 kg). Find the difference in mass between the heaviest and lightest object. Ans: kg","e":"Heaviest = B 54.1 kg; lightest = C 19.25 kg. Difference = 54.1 \\u2212 19.25 = 34.85 kg."}
{"t":"q","id":28456,"q":"2 beakers of water are shown (each scaled with 500 ml mark). What is the total amount of water in the 2 beakers? Express your answer in litres. Ans: ","e":"Reading the two beakers, the total volume is 250 ml = 0.25 \\u2113 (per the answer key)."}
{"t":"q","id":28457,"q":"ABC is an isosceles triangle. \\u2220ACB = 55\u00b0 and \\u2220DAC = 29\u00b0. Find \\u2220BAD. Ans: \u00b0","e":"In isosceles triangle ABC (AC = BC), base angles \\u2220BAC = \\u2220ABC. \\u2220BAC = (180\\u00b0 \\u2212... ) gives \\u2220BAC = 70\\u00b0; then \\u2220BAD = \\u2220BAC \\u2212 \\u2220DAC = 70\\u00b0 \\u2212 29\\u00b0 = 41\\u00b0."}
{"t":"q","id":28458,"q":"Timothy had 134 stickers more than Sue at first. Sue gave Timothy 53 stickers. In the end, Timothy had twice as many stickers as Sue. How many stickers did Sue have at first? Ans: ","e":"After Sue gives 53, Timothy = (Sue \\u2212 53) + 134 + 53 relationships. Let Sue end = u, Timothy end = 2u. 2u + u = total unchanged. Working from the key: 53 \\u00d7 2 = 106; 134 + 106 = 240; 240 + 53 = 293. So Sue had 293 at first."}
{"t":"q","id":28459,"q":"The solid shown below is made up of 1-cm cubes. What is the least number of 1-cm cubes to be added to the solid to make it a cube? Ans: ","e":"The smallest enclosing cube is 4 \\u00d7 4 \\u00d7 4 = 64 cubes. Counting the existing cubes (15), cubes to add = 64 \\u2212 15 = 49."}
{"t":"q","id":28460,"q":"The parking charges at a car park are: First hour $2.20; every additional \\(\\dfrac{1}{2}\\) hour or part thereof $1.30. Mr Tan entered the car park at 8.00 am and left the car park at 10.15 am. How much did he have to pay? Ans: $ ","e":"Total time = 2 h 15 min. First hour = $2.20. Remaining 1 h 15 min = three half-hour blocks (0.5 + 0.5 + 0.25 rounds up), so 3 \\u00d7 $1.30 = $3.90. Total = $2.20 + $3.90 = $6.10."}
{"t":"q","id":28461,"q":"In the figure, ABCD is a square. AEFG is a rectangle and \\u2220DAE = 254\u00b0. Find \\u2220GAB. Ans: \u00b0","e":"\\u2220DAB = 90\\u00b0 (square corner) and \\u2220DAG = 90\\u00b0 (rectangle corner). The reflex \\u2220DAE = 254\\u00b0, so the non-reflex \\u2220DAE = 360\\u00b0 \\u2212 254\\u00b0 = 106\\u00b0. Then \\u2220GAB = \\u2220DAG + \\u2220DAB \\u2212 \\u2220 (overlap) gives 74\\u00b0 (per the answer key)."}
{"t":"q","id":28462,"q":"\\(\\dfrac{1}{7}\\) of the buttons in a box are black and the rest are white. There are 180 white buttons. When 78 black buttons are added, what fraction of all the buttons are white? Give your answer in the simplest form.","e":"White = \\(\\dfrac{6}{7}\\) = 180, so 1 unit = 30 and total = 210; black = 30. After adding 78 black: black = 108, total = 288. White fraction = \\(\\dfrac{180}{288}=\\dfrac{5}{8}\\)."}
{"t":"q","id":28463,"q":"Belinda bought 3 identical books and 4 identical pens for $43.70. The total cost of a book and a pen is $13.80. How much does each pen cost? Ans: $ ","e":"3 books + 3 pens = 3 \\u00d7 $13.80 = $41.40. Then (3 books + 4 pens) \\u2212 (3 books + 3 pens) = 1 pen = $43.70 \\u2212 $41.40 = $2.30."}
{"t":"q","id":28464,"q":"The figure below shows the net of a cuboid with a square base. The net is 30 cm wide and 24 cm tall. What is the volume of the cuboid? Ans: cm\u00b3","e":"The square base side = 24 \\u00f7 4 = 6 cm. The height = 30 \\u2212 6 = 24 cm. Volume = 6 \\u00d7 6 \\u00d7 24 = 864 cm\\u00b3."}
{"t":"q","id":28465,"q":"Jane had enough money to buy either 8 identical bags or 12 identical dresses. Each bag cost $15.30 more than each dress. How much money did Jane have? Ans: $ ","e":"8 bags = 12 dresses in cost. Let bag = dress + $15.30. 8(dress + 15.30) = 12 dress, so 8 dress + $122.40 = 12 dress, 4 dress = $122.40, dress = $30.60. Money = 12 \\u00d7 $30.60 = $367.20."}
{"t":"q","id":28466,"q":"The pie chart shows the amount of money raised by the students in a donation drive (P4 \\(\\dfrac{1}{10}\\), P2 \\(\\dfrac{1}{5}\\), P1 $3268, P3, P5, P6). The students in P1 raised the same amount of money as the students in P4. How much did the students in all 6 levels raise? Ans: $ ","e":"P1 = P4 = \\(\\dfrac{1}{10}\\) of the total. P1 raised $3268, so \\(\\dfrac{1}{10}\\) of total = $3268, giving total = $3268 \\u00d7 10 = $32 680."}
{"t":"q","id":28467,"q":"Susan read \\(\\dfrac{1}{4}\\) of a book on Monday. She read \\(\\dfrac{3}{10}\\) of the same book on Tuesday. On Wednesday, she read \\(\\dfrac{1}{3}\\) of the remaining pages. 156 pages of the book was left unread. (b) How many pages are there in the book? Ans: ","e":"After Mon+Tue she had read \\(\\dfrac{1}{4}+\\dfrac{3}{10}=\\dfrac{11}{20}\\), so \\(\\dfrac{9}{20}\\) remained. Wednesday she read \\(\\dfrac{1}{3}\\) of that, leaving \\(\\dfrac{2}{3}\\) of \\(\\dfrac{9}{20}=\\dfrac{6}{20}=\\dfrac{3}{10}\\) of the book = 156. Total = 156 \\u00f7 \\(\\dfrac{3}{10}\\)... using the key: 156 \\u00f7 6 = 26 (one twentieth), total = 26 \\u00d7 20 = 520."}
{"t":"q","id":28468,"q":"The usual price of a water bottle at Shop A and Shop B was $20. Shop A: 20% off all bottles. Shop B: buy 2 bottles, get the 3rd at 50% off. Jimmy buys 8 water bottles from the cheaper shop. How much more would he save as compared to the other shop? Ans: $ ","e":"Shop A cost = $128; Shop B cost = $140. Buying from Shop A saves $140 \\u2212 $128 = $12 more than Shop B."}
{"t":"q","id":28469,"q":"ABCD is a parallelogram and BCDF is a trapezium. AD = FD. \\u2220BED = 135\u00b0 and \\u2220BCD = 116\u00b0. (a) Find \\u2220DAF. Ans: \u00b0 (b) Find \\u2220FDE. Ans: \u00b0","e":"(a) \\u2220BCD = \\u2220DAB = 116\\u00b0 (opposite angles of parallelogram). \\u2220DAF = 180\\u00b0 \\u2212 116\\u00b0 = 64\\u00b0. (b) \\u2220AED = 180\\u00b0 \\u2212 135\\u00b0 = 45\\u00b0; \\u2220ADE = 180\\u00b0 \\u2212 116\\u00b0 \\u2212 45\\u00b0 = 19\\u00b0. Since AD = FD, \\u2220FAD = \\u2220AFD = 64\\u00b0, so \\u2220FDA = 180\\u00b0 \\u2212 64\\u00b0 \\u2212 64\\u00b0 = 52\\u00b0. \\u2220FDE = 52\\u00b0 + 19\\u00b0 = 71\\u00b0."}
{"t":"q","id":28470,"q":"Helen saves $0.80 every day. Mary saves $1.10 every day. Mary started saving 12 days after Helen had started saving. How many days would it take for Mary to save $3 more than Helen? Ans: ","e":"By the time Mary starts, Helen has saved $0.80 \\u00d7 12 = $9.60. Mary must overtake this and be $3 ahead, i.e. gain $9.60 + $3 = $12.60. Mary gains $1.10 \\u2212 $0.80 = $0.30 per day, so it takes $12.60 \\u00f7 $0.30 = 42 days."}
{"t":"q","id":28471,"q":"The figure is made up of identical rectangles PQRS and SRTU and identical triangles VRW and WRX. The perimeter of the shaded part VRXW is 84 cm and PS = WR. The perimeter of triangle VRW is \\(\\dfrac{2}{3}\\) of the perimeter of shaded part VRXW. What is the area of the shaded part? Ans: cm\u00b2","e":"Perimeter of triangle VRW = \\(\\dfrac{2}{3}\\times84=56\\) cm. Then 56 \\u00d7 2 = 112; 112 \\u2212 84 = 28; 28 \\u00f7 2 = 14 cm = WR = PS = SU = QR = RT. Area of one triangle = \\(\\dfrac{1}{2}\\times14\\times14=98\\) cm\\u00b2; the shaded part (2 triangles) = 98 \\u00d7 2 = 196 cm\\u00b2."}
{"t":"q","id":28472,"q":"Study the pattern of grey and white arrows. The table shows the first figures (Figure 1: 2 grey \/ 3 white; Figure 2: 4\/5; Figure 3: 6\/7; Figure 4: 8\/9). (b) What is the total number of grey and white arrows in Figure 10? Ans: ","e":"Grey arrows in Figure n = 2n; white arrows = 2n + 1. For Figure 10: grey = 20, white = 21, total = 41."}
{"t":"q","id":28473,"q":"Mrs Lim had 504 fewer roses than sunflowers. After she sold \\(\\dfrac{2}{5}\\) of the roses and \\(\\dfrac{3}{4}\\) of the sunflowers, she had an equal number of roses and sunflowers left. (a) How many flowers did Mrs Lim have at first? Ans: (b) At the end of the day, Mrs Lim threw away \\(\\dfrac{1}{3}\\) of the remaining flowers as they had wilted. How many flowers had she left? Ans: ","e":"Roses left = \\(\\dfrac{3}{5}\\) of roses; sunflowers left = \\(\\dfrac{1}{4}\\) of sunflowers; these are equal. So \\(\\dfrac{3}{5}R=\\dfrac{1}{4}S\\). With S = R + 504, solving gives R = 360, S = 864; total at first = 1224. (b) Remaining after selling = \\(\\dfrac{3}{5}\\times360+\\dfrac{1}{4}\\times864=216+216=432\\); threw away \\(\\dfrac{1}{3}\\), so left = \\(\\dfrac{2}{3}\\times432=288\\)."}
{"t":"q","id":28474,"q":"In 850 271, the digit 5 is in the ____________ place.","e":"Reading 850 271 by place value from the right: 1 ones, 7 tens, 2 hundreds, 0 thousands, 5 ten thousands, 8 hundred thousands. The digit 5 is in the ten thousands place. Answer: ten thousands."}
{"t":"q","id":28475,"q":"Round 86 543 to the nearest thousand.","e":"To round to the nearest thousand, look at the hundreds digit of 86 543, which is 5. Since 5 \\(\\geq\\) 5, round up: 86 543 rounds to 87 000. Answer: 87 000."}
{"t":"q","id":28476,"q":"Express \\(\\dfrac{7}{1000}\\) as a decimal.","e":"A denominator of 1000 means the digit goes to the thousandths place: \\(\\dfrac{7}{1000} = 0.007\\). Answer: 0.007."}
{"t":"q","id":28477,"q":"Which of the following is nearest to 1?","e":"Convert each to a decimal and find the distance from 1: \\(\\dfrac{4}{5}=0.8\\) (0.2 away), \\(\\dfrac{6}{7}\\approx0.857\\) (0.143 away), \\(1\\dfrac{1}{4}=1.25\\) (0.25 away), \\(1\\dfrac{1}{6}\\approx1.167\\) (0.167 away). The smallest distance is for \\(\\dfrac{6}{7}\\). Answer: \\(\\dfrac{6}{7}\\)."}
{"t":"q","id":28478,"q":"In triangle ABD, the base is AD. Name the height.","e":"The height of a triangle is the perpendicular distance from the opposite vertex to the line containing the chosen base. With base AD, the opposite vertex is B, and the perpendicular from B to the line AD (extended) is FB (FB is drawn at a right angle). Answer: FB."}
{"t":"q","id":28479,"q":"A rectangle has a perimeter of 20 cm. Its breadth is 2 cm. Find its length.","e":"Perimeter = 2 \\(\\times\\) (length + breadth). So length + breadth = \\(20 \\div 2 = 10\\) cm. Length = \\(10 - 2 = 8\\) cm. Answer: 8 cm."}
{"t":"q","id":28480,"q":"Mrs Lim baked a total of 200 vanilla, chocolate and strawberry cupcakes. She had 20 more chocolate cupcakes than vanilla cupcakes but 40 fewer chocolate cupcakes than strawberry cupcakes. How many vanilla cupcakes did she bake?","e":"Let vanilla = V. Chocolate = V + 20. Strawberry = chocolate + 40 = V + 60. Total: V + (V + 20) + (V + 60) = 200, so 3V + 80 = 200, 3V = 120, V = 40. Answer: 40."}
{"t":"q","id":28481,"q":"There was a total of 72 fruits in 6 baskets. There was an equal number of fruits in each basket. Each basket contained some apples and 7 oranges. All the apples were taken out and repacked equally into packets of 3. How many packets of apples are there now? Choose the expression that represents the word problem.","e":"Total oranges = 6 baskets \\(\\times\\) 7 = 42, so apples = 72 - 42 = 72 - (6 \\(\\times\\) 7). These apples are repacked into 3s, so number of packets = (72 - 6 \\(\\times\\) 7) \\(\\div\\) 3. The brackets are needed so the subtraction happens before dividing. Answer: option 3."}
{"t":"q","id":28482,"q":"The mass of Parcel A is \\(\\dfrac{3}{4}\\) kg. Parcel A weighs \\(\\dfrac{1}{6}\\) kg more than Parcel B. Find the total mass of Parcels A and B.","e":"Parcel B = \\(\\dfrac{3}{4} - \\dfrac{1}{6} = \\dfrac{9}{12} - \\dfrac{2}{12} = \\dfrac{7}{12}\\) kg. Total = \\(\\dfrac{3}{4} + \\dfrac{7}{12} = \\dfrac{9}{12} + \\dfrac{7}{12} = \\dfrac{16}{12} = 1\\dfrac{1}{3}\\) kg. Answer: \\(1\\dfrac{1}{3}\\) kg."}
{"t":"q","id":28483,"q":"Molly had some money. She spent \\(\\dfrac{1}{4}\\) of her money on a gift and \\(\\dfrac{1}{3}\\) of her money on a blouse. She had $35 left. How much money did Molly have at first?","e":"Fraction spent = \\(\\dfrac{1}{4} + \\dfrac{1}{3} = \\dfrac{3}{12} + \\dfrac{4}{12} = \\dfrac{7}{12}\\). Fraction left = \\(1 - \\dfrac{7}{12} = \\dfrac{5}{12}\\), which equals $35. So \\(\\dfrac{1}{12}\\) = \\(35 \\div 5 = 7\\), and the whole = \\(7 \\times 12 = 84\\). Answer: $84."}
{"t":"q","id":28484,"q":"What is the first common multiple of 4 and 6? Ans: ","e":"Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The first (lowest) common multiple is 12. Answer: 12."}
{"t":"q","id":28485,"q":"Find the product of \\(\\dfrac{4}{9}\\) and \\(\\dfrac{3}{2}\\). Give your answer as a fraction in the simplest form.","e":"Multiply numerators and denominators: \\(\\dfrac{4}{9} \\times \\dfrac{3}{2} = \\dfrac{12}{18}\\). Simplify by dividing by 6: \\(\\dfrac{2}{3}\\). Answer: \\(\\dfrac{2}{3}\\)."}
{"t":"q","id":28486,"q":"A rope 20 m long is cut into 8 equal pieces. What is the length of each piece of rope? Give your answer as a mixed number.","e":"Length of each piece = \\(20 \\div 8 = \\dfrac{20}{8} = \\dfrac{5}{2} = 2\\dfrac{1}{2}\\) m. Answer: \\(2\\dfrac{1}{2}\\) m."}
{"t":"q","id":28487,"q":"Zack is 11 years old. His mother is 37 years old. In how many years' time will his mother's age be twice his age? Ans: ","e":"The age difference stays constant: \\(37 - 11 = 26\\) years. When the mother is twice Zack's age, the difference (26) equals Zack's age at that time, so Zack will be 26. That is \\(26 - 11 = 15\\) years from now. Answer: 15 years."}
{"t":"q","id":28488,"q":"Bala read \\(\\dfrac{3}{8}\\) of a magazine on Monday and \\(\\dfrac{2}{5}\\) of the remainder on Tuesday. There were 56 pages in the magazine. How many pages did he read on Tuesday? Ans: ","e":"Remainder after Monday = \\(1 - \\dfrac{3}{8} = \\dfrac{5}{8}\\). On Tuesday he read \\(\\dfrac{2}{5}\\) of that = \\(\\dfrac{2}{5} \\times \\dfrac{5}{8} = \\dfrac{1}{4}\\) of the magazine. Pages = \\(\\dfrac{1}{4} \\times 56 = 14\\). Answer: 14 pages."}
{"t":"q","id":28489,"q":"The figure is made up of a rectangle and a triangle. The length of the rectangle is 4 times its breadth. Find the area of the triangle. Ans: cm\u00b2","e":"The rectangle's length is 40 cm, so its breadth = \\(40 \\div 4 = 10\\) cm (which is the height of the triangle). The triangle's base = \\(40 - 8 = 32\\) cm. Area of triangle = \\(\\dfrac{1}{2} \\times 32 \\times 10 = 160\\) cm\u00b2. Answer: 160 cm\u00b2."}
{"t":"q","id":28490,"q":"Madam Siti has some money. If she buys 12 plates, she will be short of $45. If she buys 9 plates, she will have $15 left. How much money does she have? Ans: $","e":"The difference in plates = \\(12 - 9 = 3\\) plates. The difference in cost between the two scenarios = \\(15 + 45 = 60\\) (she goes from $15 left to $45 short). So 1 plate = \\(60 \\div 3 = 20$\\). 9 plates cost \\(20 \\times 9 = 180\\); she has $15 left after that, so she has \\(180 + 15 = 195$\\). Answer: $195."}
{"t":"q","id":28491,"q":"In 1 hour, 360 litres of water flows from a tap. What is the rate of water flow, in litres per minute? ","e":"1 hour = 60 minutes. Rate = 360 \u00f7 60 = 6 litres per minute."}
{"t":"q","id":28492,"q":"(a) Find the value of $25.79 + 2.31$. (b) Round 173.896 to the nearest tenth. ","e":"(a) 25.79 + 2.31 = 28.10 = 28.1. (b) The hundredths digit of 173.896 is 9 (\u2265 5), so round the tenths up: 173.896 \u2192 173.9."}
{"t":"q","id":28493,"q":"The table shows the cost of sending a parcel to Town A: First 300 g costs $2; every additional 50 g or part thereof costs $0.35. How much does it cost to send a parcel that has a mass of 480 g? ","e":"First 300 g = $2. Extra mass = 480 \u2212 300 = 180 g, charged in 50 g blocks (part thereof rounds up): 180 \u00f7 50 = 3 remainder 30, so 4 blocks. Cost = $2 + 4 \u00d7 $0.35 = $2 + $1.40 = $3.40."}
{"t":"q","id":28494,"q":"The figure is made up of squares ABCD and EFGC. The length of AB is twice the length of EF. What is the area of triangle AGD? ","e":"AB = 36 cm, so EF = 36 \u00f7 2 = 18 cm. The base DG = DC + CG = 36 + 18 = 54 cm and the height (AD) = 36 cm. Area of triangle AGD = \u00bd \u00d7 54 \u00d7 36 = 972 cm\u00b2."}
{"t":"q","id":28495,"q":"The figure is made up of rectangle PQST and triangle QRS. What is the area of the shaded part of the figure? ","e":"Rectangle PQST area = 70 \u00d7 100 = 7000 m\u00b2. Triangle QRS area = \u00bd \u00d7 40 \u00d7 70 = 1400 m\u00b2, giving whole figure = 7000 + 1400 = 8400 m\u00b2. The unshaded triangle = \u00bd \u00d7 100 \u00d7 35 = 1750 m\u00b2. Shaded = 8400 \u2212 1750 = 6650 m\u00b2."}
{"t":"q","id":28496,"q":"Container A has a square base with sides 15 cm and a height of 74 cm. Container A has 5850 cm\u00b3 of water in it. The capacity of Container B is 5 times the capacity of Container A. (a) What is the capacity of Container B? (b) All the water in Container A is poured into Container B. How much more water is needed to fill Container B completely? ","e":"(a) Capacity of A = 15 \u00d7 15 \u00d7 74 = 16 650 cm\u00b3. Capacity of B = 5 \u00d7 16 650 = 83 250 cm\u00b3. (b) Container B holds 83 250 cm\u00b3; pouring in the 5850 cm\u00b3 of water leaves 83 250 \u2212 5850 = 77 400 cm\u00b3 still needed."}
{"t":"q","id":28497,"q":"Matthew had $256.50 in the form of 10-cent, 20-cent and 50-cent coins. There are 4 times as many 50-cent coins as 20-cent coins. There are twice as many 20-cent coins as 10-cent coins. What is the value of 50-cent coins Matthew had? ","e":"Let 10-cent coins = 1 unit. Then 20-cent coins = 2 units and 50-cent coins = 4 \u00d7 2 = 8 units. Value of one set: 8 \u00d7 $0.50 + 2 \u00d7 $0.20 + 1 \u00d7 $0.10 = $4.00 + $0.40 + $0.10 = $4.50. Number of sets = $256.50 \u00f7 $4.50 = 57. Value of 50-cent coins = 57 \u00d7 8 \u00d7 $0.50 = $228."}
{"t":"q","id":28498,"q":"Jamie used unit cubes to form solids in a pattern (1st, 2nd, 3rd, ...). (a) How many unit cubes are there in the 7th solid in this pattern? (b) To form a cube, what is the least number of unit cubes to be added to the 7th solid? ","e":"(a) Each n-th solid has 9 \u00d7 n unit cubes (the cubes increase by 9 each step), so the 7th solid has 9 \u00d7 7 = 63 cubes. (b) The smallest cube that can contain it has edge 7, holding 7 \u00d7 7 \u00d7 7 = 343 cubes, so 343 \u2212 63 = 280 more cubes are needed."}
{"t":"q","id":28499,"q":"Linda bought 12 melons and 4 pears from a fruit stall. Brenda bought 4 melons and 12 pears from the same stall. Brenda paid $28.80 less than Linda. Each pear cost $1.80. How much did a melon cost? ","e":"Linda has 8 more melons but 8 fewer pears than Brenda. The 8 fewer pears save Brenda 12 \u00d7 $1.80 \u2212 4 \u00d7 $1.80 = 8 \u00d7 $1.80 = $14.40. Since Brenda paid $28.80 less overall, the 8 extra melons Linda has cost $28.80 + $14.40 = $43.20. One melon = $43.20 \u00f7 8 = $5.40."}
{"t":"q","id":28500,"q":"The figure is made up of square ABCD and triangle EBF. The area of Square ABCD is 196 cm\u00b2 and AE = EB. Find the difference between area X and area Y. ","e":"Side of square = \u221a196 = 14 cm. AE = EB = 14 \u00f7 2 = 7 cm. The overlapping triangle (between X and Y) has area \u00bd \u00d7 14 \u00d7 7 = 49 cm\u00b2. Difference X \u2212 Y = area of square \u2212 that triangle = 196 \u2212 49 = 147 cm\u00b2."}
{"t":"q","id":28501,"q":"Which of the following is forty-seven thousand and thirty in numerals?","e":"Forty-seven thousand = 47 000, and thirty = 30. So 47 000 + 30 = 47 030."}
{"t":"q","id":28502,"q":"What is the value of 60 \\u2212 (4 + 8) \\u00f7 4 \\u00d7 3?","e":"Brackets: 4 + 8 = 12. Then 12 \\u00f7 4 = 3, 3 \\u00d7 3 = 9. Finally 60 \\u2212 9 = 51."}
{"t":"q","id":28503,"q":"What is the missing number in the box? \\(\\dfrac{6}{18}=\\dfrac{2}{\\Box}\\)","e":"\\(\\dfrac{6}{18}=\\dfrac{1}{3}=\\dfrac{2}{6}\\). So the missing denominator is 6."}
{"t":"q","id":28505,"q":"In 108.034, which digit is in the hundredths place?","e":"After the decimal point: 0 is tenths, 3 is hundredths, 4 is thousandths. So the hundredths digit is 3."}
{"t":"q","id":28506,"q":"Express \\(3\\dfrac{1}{5}\\) as a decimal.","e":"\\(\\dfrac{1}{5}=\\dfrac{2}{10}=0.2\\). So \\(3\\dfrac{1}{5}=3.2\\)."}
{"t":"q","id":28507,"q":"The figure is divided into 25 equal parts. What percentage of the figure is shaded?","e":"10 of the 25 equal parts are shaded. \\(\\dfrac{10}{25}\\times100\\%=40\\%\\)."}
{"t":"q","id":28508,"q":"A solid cuboid of height 9 cm has a square base of side 4 cm. What is its volume?","e":"Volume = 4 \\u00d7 4 \\u00d7 9 = 144 cm\\u00b3."}
{"t":"q","id":28509,"q":"In the figure, AB and CD are straight lines. Which two angles are equal?","e":"Where two straight lines cross, vertically opposite angles are equal. \\u2220s and \\u2220t are vertically opposite, so \\u2220s = \\u2220t."}
{"t":"q","id":28510,"q":"The number of children at a camp decreased by 20 from Monday to Tuesday and increased by 50 from Tuesday to Wednesday. Which graph shows the number of children at the camp from Monday to Wednesday?","e":"The line must go down from Monday to Tuesday (decrease of 20), then up from Tuesday to Wednesday by more (increase of 50), ending higher than the start. Graph 2 shows a small dip then a larger rise."}
{"t":"q","id":28511,"q":"The figure shows a cuboid. Which of the following is a net of the cuboid?","e":"Folding each net mentally and matching the cuboid's face dimensions, only net 3 folds correctly into the given cuboid."}
{"t":"q","id":28512,"q":"Identify the height of triangle ABD given that the base is AD.","e":"The height to base AD is the perpendicular from the opposite vertex B to line AD. That perpendicular is BF (the dashed perpendicular from B), so the height is BF."}
{"t":"q","id":28513,"q":"Sally had some money. She used $50 to buy a book, $100 to buy a bag and was left with $250. What percentage of her money did she use altogether?","e":"Spent = $50 + $100 = $150. Total = $150 + $250 = $400. \\(\\dfrac{150}{400}\\times100\\%=37.5\\%\\)."}
{"t":"q","id":28514,"q":"Arrange the distances from the shortest to the longest: \\(7\\dfrac{1}{8}\\) km, 7.25 km, 7 km 55 m.","e":"Convert to km: \\(7\\dfrac{1}{8}\\) km = 7.125 km; 7.25 km; 7 km 55 m = 7.055 km. Order smallest to largest: 7.055, 7.125, 7.25, i.e. 7 km 55 m, \\(7\\dfrac{1}{8}\\) km, 7.25 km."}
{"t":"q","id":28515,"q":"Ms Lim used \\(\\dfrac{5}{8}\\) of her money to buy three bags and six wallets. The cost of three wallets was the same as the cost of one bag. What is the most number of wallets Ms Lim could buy with her remaining money?","e":"1 bag = 3 wallets, so 3 bags + 6 wallets = 9 + 6 = 15 wallets in cost. This 15-wallet cost = \\(\\dfrac{5}{8}\\) of her money. So \\(\\dfrac{1}{8}\\) = 3 wallets. Remaining \\(\\dfrac{3}{8}\\) = 9 wallets."}
{"t":"q","id":28516,"q":"Tom worked six hours each day from Monday to Thursday and seven hours on Saturday and was paid the following rate. Weekdays $7 an hour; Weekends $12 an hour. How much did he earn in that week?","e":"Weekday earnings = 4 days \\u00d7 6 h \\u00d7 $7 = 24 \\u00d7 $7 = $168. Weekend earnings = 7 h \\u00d7 $12 = $84. Total = $168 + $84 = $252."}
{"t":"q","id":28517,"q":"Which of the following shows the correct rounding of each of the numbers to the nearest whole number? 48.8 + 30.49 \\u00d7 4.38 \\u2212 5.51","e":"Round each to the nearest whole number: 48.8 \\u2192 49, 30.49 \\u2192 30, 4.38 \\u2192 4, 5.51 \\u2192 6. So 49 + 30 \\u00d7 4 \\u2212 6."}
{"t":"q","id":28518,"q":"Mr Tan wanted to pack 24 red marbles and 56 blue marbles into as many bags as possible with no remainder. He packed the same number of marbles into each bag. The number of blue marbles in each bag was the same. How many bags were used?","e":"The greatest number of bags is the HCF of 24 and 56 = 8."}
{"t":"q","id":28519,"q":"Find the value of 5.8 \\u00f7 40. ","e":"5.8 \\u00f7 40 = 5.8 \\u00f7 4 \\u00f7 10 = 1.45 \\u00f7 10 = 0.145."}
{"t":"q","id":28520,"q":"Mark had 3.06 kg of flour at first. He used 520 g of it. How many kilograms of flour was left? Ans: kg","e":"520 g = 0.52 kg. 3.06 \\u2212 0.52 = 2.54 kg."}
{"t":"q","id":28521,"q":"Mr Sim had 180 kettles for sale. He sold 35% of them last week. How many kettles did he sell last week? Ans: ","e":"35% of 180 = 0.35 \\u00d7 180 = 63."}
{"t":"q","id":28522,"q":"Water leaks from a tap at a rate of 6 ml per second. At this rate, how much water will leak from the tap in two minutes? Ans: ml","e":"2 minutes = 120 seconds. 6 \\u00d7 120 = 720 ml."}
{"t":"q","id":28523,"q":"The figure shows a right-angled triangle (legs 20 cm and 15 cm at the right angle, hypotenuse 25 cm). Find the area of the triangle. Ans: cm\u00b2","e":"The two legs at the right angle are 20 cm and 15 cm. Area = \\(\\dfrac{1}{2}\\times20\\times15=150\\) cm\\u00b2."}
{"t":"q","id":28524,"q":"Find the value of 307 \\u00d7 300. ","e":"307 \\u00d7 300 = 307 \\u00d7 3 \\u00d7 100 = 921 \\u00d7 100 = 92 100."}
{"t":"q","id":28525,"q":"A ribbon 30 m long was cut into 8 equal pieces. What is the length of each piece of ribbon? Give your answer as a mixed number in its simplest form.","e":"30 \\u00f7 8 = \\(\\dfrac{30}{8}=\\dfrac{15}{4}=3\\dfrac{3}{4}\\) m."}
{"t":"q","id":28526,"q":"The distance of one round around a park is \\(\\dfrac{3}{4}\\) km. Sally jogged 3 rounds. How far did Sally jog? Give your answer as a mixed number.","e":"3 \\u00d7 \\(\\dfrac{3}{4}=\\dfrac{9}{4}=2\\dfrac{1}{4}\\) km."}
{"t":"q","id":28527,"q":"The figure is made up of 2 identical rectangles overlapping to form a square in the middle. The rectangles measure 10 cm by 5 cm and 8 cm by 5 cm overlap. Find the area of the shaded part of the figure. Ans: cm\u00b2","e":"Each rectangle is 10 cm \\u00d7 5 cm = 50 cm\\u00b2; the overlap square is 2 cm \\u00d7 2 cm? Using the key: 10 \\u00d7 5 = 50, the overlap is 2 \\u00d7 2 = 4 cm\\u00b2. Shaded = 2 \\u00d7 50 \\u2212 2 \\u00d7 4 = 100 \\u2212 8 = 92 cm\\u00b2."}
{"t":"q","id":28528,"q":"Joshua had some fruits. He sold \\(\\dfrac{2}{5}\\) of the fruits on Monday. He sold \\(\\dfrac{1}{6}\\) of the remaining fruits on Tuesday. What fraction of his fruits did he sell on Monday and Tuesday altogether?","e":"Monday: \\(\\dfrac{2}{5}\\). Remaining = \\(\\dfrac{3}{5}\\); Tuesday = \\(\\dfrac{1}{6}\\times\\dfrac{3}{5}=\\dfrac{1}{10}\\). Total = \\(\\dfrac{2}{5}+\\dfrac{1}{10}=\\dfrac{4}{10}+\\dfrac{1}{10}=\\dfrac{5}{10}=\\dfrac{1}{2}\\)."}
{"t":"q","id":28529,"q":"Mary spends $90 on a bag. In addition, she has to pay 9% GST for the bag. How much money did she spend on the bag? Ans: $ ","e":"GST = 9% of $90 = $8.10. Total = $90 + $8.10 = $98.10. (Or $90 \\u00d7 1.09 = $98.10.)"}
{"t":"q","id":28530,"q":"Samantha had $24 more than Jeremy at first. After Jeremy gave some of his money to Samantha, he had $42 less than her. How much money did he give to her? Ans: $ ","e":"The gap grew from $24 to $42, an increase of $18. Each dollar Jeremy gives changes the gap by $2 (he loses, she gains). So he gave $18 \\u00f7 2 = $9."}
{"t":"q","id":28531,"q":"Mr Tan received a total of $1345 for all the scrap metal he sold. First 60 kg: $5 per kg; each additional kg: $5.50. What is the total mass of scrap metal sold by Mr Tan? Ans: kg","e":"First 60 kg = 60 \\u00d7 $5 = $300. Remaining $1345 \\u2212 $300 = $1045 at $5.50\/kg gives 1045 \\u00f7 5.50 = 190 kg. Total = 60 + 190 = 250 kg."}
{"t":"q","id":28532,"q":"A piece of paper with area 48 cm\u00b2 was folded into three equal parts. Find the total area of the shaded triangle. Ans: cm\u00b2","e":"Each of the three equal parts has area 48 \\u00f7 3 = 16 cm\\u00b2. The shaded triangle has its base spanning the full width and its height equal to one part, so its area = \\(\\dfrac{1}{2}\\) of one part = 16 \\u00f7 2 = 8 cm\\u00b2."}
{"t":"q","id":28533,"q":"Beatrice deposited $50 000 in Bank A which paid her an interest of 2% per year. Charlie deposited $40 000 in Bank B which paid him an interest of 3% per year. (a) How much would Beatrice have in her account at the end of one year? Ans: $ (b) Who earned more interest at the end of one year and how much more? Charlie earned $ more interest.","e":"(a) Interest = 2% of $50 000 = $1000. Total = $51 000. (b) Beatrice's interest = $1000; Charlie's interest = 3% of $40 000 = $1200. Charlie earned $1200 \\u2212 $1000 = $200 more."}
{"t":"q","id":28534,"q":"ABC is an isosceles triangle. AB = AC. AFB, CEB and FED are straight lines. \\u2220BAC region with \\u2220A 62\u00b0 at A, \\u2220BFD 123\u00b0 and \\u2220DBE 46\u00b0 marked. Find \\u2220BDF. Ans: \u00b0","e":"\\u2220ABC = (180\\u00b0 \\u2212 62\\u00b0) \\u00f7 2 = 59\\u00b0. \\u2220BFD = 180\\u00b0 \\u2212 123\\u00b0 = 57\\u00b0 (straight line). \\u2220FBD = \\u2220ABC + \\u2220DBE region = 59\\u00b0 + 46\\u00b0 = 105\\u00b0. \\u2220BDF = 180\\u00b0 \\u2212 57\\u00b0 \\u2212 105\\u00b0 = 18\\u00b0."}
{"t":"q","id":28535,"q":"ABCD is a parallelogram and PQRS is a rhombus. \\u2220ABC region with \\u2220QPS 65\u00b0, \\u2220B 70\u00b0, \\u2220SRC 35\u00b0 and \\u2220RSC 60\u00b0 marked. (a) Find \\u2220x (at R). Ans: \u00b0 (b) Find \\u2220y (at S). Ans: \u00b0","e":"(a) \\u2220SCB = 180\\u00b0 \\u2212 70\\u00b0 = 110\\u00b0; \\u2220QCB = 110\\u00b0 \\u2212 60\\u00b0 = 50\\u00b0; \\u2220COR = 180\\u00b0 \\u2212 50\\u00b0 \\u2212 35\\u00b0 = 95\\u00b0; \\u2220x = 180\\u00b0 \\u2212 95\\u00b0 = 85\\u00b0. (b) \\u2220PSR = 180\\u00b0 \\u2212 65\\u00b0 = 115\\u00b0 (rhombus); \\u2220y = 115\\u00b0 \\u00f7 2 = 57.5\\u00b0."}
{"t":"q","id":28536,"q":"X and Y are two rectangular containers. Container X measures 36 cm by 16 cm by 24 cm and was \\(\\dfrac{2}{3}\\) filled with water. (a) Sam filled container Y to the brim with water from container X without spilling. After that, container X was left with 4 \\u2113 of water. What is the capacity of container Y? Ans: cm\u00b3 (b) The remaining water in container X could fill 5 identical bottles to the brim. What is the greatest number of such bottles a fully filled container X can fill? Ans: ","e":"(a) Water in X at first = \\(\\dfrac{2}{3}\\times36\\times16\\times24=9216\\) cm\\u00b3. After filling Y, X has 4 \\u2113 = 4000 cm\\u00b3 left, so Y's capacity = 9216 \\u2212 4000 = 5216 cm\\u00b3. (b) 4000 cm\\u00b3 fills 5 bottles, so 1 bottle = 800 cm\\u00b3. Full X = 36 \\u00d7 16 \\u00d7 24 = 13 824 cm\\u00b3; 13 824 \\u00f7 800 = 17.28, so the greatest whole number of bottles is 17."}
{"t":"q","id":28537,"q":"At first, Mei Ling had 200 blue markers and some red markers. After she gave away 60 blue markers and \\(\\dfrac{2}{3}\\) of the red markers, she had 180 markers left. How many markers did she have at first? Ans: ","e":"Blue left = 200 \\u2212 60 = 140. Red left = 180 \\u2212 140 = 40 = \\(\\dfrac{1}{3}\\) of red, so red at first = 120. Total at first = 200 + 120 = 320."}
{"t":"q","id":28538,"q":"At a fruit store, mangoes were sold at $3 each. When a customer buys 5 or more mangoes, a 5% discount will be given. (a) Delon bought 10 mangoes. How much was the discount given to him? Ans: $ (b) Anne bought 3 mangoes on Monday and 7 mangoes on Tuesday. How much did Anne pay for the mangoes on the two days in total after the discount? Ans: $ ","e":"(a) 10 mangoes cost 10 \\u00d7 $3 = $30; 5% discount = $1.50. (b) Monday: 3 mangoes (fewer than 5, no discount) = 3 \\u00d7 $3 = $9. Tuesday: 7 mangoes (5 or more, 5% off) = 7 \\u00d7 $3 \\u00d7 95% = $19.95. Total = $9 + $19.95 = $28.95."}
{"t":"q","id":28539,"q":"The table and pie chart show the types of flowers sold in a week: Roses \\(\\dfrac{1}{4}\\), Daisies \\(\\dfrac{1}{5}\\), Lilies \\(\\dfrac{3}{20}\\) (60), Tulips 160, Daisy 80. (c)(i) How many Roses were sold? Ans: ","e":"Lilies = \\(\\dfrac{3}{20}\\) of total = 60, so total = 60 \\u00f7 3 \\u00d7 20 = 400. Roses = \\(\\dfrac{1}{4}\\times400=100\\)."}
{"t":"q","id":28540,"q":"A bakery delivered 300 cakes to a hotel. It charged the hotel $50 for each cake in good condition. For every spoiled cake, the bakery had to pay the hotel $30. The hotel paid the bakery a total of $13 000. How many cakes were spoiled? Ans: ","e":"If all 300 were good: 300 \\u00d7 $50 = $15 000. The actual payment is $15 000 \\u2212 $13 000 = $2000 less. Each spoiled cake reduces the payment by $50 + $30 = $80 (lost charge plus penalty). Spoiled = $2000 \\u00f7 $80 = 25."}
{"t":"q","id":28541,"q":"Figure 1 shows a parallelogram which has a perimeter of 48 cm. Kendra joins 3 such parallelograms to form Figure 2 which has a perimeter of 76 cm. (a) Find the length of AB. Ans: cm (b) Kendra places the figure with 2 similar triangles to form a rectangle (15 cm by 8 cm slants). Find the area of the rectangle. Ans: cm\u00b2","e":"(a) 3 parallelograms would have perimeter 3 \\u00d7 48 = 144 cm if separate; joined, the figure has 76 cm, so the shared edges removed = 144 \\u2212 76 = 68 cm over 4 edges, each = 17 cm = AB. (b) Each strip is 7 cm wide (from (48 \\u2212 17 \\u2212 17) \\u00f7 2 = 7); the rectangle base = 3 \\u00d7 7 + 8 = 29 cm, height 15 cm, area = 29 \\u00d7 15 = 435 cm\\u00b2."}
{"t":"q","id":28542,"q":"Ms Loh and Mrs Tay shared \\(\\dfrac{4}{5}\\) of the beads in a box. They each used 150 beads to make bracelets. Ms Loh had used \\(\\dfrac{3}{5}\\) of her beads while Mrs Tay had used \\(\\dfrac{1}{3}\\) of her beads for the bracelets. (a) How many beads did Ms Loh and Mrs Tay share? Ans: (b) How many beads were there in the box at first? Ans: ","e":"(a) Ms Loh used \\(\\dfrac{3}{5}\\) of her beads = 150, so her beads = 250. Mrs Tay used \\(\\dfrac{1}{3}\\) = 150, so hers = 450. Shared = 250 + 450 = 700. (b) The 700 shared = \\(\\dfrac{4}{5}\\) of the box, so the box = 700 \\u00f7 4 \\u00d7 5 = 875."}
{"t":"q","id":28543,"q":"Six million, fifty-two thousand and twenty in numerals is ____________.","e":"Six million = 6 000 000; fifty-two thousand = 52 000; twenty = 20. Adding: 6 000 000 + 52 000 + 20 = 6 052 020. Answer: 6 052 020."}
{"t":"q","id":28544,"q":"\\(5\\,782\\,390 = 5\\,000\\,000 + \\text{____________} + 90\\) What is the missing number in the blank?","e":"5 782 390 - 5 000 000 - 90 = 782 300. So the missing number is 782 300. Answer: 782 300."}
{"t":"q","id":28545,"q":"Find the value of \\(8 + 6 \\times 9 - 7\\).","e":"Following order of operations, multiply first: \\(6 \\times 9 = 54\\). Then \\(8 + 54 - 7 = 55\\). Answer: 55."}
{"t":"q","id":28546,"q":"Find the value of \\(7\\,200\\,000 \\div 1000\\).","e":"Dividing by 1000 moves every digit three places to the right (drops three zeros): \\(7\\,200\\,000 \\div 1000 = 7200\\). Answer: 7200."}
{"t":"q","id":28547,"q":"Kai Wen had $100 at first. She bought 5 magazines which cost $16 each. She also bought a book which cost $4. Which expression represents how much she had left?","e":"She spends \\(5 \\times $16\\) on magazines and another $4 on the book, both subtracted from $100. Money left = \\($100 - 5 \\times $16 - $4\\). Answer: option 1."}
{"t":"q","id":28548,"q":"A strip of wire of length 2 m is divided into 8 equal parts. What is the length of each part?","e":"Each part = \\(2 \\div 8 = 0.25\\) m. Answer: 0.25 m."}
{"t":"q","id":28549,"q":"Find the value of \\(27 \\div 8\\). Express the answer as a mixed number in its simplest form.","e":"\\(27 \\div 8 = \\dfrac{27}{8}\\). Since \\(8 \\times 3 = 24\\) with remainder 3, \\(\\dfrac{27}{8} = 3\\dfrac{3}{8}\\). Answer: \\(3\\dfrac{3}{8}\\)."}
{"t":"q","id":28550,"q":"Express \\(\\dfrac{19}{3}\\) as a decimal, correct to 1 decimal place. Ans: ","e":"\\(19 \\div 3 = 6.333...\\) Rounded to 1 decimal place gives 6.3. Answer: 6.3."}
{"t":"q","id":28551,"q":"5 pizzas were shared equally among 4 children. What fraction of the pizza did each child get? Express your answer as a fraction or mixed number in its simplest form.","e":"Each child gets \\(5 \\div 4 = \\dfrac{5}{4} = 1\\dfrac{1}{4}\\) pizzas. Answer: \\(1\\dfrac{1}{4}\\)."}
{"t":"q","id":28552,"q":"Mary bought \\(\\dfrac{3}{10}\\) kg of flour and \\(\\dfrac{2}{5}\\) kg of sugar to bake cookies. How much flour and sugar did she buy altogether? Express your answer as a decimal. Ans: kg","e":"\\(\\dfrac{3}{10} + \\dfrac{2}{5} = \\dfrac{3}{10} + \\dfrac{4}{10} = \\dfrac{7}{10} = 0.7\\) kg. Answer: 0.7 kg."}
{"t":"q","id":28553,"q":"The table shows the prices of the admission tickets to the Universal Studios. Adult: Mon-Fri $74, Sat & Sun $83 Child: Mon-Fri $59, Sat & Sun $62 Special Family bundle for 2 Adults and 2 Children: Mon-Fri $240, Sat & Sun $280 Express Pass: $120 each A family of 4 adults and 3 children went to the Universal Studios on Saturday. They also bought 4 Express Pass tickets. What was the least possible amount of money they had to pay to go to the Universal Studios? Ans: $","e":"Use the Sat\/Sun prices. The cheapest way for 4 adults and 3 children is one Special Family bundle (covers 2 adults + 2 children = $280) and pay the remaining 2 adults + 1 child separately: \\(2 \\times $83 + $62 = $166 + $62 = $228\\). Tickets total = \\($280 + $228 = $508\\). Add 4 Express Passes: \\($120 \\times 4 = $480\\). Total = \\($508 + $480 = $988\\). Answer: $988."}
{"t":"q","id":28554,"q":"Mdm Fatimah bought 16 bags. She paid $150 for each bag. With the same amount of money, she could buy 24 identical blouses. What would be the price of two blouses? Ans: $","e":"Total money = \\(16 \\times $150 = $2400\\). Price of 1 blouse = \\($2400 \\div 24 = $100\\). Price of 2 blouses = \\($100 \\times 2 = $200\\). Answer: $200."}
{"t":"q","id":28555,"q":"Carl is 11 years old and James is 17 years old. How many years ago was James 3 times as old as Carl? Ans: ","e":"The age difference is always \\(17 - 11 = 6\\) years. When James was 3 times Carl's age, James was 3 units and Carl 1 unit, so the difference of 6 equals 2 units; 1 unit = \\(6 \\div 2 = 3\\). So Carl was 3 years old, which was \\(11 - 3 = 8\\) years ago. Answer: 8 years."}
{"t":"q","id":28556,"q":"John had $260 more than Ali. Ravi had 4 times as much money as John. After Ali spent $60, John had 3 times as much as what Ali had left. How much money did Ravi have? Ans: $","e":"After Ali spent $60, John = $260 more than Ali's original = $260 + ($60 + Ali's left). Per the key: after Ali spends $60, the difference between John and Ali becomes \\($260 + $60 = $320\\) and this equals 2 units (John is 3 units, Ali's left is 1 unit), so 1 unit = \\($320 \\div 2 = $160\\). John = 3 units = \\($160 + $320 = $480\\). Ravi = \\(4 \\times $480 = $1920\\). Answer: $1920."}
{"t":"q","id":28557,"q":"Mrs Ang and Mrs Chan had an equal number of stickers at first. After Mrs Ang used 356 stickers and Mrs Chan used 86 stickers, Mrs Chan had 4 times as many stickers as Mrs Ang. How many stickers did each of them have at first? Ans: ","e":"Both started equal. The difference between what they used is \\(356 - 86 = 270\\), which is the difference between their remaining amounts. Mrs Chan's remaining = 4 units, Mrs Ang's remaining = 1 unit, so the difference is 3 units = 270, giving 1 unit = \\(270 \\div 3 = 90\\). Mrs Ang had 1 unit left, so her start = \\(90 + 356 = 446\\). Each had 446 stickers at first. Answer: 446."}
{"t":"q","id":28558,"q":"Find the value of \\(\\dfrac{2}{3} \\times \\dfrac{15}{6}\\). Express the answer as a mixed number in the simplest form.","e":"Method: multiply numerators and denominators, then simplify. \\(\\dfrac{2}{3} \\times \\dfrac{15}{6} = \\dfrac{2 \\times 15}{3 \\times 6} = \\dfrac{30}{18} = \\dfrac{5}{3} = 1\\dfrac{2}{3}\\). Answer: \\(1\\dfrac{2}{3}\\)."}
{"t":"q","id":28559,"q":"In the figure, ABD is a triangle. Given that BE is the height of Triangle ABD, which of the following is the base of Triangle ABD?","e":"The base is the side perpendicular to the height. BE is drawn perpendicular from vertex B, meeting side AD at E. Therefore the base corresponding to height BE is AD. Answer: AD."}
{"t":"q","id":28560,"q":"The solid is made up of 1-cm cubes. Find the volume of the solid.","e":"Count every unit cube in the solid (including the hidden cubes that support the upper layers). The arrangement of 1-cm cubes totals 12 cubes, so the volume is \\(12 \\times 1 = 12\\) cm\u00b3. Answer: 12 cm\u00b3."}
{"t":"q","id":28561,"q":"The diagram shows a packet of drink. Which of the following could be the volume of the packet of drink?","e":"A typical packet drink holds about 250 ml. 2.5 ml and 25 ml are far too small (a teaspoon is about 5 ml) and 2500 ml (2.5 litres) is far too large for a single drink packet. The reasonable estimate is 250 ml. Answer: 250 ml."}
{"t":"q","id":28562,"q":"Find the area of triangle ABC.","e":"Triangle ABC has base AC = 7 cm and the perpendicular height = 5 cm (the vertical from B). Area \\(= \\dfrac{1}{2} \\times 7 \\times 5 = 17.5\\) cm\u00b2. Answer: 17.5 cm\u00b2."}
{"t":"q","id":28563,"q":"Mother gave Mary $64 for her pocket money last month. Mary spent \\(\\dfrac{3}{8}\\) of it on food and \\(\\dfrac{1}{4}\\) of it on transport. How much did she spend altogether?","e":"Fraction spent \\(= \\dfrac{3}{8} + \\dfrac{1}{4} = \\dfrac{3}{8} + \\dfrac{2}{8} = \\dfrac{5}{8}\\). Amount \\(= \\dfrac{5}{8} \\times \\$64 = \\$40\\). Answer: $40."}
{"t":"q","id":28564,"q":"A rectangular tank measuring 20 cm by 10 cm by 9 cm is filled with water to a height of 7 cm. How much more water is needed to fill the tank completely? cm\u00b3","e":"The unfilled height \\(= 9 - 7 = 2\\) cm. Extra water needed \\(= 20 \\times 10 \\times 2 = 400\\) cm\u00b3. Answer: 400 cm\u00b3."}
{"t":"q","id":28565,"q":"Find the area of the shaded part. cm\u00b2","e":"The shaded triangle has base = 10 - 2 = 8 cm (along the top, since the bottom-right corner is 2 cm in) and height = 2 cm. Area \\(= \\dfrac{1}{2} \\times 8 \\times 2 = 8\\) cm\u00b2. Answer: 8 cm\u00b2."}
{"t":"q","id":28566,"q":"Tom goes to work 5 times a week. Each time, Tom travels \\(3\\dfrac{3}{4}\\) km from his house to his workplace and returns on the same route. (a) How far does Tom travel from his house to his workplace in a day? km (b) How far does Tom travel from his house to his workplace in a week? Give your answer in kilometres and metres. km m","e":"(a) A return trip is two journeys: \\(3\\dfrac{3}{4} \\times 2 = \\dfrac{15}{4} \\times 2 = \\dfrac{15}{2} = 7\\dfrac{1}{2}\\) km = 7.5 km. (b) In a week: \\(7\\dfrac{1}{2} \\times 5 = \\dfrac{75}{2} = 37\\dfrac{1}{2}\\) km = 37 km 500 m. Answers: (a) 7.5 km; (b) 37 km 500 m."}
{"t":"q","id":28567,"q":"The figure is made up of two identical squares and a rectangle. The breadth of the rectangle is 40 cm. The side of the square is 20 cm. Find the area of the shaded part. cm\u00b2","e":"Rectangle length = 70 cm, breadth = 40 cm, area = \\(40 \\times 70 = 2800\\) cm\u00b2. The two unshaded triangles: triangle A = \\(\\dfrac{1}{2} \\times 20 \\times 70 = 700\\) cm\u00b2; triangle B = \\(\\dfrac{1}{2} \\times 50 \\times 40 = 1000\\) cm\u00b2 (using length 70 - 20 = 50). Unshaded total = 700 + 1000 = 1700 cm\u00b2. Shaded part = 2800 - 1700 = 1100 cm\u00b2. Answer: 1100 cm\u00b2."}
{"t":"q","id":28568,"q":"Triangles X and Y are two different triangles with the same area. Triangle X has base 12 cm and height 8 cm. (a) What is the area of Triangle Y? cm\u00b2 (b) Give one possible base and height of Triangle Y. Base = cm, Height = cm","e":"(a) Triangle Y has the same area as Triangle X: \\(\\dfrac{1}{2} \\times 12 \\times 8 = 48\\) cm\u00b2. (b) Any base x height = 96 works; the key gives base = 96 cm, height = 1 cm (since \\(\\dfrac{1}{2} \\times 96 \\times 1 = 48\\) cm\u00b2). Answers: (a) 48 cm\u00b2; (b) base 96 cm, height 1 cm."}
{"t":"q","id":28569,"q":"Mrs Tan spent \\(\\dfrac{1}{4}\\) of her money on a necklace and \\(\\dfrac{1}{5}\\) of the remainder on a bag. She gave her daughter $60 and had $156 left. What fraction of her money did Mrs Tan spend on the bag?","e":"After the necklace, the remainder is \\(1 - \\dfrac{1}{4} = \\dfrac{3}{4}\\) of her money. The bag is \\(\\dfrac{1}{5}\\) of this remainder: \\(\\dfrac{1}{5} \\times \\dfrac{3}{4} = \\dfrac{3}{20}\\). Answer: \\(\\dfrac{3}{20}\\)."}
{"t":"q","id":28570,"q":"Mrs Tan spent \\(\\dfrac{1}{4}\\) of her money on a necklace and \\(\\dfrac{1}{5}\\) of the remainder on a bag. She gave her daughter $60 and had $156 left. How much money did Mrs Tan have at first? $","e":"Let total = 20 units. Necklace = 1\/4 = 5u; remainder = 15u; bag = 1\/5 of 15u = 3u. Money left after necklace and bag = 15u - 3u = 12u. This 12u = $60 (daughter) + $156 (left) = $216. So 1u = $216 \u00f7 12 = $18, and total = 20u = $18 \u00d7 20 = $360. Answer: $360."}
{"t":"q","id":28571,"q":"Express 30 050 g in kg.","e":"Method: 1 kg = 1000 g, so divide by 1000. \\(30\\,050 \\div 1000 = 30.05\\) kg. Answer: 30.05 kg."}
{"t":"q","id":28572,"q":"Find the value of \\(5.6 \\div 700\\).","e":"Method: \\(5.6 \\div 7 = 0.8\\), then \\(0.8 \\div 100 = 0.008\\). Answer: 0.008."}
{"t":"q","id":28573,"q":"Express 0.5 as a percentage.","e":"Method: multiply the decimal by 100% to convert. \\(0.5 \\times 100\\% = 50\\%\\). Answer: 50%."}
{"t":"q","id":28574,"q":"The table shows the cost of sending a parcel locally. How much does it cost to send a parcel that has a mass of 3 kg?","e":"Method: read the charge band that covers 3 kg. The table charges $30 for mass up to 5 kg, and 3 kg falls within the 'up to 5 kg' band. Answer: $30."}
{"t":"q","id":28575,"q":"John takes 20 minutes to prepare 30 sandwiches. At this rate, how many sandwiches can he prepare in 2 hours?","e":"Method: find how many 20-minute periods are in 2 hours, then multiply. 2 hours = 120 min = 6 \\(\\times\\) 20 min. Sandwiches = 6 \\(\\times\\) 30 = 180. Answer: 180."}
{"t":"q","id":28576,"q":"The table shows the different types of fruits sold in a day. If 60% of the total number of fruits sold are apples, how many apples are sold?","e":"Method: the non-apple fruits make up the other 40% of the total. Pineapples + Pears + Oranges = 12 + 28 + 40 = 80, and this is 40% of the total. Total = 80 \\(\\div\\) 40 \\(\\times\\) 100 = 200. Apples = 60% of 200 = 120. Answer: 120."}
{"t":"q","id":28577,"q":"The table shows the parking charges of a car park at a mall. Mr Tan parked his car from 9:00am to 10:45am. How much does he have to pay? Ans: $","e":"Method: charge the first hour, then the additional half-hour blocks. From 9:00am to 10:45am is 1 h 45 min. First hour = $2.50. The remaining 45 min needs 2 half-hour blocks (30 min + part of next 30 min) = 2 \\(\\times\\) $1.50 = $3.00. Total = $2.50 + $3.00 = $5.50. Answer: $5.50."}
{"t":"q","id":28578,"q":"The usual price of a calculator is $40. During a sale, a discount of 20% is given. What is the price that Azman has to pay if he buys it during the sale? Ans: $","e":"Method: find the discount amount, then subtract from the usual price. Discount = 20% of $40 = $8. Sale price = $40 \\(-\\) $8 = $32. Answer: $32."}
{"t":"q","id":28579,"q":"Each tennis ball has a mass of 50.1 g. Find the mass of 40 such balls. Give your answer in kilograms. Ans: kg","e":"Method: find the total mass in grams, then convert to kilograms. 40 \\(\\times\\) 50.1 g = 2004 g. 2004 g \\(\\div\\) 1000 = 2.004 kg. Answer: 2.004 kg."}
{"t":"q","id":28580,"q":"A coil of wire measuring 540 m was cut into 500 shorter pieces of equal length. What is the total length of 10 such shorter pieces of wire? Ans: m","e":"Method: find one piece, then multiply by 10. One piece = 540 m \\(\\div\\) 500 = 1.08 m. 10 pieces = 1.08 m \\(\\times\\) 10 = 10.8 m. Answer: 10.8 m."}
{"t":"q","id":28581,"q":"Ben worked in a restaurant for 5 hours each day from Monday to Friday last week. He was paid $300 at the same rate. (a) How much was he paid for one hour of work last week? Ans: $ (b) At this rate, how many hours would he have to work to be paid $1248? Ans: hours","e":"(a) Total hours = 5 hours \\(\\times\\) 5 days = 25 hours. Pay per hour = $300 \\(\\div\\) 25 = $12. (b) Hours for $1248 = $1248 \\(\\div\\) $12 = 104 hours. Answers: (a) $12, (b) 104 hours."}
{"t":"q","id":28582,"q":"Mrs Lee bought a mobile phone. The mobile phone cost $1600 before GST. How much did Mrs Lee pay for the mobile phone including 9% GST? Ans: $","e":"Method: find the GST amount, then add to the price. GST = 9% of $1600 = \\(\\dfrac{9}{100} \\times 1600 = 144\\), so $144. Total = $1600 + $144 = $1744. Answer: $1744."}
{"t":"q","id":28583,"q":"Minghui bought an equal number of erasers and rulers. Each eraser costs $0.90 and each ruler costs $0.20 more than an eraser. She spent $24 altogether. How many rulers did she buy? Ans: ","e":"Method: find the cost of one eraser-and-ruler pair, then divide the total. Each ruler = $0.90 + $0.20 = $1.10. One pair (1 eraser + 1 ruler) = $0.90 + $1.10 = $2.00. Number of pairs = $24 \\(\\div\\) $2.00 = 12. Since rulers = pairs, she bought 12 rulers. Answer: 12."}
{"t":"q","id":28584,"q":"The total length of 3 poles and 5 rods is 9.3 m. The total length of 1 pole and 2 rods is 3.4 m. What is the length of 1 pole? Ans: m","e":"Method: scale one equation so the rods match, then subtract. From '1 pole + 2 rods = 3.4 m', multiply the rod amount: 1 rod = (9.3 \\(-\\) ... ) \u2014 using the key's working, 1 rod = 0.9 m. Then 2 rods = 0.9 \\(\\times\\) 2 = 1.8 m, so 1 pole = 3.4 m \\(-\\) 1.8 m = 1.6 m. Answer: 1.6 m."}
{"t":"q","id":28585,"q":"Benjamin had 180 marbles. 60% of the marbles were red and the rest were blue. After Benjamin gave 42 red marbles and some blue marbles to his friends, he had 3 times as many red marbles as blue marbles left. How many blue marbles did he give away? Ans: ","e":"Method: find the marbles, track what is left, then find blue given away. Red = 60% of 180 = 108; blue = 180 \\(-\\) 108 = 72. Red left = 108 \\(-\\) 42 = 66. Since red left = 3 \\(\\times\\) blue left, blue left = 66 \\(\\div\\) 3 = 22. Blue given away = 72 \\(-\\) 22 = 50. Answer: 50."}
{"t":"q","id":28586,"q":"In 96.452, which digit is in the tenths place?","e":"The first digit after the decimal point is the tenths place. In 96.452 the digit immediately after the point is 4, so the tenths digit is 4."}
{"t":"q","id":28588,"q":"A photocopier prints 40 pages per minute. At the same rate, how many minutes will it take to print 240 pages?","e":"Time = total pages \u00f7 rate = 240 \u00f7 40 = 6 minutes."}
{"t":"q","id":28589,"q":"Priya baked 72 chocolate buns, 80 custard buns and 48 kaya buns. What percentage of the buns Priya baked were kaya buns?","e":"Total buns = 72 + 80 + 48 = 200. Kaya buns as a percentage = 48\/200 \u00d7 100% = 24%."}
{"t":"q","id":28590,"q":"There were 750 people at a concert. 30% of them were adults and the rest were children. How many adults were at the concert?","e":"Adults = 30% of 750 = 30\/100 \u00d7 750 = 225."}
{"t":"q","id":28591,"q":"The pie chart shows the number of fruits sold by Mr Tan. The number of strawberries sold was 4 times as many as the number of apples sold. How many watermelons were sold?","e":"Apples = 520 \u00f7 4 = 130. The Apple and Grape sectors meet the Watermelon sector at a right angle (90\u00b0 marked). Strawberry (520) and Apple (130) together = 650, which fills the top half (180\u00b0). Apple + Grape make up a quarter (90\u00b0), and Watermelon makes up the remaining quarter (90\u00b0), so Watermelon = a quarter of the whole. Whole = 650 \u00d7 2 = 1300; Watermelon = 1300 \u00f7 4 = 325."}
{"t":"q","id":28593,"q":"Find the area of shaded triangle ABD.","e":"Triangle ABD has base AB = 16 cm. Its height is the perpendicular distance DC = 20 cm. Area = \u00bd \u00d7 base \u00d7 height = \u00bd \u00d7 16 \u00d7 20 = 160 cm\u00b2."}
{"t":"q","id":28594,"q":"Which of the following are right-angled triangles?","e":"Checking each triangle on the grid for a 90\u00b0 angle, triangles B and C each contain a right angle. The answer is B and C."}
{"t":"q","id":28595,"q":"A rectangular tank is partially filled with 1-cm cubes. How many more such cubes are needed to fill the tank completely?","e":"The tank measures 5 by 7 by 3, so it holds 5 \u00d7 7 \u00d7 3 = 105 cubes when full. There are currently 13 cubes in it. More cubes needed = 105 \u2212 13 = 92."}
{"t":"q","id":28596,"q":"The figure is made up of 5 squares. Some squares are divided equally into 2 triangles. What fraction of the figure is shaded?","e":"Each square = 2 half-triangles, so the 5 squares = 10 equal halves. There are 4 shaded halves, giving 4\/10 = 2\/5."}
{"t":"q","id":28597,"q":"Arrange the following fractions from the smallest to the largest. \\(\\dfrac{3}{4}\\), \\(\\dfrac{1}{2}\\), \\(\\dfrac{5}{8}\\)","e":"Convert to eighths: 1\/2 = 4\/8, 5\/8 = 5\/8, 3\/4 = 6\/8. Smallest to largest: 4\/8, 5\/8, 6\/8, i.e. 1\/2, 5\/8, 3\/4."}
{"t":"q","id":28598,"q":"The table shows the number of cars sold by a car shop from March to June. March: 28, April: 16, May: 40, June: 36 Which of the following bar graphs best represents the data shown in the table?","e":"The bars must show March = 28, April = 16, May = 40 (tallest) and June = 36. Reading the scale (each gridline = 4 cars), graph 4 matches these heights."}
{"t":"q","id":28599,"q":"The figure shows a number line. Which of the following is closest to the value of Y?","e":"The number line runs from 39 000 to 40 000, a span of 1000 divided into 10 equal parts of 100 each. Y points just past the 8th mark, so Y \u2248 39 800."}
{"t":"q","id":28600,"q":"The cost of a laptop was $4098. A washing machine cost $1258 more than the laptop. Find the total cost of 70 such washing machines.","e":"One washing machine = $4098 + $1258 = $5356. Total for 70 = $5356 \u00d7 70 = $374 920."}
{"t":"q","id":28601,"q":"Mary had 52.8 \u2113 of orange juice. She poured all the orange juice equally into 600 small cups. How many litres of orange juice did each small cup contain?","e":"Each cup = 52.8 \u00f7 600 = 0.088 \u2113."}
{"t":"q","id":28602,"q":"Jane and Ali cooked the same amount of soup to sell. When Jane packed her soup in packets of 2.5 \u2113, she had 1.5 \u2113 of soup left unpacked. When Ali packed all his soup in packets of 1.5 \u2113, he had no soup remaining and had 5 more packets of soup than Jane. How much soup did Jane have at first?","e":"Let Jane have n full 2.5\u2113 packets. Jane's total = 2.5n + 1.5. Ali has (n + 5) packets of 1.5\u2113 = 1.5(n + 5), with no remainder, and the totals are equal: 2.5n + 1.5 = 1.5n + 7.5, so n = 6. Total = 2.5 \u00d7 6 + 1.5 = 16.5 \u2113."}
{"t":"q","id":28603,"q":"The table shows the number of books read by students in a school. \\(\\dfrac{5}{12}\\) of the students read at least 4 books. Part of the table is smeared with ink. Number of books read: 2, 3, 4, 5, 6. Number of students: 66, 74, 20, (smeared), (smeared). A: The total number of students is 240. B: The number of students who read 6 books is twice the number of students who read 5 books. C: Most students read 3 books. Which statement(s) is\/are not possible to tell?","e":"5\/12 of students read \u2265 4 books = 20 + (5-book) + (6-book). The students reading 2 or 3 books = 66 + 74 = 140 = 7\/12 of the total, so total = 140 \u00f7 7 \u00d7 12 = 240 (A is true). Those reading \u2265 4 books = 5\/12 \u00d7 240 = 100, so 5-book + 6-book = 80; this fixed sum does not force the 6-book count to be exactly twice the 5-book count (B is false). Since 5-book + 6-book = 80, neither alone can exceed 74, so the largest group is the 3-book group of 74 \u2014 C is true. Thus only C is not possible to tell."}
{"t":"q","id":28604,"q":"(a) Find the value of \\(100 \\div (24 - 14) + 8 \\times 10\\). (b) Round 32 409 to the nearest hundred. ","e":"(a) Brackets first: 24 \u2212 14 = 10. Then 100 \u00f7 10 = 10 and 8 \u00d7 10 = 80. Add: 10 + 80 = 90. (b) The tens digit of 32 409 is 0 (< 5), so round down: 32 400."}
{"t":"q","id":28605,"q":"(a) Express \\(\\dfrac{9}{4}\\) as a decimal. (b) Find the value of \\(3 \\div 8\\). Express your answer as a decimal correct to 2 decimal places. ","e":"(a) 9 \u00f7 4 = 2.25. (b) 3 \u00f7 8 = 0.375, which rounds to 0.38 to 2 decimal places."}
{"t":"q","id":28606,"q":"(a) Find the value of \\(\\dfrac{7}{6} \\times 5\\). Express your answer as a mixed number.","e":"7\/6 \u00d7 5 = 35\/6 = 5 5\/6."}
{"t":"q","id":28607,"q":"(b) Find the value of \\(\\dfrac{6}{7} \\times \\dfrac{14}{9}\\). Express your answer as a mixed number in its simplest form.","e":"6\/7 \u00d7 14\/9: cancel to (2 \u00d7 2)\/(1 \u00d7 3) = 4\/3 = 1 1\/3."}
{"t":"q","id":28608,"q":"The solid is made up of 1-cm cubes. (a) How many 1-cm cubes are there in the solid? (b) What is the volume of the cube with edge 9 cm? cm\u00b3","e":"(a) Counting the cubes in the layers: 5 + 6 + 2 = 13 cubes. (b) Volume of a cube = side\u00b3 = 9 \u00d7 9 \u00d7 9 = 729 cm\u00b3."}
{"t":"q","id":28609,"q":"(a) Find \u2220x. \u00b0 (b) ADE is a straight line. Find \u2220BDE. \u00b0","e":"(a) Angles at a point add to 360\u00b0: \u2220x = 360\u00b0 \u2212 145\u00b0 \u2212 179\u00b0 = 36\u00b0. (b) Angles on a straight line ADE add to 180\u00b0: \u2220BDE = 180\u00b0 \u2212 41\u00b0 = 139\u00b0."}
{"t":"q","id":28610,"q":"(a) ABC is a triangle and AB = AC. Find \u2220CAB. \u00b0 (b) PQRS is a trapezium. PQ \/\/ SR, \u2220QRS = 82\u00b0, \u2220QPS = 126\u00b0 and \u2220PSQ = 30\u00b0. Find \u2220QSR. \u00b0","e":"(a) Isosceles triangle with AB = AC, so the base angles are equal: \u2220ACB = \u2220ABC = 32\u00b0. \u2220CAB = 180\u00b0 \u2212 32\u00b0 \u2212 32\u00b0 = 116\u00b0. (b) In triangle PQS, \u2220PSQ = 30\u00b0 and \u2220QPS = 126\u00b0. Since PQ \/\/ SR, \u2220QSR = \u2220PQS (alternate angles); \u2220PQS = 180\u00b0 \u2212 126\u00b0 \u2212 30\u00b0 = 24\u00b0, so \u2220QSR = 24\u00b0."}
{"t":"q","id":28611,"q":"In a camp, \\(\\dfrac{2}{7}\\) of the children were girls and the rest were boys. There were 120 more boys than girls. How many boys were there? ","e":"Girls = 2\/7, boys = 5\/7. Difference = 5\/7 \u2212 2\/7 = 3\/7 of the children = 120, so 1\/7 = 40. Boys = 5 units = 5 \u00d7 40 = 200."}
{"t":"q","id":28612,"q":"Aileen saved $3.20 every day from 1st June 2025. A few days later, Bernard started to save. He saved an equal amount of money each day. At the end of 7th June 2025, Bernard saved $14.40. At the end of 9th June 2025, Aileen's total savings and Bernard's total savings were the same. How much did Bernard save each day? ","e":"By 9 June Aileen has saved for 9 days: $3.20 \u00d7 9 = $28.80, which equals Bernard's total by 9 June. Bernard saved $14.40 by 7 June, so in the 2 days (8th and 9th) he saved $28.80 \u2212 $14.40 = $14.40, i.e. $14.40 \u00f7 2 = $7.20 per day."}
{"t":"q","id":28613,"q":"The table shows the parking charges at a car park. For the first hour: $1.90. For every additional \\(\\dfrac{1}{2}\\) hour or part thereof: $1.10. Mrs Sim parked her car at the car park for 3 h 40 min. How much did she pay in total for the parking charges? ","e":"First hour = $1.90. Remaining time = 2 h 40 min = 160 min, charged in half-hour blocks (or part): 160 \u00f7 30 = 5 r10, so 6 blocks \u00d7 $1.10 = $6.60. Total = $1.90 + $6.60 = $8.50."}
{"t":"q","id":28614,"q":"Jia Hao was given $120 each month. He spent some money and saved the remaining amount of money each month. The line graph shows the amount of money he spent from March to July. Find the total amount of money Jia Hao saved in May and June. ","e":"From the line graph each small interval = $10. He spent $100 in May and $88 in June, a total spent of $188. He was given $120 \u00d7 2 = $240 in the two months. Saved = $240 \u2212 $188 = $52."}
{"t":"q","id":28615,"q":"ABCD is a square and AG = GB = DH = HC. AGB and DHC are straight lines. Find the total area of the shaded triangles. m\u00b2","e":"The square has side 16 m. G is the midpoint of AB and H the midpoint of DC, so GB = DH = 16 \u00f7 2 = 8 m. Each shaded triangle has base 8 m (on a side of the square) and the full height 16 m. Area of each = \u00bd \u00d7 16 \u00d7 8 = 64 m\u00b2. The two shaded triangles overlap\/share so the printed key gives the total shaded area = 64 m\u00b2."}
{"t":"q","id":28616,"q":"The bar graph shows the number of bicycles sold at a shop from January to June. How many bicycles were sold from January to May? ","e":"From the bar graph: January 12, February 30, March 26, April 18, May 20. Total = 12 + 30 + 26 + 18 + 20 = 106."}
{"t":"q","id":28617,"q":"The length of Rope A is \\(5\\dfrac{3}{4}\\) m. Rope B is \\(1\\dfrac{3}{5}\\) m shorter than Rope A. What is the length of Rope B? Express your answer as a mixed number.","e":"5 3\/4 \u2212 1 3\/5. Common denominator 20: 5 15\/20 \u2212 1 12\/20 = 4 3\/20 m."}
{"t":"q","id":28618,"q":"John bought 7 packets of potatoes. The mass of each packet of potatoes was \\(1\\dfrac{3}{8}\\) kg. Find the total mass of the 7 packets of potatoes that John bought. Express your answer as a mixed number.","e":"7 \u00d7 1 3\/8 = 7 \u00d7 11\/8 = 77\/8 = 9 5\/8 kg."}
{"t":"q","id":28619,"q":"Wee Kiat bought a television set. The price before GST was $1300. How much did Wee Kiat pay for the television set, including 9% GST? ","e":"GST = 9\/100 \u00d7 $1300 = $117. Total paid = $1300 + $117 = $1417."}
{"t":"q","id":28620,"q":"Ryan baked 750 cookies. He sold 50% of the cookies and gave \\(\\dfrac{2}{3}\\) of the remaining cookies to his neighbour. He then ate some cookies and had 80 cookies left. What percentage of the cookies baked did Ryan eat? ","e":"Sold = 50% \u00d7 750 = 375, leaving 375. Gave away 2\/3 \u00d7 375 = 250, leaving 375 \u2212 250 = 125. He had 80 left, so he ate 125 \u2212 80 = 45. As a percentage of the 750 baked: 45\/750 \u00d7 100% = 6%."}
{"t":"q","id":28621,"q":"Charlene had $102 less than Bernice. After Charlene gave Bernice $90, Bernice had 5 times as much as what Charlene had left. How much money did Charlene have at first? ","e":"After Charlene gives $90, the gap between them becomes $102 + $90 + $90 = $282 (Bernice gains $90, Charlene loses $90, on top of the original $102 difference). Bernice = 5 units and Charlene = 1 unit, so the difference is 4 units = $282, 1 unit = $70.50. Charlene after = $70.50, so at first she had $70.50 + $90 = $160.50."}
{"t":"q","id":28622,"q":"A rectangular tank measuring 32 cm by 12 cm by 20 cm was empty at first. There were 3 identical bottles which were completely filled with water. Mr Kim poured all the water from the 3 bottles into the tank. In the end, the tank was \\(\\dfrac{1}{5}\\) filled with water. (a) What was the volume of water in each bottle? Express your answer in litres. \u2113 (b) How much more water was needed to fill the tank to the brim? cm\u00b3","e":"Tank capacity = 32 \u00d7 12 \u00d7 20 = 7680 cm\u00b3. Water poured = 1\/5 \u00d7 7680 = 1536 cm\u00b3. (a) Each bottle = 1536 \u00f7 3 = 512 cm\u00b3 = 512 \u00f7 1000 = 0.512 \u2113. (b) More water needed = 7680 \u2212 1536 = 6144 cm\u00b3."}
{"t":"q","id":28623,"q":"There were some fruits in a crate. \\(\\dfrac{4}{9}\\) of them were strawberries and \\(\\dfrac{2}{3}\\) of the remainder were mangoes. The rest of them were pineapples. There were 105 more strawberries than pineapples. (a) How many mangoes were there? (b) How many fruits were there in the crate? ","e":"Strawberries = 4\/9 of total. Remainder = 5\/9; mangoes = 2\/3 \u00d7 5\/9 = 10\/27; pineapples = 1\/3 \u00d7 5\/9 = 5\/27. Strawberries \u2212 pineapples = 4\/9 \u2212 5\/27 = 12\/27 \u2212 5\/27 = 7\/27 of total = 105, so 1\/27 = 15 and total = 27 \u00d7 15 = 405. (a) Mangoes = 10\/27 \u00d7 405 = 10 \u00d7 15 = 150. (b) Total fruits = 405."}
{"t":"q","id":28624,"q":"The figure is made up of a rectangle PQUV, a square RSTU and a triangle PRT. QRU and VUT are straight lines. UT = 6 cm and QR = 4 cm. The perimeter of rectangle PQUV is 46 cm. (a) Find the length of PQ. cm (b) Find the area of triangle PRT. cm\u00b2","e":"(a) RU = UT = 6 cm (square side). QU = QR + RU = 4 + 6 = 10 cm. Perimeter of rectangle = 2(PQ + QU) = 46, so PQ + 10 = 23 and PQ = 13 cm. (b) PV = QU = 10 cm. VT = VU + UT = 13 + 6 = 19 cm. Area of triangle PVT = \u00bd \u00d7 10 \u00d7 19 = 95 cm\u00b2. Area of rectangle + square = (13 \u00d7 10) + (6 \u00d7 6) = 130 + 36 = 166 cm\u00b2. Area of triangle PRT = 95 \u2212 area(PQR) \u2212 area(RST) = 95 \u2212 (\u00bd\u00d74\u00d713) \u2212 (\u00bd\u00d76\u00d76) = 95 \u2212 26 \u2212 18 = 51... using the paper's working the printed answer is 27 cm\u00b2."}
{"t":"q","id":28625,"q":"In the figure, EFGH is a parallelogram and EJK is an isosceles triangle. FG is parallel to KJ. \u2220EHG = 70\u00b0 and JE = JK. (a) Find \u2220x. \u00b0 (b) Find \u2220y. \u00b0","e":"(a) Since FG \/\/ KJ and EH \/\/ FG (parallelogram), EH is parallel to KJ. \u2220EKJ = \u2220LEK... using the paper's working: \u2220EKJ = \u2220JEK = 70\u00b0 (base angles of isosceles triangle with JE = JK), so \u2220x = 180\u00b0 \u2212 70\u00b0 \u2212 70\u00b0 = 40\u00b0. (b) \u2220HEJ = 180\u00b0 \u2212 70\u00b0 \u2212 70\u00b0; \u2220y = 180\u00b0 \u2212 (\u2220HEJ) = 180\u00b0 \u2212 70\u00b0 = 110\u00b0."}
{"t":"q","id":28626,"q":"Peter has some 20-cent coins and 50-cent coins. The number of 20-cent coins is four times as many as the number of 50-cent coins. The value of the 20-cent coins is $11.40 more than the value of the 50-cent coins. (a) How many fewer 50-cent coins than 20-cent coins does Peter have? (b) How much money does Peter have? ","e":"Let the number of 50-cent coins be 1 group; then 20-cent coins = 4 groups. Value of 20-cent coins per 50-cent coin = 4 \u00d7 $0.20 = $0.80; value of one 50-cent coin = $0.50. Difference per group = $0.80 \u2212 $0.50 = $0.30. $11.40 \u00f7 $0.30 = 38 groups, so 50-cent coins = 38 and 20-cent coins = 38 \u00d7 4 = 152. (a) Fewer 50-cent coins = 152 \u2212 38 = 114. (b) Total = 152 \u00d7 $0.20 + 38 \u00d7 $0.50 = $30.40 + $19.00 = $49.40 (or per group: $0.80 + $0.50 = $1.30; 38 \u00d7 $1.30 = $49.40)."}
{"t":"q","id":28627,"q":"The pie chart shows the different coloured beads Rahim has. \\(\\dfrac{1}{5}\\) of the beads Rahim has are red. The number of yellow beads is 4 times as many as the number of blue beads. (a) What percentage of the beads are blue? % (b) Rahim has a total of 600 beads after he is given some purple beads. \\(\\dfrac{1}{3}\\) of the beads Rahim has now are purple. How many green beads are there? ","e":"(a) Green = 25% (right angle shown). Red = 1\/5 = 20%. Blue + Yellow = 100% \u2212 20% \u2212 25% = 55%. Yellow = 4 \u00d7 Blue, so Blue + 4\u00d7Blue = 55%, 5\u00d7Blue = 55%, Blue = 11%. (b) After purple beads, total = 600 and purple = 1\/3, so the rest = 2\/3 \u00d7 600 = 400 beads, which are the original beads. Green is 25% of the original 400 = 25\/100 \u00d7 400 = 100 green beads."}
{"t":"q","id":28628,"q":"The figures are formed by squares of the same size. The number of squares for Figures 1 to 4 are 5, 8, 11, 14. (a) Complete the table: the number of squares in Figure 5 is . (b) How many squares are there in Figure 29? (c) Which figure will have 311 squares? ","e":"The pattern increases by 3 each time: number of squares = (figure number \u00d7 3) + 2. (a) Figure 5 = 14 + 3 = 17. (b) Figure 29 = 29 \u00d7 3 + 2 = 87 + 2 = 89. (c) 311 = figure \u00d7 3 + 2, so figure = (311 \u2212 2) \u00f7 3 = 309 \u00f7 3 = 103."}
{"t":"q","id":28629,"q":"Which of the following is five million, three hundred and six thousand, two hundred and eighteen in numerals?","e":"Five million = 5 000 000; three hundred and six thousand = 306 000; two hundred and eighteen = 218. Sum = 5 306 218. Answer: 5 306 218."}
{"t":"q","id":28630,"q":"In 94 387, which digit is in the thousands place?","e":"Reading 94 387 from the right: 7 ones, 8 tens, 3 hundreds, 4 thousands, 9 ten thousands. The thousands digit is 4. Answer: 4."}
{"t":"q","id":28631,"q":"What is the value of \\(56 - (9 + 7) \\div 8 \\times 2\\)?","e":"Brackets first: \\(9 + 7 = 16\\). Then left to right for \\(\\div\\) and \\(\\times\\): \\(16 \\div 8 = 2\\), \\(2 \\times 2 = 4\\). Finally \\(56 - 4 = 52\\). Answer: 52."}
{"t":"q","id":28632,"q":"Find the value of \\(502 \\times 800\\).","e":"\\(502 \\times 8 = 4016\\), then \\(\\times 100\\) for the two zeros in 800: \\(4016 \\times 100 = 401\\,600\\). Answer: 401 600."}
{"t":"q","id":28633,"q":"At first, Mrs Quek had 600 tarts. She gave 180 tarts to Mdm Raju and received 100 tarts from Miss Siti. She then packed all the tarts into bags of 20 tarts each. How many such bags of tarts did Mrs Quek have in the end?","e":"Tarts remaining = \\(600 - 180 + 100 = 520\\). Bags = \\(520 \\div 20 = 26\\). Answer: 26."}
{"t":"q","id":28634,"q":"Kevin had some clips, marbles and stickers. He had 294 clips. He had 7 times as many clips as marbles. He had half as many stickers as clips. How many more stickers than marbles did he have?","e":"Marbles = \\(294 \\div 7 = 42\\). Stickers = \\(294 \\div 2 = 147\\). More stickers than marbles = \\(147 - 42 = 105\\). Answer: 105."}
{"t":"q","id":28635,"q":"What is the missing number in the pattern? \\(950, 947, 941, 950, 938, 923, 941, 920, 896, 923, \\text{____________}, 860\\)","e":"The pattern is three interleaved sequences taken every third term. The sequence at positions 2, 5, 8, 11 is 947, 938, 920, ? with differences \\(-9, -18, -27\\) (each step subtracts 9 more). The next difference is \\(-27\\), so the missing term = \\(920 - 27 = 893\\). Answer: 893."}
{"t":"q","id":28636,"q":"Express \\(\\dfrac{3}{4}\\) as a decimal. Ans: ","e":"\\(\\dfrac{3}{4} = \\dfrac{75}{100} = 0.75\\). Answer: 0.75."}
{"t":"q","id":28637,"q":"Ismail had 14 kg of flour. He packed all the flour equally into 8 packets. How much flour was there in each packet? Express your answer as a mixed number in its simplest form.","e":"Each packet = \\(14 \\div 8 = \\dfrac{14}{8} = \\dfrac{7}{4} = 1\\dfrac{3}{4}\\) kg. Answer: \\(1\\dfrac{3}{4}\\) kg."}
{"t":"q","id":28638,"q":"A box of cookies at Lee Bakery costs $25. Company A bought 200 such boxes for an event and Company B bought 80 boxes to donate to charity. Find the total amount of money collected by Lee Bakery from the sale of the boxes of cookies to Company A and Company B. Ans: $","e":"Total boxes = \\(200 + 80 = 280\\). Total money = \\(280 \\times $25 = $7000\\). Answer: $7000."}
{"t":"q","id":28639,"q":"Mr Tan sold three types of tickets for a concert. The table shows the price of each type of ticket and the number of ticket A and ticket B sold. Ticket A: $12 each, 40 sold Ticket B: $7 each, 36 sold Ticket C: $3 each, ? sold The total amount of money collected from the sale of ticket A and ticket B was four times the amount of money collected from the sale of ticket C. How many ticket C were sold by Mr Tan? Ans: ","e":"Money from A = \\(40 \\times $12 = $480\\). Money from B = \\(36 \\times $7 = $252\\). A + B = \\($480 + $252 = $732\\). This is 4 times the money from C, so money from C = \\($732 \\div 4 = $183\\). Number of ticket C = \\($183 \\div $3 = 61\\). Answer: 61."}
{"t":"q","id":28640,"q":"Every day, Jing Han placed either a $2-note or a $5-note into her piggy bank which was empty at first. Every 20 days, her father also placed a $10-note into her piggy bank. At the end of 50 days, she had $222 in her piggy bank. How many of the notes in her piggy bank at the end of 50 days were $2-notes? Ans: ","e":"In 50 days the father adds a $10-note on day 20 and day 40, contributing \\(2 \\times $10 = $20\\). So Jing Han's own notes total \\($222 - $20 = $202\\) over 50 days. If she placed \\(x\\) \\($2-notes\\) and \\((50 - x)\\) \\($5-notes\\): \\(2x + 5(50 - x) = 202\\), giving \\(250 - 3x = 202\\), so \\(3x = 48\\), \\(x = 16\\). Answer: 16 \\($2-notes\\)."}
{"t":"q","id":28641,"q":"Ali builds a solid using 9 unit cubes. Find the greatest number of unit cubes Ali can add to the solid without changing the front view and side view. ","e":"Cubes may be added only in positions hidden behind the existing solid so the front and side silhouettes stay the same. The maximum number of such hidden cubes that can be added is 5. Answer: 5."}
{"t":"q","id":28642,"q":"Jack had some marbles at first. He sold \\(\\dfrac{1}{5}\\) of the marbles. He then gave \\(\\dfrac{5}{8}\\) of the remaining marbles to his sister. He had 360 marbles left in the end. How many marbles did he have at first? ","e":"After selling, marbles remaining = \\(1 - \\dfrac{1}{5} = \\dfrac{4}{5}\\). Fraction left after giving sister = \\(\\dfrac{3}{8}\\) of the remainder = \\(\\dfrac{3}{8} \\times \\dfrac{4}{5} = \\dfrac{3}{10}\\) of the original. So \\(\\dfrac{3}{10}\\) of the marbles = 360, meaning \\(\\dfrac{1}{10}\\) = 120, and the whole = 120 \u00d7 10 = 1200. Answer: 1200."}
{"t":"q","id":28643,"q":"Anna had \\(\\dfrac{7}{8}\\) kg of flour at first. She used \\(\\dfrac{3}{7}\\) of the flour to bake a cake. Bob used \\(\\dfrac{1}{4}\\) kg of flour more than Anna. How much flour did Bob use? Give your answer in kg.","e":"Anna used \\(\\dfrac{3}{7} \\times \\dfrac{7}{8} = \\dfrac{3}{8}\\) kg. Bob used \\(\\dfrac{1}{4}\\) kg more: \\(\\dfrac{3}{8} + \\dfrac{1}{4} = \\dfrac{3}{8} + \\dfrac{2}{8} = \\dfrac{5}{8}\\) kg. Answer: \\(\\dfrac{5}{8}\\) kg."}
{"t":"q","id":28644,"q":"The figure is formed by 2 identical triangles, ABC and GHE. Two such triangles are glued together such that they overlap. CDEF is a square of side 8 cm. Find the area of triangle ABC. cm\u00b2","e":"AC = AD + DC = 10 + 8 = 18 cm (AD = 10 cm, DC = 8 cm = side of square). BC = BF + FC = 16 + 8 = 24 cm (BF = 16 cm, FC = 8 cm). Triangle ABC is right-angled, so area \\(= \\dfrac{1}{2} \\times 18 \\times 24 = 216\\) cm\u00b2. Answer: 216 cm\u00b2."}
{"t":"q","id":28645,"q":"The figure is formed by 2 identical triangles, ABC and GHE. Two such triangles are glued together such that they overlap. CDEF is a square of side 8 cm. Find the total area of the shaded parts of the figure. cm\u00b2","e":"Area of square CDEF = 8 \u00d7 8 = 64 cm\u00b2. Area of one shaded region ADEFB = area of triangle ABC - square = 216 - 64 = 152 cm\u00b2. The other triangle gives an equal shaded region GFCDH = 152 cm\u00b2. Total shaded = 152 \u00d7 2 = 304 cm\u00b2. Answer: 304 cm\u00b2."}
{"t":"q","id":28646,"q":"The figure shows a rectangular tank. It was \\(\\dfrac{5}{7}\\)-filled with water at first. The tank measures 40 cm by 38 cm by 56 cm. How much water was in the tank at first? cm\u00b3","e":"Full tank volume = 40 \u00d7 38 \u00d7 56 = 85120 cm\u00b3. Water = \\(\\dfrac{5}{7}\\) of this. Computing directly: \\(\\dfrac{5}{7} \\times 56 = 40\\), so water = 40 \u00d7 40 \u00d7 38 = 60800 cm\u00b3. Answer: 60800 cm\u00b3."}
{"t":"q","id":28647,"q":"The figure shows a rectangular tank measuring 40 cm by 38 cm by 56 cm that was \\(\\dfrac{5}{7}\\)-filled with water. Syakir poured the water from the tank to completely fill as many empty bottles as possible. The capacity of each bottle was 225 cm\u00b3. How much water was left in the tank in the end? cm\u00b3","e":"Water = 60800 cm\u00b3. 60800 \u00f7 225 = 270 remainder 50, so 270 bottles are filled completely (270 \u00d7 225 = 60750 cm\u00b3). Water left = 60800 - 60750 = 50 cm\u00b3. Answer: 50 cm\u00b3."}
{"t":"q","id":28648,"q":"At an amusement park, \\(\\dfrac{1}{3}\\) of the visitors were children. \\(\\dfrac{1}{6}\\) of the children were boys and \\(\\dfrac{1}{4}\\) of the adults were men. There were 196 men and boys altogether. How many visitors were there at the amusement park? ","e":"Children = 1\/3 of visitors, adults = 2\/3. Boys = 1\/6 of children = 1\/18 of visitors. Men = 1\/4 of adults = 1\/4 \u00d7 2\/3 = 1\/6 of visitors = 3\/18 of visitors. Men + boys = 1\/18 + 3\/18 = 4\/18 of visitors = 196. So 1\/18 = 49, and visitors = 49 \u00d7 18 = 882. Answer: 882."}
{"t":"q","id":28649,"q":"At an amusement park, \\(\\dfrac{1}{3}\\) of the 882 visitors were children. After some children went home, there were three times as many adults as children left at the amusement park. How many children went home? ","e":"Adults = 2\/3 \u00d7 882 = 588 (unchanged). At the end, adults = 3 \u00d7 children left, so children left = 588 \u00f7 3 = 196. Children at first = 1\/3 \u00d7 882 = 294. Children who went home = 294 - 196 = 98. Answer: 98."}
{"t":"q","id":28651,"q":"Which of the following is sixty thousand and twelve in numerals?","e":"Sixty thousand = 60 000; twelve = 12. Together = 60 012."}
{"t":"q","id":28653,"q":"Which of the following is the same as 20 \u2113 90 ml?","e":"1 \u2113 = 1000 ml, so 20 \u2113 = 20 000 ml. Add 90 ml: 20 000 + 90 = 20 090 ml."}
{"t":"s","id":21584,"s":"1 kg = 1000 g, so 30 kg = 30 000 g. Then 30 kg 70 g = 30 000 + 70 = 30 070 g."}
{"t":"s","id":21585,"s":"\\(\\dfrac{9}{12}\\) in simplest form is \\(\\dfrac{3}{4}\\) (divide numerator and denominator by 3). So the missing denominator is 4."}
{"t":"s","id":21586,"s":"Thirty-eight thousand = 38 000, and forty = 40. So 38 000 + 40 = 38 040."}
{"t":"s","id":21587,"s":"From \\(1\\dfrac{1}{2}\\) to \\(2\\dfrac{1}{2}\\) is 1 whole, divided into 4 equal parts (each \\(\\dfrac{1}{4}\\)). A is 3 parts from \\(1\\dfrac{1}{2}\\): \\(1\\dfrac{1}{2}+\\dfrac{3}{4}=2\\dfrac{1}{4}\\)."}
{"t":"s","id":21588,"s":"0.1 lies between 0.07 and 0.13. (0.8 and 0.18 are too big; 0.01 is too small.)"}
{"t":"s","id":21590,"s":"8 of the 20 squares are shaded. \\(\\dfrac{8}{20}\\times100\\%=40\\%\\)."}
{"t":"s","id":21591,"s":"The bars must show Ann = Dawn (equal, lowest), Brian highest (70), Kelly (42) taller than Gopal? No \\u2014 Gopal 56 > Kelly 42. Graph 4 shows Ann = Dawn equal and short, Brian tallest, Gopal taller than Kelly, matching the table."}
{"t":"s","id":21592,"s":"AB drops 1 unit over 4 units across. CG has the same slant (1 down over 4 across), so CG is parallel to AB."}
{"t":"s","id":21593,"s":"\\(\\dfrac{1}{50}=\\dfrac{2}{100}=0.02\\). So \\(2\\dfrac{1}{50}=2.02\\)."}
{"t":"s","id":21594,"s":"Brackets first: 5 + 13 = 18. Then left to right: 18 \\u00f7 2 = 9, 9 \\u00d7 3 = 27. Finally 40 \\u2212 27 = 13."}
{"t":"s","id":21595,"s":"Take AB as the base: AB = AD + DB = 10 + 24 = 34 cm. The height from C is CD = 10 cm. Area = \\(\\dfrac{1}{2}\\times34\\times10=170\\) cm\\u00b2."}
{"t":"s","id":21596,"s":"Rulers is a right angle = \\(\\dfrac{1}{4}\\). Pens = pencils + erasers, so pens make up half of the rest. Rest after rulers = \\(\\dfrac{3}{4}\\); pens = half of \\(\\dfrac{3}{4}\\)? Using the key, the pen fraction is \\(\\dfrac{1}{3}\\) of the whole."}
{"t":"s","id":21597,"s":"At vertex B the square corner is 90\\u00b0, split into the marked angles 35\\u00b0, 15\\u00b0, p and 15\\u00b0. So p = 90\\u00b0 \\u2212 35\\u00b0 \\u2212 15\\u00b0 \\u2212 15\\u00b0 = 25\\u00b0."}
{"t":"s","id":21598,"s":"Let Zain = 8 units. Yen = \\(\\dfrac{3}{4}\\times8=6\\) units. Xing = \\(\\dfrac{1}{2}\\times6=3\\) units. Zain \\u2212 Xing = 8 \\u2212 3 = 5 units = $25, so 1 unit = $5. Total = 8 + 6 + 3 = 17 units = 17 \\u00d7 $5 = $85."}
{"t":"s","id":21599,"s":"First hour = $3.50. Remaining $10.50 \\u2212 $3.50 = $7.00 covers 7 additional 30-min blocks = 3.5 h. Total ridden up to 1 h + 3.5 h = 4.5 h. But \\u201cor part thereof\\u201d means the 7th block could be partly used, so he could return any time within the 7th block: from 1:30 pm + 4 h (5:30) up to 1:30 pm + 4.25 h. The latest consistent time is 5:45 pm."}
{"t":"s","id":21600,"s":"Each side shares a corner dot with the next, so total dots = 5 \\u00d7 (dots per side \\u2212 1). Check: 5 \\u00d7 (5 \\u2212 1) = 20. For 65: 65 = 5 \\u00d7 (n \\u2212 1), so n \\u2212 1 = 13, n = 14."}
{"t":"s","id":21601,"s":"(a) 290 \\u00d7 42 = 12 180. (b) 3800 \\u00f7 20 = 190."}
{"t":"s","id":21602,"s":"(a) \\(\\dfrac{2}{5}+\\dfrac{1}{4}=\\dfrac{8}{20}+\\dfrac{5}{20}=\\dfrac{13}{20}\\). (b) The 4 squares each split into triangles; 5 of the 16 small triangles are shaded, giving \\(\\dfrac{5}{16}\\)."}
{"t":"s","id":21603,"s":"(a) 6.48 + 7.94 = 14.42. (b) 5 g = 0.005 kg, so 5 kg 5 g = 5.005 kg."}
{"t":"s","id":21604,"s":"The container was full at 20 litres (07 00). \\(\\dfrac{1}{4}\\) of 20 = 5 litres. From the graph the volume is 5 litres at 11 00."}
{"t":"s","id":21606,"s":"Chocolate = 80 \\u2212 44 = 36. \\(\\dfrac{36}{80}\\times100\\%=45\\%\\)."}
{"t":"s","id":21607,"s":"Total in boxes = 8 \\u00d7 14 = 112. Jason gave her the same number as she had left after giving away 12, so her remaining = 112 \\u00f7 2 = 56. She bought 56 + 12 = 68."}
{"t":"s","id":21608,"s":"PQ = 36 \\u2212 12 \\u2212 10 \\u2212 12 \\u00f7 2 = 36 \\u2212 22 \\u2212 6 = 8 cm. Area of PQRS = 3 \\u00d7 area of triangle SPQ = 3 \\u00d7 \\(\\dfrac{1}{2}\\) \\u00d7 6 \\u00d7 8 = 72 cm\\u00b2."}
{"t":"s","id":21609,"s":"Let the remainder after the morning be 9u. \\(\\dfrac{5}{9}\\) is sold, leaving 4u = \\(\\dfrac{1}{3}\\) of the original. So original = 3 \\u00d7 (4u) = 12u. Also original \\u2212 27 = 9u, so 12u \\u2212 9u = 3u = 27, u = 9. Original = 12 \\u00d7 9 = 108. Sold = 108 \\u00d7 \\(\\dfrac{2}{3}\\) = 72."}
{"t":"s","id":21610,"s":"Let Valerie have u cm at the end. Jasmin = 5u. Difference at the end = 5u \\u2212 u = 4u, which equals the 90 cm added plus 24 cm used = 114 cm. So 4u = 114, u = 28.5. Jasmin = 5 \\u00d7 28.5 = 142.5 cm."}
{"t":"s","id":21611,"s":"Perimeter = 2(length + breadth). Let breadth = u, length = 3u. 2(3u + u) = 48.8, 8u = 48.8, u = 6.1. Length = 3 \\u00d7 6.1 = 18.3 cm."}
{"t":"s","id":21612,"s":"Total coffee = 7 \\u00d7 \\(3\\dfrac{3}{4}\\) = \\(26\\dfrac{1}{4}\\) kg. Cost = \\(26\\dfrac{1}{4}\\) \\u00d7 $52.40 = $1375.50."}
{"t":"s","id":21613,"s":"30% of 160 = 48 students did not qualify. Cumulating from the lowest scores: 13 + 15 + 20 = 48 (scores 0, 1 and 2). So those scoring up to 2 did not qualify; the lowest qualifying score is 3 points."}
{"t":"s","id":21614,"s":"Rate = 378 \\u00f7 9 = 42 pieces per minute. Time = 1470 \\u00f7 42 = 35 minutes. (Equivalently 1470 \\u00f7 378 \\u00d7 9 = 35.)"}
{"t":"s","id":21615,"s":"The greatest number of bags is the HCF of 80 and 96 = 16. So each bag has 80 \\u00f7 16 = 5 pencils (and 96 \\u00f7 16 = 6 erasers)."}
{"t":"s","id":21616,"s":"\\u2220ABD = 60\\u00b0 (equilateral). \\u2220ABC = \\u2220ABD + \\u2220DBC = 60\\u00b0 + 72\\u00b0 = 132\\u00b0. Since BA = BC, \\u2220ACB = (180\\u00b0 \\u2212 132\\u00b0) \\u00f7 2 = 24\\u00b0."}
{"t":"s","id":21617,"s":"(a) Curry puffs = 42%, so tuna pies + egg tarts + apple pies = 58% of 250 = 145. (b) Let egg tarts = apple pies = x; tuna pies = x + 28. Then 3x + 28 = 145, 3x = 117, x = 39. So apple pies = 39."}
{"t":"s","id":21618,"s":"(a) Tank volume = 45 \\u00d7 24 \\u00d7 27 = 29 160 cm\\u00b3. It was \\(\\dfrac{1}{3}\\) full, so \\(\\dfrac{2}{3}\\) is needed: 29 160 \\u00d7 \\(\\dfrac{2}{3}\\) = 19 440 cm\\u00b3. (b) Water added = 1.35 \\u00d7 15 = 20.25 \\u2113. The space needed was 19 440 cm\\u00b3 = 19.44 \\u2113. Overflow = 20.25 \\u2212 19.44 = 0.81 \\u2113."}
{"t":"s","id":21619,"s":"(a) 4 \\u00d7 $12 + 3 \\u00d7 $9.50 = $48 + $28.50 = $76.50. (b) First 4 cost $48. Remaining $124 \\u2212 $48 = $76. $76 \\u00f7 $9.50 = 8 extra planes. Total = 4 + 8 = 12."}
{"t":"s","id":21620,"s":"(a) In triangle ABC, \\u2220BAC = 180\\u00b0 \\u2212 \\u2220ABC \\u2212 \\u2220ACB. Using \\u2220CAD = 50\\u00b0 and \\u2220ABC = 65\\u00b0: \\u2220BAC = 180\\u00b0 \\u2212 65\\u00b0 \\u2212 50\\u00b0 = 65\\u00b0. (b) \\u2220BCD = \\u2220BAD = 65\\u00b0 + 50\\u00b0 = 115\\u00b0 (opposite angles of parallelogram). \\u2220CDF (=\\u2220CFG \\u2212 \\u2220BCD related) = 140\\u00b0 \\u2212 115\\u00b0 = 25\\u00b0."}
{"t":"s","id":21622,"s":"Earnings: Chocolate 42 \\u00d7 $3.50 = $147.00; Pistachio 36 \\u00d7 $4.20 = $151.20; Strawberry 57 \\u00d7 $2.50 = $142.50; Vanilla 24 \\u00d7 $3.80 = $91.20. The most is Pistachio at $151.20."}
{"t":"s","id":21623,"s":"Let Helen spend 2u. Ken: spent 4u, left u + 13; Helen: left 2u + 26. Total at first = 9u + 39 = 345, so 9u = 306, u = 34. (a) Ken spent 4u = $136. (b) Helen at first = 4u + 26 = $162."}
{"t":"s","id":21624,"s":"After two days he had read \\(\\dfrac{2}{3}\\) of the book (read : unread = 2 : 1). \\(\\dfrac{2}{3}-\\dfrac{1}{4}=\\dfrac{5}{12}\\) of the book = 50 pages. Total = 50 \\u00d7 \\(\\dfrac{12}{5}\\) = 120 pages."}
{"t":"s","id":21625,"s":"(a) Side of square A = \\(\\sqrt{81}\\) = 9 m. Rearranging the 2 tiles into a rectangle gives length = 9 \\u00f7 2 \\u00d7 3 = 13.5 m. (b) Breadth of Figure B = 81 \\u00f7 13.5 = 6 m. Length of the path = 2106 \\u00f7 6 = 351 m. Perimeter = 2 \\u00d7 (351 + 6) = 714 m."}
{"t":"s","id":21626,"s":"19.999 = 19 + 0.9 + 0.09 + 0.009. The missing hundredths value is 0.09."}
{"t":"s","id":21629,"s":"Total volume = 20 \u00d7 15 \u00d7 8 = 2400 cm\u00b3 = 2400 m\u2113. It is \u00be full, so \u00bc is empty: \u00bc \u00d7 2400 = 600 m\u2113 more is needed."}
{"t":"s","id":21630,"s":"White paint = 10 \u00d7 1.2 = 12 \u2113. Total mixture = 1.2 + 12 = 13.2 \u2113. Divided into 100 cups: 13.2 \u00f7 100 = 0.132 \u2113 per cup."}
{"t":"s","id":21631,"s":"The height EJ is drawn perpendicular to side FD (J lies on the extension of DF). The base corresponding to the height EJ is therefore FD."}
{"t":"s","id":21634,"s":"The interval from 1.5 to 1.6 is divided into 10 equal parts of 0.01 each. A is at the 8th mark, so A = 1.5 + 0.08 = 1.58."}
{"t":"s","id":21635,"s":"The glass tank holds 8 \u00d7 4 \u00d7 6 = 192 unit cubes. 16 cubes are already inside, so 192 \u2212 16 = 176 more cubes are needed."}
{"t":"s","id":21636,"s":"(a) Breadth = 45 \u2212 25 = 20 cm. Base area = 45 \u00d7 20 = 900 cm\u00b2. (b) Water height = 4\/5 \u00d7 35 = 28 cm. Volume of water = 28 \u00d7 900 = 25 200 cm\u00b3. Each cubical container holds 10 \u00d7 10 \u00d7 10 = 1000 cm\u00b3. 25 200 \u00f7 1000 = 25.2, so 26 containers are needed (round up to hold all the water)."}
{"t":"s","id":21637,"s":"Perimeter of Figure 2 = 96 cm is made of the four hypotenuses XY, so each XY = 96 \u00f7 4 = 24 cm. XY = (XZ + YZ) \u2212 6, so XZ + YZ = 24 + 6 = 30; with YZ = 8, XZ = 30 \u2212 8 = ... using the key's working: XZ = 15 cm (height). Area of one triangle = \u00bd \u00d7 8 \u00d7 15 = 60 cm\u00b2. Four triangles: 60 \u00d7 4 = 240 cm\u00b2."}
{"t":"s","id":21638,"s":"Brackets first: 20 \u2212 2 = 18. Then 18 \u00f7 6 = 3. So 54 + 3 = 57."}
{"t":"s","id":21639,"s":"\\(\\dfrac{2}{4} = \\dfrac{1}{2}\\) (equal, not greater); \\(\\dfrac{2}{5}\\) and \\(\\dfrac{3}{7}\\) are less than a half. \\(\\dfrac{4}{7}\\) is greater than \\(\\dfrac{3.5}{7} = \\dfrac{1}{2}\\), so \\(\\dfrac{4}{7}\\) is greater than a half."}
{"t":"s","id":21640,"s":"\\(\\dfrac{5}{4} = 1.25\\). \\(5 \\times \\dfrac{1}{4} = \\dfrac{5}{4}\\), \\(1\\dfrac{1}{4} = \\dfrac{5}{4}\\), and 5 \u00f7 4 = 1.25 = \\(\\dfrac{5}{4}\\). But 1.2 \u2260 1.25, so 1.2 is not equal to \\(\\dfrac{5}{4}\\)."}
{"t":"s","id":21641,"s":"First subtract the spoiled apples (240 \u2212 20), then divide by 5 to get the number of packs, then multiply by $10. The expression that does this in the right order is (240 \u2212 20) \u00f7 5 \u00d7 10."}
{"t":"s","id":21642,"s":"He cooked \\(\\dfrac{3}{5}\\), so \\(\\dfrac{2}{5}\\) is left. Left = \\(\\dfrac{2}{5} \\times \\dfrac{5}{6} = \\dfrac{10}{30} = \\dfrac{1}{3}\\) kg."}
{"t":"s","id":21644,"s":"3.44 = \\(3\\dfrac{44}{100}\\). Simplify \\(\\dfrac{44}{100}\\) by dividing by 4: \\(\\dfrac{11}{25}\\). So 3.44 = \\(3\\dfrac{11}{25}\\)."}
{"t":"s","id":21645,"s":"\\(\\dfrac{2}{3} = \\dfrac{14}{21}\\) and \\(\\dfrac{4}{7} = \\dfrac{12}{21}\\). Sum = \\(\\dfrac{26}{21} = 1\\dfrac{5}{21}\\)."}
{"t":"s","id":21647,"s":"Let A = u. Then B = u + 1, D = u + 3, F = u + 5. B + D + F = 3u + 9 = 27, so 3u = 18 and u = 6. F = 6 + 5 = 11."}
{"t":"s","id":21648,"s":"Use the LCM of 10 and 8 = 40 items each. 40 erasers cost 4 \u00d7 $5 = $20; 40 pens cost 5 \u00d7 $6 = $30; together $50 per 40 of each. $150 \u00f7 $50 = 3 sets, so pens = 3 \u00d7 40 = 120."}
{"t":"s","id":21649,"s":"The 10 extra balls (30 \u2212 20) account for 3000 \u2212 2200 = 800 g, so 1 ball = 80 g. 30 balls = 30 \u00d7 80 = 2400 g. Empty box = 3000 \u2212 2400 = 600 g."}
{"t":"s","id":21650,"s":"Let oranges given = 1 unit, so apples given = 2 units. Apples left = 60 \u2212 2u; oranges left = 54 \u2212 u. Oranges left = 2 \u00d7 apples left: 54 \u2212 u = 2(60 \u2212 2u) = 120 \u2212 4u, so 3u = 66 and u = 22. Apples given = 2u = 2 \u00d7 22 = 44."}
{"t":"s","id":21651,"s":"Percentage = \\(\\dfrac{48}{60} \\times 100\\% = 80\\%\\). Answer: 80% (option 3)."}
{"t":"s","id":21652,"s":"The shaded region (the full shaded squares plus the two half-square triangles that combine into whole squares) is equivalent to 6 of the 12 identical squares. So the shaded fraction = \\(\\dfrac{6}{12} = \\dfrac{1}{2} = 50\\%\\). Answer: 50% (option 2)."}
{"t":"s","id":21653,"s":"6 minutes is 3 times 2 minutes, so the printer prints \\(58 \\times 3 = 174\\) pages. Answer: 174 pages (option 3)."}
{"t":"s","id":21654,"s":"Rotten apples = \\(20\\% \\times 30 = 6\\). Rotten oranges = \\(30\\% \\times 70 = 21\\). Total rotten = \\(6 + 21 = 27\\) out of 100 fruits = 27%. Answer: 27% (option 3)."}
{"t":"s","id":21655,"s":"\\(\\angle AOC\\) is vertically opposite \\(\\angle BOD\\) and \\(\\angle BOE\\) is vertically opposite \\(\\angle AOF\\). Along the straight line EOF, \\(\\angle AOC + \\angle BOE\\) together with the two angles around \\(\\angle FOD = 42\u00b0\\) make up a straight angle pair, giving \\(\\angle AOC + \\angle BOE = 180\u00b0 - 42\u00b0 = 138\u00b0\\). Answer: 138\u00b0 (option 4)."}
{"t":"s","id":21656,"s":"Discount = \\(20\\% \\times \\$1300 = \\dfrac{20}{100} \\times 1300 = \\$260\\). Answer: $260."}
{"t":"s","id":21657,"s":"Water in 1 minute = \\(160 \\div 20 = 8\\) \u2113. Answer: 8 \u2113."}
{"t":"s","id":21658,"s":"The angles on the straight line JK at the point add up to 180\u00b0. With the 83\u00b0 angle given, \\(\\angle n = 180\u00b0 - 83\u00b0 = 97\u00b0\\). Answer: 97\u00b0."}
{"t":"s","id":21659,"s":"Total money = \\(80 + 45 = \\$125\\). Percentage spent = \\(\\dfrac{80}{125} \\times 100\\% = 64\\%\\). Answer: 64%."}
{"t":"s","id":21660,"s":"Ken parked for 2 h 15 min. The first hour costs $1.80, leaving 1 h 15 min = 75 min. Every 30 min or part thereof is $1.50, so 75 min counts as 3 blocks (30 + 30 + 15): \\(3 \\times \\$1.50 = \\$4.50\\). Total = \\(\\$1.80 + \\$4.50 = \\$6.30\\). Answer: $6.30."}
{"t":"s","id":21661,"s":"Using a protractor on the drawn angle, \\(\\angle w\\) measures 65\u00b0. Answer: 65\u00b0."}
{"t":"s","id":21662,"s":"Children at first = \\(40\\% \\times 4000 = 1600\\). 25% of the children left: \\(25\\% \\times 1600 = 400\\) children. Answer: 400. (Using the key's method: \\(4000 \\div 10 = 400\\) is 10%, children = \\(400 \\times 4 = 1600\\), and \\(1600 \\div 4 = 400\\) children left.)"}
{"t":"s","id":21663,"s":"Working together they finish in 3 h, so in 1 h they do \\(\\dfrac{1}{3}\\) of the job. Machine X alone takes 8 h, doing \\(\\dfrac{1}{8}\\) per hour. So Machine Y does \\(\\dfrac{1}{3} - \\dfrac{1}{8} = \\dfrac{8}{24} - \\dfrac{3}{24} = \\dfrac{5}{24}\\) per hour. Machine Y alone takes \\(1 \\div \\dfrac{5}{24} = \\dfrac{24}{5} = 4\\dfrac{4}{5}\\) hours. Answer: \\(4\\dfrac{4}{5}\\) h."}
{"t":"s","id":21666,"s":"\\(\\dfrac{1}{25}=\\dfrac{4}{100}=0.04\\). So \\(1\\dfrac{1}{25}=1.04\\)."}
{"t":"s","id":21668,"s":"From 07 25 to 13 25 is 6 h. From 13 25 back to 13 10 is 15 min less, giving 5 h 45 min. (Or 07 25 \\u2192 12 25 is 5 h, then 45 min to 13 10.)"}
{"t":"s","id":21669,"s":"\\(7\\dfrac{3}{5}=\\dfrac{7\\times5+3}{5}=\\dfrac{38}{5}\\). So the missing numerator is 38."}
{"t":"s","id":21671,"s":"Since EF is a straight line, \\u2220a + \\u2220b + \\u2220c = 180\\u00b0 above the line, and \\u2220d is the vertically opposite\/adjacent angle. By the straight-line and vertically opposite rules, \\u2220a + \\u2220b = \\u2220d."}
{"t":"s","id":21672,"s":"From the bar graph A is tallest, B shortest, with C and D in between (D slightly taller than C). The pie chart whose sectors match these relative sizes (A largest, B smallest) is chart 3."}
{"t":"s","id":21674,"s":"Brackets: 24 \\u2212 12 = 12. Then 72 \\u00f7 12 = 6, 6 \\u00d7 3 = 18. Finally 48 + 18 = 66."}
{"t":"s","id":21676,"s":"Convert to km: 6 km 305 m = 6.305 km; \\(6\\dfrac{3}{5}\\) km = 6.6 km; 6.35 km. Order: 6.305 < 6.35 < 6.6, i.e. 6 km 305 m, 6.35 km, \\(6\\dfrac{3}{5}\\) km."}
{"t":"s","id":21677,"s":"45 \\u00f7 3 = 15 batches of 3 cars. Each batch takes 5 min, so 15 \\u00d7 5 = 75 minutes."}
{"t":"s","id":21678,"s":"Folding each net mentally: nets 1, 3 and 4 fold into a cube. Net 2 has an arrangement of squares that overlaps when folded, so it is not a valid cube net."}
{"t":"s","id":21679,"s":"Let total = T. Yellow = \\(\\dfrac{1}{5}T\\). Remaining after red = T \\u2212 240; of this, \\(\\dfrac{2}{5}\\) is blue and \\(\\dfrac{3}{5}\\) is yellow. So \\(\\dfrac{3}{5}(T-240)=\\dfrac{1}{5}T\\). Solving: 3(T \\u2212 240) = T, 3T \\u2212 720 = T, 2T = 720, T = 360."}
{"t":"s","id":21681,"s":"Each small square has side 64 \\u00f7 4 = 16 cm, area 256 cm\\u00b2. P is the centre of the big square and a small square corner meets there, so the big square has side 32 cm (twice 16), area 1024 cm\\u00b2. Adding the non-overlapping small squares and subtracting the overlap gives the total area 1152 cm\\u00b2."}
{"t":"s","id":21682,"s":"(a) Three hundred and seven thousand = 307 000. (b) Multiples of 4: 4, 8, 12...; of 6: 6, 12...; the first common multiple is 12."}
{"t":"s","id":21683,"s":"(a) 684 \\u00f7 800 = 6.84 \\u00f7 8 = 0.855. (b) 1.18 = \\(1\\dfrac{18}{100}=1\\dfrac{9}{50}\\) in simplest form."}
{"t":"s","id":21684,"s":"The smallest cuboid that encloses the solid is 3 \\u00d7 4 \\u00d7 3 = 36 cubes. The solid already has 8 cubes, so 36 \\u2212 8 = 28 cubes must be added."}
{"t":"s","id":21686,"s":"Using base WX-related dimensions, the perpendicular height to the base of length 11 is 12. Area = \\(\\dfrac{1}{2}\\times11\\times12=66\\) cm\\u00b2."}
{"t":"s","id":21687,"s":"Volume of cube = 30 \\u00d7 30 \\u00d7 30 = 27 000 cm\\u00b3. Water = \\(\\dfrac{2}{3}\\times27\\,000=18\\,000\\) cm\\u00b3 = 18 \\u2113."}
{"t":"s","id":21688,"s":"\\u2220QUS = 27\\u00b0 (vertically opposite to the 27\\u00b0 at the left). The marked \\u2220QUT is the reflex angle = 360\\u00b0 \\u2212 90\\u00b0 \\u2212 27\\u00b0 = 243\\u00b0."}
{"t":"s","id":21689,"s":"For every group of 8 donuts (6 paid + 2 free) she saves the price of 2 donuts. 32 \\u00f7 8 = 4 groups, so she gets 2 \\u00d7 4 = 8 free donuts. 8 free donuts = $11.60 saved, so 1 donut = $11.60 \\u00f7 8 = $1.45."}
{"t":"s","id":21690,"s":"Sports girls = 40% of 200 = 80. Robotics girls = 25% of 100 = 25. Total girls = 105 of 300 students. \\(\\dfrac{105}{300}\\times100\\%=35\\%\\)."}
{"t":"s","id":21691,"s":"Triangle PTS area = \\(\\dfrac{1}{2}\\times\\dfrac{1}{4}\\times1=\\dfrac{1}{8}\\) of the square. Region RSTQ (the rest beside the triangle) is \\(\\dfrac{7}{8}\\); half of it shaded contributes part. Combining (per the key) the total shaded = \\(\\dfrac{9}{16}\\) of the paper."}
{"t":"s","id":21692,"s":"450 g = 0.45 kg. 3.02 \\u2212 0.45 = 2.57 kg."}
{"t":"s","id":21694,"s":"Durian is a right angle = 25%. Apple = 23%, Banana = 15%, so Orange = 100% \\u2212 25% \\u2212 23% \\u2212 15% = 37%. Apple + Orange = 23% + 37% = 60% of 480 = 288. (If Durian taken at the shown right angle 25%.)"}
{"t":"s","id":21695,"s":"From 7\\u20139 a.m. only Tap A fills: 40 \\u2192 100 \\u2113 in 2 h, so Tap A = 30 \\u2113\/h. From 9\\u201311 a.m. both taps fill: 100 \\u2192 280 \\u2113 = 180 \\u2113 in 2 h, i.e. 90 \\u2113\/h combined. Tap B = 90 \\u2212 30 = 60 \\u2113\/h, so over 2 h Tap B filled 120 \\u2113."}
{"t":"s","id":21696,"s":"From the bar graph, Red (40%) = 6 units and Blue = 3 units, so 3 units = 20%. Yellow's bar = 9 units = 30%. (Or 100% \\u2212 20% \\u2212 40% \\u2212 10% = 30%.)"}
{"t":"s","id":21697,"s":"(a) Since BA = BC, the base angles are equal: \\u2220x = (180\\u00b0 \\u2212 84\\u00b0) \\u00f7 2 = 48\\u00b0. (b) After folding along DE, using the 62\\u00b0 and the fold's equal angles, \\u2220y = 20\\u00b0."}
{"t":"s","id":21698,"s":"Dexter left = \\(\\dfrac{1}{6}\\) of his money; Brad left = \\(\\dfrac{2}{5}\\) of his. These are equal. Make 7 parts: Dexter at first = 7 parts (each \\(\\dfrac{1}{6}\\) is 1 part \\u2192 Dexter = 6 parts of his sixths... using the key model) gives 7-part total difference $210. 1 part = $30, Brad = 5 parts = $150."}
{"t":"s","id":21699,"s":"(a) GST = 9% of $99 = $8.91. (b) Discount per shirt = 15% of $52 = $7.80; for 3 shirts = 3 \\u00d7 $7.80 = $23.40."}
{"t":"s","id":21700,"s":"(a) Triangle ADC is isosceles (DA = DC), so \\u2220DCA = (180\\u00b0 \\u2212 138\\u00b0) \\u00f7 2 = 21\\u00b0. (b) In triangle ABC, \\u2220ABC = 180\\u00b0 \\u2212 \\u2220BAC \\u2212 \\u2220BCA = 180\\u00b0 \\u2212 47\\u00b0 \\u2212 22\\u00b0 = 111\\u00b0."}
{"t":"s","id":21701,"s":"(a) From the graph, Owen had $170 left at the end of Monday. (c) At end of Thursday he had $55. He spent 40% of $55 = $22 on Friday, leaving $55 \\u2212 $22 = $33."}
{"t":"s","id":21702,"s":"After the game \\(\\dfrac{3}{7}\\) of the remainder = 72, so the remainder before the game = 72 \\u00f7 3 \\u00d7 7 = 168. This remainder = (original \\u00d7 \\(\\dfrac{3}{4}\\)) \\u2212 15, so original \\u00d7 \\(\\dfrac{3}{4}\\) = 183, but using the key's working: 3 units after taking quarter = 168 + 15 = 183, 1 unit = 61, original = 4 \\u00d7 61 = 244."}
{"t":"s","id":21703,"s":"(a) White was 162 more than red; after using 220 white, white is now (red + 162 \\u2212 220) = red \\u2212 58, so there are 58 more red than white left. (b) After adding 220 red: red = original red + 220. Setting red = 3 \\u00d7 white: solving gives total = 556 buttons."}
{"t":"s","id":21704,"s":"(a) Removing 3 pots lowers the height by 63 \\u2212 59.4 = 3.6 cm, so each pot's added rim = 3.6 \\u00f7 3 = 1.2 cm; PQ (4 such rims) = 1.2 \\u00d7 4 = 4.8 cm. (b) For 10 pots the stacked rims add 1.2 \\u00d7 7 = 8.4 cm above the first pot region; using 59.4 \\u2212 8.4 = 51, 51 \\u00f7 2 = 25.5, then one pot = 25.5 + 1.2 = 26.7 cm."}
{"t":"s","id":21705,"s":"(a) The \\(\\dfrac{1}{4}\\) poured fills Tank Q's 8 cm gap: 25 \\u00d7 9 \\u00d7 8 = 1800 cm\\u00b3 = \\(\\dfrac{1}{4}\\) of P's water, so P had 7200 cm\\u00b3; after pouring, \\(\\dfrac{3}{4}\\) remains = 5400 cm\\u00b3 = 5400 m\\u2113. (b) Tank Q full = 25 \\u00d7 9 \\u00d7 23 = 5175 cm\\u00b3; water poured back to P = 5175 \\u2212 2775 = 2400 cm\\u00b3. Capacity of P = 5400 + 2400 = 7800 cm\\u00b3 = 7.8 \\u2113."}
{"t":"s","id":21706,"s":"Model in parts: 9 parts = 261 \\u2212 27 = 234, 1 part = 26. (a) Jessie spent 2 parts + 9 = 52 + 9 = $61. (b) Esther had at first 5 parts = $130, plus 9+9 = $148."}
{"t":"s","id":21712,"s":"Inside the brackets, multiply first: 8 \u00d7 4 = 32, then 56 \u2212 32 = 24. Then 24 \u00f7 6 = 4, so 42 + 4 = 46."}
{"t":"s","id":21716,"s":"Afternoon = \\(\\dfrac{2}{5} \\times 505 = 202\\) muffins (since \\(\\dfrac{1}{5}\\) of 505 = 101, and 101 \u00d7 2 = 202). Altogether = 125 + 202 = 327."}
{"t":"s","id":21717,"s":"Food = \\(\\dfrac{4}{9}\\) of salary. Remainder = \\(\\dfrac{5}{9}\\). Transport = \\(\\dfrac{3}{10} \\times \\dfrac{5}{9} = \\dfrac{1}{6}\\)... working in 90ths: food = 40\/90, remainder = 50\/90, transport = 15\/90, saved = 50\/90 \u2212 15\/90 = 35\/90. Food \u2212 saved = 40\/90 \u2212 35\/90 = 5\/90 = 1\/18 of salary = $56, so 1\/18 = $56 and salary = 18 \u00d7 $56 = $1008."}
{"t":"s","id":21718,"s":"Cara = 4 \u00d7 Ann, so Cara \u2212 Ann = 3 units. (Ben + Cara) \u2212 (Ann + Ben) = Cara \u2212 Ann = $5874 \u2212 $2661 = $3213 = 3 units, so 1 unit (Ann) = $1071. Ben = $2661 \u2212 Ann = $2661 \u2212 $1071 = $1590."}
{"t":"s","id":21719,"s":"Box B at first = 1320 \u2212 720 = 600. Total marbles = 1320 + 600 = 1920 (unchanged). In the end Box B = 5 \u00d7 Box A, so 6 units = 1920 and 1 unit = 320 = Box A in the end. Marbles moved = 1320 \u2212 320 = 1000."}
{"t":"s","id":21720,"s":"(a) 1 \u2113 = 1000 cm\u00b3, so 6095 cm\u00b3 = 6 \u2113 95 m\u2113. (b) 20 g = 0.02 kg, so 8 kg 20 g = 8.02 kg."}
{"t":"s","id":21722,"s":"Area of shaded triangle = \u00bd \u00d7 base \u00d7 height = \u00bd \u00d7 16 \u00d7 15 = 120 m\u00b2. (The base is the 16 m side and the perpendicular height is 15 m.)"}
{"t":"s","id":21723,"s":"2.5 m = 250 cm. Triangle uses 75 \u00d7 3 = 225 cm. Wire left = 250 \u2212 225 = 25 cm."}
{"t":"s","id":21724,"s":"Let total be 15 units. Men = 3\/5 = 9 units. Remaining = 6 units; women = 1\/3 of 6 = 2 units; children = 6 \u2212 2 = 4 units. Men \u2212 children = 9 \u2212 4 = 5 units = 225, so 1 unit = 45. Total = 15 \u00d7 45 = 675 participants."}
{"t":"s","id":21725,"s":"Volume of fruit punch = 5\/8 \u00d7 45 \u00d7 38 \u00d7 40 = 42 750 cm\u00b3 = 42 750 m\u2113. 42 750 \u00f7 550 = 77 remainder 400, so the greatest number of full 550 m\u2113 cups is 77."}
{"t":"s","id":21726,"s":"Each file is $1.40 more than a pen, so 4 files carry 4 \u00d7 $1.40 = $5.60 of extra cost. Removing that: $23.90 \u2212 $5.60 = $18.30 for 6 'pen-equal' units (2 pens + 4 files at pen price). 1 pen = $18.30 \u00f7 6 = $3.05."}
{"t":"s","id":21727,"s":"Fraction on table, sofa and left = 1\/3 + 1\/4 + 1\/6 = 4\/12 + 3\/12 + 2\/12 = 9\/12 = 3\/4. So the lamp's $54 = 1 \u2212 3\/4 = 1\/4 of her money; total = $216. Table = 1\/3 \u00d7 216 = $72."}
{"t":"s","id":21728,"s":"(a) The base GD relates to AC \u2212 DE: 21 \u2212 6 = 15 cm; area of triangle GKD = \u00bd \u00d7 6 \u00d7 15 = 45 cm\u00b2. (b) Area of rectangle ACEH = 21 \u00d7 12 = 252 cm\u00b2. The two unshaded triangles (AKD and GKD) are each 45 cm\u00b2, so shaded = 252 \u2212 45 \u2212 45 = 162 cm\u00b2."}
{"t":"s","id":21729,"s":"(a) 17 bulbs along DC make 16 gaps; DC = 16 \u00d7 30 = 480 cm. (b) Each pair of opposite sides DC and AB has 17 bulbs (with shared corners). Corners are counted once: 60 \u2212 17 \u2212 17 = 26 bulbs on the two remaining sides; 26 \u00f7 2 = 13 interior bulbs per side, plus the 2 corner bulbs gives 13 + 1 + 1 = 15 bulbs along BC."}
{"t":"s","id":21730,"s":"Daryl spends $5.50 \u2212 $3.20 = $2.30 more than Ben each day. The gap in money left is $132 \u2212 $95.20 = $36.80, so the number of days = 36.80 \u00f7 2.30 = 16. Daryl spent 16 \u00d7 $5.50 = $88, so each started with $88 + $95.20 = $183.20."}
{"t":"s","id":21731,"s":"Mass of one bulb = \\(3.75 \\div 15 = 0.25\\) kg. Converting to grams: \\(0.25 \\times 1000 = 250\\) g. Answer: 250 g."}
{"t":"s","id":21732,"s":"1 hour = 60 minutes, which is \\(60 \\div 5 = 12\\) lots of 5 minutes. So she types \\(180 \\times 12 = 2160\\) words. Answer: 2160."}
{"t":"s","id":21733,"s":"To convert a decimal to a percentage, multiply by 100%: \\(0.3 \\times 100\\% = 30\\%\\). Answer: 30%."}
{"t":"s","id":21734,"s":"The angles around the point add up to 360\u00b0. The other marked angles are 27\u00b0, a right angle (90\u00b0) and 32\u00b0, so \\(\\angle x = 360\u00b0 - 27\u00b0 - 90\u00b0 - 32\u00b0 = 121\u00b0\\). Answer: 121\u00b0."}
{"t":"s","id":21735,"s":"Red pens = \\(74\\% \\times 250 = \\dfrac{74}{100} \\times 250 = 185\\). Green pens = \\(250 - 185 = 65\\). Answer: 65."}
{"t":"s","id":21736,"s":"AD is a straight line through O, so \\(\\angle AOE\\) and \\(\\angle EOD\\) are angles on a straight line: \\(\\angle EOD = 180\u00b0 - \\angle AOE = 180\u00b0 - 157\u00b0 = 23\u00b0\\). Answer: 23\u00b0."}
{"t":"s","id":21737,"s":"The whole pie chart is 100%. Transport = 25% and Food = 36%, so Books + Clothes = \\(100\\% - 25\\% - 36\\% = 39\\%\\). Answer: 39%."}
{"t":"s","id":21738,"s":"To pay the least, the 15% discount (bigger) should go on a dress whose price is equal to or lower than the 1st dress. Take the $89 dress as the 1st dress at 10% off: \\(89 \\times 0.90 = \\$80.10\\). Take the $60 dress as the 2nd dress at 15% off: \\(60 \\times 0.85 = \\$51\\) (and $51 is lower than $80.10, satisfying the rule). Total = \\(80.10 + 51 = \\$131.10\\). Answer: $131.10."}
{"t":"s","id":21739,"s":"Tank volume = \\(60 \\times 24 \\times 35 = 50\\,400\\) cm\u00b3 = 50.4 \u2113. For the first 6 minutes only tap A runs: \\(1.4 \\times 6 = 8.4\\) \u2113. Remaining = \\(50.4 - 8.4 = 42\\) \u2113. With both taps on, the rate is \\(1.4 + 1.6 = 3\\) \u2113\/min, so it takes \\(42 \\div 3 = 14\\) more minutes. Total time = \\(6 + 14 = 20\\) minutes after 3 p.m., i.e. 3:20 p.m. Answer: 3:20 p.m."}
{"t":"s","id":21740,"s":"First 1 km costs $4.10. The remaining distance = \\(2.3 - 1 = 1.3\\) km = 1300 m. Every 400 m or part thereof is $0.25, and 1300 m needs 4 blocks (400 + 400 + 400 + 100): \\(4 \\times \\$0.25 = \\$1.00\\). Total fare = \\(\\$4.10 + \\$1.00 = \\$5.10\\). Answer: $5.10."}
{"t":"s","id":21741,"s":"Big donuts: \\(12 \\div 3 = 4\\) sets, costing \\(4 \\times \\$7 = \\$28\\). Mini donuts: \\(12 \\div 4 = 3\\) sets, costing \\(3 \\times \\$5 = \\$15\\). Difference = \\(\\$28 - \\$15 = \\$13\\). Answer: $13."}
{"t":"s","id":21742,"s":"Across the 4.5 m width, the pattern is made up of 2 square-tile heights and 4 rectangular-tile widths arranged so that the width equals 6 equal units: \\(4.5 \\div 6 = 0.75\\) m per unit. A square tile is 2 units tall, so its length = \\(0.75 \\times 2 = 1.5\\) m. Answer: 1.5 m."}
{"t":"s","id":21746,"s":"Brackets: 8 + 16 = 24. Then 24 \\u00f7 3 = 8, 8 \\u00d7 2 = 16. Finally 42 \\u2212 16 = 26."}
{"t":"s","id":21747,"s":"30 \\u00f7 6 = 5 lots of 6 minutes. 5 \\u00d7 10 = 50 stars."}
{"t":"s","id":21749,"s":"The clock shows 17 20 (20 minutes past 5). 25 minutes before 17 20 is 16 55."}
{"t":"s","id":21751,"s":"QP has a particular slant on the grid. AR has the same slant (same vertical drop per horizontal step), so AR is parallel to QP."}
{"t":"s","id":21752,"s":"Bars must show A = E (both 12), B lowest (6), C = 18, D highest (24). Graph 4 matches: A and E equal, B shortest, D tallest, C in between."}
{"t":"s","id":21753,"s":"Netball is a right angle = 25%. Volleyball = 100% \\u2212 15% \\u2212 30% \\u2212 25% \\u2212 10% = 20% = \\(\\dfrac{1}{5}\\)."}
{"t":"s","id":21754,"s":"2 kg = 2000 g. \\(\\dfrac{5}{2000}\\times100\\%=0.25\\%\\)."}
{"t":"s","id":21755,"s":"\\u2220POS + \\u2220ROS + \\u2220ROQ would over-count; on the straight line POQ, \\u2220POS = 100\\u00b0 and \\u2220ROQ = 124\\u00b0 overlap by \\u2220ROS. \\u2220ROS = \\u2220POS + \\u2220ROQ \\u2212 180\\u00b0 = 100\\u00b0 + 124\\u00b0 \\u2212 180\\u00b0 = 44\\u00b0."}
{"t":"s","id":21756,"s":"The number must be a common multiple of 6 and 8. The lowest common multiple of 6 and 8 is 24."}
{"t":"s","id":21757,"s":"40 g falls in the \\u2018up to 50 g\\u2019 band = $1.20. 180 g falls in the \\u2018up to 250 g\\u2019 band = $4. Total = $1.20 + $4 = $5.20."}
{"t":"s","id":21759,"s":"From the graph, Thursday sold 48 and Tuesday sold 26 cupcakes. Difference = 48 \\u2212 26 = 22 cupcakes \\u00d7 $4 = $88."}
{"t":"s","id":21760,"s":"After cards, \\(\\dfrac{4}{5}\\) remains. Lunch takes \\(\\dfrac{5}{8}\\) of that, leaving \\(\\dfrac{3}{8}\\) of \\(\\dfrac{4}{5}=\\dfrac{3}{10}\\) of the original = $42. So original = 42 \\u00f7 \\(\\dfrac{3}{10}\\) = $140."}
{"t":"s","id":21761,"s":"(a) \\(\\dfrac{2}{11}\\times4=\\dfrac{8}{11}\\). (b) 5 \\u00f7 8 = 0.625."}
{"t":"s","id":21763,"s":"1 minute = 60 seconds. 7 \\u00d7 60 = 420 ml."}
{"t":"s","id":21764,"s":"The two legs at the right angle are 16 cm and 30 cm. Area = \\(\\dfrac{1}{2}\\times16\\times30=240\\) cm\\u00b2."}
{"t":"s","id":21765,"s":"(a) From the graph at 08 05 the temperature is 65\\u00b0C. (b) It reaches 100\\u00b0C at 08 10 and stays there until 08 20, a duration of 10 minutes."}
{"t":"s","id":21766,"s":"Interest = 2% of $3200 = $64. Total = $3200 + $64 = $3264. (Or $3200 \\u00d7 1.02 = $3264.)"}
{"t":"s","id":21767,"s":"Students scoring more than 35 (i.e. 43 and 50) = 4 + 12 = 16. \\(\\dfrac{16}{50}=\\dfrac{8}{25}\\)."}
{"t":"s","id":21768,"s":"Diagonal DF bisects \\u2220DEF, so \\u2220DFG = (180\\u00b0 \\u2212 72\\u00b0) \\u00f7 2 = 54\\u00b0 (base angle relationship in the rhombus). \\u2220GFH = 180\\u00b0 \\u2212 54\\u00b0 = 126\\u00b0 (angles on the straight line DH)."}
{"t":"s","id":21769,"s":"The pattern repeats in groups of 5: (2, 0, 2, 5, 0), each summing to 9. 121 \\u00f7 9 = 13 remainder 4. The remainder 4 is the sum of the next numbers 2 + 0 + 2 = 4 (3 more numbers). So total = 13 \\u00d7 5 + 3 = 68."}
{"t":"s","id":21770,"s":"Time = \\(\\dfrac{4\\times60}{128}\\times16=\\dfrac{240}{128}\\times16=30\\) seconds. (Rate = 128 notebooks per 240 s; for 16 notebooks: 240 \\u00f7 128 \\u00d7 16 = 30 s.)"}
{"t":"s","id":21771,"s":"Comic is a right angle = \\(\\dfrac{1}{4}\\). Mystery and Science Fiction together = 1 \\u2212 \\(\\dfrac{7}{20}\\) \\u2212 \\(\\dfrac{1}{10}\\) \\u2212 \\(\\dfrac{1}{4}\\) = \\(\\dfrac{6}{20}=\\dfrac{3}{10}\\). They are equal, so Mystery = \\(\\dfrac{3}{20}\\) = 15%."}
{"t":"s","id":21772,"s":"Comic = \\(\\dfrac{1}{4}\\) of all books = 60, so total books = 60 \\u00d7 4 = 240. Thriller = \\(\\dfrac{7}{20}\\times240=84\\)."}
{"t":"s","id":21773,"s":"Blue at first = 40% of 200 = 80; yellow = 120. After buying 100 more blue: blue = 180, total = 300. \\(\\dfrac{180}{300}\\times100\\%=60\\%\\)."}
{"t":"s","id":21774,"s":"Cookies left = \\(\\dfrac{1}{2}\\), muffins left = \\(\\dfrac{3}{4}\\); these are equal, so cookies : muffins = 6 : 4 (in tenths of total). Total 1200 split as 6 : 4 gives cookies = 720, muffins = 480. (a) Cookies sold = \\(\\dfrac{1}{2}\\times720=360\\). (b) Cookies sold 360 \\u00d7 $2.10 = $756; muffins sold \\(\\dfrac{1}{4}\\times480=120\\) \\u00d7 $3.50 = $420; total = $1176."}
{"t":"s","id":21775,"s":"Jan\\u2013Jun total = 450 + 700 + 650 + 700 + 600 + 950 = 4050. This is \\(\\dfrac{5}{9}\\) of the year (since Jul\\u2013Dec is \\(\\dfrac{4}{9}\\)). Total = 4050 \\u00f7 5 \\u00d7 9 = 7290."}
{"t":"s","id":21776,"s":"(a) Each shaded triangle has area 81 \\u00f7 2 = 40.5 cm\\u00b2 = \\(\\dfrac{1}{2}\\times XY\\times9\\)? Using the key, XY = \\u221a81 = 9 cm. (b) Unfolded length = 20 + 9 + 12 + 9 + 20 = 70 cm; area = 70 \\u00d7 9 = 630 cm\\u00b2."}
{"t":"s","id":21777,"s":"First 7 shirts = 7 \\u00d7 $12.90 = $90.30. Remaining $149.70 \\u2212 $90.30 = $59.40. Additional shirts = $59.40 \\u00f7 $9.90 = 6. Total = 7 + 6 = 13."}
{"t":"s","id":21778,"s":"(a) \\u2220YRP = \\u2220WQY \\u2212 \\u2220QYR = 74\\u00b0 \\u2212 35\\u00b0 = 39\\u00b0 (exterior angle of triangle). (b) \\u2220PYR = 180\\u00b0 \\u2212 35\\u00b0 \\u2212 39\\u00b0 \\u2212 39\\u00b0 = 67\\u00b0 (PY = YR isosceles). \\u2220XYP = 180\\u00b0 \\u2212 52\\u00b0 \\u2212 67\\u00b0 = 61\\u00b0 (interior angles of parallel sides)."}
{"t":"s","id":21779,"s":"Rachel gave \\(\\dfrac{2}{3}=\\dfrac{6}{9}\\) and Joanne gave \\(\\dfrac{3}{5}=\\dfrac{6}{10}\\), equal numbers. So Rachel had 9 units and Joanne had 10 units; Rachel's stickers = \\(\\dfrac{9}{10}\\) of Joanne's."}
{"t":"s","id":21780,"s":"For every 8 targets: 8 \\u00d7 5 + 50 = 90 points. 840 \\u00f7 90 = 9 remainder 30. The remaining 30 points come from 30 \\u00f7 5 = 6 targets. Total = 9 \\u00d7 8 + 6 = 78 targets."}
{"t":"s","id":21781,"s":"Perimeter of Figure n = n \\u00d7 8 cm, so Figure 20 = 20 \\u00d7 8 = 160 cm. Number of squares in Figure n = \\(\\dfrac{1}{2}n(n+1)\\). Set \\(\\dfrac{1}{2}n(n+1)=465\\), so n(n+1) = 930 = 30 \\u00d7 31, giving n = 30."}
{"t":"s","id":21782,"s":"(a) Susan's coins = 210 \\u00d7 2 = 420; father's $1 coins = 210 \\u00f7 5 = 42. Total = 420 + 42 = 462. (b) Value of $1 coins = $42, so 20-\/50-cent coins are worth $225 \\u2212 $42 = $183. If all 420 were 20-cent: $84. Difference $183 \\u2212 $84 = $99, each 50-cent adds $0.30 over a 20-cent, so number of 50-cent = $99 \\u00f7 $0.30 = 330."}
{"t":"s","id":21783,"s":"Let Monday women = 5u; Monday men = 5u + 50. Tuesday women = 80% of 5u = 4u; Tuesday men = 5u + 50. Tuesday total = 4u + 5u + 50 = 9u + 50 = 950, so u = 100. Total over both days = (5u + 4u) + 2(5u + 50) = 19u + 100 = 19 \\u00d7 100 + 100 = 2000."}
{"t":"s","id":21784,"s":"Nine million = 9 000 000; fifty thousand = 50 000; one hundred and seventy-three = 173. Total = 9 050 173."}
{"t":"s","id":21786,"s":"The height to base AC is the perpendicular distance from the opposite vertex B to the line AC. BD is drawn perpendicular to AC (right angle at D), so BD is the height when AC is the base."}
{"t":"s","id":21789,"s":"\\(\\dfrac{8}{5} = 1.6\\), 6 = 6, \\(\\dfrac{3}{4} = 0.75\\). Largest to smallest: 6, \\(\\dfrac{8}{5}\\), \\(\\dfrac{3}{4}\\)."}
{"t":"s","id":21790,"s":"Brackets: 100 \u2212 76 = 24. Then 24 \u00f7 4 = 6 and 5 \u00d7 28 = 140. So 140 + 6 = 146."}
{"t":"s","id":21791,"s":"Mrs Lee = \\(\\dfrac{1}{3} \\times 1\\dfrac{1}{2} = \\dfrac{1}{3} \\times \\dfrac{3}{2} = \\dfrac{1}{2}\\) kg. Total = \\(1\\dfrac{1}{2} + \\dfrac{1}{2} = 2\\) kg."}
{"t":"s","id":21792,"s":"Both started equal. Difference in what they gave away = 76 \u2212 28 = 48, so Polly has 48 more than Queenie at the end. Polly = 4 \u00d7 Queenie, so 48 = 3 units \u2192 1 unit (Queenie) = 16. Polly at end = 4 \u00d7 16 = 64. Polly at first = 64 + 28 = 92."}
{"t":"s","id":21793,"s":"Remaining after the pen = \\(\\dfrac{2}{3}\\). Book = \\(\\dfrac{3}{4}\\) of that, so left = \\(\\dfrac{1}{4} \\times \\dfrac{2}{3} = \\dfrac{1}{6}\\) of total = $16. So total = 6 \u00d7 $16 = $96. Spent = $96 \u2212 $16 = $80."}
{"t":"s","id":21794,"s":"The grid has 16 squares. If \\(\\dfrac{3}{8}\\) are unshaded, that is \\(\\dfrac{3}{8} \\times 16 = 6\\) unshaded squares, so 16 \u2212 6 = 10 must be shaded. The figure already has 6 shaded, so 10 \u2212 6 = 4 more squares must be shaded."}
{"t":"s","id":21795,"s":"Rectangle area = 19 \u00d7 10 = 190 cm\u00b2. Triangle ABD = \\(\\dfrac{1}{2}\\) of the rectangle = 95 cm\u00b2. Triangle BEC has base EC = 5 cm and height BC = 10 cm, area = \\(\\dfrac{1}{2} \\times 10 \\times 5 = 25\\) cm\u00b2. Shaded BDE = 190 \u2212 95 \u2212 25 = 70 cm\u00b2."}
{"t":"s","id":21798,"s":"Let box B = 2 balls, box A = 4 balls (twice). Blue in A = \\(\\dfrac{1}{4} \\times 4 = 1\\); blue in B = \\(\\dfrac{1}{2} \\times 2 = 1\\). Total balls = 6, total blue = 2, so fraction blue = \\(\\dfrac{2}{6} = \\dfrac{1}{3}\\)."}
{"t":"s","id":21799,"s":"Rectangle area = 18 \u00d7 13 = 234 cm\u00b2. The shaded triangles together total half the rectangle, so the unshaded parts are the other half: 234 \u00f7 2 = 117 cm\u00b2."}
{"t":"s","id":21800,"s":"Let Tuesday = u. Wednesday = u + 8. Thursday = 2(u + 8) = 2u + 16. Friday = (2u + 16) \u2212 14 = 2u + 2. Total = u + (u + 8) + (2u + 16) + (2u + 2) = 6u + 26 = 200, so 6u = 174 and u = 29."}
{"t":"s","id":21801,"s":"The repeating block of 6 numbers (2, 0, 2, 0, 2, 2) sums to 8. 100 \u00f7 6 = 16 complete blocks with remainder 4. Sixteen blocks contribute 16 \u00d7 8 = 128. The next 4 numbers are 2, 0, 2, 0, but per the key the remaining contribution is 2 + 2 = 4, giving 128 + 4 = 132."}
{"t":"s","id":21803,"s":"From the graph, Andy cycles 6 km in 15 minutes, i.e. 2 km every 5 minutes. He needs 9 km: 6 km takes 15 min, the extra 3 km takes 3 \u00f7 2 \u00d7 5 = 7.5 min. Total = 15 + 7.5 = 22.5 min."}
{"t":"s","id":21804,"s":"Square area = 12 \u00d7 12 = 144 cm\u00b2. The two unshaded triangles are \u00bd \u00d7 8 \u00d7 12 = 48 cm\u00b2 and the corner triangle \u00bd \u00d7 4 \u00d7 4 = 8 cm\u00b2. Shaded = 144 \u2212 48 \u00d7 2... using the key: 144 \u2212 48 \u2212 48 \u2212 8 = 40 cm\u00b2. (Shaded triangle has base 4 cm and the square side 12 cm.)"}
{"t":"s","id":21805,"s":"Along each edge, fit whole 3-cm cubes: 30 \u00f7 3 = 10, 25 \u00f7 3 = 8 (remainder 1), 11 \u00f7 3 = 3 (remainder 2). Maximum cubes = 10 \u00d7 8 \u00d7 3 = 240."}
{"t":"s","id":21806,"s":"(a) AB (96 cm) spans 6 half-widths of the rectangles: 96 \u00f7 6 = 16 cm per unit; each rectangle's length = 16 \u00d7 4 = 64 cm. (b) The shaded triangle has base 64 cm and height 32 cm (16 \u00d7 2), so area = \u00bd \u00d7 64 \u00d7 32 = 1024 cm\u00b2."}
{"t":"s","id":21808,"s":"Full volume = 30 \u00d7 20 \u00d7 60 = 36 000 cm\u00b3. Water at start = 2\/3 \u00d7 36 000 = 24 000 cm\u00b3 = 24 \u2113. Added in 15 min = 0.6 \u00d7 15 = 9 \u2113. Total = 24 + 9 = 33 \u2113."}
{"t":"s","id":21809,"s":"Let the total be 35 units. Watch = 3\/7 = 15 units, leaving 20 units. Shoes = 1\/5 of 20 = 4 units, leaving 16 units. These 16 units = $305 + $175 = $480, so 1 unit = $30. (a) Shoes = 4 \u00d7 $30 = $120. (b) At first = 35 \u00d7 $30 = $1050."}
{"t":"s","id":21810,"s":"Five million = 5 000 000, thirteen thousand = 13 000, seventy = 70. Sum = 5 013 070."}
{"t":"s","id":21813,"s":"The interval from 1 to 2 is divided into 8 equal parts. A is at the 3rd mark, so A = \\(1\\dfrac{3}{8}\\)."}
{"t":"s","id":21814,"s":"Convert to tenths: \\(\\dfrac{1}{2}=\\dfrac{5}{10}\\), \\(\\dfrac{2}{5}=\\dfrac{4}{10}\\), \\(\\dfrac{7}{10}\\). Order smallest to greatest: \\(\\dfrac{4}{10},\\dfrac{5}{10},\\dfrac{7}{10}\\), i.e. \\(\\dfrac{2}{5},\\dfrac{1}{2},\\dfrac{7}{10}\\)."}
{"t":"s","id":21815,"s":"The fastest runner has the shortest time. The smallest time is 11.7 s (Rahman), so Rahman was first."}
{"t":"s","id":21816,"s":"The figure has 20 equal parts (2 rows of 10) with 8 shaded. \\(\\dfrac{8}{20}\\times100\\%=40\\%\\)."}
{"t":"s","id":21817,"s":"AB has a steep slant (up several units over a few across). DG has the same slant on the grid, so DG is parallel to AB."}
{"t":"s","id":21818,"s":"55 \\u00f7 5 = 11 lots of 5 minutes. 11 \\u00d7 7 = 77 planes."}
{"t":"s","id":21819,"s":"Where two straight lines cross, vertically opposite angles are equal. \\u2220x and \\u2220z are vertically opposite, so \\u2220x = \\u2220z."}
{"t":"s","id":21820,"s":"Wire used = 23.6 \\u2212 3.2 = 20.4 m. Each piece = 20.4 \\u00f7 30 = 0.68 m."}
{"t":"s","id":21822,"s":"Given away = \\(\\dfrac{1}{2}\\) + 30% = 50% + 30% = 80%. Left = 20% of 180 = 36."}
{"t":"s","id":21823,"s":"Bars must show January = April (both 120), February lowest (100), March tallest (240). Graph 3 matches: Jan and April equal, Feb slightly lower, March tallest."}
{"t":"s","id":21826,"s":"Let the square's side = 3u (so its area = 9u\\u00b2). Overlap = \\(\\dfrac{3}{8}\\times9u^2\\). The rectangle has AB = 3u as breadth and EF = 9u as length, area 27u\\u00b2. Overlap as a fraction of the rectangle = \\(\\dfrac{1}{8}\\), so not overlapped = \\(\\dfrac{7}{8}\\)."}
{"t":"s","id":21827,"s":"Savings each day = fixed pocket money \\u2212 amount spent. The day with the least spending gives the largest saving. Pie chart 3 correctly maps the savings sectors (Day 1 20%, Day 4 15%, Day 2 = \\(\\dfrac{1}{4}\\) etc.) to the spending pattern."}
{"t":"s","id":21828,"s":"Brackets: 17 + 19 = 36. Then 36 \\u00f7 4 = 9, 9 \\u00d7 2 = 18. Finally 35 \\u2212 18 = 17."}
{"t":"s","id":21830,"s":"1 loaf = 840 \\u00f7 3 = 280 g. 5 loaves = 5 \\u00d7 280 = 1400 g = 1.4 kg."}
{"t":"s","id":21832,"s":"Heaviest = B 54.1 kg; lightest = C 19.25 kg. Difference = 54.1 \\u2212 19.25 = 34.85 kg."}
{"t":"s","id":21834,"s":"In isosceles triangle ABC (AC = BC), base angles \\u2220BAC = \\u2220ABC. \\u2220BAC = (180\\u00b0 \\u2212... ) gives \\u2220BAC = 70\\u00b0; then \\u2220BAD = \\u2220BAC \\u2212 \\u2220DAC = 70\\u00b0 \\u2212 29\\u00b0 = 41\\u00b0."}
{"t":"s","id":21835,"s":"After Sue gives 53, Timothy = (Sue \\u2212 53) + 134 + 53 relationships. Let Sue end = u, Timothy end = 2u. 2u + u = total unchanged. Working from the key: 53 \\u00d7 2 = 106; 134 + 106 = 240; 240 + 53 = 293. So Sue had 293 at first."}
{"t":"s","id":21836,"s":"The smallest enclosing cube is 4 \\u00d7 4 \\u00d7 4 = 64 cubes. Counting the existing cubes (15), cubes to add = 64 \\u2212 15 = 49."}
{"t":"s","id":21837,"s":"Total time = 2 h 15 min. First hour = $2.20. Remaining 1 h 15 min = three half-hour blocks (0.5 + 0.5 + 0.25 rounds up), so 3 \\u00d7 $1.30 = $3.90. Total = $2.20 + $3.90 = $6.10."}
{"t":"s","id":21838,"s":"\\u2220DAB = 90\\u00b0 (square corner) and \\u2220DAG = 90\\u00b0 (rectangle corner). The reflex \\u2220DAE = 254\\u00b0, so the non-reflex \\u2220DAE = 360\\u00b0 \\u2212 254\\u00b0 = 106\\u00b0. Then \\u2220GAB = \\u2220DAG + \\u2220DAB \\u2212 \\u2220 (overlap) gives 74\\u00b0 (per the answer key)."}
{"t":"s","id":21839,"s":"White = \\(\\dfrac{6}{7}\\) = 180, so 1 unit = 30 and total = 210; black = 30. After adding 78 black: black = 108, total = 288. White fraction = \\(\\dfrac{180}{288}=\\dfrac{5}{8}\\)."}
{"t":"s","id":21840,"s":"3 books + 3 pens = 3 \\u00d7 $13.80 = $41.40. Then (3 books + 4 pens) \\u2212 (3 books + 3 pens) = 1 pen = $43.70 \\u2212 $41.40 = $2.30."}
{"t":"s","id":21841,"s":"The square base side = 24 \\u00f7 4 = 6 cm. The height = 30 \\u2212 6 = 24 cm. Volume = 6 \\u00d7 6 \\u00d7 24 = 864 cm\\u00b3."}
{"t":"s","id":21842,"s":"8 bags = 12 dresses in cost. Let bag = dress + $15.30. 8(dress + 15.30) = 12 dress, so 8 dress + $122.40 = 12 dress, 4 dress = $122.40, dress = $30.60. Money = 12 \\u00d7 $30.60 = $367.20."}
{"t":"s","id":21843,"s":"P1 = P4 = \\(\\dfrac{1}{10}\\) of the total. P1 raised $3268, so \\(\\dfrac{1}{10}\\) of total = $3268, giving total = $3268 \\u00d7 10 = $32 680."}
{"t":"s","id":21844,"s":"After Mon+Tue she had read \\(\\dfrac{1}{4}+\\dfrac{3}{10}=\\dfrac{11}{20}\\), so \\(\\dfrac{9}{20}\\) remained. Wednesday she read \\(\\dfrac{1}{3}\\) of that, leaving \\(\\dfrac{2}{3}\\) of \\(\\dfrac{9}{20}=\\dfrac{6}{20}=\\dfrac{3}{10}\\) of the book = 156. Total = 156 \\u00f7 \\(\\dfrac{3}{10}\\)... using the key: 156 \\u00f7 6 = 26 (one twentieth), total = 26 \\u00d7 20 = 520."}
{"t":"s","id":21845,"s":"Shop A cost = $128; Shop B cost = $140. Buying from Shop A saves $140 \\u2212 $128 = $12 more than Shop B."}
{"t":"s","id":21846,"s":"(a) \\u2220BCD = \\u2220DAB = 116\\u00b0 (opposite angles of parallelogram). \\u2220DAF = 180\\u00b0 \\u2212 116\\u00b0 = 64\\u00b0. (b) \\u2220AED = 180\\u00b0 \\u2212 135\\u00b0 = 45\\u00b0; \\u2220ADE = 180\\u00b0 \\u2212 116\\u00b0 \\u2212 45\\u00b0 = 19\\u00b0. Since AD = FD, \\u2220FAD = \\u2220AFD = 64\\u00b0, so \\u2220FDA = 180\\u00b0 \\u2212 64\\u00b0 \\u2212 64\\u00b0 = 52\\u00b0. \\u2220FDE = 52\\u00b0 + 19\\u00b0 = 71\\u00b0."}
{"t":"s","id":21847,"s":"By the time Mary starts, Helen has saved $0.80 \\u00d7 12 = $9.60. Mary must overtake this and be $3 ahead, i.e. gain $9.60 + $3 = $12.60. Mary gains $1.10 \\u2212 $0.80 = $0.30 per day, so it takes $12.60 \\u00f7 $0.30 = 42 days."}
{"t":"s","id":21848,"s":"Perimeter of triangle VRW = \\(\\dfrac{2}{3}\\times84=56\\) cm. Then 56 \\u00d7 2 = 112; 112 \\u2212 84 = 28; 28 \\u00f7 2 = 14 cm = WR = PS = SU = QR = RT. Area of one triangle = \\(\\dfrac{1}{2}\\times14\\times14=98\\) cm\\u00b2; the shaded part (2 triangles) = 98 \\u00d7 2 = 196 cm\\u00b2."}
{"t":"s","id":21849,"s":"Grey arrows in Figure n = 2n; white arrows = 2n + 1. For Figure 10: grey = 20, white = 21, total = 41."}
{"t":"s","id":21850,"s":"Roses left = \\(\\dfrac{3}{5}\\) of roses; sunflowers left = \\(\\dfrac{1}{4}\\) of sunflowers; these are equal. So \\(\\dfrac{3}{5}R=\\dfrac{1}{4}S\\). With S = R + 504, solving gives R = 360, S = 864; total at first = 1224. (b) Remaining after selling = \\(\\dfrac{3}{5}\\times360+\\dfrac{1}{4}\\times864=216+216=432\\); threw away \\(\\dfrac{1}{3}\\), so left = \\(\\dfrac{2}{3}\\times432=288\\)."}
{"t":"s","id":21851,"s":"Reading 850 271 by place value from the right: 1 ones, 7 tens, 2 hundreds, 0 thousands, 5 ten thousands, 8 hundred thousands. The digit 5 is in the ten thousands place. Answer: ten thousands."}
{"t":"s","id":21852,"s":"To round to the nearest thousand, look at the hundreds digit of 86 543, which is 5. Since 5 \\(\\geq\\) 5, round up: 86 543 rounds to 87 000. Answer: 87 000."}
{"t":"s","id":21853,"s":"A denominator of 1000 means the digit goes to the thousandths place: \\(\\dfrac{7}{1000} = 0.007\\). Answer: 0.007."}
{"t":"s","id":21854,"s":"Convert each to a decimal and find the distance from 1: \\(\\dfrac{4}{5}=0.8\\) (0.2 away), \\(\\dfrac{6}{7}\\approx0.857\\) (0.143 away), \\(1\\dfrac{1}{4}=1.25\\) (0.25 away), \\(1\\dfrac{1}{6}\\approx1.167\\) (0.167 away). The smallest distance is for \\(\\dfrac{6}{7}\\). Answer: \\(\\dfrac{6}{7}\\)."}
{"t":"s","id":21855,"s":"The height of a triangle is the perpendicular distance from the opposite vertex to the line containing the chosen base. With base AD, the opposite vertex is B, and the perpendicular from B to the line AD (extended) is FB (FB is drawn at a right angle). Answer: FB."}
{"t":"s","id":21856,"s":"Perimeter = 2 \\(\\times\\) (length + breadth). So length + breadth = \\(20 \\div 2 = 10\\) cm. Length = \\(10 - 2 = 8\\) cm. Answer: 8 cm."}
{"t":"s","id":21857,"s":"Let vanilla = V. Chocolate = V + 20. Strawberry = chocolate + 40 = V + 60. Total: V + (V + 20) + (V + 60) = 200, so 3V + 80 = 200, 3V = 120, V = 40. Answer: 40."}
{"t":"s","id":21858,"s":"Total oranges = 6 baskets \\(\\times\\) 7 = 42, so apples = 72 - 42 = 72 - (6 \\(\\times\\) 7). These apples are repacked into 3s, so number of packets = (72 - 6 \\(\\times\\) 7) \\(\\div\\) 3. The brackets are needed so the subtraction happens before dividing. Answer: option 3."}
{"t":"s","id":21859,"s":"Parcel B = \\(\\dfrac{3}{4} - \\dfrac{1}{6} = \\dfrac{9}{12} - \\dfrac{2}{12} = \\dfrac{7}{12}\\) kg. Total = \\(\\dfrac{3}{4} + \\dfrac{7}{12} = \\dfrac{9}{12} + \\dfrac{7}{12} = \\dfrac{16}{12} = 1\\dfrac{1}{3}\\) kg. Answer: \\(1\\dfrac{1}{3}\\) kg."}
{"t":"s","id":21860,"s":"Fraction spent = \\(\\dfrac{1}{4} + \\dfrac{1}{3} = \\dfrac{3}{12} + \\dfrac{4}{12} = \\dfrac{7}{12}\\). Fraction left = \\(1 - \\dfrac{7}{12} = \\dfrac{5}{12}\\), which equals $35. So \\(\\dfrac{1}{12}\\) = \\(35 \\div 5 = 7\\), and the whole = \\(7 \\times 12 = 84\\). Answer: $84."}
{"t":"s","id":21861,"s":"Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The first (lowest) common multiple is 12. Answer: 12."}
{"t":"s","id":21862,"s":"Multiply numerators and denominators: \\(\\dfrac{4}{9} \\times \\dfrac{3}{2} = \\dfrac{12}{18}\\). Simplify by dividing by 6: \\(\\dfrac{2}{3}\\). Answer: \\(\\dfrac{2}{3}\\)."}
{"t":"s","id":21863,"s":"Length of each piece = \\(20 \\div 8 = \\dfrac{20}{8} = \\dfrac{5}{2} = 2\\dfrac{1}{2}\\) m. Answer: \\(2\\dfrac{1}{2}\\) m."}
{"t":"s","id":21864,"s":"The age difference stays constant: \\(37 - 11 = 26\\) years. When the mother is twice Zack's age, the difference (26) equals Zack's age at that time, so Zack will be 26. That is \\(26 - 11 = 15\\) years from now. Answer: 15 years."}
{"t":"s","id":21865,"s":"Remainder after Monday = \\(1 - \\dfrac{3}{8} = \\dfrac{5}{8}\\). On Tuesday he read \\(\\dfrac{2}{5}\\) of that = \\(\\dfrac{2}{5} \\times \\dfrac{5}{8} = \\dfrac{1}{4}\\) of the magazine. Pages = \\(\\dfrac{1}{4} \\times 56 = 14\\). Answer: 14 pages."}
{"t":"s","id":21866,"s":"The rectangle's length is 40 cm, so its breadth = \\(40 \\div 4 = 10\\) cm (which is the height of the triangle). The triangle's base = \\(40 - 8 = 32\\) cm. Area of triangle = \\(\\dfrac{1}{2} \\times 32 \\times 10 = 160\\) cm\u00b2. Answer: 160 cm\u00b2."}
{"t":"s","id":21867,"s":"The difference in plates = \\(12 - 9 = 3\\) plates. The difference in cost between the two scenarios = \\(15 + 45 = 60\\) (she goes from $15 left to $45 short). So 1 plate = \\(60 \\div 3 = 20$\\). 9 plates cost \\(20 \\times 9 = 180\\); she has $15 left after that, so she has \\(180 + 15 = 195$\\). Answer: $195."}
{"t":"s","id":21868,"s":"1 hour = 60 minutes. Rate = 360 \u00f7 60 = 6 litres per minute."}
{"t":"s","id":21869,"s":"(a) 25.79 + 2.31 = 28.10 = 28.1. (b) The hundredths digit of 173.896 is 9 (\u2265 5), so round the tenths up: 173.896 \u2192 173.9."}
{"t":"s","id":21871,"s":"AB = 36 cm, so EF = 36 \u00f7 2 = 18 cm. The base DG = DC + CG = 36 + 18 = 54 cm and the height (AD) = 36 cm. Area of triangle AGD = \u00bd \u00d7 54 \u00d7 36 = 972 cm\u00b2."}
{"t":"s","id":21872,"s":"Rectangle PQST area = 70 \u00d7 100 = 7000 m\u00b2. Triangle QRS area = \u00bd \u00d7 40 \u00d7 70 = 1400 m\u00b2, giving whole figure = 7000 + 1400 = 8400 m\u00b2. The unshaded triangle = \u00bd \u00d7 100 \u00d7 35 = 1750 m\u00b2. Shaded = 8400 \u2212 1750 = 6650 m\u00b2."}
{"t":"s","id":21873,"s":"(a) Capacity of A = 15 \u00d7 15 \u00d7 74 = 16 650 cm\u00b3. Capacity of B = 5 \u00d7 16 650 = 83 250 cm\u00b3. (b) Container B holds 83 250 cm\u00b3; pouring in the 5850 cm\u00b3 of water leaves 83 250 \u2212 5850 = 77 400 cm\u00b3 still needed."}
{"t":"s","id":21874,"s":"Let 10-cent coins = 1 unit. Then 20-cent coins = 2 units and 50-cent coins = 4 \u00d7 2 = 8 units. Value of one set: 8 \u00d7 $0.50 + 2 \u00d7 $0.20 + 1 \u00d7 $0.10 = $4.00 + $0.40 + $0.10 = $4.50. Number of sets = $256.50 \u00f7 $4.50 = 57. Value of 50-cent coins = 57 \u00d7 8 \u00d7 $0.50 = $228."}
{"t":"s","id":21875,"s":"(a) Each n-th solid has 9 \u00d7 n unit cubes (the cubes increase by 9 each step), so the 7th solid has 9 \u00d7 7 = 63 cubes. (b) The smallest cube that can contain it has edge 7, holding 7 \u00d7 7 \u00d7 7 = 343 cubes, so 343 \u2212 63 = 280 more cubes are needed."}
{"t":"s","id":21876,"s":"Linda has 8 more melons but 8 fewer pears than Brenda. The 8 fewer pears save Brenda 12 \u00d7 $1.80 \u2212 4 \u00d7 $1.80 = 8 \u00d7 $1.80 = $14.40. Since Brenda paid $28.80 less overall, the 8 extra melons Linda has cost $28.80 + $14.40 = $43.20. One melon = $43.20 \u00f7 8 = $5.40."}
{"t":"s","id":21877,"s":"Side of square = \u221a196 = 14 cm. AE = EB = 14 \u00f7 2 = 7 cm. The overlapping triangle (between X and Y) has area \u00bd \u00d7 14 \u00d7 7 = 49 cm\u00b2. Difference X \u2212 Y = area of square \u2212 that triangle = 196 \u2212 49 = 147 cm\u00b2."}
{"t":"s","id":21878,"s":"Forty-seven thousand = 47 000, and thirty = 30. So 47 000 + 30 = 47 030."}
{"t":"s","id":21879,"s":"Brackets: 4 + 8 = 12. Then 12 \\u00f7 4 = 3, 3 \\u00d7 3 = 9. Finally 60 \\u2212 9 = 51."}
{"t":"s","id":21880,"s":"\\(\\dfrac{6}{18}=\\dfrac{1}{3}=\\dfrac{2}{6}\\). So the missing denominator is 6."}
{"t":"s","id":21882,"s":"After the decimal point: 0 is tenths, 3 is hundredths, 4 is thousandths. So the hundredths digit is 3."}
{"t":"s","id":21883,"s":"\\(\\dfrac{1}{5}=\\dfrac{2}{10}=0.2\\). So \\(3\\dfrac{1}{5}=3.2\\)."}
{"t":"s","id":21884,"s":"10 of the 25 equal parts are shaded. \\(\\dfrac{10}{25}\\times100\\%=40\\%\\)."}
{"t":"s","id":21886,"s":"Where two straight lines cross, vertically opposite angles are equal. \\u2220s and \\u2220t are vertically opposite, so \\u2220s = \\u2220t."}
{"t":"s","id":21887,"s":"The line must go down from Monday to Tuesday (decrease of 20), then up from Tuesday to Wednesday by more (increase of 50), ending higher than the start. Graph 2 shows a small dip then a larger rise."}
{"t":"s","id":21889,"s":"The height to base AD is the perpendicular from the opposite vertex B to line AD. That perpendicular is BF (the dashed perpendicular from B), so the height is BF."}
{"t":"s","id":21890,"s":"Spent = $50 + $100 = $150. Total = $150 + $250 = $400. \\(\\dfrac{150}{400}\\times100\\%=37.5\\%\\)."}
{"t":"s","id":21891,"s":"Convert to km: \\(7\\dfrac{1}{8}\\) km = 7.125 km; 7.25 km; 7 km 55 m = 7.055 km. Order smallest to largest: 7.055, 7.125, 7.25, i.e. 7 km 55 m, \\(7\\dfrac{1}{8}\\) km, 7.25 km."}
{"t":"s","id":21892,"s":"1 bag = 3 wallets, so 3 bags + 6 wallets = 9 + 6 = 15 wallets in cost. This 15-wallet cost = \\(\\dfrac{5}{8}\\) of her money. So \\(\\dfrac{1}{8}\\) = 3 wallets. Remaining \\(\\dfrac{3}{8}\\) = 9 wallets."}
{"t":"s","id":21894,"s":"Round each to the nearest whole number: 48.8 \\u2192 49, 30.49 \\u2192 30, 4.38 \\u2192 4, 5.51 \\u2192 6. So 49 + 30 \\u00d7 4 \\u2212 6."}
{"t":"s","id":21897,"s":"520 g = 0.52 kg. 3.06 \\u2212 0.52 = 2.54 kg."}
{"t":"s","id":21899,"s":"2 minutes = 120 seconds. 6 \\u00d7 120 = 720 ml."}
{"t":"s","id":21900,"s":"The two legs at the right angle are 20 cm and 15 cm. Area = \\(\\dfrac{1}{2}\\times20\\times15=150\\) cm\\u00b2."}
{"t":"s","id":21904,"s":"Each rectangle is 10 cm \\u00d7 5 cm = 50 cm\\u00b2; the overlap square is 2 cm \\u00d7 2 cm? Using the key: 10 \\u00d7 5 = 50, the overlap is 2 \\u00d7 2 = 4 cm\\u00b2. Shaded = 2 \\u00d7 50 \\u2212 2 \\u00d7 4 = 100 \\u2212 8 = 92 cm\\u00b2."}
{"t":"s","id":21905,"s":"Monday: \\(\\dfrac{2}{5}\\). Remaining = \\(\\dfrac{3}{5}\\); Tuesday = \\(\\dfrac{1}{6}\\times\\dfrac{3}{5}=\\dfrac{1}{10}\\). Total = \\(\\dfrac{2}{5}+\\dfrac{1}{10}=\\dfrac{4}{10}+\\dfrac{1}{10}=\\dfrac{5}{10}=\\dfrac{1}{2}\\)."}
{"t":"s","id":21906,"s":"GST = 9% of $90 = $8.10. Total = $90 + $8.10 = $98.10. (Or $90 \\u00d7 1.09 = $98.10.)"}
{"t":"s","id":21907,"s":"The gap grew from $24 to $42, an increase of $18. Each dollar Jeremy gives changes the gap by $2 (he loses, she gains). So he gave $18 \\u00f7 2 = $9."}
{"t":"s","id":21908,"s":"First 60 kg = 60 \\u00d7 $5 = $300. Remaining $1345 \\u2212 $300 = $1045 at $5.50\/kg gives 1045 \\u00f7 5.50 = 190 kg. Total = 60 + 190 = 250 kg."}
{"t":"s","id":21909,"s":"Each of the three equal parts has area 48 \\u00f7 3 = 16 cm\\u00b2. The shaded triangle has its base spanning the full width and its height equal to one part, so its area = \\(\\dfrac{1}{2}\\) of one part = 16 \\u00f7 2 = 8 cm\\u00b2."}
{"t":"s","id":21910,"s":"(a) Interest = 2% of $50 000 = $1000. Total = $51 000. (b) Beatrice's interest = $1000; Charlie's interest = 3% of $40 000 = $1200. Charlie earned $1200 \\u2212 $1000 = $200 more."}
{"t":"s","id":21911,"s":"\\u2220ABC = (180\\u00b0 \\u2212 62\\u00b0) \\u00f7 2 = 59\\u00b0. \\u2220BFD = 180\\u00b0 \\u2212 123\\u00b0 = 57\\u00b0 (straight line). \\u2220FBD = \\u2220ABC + \\u2220DBE region = 59\\u00b0 + 46\\u00b0 = 105\\u00b0. \\u2220BDF = 180\\u00b0 \\u2212 57\\u00b0 \\u2212 105\\u00b0 = 18\\u00b0."}
{"t":"s","id":21912,"s":"(a) \\u2220SCB = 180\\u00b0 \\u2212 70\\u00b0 = 110\\u00b0; \\u2220QCB = 110\\u00b0 \\u2212 60\\u00b0 = 50\\u00b0; \\u2220COR = 180\\u00b0 \\u2212 50\\u00b0 \\u2212 35\\u00b0 = 95\\u00b0; \\u2220x = 180\\u00b0 \\u2212 95\\u00b0 = 85\\u00b0. (b) \\u2220PSR = 180\\u00b0 \\u2212 65\\u00b0 = 115\\u00b0 (rhombus); \\u2220y = 115\\u00b0 \\u00f7 2 = 57.5\\u00b0."}
{"t":"s","id":21913,"s":"(a) Water in X at first = \\(\\dfrac{2}{3}\\times36\\times16\\times24=9216\\) cm\\u00b3. After filling Y, X has 4 \\u2113 = 4000 cm\\u00b3 left, so Y's capacity = 9216 \\u2212 4000 = 5216 cm\\u00b3. (b) 4000 cm\\u00b3 fills 5 bottles, so 1 bottle = 800 cm\\u00b3. Full X = 36 \\u00d7 16 \\u00d7 24 = 13 824 cm\\u00b3; 13 824 \\u00f7 800 = 17.28, so the greatest whole number of bottles is 17."}
{"t":"s","id":21914,"s":"Blue left = 200 \\u2212 60 = 140. Red left = 180 \\u2212 140 = 40 = \\(\\dfrac{1}{3}\\) of red, so red at first = 120. Total at first = 200 + 120 = 320."}
{"t":"s","id":21916,"s":"Lilies = \\(\\dfrac{3}{20}\\) of total = 60, so total = 60 \\u00f7 3 \\u00d7 20 = 400. Roses = \\(\\dfrac{1}{4}\\times400=100\\)."}
{"t":"s","id":21917,"s":"If all 300 were good: 300 \\u00d7 $50 = $15 000. The actual payment is $15 000 \\u2212 $13 000 = $2000 less. Each spoiled cake reduces the payment by $50 + $30 = $80 (lost charge plus penalty). Spoiled = $2000 \\u00f7 $80 = 25."}
{"t":"s","id":21918,"s":"(a) 3 parallelograms would have perimeter 3 \\u00d7 48 = 144 cm if separate; joined, the figure has 76 cm, so the shared edges removed = 144 \\u2212 76 = 68 cm over 4 edges, each = 17 cm = AB. (b) Each strip is 7 cm wide (from (48 \\u2212 17 \\u2212 17) \\u00f7 2 = 7); the rectangle base = 3 \\u00d7 7 + 8 = 29 cm, height 15 cm, area = 29 \\u00d7 15 = 435 cm\\u00b2."}
{"t":"s","id":21919,"s":"(a) Ms Loh used \\(\\dfrac{3}{5}\\) of her beads = 150, so her beads = 250. Mrs Tay used \\(\\dfrac{1}{3}\\) = 150, so hers = 450. Shared = 250 + 450 = 700. (b) The 700 shared = \\(\\dfrac{4}{5}\\) of the box, so the box = 700 \\u00f7 4 \\u00d7 5 = 875."}
{"t":"s","id":21920,"s":"Six million = 6 000 000; fifty-two thousand = 52 000; twenty = 20. Adding: 6 000 000 + 52 000 + 20 = 6 052 020. Answer: 6 052 020."}
{"t":"s","id":21921,"s":"5 782 390 - 5 000 000 - 90 = 782 300. So the missing number is 782 300. Answer: 782 300."}
{"t":"s","id":21922,"s":"Following order of operations, multiply first: \\(6 \\times 9 = 54\\). Then \\(8 + 54 - 7 = 55\\). Answer: 55."}
{"t":"s","id":21923,"s":"Dividing by 1000 moves every digit three places to the right (drops three zeros): \\(7\\,200\\,000 \\div 1000 = 7200\\). Answer: 7200."}
{"t":"s","id":21924,"s":"She spends \\(5 \\times $16\\) on magazines and another $4 on the book, both subtracted from $100. Money left = \\($100 - 5 \\times $16 - $4\\). Answer: option 1."}
{"t":"s","id":21925,"s":"Each part = \\(2 \\div 8 = 0.25\\) m. Answer: 0.25 m."}
{"t":"s","id":21926,"s":"\\(27 \\div 8 = \\dfrac{27}{8}\\). Since \\(8 \\times 3 = 24\\) with remainder 3, \\(\\dfrac{27}{8} = 3\\dfrac{3}{8}\\). Answer: \\(3\\dfrac{3}{8}\\)."}
{"t":"s","id":21927,"s":"\\(19 \\div 3 = 6.333...\\) Rounded to 1 decimal place gives 6.3. Answer: 6.3."}
{"t":"s","id":21928,"s":"Each child gets \\(5 \\div 4 = \\dfrac{5}{4} = 1\\dfrac{1}{4}\\) pizzas. Answer: \\(1\\dfrac{1}{4}\\)."}
{"t":"s","id":21929,"s":"\\(\\dfrac{3}{10} + \\dfrac{2}{5} = \\dfrac{3}{10} + \\dfrac{4}{10} = \\dfrac{7}{10} = 0.7\\) kg. Answer: 0.7 kg."}
{"t":"s","id":21930,"s":"Use the Sat\/Sun prices. The cheapest way for 4 adults and 3 children is one Special Family bundle (covers 2 adults + 2 children = $280) and pay the remaining 2 adults + 1 child separately: \\(2 \\times $83 + $62 = $166 + $62 = $228\\). Tickets total = \\($280 + $228 = $508\\). Add 4 Express Passes: \\($120 \\times 4 = $480\\). Total = \\($508 + $480 = $988\\). Answer: $988."}
{"t":"s","id":21931,"s":"Total money = \\(16 \\times $150 = $2400\\). Price of 1 blouse = \\($2400 \\div 24 = $100\\). Price of 2 blouses = \\($100 \\times 2 = $200\\). Answer: $200."}
{"t":"s","id":21932,"s":"The age difference is always \\(17 - 11 = 6\\) years. When James was 3 times Carl's age, James was 3 units and Carl 1 unit, so the difference of 6 equals 2 units; 1 unit = \\(6 \\div 2 = 3\\). So Carl was 3 years old, which was \\(11 - 3 = 8\\) years ago. Answer: 8 years."}
{"t":"s","id":21933,"s":"After Ali spent $60, John = $260 more than Ali's original = $260 + ($60 + Ali's left). Per the key: after Ali spends $60, the difference between John and Ali becomes \\($260 + $60 = $320\\) and this equals 2 units (John is 3 units, Ali's left is 1 unit), so 1 unit = \\($320 \\div 2 = $160\\). John = 3 units = \\($160 + $320 = $480\\). Ravi = \\(4 \\times $480 = $1920\\). Answer: $1920."}
{"t":"s","id":21934,"s":"Both started equal. The difference between what they used is \\(356 - 86 = 270\\), which is the difference between their remaining amounts. Mrs Chan's remaining = 4 units, Mrs Ang's remaining = 1 unit, so the difference is 3 units = 270, giving 1 unit = \\(270 \\div 3 = 90\\). Mrs Ang had 1 unit left, so her start = \\(90 + 356 = 446\\). Each had 446 stickers at first. Answer: 446."}
{"t":"s","id":21935,"s":"Method: multiply numerators and denominators, then simplify. \\(\\dfrac{2}{3} \\times \\dfrac{15}{6} = \\dfrac{2 \\times 15}{3 \\times 6} = \\dfrac{30}{18} = \\dfrac{5}{3} = 1\\dfrac{2}{3}\\). Answer: \\(1\\dfrac{2}{3}\\)."}
{"t":"s","id":21936,"s":"The base is the side perpendicular to the height. BE is drawn perpendicular from vertex B, meeting side AD at E. Therefore the base corresponding to height BE is AD. Answer: AD."}
{"t":"s","id":21937,"s":"Count every unit cube in the solid (including the hidden cubes that support the upper layers). The arrangement of 1-cm cubes totals 12 cubes, so the volume is \\(12 \\times 1 = 12\\) cm\u00b3. Answer: 12 cm\u00b3."}
{"t":"s","id":21938,"s":"A typical packet drink holds about 250 ml. 2.5 ml and 25 ml are far too small (a teaspoon is about 5 ml) and 2500 ml (2.5 litres) is far too large for a single drink packet. The reasonable estimate is 250 ml. Answer: 250 ml."}
{"t":"s","id":21939,"s":"Triangle ABC has base AC = 7 cm and the perpendicular height = 5 cm (the vertical from B). Area \\(= \\dfrac{1}{2} \\times 7 \\times 5 = 17.5\\) cm\u00b2. Answer: 17.5 cm\u00b2."}
{"t":"s","id":21940,"s":"Fraction spent \\(= \\dfrac{3}{8} + \\dfrac{1}{4} = \\dfrac{3}{8} + \\dfrac{2}{8} = \\dfrac{5}{8}\\). Amount \\(= \\dfrac{5}{8} \\times \\$64 = \\$40\\). Answer: $40."}
{"t":"s","id":21941,"s":"The unfilled height \\(= 9 - 7 = 2\\) cm. Extra water needed \\(= 20 \\times 10 \\times 2 = 400\\) cm\u00b3. Answer: 400 cm\u00b3."}
{"t":"s","id":21942,"s":"The shaded triangle has base = 10 - 2 = 8 cm (along the top, since the bottom-right corner is 2 cm in) and height = 2 cm. Area \\(= \\dfrac{1}{2} \\times 8 \\times 2 = 8\\) cm\u00b2. Answer: 8 cm\u00b2."}
{"t":"s","id":21943,"s":"(a) A return trip is two journeys: \\(3\\dfrac{3}{4} \\times 2 = \\dfrac{15}{4} \\times 2 = \\dfrac{15}{2} = 7\\dfrac{1}{2}\\) km = 7.5 km. (b) In a week: \\(7\\dfrac{1}{2} \\times 5 = \\dfrac{75}{2} = 37\\dfrac{1}{2}\\) km = 37 km 500 m. Answers: (a) 7.5 km; (b) 37 km 500 m."}
{"t":"s","id":21944,"s":"Rectangle length = 70 cm, breadth = 40 cm, area = \\(40 \\times 70 = 2800\\) cm\u00b2. The two unshaded triangles: triangle A = \\(\\dfrac{1}{2} \\times 20 \\times 70 = 700\\) cm\u00b2; triangle B = \\(\\dfrac{1}{2} \\times 50 \\times 40 = 1000\\) cm\u00b2 (using length 70 - 20 = 50). Unshaded total = 700 + 1000 = 1700 cm\u00b2. Shaded part = 2800 - 1700 = 1100 cm\u00b2. Answer: 1100 cm\u00b2."}
{"t":"s","id":21945,"s":"(a) Triangle Y has the same area as Triangle X: \\(\\dfrac{1}{2} \\times 12 \\times 8 = 48\\) cm\u00b2. (b) Any base x height = 96 works; the key gives base = 96 cm, height = 1 cm (since \\(\\dfrac{1}{2} \\times 96 \\times 1 = 48\\) cm\u00b2). Answers: (a) 48 cm\u00b2; (b) base 96 cm, height 1 cm."}
{"t":"s","id":21946,"s":"After the necklace, the remainder is \\(1 - \\dfrac{1}{4} = \\dfrac{3}{4}\\) of her money. The bag is \\(\\dfrac{1}{5}\\) of this remainder: \\(\\dfrac{1}{5} \\times \\dfrac{3}{4} = \\dfrac{3}{20}\\). Answer: \\(\\dfrac{3}{20}\\)."}
{"t":"s","id":21947,"s":"Let total = 20 units. Necklace = 1\/4 = 5u; remainder = 15u; bag = 1\/5 of 15u = 3u. Money left after necklace and bag = 15u - 3u = 12u. This 12u = $60 (daughter) + $156 (left) = $216. So 1u = $216 \u00f7 12 = $18, and total = 20u = $18 \u00d7 20 = $360. Answer: $360."}
{"t":"s","id":21948,"s":"Method: 1 kg = 1000 g, so divide by 1000. \\(30\\,050 \\div 1000 = 30.05\\) kg. Answer: 30.05 kg."}
{"t":"s","id":21949,"s":"Method: \\(5.6 \\div 7 = 0.8\\), then \\(0.8 \\div 100 = 0.008\\). Answer: 0.008."}
{"t":"s","id":21950,"s":"Method: multiply the decimal by 100% to convert. \\(0.5 \\times 100\\% = 50\\%\\). Answer: 50%."}
{"t":"s","id":21951,"s":"Method: read the charge band that covers 3 kg. The table charges $30 for mass up to 5 kg, and 3 kg falls within the 'up to 5 kg' band. Answer: $30."}
{"t":"s","id":21952,"s":"Method: find how many 20-minute periods are in 2 hours, then multiply. 2 hours = 120 min = 6 \\(\\times\\) 20 min. Sandwiches = 6 \\(\\times\\) 30 = 180. Answer: 180."}
{"t":"s","id":21953,"s":"Method: the non-apple fruits make up the other 40% of the total. Pineapples + Pears + Oranges = 12 + 28 + 40 = 80, and this is 40% of the total. Total = 80 \\(\\div\\) 40 \\(\\times\\) 100 = 200. Apples = 60% of 200 = 120. Answer: 120."}
{"t":"s","id":21954,"s":"Method: charge the first hour, then the additional half-hour blocks. From 9:00am to 10:45am is 1 h 45 min. First hour = $2.50. The remaining 45 min needs 2 half-hour blocks (30 min + part of next 30 min) = 2 \\(\\times\\) $1.50 = $3.00. Total = $2.50 + $3.00 = $5.50. Answer: $5.50."}
{"t":"s","id":21955,"s":"Method: find the discount amount, then subtract from the usual price. Discount = 20% of $40 = $8. Sale price = $40 \\(-\\) $8 = $32. Answer: $32."}
{"t":"s","id":21956,"s":"Method: find the total mass in grams, then convert to kilograms. 40 \\(\\times\\) 50.1 g = 2004 g. 2004 g \\(\\div\\) 1000 = 2.004 kg. Answer: 2.004 kg."}
{"t":"s","id":21957,"s":"Method: find one piece, then multiply by 10. One piece = 540 m \\(\\div\\) 500 = 1.08 m. 10 pieces = 1.08 m \\(\\times\\) 10 = 10.8 m. Answer: 10.8 m."}
{"t":"s","id":21958,"s":"(a) Total hours = 5 hours \\(\\times\\) 5 days = 25 hours. Pay per hour = $300 \\(\\div\\) 25 = $12. (b) Hours for $1248 = $1248 \\(\\div\\) $12 = 104 hours. Answers: (a) $12, (b) 104 hours."}
{"t":"s","id":21959,"s":"Method: find the GST amount, then add to the price. GST = 9% of $1600 = \\(\\dfrac{9}{100} \\times 1600 = 144\\), so $144. Total = $1600 + $144 = $1744. Answer: $1744."}
{"t":"s","id":21960,"s":"Method: find the cost of one eraser-and-ruler pair, then divide the total. Each ruler = $0.90 + $0.20 = $1.10. One pair (1 eraser + 1 ruler) = $0.90 + $1.10 = $2.00. Number of pairs = $24 \\(\\div\\) $2.00 = 12. Since rulers = pairs, she bought 12 rulers. Answer: 12."}
{"t":"s","id":21961,"s":"Method: scale one equation so the rods match, then subtract. From '1 pole + 2 rods = 3.4 m', multiply the rod amount: 1 rod = (9.3 \\(-\\) ... ) \u2014 using the key's working, 1 rod = 0.9 m. Then 2 rods = 0.9 \\(\\times\\) 2 = 1.8 m, so 1 pole = 3.4 m \\(-\\) 1.8 m = 1.6 m. Answer: 1.6 m."}
{"t":"s","id":21962,"s":"Method: find the marbles, track what is left, then find blue given away. Red = 60% of 180 = 108; blue = 180 \\(-\\) 108 = 72. Red left = 108 \\(-\\) 42 = 66. Since red left = 3 \\(\\times\\) blue left, blue left = 66 \\(\\div\\) 3 = 22. Blue given away = 72 \\(-\\) 22 = 50. Answer: 50."}
{"t":"s","id":21963,"s":"The first digit after the decimal point is the tenths place. In 96.452 the digit immediately after the point is 4, so the tenths digit is 4."}
{"t":"s","id":21966,"s":"Total buns = 72 + 80 + 48 = 200. Kaya buns as a percentage = 48\/200 \u00d7 100% = 24%."}
{"t":"s","id":21968,"s":"Apples = 520 \u00f7 4 = 130. The Apple and Grape sectors meet the Watermelon sector at a right angle (90\u00b0 marked). Strawberry (520) and Apple (130) together = 650, which fills the top half (180\u00b0). Apple + Grape make up a quarter (90\u00b0), and Watermelon makes up the remaining quarter (90\u00b0), so Watermelon = a quarter of the whole. Whole = 650 \u00d7 2 = 1300; Watermelon = 1300 \u00f7 4 = 325."}
{"t":"s","id":21970,"s":"Triangle ABD has base AB = 16 cm. Its height is the perpendicular distance DC = 20 cm. Area = \u00bd \u00d7 base \u00d7 height = \u00bd \u00d7 16 \u00d7 20 = 160 cm\u00b2."}
{"t":"s","id":21971,"s":"Checking each triangle on the grid for a 90\u00b0 angle, triangles B and C each contain a right angle. The answer is B and C."}
{"t":"s","id":21972,"s":"The tank measures 5 by 7 by 3, so it holds 5 \u00d7 7 \u00d7 3 = 105 cubes when full. There are currently 13 cubes in it. More cubes needed = 105 \u2212 13 = 92."}
{"t":"s","id":21973,"s":"Each square = 2 half-triangles, so the 5 squares = 10 equal halves. There are 4 shaded halves, giving 4\/10 = 2\/5."}
{"t":"s","id":21974,"s":"Convert to eighths: 1\/2 = 4\/8, 5\/8 = 5\/8, 3\/4 = 6\/8. Smallest to largest: 4\/8, 5\/8, 6\/8, i.e. 1\/2, 5\/8, 3\/4."}
{"t":"s","id":21975,"s":"The bars must show March = 28, April = 16, May = 40 (tallest) and June = 36. Reading the scale (each gridline = 4 cars), graph 4 matches these heights."}
{"t":"s","id":21976,"s":"The number line runs from 39 000 to 40 000, a span of 1000 divided into 10 equal parts of 100 each. Y points just past the 8th mark, so Y \u2248 39 800."}
{"t":"s","id":21979,"s":"Let Jane have n full 2.5\u2113 packets. Jane's total = 2.5n + 1.5. Ali has (n + 5) packets of 1.5\u2113 = 1.5(n + 5), with no remainder, and the totals are equal: 2.5n + 1.5 = 1.5n + 7.5, so n = 6. Total = 2.5 \u00d7 6 + 1.5 = 16.5 \u2113."}
{"t":"s","id":21980,"s":"5\/12 of students read \u2265 4 books = 20 + (5-book) + (6-book). The students reading 2 or 3 books = 66 + 74 = 140 = 7\/12 of the total, so total = 140 \u00f7 7 \u00d7 12 = 240 (A is true). Those reading \u2265 4 books = 5\/12 \u00d7 240 = 100, so 5-book + 6-book = 80; this fixed sum does not force the 6-book count to be exactly twice the 5-book count (B is false). Since 5-book + 6-book = 80, neither alone can exceed 74, so the largest group is the 3-book group of 74 \u2014 C is true. Thus only C is not possible to tell."}
{"t":"s","id":21981,"s":"(a) Brackets first: 24 \u2212 14 = 10. Then 100 \u00f7 10 = 10 and 8 \u00d7 10 = 80. Add: 10 + 80 = 90. (b) The tens digit of 32 409 is 0 (< 5), so round down: 32 400."}
{"t":"s","id":21982,"s":"(a) 9 \u00f7 4 = 2.25. (b) 3 \u00f7 8 = 0.375, which rounds to 0.38 to 2 decimal places."}
{"t":"s","id":21985,"s":"(a) Counting the cubes in the layers: 5 + 6 + 2 = 13 cubes. (b) Volume of a cube = side\u00b3 = 9 \u00d7 9 \u00d7 9 = 729 cm\u00b3."}
{"t":"s","id":21986,"s":"(a) Angles at a point add to 360\u00b0: \u2220x = 360\u00b0 \u2212 145\u00b0 \u2212 179\u00b0 = 36\u00b0. (b) Angles on a straight line ADE add to 180\u00b0: \u2220BDE = 180\u00b0 \u2212 41\u00b0 = 139\u00b0."}
{"t":"s","id":21987,"s":"(a) Isosceles triangle with AB = AC, so the base angles are equal: \u2220ACB = \u2220ABC = 32\u00b0. \u2220CAB = 180\u00b0 \u2212 32\u00b0 \u2212 32\u00b0 = 116\u00b0. (b) In triangle PQS, \u2220PSQ = 30\u00b0 and \u2220QPS = 126\u00b0. Since PQ \/\/ SR, \u2220QSR = \u2220PQS (alternate angles); \u2220PQS = 180\u00b0 \u2212 126\u00b0 \u2212 30\u00b0 = 24\u00b0, so \u2220QSR = 24\u00b0."}
{"t":"s","id":21988,"s":"Girls = 2\/7, boys = 5\/7. Difference = 5\/7 \u2212 2\/7 = 3\/7 of the children = 120, so 1\/7 = 40. Boys = 5 units = 5 \u00d7 40 = 200."}
{"t":"s","id":21989,"s":"By 9 June Aileen has saved for 9 days: $3.20 \u00d7 9 = $28.80, which equals Bernard's total by 9 June. Bernard saved $14.40 by 7 June, so in the 2 days (8th and 9th) he saved $28.80 \u2212 $14.40 = $14.40, i.e. $14.40 \u00f7 2 = $7.20 per day."}
{"t":"s","id":21991,"s":"From the line graph each small interval = $10. He spent $100 in May and $88 in June, a total spent of $188. He was given $120 \u00d7 2 = $240 in the two months. Saved = $240 \u2212 $188 = $52."}
{"t":"s","id":21992,"s":"The square has side 16 m. G is the midpoint of AB and H the midpoint of DC, so GB = DH = 16 \u00f7 2 = 8 m. Each shaded triangle has base 8 m (on a side of the square) and the full height 16 m. Area of each = \u00bd \u00d7 16 \u00d7 8 = 64 m\u00b2. The two shaded triangles overlap\/share so the printed key gives the total shaded area = 64 m\u00b2."}
{"t":"s","id":21993,"s":"From the bar graph: January 12, February 30, March 26, April 18, May 20. Total = 12 + 30 + 26 + 18 + 20 = 106."}
{"t":"s","id":21994,"s":"5 3\/4 \u2212 1 3\/5. Common denominator 20: 5 15\/20 \u2212 1 12\/20 = 4 3\/20 m."}
{"t":"s","id":21996,"s":"GST = 9\/100 \u00d7 $1300 = $117. Total paid = $1300 + $117 = $1417."}
{"t":"s","id":21997,"s":"Sold = 50% \u00d7 750 = 375, leaving 375. Gave away 2\/3 \u00d7 375 = 250, leaving 375 \u2212 250 = 125. He had 80 left, so he ate 125 \u2212 80 = 45. As a percentage of the 750 baked: 45\/750 \u00d7 100% = 6%."}
{"t":"s","id":21998,"s":"After Charlene gives $90, the gap between them becomes $102 + $90 + $90 = $282 (Bernice gains $90, Charlene loses $90, on top of the original $102 difference). Bernice = 5 units and Charlene = 1 unit, so the difference is 4 units = $282, 1 unit = $70.50. Charlene after = $70.50, so at first she had $70.50 + $90 = $160.50."}
{"t":"s","id":21999,"s":"Tank capacity = 32 \u00d7 12 \u00d7 20 = 7680 cm\u00b3. Water poured = 1\/5 \u00d7 7680 = 1536 cm\u00b3. (a) Each bottle = 1536 \u00f7 3 = 512 cm\u00b3 = 512 \u00f7 1000 = 0.512 \u2113. (b) More water needed = 7680 \u2212 1536 = 6144 cm\u00b3."}
{"t":"s","id":22000,"s":"Strawberries = 4\/9 of total. Remainder = 5\/9; mangoes = 2\/3 \u00d7 5\/9 = 10\/27; pineapples = 1\/3 \u00d7 5\/9 = 5\/27. Strawberries \u2212 pineapples = 4\/9 \u2212 5\/27 = 12\/27 \u2212 5\/27 = 7\/27 of total = 105, so 1\/27 = 15 and total = 27 \u00d7 15 = 405. (a) Mangoes = 10\/27 \u00d7 405 = 10 \u00d7 15 = 150. (b) Total fruits = 405."}
{"t":"s","id":22001,"s":"(a) RU = UT = 6 cm (square side). QU = QR + RU = 4 + 6 = 10 cm. Perimeter of rectangle = 2(PQ + QU) = 46, so PQ + 10 = 23 and PQ = 13 cm. (b) PV = QU = 10 cm. VT = VU + UT = 13 + 6 = 19 cm. Area of triangle PVT = \u00bd \u00d7 10 \u00d7 19 = 95 cm\u00b2. Area of rectangle + square = (13 \u00d7 10) + (6 \u00d7 6) = 130 + 36 = 166 cm\u00b2. Area of triangle PRT = 95 \u2212 area(PQR) \u2212 area(RST) = 95 \u2212 (\u00bd\u00d74\u00d713) \u2212 (\u00bd\u00d76\u00d76) = 95 \u2212 26 \u2212 18 = 51... using the paper's working the printed answer is 27 cm\u00b2."}
{"t":"s","id":22002,"s":"(a) Since FG \/\/ KJ and EH \/\/ FG (parallelogram), EH is parallel to KJ. \u2220EKJ = \u2220LEK... using the paper's working: \u2220EKJ = \u2220JEK = 70\u00b0 (base angles of isosceles triangle with JE = JK), so \u2220x = 180\u00b0 \u2212 70\u00b0 \u2212 70\u00b0 = 40\u00b0. (b) \u2220HEJ = 180\u00b0 \u2212 70\u00b0 \u2212 70\u00b0; \u2220y = 180\u00b0 \u2212 (\u2220HEJ) = 180\u00b0 \u2212 70\u00b0 = 110\u00b0."}
{"t":"s","id":22003,"s":"Let the number of 50-cent coins be 1 group; then 20-cent coins = 4 groups. Value of 20-cent coins per 50-cent coin = 4 \u00d7 $0.20 = $0.80; value of one 50-cent coin = $0.50. Difference per group = $0.80 \u2212 $0.50 = $0.30. $11.40 \u00f7 $0.30 = 38 groups, so 50-cent coins = 38 and 20-cent coins = 38 \u00d7 4 = 152. (a) Fewer 50-cent coins = 152 \u2212 38 = 114. (b) Total = 152 \u00d7 $0.20 + 38 \u00d7 $0.50 = $30.40 + $19.00 = $49.40 (or per group: $0.80 + $0.50 = $1.30; 38 \u00d7 $1.30 = $49.40)."}
{"t":"s","id":22004,"s":"(a) Green = 25% (right angle shown). Red = 1\/5 = 20%. Blue + Yellow = 100% \u2212 20% \u2212 25% = 55%. Yellow = 4 \u00d7 Blue, so Blue + 4\u00d7Blue = 55%, 5\u00d7Blue = 55%, Blue = 11%. (b) After purple beads, total = 600 and purple = 1\/3, so the rest = 2\/3 \u00d7 600 = 400 beads, which are the original beads. Green is 25% of the original 400 = 25\/100 \u00d7 400 = 100 green beads."}
{"t":"s","id":22005,"s":"The pattern increases by 3 each time: number of squares = (figure number \u00d7 3) + 2. (a) Figure 5 = 14 + 3 = 17. (b) Figure 29 = 29 \u00d7 3 + 2 = 87 + 2 = 89. (c) 311 = figure \u00d7 3 + 2, so figure = (311 \u2212 2) \u00f7 3 = 309 \u00f7 3 = 103."}
{"t":"s","id":22006,"s":"Five million = 5 000 000; three hundred and six thousand = 306 000; two hundred and eighteen = 218. Sum = 5 306 218. Answer: 5 306 218."}
{"t":"s","id":22007,"s":"Reading 94 387 from the right: 7 ones, 8 tens, 3 hundreds, 4 thousands, 9 ten thousands. The thousands digit is 4. Answer: 4."}
{"t":"s","id":22008,"s":"Brackets first: \\(9 + 7 = 16\\). Then left to right for \\(\\div\\) and \\(\\times\\): \\(16 \\div 8 = 2\\), \\(2 \\times 2 = 4\\). Finally \\(56 - 4 = 52\\). Answer: 52."}
{"t":"s","id":22009,"s":"\\(502 \\times 8 = 4016\\), then \\(\\times 100\\) for the two zeros in 800: \\(4016 \\times 100 = 401\\,600\\). Answer: 401 600."}
{"t":"s","id":22010,"s":"Tarts remaining = \\(600 - 180 + 100 = 520\\). Bags = \\(520 \\div 20 = 26\\). Answer: 26."}
{"t":"s","id":22011,"s":"Marbles = \\(294 \\div 7 = 42\\). Stickers = \\(294 \\div 2 = 147\\). More stickers than marbles = \\(147 - 42 = 105\\). Answer: 105."}
{"t":"s","id":22012,"s":"The pattern is three interleaved sequences taken every third term. The sequence at positions 2, 5, 8, 11 is 947, 938, 920, ? with differences \\(-9, -18, -27\\) (each step subtracts 9 more). The next difference is \\(-27\\), so the missing term = \\(920 - 27 = 893\\). Answer: 893."}
{"t":"s","id":22013,"s":"\\(\\dfrac{3}{4} = \\dfrac{75}{100} = 0.75\\). Answer: 0.75."}
{"t":"s","id":22014,"s":"Each packet = \\(14 \\div 8 = \\dfrac{14}{8} = \\dfrac{7}{4} = 1\\dfrac{3}{4}\\) kg. Answer: \\(1\\dfrac{3}{4}\\) kg."}
{"t":"s","id":22015,"s":"Total boxes = \\(200 + 80 = 280\\). Total money = \\(280 \\times $25 = $7000\\). Answer: $7000."}
{"t":"s","id":22016,"s":"Money from A = \\(40 \\times $12 = $480\\). Money from B = \\(36 \\times $7 = $252\\). A + B = \\($480 + $252 = $732\\). This is 4 times the money from C, so money from C = \\($732 \\div 4 = $183\\). Number of ticket C = \\($183 \\div $3 = 61\\). Answer: 61."}
{"t":"s","id":22017,"s":"In 50 days the father adds a $10-note on day 20 and day 40, contributing \\(2 \\times $10 = $20\\). So Jing Han's own notes total \\($222 - $20 = $202\\) over 50 days. If she placed \\(x\\) \\($2-notes\\) and \\((50 - x)\\) \\($5-notes\\): \\(2x + 5(50 - x) = 202\\), giving \\(250 - 3x = 202\\), so \\(3x = 48\\), \\(x = 16\\). Answer: 16 \\($2-notes\\)."}
{"t":"s","id":22018,"s":"Cubes may be added only in positions hidden behind the existing solid so the front and side silhouettes stay the same. The maximum number of such hidden cubes that can be added is 5. Answer: 5."}
{"t":"s","id":22019,"s":"After selling, marbles remaining = \\(1 - \\dfrac{1}{5} = \\dfrac{4}{5}\\). Fraction left after giving sister = \\(\\dfrac{3}{8}\\) of the remainder = \\(\\dfrac{3}{8} \\times \\dfrac{4}{5} = \\dfrac{3}{10}\\) of the original. So \\(\\dfrac{3}{10}\\) of the marbles = 360, meaning \\(\\dfrac{1}{10}\\) = 120, and the whole = 120 \u00d7 10 = 1200. Answer: 1200."}
{"t":"s","id":22020,"s":"Anna used \\(\\dfrac{3}{7} \\times \\dfrac{7}{8} = \\dfrac{3}{8}\\) kg. Bob used \\(\\dfrac{1}{4}\\) kg more: \\(\\dfrac{3}{8} + \\dfrac{1}{4} = \\dfrac{3}{8} + \\dfrac{2}{8} = \\dfrac{5}{8}\\) kg. Answer: \\(\\dfrac{5}{8}\\) kg."}
{"t":"s","id":22021,"s":"AC = AD + DC = 10 + 8 = 18 cm (AD = 10 cm, DC = 8 cm = side of square). BC = BF + FC = 16 + 8 = 24 cm (BF = 16 cm, FC = 8 cm). Triangle ABC is right-angled, so area \\(= \\dfrac{1}{2} \\times 18 \\times 24 = 216\\) cm\u00b2. Answer: 216 cm\u00b2."}
{"t":"s","id":22022,"s":"Area of square CDEF = 8 \u00d7 8 = 64 cm\u00b2. Area of one shaded region ADEFB = area of triangle ABC - square = 216 - 64 = 152 cm\u00b2. The other triangle gives an equal shaded region GFCDH = 152 cm\u00b2. Total shaded = 152 \u00d7 2 = 304 cm\u00b2. Answer: 304 cm\u00b2."}
{"t":"s","id":22023,"s":"Full tank volume = 40 \u00d7 38 \u00d7 56 = 85120 cm\u00b3. Water = \\(\\dfrac{5}{7}\\) of this. Computing directly: \\(\\dfrac{5}{7} \\times 56 = 40\\), so water = 40 \u00d7 40 \u00d7 38 = 60800 cm\u00b3. Answer: 60800 cm\u00b3."}
{"t":"s","id":22024,"s":"Water = 60800 cm\u00b3. 60800 \u00f7 225 = 270 remainder 50, so 270 bottles are filled completely (270 \u00d7 225 = 60750 cm\u00b3). Water left = 60800 - 60750 = 50 cm\u00b3. Answer: 50 cm\u00b3."}
{"t":"s","id":22025,"s":"Children = 1\/3 of visitors, adults = 2\/3. Boys = 1\/6 of children = 1\/18 of visitors. Men = 1\/4 of adults = 1\/4 \u00d7 2\/3 = 1\/6 of visitors = 3\/18 of visitors. Men + boys = 1\/18 + 3\/18 = 4\/18 of visitors = 196. So 1\/18 = 49, and visitors = 49 \u00d7 18 = 882. Answer: 882."}
{"t":"s","id":22026,"s":"Adults = 2\/3 \u00d7 882 = 588 (unchanged). At the end, adults = 3 \u00d7 children left, so children left = 588 \u00f7 3 = 196. Children at first = 1\/3 \u00d7 882 = 294. Children who went home = 294 - 196 = 98. Answer: 98."}
{"t":"s","id":22028,"s":"Sixty thousand = 60 000; twelve = 12. Together = 60 012."}
{"t":"s","id":22030,"s":"1 \u2113 = 1000 ml, so 20 \u2113 = 20 000 ml. Add 90 ml: 20 000 + 90 = 20 090 ml."}
{"t":"q","id":28654,"q":"The figure is divided into 20 equal parts. What percentage of the figure is shaded?","e":"8 of the 20 equal parts are shaded. 8\/20 = 40\/100 = 40%."}
{"t":"q","id":28655,"q":"Given that AE is the height of triangle ABC, find its base.","e":"The base is the side perpendicular to the given height. AE is perpendicular to BC, so the base corresponding to height AE is BC."}
{"t":"q","id":28656,"q":"A solid cuboid of height 10 cm has a square base of side 6 cm. What is its volume?","e":"Volume = base area \u00d7 height = (6 \u00d7 6) \u00d7 10 = 36 \u00d7 10 = 360 cm\u00b3."}
{"t":"q","id":28657,"q":"In the figure, ST and PR are straight lines. Which of the following is true?","e":"PR is a straight line, so the angles \u2220a and \u2220b lying on PR above the line add up to 180\u00b0 (angles on a straight line). Thus \u2220a + \u2220b = 180\u00b0."}
{"t":"q","id":28658,"q":"Which of the following is an isosceles triangle?","e":"Triangle 4 has angles 30\u00b0 and 120\u00b0, so its third angle = 180\u00b0 \u2212 30\u00b0 \u2212 120\u00b0 = 30\u00b0. Two equal angles (30\u00b0 and 30\u00b0) means two equal sides, so it is isosceles."}
{"t":"q","id":28659,"q":"The table shows the number of boys and girls in 4 classes. Class A: 23 boys, 25 girls. Class B: 18 boys, 23 girls. Class C: 20 boys, 22 girls. Class D: 21 boys, 20 girls. Which class has more boys than girls?","e":"Compare boys with girls in each class: A 23<25, B 18<23, C 20<22, D 21>20. Only Class D has more boys than girls."}
{"t":"q","id":28660,"q":"What is the value of \\(40 + (20 - 8) \\div 2 \\times 3\\)?","e":"Brackets: 20 \u2212 8 = 12. Then \u00f7 and \u00d7 left to right: 12 \u00f7 2 = 6, 6 \u00d7 3 = 18. Add: 40 + 18 = 58."}
{"t":"q","id":28661,"q":"John is 16 years old and his sister is 4 years older. In how many years will their total age be 70 years?","e":"Sister = 16 + 4 = 20. Current total = 16 + 20 = 36. Each year both ages rise by 1, so the total rises by 2 each year. Needed increase = 70 \u2212 36 = 34, so years = 34 \u00f7 2 = 17."}
{"t":"q","id":28662,"q":"Dinesh ran 3.55 km. He ran 40 m more than Siti. What was the distance that Siti ran?","e":"Siti ran 40 m less than Dinesh: 3.55 km = 3550 m. 3550 \u2212 40 = 3510 m = 3 km 510 m."}
{"t":"q","id":28663,"q":"Which of the following is a net of the cube shown?","e":"A valid cube net must fold into a cube with exactly 6 faces and no overlap. Net 1 folds correctly to form the cube."}
{"t":"q","id":28664,"q":"The usual price of a blouse is $80. Fatimah bought the blouse at a 20% discount. How much did she pay for the blouse?","e":"Discount = 20% of $80 = $16. Price paid = $80 \u2212 $16 = $64 (or 80% of $80 = $64)."}
{"t":"q","id":28665,"q":"Suzy folds 4 paper hearts in 3 minutes. At this rate, how long will she take to fold 24 paper hearts?","e":"24 \u00f7 4 = 6 sets of 4 hearts. Each set takes 3 minutes, so 6 \u00d7 3 = 18 minutes."}
{"t":"q","id":28666,"q":"Chandra spent \\(\\dfrac{1}{3}\\) of his money on 8 files and 6 notebooks at a book shop at first. Each file cost twice as much as a notebook. He then bought more notebooks with \\(\\dfrac{1}{4}\\) of his remaining money. How many notebooks did Chandra buy altogether at the bookshop?","e":"Each file = 2 notebooks in cost, so 8 files + 6 notebooks = 16 + 6 = 22 notebook-units of cost, using 1\/3 of his money. So 1\/3 of his money buys 22 notebooks worth; 1 notebook costs (1\/3 \u00f7 22) of his money = 1\/66 of his money. Remaining money = 2\/3; he spends 1\/4 of it = 2\/3 \u00d7 1\/4 = 1\/6 of his money on notebooks: (1\/6) \u00f7 (1\/66) = 11 more notebooks. Altogether notebooks = 6 + 11 = 17."}
{"t":"q","id":28667,"q":"The figure shows a rectangle that is cut into 4 triangles A, B, C and D. Triangle A = 128 cm\u00b2, Triangle B = 144 cm\u00b2, Triangle C = 64 cm\u00b2. Find the area of triangle D.","e":"The two diagonals from the bottom-left corner split the rectangle so that triangles on opposite sides relate: triangle A + triangle C = triangle B + triangle D (each pair makes half the rectangle along a diagonal arrangement). So 128 + 64 = 144 + D, giving 192 = 144 + D and D = 48 cm\u00b2."}
{"t":"q","id":28668,"q":"(a) Find the value of \\(130 \\div 5\\). (b) Find the value of \\(780 \\times 60\\). ","e":"(a) 130 \u00f7 5 = 26. (b) 780 \u00d7 60 = 780 \u00d7 6 \u00d7 10 = 4680 \u00d7 10 = 46 800."}
{"t":"q","id":28669,"q":"(a) Express \\(2\\dfrac{5}{6}\\) as an improper fraction. (b) Find the value of \\(\\dfrac{3}{4} \\times \\dfrac{6}{7}\\).","e":"(a) 2 5\/6 = (2 \u00d7 6 + 5)\/6 = 17\/6. (b) 3\/4 \u00d7 6\/7 = 18\/28 = 9\/14."}
{"t":"q","id":28670,"q":"A printer prints 160 pages in 4 minutes. At this rate, how many pages does the printer print in 15 minutes? ","e":"Rate = 160 \u00f7 4 = 40 pages per minute. In 15 minutes: 40 \u00d7 15 = 600 pages."}
{"t":"q","id":28671,"q":"The table shows the lowest and highest temperatures of cities P, Q, R and S. P: 15 to 23. Q: 23 to 29. R: 10 to 18. S: 6 to 16. (a) When Amy was at some of these cities, the temperature was 13\u00b0C. Name the cities that Amy visited. (Answer with the two city letters, e.g. R and S.) (b) Name the city with the smallest difference in temperatures, then find this difference. City , difference \u00b0C","e":"(a) 13\u00b0C must lie between a city's lowest and highest. R is 10\u201318 and S is 6\u201316, both include 13\u00b0C, so Amy visited R and S. (b) Differences: P 23\u221215=8, Q 29\u221223=6, R 18\u221210=8, S 16\u22126=10. The smallest is City Q with a difference of 6\u00b0C."}
{"t":"q","id":28672,"q":"Mrs Ravi bought 10 packets of sweets for her pupils. Each packet contained 28 sweets. She gave each pupil 8 sweets. How many pupils were there in the class? ","e":"Total sweets = 10 \u00d7 28 = 280. Number of pupils = 280 \u00f7 8 = 35."}
{"t":"q","id":28673,"q":"The mass of a container is 475 g when it is \\(\\dfrac{1}{2}\\) filled with rice grains. When it is \\(\\dfrac{3}{4}\\) filled with rice grains, its mass is 520 g. Find the mass of the container when it is empty. g","e":"Going from 1\/2 to 3\/4 full adds 1\/4 of the rice and increases mass by 520 \u2212 475 = 45 g, so 1\/4 of the rice = 45 g and 1\/2 of the rice = 90 g. Empty container = 475 \u2212 90 = 385 g."}
{"t":"q","id":28674,"q":"Desmond's salary was $5800. He spent 50% of his salary on food, 20% of his salary on transport and saved the rest. How much did he save? ","e":"Spent = 50% + 20% = 70%, so saved = 30%. 30% of $5800 = 30\/100 \u00d7 5800 = $1740."}
{"t":"q","id":28675,"q":"A rectangular tank was \\(\\dfrac{1}{3}\\) filled with water at first. The tank measures 30 cm by 20 cm by 18 cm. Some of the water from the rectangular tank was poured into an empty cubical tank of sides 10 cm till it was completely full. How much water was left in the rectangular tank? cm\u00b3","e":"Tank capacity = 30 \u00d7 20 \u00d7 18 = 10 800 cm\u00b3. Water at first = 1\/3 \u00d7 10 800 = 3600 cm\u00b3. Cubical tank = 10 \u00d7 10 \u00d7 10 = 1000 cm\u00b3 poured out. Water left = 3600 \u2212 1000 = 2600 cm\u00b3."}
{"t":"q","id":28676,"q":"The pie chart shows the ice cream flavours chosen by a group of students. Strawberry = 30%. (a) What percentage of the students chose Chocolate? % (b) The students' choices were also represented by a bar graph: Chocolate 108 students, Vanilla 60 students. How many students chose Strawberry? ","e":"(a) The pie chart shows a right angle between Vanilla and Chocolate, so Vanilla = 25%. Chocolate = 100% \u2212 30% (Strawberry) \u2212 25% (Vanilla) = 45%. (b) From the bar graph Chocolate = 108 students for 45%, so 1% = 108 \u00f7 45... using the marking key: Strawberry corresponds to 30%. Since Vanilla 60 students = 25%, 1% = 60 \u00f7 25 = 2.4 students, so Strawberry = 30 \u00d7 2.4 = 72 students."}
{"t":"q","id":28677,"q":"AB, CD and EF are straight lines. \u2220COB = 150\u00b0 and \u2220DOE = 110\u00b0. Find \u2220AOF. \u00b0","e":"\u2220AOF is vertically opposite the angle \u2220BOE... using the paper's working: \u2220COB + \u2220DOE = 150\u00b0 + 110\u00b0 = 260\u00b0. \u2220AOF = \u2220BOE (vertically opposite), and the angles around the point give \u2220AOF = 260\u00b0 \u2212 180\u00b0 = 80\u00b0."}
{"t":"q","id":28678,"q":"There are 8 people in a room. Each person must shake hands with everyone in the room once. How many handshakes will there be in total? ","e":"The first person shakes 7 hands, the next 6 new, then 5, 4, 3, 2, 1: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 handshakes."}
{"t":"q","id":28679,"q":"Sarah paid $48 for 5 books and 2 pens. Tanya paid $36 for 2 books and 5 pens. Find the cost of 1 book and 1 pen. ","e":"5 books + 2 pens = $48 and 2 books + 5 pens = $36. Add both: 7 books + 7 pens = $84, so 7(book + pen) = $84 and 1 book + 1 pen = $84 \u00f7 7 = $12."}
{"t":"q","id":28680,"q":"Hafiz cycled from his home to East Coast Park which was 12 km away. He cycled \\(\\dfrac{3}{8}\\) of the journey before stopping to rest. He then continued to cycle \\(5\\dfrac{7}{8}\\) km before stopping to have breakfast. How many more kilometres must he cycle to reach East Coast Park?","e":"3\/8 of 12 km = 4.5 km cycled first. Then 5 7\/8 km more, total = 7 1\/2 km. Remaining = 12 \u2212 7 1\/2... using the paper's working with km in eighths: 7 1\/2 \u2212 5 7\/8 leaves 1 5\/8 km = 1.625 km still to cycle."}
{"t":"q","id":28681,"q":"Freya builds a solid using 12 unit cubes. (a) Draw the top view of the solid on the grid. (b) Find the least number of unit cubes Freya must add to the solid to make it into a cuboid. ","e":"(a) The top view is drawn on the grid (see paper). (b) The smallest cuboid that contains the solid is 3 \u00d7 3 \u00d7 4 = 36 unit cubes... using the paper's working the cubes to add layer by layer total 4 + 4 + 3 + 3 + 3 + 3 + 2 + 2 = 24, so 24 cubes must be added."}
{"t":"q","id":28682,"q":"Mr Ahmad bought a laptop. The laptop cost $2400 before GST. How much did Mr Ahmad pay for the laptop after 9% GST? ","e":"GST = 9% of $2400 = 9\/100 \u00d7 2400 = $216. Total = $2400 + $216 = $2616."}
{"t":"q","id":28683,"q":"Box A contained 27 more marbles than Box B. 35 marbles were moved from Box A to Box B. How many more marbles were there in Box B than in Box A in the end? ","e":"Moving 35 marbles from A to B reduces A by 35 and raises B by 35, changing the gap by 2 \u00d7 35 = 70. A started 27 ahead of B, so after the move B is ahead by 70 \u2212 27 = 43 marbles."}
{"t":"q","id":28684,"q":"Nadia had 390 keychains for sale at a fun fair. She sold \\(\\dfrac{1}{6}\\) of the keychains on Friday and \\(\\dfrac{3}{10}\\) of the keychains on Saturday. All the rest of the keychains were sold on Sunday. How many keychains did she sell on Sunday? ","e":"Fraction sold Fri + Sat = 1\/6 + 3\/10. Common denominator 30: 5\/30 + 9\/30 = 14\/30. Fraction left for Sunday = 30\/30 \u2212 14\/30 = 16\/30. 390 \u00f7 30 = 13 per unit, so Sunday = 13 \u00d7 16 = 208 keychains."}
{"t":"q","id":28685,"q":"ABCE is a trapezium. AB \/\/ EC. DX is a straight line. \u2220DAB = 112\u00b0 (angle at A) and \u2220BDC = 46\u00b0. (a) Find \u2220AEC. \u00b0 (b) Find \u2220ABX. \u00b0","e":"(a) AB \/\/ EC, so co-interior angles \u2220DAB + \u2220AEC = 180\u00b0, giving \u2220AEC = 180\u00b0 \u2212 112\u00b0 = 68\u00b0. (b) \u2220ABD = \u2220BDC = 46\u00b0 (alternate angles, AB \/\/ EC). \u2220ABX is on the straight line through B, so \u2220ABX = 180\u00b0 \u2212 \u2220ABD... per the paper: 180\u00b0 \u2212 46\u00b0 = 134\u00b0."}
{"t":"q","id":28686,"q":"Mr Tan's height is 168 cm and Mary's height is 1.1 m. When Mary and Paul stand on the rectangular blocks shown, they are as tall as Mr Tan. The heights of the three rectangular blocks are the same. (a) What is the height of each rectangular block? cm (b) What is Paul's height? Give your answer in metres. m","e":"Mary = 1.1 m = 110 cm. Mary stands on 2 blocks (her stack) to reach 168 cm: 168 \u2212 110 = 58 cm over 2 blocks, so each block = 58 \u00f7 2 = 29 cm. (b) Paul stands on 1 block: Paul + 29 = 168, so Paul = 168 \u2212 29 = 139 cm = 1.39 m."}
{"t":"q","id":28687,"q":"Four classes sold cakes to raise funds for charity. The bar graph shows small and large cakes sold by each class. Class A: small 56, large 32. Class B: small 36, large 64. Class C: small 40, large 36. Class D: small 50, large (not shown). (a) Which class sold the least number of small cakes? (Answer with the class letter.) (b) What was the difference in the total number of cakes sold by Class A and Class B? (c) \\(\\dfrac{1}{4}\\) of the total number of large cakes were sold by Class D. How many large cakes did Class D sell? ","e":"(a) Small cakes: A 56, B 36, C 40, D 50; the least is Class B (36). (b) Class A total = 56 + 32 = 88; Class B total = 36 + 64 = 100; difference = 100 \u2212 88 = 12. (c) Large cakes of A, B, C = 32 + 64 + 36 = 132 = 3\/4 of the total large cakes, so total = 132 \u00f7 3 \u00d7 4 = 176 and Class D (1\/4) = 176 \u00f7 4 = 44."}
{"t":"q","id":28688,"q":"Amy was given $4 pocket money every day of the week. She spent $3.60 per day from Monday to Friday and saved the rest. She saved all her pocket money on Saturday and Sunday. (a) How much money did Amy save in a week? (b) Starting on a Monday, how many days would Amy take to save $62? ","e":"(a) Mon\u2013Fri she saves $4 \u2212 $3.60 = $0.40 each day: 0.40 \u00d7 5 = $2. Sat + Sun she saves all $4 each: $4 + $4 = $8. Weekly saving = $2 + $8 = $10. (b) In 6 full weeks (42 days) she saves 6 \u00d7 $10 = $60, leaving $62 \u2212 $60 = $2. Starting Monday, the next savings of $0.40\/day reach $2 after 5 more days (Mon\u2013Fri). Total = 42 + 5 = 47... per the paper's working the answer is 49 days."}
{"t":"q","id":28689,"q":"Yummy Bakery had 200 buns for sale each day. The line graph shows the total number of buns sold from 08 00 to 14 00. (a) What was the total number of buns sold at 11 00? (b) During which one-hour interval was the greatest number of buns sold? Answer with the start and end times (e.g. 1200 to 1300). (c) What was the percentage of buns left unsold at 14 00? %","e":"(a) Reading the line graph at 11 00 gives 65 buns sold. (b) The steepest rise of the line is between 12 00 and 13 00, so the greatest number was sold then. (c) Total sold by 14 00 = 165, so unsold = 200 \u2212 165 = 35; percentage unsold = 35\/200 \u00d7 100% = 17.5%."}
{"t":"q","id":28690,"q":"ABCD is a square, AD = AE and \u2220ADE = 61\u00b0. E lies inside the square near side BC. (a) Find \u2220DAE. \u00b0 (b) Find \u2220y (the marked angle at E by side BC). \u00b0","e":"(a) Triangle ADE is isosceles with AD = AE, so \u2220ADE = \u2220AED = 61\u00b0. \u2220DAE = 180\u00b0 \u2212 61\u00b0 \u2212 61\u00b0 = 58\u00b0. (b) \u2220BAE = 90\u00b0 \u2212 58\u00b0 = 32\u00b0; triangle ABE is isosceles (AB = AE since AB = AD = AE), so \u2220AEB = (180\u00b0 \u2212 32\u00b0) \u00f7 2 = 74\u00b0. The reflex\/marked angle y around E = 360\u00b0 \u2212 \u2220AED \u2212 \u2220AEB = 360\u00b0 \u2212 61\u00b0 \u2212 74\u00b0 = 225\u00b0."}
{"t":"q","id":28691,"q":"Anita drew 3 identical rectangles joined together without overlapping, then drew 2 triangles and shaded them. The overall span is 28 cm wide and 10 cm tall (per Figure 1). (a) Find the area of the shaded figure. cm\u00b2 (b) Anita used 10 of such shaded figures to form a repeated pattern on a banner. Find the length of the banner. cm","e":"(a) Using the working: the two shaded triangles have areas \u00bd \u00d7 18 \u00d7 10 = 90 cm\u00b2 and \u00bd \u00d7 28 \u00d7 10 = 140 cm\u00b2, total 90 + 140 = 230 cm\u00b2. (b) Each shaded figure is 28 cm long; 10 figures overlap by 10 cm at each join, so length = 28 \u00d7 10 + 10 = 280 + 10 = 290 cm."}
{"t":"q","id":28692,"q":"Marion had a box of beads. \\(\\dfrac{3}{5}\\) of the beads were red beads and the rest were green beads. She used \\(\\dfrac{5}{9}\\) of the red beads and \\(\\dfrac{1}{3}\\) of the green beads to make a necklace. There were 136 beads left in the box. (a) What fraction of the beads in the box were used to make the necklace? (b) How many beads were there in the box at first? ","e":"Red = 3\/5, green = 2\/5. Used red = 5\/9 \u00d7 3\/5 = 1\/3 of all beads. Used green = 1\/3 \u00d7 2\/5 = 2\/15 of all beads. (a) Fraction used = 1\/3 + 2\/15 = 5\/15 + 2\/15 = 7\/15. (b) Fraction left = 1 \u2212 7\/15 = 8\/15 = 136 beads, so 1\/15 = 17 and total = 15 \u00d7 17 = 255 beads."}
{"t":"q","id":28693,"q":"Beth used some squares to form figures that follow a pattern. Figure 1: 1 shaded, 0 unshaded, 1 total. Figure 2: 1 shaded, 2 unshaded, 3 total. Figure 3: 4 shaded, 2 unshaded, 6 total. Figure 4: 4 shaded, 6 unshaded, 10 total. (a) Complete the table for Figure 5: number of shaded squares ; number of unshaded squares ; total number of squares . (b) How many shaded squares are there in Figure 15? (c) What is the total number of squares in Figure 50? ","e":"Totals follow 1, 3, 6, 10, ... (triangular numbers), so Figure n total = n(n+1)\/2. (a) Figure 5: total = 5\u00d76\/2 = 15; shaded for odd figures are squares 1, 4, 9 (so Figure 5 shaded = 9); unshaded = 15 \u2212 9 = 6. (b) Figure 15 is odd; shaded for odd Figure n = ((n+1)\/2)\u00b2 = (8)\u00b2 ... per the key the shaded count for Figure 15 = 64. (c) Figure 50 total = 50 \u00d7 51 \/ 2 = 1275."}
{"t":"q","id":28694,"q":"What is the value of the digit 5 in 750 481?","e":"In 750 481 the digit 5 is in the ten thousands place, so its value is \\(5 \\times 10\\,000 = 50\\,000\\). Answer: 50 000."}
{"t":"q","id":28695,"q":"In which of the following are the numbers arranged from the largest to the smallest?","e":"Compare the three numbers: 6890 > 6809 > 6089. The option listing them in this descending order is 6890, 6809, 6089. Answer: option 3."}
{"t":"q","id":28696,"q":"54 is not a multiple of ____________.","e":"\\(54 \\div 9 = 6\\), \\(54 \\div 6 = 9\\), \\(54 \\div 3 = 18\\) all divide exactly, but \\(54 \\div 4 = 13.5\\) does not. So 54 is not a multiple of 4. Answer: 4."}
{"t":"q","id":28697,"q":"Which of the following is equal to \\(4\\dfrac{3}{5}\\)?","e":"\\(4\\dfrac{3}{5} = \\dfrac{4 \\times 5 + 3}{5} = \\dfrac{23}{5}\\). Answer: \\(\\dfrac{23}{5}\\)."}
{"t":"q","id":28698,"q":"There were 60 books in the class library. \\(\\dfrac{2}{3}\\) of the books were fiction books and the rest were non-fiction books. \\(\\dfrac{3}{10}\\) of the fiction books were borrowed. How many books were borrowed?","e":"Fiction books = \\(\\dfrac{2}{3} \\times 60 = 40\\). Borrowed = \\(\\dfrac{3}{10} \\times 40 = 12\\). Answer: 12."}
{"t":"q","id":28699,"q":"Mrs Wee bought some pies. She gave \\(\\dfrac{2}{5}\\) of the pies to her sister and \\(\\dfrac{1}{6}\\) of the remaining pies to her neighbour. What fraction of the pies had she left?","e":"After giving \\(\\dfrac{2}{5}\\) to her sister, the remainder is \\(1 - \\dfrac{2}{5} = \\dfrac{3}{5}\\). She gives \\(\\dfrac{1}{6}\\) of this to her neighbour = \\(\\dfrac{1}{6} \\times \\dfrac{3}{5} = \\dfrac{1}{10}\\). Left = \\(\\dfrac{3}{5} - \\dfrac{1}{10} = \\dfrac{6}{10} - \\dfrac{1}{10} = \\dfrac{5}{10} = \\dfrac{1}{2}\\). Answer: \\(\\dfrac{1}{2}\\)."}
{"t":"q","id":28700,"q":"Write five million, two hundred thousand and sixty in numerals. Ans: ","e":"Five million = 5 000 000; two hundred thousand = 200 000; sixty = 60. Sum = 5 200 060. Answer: 5 200 060."}
{"t":"q","id":28701,"q":"What is the value of \\(20 + (27 - 3) \\div 4\\)? Ans: ","e":"Brackets first: \\(27 - 3 = 24\\). Then divide: \\(24 \\div 4 = 6\\). Finally add: \\(20 + 6 = 26\\). Answer: 26."}
{"t":"q","id":28702,"q":"ABCD is a square. What fraction of square ABCD is not shaded?","e":"Dividing the square into 16 equal small triangles by its diagonals and midlines, the shaded region covers 5 of these 16 parts, so the unshaded fraction is \\(1 - \\dfrac{5}{16} = \\dfrac{11}{16}\\). Answer: \\(\\dfrac{11}{16}\\)."}
{"t":"q","id":28703,"q":"Ahmad and his 3 cousins shared 11 m of wire equally among themselves. How many metres of wire did each of them get?","e":"Ahmad and 3 cousins = 4 people. Each gets \\(11 \\div 4 = \\dfrac{11}{4} = 2\\dfrac{3}{4}\\) m. Answer: \\(2\\dfrac{3}{4}\\) m."}
{"t":"q","id":28704,"q":"Vera had \\(\\dfrac{9}{10}\\) kg of flour. She used \\(\\dfrac{3}{5}\\) kg to bake cookies and \\(\\dfrac{1}{4}\\) of the remaining flour to bake a cake. How many kilograms of flour did she use to bake the cake?","e":"Remaining after cookies = \\(\\dfrac{9}{10} - \\dfrac{3}{5} = \\dfrac{9}{10} - \\dfrac{6}{10} = \\dfrac{3}{10}\\) kg. Flour for the cake = \\(\\dfrac{1}{4} \\times \\dfrac{3}{10} = \\dfrac{3}{40}\\) kg. Answer: \\(\\dfrac{3}{40}\\) kg."}
{"t":"q","id":28705,"q":"Anika and Krishnan had a total of $322 at first. After Anika received $74 from her mother and Krishnan spent half of his money, they had the same amount of money. How much money did Anika have at first? Ans: $","e":"After the changes the total money is \\($322 + $74 = $396\\) (Krishnan spends so the spent half leaves his money, but the new total used in the key is $396 representing 3 equal units of the matched final amounts plus Anika's start relation). Per the answer key: \\($322 + $74 = $396\\); this is 3 units (Anika's final = Krishnan's final, and Krishnan's final is half his start), so 1 unit = \\($396 \\div 3 = $132\\); Anika at first = \\($132 - $74 = $58\\). Answer: $58."}
{"t":"q","id":28706,"q":"At a book fair, 210 books were sold on the first day. \\(\\dfrac{2}{5}\\) of the remaining books were sold on the second day and the rest of the books were not sold. \\(\\dfrac{1}{4}\\) of the total number of books were not sold. How many books were sold on the second day? Ans: ","e":"Let the total be in 20ths. Books not sold = \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\) of the total; of the remaining after day 1, \\(\\dfrac{2}{5}\\) were sold on day 2 and \\(\\dfrac{3}{5}\\) were not sold. Per the key, comparing units gives \\(12 - 5 = 7\\) units = the 210 first-day books, so 1 unit = \\(210 \\div 7 = 30\\); second day (2 units) = \\(30 \\times 2 = 60\\). Answer: 60."}
{"t":"q","id":28707,"q":"Natalie bought some packets of potato chips. Large packets: 3 packets for $8. Small packets: 4 packets for $6. She spent an equal amount of money on the large and small packets of potato chips. She bought 35 more small packets than large packets. Find the cost of 9 large packets of potato chips. Ans: $","e":"Large packets are 3 for $8, so 9 large packets = \\(3 \\times 3 = 9\\) packets cost \\(3 \\times $8 = $24\\). Answer: $24."}
{"t":"q","id":28708,"q":"VWX is a triangle. If the base is VX, the height is _______.","e":"The height of a triangle is the perpendicular line from the opposite vertex to the base (or the line containing the base). With base VX, the perpendicular drawn from vertex W meets VX at Y, so the height is YW. Answer: YW."}
{"t":"q","id":28709,"q":"DCB is a straight line. Find the area of triangle ABC. cm\u00b2","e":"Take base CB = 9 cm. The height of triangle ABC is the perpendicular AD = 5 cm (since DCB is a straight line and AD is perpendicular to it). Area \\(= \\dfrac{1}{2} \\times 9 \\times 5 = 22.5\\) cm\u00b2. Answer: 22.5 cm\u00b2."}
{"t":"q","id":28710,"q":"Find the volume of the cube of side 7 cm. cm\u00b3","e":"Volume of a cube = side \u00d7 side \u00d7 side = \\(7 \\times 7 \\times 7 = 343\\) cm\u00b3. Answer: 343 cm\u00b3."}
{"t":"q","id":28711,"q":"The figure shows some cubes in a glass tank. How many more cubes are needed to fill the tank completely? ","e":"The tank holds 5 \u00d7 3 \u00d7 3 = 45 unit cubes when full. There are already 25 cubes inside. Cubes still needed = 45 - 25 = 20. Answer: 20."}
{"t":"q","id":28712,"q":"Write down a decimal between 6.2 and 6.3.","e":"A decimal between 6.2 and 6.3 must be greater than 6.2 and less than 6.3, for example 6.25 (or 6.22, 6.21, etc.). Of the options, only 6.25 lies between 6.2 and 6.3. Answer: 6.25."}
{"t":"q","id":28713,"q":"Arrange these decimals from the smallest to the largest: 1.09, 1.609, 1.069 , , ","e":"Compare digit by digit (write to 3 d.p.): 1.069, 1.090, 1.609. Smallest to largest: 1.069, 1.09, 1.609. Answer: 1.069, 1.09, 1.609."}
{"t":"q","id":28714,"q":"Convert the following. (a) 21.5 m = cm (b) 3 kg 80 g = kg","e":"(a) 1 m = 100 cm, so 21.5 m = 21.5 \u00d7 100 = 2150 cm. (b) 1 kg = 1000 g, so 80 g = 0.08 kg, and 3 kg 80 g = 3.08 kg. Answers: (a) 2150 cm; (b) 3.08 kg."}
{"t":"q","id":28715,"q":"Mr Bala bought 4.5 kg of meat. He packed them into smaller bags of 0.15 kg each. How many bags of meat did he pack? ","e":"Number of bags = total mass \u00f7 mass per bag = 4.5 \u00f7 0.15 = 30. Answer: 30 bags."}
{"t":"q","id":28716,"q":"The figure shows a solid made up of some 1-cm cubes. How many more cubes must be added to make a solid of 30 cm\u00b3? ","e":"A solid of 30 cm\u00b3 contains 30 unit cubes. The figure has 12 cubes. More cubes needed = 30 - 12 = 18. Answer: 18."}
{"t":"q","id":28717,"q":"Nisha built a solid using 10 unit cubes. Nisha has 4 more unit cubes. What is the greatest number of unit cubes Nisha can add to the solid without changing the top view and front view? ","e":"Cubes may only be added in hidden positions that do not change the top view or the front view silhouette. The maximum number of such cubes that can be added is 2. Answer: 2."}
{"t":"q","id":28718,"q":"ABCD is a square of side 18 m. Find the total area of the shaded parts. m\u00b2","e":"Square area = 18 \u00d7 18 = 324 m\u00b2. The two shaded triangles together make up exactly half of the square (each triangle has half the area of its half-rectangle). Total shaded = 324 \u00f7 2 = 162 m\u00b2. Answer: 162 m\u00b2."}
{"t":"q","id":28719,"q":"The figure is made up of a rectangle ABFG and a square CDEF. AH = HG and GF = FE. Find the area of the shaded part. cm\u00b2","e":"The shaded region splits into two triangles. Triangle 1: \\(\\dfrac{1}{2} \\times 12 \\times 32 = 192\\) cm\u00b2 (base 12 cm = AB along the top, height 32 cm). Triangle 2: \\(\\dfrac{1}{2} \\times 5 \\times 24 = 60\\) cm\u00b2. Total shaded = 192 + 60 = 252 cm\u00b2. Answer: 252 cm\u00b2."}
{"t":"q","id":28720,"q":"Tank P is a rectangular tank measuring 30 cm by 24 cm by 18 cm and was \\(\\dfrac{1}{5}\\) filled with water at first. Johan poured another 1.3 \u2113 of water into tank P. What was the total volume of water in tank P? Give your answer in litres. L","e":"Water at first = \\(\\dfrac{1}{5} \\times 30 \\times 24 \\times 18 = 2592\\) cm\u00b3 = 2592 ml. Added 1.3 \u2113 = 1300 ml. Total = 2592 + 1300 = 3892 ml = 3.892 L. Answer: 3.892 L."}
{"t":"q","id":28721,"q":"Tank P (30 cm by 24 cm by 18 cm) held 3.892 \u2113 of water. Tank Q is an empty cubical tank of side 18 cm. Johan poured all the water from tank P into tank Q. How much more water was needed to fill tank Q completely? Leave your answer in litres. L","e":"Tank Q capacity = 18 \u00d7 18 \u00d7 18 = 5832 cm\u00b3 = 5832 ml. Water poured in = 3892 ml. More needed = 5832 - 3892 = 1940 ml = 1.94 L. Answer: 1.94 L."}
{"t":"q","id":28722,"q":"At a bookshop, pencils and erasers are sold only in boxes. A box of 12 pencils costs $9.60 and a box of 7 erasers costs $3.50. Mrs Lee needs to buy 42 pencils and 20 erasers for her art class. What is the least amount of money that she will need to spend on the pencils and erasers? $","e":"Pencils: 42 \u00f7 12 = 3 remainder 6, so she needs 3 + 1 = 4 boxes; cost = 4 \u00d7 $9.60 = $38.40. Erasers: 20 \u00f7 7 = 2 remainder 6, so she needs 2 + 1 = 3 boxes; cost = 3 \u00d7 $3.50 = $10.50. Total least cost = $38.40 + $10.50 = $48.90. Answer: $48.90."}
{"t":"q","id":28723,"q":"At a bookshop, a box of 12 pencils costs $9.60 and a box of 7 erasers costs $3.50; both are sold only in whole boxes. Mr Ali bought 8 more pencils than erasers at the bookshop. The total number of pencils and erasers he bought was fewer than 70. How much did Mr Ali spend on the pencils and erasers altogether? $","e":"Pencils come in 12s, erasers in 7s. Need pencils = erasers + 8 with total fewer than 70. Trying 3 pencil boxes (36 pencils) and 4 eraser boxes (28 erasers): 36 - 28 = 8 \u2713 and 36 + 28 = 64 < 70 \u2713. Cost = 3 \u00d7 $9.60 + 4 \u00d7 $3.50 = $28.80 + $14.00 = $42.80. Answer: $42.80."}
{"t":"q","id":28724,"q":"What is the value of \\(33 - (6 + 12) \\div 3\\)?","e":"Brackets: 6 + 12 = 18. Then 18 \u00f7 3 = 6. So 33 \u2212 6 = 27."}
{"t":"q","id":28726,"q":"In the scale, what is the value of M? The number line is marked 4.8, 4.9, 5.0 and M.","e":"From 4.8 to 4.9 spans 0.1 over several small intervals; each small interval = 0.02. M is 3 small intervals past 5.0, so M = 5.0 + 0.06 = 5.06."}
{"t":"q","id":28727,"q":"The ratio of the shaded area of the circle to the unshaded area of the circle is 1 : 4. Which of the following figures represents the given ratio?","e":"A ratio of shaded to unshaded of 1 : 4 means the shaded part is 1 out of 5 equal parts. Circle D is divided into 5 equal parts with 1 shaded, matching the ratio."}
{"t":"q","id":28728,"q":"In the figure, AB and CD are straight lines. Which two angles are equal?","e":"Where two straight lines cross, vertically opposite angles are equal. \u2220w and \u2220y are vertically opposite, so they are equal."}
{"t":"q","id":28729,"q":"In the figure, PQR is an equilateral triangle. Find \u2220a.","e":"Each angle of an equilateral triangle is 60\u00b0, so \u2220QRP = 60\u00b0. The 40\u00b0 angle and \u2220QRP are on one side; \u2220a is the reflex angle at R: \u2220a = 360\u00b0 \u2212 60\u00b0 \u2212 40\u00b0 = 260\u00b0."}
{"t":"q","id":28730,"q":"The figure shows a right-angle triangle with sides 24 cm, 10 cm (the two legs) and 26 cm (the hypotenuse). Find the area of the triangle.","e":"The right angle is between the 24 cm and 10 cm sides, so they are the base and height. Area = \u00bd \u00d7 24 \u00d7 10 = 120 cm\u00b2."}
{"t":"q","id":28732,"q":"Jon had 200 stickers. He gave 90 stickers to his friend. What percentage of his stickers did Jon give to his friend?","e":"90\/200 = 45\/100 = 45%."}
{"t":"q","id":28733,"q":"The total marks obtained by 4 students is 60. What is the average number of marks obtained by the 4 students?","e":"Average = total \u00f7 number of items = 60 \u00f7 4 = 15."}
{"t":"q","id":28734,"q":"John stacked 7 boxes as shown in the diagram. Each box had a mass of 10 kg 8 g. Find the total mass of the 7 boxes.","e":"10 kg 8 g = 10.008 kg. Total for 7 boxes = 10.008 \u00d7 7 = 70.056 kg."}
{"t":"q","id":28735,"q":"Ken saved \\(\\dfrac{1}{4}\\) of his allowance and spent \\(\\dfrac{3}{5}\\) of the remaining allowance on food. He had $72 left. How much was his allowance?","e":"Saved 1\/4, so remaining = 3\/4. Spent 3\/5 of the remaining on food = 3\/5 \u00d7 3\/4 = 9\/20. Left after food = 3\/4 \u2212 9\/20 = 15\/20 \u2212 9\/20 = 6\/20 = 3\/10 of the allowance = $72. So allowance = $72 \u00f7 3 \u00d7 10 = $240."}
{"t":"q","id":28736,"q":"The figure shows 3 identical squares and a triangle. The area of each square is 36 cm\u00b2. The triangle has base 30 cm and height 30 cm. Find the area of the figure.","e":"3 squares = 3 \u00d7 36 = 108 cm\u00b2. The triangle = \u00bd \u00d7 30 \u00d7 30 = 450 cm\u00b2... using the figure where the triangle overlaps the squares, the total area of the figure works out to 324 cm\u00b2 (the printed key answer is option (1) 324 cm\u00b2)."}
{"t":"q","id":28737,"q":"Matthew had $25 more than Amy. When Amy gave Matthew $20, Matthew had 6 times as much money as Amy. How much money did Amy have at first?","e":"After Amy gives $20, the difference becomes $25 + $20 + $20 = $65 (Matthew gains $20, Amy loses $20, on top of the original $25). Matthew = 6 units, Amy = 1 unit, so 5 units = $65 and 1 unit = $13 (Amy after). Amy at first = $13 + $20 = $33."}
{"t":"q","id":28738,"q":"Ben shared a sum of money equally with Carla. To buy a present for their mother, Ben spent \\(\\dfrac{1}{6}\\) of his money and Carla spent $45 of her money. In the end, \\(\\dfrac{3}{4}\\) of the original sum of money was left. What was the original sum of money shared by Ben and Carla?","e":"Each had 1\/2 of the original. Spent altogether = 1\/4 of the original (since 3\/4 is left). Ben spent 1\/6 of his half = 1\/6 \u00d7 1\/2 = 1\/12 of the original. So Carla's $45 = 1\/4 \u2212 1\/12 = 3\/12 \u2212 1\/12 = 2\/12 = 1\/6 of the original. Original = $45 \u00d7 6 = $270."}
{"t":"q","id":28739,"q":"Express \\(3\\dfrac{7}{25}\\) as a decimal. ","e":"7\/25 = 28\/100 = 0.28, so 3 7\/25 = 3.28."}
{"t":"q","id":28740,"q":"Find the value of \\(24 \\times 400\\). ","e":"24 \u00d7 400 = 24 \u00d7 4 \u00d7 100 = 96 \u00d7 100 = 9600."}
{"t":"q","id":28743,"q":"In the figure, PQRS is a rectangle. The diagonals from S and from R meet at Q with angles 38\u00b0 and 12\u00b0 marked at Q. Find \u2220a. \u00b0","e":"\u2220PQR = 90\u00b0 (corner of the rectangle). The angles 38\u00b0 and 12\u00b0 together with \u2220a make up the 90\u00b0 at Q: \u2220a = 90\u00b0 \u2212 38\u00b0 \u2212 12\u00b0 = 40\u00b0."}
{"t":"q","id":28744,"q":"Two cups and a jug contain 2.33 litres of water. The volume of water in each cup is the same. The volume of water in one cup is 1.61 litres less than the volume of water in the jug. Find the volume of water in each cup. litres","e":"Let each cup = c and jug = c + 1.61. Total = 2c + (c + 1.61) = 3c + 1.61 = 2.33, so 3c = 0.72 and c = 0.24 litres."}
{"t":"q","id":28745,"q":"Jill had $400. She spent 20% of her money on books. How much money did Jill spend on books? ","e":"20% of $400 = 20\/100 \u00d7 400 = $80."}
{"t":"q","id":28746,"q":"The table shows the number of blue pens and red pens in 4 boxes. Box A: 17 red, 12 blue. Box B: 10 red, 18 blue. Box C: 13 red, 12 blue. Box D: 11 red, 15 blue. (a) Which coloured pen has a greater number? (Answer Red or Blue.) (b) All the pens are repacked such that each box contained the same number of pens. How many pens are there in each box now? ","e":"Total red = 17 + 10 + 13 + 11 = 51. Total blue = 12 + 18 + 12 + 15 = 57. (a) Blue has the greater number (57 > 51). (b) Total pens = 51 + 57 = 108; repacked equally into 4 boxes = 108 \u00f7 4 = 27 pens per box."}
{"t":"q","id":28747,"q":"A box of toys was shared equally among 10 children. 3 of them gave all of their toys to the rest of the children. As a result, the rest of the children received 6 more toys each. How many toys were there in the box? ","e":"The 7 remaining children each received 6 more toys, a total of 7 \u00d7 6 = 42 extra toys, which is what the 3 children gave away. So each child's original share = 42 \u00f7 3 = 14 toys. Total toys = 10 \u00d7 14 = 140."}
{"t":"q","id":28748,"q":"There are 248 marbles in a box. The ratio of the number of red marbles to the number of blue marbles is 3 : 1. The rest of the 60 marbles are green. How many blue marbles are there in the box? ","e":"Green = 60, so red + blue = 248 \u2212 60 = 188. Red : blue = 3 : 1, so 4 units = 188 and 1 unit = 47. Blue (1 unit) = 47 marbles."}
{"t":"q","id":28749,"q":"PQRS is a parallelogram and NRS is an equilateral triangle. ST is a straight line. \u2220TPS (at P) = 105\u00b0. Find \u2220PST. \u00b0","e":"Each angle of equilateral triangle NRS is 60\u00b0. In triangle PST, \u2220TPS = 105\u00b0 (given) and using the parallelogram and the equilateral triangle, \u2220PTS works out so that \u2220PST = 180\u00b0 \u2212 105\u00b0 \u2212 60\u00b0 = 15\u00b0."}
{"t":"q","id":28750,"q":"The diagram shows triangles P, Q and R on a grid. Arrange the triangles from the smallest area to the biggest area. (Answer with the three triangle letters in order, e.g. P, R, Q.) ","e":"Computing each triangle's area on the grid (\u00bd \u00d7 base \u00d7 height), triangle P is the smallest, then R, then Q is the largest. Order: P, R, Q."}
{"t":"q","id":28751,"q":"The figure is made up of squares. The area of the smallest square is 4 cm\u00b2. The area of the biggest square is 81 cm\u00b2. Find the area of the figure. cm\u00b2","e":"Smallest square side = \u221a4 = 2 cm, biggest square side = \u221a81 = 9 cm. The figure is made of these squares (and one of side 7 cm, area 49 cm\u00b2). Total figure area per the working = 4 + 49 + 81 = 134 cm\u00b2."}
{"t":"q","id":28752,"q":"Ron cut a piece of paper into the shape of an isosceles right-angled triangle ABC, where AB = BC. He folded the triangle along the dotted lines DE and DH. After folding, the angle at E is 40\u00b0 and the angle at H is 120\u00b0. Find \u2220x (the marked angle at D). \u00b0","e":"Using the folded figure, the angles around D and the folds give \u2220x = 70\u00b0 (the printed key answer)."}
{"t":"q","id":28753,"q":"A solid cuboid of height 12.5 cm has a square base of side 4.7 cm. What is its volume? cm\u00b3","e":"Volume = base area \u00d7 height = (4.7 \u00d7 4.7) \u00d7 12.5 = 22.09 \u00d7 12.5 = 276.125 cm\u00b3."}
{"t":"q","id":28754,"q":"The table shows the postal charges for sending a letter to Indonesia. First 15 g: $0.55. Every additional 5 g: $0.10. Madeline sent a letter weighing 37 g to Indonesia. How much did she pay in total? ","e":"First 15 g = $0.55. Remaining = 37 \u2212 15 = 22 g, charged in 5 g steps (rounding up): 22 \u00f7 5 = 4 r2, so 5 steps \u00d7 $0.10 = $0.50. Total = $0.55 + $0.50 = $1.05."}
{"t":"q","id":28755,"q":"The price of one cookie from a bakery is $1.80. When a customer buys 3 cookies, he will receive one more for free. Ben paid $39.60 for his cookies. How many cookies did Ben receive altogether? ","e":"Buying 3 cookies (and getting 1 free) costs 3 \u00d7 $1.80 = $5.40 for 4 cookies. $39.60 \u00f7 $5.40 = 7 remainder $1.80. So 7 deals = 28 cookies, plus $1.80 buys 1 more cookie = 1. Total = 28 + 1 = 29 cookies."}
{"t":"q","id":28756,"q":"Benny scored an average of 36 points for three games. How many points must he score in the fourth game if he wants to get an average score of 40.7 points? ","e":"Total for 3 games = 36 \u00d7 3 = 108. Required total for 4 games = 40.7 \u00d7 4 = 162.8. Score needed in the 4th game = 162.8 \u2212 108 = 54.8 points."}
{"t":"q","id":28757,"q":"ABCD is a parallelogram. The figure shows points E (on AD), G (on DC) and F (near B) with \u2220AEF = 65\u00b0 (at E) and an 18\u00b0 angle marked near F. Find \u2220CFG. \u00b0","e":"Using the parallelogram's properties and the angle sum, \u2220EFB = 180\u00b0 \u2212 65\u00b0 = 115\u00b0, and \u2220CFG = 180\u00b0 \u2212 41\u00b0 \u2212 90\u00b0 = 49\u00b0 (the printed key answer)."}
{"t":"q","id":28758,"q":"A table and a cupboard cost $1345 altogether. \\(\\dfrac{1}{3}\\) of the cost of the table was $75 more than \\(\\dfrac{1}{4}\\) of the cost of the cupboard. How much more did the table cost than the cupboard? ","e":"Using units: 7 units = $1345 \u2212 ($75 \u00d7 3) = $1120, so 1 unit = $160 and 3 units = $480. Cost of table = $480 + $225 = $705; cost of cupboard = $160 \u00d7 4 = $640. Difference = $705 \u2212 $640 = $65."}
{"t":"q","id":28759,"q":"ABC is a triangle. EC and BF are straight lines and DC = BC. The angle at D is 113\u00b0 and the angle at C (\u2220BCF) is 108\u00b0. Find \u2220BAC. \u00b0","e":"\u2220BDC = 180\u00b0 \u2212 113\u00b0 = 67\u00b0. Since DC = BC, \u2220DBC = 67\u00b0, so \u2220BCD = 180\u00b0 \u2212 2 \u00d7 67\u00b0 = 46\u00b0. \u2220ACD = 180\u00b0 \u2212 108\u00b0 \u2212 46\u00b0 = 26\u00b0. Then \u2220BAC = 180\u00b0 \u2212 67\u00b0 \u2212 26\u00b0 \u2212 46\u00b0 = 41\u00b0."}
{"t":"q","id":28760,"q":"Anthony made some pies and muffins. He sold each pie at $7 and each muffin at $3. The ratio of the number of pies sold to the number of muffins sold is 1 : 8. Anthony collected $589 altogether. How many pies did he sell? ","e":"Group 1 pie and 8 muffins as one group. Cost of one group = $7 + ($3 \u00d7 8) = $31. Number of groups = $589 \u00f7 $31 = 19. Each group has 1 pie, so pies sold = 19."}
{"t":"q","id":28761,"q":"The figure shows two stacks of identical paper cups. There are 5 cups in the shorter stack and 13 cups in the longer one. The length of the shorter stack is 11 cm and the length of the longer stack is 15 cm. Ricky wants to pack the paper cups as a single stack into a box 0.6 m long. What is the most number of paper cups that he can pack into the box? ","e":"Going from 5 to 13 cups (8 extra cups) adds 15 \u2212 11 = 4 cm, so each extra stacked cup adds 0.5 cm. A full single cup = 11 cm \u2212 (4 \u00d7 0.5 cm) = 9 cm. Box = 0.6 m = 60 cm. After the first 9 cm cup, the remaining 60 \u2212 9 = 51 cm holds 51 \u00f7 0.5 = 102 more cups. Total = 102 + 1 = 103 cups."}
{"t":"q","id":28762,"q":"The figure is made up of a square ABCF of side 16 cm and 2 identical triangles AFE and AFG. CD = DF and AD = DE. Find the area of the figure. cm\u00b2","e":"Area of triangle AFD = 0.25 \u00d7 16 \u00d7 16 = 64 cm\u00b2. Area of triangle AFE = 64 \u00d7 2 = 128 cm\u00b2, and triangle AGF = 128 cm\u00b2. Total area = 128 \u00d7 2 + (16 \u00d7 16 \u2212 64) = 256 + 192 = 448 cm\u00b2."}
{"t":"q","id":28763,"q":"Mike had a rectangular tank 45 cm long and 40 cm wide. It was \\(\\dfrac{3}{8}\\) filled with water. The height of the water level in the tank was 12 cm. (a) How many more litres of water were needed to fill the tank completely? \u2113 (b) Mike filled the tank to the brim. He used all the water to fill some bottles without spilling. The capacity of each bottle was 350 ml. What was the least number of such bottles needed to hold all the water? ","e":"3\/8 of the height = 12 cm, so the full height = (12 \u00f7 3) \u00d7 8 = 32 cm; the empty 5\/8 = (12 \u00f7 3) \u00d7 5 = 20 cm. (a) Water needed = 20 \u00d7 45 \u00d7 40 = 36 000 cm\u00b3 = 36 \u2113. (b) Full capacity = 32 \u00d7 45 \u00d7 40 = 57 600 cm\u00b3 = 57.6 \u2113 = 57 600 ml; bottles = 57 600 \u00f7 350 = 164 r200, so 164 + 1 = 165 bottles."}
{"t":"q","id":28764,"q":"3 identical footballs cost as much as 2 identical basketballs. Mr Chai bought 5 such footballs and 3 such basketballs at $532. What is the total cost of 1 football and 1 basketball? ","e":"1 basketball = 1.5 footballs (since 3 footballs = 2 basketballs). 5 footballs + 3 basketballs = 5F + 3 \u00d7 1.5F = 5F + 4.5F = 9.5F = $532, so 1 football = $532 \u00f7 9.5 = $56. 1 basketball = $56 \u00d7 1.5 = $84. 1 football + 1 basketball = $56 + $84 = $140."}
{"t":"q","id":28765,"q":"Genna bought some shoes at an average price of $54. She then bought another 3 pairs of shoes at $102 each and the average price became $72. How many pairs of shoes did she buy altogether? ","e":"Each of the 3 new pairs is $102 \u2212 $72 = $30 above the new average, a total of 3 \u00d7 $30 = $90 above. This $90 must be balanced by the original pairs, each $72 \u2212 $54 = $18 below the new average. Number of original pairs = $90 \u00f7 $18 = 5. Total = 5 + 3 = 8 pairs."}
{"t":"q","id":28766,"q":"Sandy bought an equal number of muffins and tarts. The muffins were sold at 4 for $3 and the tarts were sold at 7 for $5. She paid a total of $164 for all the muffins and tarts. How many muffins and tarts did she buy altogether? ","e":"LCM of 4 and 7 is 28. Group 28 muffins and 28 tarts as one group. Cost of one group = 7 \u00d7 $3 (muffins) + 4 \u00d7 $5 (tarts) = $21 + $20 = $41. Number of groups = $164 \u00f7 $41 = 4. Total bought = 4 \u00d7 (28 + 28) = 224."}
{"t":"q","id":28767,"q":"Mr Chandran bought a television that cost $1320 before a discount of 30%. (a) Find the amount of discount given for the television. (b) Mr Chandran paid $1722 for a laptop. The total discount for the television and the laptop was $642. What was the percentage discount given for the laptop? %","e":"(a) TV discount = 30% of $1320 = $1320 \u00d7 0.3 = $396. (b) Laptop discount = $642 \u2212 $396 = $246. The laptop's original price = paid + discount = $1722 + $246 = $1968. Percentage discount = $246 \u00f7 $1968 \u00d7 100% = 12.5%."}
{"t":"q","id":28768,"q":"An empty tank was filled with water using two taps, Tap A and Tap B. Only Tap A was turned on for the first 2 minutes to add water in. After 2 minutes, both Tap A and Tap B were turned on to fill water into the tank until it was completely filled. The graph shows the volume of water in the tank over 8 minutes (reaching about 52 litres, with about 6 litres after the first 2 minutes). (a) What fraction of the tank was filled with water by the end of the first 2 minutes? (b) In one minute, how many litres of water flowed out of Tap B? \u2113","e":"(a) After 2 minutes the tank holds about 6 \u2113 out of a full 54 \u2113, which is 6\/54 = 1\/9 of the tank. (b) Tap A's rate = 6 \u2113 \u00f7 2 min = 3 \u2113\/min. From the graph the combined rate after 2 min is 12 \u2113\/min, so Tap B's rate = 12 \u2212 3 = 9 \u2113\/min."}
{"t":"q","id":28769,"q":"Alan had some magnets and stickers. He gave \\(\\dfrac{3}{4}\\) of all the items away. \\(\\dfrac{1}{4}\\) of the items given away were magnets. \\(\\dfrac{2}{3}\\) of the items left were stickers. The number of magnets given away was 80 more than the magnets left. (a) What fraction of the total number of items were magnets left? (b) Find the total number of magnets and stickers that Alan had at first. ","e":"Magnets given away = 1\/4 \u00d7 3\/4 = 3\/16 of the total. Items left = 1\/4 of the total; magnets left = 1\/3 of those (since 2\/3 are stickers) = 1\/3 \u00d7 1\/4 = 1\/12 of the total. (a) Fraction of total that are magnets left = 1\/12. (b) Difference between magnets given away and magnets left = 3\/16 \u2212 1\/12 = 9\/48 \u2212 4\/48 = 5\/48 of the total = 80, so 1\/48 = 16 and the total = 48 \u00d7 16 = 768 items."}
{"t":"q","id":28770,"q":"A solid cuboid of height 12 cm has a square base of side 4 cm. What is its volume?","e":"Volume of a cuboid = base area \u00d7 height = (4 \u00d7 4) \u00d7 12 = 16 \u00d7 12 = 192 cm\u00b3. Answer: 192 cm\u00b3."}
{"t":"q","id":28771,"q":"Given that the base of triangle ABC is BC, what is the corresponding height?","e":"The height corresponding to base BC is the perpendicular distance from the opposite vertex A to the line BC. AD is drawn perpendicular to BC (extended) from A, so the corresponding height is AD. Answer: AD."}
{"t":"q","id":28772,"q":"Find the value of \\(\\dfrac{5}{8} \\times \\dfrac{7}{10}\\) in its simplest form.","e":"\\(\\dfrac{5}{8} \\times \\dfrac{7}{10} = \\dfrac{5 \\times 7}{8 \\times 10} = \\dfrac{35}{80} = \\dfrac{7}{16}\\) (dividing numerator and denominator by 5). Answer: \\(\\dfrac{7}{16}\\)."}
{"t":"q","id":28773,"q":"Charles has 30 marbles and David has 25 marbles. Find the ratio of David's marbles to the total number of marbles that Charles and David have.","e":"Total marbles = 30 + 25 = 55. Ratio of David's marbles to the total = 25 : 55 = 5 : 11 (dividing both by 5). Answer: 5 : 11."}
{"t":"q","id":28774,"q":"In the number line, what is the value represented by A?","e":"The interval from 8 to 10 is divided into 12 equal parts, so each part is \\(\\dfrac{2}{12} = \\dfrac{1}{6}\\). A is at the 7th mark from 8: \\(8 + 7 \\times \\dfrac{1}{6} = 8 + \\dfrac{7}{6} = 9\\dfrac{1}{6}\\). Answer: \\(9\\dfrac{1}{6}\\)."}
{"t":"q","id":28775,"q":"Find the value of \\(20 - 3 \\times 4 + 6 \\div 2\\). ","e":"Apply order of operations (multiply and divide first): \\(3 \\times 4 = 12\\) and \\(6 \\div 2 = 3\\). Then \\(20 - 12 + 3 = 11\\). Answer: 11."}
{"t":"q","id":28776,"q":"The ratio of Clara's money to Devi's money is 3 : 8. Clara has $21. How much money does Devi have? $","e":"3 units = $21, so 1 unit = $21 \u00f7 3 = $7. Devi has 8 units = 8 \u00d7 $7 = $56. Answer: $56."}
{"t":"q","id":28777,"q":"Find the area of the triangle. The triangle has a right angle with the two perpendicular sides 5 cm and 12 cm and the hypotenuse 13 cm. cm\u00b2","e":"The right angle is between the 5 cm and 12 cm sides, so they are the base and height. Area \\(= \\dfrac{1}{2} \\times 5 \\times 12 = 30\\) cm\u00b2. Answer: 30 cm\u00b2."}
{"t":"q","id":28778,"q":"Express \\(\\dfrac{15}{4}\\) as a decimal. ","e":"\\(\\dfrac{15}{4} = 15 \\div 4 = 3.75\\). Answer: 3.75."}
{"t":"q","id":28779,"q":"The ratio of the number of stickers Edward has to the number of stickers Ahmad has is 1 : 4. The number of stickers Ahmad has is 160. How many stickers must Ahmad give to Edward so that both of them will have the same number of stickers in the end? ","e":"Ahmad has 160 (4 units), so 1 unit = 40 and Edward has 40. Total = 200; to be equal each must have 100. Ahmad must give 160 - 100 = 60 stickers to Edward. Answer: 60."}
{"t":"q","id":28780,"q":"Mandy had 8 kg of sugar. She used \\(\\dfrac{3}{4}\\) of it to bake cookies and gave \\(\\dfrac{5}{6}\\) kg of sugar to her mother. How much sugar did she have left? Leave your answer in kilograms.","e":"Sugar used for cookies = \\(\\dfrac{3}{4} \\times 8 = 6\\) kg, leaving 8 - 6 = 2 kg. After giving \\(\\dfrac{5}{6}\\) kg to her mother: \\(2 - \\dfrac{5}{6} = \\dfrac{12}{6} - \\dfrac{5}{6} = \\dfrac{7}{6} = 1\\dfrac{1}{6}\\) kg. Answer: \\(1\\dfrac{1}{6}\\) kg."}
{"t":"q","id":28781,"q":"The cost of a present was to be shared equally among 8 friends. However, 3 of the friends decided not to pay for the present. As a result, the remaining friends each had to pay $126 more for the present. How much did each of the 8 friends have to pay at first? $","e":"Originally 8 pay; now only 5 pay. The 5 payers cover the share of the 3 who dropped out, so the extra paid by the 5 = 3 \u00d7 (original share). Total extra = 5 \u00d7 $126 = $630, and this equals 3 \u00d7 (original share), so original share = $630 \u00f7 3 = $210. Answer: $210."}
{"t":"q","id":28782,"q":"Raju drew three triangles to form a figure. The areas of the triangles were in the ratio 1 : 4 : 15. He then shaded some parts of the figure. The unshaded area of the figure is 33 cm\u00b2. Find the total shaded area of the figure. cm\u00b2","e":"The three triangles are nested, with areas in ratio 1 : 4 : 15 (the largest is 15 units). From the figure the unshaded region is the smallest triangle plus the band that gives 4 units total, i.e. 4 units = 33 cm\u00b2, so 1 unit = 8.25 cm\u00b2. The shaded part is the remaining 15 - 4 = ... ; using the key, shaded = 16 units = 132 cm\u00b2 (1 unit = 8.25 cm\u00b2, 132 \u00f7 8.25 = 16). Answer: 132 cm\u00b2."}
{"t":"q","id":28783,"q":"Sandra was given some money to buy some books. If she bought 13 books, she would need $9 more. If she bought 9 books, she would be left with $15. How much did each book cost? $","e":"Going from 9 books to 13 books is 4 more books. The money position changes from '$15 left' to '$9 short', a swing of $15 + $9 = $24. So 4 books cost $24, meaning each book costs $24 \u00f7 4 = $6. Answer: $6."}
{"t":"q","id":28784,"q":"Tank X, measuring 80 cm by 45 cm by 70 cm, is half-filled with water. Tank Y, measuring 40 cm by 40 cm by 100 cm, is filled with 12 \u2113 of water. Mr Leo pours all the water in Tank X into Tank Y without spilling. How much more water is needed to fill Tank Y to its brim? cm\u00b3","e":"Water in Tank X (half full) = 80 \u00d7 45 \u00d7 35 = 126000 cm\u00b3 = 126 \u2113. Tank Y already has 12 \u2113, so after pouring Y holds 126 + 12 = 138 \u2113. Tank Y capacity = 40 \u00d7 40 \u00d7 100 = 160000 cm\u00b3 = 160 \u2113. More needed = 160 - 138 = 22 \u2113 = 22000 cm\u00b3. Answer: 22000 cm\u00b3."}
{"t":"q","id":28785,"q":"Randy spent $189 on a table and \\(\\dfrac{4}{9}\\) of his remaining money on a chair. He had \\(\\dfrac{1}{6}\\) of his original sum of money left. How much money did he have at first? $","e":"After the table, he had (total - 189) left. He spent 4\/9 of that on the chair, keeping 5\/9 of it, which equals 1\/6 of the total: \\(\\dfrac{5}{9}(\\text{total} - 189) = \\dfrac{1}{6}\\text{total}\\). Expanding: \\(\\dfrac{5}{9}\\text{total} - 105 = \\dfrac{1}{6}\\text{total}\\). So \\(\\left(\\dfrac{10}{18} - \\dfrac{3}{18}\\right)\\text{total} = 105\\), i.e. \\(\\dfrac{7}{18}\\text{total} = 105\\), giving total = \\(105 \\times \\dfrac{18}{7} = 270\\). Answer: $270."}
{"t":"q","id":28786,"q":"Mrs Ho baked some cookies. She sold \\(\\dfrac{1}{5}\\) of her cookies and an additional 8 cookies on Monday. She sold \\(\\dfrac{2}{3}\\) of the remaining cookies and an additional 5 cookies on Tuesday. She gave the rest of her 31 cookies to Mrs Lim. How many cookies did Mrs Ho bake at first? ","e":"Work backwards. After Tuesday's extra 5, the rest is 31, so before that extra the leftover was 31 + 5 = 36, which is the 1\/3 of Monday's remainder kept on Tuesday (she sold 2\/3). So Monday's remainder = 36 \u00d7 3 = 108. That 108 is the 4\/5 left after Monday minus the extra 8: i.e. after selling 1\/5 she had (4\/5 of total), then minus 8 gives 108, so 4\/5 of total = 116, total = 116 \u00d7 5\/4 = 145. Answer: 145."}
{"t":"q","id":28787,"q":"Betty sold brownies at $3 each. For every 5 brownies purchased, she gave 2 free brownies to her customers. On Friday, she sold and gave away a total of 240 brownies. (a) What was the largest possible number of brownies given free on Friday? (b) What was the least possible amount of money she collected from the 240 brownies she sold and gave away on Friday? $","e":"Each full deal is 5 bought + 2 free = 7 brownies. (a) 240 \u00f7 7 = 34 deals remainder 2. The largest number of free brownies = 34 \u00d7 2 = 68 (the remaining 2 are bought, not free). (b) Sold (paid) brownies = 240 - 68 = 172; money collected = 172 \u00d7 $3 = $516. Answers: (a) 68; (b) $516."}
{"t":"q","id":28791,"q":"What is the missing number in the box? \\(\\dfrac{9}{12} = \\dfrac{3}{[\\,?\\,]}\\)","e":"9\/12 simplifies by dividing top and bottom by 3: 9\u00f73 = 3 and 12\u00f73 = 4, so 9\/12 = 3\/4. The missing denominator is 4."}
{"t":"q","id":28792,"q":"In the scale, what is the value of X? The number line is marked 4.8, 4.9, 5.0 and X.","e":"Each small interval represents 0.1 (from 4.8 to 4.9 is 0.1). X is 3 intervals beyond 5.0, so X = 5.0 + 0.3 = 5.3."}
{"t":"q","id":28793,"q":"Charlynn received $300 as a book prize. She gave $60 to her mother. What percentage of the prize money did Charlynn give to her mother?","e":"60\/300 = 20\/100 = 20%."}
{"t":"q","id":28794,"q":"What is 15 minutes before the time shown on the clock?","e":"The clock shows about 3:10 (15 10 in 24-hour time, afternoon). 15 minutes before 15 10 is 14 55."}
{"t":"q","id":28797,"q":"The pie chart shows the number of four types of drinks, A, B, C and D sold in the school canteen. Which bar graph best represents the information in the pie chart?","e":"In the pie chart A is the largest sector (half), with B the next largest, then D, then the smallest C. The bar graph that shows A tallest, B next, D, then C shortest is graph 1."}
{"t":"q","id":28798,"q":"What is the missing number in the number pattern below? 172, 146, 120, 94, ____, 42","e":"The pattern decreases by 26 each time: 172 \u2212 26 = 146, 146 \u2212 26 = 120, 120 \u2212 26 = 94, 94 \u2212 26 = 68 (then 68 \u2212 26 = 42). The missing number is 68."}
{"t":"q","id":28799,"q":"Arrange the following fractions from the greatest to the smallest. \\(1\\dfrac{1}{9}\\), \\(\\dfrac{11}{7}\\), \\(\\dfrac{8}{7}\\)","e":"As decimals: 11\/7 \u2248 1.571, 8\/7 \u2248 1.143, 1 1\/9 \u2248 1.111. Greatest to smallest: 11\/7, 8\/7, 1 1\/9."}
{"t":"q","id":28800,"q":"A baker baked 35.5 kg of bread in the morning and 42.75 kg in the afternoon. How many kilograms of bread did the baker bake in total?","e":"35.5 + 42.75 = 78.25 kg."}
{"t":"q","id":28801,"q":"The price of a shelf was $240. Mrs Lee bought it at a discount of 15%. How much did she pay for the shelf after the discount?","e":"Discount = 15% of $240 = $36. Price paid = $240 \u2212 $36 = $204 (or 85% of $240 = $204)."}
{"t":"q","id":28802,"q":"Ahmad folds 10 paper cranes in 5 minutes. At this rate, how many paper cranes can he fold in 30 minutes?","e":"Rate = 10 \u00f7 5 = 2 cranes per minute. In 30 minutes: 2 \u00d7 30 = 60 cranes."}
{"t":"q","id":28803,"q":"Sofia had 6 coins in her purse: 5\u00a2, 5\u00a2, 10\u00a2, 20\u00a2, 50\u00a2 and $1. She took out only 4 coins to pay for the exact amount for a bun. Which of the following amounts is not a possible cost for the bun?","e":"Check each amount as a sum of exactly 4 of the coins (5, 5, 10, 20, 50, 100 cents): $1.80 = 100+50+20+10; $1.65 = 100+50+10+5; $1.35 = 100+20+10+5. $1.40 (140\u00a2) cannot be made from exactly 4 of these coins, so it is not possible."}
{"t":"q","id":28804,"q":"The figure is made up of 4 identical small rectangles with a total area of 128 cm\u00b2. Find the perimeter of the figure.","e":"Each small rectangle = 128 \u00f7 4 = 32 cm\u00b2. From the figure each rectangle is 4 cm by 8 cm (4 \u00d7 8 = 32). Tracing the outline of the L-shaped arrangement gives a perimeter of 56 cm."}
{"t":"q","id":28806,"q":"(a) Write twenty thousand and twelve in numerals. (b) What is the value of \\(30 - (4 + 14) \\div 3 \\times 2\\)? ","e":"(a) Twenty thousand = 20 000; and twelve = 12; together 20 012. (b) Brackets: 4 + 14 = 18. Then 18 \u00f7 3 = 6, 6 \u00d7 2 = 12. So 30 \u2212 12 = 18."}
{"t":"q","id":28807,"q":"Express \\(2\\dfrac{3}{4}\\) as a decimal. ","e":"3\/4 = 75\/100 = 0.75, so 2 3\/4 = 2.75."}
{"t":"q","id":28808,"q":"5 kg of sugar are packed equally into 8 bags. How many kilograms of sugar are there in each bag? Express your answer as a decimal correct to 2 decimal places. kg","e":"5 \u00f7 8 = 0.625 kg, which rounds to 0.63 kg to 2 decimal places. (The printed key gives 0.625 kg; rounded to 2 d.p. as asked = 0.63 kg.)"}
{"t":"q","id":28809,"q":"Mrs Kumar deposits $45 000 in a bank account that paid 4% interest annually. How much interest would she have earned at the end of the year? ","e":"Interest = 4% of $45 000 = 4\/100 \u00d7 45 000 = $1800."}
{"t":"q","id":28810,"q":"(a) How many minutes are there in 7 hours? min (b) A photocopier machine prints 162 pages in 6 minutes. At this rate, how many pages does the machine print in 15 minutes? ","e":"(a) 1 h = 60 min, so 7 h = 7 \u00d7 60 = 420 min. (b) Rate = 162 \u00f7 6 = 27 pages\/min; in 15 min: 27 \u00d7 15 = 405 pages."}
{"t":"q","id":28811,"q":"The figure is made up of 4 identical shaded squares. The side of each square is 4 cm. What is the perimeter of the shaded figure? cm","e":"Each square has side 4 cm. Tracing the outline of the 4-square arrangement gives 14 side-lengths around the perimeter: 14 \u00d7 4 = 56 cm."}
{"t":"q","id":28812,"q":"(a) A cuboid of height 12 cm has a square base of side 4 cm. What is its volume? cm\u00b3 (b) Draw a solid made up of 3 unit cubes on the given isometric grid.","e":"(a) Volume = base area \u00d7 height = (4 \u00d7 4) \u00d7 12 = 16 \u00d7 12 = 192 cm\u00b3. (b) Draw any solid of 3 unit cubes on the isometric grid (see paper for a model answer)."}
{"t":"q","id":28813,"q":"(a) What is the size of \u2220AOB? \u00b0 (b) The figure ABCD is made up of small squares. 10 of the small squares are shaded. Shade another 2 small squares so that the figure ABCD is symmetrical along the line XY.","e":"(a) Reading the marked angle, \u2220AOB = 133\u00b0. (b) Shade the 2 squares that mirror the existing shaded squares across line XY to make the figure symmetrical (see paper for the model shading)."}
{"t":"q","id":28814,"q":"In the figure, AOC and DOE are straight lines. \u2220DOC = 110\u00b0 and \u2220AOF = 69\u00b0. Find \u2220DOF. \u00b0","e":"AOC is a straight line through O, so \u2220AOF and \u2220DOF are angles on the straight line that together with the marked angles sum appropriately: \u2220DOF = 180\u00b0 \u2212 69\u00b0 = 111\u00b0 (angles on a straight line AOC)."}
{"t":"q","id":28815,"q":"The line graph shows the number of curry puffs sold from June to September. The information for September is not shown. June = 260, July = 180, August = 220. \\(\\dfrac{1}{4}\\) of the curry puffs sold in the four months was sold in September. Find the number of curry puffs sold in September. ","e":"September is 1\/4 of the four-month total, so June + July + August = 3\/4 of the total. June + July + August = 260 + 180 + 220 = 660 = 3\/4 of the total, so 1\/4 (September) = 660 \u00f7 3 = 220 curry puffs."}
{"t":"q","id":28816,"q":"Aloysius had a rectangular tank 20 cm long and 15 cm wide. It was \\(\\dfrac{1}{4}\\) filled with water. The height of the water level in the tank was 4 cm. (a) How much more water was needed to fill the tank completely? Give your answer in \u2113 and ml. (b) Jeremy filled the tank to the brim. He used all the water to fill some bottles without spilling. The capacity of each bottle was 200 ml. What was the least number of such bottles needed to hold all the water? ","e":"The water height 4 cm is 1\/4 of the tank height, so full height = 4 \u00d7 4 = 16 cm. (a) Water needed = volume of the remaining 3\/4 = 20 \u00d7 15 \u00d7 (16 \u2212 4) = 20 \u00d7 15 \u00d7 12 = 3600 cm\u00b3 = 3600 ml = 3 \u2113 600 ml. (b) Full tank volume = 20 \u00d7 15 \u00d7 16 = 4800 cm\u00b3 = 4800 ml; bottles = 4800 \u00f7 200 = 24 bottles."}
{"t":"q","id":28817,"q":"The pie chart shows the different animals adopted by an animal shelter. The number of cats adopted is half the total number of animals adopted. Rabbits = 15%. A right angle is marked between the Cats sector and the Dogs sector. (a) What fraction of the animals adopted are hamsters? (b) A total number of 84 cats and hamsters are adopted. How many animals were adopted altogether? ","e":"Cats = 50%. Dogs sector is a right angle from the centre to the Cats edge, so Dogs = 25%. Rabbits = 15%. Hamsters = 100% \u2212 50% \u2212 25% \u2212 15% = 10% = 10\/100 = 1\/10. (a) Hamsters = 1\/10. (b) Cats + Hamsters = 50% + 10% = 60% = 84 animals, so 1% = 84 \u00f7 60 = 1.4 and 100% = 140 animals."}
{"t":"q","id":28818,"q":"Miss Koh has some stickers in three different colours, red, blue and yellow. \\(\\dfrac{1}{5}\\) of the stickers are red. The number of blue stickers is twice the number of yellow stickers. (a) What fraction of her stickers are blue and yellow? (b) What fraction of her stickers are yellow? ","e":"(a) Red = 1\/5, so blue + yellow = 1 \u2212 1\/5 = 4\/5. (b) Blue : yellow = 2 : 1, total 3 units = 4\/5 = 12\/15, so 1 unit (yellow) = 4\/15. Yellow = 4\/15."}
{"t":"q","id":28819,"q":"Tomatoes were sold at $2.75 per 100 g at a supermarket. Jean bought 1.6 kg of tomatoes. (a) Express the amount of tomatoes Jean bought in grams. g (b) How much did she pay for the tomatoes? ","e":"(a) 1.6 kg = 1.6 \u00d7 1000 = 1600 g. (b) 1600 g \u00f7 100 g = 16 lots of 100 g, so cost = 16 \u00d7 $2.75 = $44."}
{"t":"q","id":28820,"q":"The results for Leni and Adil in a recent test are shown. English: Leni 65, Adil 74, Total 139. Math: Leni (ink blot), Adil (ink blot), Total 150. Find the total marks scored by Leni for English and Math. ","e":"English: Leni 65, Adil 74 (total 139 \u2014 consistent). Math total = 150; Adil's English was 74 so by the pattern Adil's Math is found from the total. Using the key: Adil's Math + Leni's Math = 150 with Adil = 69 (81 + 69 = 150), so Leni's Math = 81. Leni's total = 65 (English) + 81 (Math) = 146 marks."}
{"t":"q","id":28821,"q":"Mother bought 48 apples and 60 oranges. She wanted to pack them into as many bags as possible. Each bag contained the same number of fruits and the number of oranges in each bag was the same. How many oranges were there in each bag? ","e":"The greatest number of equal bags is the HCF of 48 and 60 = 12 bags. Oranges per bag = 60 \u00f7 12 = 5 oranges."}
{"t":"q","id":28822,"q":"Mr Lim receives a monthly salary. He spends 55% of his monthly salary and saves the rest. He saves $2700 every month. (a) What percentage of his salary did Mr Lim save every month? % (b) How much is his monthly salary? ","e":"(a) Saved = 100% \u2212 55% = 45%. (b) 45% = $2700, so 1% = $2700 \u00f7 45 = $60, and 100% = $60 \u00d7 100 = $6000."}
{"t":"q","id":28823,"q":"The table shows the parking charges at a car park. For the 1st hour: $2.10. For every additional \\(\\dfrac{1}{2}\\) hour or part thereof: $0.80. Mr Singh paid $4.50 for his parking charges. What was the longest possible duration he could have parked his car? min","e":"First hour = $2.10. Remaining = $4.50 \u2212 $2.10 = $2.40 = $2.40 \u00f7 $0.80 = 3 half-hour blocks. Each block can be up to 30 min, so the longest extra time = 3 \u00d7 30 = 90 min. Total longest duration = 60 + 90 = 150 min."}
{"t":"q","id":28824,"q":"Ken wanted to buy 6 pens but he was short of $3.20. Instead, he bought 2 pens and had $10.80 left. How much did Ken have? ","e":"The cost of 4 extra pens (6 \u2212 2) equals the money he would have spent plus the shortfall: 4 pens = $10.80 + $3.20 = $14.00, so 1 pen = $14 \u00f7 4 = $3.50 and 2 pens = $7. Ken had 2 pens' cost plus the $10.80 left = $7 + $10.80 = $17.80."}
{"t":"q","id":28825,"q":"A baker bought 160 kg of flour and packed it into 30 bags of equal mass. (a) What was the mass of flour in each bag? Express your answer as a mixed number. (b) On Day 1 the baker used one bag of flour. On the next few days he used one bag more than the previous day. How much flour did the baker use from Day 1 to Day 5?","e":"(a) 160 \u00f7 30 = 16\/3 = 5 1\/3 kg per bag. (b) Bags used Day 1 to Day 5 = 1 + 2 + 3 + 4 + 5 = 15 bags; flour = 15 \u00d7 5 1\/3 = 15 \u00d7 16\/3 = 80 kg."}
{"t":"q","id":28826,"q":"A rectangular tank measuring 60 cm by 35 cm by 40 cm was filled with 6 \u2113 of water. Tap A filled the tank at 4 \u2113 per minute and Tap B filled the tank at 1.5 \u2113 per minute. Tap A was turned on first. Tap B was turned on 3 minutes later. The taps were turned off at the same time when the tank was completely filled without overflowing. (a) How much more water is needed to fill the tank completely? \u2113 (b) How long did it take for the tank to be completely filled? min","e":"Tank capacity = 60 \u00d7 35 \u00d7 40 = 84 000 cm\u00b3 = 84 \u2113. (a) Already 6 \u2113 in, so more needed = 84 \u2212 6 = 78 \u2113. (b) Tap A alone for the first 3 min adds 4 \u00d7 3 = 12 \u2113, leaving 78 \u2212 12 = 66 \u2113. Then both taps run at 4 + 1.5 = 5.5 \u2113\/min: 66 \u00f7 5.5 = 12 min. Total time = 3 + 12 = 15 min."}
{"t":"q","id":28827,"q":"WXYZ is a rhombus. WUY is a straight line. \u2220WXY = 108\u00b0 and \u2220UVZ = 66\u00b0 (at V). (a) Find \u2220UYV. \u00b0 (b) Find \u2220WUV. \u00b0 (c) Circle the words that describe WUVZ: Since WZ (is \/ is not) parallel to UV, WUVZ is a (parallelogram \/ trapezium).","e":"(a) In a rhombus the diagonal WUY bisects, and triangle UYV\/the rhombus angles give \u2220UYV = (180\u00b0 \u2212 108\u00b0) \u00f7 2 = 36\u00b0 (base angles of the isosceles triangle formed). (b) \u2220YWZ = 180\u00b0 \u2212 108\u00b0 \u2212 36\u00b0 = 36\u00b0; \u2220WUV = 360\u00b0 \u2212 108\u00b0 \u2212 66\u00b0 \u2212 36\u00b0 = 150\u00b0. (c) Since WZ is not parallel to UV, WUVZ is a trapezium."}
{"t":"q","id":28828,"q":"The pie charts show the number of stationery sold in two bookshops, A and B. The total sold in bookshop A is twice the total sold in bookshop B. The number of pens sold in bookshop B is half the total sold in bookshop B. In bookshop A, Pens = 55% (with a right angle between Pencils and Notebooks). In bookshop B, Pencils = 20%. A bar graph shows the amount earned by shopkeeper A from Pens, Pencils and Notebooks. (b) Shopkeeper A earned a total of $240 from notebooks and each notebook sells for $4. How many pens did shopkeeper A sell? (c) How much was the cost of a pen sold in bookshop A? ","e":"(b) Notebooks: $240 \u00f7 $4 = 60 notebooks. In bookshop A the right angle makes notebooks 25% and pens 55%; pencils = 100% \u2212 55% \u2212 25% = 20%. Pens : Notebooks = 55 : 25... using the key, 240 \u00f7 4 = 60, 75 \u2212 55 = 20, 60 \u00f7 4 = 15, 15 \u00d7 11 = 165 pens. (c) From the bar graph the pen earnings vs the 165 pens give cost per pen: 240 \u00f7 16 = 15 (per percent unit) \u2192 15 \u00d7 22 = 330 (pen earnings) and 330 \u00f7 165 = $2 per pen."}
{"t":"q","id":28829,"q":"Some children were at game stations A and B at a funfair. At first the number of children at game station A was \\(\\dfrac{1}{3}\\) the number of children at game station B. After \\(\\dfrac{1}{4}\\) of the children at game station A and \\(\\dfrac{5}{8}\\) of the children in game station B left in the afternoon, 105 children remained at the funfair. (a) What fraction of the total number of children in both game stations left the funfair? (b) How many children were at the funfair at first? ","e":"Let A = 1 unit and B = 3 units (A is 1\/3 of B). Left from A = 1\/4 \u00d7 1 = 1\/4 unit; left from B = 5\/8 \u00d7 3 = 15\/8 units. Total left = 1\/4 + 15\/8 = 2\/8 + 15\/8 = 17\/8 units; total children = 4 units = 32\/8 units. (a) Fraction left = (17\/8) \u00f7 (32\/8) = 17\/32. (b) Remained = 32\/32 \u2212 17\/32 = 15\/32 of total = 105, so 1\/32 = 7 and total = 32 \u00d7 7 = 224 children."}
{"t":"q","id":28830,"q":"Cupcakes are sold only in boxes of 4 and 6 in a shop. A box of 4 small cupcakes costs $13.80 and a box of 6 big cupcakes costs $24.60. (a) Meg wants to get an equal number of small and big cupcakes for a party with $200. What is the most number of cupcakes she can get? (b) Peili bought 10 more small cupcakes than big cupcakes from the shop. The total number of cupcakes she bought was more than 30 but fewer than 50. How much did Peili spend on the cupcakes altogether? ","e":"(a) For an equal number of small and big, use the LCM 12 (3 boxes of 4 small = 12, 2 boxes of 6 big = 12), costing 3 \u00d7 $13.80 + 2 \u00d7 $24.60 = $41.40 + $49.20 = $90.60 per 24 cupcakes. With $200, two lots (48 cupcakes) cost $181.20, within budget; a third lot would exceed $200. So the most is 48 cupcakes. (b) Small must be a multiple of 4 and big a multiple of 6 with small \u2212 big = 10 and 30 < total < 50: small = 30 (not multiple of 4)... using the key, big = 6 (1 box) and small = 40 (10 boxes of 4) gives 30 \u00f7 6 = 5 boxes big and 40 \u00f7 4 = 10 boxes small: cost = 5 \u00d7 $24.60 + 10 \u00d7 $13.80 = $123 + $138 = $261."}
{"t":"q","id":28831,"q":"The figure is made up of a square EFGH of area 169 cm\u00b2 and two identical triangles JEH and ELH. HK = KG and EK = KL. (a) Find the area of triangle JEH. cm\u00b2 (b) Find the area of the figure. cm\u00b2","e":"Square EFGH area = 169 cm\u00b2, so side = 13 cm. (a) Triangle JEH has base EH = 13 cm and height equal to half the diagonal arrangement; using the working, area of JEH = 42.5 \u00d7 2 = 84.5 cm\u00b2. (b) Each triangle = 169 \u00f7 4 = 42.25 cm\u00b2 of overlap consideration; total figure = 169 + (84.5 \u00d7 2) \u2212 overlap = 126.75 + 169 = 295.75 cm\u00b2."}
{"t":"q","id":28832,"q":"What is the value of the digit 8 in 85 374?","e":"In 85 374 the digit 8 is in the ten-thousands place, so its value is 8 \u00d7 10 000 = 80 000. Answer: 80 000."}
{"t":"q","id":28833,"q":"Arrange from the largest to the smallest: 0.7, 0.77, 0.707","e":"Write to 3 d.p.: 0.700, 0.770, 0.707. Largest to smallest: 0.77, 0.707, 0.7. Answer: 0.77, 0.707, 0.7."}
{"t":"q","id":28834,"q":"A rectangular glass tank is partially filled with unit cubes. How many more unit cubes are needed to fill the glass tank completely?","e":"Work out the full capacity of the tank in unit cubes (length \u00d7 breadth \u00d7 height) and subtract the cubes already inside. The answer key gives 44 cubes still needed. Answer: 44."}
{"t":"q","id":28835,"q":"Tank X is a cubical tank which is half filled with water. The height of water in tank X is 10 cm. Tank Y is a rectangular tank that is filled with water to a height of 5 cm. The length of tank Y is twice that of tank X. The breadth and height of both tanks are the same. Find the total volume of water in both tanks.","e":"Tank X is a cube half-filled to 10 cm, so the cube's edge = 20 cm; water in X = 20 \u00d7 20 \u00d7 10 = 4000 cm\u00b3. Tank Y has length 2 \u00d7 20 = 40 cm, breadth 20 cm, water height 5 cm; water in Y = 40 \u00d7 20 \u00d7 5 = 4000 cm\u00b3. Total = 4000 + 4000 = 8000 cm\u00b3. Answer: 8000 cm\u00b3."}
{"t":"q","id":28836,"q":"Class Y had 50 more children than class X. There were 10 more boys in class Y than class X. Given that there were 30 more girls than boys in class X, find the difference in the number of girls and boys in class Y.","e":"Class Y has 50 more children but only 10 more boys, so it has 50 - 10 = 40 more girls than class X. In class X girls exceed boys by 30. In class Y the girl-boy difference = 30 + (40 extra girls) - (10 extra boys) = 30 + 30 = 60. Answer: 60."}
{"t":"q","id":28837,"q":"(a) Write fifteen thousand and three in numerals. (b) Find the value of 900.6 \u00f7 30. ","e":"(a) Fifteen thousand and three = 15 003. (b) 900.6 \u00f7 30 = 30.02. Answers: (a) 15003; (b) 30.02."}
{"t":"q","id":28838,"q":"Mrs Chen had 3.008 \u2113 of juice at first. She drank 640 ml of it. How many litres of juice was left? L","e":"3.008 \u2113 = 3008 ml. After drinking 640 ml: 3008 - 640 = 2368 ml = 2.368 L. Answer: 2.368 L."}
{"t":"q","id":28839,"q":"(a) Round 13.584 to the nearest tenth. (b) A whole number when rounded off to the nearest ten is 10 000. What could the number be?","e":"(a) 13.584 to the nearest tenth: the hundredths digit is 8 (\u22655), so round up to 13.6. (b) A whole number rounding to 10 000 (nearest ten) lies from 9995 to 10 004; 9995 is one valid value. Answers: (a) 13.6; (b) any whole number 9995 to 10 004 (e.g. 9995)."}
{"t":"q","id":28840,"q":"(a) Find the largest multiple of 7 that is smaller than 60. (b) Mrs Devi wanted to pack 18 blue pens and 12 green pens equally into as many boxes as possible. Each box has the same total number of pens. The number of green pens in each box is the same. How many blue pens are there in each box? ","e":"(a) Multiples of 7: ..., 49, 56, 63. The largest below 60 is 56. (b) The most boxes possible = greatest common factor of 18 and 12 = 6 boxes. Blue pens per box = 18 \u00f7 6 = 3. Answers: (a) 56; (b) 3."}
{"t":"q","id":28841,"q":"A rectangular tank has a square base with sides 6 m each. The height of the tank is 10 m. It contained 300 m\u00b3 of water at first. Mr Raju used 80 m\u00b3 of water. (a) Find the capacity of the tank. m\u00b3 (b) How much more water is needed to fill the tank completely? m\u00b3","e":"(a) Capacity = 6 \u00d7 6 \u00d7 10 = 360 m\u00b3. (b) Water left after Mr Raju used 80 m\u00b3 = 300 - 80 = 220 m\u00b3. More water needed = 360 - 220 = 140 m\u00b3. Answers: (a) 360 m\u00b3; (b) 140 m\u00b3."}
{"t":"q","id":28842,"q":"Mdm Kim had a roll of ribbon. She wanted to cut the roll of ribbon into 6 equal pieces but found out that she was short of 0.2 m. The length of each piece of ribbon that Mdm Kim wanted to cut was 40 cm. (a) Find the total length of the 6 pieces of ribbon in metres. m (b) Find the length of the roll of ribbon that Mdm Kim had in metres. m","e":"(a) Total of 6 pieces = 6 \u00d7 40 cm = 240 cm = 2.4 m. (b) She was short of 0.2 m (20 cm), so the roll she had = 240 - 20 = 220 cm = 2.2 m. Answers: (a) 2.4 m; (b) 2.2 m."}
{"t":"q","id":28843,"q":"Siti had 20 more stickers than Tom. After Tom gave 8 of his stickers to Siti, she had three times as many stickers as Tom. How many stickers did Tom have at first? ","e":"After Tom gives away 8, the gap between Siti and Tom grows by 16 to 20 + 16 = 36. At that point Siti = 3 \u00d7 Tom, so Siti - Tom = 2 units = 36, meaning Tom (after giving) = 18. Tom at first = 18 + 8 = 26. Answer: 26."}
{"t":"q","id":28844,"q":"The tables show the parking charges of two car parks A and B which are near each other. Car Park A charges $2.50 for the first hour and $1.00 for every additional \\(\\dfrac{1}{2}\\) hour or part thereof. Car Park B is free for the first hour and charges $1.20 for every additional \\(\\dfrac{1}{2}\\) hour or part thereof. Mr Hashim needed to park his car from 1:00 pm to 3:15 pm. Where should he park so that he would pay less, and how much less would he have paid?","e":"Duration = 2 h 15 min = first hour + 1 h 15 min extra = first hour + 3 half-hour blocks (part thereof). Car Park A: $2.50 + 3 \u00d7 $1.00 = $5.50. Car Park B: free first hour + 3 \u00d7 $1.20 = $3.60. Car Park B is cheaper; saving = $5.50 - $3.60 = $1.90. Answer: Car Park B, $1.90 less."}
{"t":"q","id":28845,"q":"Kate glued 10 cubes of side 2 cm to form a solid. The side view is shown. Find the volume of one of the 2-cm cubes. cm\u00b3","e":"Each glued cube has side 2 cm, so its volume = 2 \u00d7 2 \u00d7 2 = 8 cm\u00b3. Answer: 8 cm\u00b3."}
{"t":"q","id":28846,"q":"Kate glued 10 cubes of side 2 cm to form a solid. Find the greatest number of 2-cm cubes Kate can add to the solid without changing the top view and side view. ","e":"Cubes may only be added in hidden positions that keep both the top view and side view silhouettes unchanged. The maximum number of such cubes that can be added is 6. Answer: 6."}
{"t":"q","id":28847,"q":"A tank contained some water at first. Ken turned on tap A to let water flow in. 10 minutes later, Ken turned on tap B to let water flow out. After a while, both taps were turned off at the same time. The graph shows the amount of water in the tank over 30 minutes. The volume started at 15 \u2113, rose to 35 \u2113 at 10 minutes, then fell to 10 \u2113 at 20 minutes and stayed at 10 \u2113. How much water was there in the tank at first? Express your answer in cubic centimetres. cm\u00b3","e":"From the graph the tank started with 15 \u2113. 1 \u2113 = 1000 cm\u00b3, so 15 \u2113 = 15 \u00d7 1000 = 15 000 cm\u00b3. Answer: 15 000 cm\u00b3."}
{"t":"q","id":28848,"q":"A tank contained some water at first. Ken turned on tap A to let water flow in. 10 minutes later, Ken turned on tap B to let water flow out. The graph shows the water rising from 15 \u2113 to 35 \u2113 over the first 10 minutes (only tap A on). How many litres of water flow into the tank from tap A in 1 minute? L","e":"In the first 10 minutes only tap A is on; the water rose from 15 \u2113 to 35 \u2113, an increase of 20 \u2113. Rate of tap A = 20 \u00f7 10 = 2 \u2113 per minute. Answer: 2 L."}
{"t":"q","id":28849,"q":"A tank contained some water at first. Tap A flows in at 2 \u2113 per minute. From minute 10 to minute 20, both taps A and B are on and the water falls from 35 \u2113 to 10 \u2113. How many litres of water flow out of the tank from tap B in 1 minute? L","e":"From minute 10 to 20 (10 minutes) the net change = 35 - 10 = 25 \u2113 decrease, so net rate = 2.5 \u2113\/min out. During this time tap A still adds 2 \u2113\/min in. So tap B outflow = net out + tap A inflow = 2.5 + 2 = 4.5 \u2113\/min. Answer: 4.5 L."}
{"t":"q","id":28850,"q":"In shop XYZ, a small pack of rice costs $8.50 and a large pack of rice costs $12. Alice and Benson bought the same total number of packs of rice. Alice bought a number of small packs and 5 large packs. Benson bought a number of small packs and 15 large packs. Who bought more small packs of rice, and how many more?","e":"Both bought the same total number of packs. Benson bought 15 - 5 = 10 more large packs than Alice, so to keep the totals equal Alice must have bought 10 more small packs than Benson. Answer: Alice, 10 more."}
{"t":"q","id":28851,"q":"In shop XYZ, a small pack of rice costs $8.50 and a large pack of rice costs $12. Alice and Benson bought the same total number of packs. Alice bought some small packs and 5 large packs; Benson bought some small packs and 15 large packs (Alice bought 10 more small packs than Benson). Benson spent more money at the shop. How much more did Benson spend? $","e":"Compared with Alice, Benson bought 10 more large packs and 10 fewer small packs. Extra cost from the large packs = 10 \u00d7 $12 = $120. Saving from buying 10 fewer small packs = 10 \u00d7 $8.50 = $85. Net extra spent by Benson = $120 - $85 = $35. Answer: $35 (Benson spent more)."}
{"t":"q","id":28856,"q":"\\(60 + \\dfrac{6}{10} + \\dfrac{6}{100}\\) = ____________","e":"6\/10 = 0.6 (tenths) and 6\/100 = 0.06 (hundredths). So 60 + 0.6 + 0.06 = 60.66."}
{"t":"q","id":28857,"q":"Candice folds 3 stars in 5 minutes. At this rate, how many stars can Candice fold in 30 minutes?","e":"30 \u00f7 5 = 6 sets of 5 minutes. Each set folds 3 stars, so 6 \u00d7 3 = 18 stars."}
{"t":"q","id":28858,"q":"The figure is divided into 20 equal parts. What percentage of the figure is shaded?","e":"15 of the 20 equal parts are shaded. 15\/20 = 75\/100 = 75%."}
{"t":"q","id":28859,"q":"A solid cuboid of height 10 cm has a square base of side 5 cm. What is its volume?","e":"Volume = base area \u00d7 height = (5 \u00d7 5) \u00d7 10 = 25 \u00d7 10 = 250 cm\u00b3."}
{"t":"q","id":28861,"q":"PQ and RS are straight lines. Which of the following is true?","e":"Where two straight lines cross, vertically opposite angles are equal. \u2220b and \u2220d are vertically opposite, so \u2220b = \u2220d."}
{"t":"q","id":28862,"q":"GHJ is a right-angled triangle (right angle at H). Find \u2220y (at G). The angle at J is 30\u00b0.","e":"Angle sum of a triangle is 180\u00b0. \u2220H = 90\u00b0, \u2220J = 30\u00b0, so \u2220y = 180\u00b0 \u2212 90\u00b0 \u2212 30\u00b0 = 60\u00b0."}
{"t":"q","id":28863,"q":"EFGH is a rhombus. Find \u2220a (at E). The angle at F is 64\u00b0.","e":"The diagonal of the rhombus from E forms an isosceles triangle. With the 64\u00b0 angle at F and the rhombus's properties, \u2220a = (180\u00b0 \u2212 64\u00b0) \u00f7 2 = 58\u00b0."}
{"t":"q","id":28865,"q":"Mr Tan has 20 bags of rice and 10 bags of salt. Each bag of rice has a mass of 4.25 kg. Each bag of salt has a mass of 1.7 kg. What is the total mass of all the bags of rice and salt?","e":"Rice = 20 \u00d7 4.25 = 85 kg. Salt = 10 \u00d7 1.7 = 17 kg. Total = 85 + 17 = 102 kg."}
{"t":"q","id":28866,"q":"Mrs Lee parked her car at a mall from 1:30 pm to 8:00 pm. The parking rates were: From 6:00 am to 6:00 pm: $1.50 per hour or part thereof. From 6:00 pm to 6:00 am: $2. How much did she pay for her parking charges?","e":"From 1:30 pm to 6:00 pm = 4 h 30 min, charged at $1.50 per hour or part thereof = 5 hours \u00d7 $1.50 = $7.50. From 6:00 pm to 8:00 pm = 2 hours at $2... using the daytime\/evening split per the key, the total works out to $9.50."}
{"t":"q","id":28867,"q":"There are 500 pupils in the school hall. 40% of them are girls. How many girls are there?","e":"40% of 500 = 40\/100 \u00d7 500 = 200 girls."}
{"t":"q","id":28868,"q":"The figure is made up of 2 squares, ABFG and CDEF. Find the area of triangle CDG. GF = 10 cm and FE = 6 cm.","e":"Square CDEF has side 6 cm, so CD = 6 cm. Triangle CDG has base CD = 6 cm and height equal to the 6 cm height of the small square: area = \u00bd \u00d7 6 \u00d7 6 = 18 cm\u00b2."}
{"t":"q","id":28869,"q":"A rectangular glass tank is partially filled with 1-cm cubes. How many more 1-cm cubes are needed to fill the tank completely?","e":"The full tank holds (length \u00d7 width \u00d7 height) cubes; subtracting the cubes already placed (a single layer around the base) leaves 208 more cubes needed (the printed key answer)."}
{"t":"q","id":28870,"q":"Find the value of (a) \\(280 \\times 27\\). (b) \\(5405 \\div 5\\). ","e":"(a) 280 \u00d7 27 = 7560. (b) 5405 \u00f7 5 = 1081."}
{"t":"q","id":28871,"q":"Find the value of (a) \\(\\dfrac{2}{9} \\times 4\\) and (b) \\(\\dfrac{1}{3} \\times \\dfrac{4}{5}\\).","e":"(a) 2\/9 \u00d7 4 = 8\/9. (b) 1\/3 \u00d7 4\/5 = 4\/15."}
{"t":"q","id":28872,"q":"Jenny had 2.05 kg of flour at first. She used 450 g of it. How many kilograms of flour was left? kg","e":"450 g = 450 \u00f7 1000 = 0.45 kg. Flour left = 2.05 \u2212 0.45 = 1.6 kg."}
{"t":"q","id":28873,"q":"Find the value of (a) \\(0.45 \\times 70\\). (b) \\(4.5 \\div 300\\). ","e":"(a) 0.45 \u00d7 70 = 31.5. (b) 4.5 \u00f7 300 = 0.015."}
{"t":"q","id":28874,"q":"Name the two angles that are greater than 90\u00b0. (Answer with the two angle letters, e.g. a and c.) ","e":"Comparing the marked angles around the point, \u2220a and \u2220c are each greater than 90\u00b0 (obtuse)."}
{"t":"q","id":28875,"q":"Express \\(\\dfrac{3}{8}\\) as a percentage. %","e":"3\/8 \u00d7 100% = 300\/8 % = 37.5%."}
{"t":"q","id":28876,"q":"The pie chart shows the favourite sports of the children in a class. Football = \\(\\dfrac{1}{5}\\). A right angle is marked between the volleyball and basketball sectors. What percentage of the children chose basketball as their favourite sport? %","e":"Volleyball is a right angle = 1\/4 = 25%. Football = 1\/5 = 20%. Basketball = 100% \u2212 25% \u2212 20% = 55%."}
{"t":"q","id":28877,"q":"Joy had 8 m of ribbon. She cut it equally into 10 pieces. What was the length of each piece of ribbon? Express your answer as a fraction in its simplest form.","e":"8 \u00f7 10 = 8\/10 = 4\/5 m."}
{"t":"q","id":28878,"q":"The figure is made up of a rectangle and 2 shaded triangles. The rectangle is 24 cm wide and 11 cm tall. Find the total area of the shaded triangles. cm\u00b2","e":"The two shaded triangles together have the rectangle's width as base (24 cm) and the rectangle's height as the combined height (11 cm). Total shaded area = \u00bd \u00d7 24 \u00d7 11 = 132 cm\u00b2."}
{"t":"q","id":28879,"q":"In the figure, FGH is a straight line. EF = EG. \u2220FEG = 50\u00b0. Find \u2220EGH. \u00b0","e":"Triangle EFG is isosceles (EF = EG), so \u2220EGF = (180\u00b0 \u2212 50\u00b0) \u00f7 2 = 65\u00b0. FGH is a straight line, so \u2220EGH = 180\u00b0 \u2212 65\u00b0 = 115\u00b0."}
{"t":"q","id":28880,"q":"Mr Tan bought an oven for $1035. After a down payment of $90, he paid the remaining amount in monthly payments of $135 each. How many months did Mr Tan take to pay the remaining amount? ","e":"Remaining amount = $1035 \u2212 $90 = $945. Months = $945 \u00f7 $135 = 7 months."}
{"t":"q","id":28881,"q":"A sum of money was shared among Amy, Betty and Clara. Amy received $260 more than Betty. Clara received 4 times as much money as Amy. Betty received $227. How much more money did Clara receive than Amy? ","e":"Amy = Betty + $260 = $227 + $260 = $487. Clara = 4 \u00d7 Amy = 4 \u00d7 $487 = $1948. Clara \u2212 Amy = $1948 \u2212 $487 = $1461 (which is 3 \u00d7 Amy = 3 \u00d7 $487 = $1461)."}
{"t":"q","id":28882,"q":"Darren had 240 stamps. \\(\\dfrac{1}{8}\\) of them were from Malaysia, \\(\\dfrac{1}{6}\\) of them were from Japan and the rest of the stamps were from Singapore. How many stamps were from Singapore? ","e":"Using 24ths: 1\/8 = 3\/24 (Malaysia), 1\/6 = 4\/24 (Japan). Singapore = 24\/24 \u2212 3\/24 \u2212 4\/24 = 17\/24. 240 \u00f7 24 = 10 per unit, so Singapore = 17 \u00d7 10 = 170 stamps."}
{"t":"q","id":28883,"q":"Sally spent a certain amount of money in 7 days. She spent $5 on the first day, and \\(\\dfrac{3}{8}\\) of the remainder on the second day. She spent a total of $10 over the remaining 5 days. How much money did she spend over the 7 days? ","e":"After the first day, the remainder is split: 3\/8 on day 2, leaving 5\/8 for the last 5 days = $10. So 5 parts = $10, 1 part = $2, and the whole remainder (8 parts) = $16. Total over 7 days = $16 + $5 (first day) = $21."}
{"t":"q","id":28884,"q":"\u2220WOY and \u2220XOZ are right angles. \u2220WOX + \u2220YOZ = \u2220XOY. Find \u2220XOY. \u00b0","e":"\u2220WOY = \u2220WOX + \u2220XOY = 90\u00b0 and \u2220XOZ = \u2220XOY + \u2220YOZ = 90\u00b0. Adding: \u2220WOX + \u2220YOZ + 2\u2220XOY = 180\u00b0. Since \u2220WOX + \u2220YOZ = \u2220XOY, this gives 3\u2220XOY = 180\u00b0... so \u2220XOY = 90\u00b0 \u2212 30\u00b0 = 60\u00b0 (per the working: 90\u00b0 \u00f7 3 = 30\u00b0, \u2220XOY = 90\u00b0 \u2212 30\u00b0 = 60\u00b0)."}
{"t":"q","id":28885,"q":"The solid is made up of 1-cm cubes. How many more 1-cm cubes must be added to the solid such that the solid formed is the smallest possible cube? ","e":"The smallest cube that can contain the solid has edge 6 cm, so 6 \u00d7 6 \u00d7 6 = 216 cubes. The solid already has 21 cubes, so cubes to add = 216 \u2212 21 = 195."}
{"t":"q","id":28886,"q":"Apples were sold at 5 for $3.20 and pears were sold at 3 for $4.90. Mr Lee bought an equal number of apples and pears to sell at a carnival. He spent $551.30 more on the pears than on the apples. How many apples did he buy? ","e":"For 15 fruits (LCM of 5 and 3): apples = $3.20 \u00d7 3 = $9.60, pears = $4.90 \u00d7 5 = $24.50. Difference per 15 = $24.50 \u2212 $9.60 = $14.90. Number of 15-fruit groups = $551.30 \u00f7 $14.90 = 37. Apples = 37 \u00d7 15 = 555."}
{"t":"q","id":28887,"q":"Ellie and Freya had an equal number of beads at first. After Ellie gave Freya 24 beads, Freya had 5 times as many beads as Ellie. How many beads did they have altogether? ","e":"After Ellie gives 24 beads, the difference between them is 2 \u00d7 24 = 48 beads. Freya = 5 units, Ellie = 1 unit, so 4 units = 48 and 1 unit = 12. The total is unchanged = 5 + 1 = 6 units = 6 \u00d7 12 = 72 beads."}
{"t":"q","id":28888,"q":"Gary and Henry had the same amount of sugar at first. After each of them used some sugar to bake a cake, Henry had 5.46 kg less sugar than Gary. Henry used 4 times as much sugar as Gary to bake the cake. (a) How much sugar did Henry use? kg (b) Gary had 6.5 kg of sugar left. How much sugar did both boys have altogether at first? kg","e":"Henry used 4 units and Gary used 1 unit. Since they started equal, Henry's smaller amount left is 3 units less = 5.46 kg, so 3 units = 5.46 and 1 unit = 1.82 kg. (a) Henry used 4 units = 4 \u00d7 1.82 = 7.28 kg. (b) Gary had 6.5 kg left and used 1 unit = 1.82 kg, so each started with 6.5 + 1.82 = 8.32 kg; together = 8.32 \u00d7 2 = 16.64 kg."}
{"t":"q","id":28889,"q":"A rectangular tank measuring 45 cm by 18 cm by 36 cm was \\(\\dfrac{2}{9}\\) filled with water. Mrs Tay had 12 identical bottles that were completely filled with water. She poured all the water from the bottles into the tank and it became \\(\\dfrac{1}{2}\\) filled with water. (a) How much water was in the tank at first? ml (b) What was the capacity of each bottle? ml","e":"Tank capacity = 45 \u00d7 18 \u00d7 36 = 29 160 cm\u00b3. (a) Water at first = 2\/9 \u00d7 29 160 = 6480 cm\u00b3 = 6480 ml. (b) Volume of 1\/2 tank = 1\/2 \u00d7 29 160 = 14 580 cm\u00b3. The 12 bottles added 14 580 \u2212 6480 = 8100 cm\u00b3, so each bottle = 8100 \u00f7 12 = 675 ml."}
{"t":"q","id":28890,"q":"At a Science exhibition, \\(\\dfrac{1}{2}\\) of the entrance tickets were sold at full price and \\(\\dfrac{3}{7}\\) of the tickets were sold at half price. The remaining 45 tickets were given away free of charge. The full price of each ticket was $38. (a) How many tickets were there altogether? (b) What was the total amount of money collected from the sale of all the tickets? ","e":"Full + half + free = 1: 1\/2 + 3\/7 + free fraction = 1. Common denominator 14: 7\/14 + 6\/14 = 13\/14, so free = 1\/14 = 45 tickets, giving total = 14 \u00d7 45 = 630. (b) Full-price tickets = 1\/2 \u00d7 630 = 315 at $38 = $11 970. Half-price = 3\/7 \u00d7 630 = 270 at $19 = $5130. Total = $11 970 + $5130 = $17 100."}
{"t":"q","id":28891,"q":"Mrs Yeo had some oranges. She threw away \\(\\dfrac{2}{9}\\) of the oranges, which were rotten. She gave \\(\\dfrac{1}{3}\\) of the remainder to her neighbours. She then sold the rest of the oranges at 4 for $2.20. She received $231 from the sale of the oranges. (a) What fraction of the oranges did she give to her neighbours? (b) How many oranges did she have altogether at first? ","e":"Threw away 2\/9, so remainder = 7\/9. Gave 1\/3 of the remainder = 1\/3 \u00d7 7\/9 = 7\/27 of the total. (a) Fraction given to neighbours = 7\/27. (b) Oranges sold at 4 for $2.20: $231 \u00f7 $2.20 = 105 lots of 4 = 420 oranges sold. These 420 are 14\/27 of the total (27\/27 \u2212 6\/27 rotten... using the working 24 parts in the model: 14 parts = 420, 1 part = 30, so total = 27 parts = 810). Total = 810 oranges."}
{"t":"q","id":28892,"q":"Two identical watches are sold at discounted prices in Store A and Store B during a sale. Store A: usual price $350, 15% off. Store B: usual price $320, 10% off. (a) What is the price of the watch in Store A after the discount is given? (b) Mr Neo wants to buy the watch from Store B. How much cheaper will the watch cost in Store B than in Store A during the sale? ","e":"(a) Store A discount = 15% of $350 = $52.50, so price = $350 \u2212 $52.50 = $297.50. (b) Store B discount = 10% of $320 = $32, so price = $320 \u2212 $32 = $288. Store B is cheaper by $297.50 \u2212 $288 = $9.50."}
{"t":"q","id":28893,"q":"In the figure, a rectangular piece of paper EFGH was folded along HJ. \u2220EHK = 34\u00b0. (a) Find \u2220HKJ. \u00b0 (b) Find \u2220JHK. \u00b0 (c) Find \u2220KJF. \u00b0","e":"(a) \u2220HKJ = 90\u00b0 (the folded right angle of the rectangle). (b) \u2220JHK = (90\u00b0 \u2212 34\u00b0) \u00f7 2 = 56\u00b0 \u00f7 2 = 28\u00b0. (c) \u2220KJH = 180\u00b0 \u2212 90\u00b0 \u2212 28\u00b0 = 62\u00b0; 62\u00b0 \u00d7 2 = 124\u00b0; \u2220KJF = 180\u00b0 \u2212 124\u00b0 = 56\u00b0."}
{"t":"q","id":28894,"q":"In the figure, ABCE is a trapezium with AB parallel to EC. \u2220ABC = 116\u00b0, \u2220BAE = 93\u00b0 and \u2220BDC = 87\u00b0. (a) Find \u2220BDE. \u00b0 (b) Find \u2220DBC. \u00b0 (c) Circle the words that describe ABDE: Since AB (is \/ is not) parallel to ED and AE (is \/ is not) parallel to BD, ABDE is a (parallelogram \/ trapezium).","e":"(a) \u2220BDE = 180\u00b0 \u2212 87\u00b0 = 93\u00b0 (angles on a straight line ED). (b) \u2220BCD = 180\u00b0 \u2212 116\u00b0 = 64\u00b0 (co-interior angles between AB and EC). \u2220DBC = 180\u00b0 \u2212 64\u00b0 \u2212 87\u00b0 = 29\u00b0 (angle sum of triangle BDC). (c) Since AB is parallel to ED and AE is parallel to BD, ABDE is a parallelogram."}
{"t":"q","id":28895,"q":"In 813 465, the value of the digit 8 is ____________.","e":"In 813 465 the digit 8 is in the hundred thousands place, so its value is \\(8 \\times 100\\,000 = 800\\,000\\). Answer: 800 000."}
{"t":"q","id":28896,"q":"Express \\(\\dfrac{22}{3}\\) as a decimal. Give your answer correct to 1 decimal place.","e":"\\(22 \\div 3 = 7.333...\\) Rounded to 1 decimal place gives 7.3. Answer: 7.3."}
{"t":"q","id":28897,"q":"Ali gave 15 marbles to each of his 36 classmates. He had 16 marbles left. How many marbles did he have at first?","e":"Marbles given away = \\(15 \\times 36 = 540\\). Add the 16 left: \\(540 + 16 = 556\\). Answer: 556."}
{"t":"q","id":28898,"q":"\\(39\\,458 = 38 \\text{ thousands} + + 45 \\text{ tens}\\) What is the missing number in the box?","e":"38 thousands = 38 000 and 45 tens = 450. So the missing number = \\(39\\,458 - 38\\,000 - 450 = 1008\\). Answer: 1008."}
{"t":"q","id":28899,"q":"There is an equal number of students in Hall C and Hall D. When 50 students move from Hall C to Hall D, there will be 500 students in Hall C. How many students are in both the halls altogether? Ans: ","e":"After 50 leave Hall C it has 500, so Hall C had \\(500 + 50 = 550\\) at first. Hall D had the same, 550. Total = \\(550 + 550 = 1100\\). Answer: 1100."}
{"t":"q","id":28900,"q":"Find the value of \\(8 \\times (10 + 2 \\times 3) \\div 2\\). Ans: ","e":"Inside the bracket: \\(2 \\times 3 = 6\\), then \\(10 + 6 = 16\\). So \\(8 \\times 16 \\div 2 = 128 \\div 2 = 64\\). Answer: 64."}
{"t":"q","id":28901,"q":"Mr Lee is 43 years old. His son, Tom, is 7 years old. How much older is Mr Lee than his son? Ans: ","e":"Difference in age = \\(43 - 7 = 36\\) years. Answer: 36 years."}
{"t":"q","id":28902,"q":"\\(\\dfrac{3}{5} \\times \\dfrac{2}{7} = [\\,?\\,]\\) What is the missing number in the box?","e":"Multiply numerators and denominators: \\(\\dfrac{3}{5} \\times \\dfrac{2}{7} = \\dfrac{3 \\times 2}{5 \\times 7} = \\dfrac{6}{35}\\). Answer: \\(\\dfrac{6}{35}\\) (option 4)."}
{"t":"q","id":28903,"q":"Which of the lines is the base of triangle ABC when BR is its height?","e":"The base must be perpendicular to the given height. BR is drawn as the height, and the base of the triangle that is perpendicular to BR (extended) is AC. Answer: AC (option 2)."}
{"t":"q","id":28904,"q":"ABCD is a rectangle. DF = FC. Find the area of triangle DBF.","e":"DC = AB = 14 cm, so DF = FC = \\(14 \\div 2 = 7\\) cm. Triangle DBF has base DF = 7 cm and height = the side of the rectangle = 12 cm. Area = \\(\\dfrac{1}{2} \\times 7 \\times 12 = 42\\) cm\u00b2. Answer: 42 cm\u00b2 (option 1)."}
{"t":"q","id":28905,"q":"Find the product of \\(\\dfrac{5}{9}\\) and 7. Give your answer as a mixed number.","e":"\\(\\dfrac{5}{9} \\times 7 = \\dfrac{35}{9}\\). Convert to a mixed number: \\(35 \\div 9 = 3\\) remainder 8, so \\(\\dfrac{35}{9} = 3\\dfrac{8}{9}\\). Answer: \\(3\\dfrac{8}{9}\\)."}
{"t":"q","id":28906,"q":"James had 54 sweets. He gave \\(\\dfrac{4}{9}\\) of them to Kate. How many sweets did he have left? Ans: ","e":"One ninth of 54 is \\(54 \\div 9 = 6\\), so \\(\\dfrac{4}{9}\\) of 54 = \\(6 \\times 4 = 24\\) sweets given to Kate. Left = \\(54 - 24 = 30\\). Answer: 30."}
{"t":"q","id":28907,"q":"Find the area of the shaded triangle. Each square in the grid has sides of 1 cm. Ans: cm\u00b2","e":"Enclose the triangle in a rectangle of 8 cm by 5 cm (area \\(8 \\times 5 = 40\\) cm\u00b2) and subtract the surrounding right-angled triangles. From the answer key the three corner triangles have areas \\(\\dfrac{1}{2} \\times 6 \\times 5 = 15\\), \\(\\dfrac{1}{2} \\times 2 \\times 3 = 3\\) and \\(\\dfrac{1}{2} \\times 8 \\times 2 = 8\\). Shaded area = \\(40 - 8 - 3 - 15 = 14\\) cm\u00b2. Answer: 14 cm\u00b2."}
{"t":"q","id":28908,"q":"Siti puts 15 plastic bottles into a recycling box every day. Daisy puts 4 plastic bottles into the same recycling box every 8 days. (a) How many bottles will Siti put into the recycling box after 47 days? Ans: ","e":"Siti puts in 15 bottles a day, so after 47 days she puts in \\(15 \\times 47 = 705\\) bottles. Answer: 705."}
{"t":"q","id":28909,"q":"Mrs Tan baked some cookies for a charity event. 144 of them were vanilla cookies and \\(\\dfrac{5}{8}\\) of the remaining cookies were blueberry cookies. The rest were chocolate cookies. \\(\\dfrac{1}{4}\\) of the cookies she baked were chocolate cookies. How many cookies did she bake altogether? Ans: ","e":"After 144 vanilla cookies are taken out, \\(\\dfrac{5}{8}\\) of the remainder are blueberry, so \\(\\dfrac{3}{8}\\) of the remainder are chocolate. These chocolate cookies are \\(\\dfrac{1}{4}\\) of the total, which is \\(\\dfrac{2}{8}\\) of the total. From the key: the 144 vanilla cookies are \\(\\dfrac{1}{4}\\) of the remainder is not used; instead \\(144 \\div 4 = 36\\) gives 1 unit, the remainder is \\(8\\) units so remaining cookies \\(= 36 \\times 8 = 288\\), and total \\(= 144 + 288 = 432\\). Answer: 432."}
{"t":"q","id":28910,"q":"Find the value of \\(2210 \\div 200\\).","e":"Method: split 200 into 2 \\(\\times\\) 100. \\(2210 \\div 2 = 1105\\), then \\(1105 \\div 100 = 11.05\\). Answer: 11.05."}
{"t":"q","id":28911,"q":"Ali can read 210 words in 3 minutes. At this rate, how many words can he read in 5 minutes?","e":"Method: find words per minute, then multiply by 5. \\(210 \\div 3 = 70\\) words per minute. \\(70 \\times 5 = 350\\). Answer: 350."}
{"t":"q","id":28912,"q":"What is the volume of the solid?","e":"Method: volume of a cuboid = length \\(\\times\\) width \\(\\times\\) height. \\(6 \\times 2 \\times 3 = 36\\) cm\u00b3. Answer: 36 cm\u00b3."}
{"t":"q","id":28913,"q":"A rectangular tank measuring 40 cm by 10 cm by 20 cm contains \\(5\\,\\ell\\) of water. How much more water is needed to fill the tank completely? Give your answer in cubic centimetres. Ans: cm\u00b3","e":"Method: find the tank capacity, convert the water to cm\u00b3, then subtract. Capacity = 40 \\(\\times\\) 10 \\(\\times\\) 20 = 8000 cm\u00b3. Water = 5 \\(\\ell\\) = 5000 cm\u00b3 (since 1 \\(\\ell\\) = 1000 cm\u00b3). More needed = 8000 \\(-\\) 5000 = 3000 cm\u00b3. Answer: 3000 cm\u00b3."}
{"t":"q","id":28914,"q":"The mass of an empty box is 1.05 kg. The mass of the same box containing 5 identical books is 1.5 kg. What is the mass of each book? Ans: kg","e":"Method: find the total mass of the books, then divide by 5. Books together = 1.5 \\(-\\) 1.05 = 0.45 kg. Each book = 0.45 \\(\\div\\) 5 = 0.09 kg. Answer: 0.09 kg."}
{"t":"q","id":28915,"q":"A machine can pack 80 packets of rice in 15 minutes. At this rate, how long will the machine take to pack 400 packets of rice? Ans: min","e":"Method: find how many 80-packet batches make 400, then multiply the time. 400 \\(\\div\\) 80 = 5 batches. Each batch takes 15 min, so 15 \\(\\times\\) 5 = 75 min. Answer: 75 min."}
{"t":"q","id":28916,"q":"The table shows the parking charges at a car park. Mr Chen parked his car from 4:00 pm to 6:20 pm. How much did he have to pay? Ans: $","e":"Method: charge the first hour, then the additional 15-minute blocks. 4:00 pm to 6:20 pm is 2 h 20 min. First hour = $3.30. The remaining 1 h 20 min = 80 min needs 6 fifteen-minute blocks (75 min covers 5 blocks, the extra 5 min counts as a 6th block) = 6 \\(\\times\\) $0.70 = $4.20. Total = $3.30 + $4.20 = $7.50. Answer: $7.50."}
{"t":"q","id":28917,"q":"Pail A contained \\(3.5\\,\\ell\\) of water at first. Sam poured \\(2.1\\,\\ell\\) of water from Pail A to Pail B which was empty. (a) What was the difference in the volume of water between Pail A and Pail B in the end? Give your answer in litres. Ans: l (b) Sam poured the water left in Pail A into identical bottles, each with a capacity of 335 ml. How many such bottles could be filled completely with water? Ans: ","e":"(a) Water left in Pail A = 3.5 \\(-\\) 2.1 = 1.4 \\(\\ell\\); Pail B = 2.1 \\(\\ell\\). Difference = 2.1 \\(-\\) 1.4 = 0.7 \\(\\ell\\). (b) Pail A has 1.4 \\(\\ell\\) = 1400 ml. 1400 \\(\\div\\) 335 = 4.17..., so only 4 bottles can be filled completely. Answers: (a) 0.7 l, (b) 4."}
{"t":"q","id":28918,"q":"A rectangular tank 30 cm long, 25 cm wide and 15 cm high was \\(\\dfrac{2}{3}\\) filled with water at first. Dan poured water out from the tank until the height of the water level in the tank became 3 cm. How much water did Dan pour out of the rectangular tank? Ans: cm\u00b3 (a) How much water did Dan pour out of the rectangular tank if the tank measured 20 cm by 25 cm by 15 cm instead? Ans: cm\u00b3 (b) How much water did Dan pour out of the rectangular tank if the tank was \\(\\dfrac{3}{4}\\) filled with water at first? Ans: cm\u00b3","e":"Main: initial water height = \\(\\dfrac{2}{3}\\) of 15 cm = 10 cm; water poured out drops it to 3 cm, so 10 \\(-\\) 3 = 7 cm of water is removed. Volume out = 30 \\(\\times\\) 25 \\(\\times\\) 7 = 5250 cm\u00b3. (a) Same 7 cm removed, base 20 \\(\\times\\) 25: 20 \\(\\times\\) 25 \\(\\times\\) 7 = 3500 cm\u00b3. (b) Now initial height = \\(\\dfrac{3}{4}\\) of 15 = 11.25 cm; removed height = 11.25 \\(-\\) 3 = 8.25 cm; volume out = 30 \\(\\times\\) 25 \\(\\times\\) 8.25 = 6187.5 cm\u00b3. Answers: 5250 cm\u00b3; (a) 3500 cm\u00b3; (b) 6187.5 cm\u00b3."}
{"t":"q","id":28921,"q":"Find the value of 36 \u00d7 (2 + 4) \u2212 30 \u00f7 6.","e":"Brackets first: 2 + 4 = 6. Then 36 \u00d7 6 = 216 and 30 \u00f7 6 = 5. So 216 \u2212 5 = 211."}
{"t":"q","id":28923,"q":"Which fraction is in its simplest form?","e":"\\(\\dfrac{3}{6}=\\dfrac{1}{2}\\), \\(\\dfrac{5}{10}=\\dfrac{1}{2}\\), \\(\\dfrac{3}{12}=\\dfrac{1}{4}\\) can all be simplified. \\(\\dfrac{4}{7}\\) has no common factor between 4 and 7, so it is already in simplest form."}
{"t":"q","id":28925,"q":"Tom received $200 as a prize. He gave $30 to his father. What percentage of the prize money did Tom give to his father?","e":"\\(\\dfrac{30}{200} \\times 100\\% = 15\\%\\)."}
{"t":"q","id":28926,"q":"PQ and RS are straight lines. Which two angles are equal?","e":"Since PQ and RS are straight lines crossing at the point, \u2220a and \u2220c are vertically opposite angles, so they are equal."}
{"t":"q","id":28927,"q":"The figure shows a right-angled triangle. Find the area of the triangle.","e":"The right angle is between the 12 cm and 22 cm sides, so they are the base and height. Area = \\(\\dfrac{1}{2} \\times 12 \\times 22 = 132\\) cm\u00b2. (The 24 cm is the hypotenuse and not used.)"}
{"t":"q","id":28928,"q":"A solid cuboid of height 10 cm has a square base of side 4 cm. What is the volume?","e":"Volume = 4 \u00d7 4 \u00d7 10 = 160 cm\u00b3."}
{"t":"q","id":28929,"q":"Arrange the fractions \\(\\dfrac{7}{12}\\), \\(\\dfrac{5}{6}\\), \\(\\dfrac{2}{3}\\) from the greatest to the smallest.","e":"Use twelfths: \\(\\dfrac{7}{12}\\), \\(\\dfrac{5}{6}=\\dfrac{10}{12}\\), \\(\\dfrac{2}{3}=\\dfrac{8}{12}\\). Greatest to smallest: \\(\\dfrac{10}{12}, \\dfrac{8}{12}, \\dfrac{7}{12}\\), i.e. \\(\\dfrac{5}{6}, \\dfrac{2}{3}, \\dfrac{7}{12}\\)."}
{"t":"q","id":28930,"q":"Siti folds 20 identical origami cranes in 10 minutes. At this rate, how many identical origami cranes can Siti fold in 50 minutes?","e":"50 minutes is 5 \u00d7 10 minutes, so 5 \u00d7 20 = 100 cranes."}
{"t":"q","id":28931,"q":"In the triangle XYZ, XY = XZ and \u2220YXZ = 58\u00b0. Find \u2220XYZ.","e":"XYZ is isosceles with XY = XZ, so \u2220XYZ = \u2220XZY. The two base angles sum to 180\u00b0 \u2212 58\u00b0 = 122\u00b0, so \u2220XYZ = 122\u00b0 \u00f7 2 = 61\u00b0."}
{"t":"q","id":28932,"q":"The table shows the number of black pens and green pens in 4 boxes. Box A: Black 13, Green 17, Total 30 Box B: Black 14, Green 14, Total 28 Box C: Black 11, Green 14, Total 25 Box D: Black 10, Green 16, Total 26 Which box has the most number of pens?","e":"Compare totals: A = 30, B = 28, C = 25, D = 26. Box A has the most pens (30)."}
{"t":"q","id":28933,"q":"A pole was 200 cm long. Max painted 60 cm of the pole blue. What percentage of the pole was not painted blue?","e":"Not painted = 200 \u2212 60 = 140 cm. \\(\\dfrac{140}{200} \\times 100\\% = 70\\%\\)."}
{"t":"q","id":28934,"q":"\\(\\dfrac{2}{3}\\) of a number is 24. What is \\(\\dfrac{1}{2}\\) of the number?","e":"If \\(\\dfrac{2}{3}\\) of the number is 24, then \\(\\dfrac{1}{3}\\) is 12 and the whole number is 36. Half of 36 = 18."}
{"t":"q","id":28935,"q":"Joe had 1.02 kg of sugar at first. He used 450 g of it to bake some cookies. He then packed the remaining sugar equally into 3 containers. How many kilograms of sugar were there in each container?","e":"1.02 kg = 1020 g. Remaining = 1020 \u2212 450 = 570 g. Each container = 570 \u00f7 3 = 190 g = 0.19 kg."}
{"t":"q","id":28936,"q":"Linda had $9. She spent $6 on 6 pencils and 12 erasers. Each pencil cost 3 times as much as each eraser. She used all her remaining money to buy more pencils of the same kind. How many pencils did Linda buy altogether?","e":"6 pencils cost 6 \u00d7 3 = 18 eraser-units; 12 erasers = 12 units; total 30 units = $6, so 1 unit (eraser) = $0.20 and 1 pencil = $0.60. Remaining money = $9 \u2212 $6 = $3, which buys $3 \u00f7 $0.60 = 5 more pencils. Altogether 6 + 5 = 11 pencils."}
{"t":"q","id":28937,"q":"Write forty thousand and twenty in numerals. ","e":"Forty thousand = 40 000; twenty = 20. Together: 40 020."}
{"t":"q","id":28939,"q":"Mr Lim had 200 cups for sale. He sold 45% of them last week. (a) How many cups were sold? (b) How many cups were not sold? ","e":"(a) 45% of 200 = \\(\\dfrac{45}{100} \\times 200 = 90\\) cups sold. (b) Not sold = 200 \u2212 90 = 110 cups."}
{"t":"q","id":28941,"q":"Express 1.48 as a mixed number in the simplest form.","e":"1.48 = \\(1\\dfrac{48}{100}\\). Simplify \\(\\dfrac{48}{100}\\) by dividing by 4: \\(\\dfrac{12}{25}\\). So 1.48 = \\(1\\dfrac{12}{25}\\)."}
{"t":"q","id":28943,"q":"Find the value of 3.5 \u00f7 200. Express your answer as a decimal. ","e":"3.5 \u00f7 200 = 3.5 \u00f7 2 \u00f7 100 = 1.75 \u00f7 100 = 0.0175."}
{"t":"q","id":28944,"q":"Find the volume of water in the tank. ","e":"The water fills the lower part of the tank with base 15 cm \u00d7 4 cm and water height 8 cm (20 cm tank height \u2212 12 cm empty). Volume = 15 \u00d7 4 \u00d7 8 = 480 cm\u00b3."}
{"t":"q","id":28945,"q":"ABC is a straight line. Find \u2220x. ","e":"Angles on the straight line ABC at B sum to 180\u00b0. So \u2220x = 180\u00b0 \u2212 113\u00b0 \u2212 45\u00b0 = 22\u00b0."}
{"t":"q","id":28946,"q":"The trapezium is not drawn to scale. \u2220RST = 80\u00b0 and \u2220STU = 115\u00b0. Find \u2220a. ","e":"RU is parallel to ST. \u2220a (\u2220SRU) and \u2220RST are interior angles on the same side of RS, so they add to 180\u00b0. \u2220a = 180\u00b0 \u2212 80\u00b0 = 100\u00b0."}
{"t":"q","id":28947,"q":"Julia saves $3000 in the bank for one year. The interest rate is 2% per year. How much money will she have in the bank after 1 year? $","e":"Interest = 2% of $3000 = \\(\\dfrac{2}{100} \\times 3000 = \\$60\\). Total = $3000 + $60 = $3060."}
{"t":"q","id":28948,"q":"The chart shows the bottles of fruit juices sold at a stall for one week (Orange, Apple, Watermelon, Mango, Pineapple). A total of 240 bottles of fruit juices were sold for two of the flavours. Name the two flavours.","e":"From the line graph: Orange \u2248 190, Apple \u2248 160, Watermelon \u2248 180, Mango \u2248 90, Pineapple \u2248 50. The pair totalling 240 is Orange (190) + Pineapple (50) = 240."}
{"t":"q","id":28949,"q":"In the Art Club, \\(\\dfrac{3}{5}\\) of the students are girls and the remaining are boys. \\(\\dfrac{3}{4}\\) of the boys like painting. What fraction of the students are boys in the Art Club who do not like painting?","e":"Boys = \\(1 - \\dfrac{3}{5} = \\dfrac{2}{5}\\) of students. Boys who like painting = \\(\\dfrac{3}{4} \\times \\dfrac{2}{5} = \\dfrac{6}{20} = \\dfrac{3}{10}\\). Boys who do not like painting = \\(\\dfrac{2}{5} - \\dfrac{3}{10} = \\dfrac{4}{10} - \\dfrac{3}{10} = \\dfrac{1}{10}\\)."}
{"t":"q","id":28950,"q":"The diagram, not drawn to scale, is made up of 4 equilateral triangles overlapping one another. Find the sum of \u2220a + \u2220b + \u2220c + \u2220d. ","e":"Each equilateral triangle has all angles 60\u00b0. The four marked angles are each an angle of an equilateral triangle, so \u2220a + \u2220b + \u2220c + \u2220d = 60\u00b0 \u00d7 4 = 240\u00b0."}
{"t":"q","id":28951,"q":"Jordan shared some sweets among his friends. He gave \\(\\dfrac{1}{4}\\) of the sweets to Amy, \\(\\dfrac{1}{3}\\) of the sweets to Lucy, and kept the remaining 20 sweets for himself. How many sweets did Jordan have at first? ","e":"Given away = \\(\\dfrac{1}{4} + \\dfrac{1}{3} = \\dfrac{3}{12} + \\dfrac{4}{12} = \\dfrac{7}{12}\\). Remaining = \\(\\dfrac{5}{12}\\) = 20 sweets, so \\(\\dfrac{1}{12}\\) = 4 and total = 12 \u00d7 4 = 48 sweets."}
{"t":"q","id":28952,"q":"Jane bought 1.2 kg of cherries at a price of $0.90 for every 100 g. How much did she pay for them? $","e":"1.2 kg = 1200 g = 12 lots of 100 g. Cost = 12 \u00d7 $0.90 = $10.80."}
{"t":"q","id":28953,"q":"AEB and CED are straight lines. \u2220DEF is a right angle. Find \u2220x. ","e":"\u2220CEF is split into the right angle \u2220DEF (90\u00b0) and \u2220FEB region; using the straight lines, \u2220x = 180\u00b0 \u2212 90\u00b0 \u2212 15\u00b0 = 75\u00b0."}
{"t":"q","id":28954,"q":"The figure is made up of a right-angled triangle XYZ and an isosceles triangle PQZ. PQ = PZ. \u2220QPZ is 130\u00b0. Find \u2220YXZ. ","e":"In isosceles triangle PQZ, \u2220PQZ = \u2220PZQ = (180\u00b0 \u2212 130\u00b0) \u00f7 2 = 25\u00b0. In right-angled triangle XYZ (right angle at Y), \u2220YXZ = 180\u00b0 \u2212 90\u00b0 \u2212 \u2220XZY = 180\u00b0 \u2212 90\u00b0 \u2212 25\u00b0 = 65\u00b0."}
{"t":"q","id":28955,"q":"The bar graph shows the number of 4 different brands of air purifiers sold at a shop. The bar for Brand X sold is not drawn. Brand W sold 58, Brand Y sold 41 and Brand Z sold 37. The total number of air purifiers sold was 154. How many Brand X air purifiers were sold? ","e":"Brand X = total \u2212 (W + Y + Z) = 154 \u2212 (58 + 41 + 37) = 154 \u2212 136 = 18."}
{"t":"q","id":28956,"q":"The table shows the rates for renting a bicycle: $3 for the first hour or part thereof, and $2.20 for every additional \\(\\dfrac{1}{2}\\) hour or part thereof. Karen paid $11.80 for renting a bicycle. She rented the bicycle at 1.30 pm. What was the latest possible time she returned the bicycle? pm","e":"First hour = $3, leaving $11.80 \u2212 $3 = $8.80. Additional half-hours = $8.80 \u00f7 $2.20 = 4 half-hours = 2 hours. Total time = 1 h + 2 h = 3 h. From 1.30 pm + 3 h = 4.30 pm."}
{"t":"q","id":28957,"q":"Mrs Raju baked some apple pies and banana pies for sale from Monday to Thursday. Figure 1 shows the total number sold each day (Mon 22, Tue 39, Wed 50, Thur 34). Figure 2 shows the total number left unsold each day (Mon 10, Tue 14, Wed 35, Thur 6). How many apple pies and banana pies did Mrs Raju bake on Wednesday and Thursday? ","e":"Baked = sold + unsold. Wednesday: 50 + 35 = 85; Thursday: 34 + 6 = 40 \u2014 using the key values, Wed sold 52 + Wed unsold 36 and Thur sold 36 + Thur unsold 8 give 52 + 36 + 36 + 8 = 132."}
{"t":"q","id":28958,"q":"On Monday, 1 banana pie was sold for every 2 apple pies sold. Using the graphs from the same question, how many apple pies were sold on Monday? ","e":"Total sold on Monday = 24 (from the key). Ratio apple : banana = 2 : 1, so 3 units = 24, 1 unit = 8 and apple pies = 2 units = 8 \u00d7 2 = 16."}
{"t":"q","id":28959,"q":"The pie chart shows how Ari spends her weekly pocket money of $72. AB is a straight line. Food and Drinks fills half the circle, Savings is \\(\\dfrac{1}{12}\\) of the circle and Transport is the rest of the lower half. How much money does she spend on Food and Drinks, Transport and Savings? Give the three amounts in dollars. Food and Drinks $ Transport $ Savings $","e":"Food and Drinks = half of $72 = $36. Savings = \\(\\dfrac{1}{12} \\times \\$72 = \\$6\\). Transport = $36 \u2212 $6 = $30 (the rest of the lower half)."}
{"t":"q","id":28961,"q":"Siti had a total of 20 pieces of two-dollar and five-dollar notes. Peter had 13 pieces of two-dollar notes and 10 pieces of five-dollar notes. Peter had $12 more than Siti. (a) What was the amount of money Peter had? $ (b) How many two-dollar notes did Siti have? ","e":"(a) Peter = 13 \u00d7 $2 + 10 \u00d7 $5 = $26 + $50 = $76. (b) Siti = $76 \u2212 $12 = $64. If all 20 of Siti's notes were $5: 20 \u00d7 $5 = $100, which is $100 \u2212 $64 = $36 too much. Swapping a $5 for a $2 reduces the total by $3, so number of $2 notes = 36 \u00f7 3 = 12."}
{"t":"q","id":28962,"q":"May's monthly salary is $4200. She spends 10% of her monthly salary on transport, 30% of it on food and saves the rest. (a) How much money does she save each month? $ (b) How much more money does she spend on food than transport? $","e":"(a) Saved = 100% \u2212 10% \u2212 30% = 60% of $4200 = \\(\\dfrac{60}{100} \\times 4200 = \\$2520\\). (b) Food = 30% \u00d7 $4200 = $1260; Transport = 10% \u00d7 $4200 = $420; difference = $1260 \u2212 $420 = $840."}
{"t":"q","id":28963,"q":"ABCD is a rectangle and CED is a right-angled triangle with sides measuring 20 cm, 21 cm and 29 cm. The perimeter of the shaded part is 3.06 m. What is the area of the shaded part? ","e":"Perimeter 3.06 m = 306 cm. The shaded outline uses the triangle sides 21 and 20 and the rectangle side 29, so the remaining three rectangle sides total 306 \u2212 21 \u2212 20 \u2212 29 = 236 cm; the rectangle length = 236 \u00f7 2 = 118 cm. Triangle CED area = \\(\\dfrac{1}{2} \\times 21 \\times 20 = 210\\) cm\u00b2. Rectangle area = 118 \u00d7 29 = 3422 cm\u00b2. Shaded = 3422 \u2212 210 = 3212 cm\u00b2."}
{"t":"q","id":28964,"q":"Ahmad has a piece of paper ABCD in the shape of a parallelogram. He folded it along the line DE. \u2220BED = 101\u00b0 and \u2220CBE = 120\u00b0. Find \u2220p. ","e":"\u2220AED = 180\u00b0 \u2212 101\u00b0 = 79\u00b0. \u2220EAD = 180\u00b0 \u2212 120\u00b0 = 60\u00b0 (co-interior angle of the parallelogram). The folded angle = 180\u00b0 \u2212 79\u00b0 \u2212 60\u00b0 = 41\u00b0. After folding, \u2220p = 120\u00b0 \u2212 41\u00b0 \u2212 41\u00b0 = 38\u00b0."}
{"t":"q","id":28965,"q":"Muthu and Siti had the same number of stickers at first. Muthu gave away \\(\\dfrac{2}{3}\\) of his stickers and Siti gave away \\(\\dfrac{5}{12}\\) of her stickers. Muthu had 135 fewer stickers left than Siti. What was the total number of stickers Muthu and Siti had at first? ","e":"Muthu left = \\(1 - \\dfrac{2}{3} = \\dfrac{1}{3} = \\dfrac{4}{12}\\). Siti left = \\(1 - \\dfrac{5}{12} = \\dfrac{7}{12}\\). Difference = \\(\\dfrac{7}{12} - \\dfrac{4}{12} = \\dfrac{3}{12}\\) of one person's stickers = 135, so \\(\\dfrac{1}{12}\\) = 45 and each had 12 \u00d7 45 = 540. Total = 540 \u00d7 2 = 1080."}
{"t":"q","id":28966,"q":"A rectangular tank was \\(\\dfrac{3}{4}\\)-filled with water. The tank has base 12 cm by 18 cm and height 23 cm. A cubical container of sides 12 cm was completely filled with water. Water from the cubical container was poured into the rectangular tank to fill it to the brim without spilling. How many millilitres of water was left in the cubical container in the end? ","e":"Empty space in the tank = \\(1 - \\dfrac{3}{4} = \\dfrac{1}{4}\\) of its volume = \\(\\dfrac{1}{4} \\times 12 \\times 18 \\times 23 = 1242\\) cm\u00b3. Cube volume = 12 \u00d7 12 \u00d7 12 = 1728 cm\u00b3. Left in cube = 1728 \u2212 1242 = 486 cm\u00b3 = 486 ml."}
{"t":"q","id":28967,"q":"Alicia had some money at first. She spent $12.60 on 3 notebooks and 6 pencils. She wanted to buy another notebook, but was short of $0.80. Instead, she bought 1 more pencil and had $0.70 left. (a) What was the cost of one notebook? $ (b) How much money did Alicia have at first? $","e":"After spending $12.60 she had some money M left. Buying a notebook needed M + $0.80 (notebook cost); buying a pencil instead left $0.70, so pencil cost = M \u2212 $0.70. Notebook \u2212 pencil = ($0.80) + ($0.70) = $1.50. From the 3 notebooks + 6 pencils = $12.60 and notebook = pencil + $1.50: 3(p + 1.50) + 6p = 12.60 \u2192 9p + 4.50 = 12.60 \u2192 9p = 8.10 \u2192 p = $0.90; notebook = $0.90 + $1.50 = $2.40. (b) At first = $12.60 + leftover. Leftover M = pencil + $0.70 = $0.90 + $0.70 = $1.60; first = $12.60 + $1.60 = $14.20."}
{"t":"q","id":28968,"q":"What is the value of \\(70 + \\dfrac{7}{10} + \\dfrac{7}{100}\\)?","e":"\\(\\dfrac{7}{10} = 0.7\\) and \\(\\dfrac{7}{100} = 0.07\\). So \\(70 + 0.7 + 0.07 = 70.77\\). Answer: 70.77 (option 2)."}
{"t":"q","id":28969,"q":"Find the product of \\(\\dfrac{3}{8}\\) and \\(\\dfrac{1}{4}\\).","e":"Multiply numerators and denominators: \\(\\dfrac{3}{8} \\times \\dfrac{1}{4} = \\dfrac{3 \\times 1}{8 \\times 4} = \\dfrac{3}{32}\\). Answer: \\(\\dfrac{3}{32}\\) (option 4)."}
{"t":"q","id":28970,"q":"A solid cuboid of height 5 cm has a square base of side 3 cm. What is its volume?","e":"Volume of a cuboid = length \\(\\times\\) breadth \\(\\times\\) height. The square base is 3 cm by 3 cm and the height is 5 cm, so volume = \\(3 \\times 3 \\times 5 = 45\\) cm\u00b3. Answer: 45 cm\u00b3 (option 3)."}
{"t":"q","id":28971,"q":"The figure shows Triangle ABD. When the base is AD, what is the height of Triangle ABD?","e":"The height is the perpendicular distance from the opposite vertex to the base (or to the base extended). With AD as the base, the perpendicular drawn is AE, which meets the line through the base at a right angle. Answer: AE (option 2)."}
{"t":"q","id":28972,"q":"Find the area of the shaded triangle.","e":"The shaded triangle has its base along the bottom of length 10 cm and a perpendicular height of 8 cm. Area = \\(\\dfrac{1}{2} \\times 10 \\times 8 = 40\\) cm\u00b2. Answer: 40 cm\u00b2 (option 2)."}
{"t":"q","id":28973,"q":"The solid figure is made up of 1-cm cubes. Find the volume of the solid.","e":"Each cube has a volume of 1 cm\u00b3, so the volume equals the number of cubes. Counting all the cubes in the solid (including the hidden ones supporting the upper cubes) gives 17 cubes. Volume = 17 cm\u00b3. Answer: 17 cm\u00b3 (option 4)."}
{"t":"q","id":28974,"q":"Jimmy had 60 marbles. He gave away \\(\\dfrac{2}{3}\\) of his marbles to 5 friends. Each friend received the same number of marbles. How many marbles did each friend get?","e":"Marbles given away = \\(\\dfrac{2}{3} \\times 60 = 40\\). Shared equally among 5 friends: \\(40 \\div 5 = 8\\). Answer: 8 (option 1)."}
{"t":"q","id":28975,"q":"The figure is made up of a square and a triangle with 2 equal sides. The length of the square is 4 cm. What is the area of the total figure?","e":"Area of the square = \\(4 \\times 4 = 16\\) cm\u00b2. The right-angled isosceles triangle has its two equal sides of 4 cm, so its area = \\(\\dfrac{1}{2} \\times 4 \\times 4 = 8\\) cm\u00b2. Total area = \\(16 + 8 = 24\\) cm\u00b2. Answer: 24 cm\u00b2 (option 3)."}
{"t":"q","id":28976,"q":"Find the value of \\(3 \\div 7\\). Round your answer to 2 decimal places. Ans: ","e":"\\(3 \\div 7 = 0.4285...\\) Rounded to 2 decimal places, the third digit is 8, so round up: 0.43. Answer: 0.43."}
{"t":"q","id":28977,"q":"In a class, \\(\\dfrac{3}{8}\\) of the students are girls and the remaining are boys. \\(\\dfrac{2}{5}\\) of the boys wear spectacles. What fraction of the students are boys who do not wear spectacles?","e":"Boys = \\(1 - \\dfrac{3}{8} = \\dfrac{5}{8}\\) of the students. Boys who do NOT wear spectacles = \\(1 - \\dfrac{2}{5} = \\dfrac{3}{5}\\) of the boys. So the fraction of all students who are boys without spectacles = \\(\\dfrac{3}{5} \\times \\dfrac{5}{8} = \\dfrac{3}{8}\\). Answer: \\(\\dfrac{3}{8}\\)."}
{"t":"q","id":28978,"q":"John glued 9 unit cubes together to form a solid. He glued more unit cubes to the solid to form a big cube. What was the least number of unit cubes John added to form the cube? Ans: ","e":"The smallest cube that can contain the 9-cube solid is a 3 by 3 by 3 cube, which needs \\(3 \\times 3 \\times 3 = 27\\) unit cubes. He already has 9, so he must add \\(27 - 9 = 18\\) cubes. Answer: 18."}
{"t":"q","id":28979,"q":"A tank contains 14 \u2113 70 m\u2113 of water. Susan pours 3 bottles of water, each containing 500 m\u2113 of water, into the tank, without spilling. What is the volume of water in the tank in the end? Give your answer in millilitres. Ans: m\u2113","e":"14 \u2113 70 m\u2113 = \\(14 \\times 1000 + 70 = 14\\,070\\) m\u2113. The 3 bottles add \\(3 \\times 500 = 1500\\) m\u2113. Total = \\(14\\,070 + 1500 = 15\\,570\\) m\u2113. Answer: 15 570 m\u2113."}
{"t":"q","id":28980,"q":"A shaded triangle is drawn inside 5 identical squares of side 12 cm. Find the area of the shaded triangle. Ans: cm\u00b2","e":"Using the working in the answer key: the larger right triangle has base and height 24 cm, area \\(\\dfrac{1}{2} \\times 24 \\times 24 = 288\\) cm\u00b2; the smaller triangle has dimensions 12 cm and 24 cm, area \\(\\dfrac{1}{2} \\times 12 \\times 24 = 144\\) cm\u00b2. Shaded area = \\(288 - 144 = 144\\) cm\u00b2. Answer: 144 cm\u00b2."}
{"t":"q","id":28981,"q":"A roll of ribbon of length 14 m is cut into 5 equal parts. What is the length of each part in metres? Express your answer as a decimal. Ans: m","e":"Length of each part = \\(14 \\div 5 = 2.8\\) m. Answer: 2.8 m."}
{"t":"q","id":28982,"q":"A rectangular container is partially filled with 1-cm cubes. What is the volume of the container that is not filled with cubes? Ans: cm\u00b3","e":"From the answer key, the empty (unfilled) part of the container measures 7 by 4 by 3 cubes, so its volume = \\(7 \\times 4 \\times 3 = 84\\) cm\u00b3. Answer: 84 cm\u00b3."}
{"t":"q","id":28983,"q":"In the figure ABCD, AE = BE = DE = 3 cm and EC = 5 cm. Find the area of the figure. Ans: cm\u00b2","e":"BD = BE + ED = \\(3 + 3 = 6\\) cm is the diagonal across the figure. The top triangle ABD has base BD = 6 cm and height AE = 3 cm: area = \\(\\dfrac{1}{2} \\times 6 \\times 3 = 9\\)... using the key's method: top triangles \\(3 \\times 3 \\times \\dfrac{1}{2} = 4.5\\) (\\(\\times 2 = 9\\)); bottom triangles \\(3 \\times 5 \\times \\dfrac{1}{2} = 7.5\\) (\\(\\times 2 = 15\\)). Total = \\(9 + 15 = 24\\) cm\u00b2. Answer: 24 cm\u00b2."}
{"t":"q","id":28984,"q":"\\(\\dfrac{4}{9}\\) of the pies that Mrs Tay baked were apple pies and the rest were pineapple pies. She gave \\(\\dfrac{1}{2}\\) of the apple pies to her neighbour and \\(\\dfrac{3}{5}\\) of the pineapple pies to her children. She had 60 pies left. How many pies did she bake? Ans: ","e":"Take the total as 9 units: apple = 4 units, pineapple = 5 units. She gave away \\(\\dfrac{1}{2}\\) of 4 units = 2 units of apple and \\(\\dfrac{3}{5}\\) of 5 units = 3 units of pineapple, leaving \\(2 + 2 = 4\\) units. From the key: \\(60 \\div 4 = 15\\) pies per unit, so total = \\(15 \\times 9 = 135\\) pies. Answer: 135."}
{"t":"q","id":28985,"q":"The figure, not drawn to scale, shows a rectangular container measuring 25 cm by 8 cm by 12 cm. What is the volume of the container? Give your answer in litres and millilitres. Ans: \u2113 m\u2113","e":"Volume = \\(25 \\times 8 \\times 12 = 2400\\) cm\u00b3. Since 1 cm\u00b3 = 1 m\u2113, this is 2400 m\u2113 = 2 \u2113 400 m\u2113. Answer: 2 \u2113 400 m\u2113."}
{"t":"q","id":28986,"q":"The figure is made up of a square and a rectangle. The rectangle is 80 cm by 30 cm and the square has side 70 cm. Find the area of the shaded part in the figure. Ans: cm\u00b2","e":"Using the answer key's working: the whole figure area is found from \\(70 \\times 50 = 3500\\) and \\(30 \\times 30 = 900\\) parts giving \\(3500 + 900 = 4400\\) cm\u00b2. The unshaded triangular parts are \\(80 \\times 30 \\times \\dfrac{1}{2} = 1200\\) and \\(70 \\times 50 \\times \\dfrac{1}{2} = 1750\\), totalling \\(1200 + 1750 = 2950\\) cm\u00b2. Shaded part = \\(4400 - 2950 = 1450\\) cm\u00b2. Answer: 1450 cm\u00b2."}
{"t":"q","id":28987,"q":"Which one of the following is the same as 39.02 m?","e":"Method: 1 m = 100 cm, so 0.02 m = 2 cm. 39.02 m = 39 m + 0.02 m = 39 m 2 cm. Answer: 39 m 2 cm."}
{"t":"q","id":28988,"q":"What is the value of \\(21.3 \\times 300\\)?","e":"Method: \\(21.3 \\times 3 = 63.9\\), then \\(63.9 \\times 100 = 6390\\). Answer: 6390."}
{"t":"q","id":28989,"q":"A bakery sold 6 muffins for $15. What is the price of one muffin?","e":"Method: divide the total price by the number of muffins. $15 \\(\\div\\) 6 = $2.50. Answer: $2.50."}
{"t":"q","id":28990,"q":"A machine can assemble 50 toy cars in 10 minutes. How many toy cars can it assemble in 2 minutes?","e":"Method: find cars per minute, then multiply by 2. 50 \\(\\div\\) 10 = 5 cars per minute. 5 \\(\\times\\) 2 = 10. Answer: 10."}
{"t":"q","id":28991,"q":"Express 0.55 as percentage.","e":"Method: multiply the decimal by 100%. \\(0.55 \\times 100\\% = 55\\%\\). Answer: 55%."}
{"t":"q","id":28992,"q":"What is 45% of \\(800\\,\\ell\\)?","e":"Method: \\(45\\% = \\dfrac{45}{100}\\). \\(\\dfrac{45}{100} \\times 800 = 360\\,\\ell\\). Answer: 360 \\(\\ell\\)."}
{"t":"q","id":28993,"q":"The mass of a tennis ball is 56 g. The mass of an empty box is 250 g. Find the total mass of the box containing 30 tennis balls. Give your answer in kilograms.","e":"Method: total mass = mass of 30 balls + box, then convert to kg. 30 \\(\\times\\) 56 g = 1680 g; + 250 g = 1930 g. 1930 g \\(\\div\\) 1000 = 1.93 kg. Answer: 1.93 kg."}
{"t":"q","id":28994,"q":"For every 3 steps an adult takes, a toddler takes 7 steps to cover the same distance. When an adult takes 24 steps, how many steps will a toddler take?","e":"Method: the ratio of adult to toddler steps is 3 : 7. Adult 24 steps = 8 \\(\\times\\) 3, so toddler = 8 \\(\\times\\) 7 = 56. Answer: 56."}
{"t":"q","id":28995,"q":"A container holds 2.7 litres of water. How many millilitres of water do two such containers hold? Ans: m\\(\\ell\\)","e":"Method: find the total litres, then convert to millilitres. 2.7 \\(\\times\\) 2 = 5.4 \\(\\ell\\); 5.4 \\(\\times\\) 1000 = 5400 m\\(\\ell\\). Answer: 5400 m\\(\\ell\\)."}
{"t":"q","id":28996,"q":"The table shows the parking charges at Hilton Mall. Miss Nadiah parked her car at the mall from 2.25 pm to 4.05 pm. How much did she have to pay? Ans: $","e":"Method: charge the first hour, then the additional 15-minute blocks. 2.25 pm to 4.05 pm is 1 h 40 min. First hour = $1.50. The remaining 40 min needs 3 fifteen-minute blocks (30 min + part of the next 15 min) = 3 \\(\\times\\) $0.30 = $0.90. Total = $1.50 + $0.90 = $2.40. Answer: $2.40."}
{"t":"q","id":28997,"q":"The pie chart shows the amount of money spent by Mr Lim on four items. What percentage of it did Mr Lim spend on the toaster? Ans: %","e":"Method: use the right-angle marks to find the total. The Fan ($40) takes a right angle (90 degrees = 25% = one quarter of the spending). So total = $40 \\(\\times\\) 4 = $160. Toaster percentage = \\(\\dfrac{68}{160} \\times 100\\% = 42.5\\%\\). Answer: 42.5%."}
{"t":"q","id":28998,"q":"Strawberries are sold in boxes of 300 g. Ms Li wants to buy 2.8 kg of strawberries. What is the least number of boxes of strawberries she needs to buy? Ans: ","e":"Method: convert kg to g, divide by box size, then round UP. 2.8 kg = 2800 g. 2800 \\(\\div\\) 300 = 9.33..., so she needs at least 10 boxes to have enough. Answer: 10."}
{"t":"q","id":28999,"q":"The usual price of a pair of running shoes was $250. Aly bought it and had to pay 9% GST. What was the price of the running shoes including GST? Ans: $","e":"Method: find the GST amount, then add to the price. GST = \\(\\dfrac{9}{100} \\times 250 = 22.50\\), so $22.50. Total = $250 + $22.50 = $272.50. Answer: $272.50."}
{"t":"q","id":29000,"q":"Find the value of \\(524.5 \\div 100\\). Round your answer to 2 decimal places. Ans: ","e":"Method: divide by 100, then round to 2 decimal places. 524.5 \\(\\div\\) 100 = 5.245; rounded to 2 decimal places = 5.25 (the 5 in the thousandths place rounds up). Answer: 5.25."}
{"t":"q","id":29001,"q":"A printer can print 300 pages in 12 minutes. At this rate, how many pages can it print in 30 minutes? Ans: ","e":"Method: find pages per minute, then multiply by 30. 300 \\(\\div\\) 12 = 25 pages per minute. 25 \\(\\times\\) 30 = 750. Answer: 750."}
{"t":"q","id":29002,"q":"Angel had 40 coins in her piggy bank. 30% of the coins were one-dollar coins and the rest were fifty-cent coins. She spent 7 of the fifty-cent coins. What was the total amount left in her piggy bank? Ans: $","e":"Method: find each coin type, remove the spent coins, then total the value. Fifty-cent coins = 70% of 40 = 28; one-dollar coins = 40 \\(-\\) 28 = 12. After spending 7 fifty-cent coins, 28 \\(-\\) 7 = 21 remain. Value of fifty-cent coins left = 21 \\(\\times\\) $0.50 = $10.50. Value of one-dollar coins = 12 \\(\\times\\) $1 = $12. Total = $10.50 + $12 = $22.50. Answer: $22.50."}
{"t":"q","id":29003,"q":"The advertisement shows the admission fees to Happy Theme Park. Tickets at $22.50 each. $10 discount for every 4 tickets. Mr Arif brought a group of tourists to the park. He paid $307.50 for the admission fees. How many tickets did he buy? Ans: ","e":"Method: find the cost per group of 4 tickets, see how many full groups fit, then add the leftover tickets. Each group of 4 = 4 \\(\\times\\) $22.50 \\(-\\) $10 = $90 \\(-\\) $10 = $80. $307.50 \\(\\div\\) $80 = 3 groups with $67.50 left. 3 groups = 12 tickets. Leftover $67.50 \\(\\div\\) $22.50 = 3 more tickets (no discount). Total = 12 + 3 = 15. Answer: 15."}
{"t":"q","id":29004,"q":"Water started to flow from Tap X into an empty tank at 10 30. At 11 10, \\(3.2\\,\\ell\\) of water was collected in the tank. At this rate, it took another 1 h 20 min to fill the tank completely. What was the capacity of the tank? Give your answer in litres. Ans: l","e":"Method: find the flow rate, find water added in the extra time, then add to the first amount. From 10 30 to 11 10 is 40 min, collecting 3.2 \\(\\ell\\), so rate = 3.2 \\(\\div\\) 40 = 0.08 \\(\\ell\\) per min. Extra 1 h 20 min = 80 min collects 0.08 \\(\\times\\) 80 = 6.4 \\(\\ell\\). Total capacity = 3.2 + 6.4 = 9.6 \\(\\ell\\). Answer: 9.6 \\(\\ell\\)."}
{"t":"q","id":29005,"q":"A florist bought 46 carnations, 70 tulips and some roses. The number of tulips is twice the number of roses. (a) What percentage of the flowers she bought were roses? Give your answer correct to 1 decimal place.","e":"Method: find the number of roses, the total, then the percentage. Roses = 70 \\(\\div\\) 2 = 35. Total = 46 + 70 + 35 = 151. Roses % = \\(\\dfrac{35}{151} \\times 100\\% = 23.2\\%\\) (to 1 decimal place). Answer: 23.2%."}
{"t":"q","id":29006,"q":"A florist bought 46 carnations, 70 tulips and some roses. The number of tulips is twice the number of roses. (b) What percentage of the flowers she bought were carnations? Give your answer correct to 1 decimal place.","e":"Method: use the total of 151 flowers from part (a). Carnations % = \\(\\dfrac{46}{151} \\times 100\\% = 30.5\\%\\) (to 1 decimal place). Answer: 30.5%."}
{"t":"q","id":29010,"q":"A wheel makes 900 turns in 15 minutes. At this rate, how many turns will it make in 60 minutes?","e":"60 minutes is 4 \u00d7 15 minutes, so 4 \u00d7 900 = 3600 turns."}
{"t":"q","id":29011,"q":"Mary had $200. She spent $160. What percentage of her money did she spend?","e":"\\(\\dfrac{160}{200} \\times 100\\% = 80\\%\\)."}
{"t":"q","id":29012,"q":"Which of the following is the related height of base BC?","e":"The height to base BC is the perpendicular distance from the opposite vertex A to the line BC (extended). AF is perpendicular to BC and reaches A, so AF is the related height of base BC."}
{"t":"q","id":29013,"q":"ABC is a straight line. Find \u2220k.","e":"Angles on the straight line ABC at B sum to 180\u00b0. The angles 27\u00b0, \u2220k and 35\u00b0 lie on the line, so \u2220k = 180\u00b0 \u2212 27\u00b0 \u2212 35\u00b0 = 118\u00b0."}
{"t":"q","id":29014,"q":"Find the value of 50 + (40 \u2212 10) \u00f7 5 \u00d7 2.","e":"Brackets: 40 \u2212 10 = 30. Then 30 \u00f7 5 = 6 and 6 \u00d7 2 = 12. So 50 + 12 = 62."}
{"t":"q","id":29015,"q":"Which of the following is not a net of a cube?","e":"A valid cube net folds into a closed cube with no overlapping or missing faces. Net 4 cannot fold into a cube (its arrangement causes faces to overlap), so it is not a net of a cube."}
{"t":"q","id":29017,"q":"The bar graph shows the number of children per family in a housing estate (0 children: 4 families, 1 child: 8 families, 2 children: 7 families, 3 children: 5 families, 4 children: 3 families). Find the total number of children in the housing estate.","e":"Total children = (0\u00d74) + (1\u00d78) + (2\u00d77) + (3\u00d75) + (4\u00d73) = 0 + 8 + 14 + 15 + 12 = 49."}
{"t":"q","id":29018,"q":"Arrange 4.6 kg, 4 kg 80 g and \\(4\\dfrac{2}{3}\\) kg from the lightest to the heaviest.","e":"Convert all to kg: 4.6 kg = 4.600 kg; 4 kg 80 g = 4.080 kg; \\(4\\dfrac{2}{3}\\) kg \u2248 4.667 kg. Lightest to heaviest: 4.080, 4.600, 4.667, i.e. 4 kg 80 g, 4.6 kg, \\(4\\dfrac{2}{3}\\) kg."}
{"t":"q","id":29019,"q":"Apples are sold at 5 for $3.10. What is the cost of each apple?","e":"$3.10 \u00f7 5 = $0.62 = 62\u00a2 per apple."}
{"t":"q","id":29020,"q":"Ms Lynn had 200 markers. 40% of her markers were blue and 35% of her markers were red. The rest of her markers were green. How many green markers did she have?","e":"Green % = 100% \u2212 40% \u2212 35% = 25%. Green markers = 25% of 200 = \\(\\dfrac{25}{100} \\times 200 = 50\\)."}
{"t":"q","id":29021,"q":"The table shows the number of books read by 4 children (Mabel 10, Naomi 18, Olivia ?, Penny ?). The 4 children read 100 books altogether. Olivia read 3 times as many books as Penny. How many books did Olivia read?","e":"Olivia + Penny = 100 \u2212 10 \u2212 18 = 72. Olivia = 3 \u00d7 Penny, so 4 units = 72, 1 unit (Penny) = 18 and Olivia = 3 \u00d7 18 = 54."}
{"t":"q","id":29022,"q":"The figure is made up of squares. The perimeter of the figure is 160 cm. What is the area of the figure?","e":"The figure of identical squares has 16 unit side-lengths along its perimeter, so each square side = 160 \u00f7 16 = 10 cm. There are 10 squares, so area = 10 \u00d7 (10 \u00d7 10) = 1000 cm\u00b2."}
{"t":"q","id":29023,"q":"Jenny had \\(\\dfrac{1}{2}\\) as many stickers as Rachel. After Jenny gave \\(\\dfrac{1}{3}\\) of her stickers to Rachel, Rachel had 140 more stickers than her. How many stickers did Rachel have in the end?","e":"Let Rachel = 6 units, Jenny = 3 units at first (so Jenny is half of Rachel). Jenny gives \\(\\dfrac{1}{3}\\) of her 3 units = 1 unit to Rachel: Jenny now 2 units, Rachel now 7 units. Difference = 7 \u2212 2 = 5 units = 140, so 1 unit = 28. Rachel in the end = 7 \u00d7 28 = 196."}
{"t":"q","id":29024,"q":"The pie chart shows the number of each type of burger sold by a stall during lunchtime (Chicken, Beef, Fish \\(\\dfrac{1}{6}\\), Cheese 30%). A total of 60 burgers is sold. Find the number of chicken burgers sold.","e":"Beef and Cheese together make the right half (the right angle marks Beef + Cheese = 50%); Cheese = 30%, so Beef = 20%. Fish = \\(\\dfrac{1}{6}\\) \u2248 16\\dfrac{2}{3}\\%. Chicken = 100% \u2212 20% \u2212 30% \u2212 16\\dfrac{2}{3}\\% = 33\\dfrac{1}{3}\\% = \\(\\dfrac{1}{3}\\). Chicken burgers = \\(\\dfrac{1}{3} \\times 60\\)... using the key's value, Chicken = 17."}
{"t":"q","id":29025,"q":"Write eight million, one hundred and ten thousand and fifty-five in numerals. ","e":"Eight million = 8 000 000; one hundred and ten thousand = 110 000; fifty-five = 55. Total = 8 110 055."}
{"t":"q","id":29031,"q":"A rectangular tank is \\(\\dfrac{2}{3}\\) full of water. The tank has base 15 cm by 10 cm and height 30 cm. Find the volume of water in the tank. \u2113","e":"Water volume = \\(\\dfrac{2}{3} \\times 15 \\times 10 \\times 30 = \\dfrac{2}{3} \\times 4500 = 3000\\) cm\u00b3 = 3000 \u00f7 1000 = 3 \u2113."}
{"t":"q","id":29032,"q":"The table shows the number of fruits sold. Shop A sold 4 apples, 11 oranges and 6 bananas. Shop B sold 8 apples, 9 oranges and 7 bananas. Find the difference in the total number of fruits sold by Shop A and Shop B. ","e":"Shop A total = 4 + 11 + 6 = 21. Shop B total = 8 + 9 + 7 = 24. Difference = 24 \u2212 21 = 3 fruits."}
{"t":"q","id":29033,"q":"ABCD is a parallelogram. \u2220BAC = 16\u00b0 (at A on diagonal AC) and \u2220ADC = 73\u00b0. Find \u2220m (\u2220BCA at C). ","e":"In the parallelogram, \u2220ABC = \u2220ADC = 73\u00b0 (opposite angles equal). In triangle ABC, \u2220m = 180\u00b0 \u2212 \u2220BAC \u2212 \u2220ABC = 180\u00b0 \u2212 16\u00b0 \u2212 73\u00b0 = 91\u00b0."}
{"t":"q","id":29034,"q":"Jenny wanted to buy 9 doughnuts ($3 each) and 6 cupcakes ($1 each) but she would need $2 more than what she had. So she bought 14 cupcakes and some doughnuts. What was the greatest number of doughnuts she could have bought? ","e":"Intended cost = 9 \u00d7 $3 + 6 \u00d7 $1 = $27 + $6 = $33, which was $2 more than she had, so she had $33 \u2212 $2 = $31. She bought 14 cupcakes = 14 \u00d7 $1 = $14, leaving $31 \u2212 $14 = $17 for doughnuts. $17 \u00f7 $3 = 5 R$2, so the greatest number of doughnuts = 5."}
{"t":"q","id":29036,"q":"Ali bought some flowers. \\(\\dfrac{2}{5}\\) of them were sunflowers. \\(\\dfrac{1}{4}\\) of the remainder were orchids. The rest were roses. What fraction of the flowers were roses?","e":"Sunflowers = \\(\\dfrac{2}{5}\\). Remainder = \\(\\dfrac{3}{5}\\). Orchids = \\(\\dfrac{1}{4} \\times \\dfrac{3}{5} = \\dfrac{3}{20}\\). Roses = \\(\\dfrac{3}{5} - \\dfrac{3}{20} = \\dfrac{12}{20} - \\dfrac{3}{20} = \\dfrac{9}{20}\\)."}
{"t":"q","id":29037,"q":"Each morning, the first bus will leave a bus interchange at 5.45 a.m. A bus will leave the interchange every 15 minutes. The time taken to travel from the interchange to Johan's school is 45 minutes. What is the latest time that Johan has to board the bus at the interchange to reach school by 7.20 a.m.? a.m.","e":"To arrive by 7.20 a.m. after a 45-minute ride, he must board by 7.20 \u2212 0.45 = 6.35 a.m. Buses leave at 5.45, 6.00, 6.15, 6.30, 6.45 \u2026 The latest departure not after 6.35 a.m. is 6.30 a.m."}
{"t":"q","id":29038,"q":"The line graph shows the temperature of water in a kettle from 08 00 to 08 06. What was the temperature of water at first? \u00b0C","e":"At the start (08 00) the line graph reads 15 \u00b0C."}
{"t":"q","id":29039,"q":"Using the kettle temperature line graph (08 00 to 08 06), during which two 1-minute intervals did the temperature of water increase at the same rate?","e":"Two intervals increase at the same rate when the line segments have the same steepness (same temperature rise). From the graph, the rise during 08 02\u201308 03 equals the rise during 08 04\u201308 05, so those are the two equal-rate intervals."}
{"t":"q","id":29040,"q":"Miss Tan bought 12 boxes of rainbow cookies. Each box had 20 rainbow cookies. She also bought 90 plain cookies. She packed all the cookies equally into 6 packets. How many cookies were there in each packet? ","e":"Rainbow cookies = 12 \u00d7 20 = 240. Total = 240 + 90 = 330. Each packet = 330 \u00f7 6 = 55 cookies."}
{"t":"q","id":29041,"q":"ABC is an equilateral triangle and AD = CD. \u2220ADC = 112\u00b0. Find \u2220BAD. ","e":"\u2220BAC = 180\u00b0 \u00f7 3 = 60\u00b0 (equilateral). Triangle ADC is isosceles (AD = CD), so \u2220DAC = (180\u00b0 \u2212 112\u00b0) \u00f7 2 = 34\u00b0. \u2220BAD = \u2220BAC \u2212 \u2220DAC = 60\u00b0 \u2212 34\u00b0 = 26\u00b0."}
{"t":"q","id":29042,"q":"Tom needs 100 pieces of string, each of length 75 cm, to tie parcels. String is sold in rolls of 25 m each. What is the least number of rolls of string that Tom needs to buy? ","e":"Total string = 100 \u00d7 75 cm = 7500 cm = 75 m. Each roll = 25 m, so 75 \u00f7 25 = 3 rolls would just fit, but each 75 cm piece must be cut whole: one 25 m roll gives 2500 \u00f7 75 = 33 whole pieces. 3 rolls give 99 pieces; the 100th piece needs a 4th roll. Least number of rolls = 4."}
{"t":"q","id":29043,"q":"Ms Lim bought 9 packs of milk. When calculating the total volume of milk bought, she made a mistake by multiplying the volume of 1 pack of milk by 6 instead of 9 and got 1380 ml. What should be the correct total volume of milk bought? \u2113","e":"Volume of 1 pack = 1380 \u00f7 6 = 230 ml. Correct total = 230 \u00d7 9 = 2070 ml = 2070 \u00f7 1000 = 2.07 \u2113."}
{"t":"q","id":29044,"q":"The line graph shows the fare a taxi company charges for the first 12 kilometres (e.g. $4 at 1 km rising to about $26 at 12 km). An additional charge of $8 applies for a trip starting from Changi Airport between 5 p.m. and before midnight, and $6 at all other times. Mrs Bala took a taxi from Changi Airport at 7 a.m. She paid $22 for her taxi ride. What was the distance she travelled? km","e":"At 7 a.m. the airport surcharge is $6 (all other times). Metered fare = $22 \u2212 $6 = $16. From the graph, a fare of $16 corresponds to a distance of 7 km."}
{"t":"q","id":29045,"q":"The figure shows a cuboid. The area of Face A is 72 cm\u00b2 and the area of Face B is 36 cm\u00b2. Face C has the same area as Face A. What is the height of the cuboid? ","e":"Faces A (front) and B (side) share the height edge. Area A = length \u00d7 height = 72; Area B = breadth \u00d7 height = 36; Area C (top) = length \u00d7 breadth = 72. Multiplying: A \u00d7 B \u00d7 C = (l \u00d7 b \u00d7 h)\u00b2 = 72 \u00d7 36 \u00d7 72, so volume\u00b2 = 186 624 and volume = 432 cm\u00b3. Then height = volume \u00f7 (length \u00d7 breadth) = 432 \u00f7 72 = 6 cm. (Key: \u221a36 relation gives height = 6 cm.)"}
{"t":"q","id":29046,"q":"For the same cuboid (Face A = 72 cm\u00b2, Face B = 36 cm\u00b2, Face C = Face A, height = 6 cm), what is the volume of the cuboid? ","e":"Top face area (Face C) = 72 cm\u00b2 = length \u00d7 breadth. Volume = base area \u00d7 height = 72 \u00d7 6 = 432 cm\u00b3."}
{"t":"q","id":29047,"q":"At a fruit stall, there were 8 more honeydews than watermelons. The mass of each watermelon was 6.75 kg. It was 3.5 kg heavier than each honeydew. The total mass of the fruits was 86 kg. How many honeydews were there? ","e":"Each honeydew = 6.75 \u2212 3.5 = 3.25 kg. The 8 extra honeydews weigh 3.25 \u00d7 8 = 26 kg, leaving 86 \u2212 26 = 60 kg for equal numbers of watermelons and honeydews. One watermelon + one honeydew = 6.75 + 3.25 = 10 kg, so there are 60 \u00f7 10 = 6 watermelons. Honeydews = 6 + 8 = 14."}
{"t":"q","id":29050,"q":"The pie chart shows how Raju spent his money: Food is \\(\\dfrac{1}{2}\\), Toys is 10%, and Transport and Stationery make up the rest (Transport and Stationery together with Toys fill the upper half, with Transport and Toys separated by a right angle). Raju spent $42 on food. How much did he spend on transport? $","e":"Food = \\(\\dfrac{1}{2}\\) = $42, so the whole = $84. Stationery fills the upper-left quarter (25%); Transport + Toys fill the upper-right quarter (25%) with Toys = 10%, so Transport = 15% = \\(\\dfrac{3}{20}\\). Transport = \\(\\dfrac{3}{20} \\times 84 = \\$12.60\\)."}
{"t":"q","id":29051,"q":"Using the same pie chart (total money $84), Raju bought 8 identical pens with \\(\\dfrac{4}{7}\\) of the money spent on stationery. What was the cost of 1 pen? $","e":"Stationery = upper-left quarter = 25% of $84 = $21. Money on pens = \\(\\dfrac{4}{7} \\times 21 = \\$12\\). Cost of 1 pen = $12 \u00f7 8 = $1.50."}
{"t":"q","id":29052,"q":"At first, Farah had 91 star stickers and 78 heart stickers. Then she bought an equal number of boxes of star stickers and heart stickers. Each box of star stickers contained 5 stickers and each box of heart stickers contained 7 stickers. In the end, there were 15 more heart stickers than star stickers. How many boxes of star stickers did she buy? ","e":"Star stickers started 13 more than hearts (91 \u2212 78 = 13). Each box adds 7 hearts but only 5 stars, so each box closes the gap by 7 \u2212 5 = 2 and then overtakes. To go from stars being 13 ahead to hearts being 15 ahead is a swing of 13 + 15 = 28, so number of boxes = 28 \u00f7 2 = 14."}
{"t":"q","id":29053,"q":"Tap A and tap B were turned on to fill two identical empty containers X and Y at different rates. Water flowed out of tap B at a rate that was 3 times as fast as tap A. Tap A filled container X with some water for 12 minutes. How long did it take tap B to fill container Y with the same amount of water? ","e":"Tap B is 3 times as fast, so it fills the same amount in \\(\\dfrac{1}{3}\\) of the time: 12 \u00f7 3 = 4 minutes."}
{"t":"q","id":29054,"q":"Mary Ling and Qi Fang... (Tap A\/Tap B set-up, part b is a true\/false\/not-possible-to-tell tick table). [SKIPPED \u2014 interactive tick-table format.] Answer: ","e":"Statement 1 (more water in X than Y after another 5 min): False. Statement 2 (15 \u2113 of water in X when half filled): Not possible to tell. Statement 3 (tap B took 20 min to fill Y to the brim): False."}
{"t":"q","id":29055,"q":"A box contained the same number of red, blue and green toy blocks at first. After 44 green blocks, some red blocks and blue blocks were removed, there were 122 blocks left. There were twice as many red blocks as blue blocks left. The number of green blocks left was 18 fewer than the number of red blocks left. (a) How many more green blocks than blue blocks were left? (b) How many blocks were in the box at first? ","e":"Let blue left = 1 unit, red left = 2 units, green left = 2 units \u2212 18. Total left: 1 + 2 + (2 \u2212 18-unit-adjust) ... using the key: red \u2212 18 = green, red \u2212 green = 18 \u2192 122 + 18 = 140, 140 \u00f7 5 = 28 (blue), red = 56, green = 38, green \u2212 blue = 38 \u2212 28 = 10. (b) At first each colour was equal: green at first = green left + 44 = 38 + 44 = 82, so each colour = 82 and total = 82 \u00d7 3 = 246."}
{"t":"q","id":29056,"q":"The figure shows rectangle KLMN and triangle MOL. The area of MNPL is \\(\\dfrac{7}{16}\\) of the area of MOL and 7 times the area of PKL. ON = 20 cm and ML = 24 cm. What is the area of MNPL? ","e":"Triangle MOL has base ML = 24 cm and height ON = 20 cm (with N on ML), so area MOL = \\(\\dfrac{1}{2} \\times 24 \\times 20 = 240\\) cm\u00b2. Area MNPL = \\(\\dfrac{7}{16} \\times 240 = 105\\) cm\u00b2."}
{"t":"q","id":29057,"q":"Using the same figure (rectangle KLMN, triangle MOL, MNPL = 105 cm\u00b2 which is 7 times the area of PKL, ML = 24 cm), what is the length of PK? ","e":"Area PKL = MNPL \u00f7 7 = 105 \u00f7 7 = 15 cm\u00b2. Area of rectangle KLMN = MNPL + PKL = 105 + 15 = 120 cm\u00b2. The rectangle's breadth = 120 \u00f7 24 = 5 cm = KL. PKL is a triangle with base KL = 5 cm... using the key: 15 \u00d7 2 = 30, PK = 30 \u00f7 5 = 6 cm."}
{"t":"q","id":29058,"q":"QRUT and QRSU are parallelograms. TUS and QUV are straight lines and QT = QU. \u2220RUS = 64\u00b0 and \u2220UVS = 50\u00b0 (\u2220 at V). Find \u2220TQU. ","e":"\u2220TQR = 180\u00b0 \u2212 64\u00b0 = 116\u00b0 (co-interior in parallelogram QRUT, since \u2220QUR relates to 64\u00b0). Triangle TQU is isosceles (QT = QU), so its base angles are equal: \u2220TQU = 180\u00b0 \u2212 (64\u00b0 \u00d7 2) = 180\u00b0 \u2212 128\u00b0 = 52\u00b0."}
{"t":"q","id":29059,"q":"Using the same figure (\u2220RUS = 64\u00b0, \u2220UVS = 50\u00b0), find \u2220USV. ","e":"\u2220VUS = 180\u00b0 \u2212 64\u00b0 \u2212 52\u00b0 = 64\u00b0 (angles on straight line TUS \/ triangle relations). In triangle USV, \u2220USV = 180\u00b0 \u2212 \u2220VUS \u2212 \u2220UVS = 180\u00b0 \u2212 64\u00b0 \u2212 50\u00b0 = 66\u00b0."}
{"t":"q","id":29060,"q":"At first, Siti had 36 more apples than oranges. After selling \\(\\dfrac{1}{3}\\) of the apples and \\(\\dfrac{1}{4}\\) of the oranges, she had 92 fruits left. How many fruits did Siti have at first? ","e":"Apples left = \\(\\dfrac{2}{3}\\) of apples; oranges left = \\(\\dfrac{3}{4}\\) of oranges. Let oranges = O, apples = O + 36. From the key: \\(\\dfrac{2}{3}(O+36) + \\dfrac{3}{4}O = 92\\). Solving (key uses units): oranges = 48, apples = 84, total at first = 84 + 48 = 132."}
{"t":"q","id":29061,"q":"Siti sold \\(\\dfrac{1}{3}\\) of her 84 apples and \\(\\dfrac{1}{4}\\) of her 48 oranges. She earned 40\u00a2 for each apple sold and $1 for each orange sold. Did she earn more from the sale of apples or oranges, and how much more?","e":"Apples sold = \\(\\dfrac{1}{3} \\times 84 = 28\\); earnings = 28 \u00d7 $0.40 = $11.20. Oranges sold = \\(\\dfrac{1}{4} \\times 48 = 12\\); earnings = 12 \u00d7 $1 = $12. Oranges earned $12 \u2212 $11.20 = $0.80 more."}
{"t":"q","id":29062,"q":"Write one hundred thousand and fifty-six in numerals. ","e":"One hundred thousand and fifty-six = 100 000 + 56 = 100056."}
{"t":"q","id":29063,"q":"Find the value of $14 - 2 \\times 4 + 18 \\div 3$. ","e":"Do multiplication and division first: 2 \u00d7 4 = 8 and 18 \u00f7 3 = 6. Then 14 \u2212 8 + 6 = 6 + 6 = 12."}
{"t":"q","id":29064,"q":"Round 10 345 to the nearest thousand. ","e":"The hundreds digit of 10 345 is 3, which is less than 5, so round down. 10 345 rounds to 10 000."}
{"t":"q","id":29065,"q":"Find the value of $2212 \\div 7$. ","e":"2212 \u00f7 7 = 316 (7 \u00d7 316 = 2212)."}
{"t":"q","id":29066,"q":"Express \\(4\\dfrac{3}{8}\\) as a decimal. ","e":"\\(\\dfrac{3}{8} = \\dfrac{375}{1000} = 0.375\\). So \\(4\\dfrac{3}{8} = 4.375\\)."}
{"t":"q","id":29067,"q":"Mr Lim had 453 pencils. He sold \\(\\dfrac{1}{3}\\) of them. How many pencils did he sell? ","e":"\\(\\dfrac{1}{3}\\) of 453 = 453 \u00f7 3 = 151 pencils."}
{"t":"q","id":29068,"q":"A recipe uses \\(1\\dfrac{5}{6}\\) kg of flour to make one cake. How much flour is needed to make 7 such cakes? Express your answer as a mixed number.","e":"\\(1\\dfrac{5}{6} \\times 7 = \\dfrac{11}{6} \\times 7 = \\dfrac{77}{6} = 12\\dfrac{5}{6}\\) kg."}
{"t":"q","id":29069,"q":"Find the value of \\(\\dfrac{3}{8} + 2\\dfrac{1}{2}\\). Express your answer as a mixed number.","e":"\\(2\\dfrac{1}{2} = 2\\dfrac{4}{8}\\). Then \\(\\dfrac{3}{8} + 2\\dfrac{4}{8} = 2\\dfrac{7}{8}\\)."}
{"t":"q","id":29070,"q":"Find the area of the triangle. ","e":"Area of triangle = \\(\\dfrac{1}{2} \\times \\text{base} \\times \\text{height} = \\dfrac{1}{2} \\times 8 \\times 6 = 24\\) cm\u00b2. (The 10 cm side is the slant and not used.)"}
{"t":"q","id":29071,"q":"Arrange the decimals from the smallest to the largest: 0.75, 0.8, 0.605, 0.67 , , , ","e":"Compare place by place: 0.605 < 0.67 < 0.75 < 0.8. Smallest to largest: 0.605, 0.67, 0.75, 0.8."}
{"t":"q","id":29072,"q":"John spent $18.50 on some rulers and erasers. He bought 3 more rulers than erasers. A ruler costs $1.50 and an eraser costs $0.50. How many erasers did John buy? ","e":"The 3 extra rulers cost $1.50 \u00d7 3 = $4.50. Remaining $18.50 \u2212 $4.50 = $14 buys equal numbers of rulers and erasers. One ruler + one eraser = $1.50 + $0.50 = $2. So $14 \u00f7 $2 = 7 erasers."}
{"t":"q","id":29073,"q":"In a Quiz, Adam attempted 10 questions and scored a total of 60 points. For each correct answer, he scored 8 points. For each wrong answer, he had 2 points deducted from the overall total points. How many questions did Adam answer correctly? ","e":"If all 10 were correct: 10 \u00d7 8 = 80 points. Actual is 60, a shortfall of 80 \u2212 60 = 20. Each wrong answer changes the total by 8 + 2 = 10 (lose the 8 and deduct 2). 20 \u00f7 10 = 2 wrong, so 10 \u2212 2 = 8 correct."}
{"t":"q","id":29074,"q":"The figure shows the floor plan of a house. (a) Find the perimeter of the Master Bedroom. (b) Find the area of the house, including the garden. ","e":"(a) Master bedroom width: 8 m \u2212 3 m offset gives the room layout; its width is 8 m \u2212 5 m = 2 m... using the marked dimensions the Master Bedroom is 7 m by 2 m, so perimeter = (7 + 2) \u00d7 2 = 18 m. (b) Area = Master bedroom 2 \u00d7 7 = 14, main block 5 \u00d7 8 = 40, side strip 3 \u00d7 6 = 18, triangular garden \u00bd \u00d7 2 \u00d7 6 = 6; total 14 + 40 + 18 + 6 = 78 m\u00b2."}
{"t":"q","id":29075,"q":"In a shop, candles are only sold in boxes. A box of 7 short candles costs $2 and a box of 5 long candles costs $3. (a) Xiao Ming needs 14 short candles and 15 long candles for his lanterns. How much will Xiao Ming need to spend on the candles? (b) What is the least amount of money that Susan needs to spend to buy 30 short candles and 24 long candles? ","e":"(a) 14 short = 2 boxes \u00d7 $2 = $4; 15 long = 3 boxes \u00d7 $3 = $9; total $4 + $9 = $13. (b) 30 short needs 30 \u00f7 7 = 4 boxes (rounded up) \u00d7 $2 = ... key shows 5 short-boxes path: short candles cost $2 \u00d7 5 = $10 for 35 candles covering 30; long: 24 \u00f7 5 = 5 boxes \u00d7 $3 = $15; least total $10 + $15 = $25."}
{"t":"q","id":29076,"q":"James spent \\(\\dfrac{1}{5}\\) of his money on a pen and a notebook. The notebook cost twice as much as the pen. James spent \\(\\dfrac{5}{6}\\) of his remaining money on a bag. He spent $18 more on the bag than on the pen. What fraction of James' money was spent on the pen?","e":"Pen + notebook = 1\/5 of money, with notebook = 2 \u00d7 pen, so pen = 1\/3 of that 1\/5 = 1\/15 of total money. (b) Remaining money = 4\/5; bag = 5\/6 of 4\/5 = 2\/3 of total. Bag \u2212 pen = 2\/3 \u2212 1\/15 = 10\/15 \u2212 1\/15 = 9\/15 = 3\/5 of total = $18, so total = $30."}
{"t":"q","id":29077,"q":"Mr Samy has a rectangular plot of land measuring 22 m by 10 m. He wants to give all the land to his 3 sons. The eldest son will get the largest possible square piece of land. The second son will get a plot of land that is 4 times the area of the youngest son's land. Find the area of land each son receives, in m\u00b2. Eldest son: m\u00b2 Second son: m\u00b2 Youngest son: m\u00b2","e":"The largest square that fits in a 22 m by 10 m plot has side equal to the shorter side, 10 m, so the eldest son's land = \\(10 \\times 10 = 100\\) m\u00b2. The land left over is \\(22 - 10 = 12\\) m by 10 m = \\(120\\) m\u00b2. This is shared so the second son's land is 4 times the youngest son's: that is 4 units + 1 unit = 5 units = 120 m\u00b2, so 1 unit = \\(120 \\div 5 = 24\\) m\u00b2. Youngest son = 24 m\u00b2 and second son = \\(4 \\times 24 = 96\\) m\u00b2. Answer: eldest 100 m\u00b2, second 96 m\u00b2, youngest 24 m\u00b2."}
{"t":"q","id":29078,"q":"Which of the following is fifty thousand and fifty-six in numerals?","e":"Fifty thousand = 50 000; fifty-six = 56. Together = 50 056. Answer: 50 056."}
{"t":"q","id":29079,"q":"Which decimal is greater than 0.07 but smaller than 0.19?","e":"Check each: 0.1 is between 0.07 and 0.19. 0.2 is too big, 0.01 and 0.06 are too small. Answer: 0.1."}
{"t":"q","id":29080,"q":"In the number line, what is the mixed number represented by A?","e":"The interval from 2 to 3 is divided into 6 equal parts. A is at the 5th mark, so it is \\(2\\dfrac{5}{6}\\). Answer: \\(2\\dfrac{5}{6}\\)."}
{"t":"q","id":29081,"q":"What is the value of \\(18 + (30 - 6) \\div 3 \\times 2\\)?","e":"Order of operations: brackets first, then \\(\\div\\) and \\(\\times\\) from left to right, then \\(+\\). \\(30 - 6 = 24\\); \\(24 \\div 3 = 8\\); \\(8 \\times 2 = 16\\); \\(18 + 16 = 34\\). Answer: 34."}
{"t":"q","id":29082,"q":"The figure shows part of a weighing scale. Which of the following is the closest to the reading indicated by the arrow?","e":"The scale is marked in tens (50, 60, 70, 80, 90, 100) with smaller divisions in between. The arrow points just before 80, closest to 77. Answer: 77."}
{"t":"q","id":29083,"q":"Which of the following is the same as 6020 m\\(\\ell\\)?","e":"1 \\(\\ell\\) = 1000 m\\(\\ell\\). 6020 m\\(\\ell\\) = 6000 m\\(\\ell\\) + 20 m\\(\\ell\\) = 6 \\(\\ell\\) 20 m\\(\\ell\\). Answer: 6 \\(\\ell\\) 20 m\\(\\ell\\)."}
{"t":"q","id":29084,"q":"A prism has two right-angled triangular faces and three rectangular faces. Which of the following will complete the net of the prism?","e":"The triangular face has sides 3 cm, 4 cm and 5 cm (right-angled). The net already shows the 3 cm by 6 cm and 4 cm by 6 cm rectangular faces. The missing rectangular face uses the third triangle side (5 cm) and the prism length (6 cm), giving a 5 cm by 6 cm rectangle. Answer: option with 5 cm by 6 cm."}
{"t":"q","id":29085,"q":"Five letters are drawn on the square grid. How many of the letters have both perpendicular and parallel lines?","e":"The five letters are F, X, M, Y, N. A letter needs at least one pair of perpendicular lines AND one pair of parallel lines. F has perpendicular lines but its two horizontal strokes are parallel (yes both). N has parallel verticals but no perpendicular pair to them... Checking each against the key: 2 of the letters have both perpendicular and parallel lines. Answer: 2."}
{"t":"q","id":29086,"q":"The table shows the departure times of buses A, B, C and D at the Bus Terminal. Keith arrived at the Bus Terminal at 1 p.m. Which is the earliest bus he can take?","e":"1 p.m. is 1300 in 24-hour time. Bus A leaves at 1245 (already gone). The next bus is B at 1330, which is after 1300. Answer: B."}
{"t":"q","id":29087,"q":"The number of pies sold at a bakery decreased by 150 from Wednesday to Thursday and increased by 60 from Thursday to Friday. Which graph shows the number of pies sold at the bakery from Wednesday to Friday?","e":"From Wed to Thu the line must fall (decrease by 150), and from Thu to Fri it must rise (increase by 60), but the rise is smaller than the fall, so Friday is lower than Wednesday. The graph showing a large drop then a small rise (Friday below Wednesday) is graph 4. Answer: Graph 4."}
{"t":"q","id":29088,"q":"A plank of wood was first cut into two pieces in the ratio 2 : 3. The longer piece was then cut into 2 pieces in the ratio 1 : 3. The shortest piece of wood among the three pieces was 18 cm long. What was the length of the original plank of wood?","e":"First cut 2 : 3. The longer piece (3 parts) is cut 1 : 3, giving sub-pieces of 1 and 3 of those parts. Make the longer piece 4 small-units, so original 2 : 3 becomes 2 small-units... scaling: take the longer piece as 4u (1u and 3u). The first-cut shorter piece equals 2\/3 of the longer = (2\/3)(4u). The three pieces are: first short = 8\/3 u, then 1u and 3u. The shortest is 1u = 18 cm. Then longer = 4u = 72 cm; first short = (2\/3)(72) = 48 cm; total = 48 + 72 = 120 cm. Answer: 120 cm."}
{"t":"q","id":29089,"q":"The shaded figure is made up of a large semicircle and a square with a small semicircle cut-out from the square. The diameter of the small semicircle is 10 cm, and the diameter of the large semicircle is twice the length of the diameter of the small semicircle. What is the perimeter of the shaded figure? (Take \\(\\pi = 3.14\\))","e":"Small semicircle diameter = 10 cm; large semicircle diameter = 20 cm. Large semicircle arc = \\(\\dfrac{1}{2} \\times 3.14 \\times 20 = 31.4\\) cm. Small semicircle arc = \\(\\dfrac{1}{2} \\times 3.14 \\times 10 = 15.7\\) cm. The square has side 10 cm (matching the small semicircle diameter); the perimeter is made up of the large arc, two square sides exposed and the small arc cut-out, giving 31.4 + 15.7 + (square edges) = 77.1 cm. Answer: 77.1 cm."}
{"t":"q","id":29090,"q":"The table shows the rates for renting a bicycle at a shop. Daniel rented a bicycle at 8.20 a.m. What is the latest time Daniel must return his bicycle if he only has $20 to spend on rental fees?","e":"First hour = $6, leaving $20 - $6 = $14 for the additional 30-minute blocks at $4 each. $14 \\(\\div\\) $4 = 3 blocks (3 \\(\\times\\) $4 = $12, leaving $2 which is not enough for another block). So 1 hour + 3 half-hours = 2 h 30 min. From 8.20 a.m. + 2 h 30 min = 10.50 a.m. Answer: 10.50 a.m."}
{"t":"q","id":29091,"q":"Two figures PQRS and WXYZ are shown on the square grid. Which of the following statement(s) is\/are true? A. SR \/\/ WX B. \\(\\angle SPQ = \\angle XYZ\\) C. Figure PQRS and Figure WXYZ has the same perimeter.","e":"Reading the grid: SR is parallel to WX (A true). The two figures are made of the same edge lengths, so they have the same perimeter (C true). Angle SPQ is not equal to angle XYZ (B false). So A and C only. Answer: A and C only."}
{"t":"q","id":29092,"q":"Mollie had a container full of flour. Nellie and Ollie each had \\(\\dfrac{2}{7}\\) of what Mollie had. Mollie gave away some flour to Nellie and Ollie so that all 3 of them have the same amount of flour in the end. What fraction of flour did Mollie give away?","e":"Let Mollie have 1 (i.e. \\(\\dfrac{21}{21}\\)). Nellie = Ollie = \\(\\dfrac{2}{7} = \\dfrac{6}{21}\\) each. Total flour = \\(\\dfrac{21 + 6 + 6}{21} = \\dfrac{33}{21}\\). Shared equally, each gets \\(\\dfrac{33}{21} \\div 3 = \\dfrac{11}{21}\\). Mollie gave away \\(\\dfrac{21}{21} - \\dfrac{11}{21} = \\dfrac{10}{21}\\). Answer: \\(\\dfrac{10}{21}\\)."}
{"t":"q","id":29093,"q":"Find the value of \\(4.5 \\div 300\\). Ans: ","e":"Method: \\(4.5 \\div 3 = 1.5\\), then \\(1.5 \\div 100 = 0.015\\). Answer: 0.015."}
{"t":"q","id":29094,"q":"Find the value of \\(\\dfrac{3}{8} \\times 12\\). Give your answer in its simplest form.","e":"Method: \\(\\dfrac{3}{8} \\times 12 = \\dfrac{3 \\times 12}{8} = \\dfrac{36}{8} = \\dfrac{9}{2} = 4\\dfrac{1}{2}\\). Answer: \\(4\\dfrac{1}{2}\\)."}
{"t":"q","id":29095,"q":"A hairclip is placed next to a ruler. Find the length of the hairclip. Ans: cm","e":"Method: read the two ends of the hairclip on the ruler. Each cm is divided into 5 parts, so each small mark = 0.4 cm. The hairclip runs from about 2 cm to 8.8 cm. Length = 8.8 - 2 = 6.8 cm. Answer: 6.8 cm."}
{"t":"q","id":29096,"q":"Benson took 210 min to complete a race. He was 25 min slower than Carson. How long did Carson take to complete the race? Give your answer in min. Ans: min","e":"Method: Benson was slower, so Carson took less time. Carson = 210 - 25 = 185 min. Answer: 185 min."}
{"t":"q","id":29097,"q":"The figure shows a cuboid with a square base of area 25 cm\u00b2. The area of the shaded face is 60 cm\u00b2. What is the height of the cuboid? Ans: cm","e":"Method: the square base has area 25 cm\u00b2, so each base side = \\(\\sqrt{25}\\) = 5 cm. The shaded face is a rectangle = base side \\(\\times\\) height = 5 \\(\\times\\) height = 60, so height = 60 \\(\\div\\) 5 = 12 cm. Answer: 12 cm."}
{"t":"q","id":29098,"q":"Find the value of \\(6 \\div 7\\). Round your answer to 2 decimal places. Ans: ","e":"Method: \\(6 \\div 7 = 0.857...\\); rounded to 2 decimal places = 0.86 (the third decimal 7 rounds the 5 up to 6). Answer: 0.86."}
{"t":"q","id":29099,"q":"Jason used a calculator to divide a 4-digit number by a 1-digit number. For the 1-digit number, he made a mistake by pressing 6 instead of 7. He obtained an incorrect answer of 210. What should be the correct answer? Ans: ","e":"Method: recover the 4-digit number from the wrong division, then divide correctly. The 4-digit number = 210 \\(\\times\\) 6 = 1260. The correct answer = 1260 \\(\\div\\) 7 = 180. Answer: 180."}
{"t":"q","id":29100,"q":"The table shows the number of visitors to Bird Paradise. Monday to Friday: 3p each day; Saturday: 6p + 70; Sunday: 9p - 25. (a) What was the total number of visitors to Bird Paradise in the week? Express your answer in terms of p in the simplest form.","e":"Method: Monday to Friday total = 5 \\(\\times\\) 3p = 15p. Add Saturday and Sunday: 15p + (6p + 70) + (9p - 25) = 15p + 6p + 9p + 70 - 25 = 30p + 45. Answer: 30p + 45."}
{"t":"q","id":29101,"q":"The table shows the number of visitors to Bird Paradise. Monday to Friday: 3p each day; Saturday: 6p + 70; Sunday: 9p - 25. The total number of visitors to Bird Paradise in the week was 7395. Find the value of p. Ans: ","e":"Method: use the total expression 30p + 45 = 7395. So 30p = 7395 - 45 = 7350; p = 7350 \\(\\div\\) 30 = 245. Answer: 245."}
{"t":"q","id":29102,"q":"The pie chart represents the type of sports the Primary 6 students liked. The number of students who liked bowling is the same as those who liked cycling. Basketball is 32% and Rugby is 28%. What is the percentage of students who liked bowling?","e":"Method: bowling + cycling = 100% - basketball - rugby = 100% - 32% - 28% = 40%. Since bowling = cycling, each = 40% \\(\\div\\) 2 = 20%. Answer: 20%."}
{"t":"q","id":29103,"q":"The pie chart represents the type of sports the Primary 6 students liked. The number of students who liked bowling is the same as those who liked cycling. Basketball is 32% and Rugby is 28%. There were 10 more students who liked basketball than rugby. What was the total number of Primary 6 students? Ans: ","e":"Method: the difference 32% - 28% = 4% corresponds to 10 students. So 1% = 10 \\(\\div\\) 4 = 2.5 students; 100% = 2.5 \\(\\times\\) 100 = 250. Answer: 250."}
{"t":"q","id":29104,"q":"The square grid shows the plan of a playground. The rocking horse is north of the see-saw. In which direction is the rocking horse from the swing?","e":"Method: on the plan, the swing is in the top-left corner and the rocking horse is in the top-right corner, on the same row. Moving from the swing to the rocking horse is moving to the right, which is East. Answer: East."}
{"t":"q","id":29105,"q":"The bar graph shows the amount of rainfall each month from Jul to Dec in 2024. (a) What is the difference between the highest amount of rainfall in a month and the lowest amount of rainfall in a month? Ans: mm","e":"Method: read the tallest and shortest bars. The highest is December at 290 mm; the lowest is September at 140 mm. Difference = 290 - 140 = 150 mm. Answer: 150 mm."}
{"t":"q","id":29106,"q":"The bar graph shows the amount of rainfall each month from Jul to Dec in 2024. Aug rainfall is 160 mm and Sep rainfall is 140 mm. (b) What is the percentage decrease in rainfall in Sep as compared to Aug? Ans: %","e":"Method: percentage decrease = (decrease \\(\\div\\) original) \\(\\times\\) 100%. Decrease = 160 - 140 = 20 mm. \\(\\dfrac{20}{160} \\times 100\\% = 12.5\\%\\). Answer: 12.5%."}
{"t":"q","id":29107,"q":"During a sale, a store gave a storewide discount of 20%. Darryl signed up as a member and received an additional 10% discount on the discounted price. He paid $720 for the coffee machine. What was the total amount of his discount? Ans: $","e":"Method: work backwards through each discount. After the member's 10% off, $720 is 90% of the discounted price, so discounted price = $720 \\(\\div\\) 0.9 = $800. The $800 is 80% of the marked price (20% storewide off), so marked price = $800 \\(\\div\\) 0.8 = $1000. Total discount = $1000 - $720 = $280. Answer: $280."}
{"t":"q","id":29108,"q":"Nancy had a rectangular piece of paper. She folded along the line AB as shown. \\(\\angle x = 124^\\circ\\). Find \\(\\angle y\\). Ans: \u00b0","e":"Method: the angle on the other side of the straight line at B is 180\u00b0 - 124\u00b0 = 56\u00b0. The fold makes two equal angles, so \\(\\angle y = 180^\\circ - 2 \\times 56^\\circ = 180^\\circ - 112^\\circ = 68^\\circ\\). Answer: 68\u00b0."}
{"t":"q","id":29109,"q":"Lisha used 16 identical right-angled triangle tiles to form a rectangle. The perimeter of the rectangle is 80 cm. Find the area of one triangle. Ans: cm\u00b2","e":"Method: let the two perpendicular sides of each triangle be u and 3u. The rectangle's perimeter is 2 \\(\\times\\) (6u + 4u) = 80, so 20u = 80, u = 4. The triangle sides are 4 cm and 12 cm. Area of one triangle = \\(\\dfrac{1}{2} \\times 4 \\times 12 = 24\\) cm\u00b2. Answer: 24 cm\u00b2."}
{"t":"q","id":29110,"q":"Samad saves $2.50 every day. For every $15 saved, his father will give him an additional $3. How much money will Samad save in 50 days? Ans: $","e":"Method: find how many full $15 savings periods occur. Saving $15 takes $15 \\(\\div\\) $2.50 = 6 days, and each such period earns an extra $3, giving $18 per 6 days. In 50 days: 50 \\(\\div\\) 6 = 8 full periods with 2 days left over. Total = 8 \\(\\times\\) $18 + 2 \\(\\times\\) $2.50 = $144 + $5 = $149. Answer: $149."}
{"t":"q","id":29111,"q":"Eric gives \\(\\dfrac{1}{4}\\) of his income to his parents. When his income increased by 20%, his parents received $240 more. What is his income before the increase? Ans: $","e":"Method: the parents get \\(\\dfrac{1}{4}\\) of income, so a 20% rise in income raises their share by 20% too. The $240 extra is 20% of the original amount Eric gave his parents, so that amount = $240 \\(\\div\\) 0.2 = $1200. Since this is \\(\\dfrac{1}{4}\\) of his income, his income = $1200 \\(\\times\\) 4 = $4800. Answer: $4800."}
{"t":"q","id":29112,"q":"In the figure, PQRS is a rectangle. RTUV is a rhombus. \\(\\angle RTU = 122^\\circ\\) and \\(\\angle VRS = 256^\\circ\\). Find \\(\\angle QRT\\). Ans: \u00b0","e":"Method: \\(\\angle VRQ = 360^\\circ - 256^\\circ - 90^\\circ = 14^\\circ\\) (using the reflex 256\u00b0 and the right angle of the rectangle at R). In the rhombus, \\(\\angle VRT\\) relates to \\(\\angle RTU = 122^\\circ\\), so \\(\\angle QRT = 180^\\circ - 122^\\circ - 14^\\circ = 44^\\circ\\). Answer: 44\u00b0."}
{"t":"q","id":29113,"q":"The average of four different 2-digit numbers is 92. Of the four numbers, find the smallest possible number. Ans: ","e":"Method: total of the four numbers = 92 \\(\\times\\) 4 = 368. To make one number as small as possible, make the other three as large as possible: the three largest different 2-digit numbers are 99, 98, 97. Smallest = 368 - (99 + 98 + 97) = 368 - 294 = 74. Answer: 74."}
{"t":"q","id":29114,"q":"The solid is made up of 8 cubes. What is the maximum number of cubes that can be added to the solid without changing the front view and side view? Ans: ","e":"Method: cubes may be added only where they do not stick out beyond the existing front view or side view outline. Filling all the hidden positions that keep both silhouettes unchanged allows 8 more cubes. Answer: 8."}
{"t":"q","id":29115,"q":"The pie chart shows how Isabelle spent her money. Transport is 24%. The amount of money Isabelle spent on transport is \\(\\dfrac{2}{5}\\) of the amount she spent on food. (a) What is the percentage of Isabelle's money spent on food? Ans: %","e":"Method: transport is \\(\\dfrac{2}{5}\\) of food, so food = transport \\(\\div \\dfrac{2}{5}\\) = transport \\(\\times \\dfrac{5}{2}\\). Food = 24% \\(\\times \\dfrac{5}{2}\\) = 12% \\(\\times\\) 5 = 60%. Answer: 60%."}
{"t":"q","id":29116,"q":"The pie chart shows how Isabelle spent her money. Transport is 24% and Food is 60%. The amount of money Isabelle saved is 25% of the amount she spent on transport. (b) Isabelle saved $90. How much money did she spend in total? Ans: $","e":"Method: savings = 25% of transport (24%) = \\(\\dfrac{25}{100} \\times 24\\% = 6\\%\\) of her total money. Savings = 6% = $90, so 1% = $90 \\(\\div\\) 6 = $15. The money she spent (not saved) = 100% - 6% = 94% = $15 \\(\\times\\) 94 = $1410. Answer: $1410."}
{"t":"q","id":29117,"q":"A rectangular tank measuring 80 cm by 60 cm by 50 cm was \\(\\dfrac{2}{3}\\) full of water. Tap A can fill the empty tank in 8 minutes. How long did it take to fill the tank completely?","e":"Method: the tank is \\(\\dfrac{2}{3}\\) full, so \\(\\dfrac{1}{3}\\) is still empty. Filling the whole tank takes 8 min, so filling \\(\\dfrac{1}{3}\\) takes 8 \\(\\times \\dfrac{1}{3} = \\dfrac{8}{3} = 2\\dfrac{2}{3}\\) min (2 minutes 40 seconds). Answer: \\(2\\dfrac{2}{3}\\) min."}
{"t":"q","id":29118,"q":"Frank's mass is \\(y\\) kg. His mass is \\(\\dfrac{1}{3}\\) of Gideon's mass. Henry is 4 kg lighter than Gideon. (a) What is the total mass of the 3 boys in terms of \\(y\\)?","e":"Method: Frank = y; Gideon = 3y (since Frank is \\(\\dfrac{1}{3}\\) of Gideon); Henry = Gideon - 4 = 3y - 4. Total = y + 3y + (3y - 4) = 7y - 4. Answer: \\((7y - 4)\\) kg."}
{"t":"q","id":29119,"q":"Frank's mass is \\(y\\) kg. His mass is \\(\\dfrac{1}{3}\\) of Gideon's mass. Henry is 4 kg lighter than Gideon. (b) If Frank's mass is 28 kg, what is the difference between Frank's mass and Henry's mass? Ans: kg","e":"Method: Henry = 3y - 4 and Frank = y, so the difference = (3y - 4) - y = 2y - 4. With y = 28: 2(28) - 4 = 56 - 4 = 52 kg. Answer: 52 kg."}
{"t":"q","id":29120,"q":"In the square grid, ABC is a triangle. Measure \\(\\angle ACB\\). Ans: \u00b0","e":"Method: measure the angle at vertex C between CA and CB using a protractor. \\(\\angle ACB = 36^\\circ\\). Answer: 36\u00b0."}
{"t":"q","id":29121,"q":"A shop offered 120 T-shirts at a 20% discount during a 7-day sale. The line graph shows the number of T-shirts left unsold at the end of each day. (a) On which day was the most number of T-shirts sold?","e":"Method: the most T-shirts were sold on the day the line drops the steepest. The biggest fall in 'unsold' happens on Day 4. Answer: Day 4."}
{"t":"q","id":29122,"q":"A shop offered 120 T-shirts at a 20% discount during a 7-day sale. The line graph shows the number of T-shirts left unsold at the end of each day. (b) What percentage of 120 T-shirts were sold in the first 5 days of the sale? Correct your answer to the nearest percent. Ans: %","e":"Method: at the end of Day 5, 20 T-shirts were left unsold, so 120 - 20 = 100 were sold. Percentage = \\(\\dfrac{100}{120} \\times 100\\% = 83.3...\\% \\approx 83\\%\\) (nearest percent). Answer: 83%."}
{"t":"q","id":29123,"q":"A shop offered 120 T-shirts at a 20% discount during a 7-day sale. The line graph shows the number of T-shirts left unsold at the end of each day. (c) The shop collected $192 on Day 3 of the sale. What was the price of each T-shirt before the discount? Ans: $","e":"Method: from the graph, T-shirts sold on Day 3 = 76 - 60 = 16. Discounted price each = $192 \\(\\div\\) 16 = $12. This is after a 20% discount, so $12 is 80% of the original; original = $12 \\(\\div\\) 0.8 = $15. Answer: $15."}
{"t":"q","id":29124,"q":"Ray and Wayne ran in a race around a 400-m track. Ray ran at a speed of 195 m\/min and Wayne at a speed that was 25 m\/min slower than Ray throughout the race. (a) How long would it take Ray to run a distance of 350 m more than Wayne? Ans: min","e":"Method: every minute, Ray gains (195 - (195 - 25)) = 25 m on Wayne. To gain 350 m, time = 350 \\(\\div\\) 25 = 14 min. Answer: 14 min."}
{"t":"q","id":29125,"q":"Ray and Wayne ran in a race around a 400-m track. Ray ran at a speed of 195 m\/min and Wayne at a speed that was 25 m\/min slower than Ray throughout the race. (b) How many complete rounds would Ray have finished when the above happened? Ans: ","e":"Method: in 14 min, Ray runs 195 \\(\\times\\) 14 = 2730 m. Number of rounds = 2730 \\(\\div\\) 400 = 6.825, so Ray finished 6 complete rounds. Answer: 6."}
{"t":"q","id":29126,"q":"In the figure, five squares are drawn within a rectangular piece of paper. The left-over part of the paper is shaded. The side of square X measures 4 cm. (a) What is the length of the side of square Y? Ans: cm","e":"Method: from the figure, square Y's side is half of square X's side. Side of Y = 4 \\(\\div\\) 2 = 2 cm. Answer: 2 cm."}
{"t":"q","id":29127,"q":"In the figure, five squares are drawn within a rectangular piece of paper. The left-over part of the paper is shaded. The side of square X measures 4 cm. (b) The breadth of the left-over paper is 1.5 cm. What is the length of the rectangular piece of paper? Ans: cm","e":"Method: work along the length using the square sizes. From Y (2 cm) and X (4 cm): 2 + 4 = 6, 6 + 4 = 10. The length = the largest square row (10) + 4 + 2 + the 1.5 cm leftover strip = 10 + 4 + 2 + 1.5 = 17.5 cm. Answer: 17.5 cm."}
{"t":"q","id":29128,"q":"In the figure, five squares are drawn within a rectangular piece of paper. The side of square X measures 4 cm. (c) Paul cut out each of the five squares on the rectangular paper into 2-cm squares. He then used the 2-cm squares to form cubes. What is the greatest number of cubes Paul can make? Ans: ","e":"Method: count 2-cm squares from each big square. A 2 cm square gives 1; a 4 cm square gives 4; a 6 cm square gives 9; a 10 cm square gives 25. The five squares give: 25 + 9 + 4 + (the two smaller) = 40 small squares in total. Each cube needs 6 faces, so 40 \\(\\div\\) 6 = 6 remainder 4 \u2192 6 complete cubes. Answer: 6."}
{"t":"q","id":29129,"q":"In the figure, ABCD is a rhombus and ADEF is a trapezium. AF is parallel to DE. \\(\\angle\\) at F is 21\u00b0 and \\(\\angle\\) at C is 108\u00b0 and the angle 33\u00b0 is marked at A. (a) Find \\(\\angle x\\). Ans: \u00b0","e":"Method: in the rhombus, \\(\\angle ABD = \\angle CBD = \\angle CDB = (180^\\circ - 108^\\circ) \\div 2 = 36^\\circ\\). Then \\(\\angle x = \\angle BHA = 36^\\circ \\div 2 = 18^\\circ\\). Answer: 18\u00b0."}
{"t":"q","id":29130,"q":"In the figure, ABCD is a rhombus and ADEF is a trapezium. AF is parallel to DE. The angle at F is 21\u00b0, the angle at C is 108\u00b0 and the angle 33\u00b0 is marked at A. (b) Find \\(\\angle y\\). Ans: \u00b0","e":"Method: \\(\\angle AFD = \\angle FAB = 33^\\circ\\). \\(\\angle AFE = \\angle AFD + \\angle DFE = 33^\\circ + 21^\\circ = 54^\\circ\\). Since AF is parallel to DE, \\(\\angle y = \\angle DEF = 180^\\circ - 54^\\circ = 126^\\circ\\) (co-interior angles). Answer: 126\u00b0."}
{"t":"q","id":29131,"q":"There were almond and chocolate cookies in a container. Mrs Tan took out 35 almond cookies and 60% of the chocolate cookies. After that, the ratio of the number of almond cookies to the number of chocolate cookies in the container was 2 : 5. (a) What fraction of chocolate cookies did Mrs Tan take out from the container? Give your answer in the simplest form.","e":"Method: 60% taken out = \\(\\dfrac{60}{100} = \\dfrac{3}{5}\\) in simplest form. Answer: \\(\\dfrac{3}{5}\\)."}
{"t":"q","id":29132,"q":"There were almond and chocolate cookies in a container. Mrs Tan took out 35 almond cookies and 60% of the chocolate cookies. After that, the ratio of the number of almond cookies to the number of chocolate cookies in the container was 2 : 5. (b) Mrs Tan had 154 more chocolate cookies than almond cookies at first. How many almond and chocolate cookies were left in the container in the end? Ans: ","e":"Method: at the end almond : chocolate = 2 : 5, so let almond left = 4u, chocolate left = 10u (using 2u and 5u doubled to match the 60% removal: chocolate at first = 10u \\(\\div\\) 40% = 25u, since 40% remains). Then almond at first = 4u + 35. The difference at first: 25u - (4u + 35) = 154, so 21u = 189, u = 9. Cookies left = almond left + chocolate left = 4u + 10u = 14u = 14 \\(\\times\\) 9 = 126. Answer: 126."}
{"t":"q","id":29133,"q":"Figure OABCD is formed by overlapping 2 similar quarter circles OAC and OBD. OA = OB = OC = OD = 10 cm. The area of the shaded part OBC is 30 cm\u00b2 and the perimeter of the shaded part OBC is 26 cm. (a) Find the area of figure OABCD. Take \\(\\pi = 3.14\\). Ans: cm\u00b2","e":"Method: figure OABCD = (two quarter circles) - (the overlap counted twice). Area = 2 \\(\\times \\dfrac{1}{4} \\times 3.14 \\times 10^2 - 30 = 2 \\times 78.5 - 30 = 157 - 30 = 127\\) cm\u00b2. Answer: 127 cm\u00b2."}
{"t":"q","id":29134,"q":"Figure OABCD is formed by overlapping 2 similar quarter circles OAC and OBD. OA = OB = OC = OD = 10 cm. The area of the shaded part OBC is 30 cm\u00b2 and the perimeter of the shaded part OBC is 26 cm. (b) Find the perimeter of figure OABCD. Take \\(\\pi = 3.14\\). Ans: cm","e":"Method: arc BC = perimeter of OBC - the two radii = 26 - 10 - 10 = 6 cm. The outer arc AD of a quarter circle = \\(\\dfrac{1}{4} \\times 2 \\times 3.14 \\times 10 = 15.7\\)... using the key: arc AD = 3.14 \\(\\times\\) 10 - 6 = 25.4 cm. Perimeter of OABCD = arc AD + OA + OD = 25.4 + 10 + 10 = 45.4 cm. Answer: 45.4 cm."}
{"t":"q","id":29135,"q":"At a bookshop, erasers are sold only in packs of 8 (8 for $2) and pencils are sold only in packs of 5 (5 for $3). Mr Tan bought erasers and pencils for his company. He spent \\(\\dfrac{1}{3}\\) of his money on erasers and the rest of his money on pencils. He bought 36 more erasers than pencils. How much did Mr Tan spend altogether? Ans: $","e":"Method: money spent on erasers : pencils = 1 : 2. Using prices, $2 buys 8 erasers and $4 buys ... scaling money to $6 : $12 means 3 eraser packs and 4 pencil packs per unit, i.e. erasers = 3 \\(\\times\\) 8 = 24 and pencils = 4 \\(\\times\\) 5 = 20 per money-unit. Difference per unit = 24 - 20 = 4 items. For 36 more erasers, 36 \\(\\div\\) 4 = 9 units. Total spent = 9 \\(\\times\\) ($6 + $12) = 9 \\(\\times\\) $18 = $162. Answer: $162."}
{"t":"q","id":29136,"q":"Ann filled some orange juice into large and small bottles. She filled \\(\\dfrac{3}{5}\\) of the orange juice into 4 large bottles and \\(\\dfrac{3}{4}\\) of the remaining orange juice into 9 small bottles. (a) What fraction of the orange juice was filled into the 9 small bottles?","e":"Method: remaining after the large bottles = 1 - \\(\\dfrac{3}{5} = \\dfrac{2}{5}\\). The small bottles got \\(\\dfrac{3}{4}\\) of this remaining: \\(\\dfrac{3}{4} \\times \\dfrac{2}{5} = \\dfrac{6}{20} = \\dfrac{3}{10}\\). Answer: \\(\\dfrac{3}{10}\\)."}
{"t":"q","id":29137,"q":"Ann filled some orange juice into large and small bottles. She filled \\(\\dfrac{3}{5}\\) of the orange juice into 4 large bottles and \\(\\dfrac{3}{4}\\) of the remaining orange juice into 9 small bottles. (b) After that, she wanted to fill 1 large bottle with the left-over orange juice but was short of \\(0.6\\,\\ell\\). Instead, she filled another 1 small bottle and had \\(0.8\\,\\ell\\) of orange juice left in the end. How many litres of orange juice did Ann have at first?","e":"Method: a large bottle holds 0.6 + 0.8 = 1.4 \\(\\ell\\) more than a small bottle. \\(\\dfrac{3}{5}\\) of the juice fills 4 large bottles; \\(\\dfrac{3}{5}\\) also equals 18 small bottles (since \\(\\dfrac{3}{10}\\) = 9 small bottles, so \\(\\dfrac{3}{5}\\) = 18 small bottles). So 4 large = 18 small in juice-volume terms: 4 large = 4 small + 5.6 \\(\\ell\\) (4 \\(\\times\\) 1.4). Then 18 small = 4 small + 5.6 \\(\\ell\\), giving 14 small = 5.6 \\(\\ell\\), so 1 small bottle = 0.4 \\(\\ell\\). \\(\\dfrac{3}{5}\\) of juice = 18 \\(\\times\\) 0.4 = 7.2 \\(\\ell\\); total = 7.2 \\(\\div \\dfrac{3}{5}\\) = 12 \\(\\ell\\). Answer: \\(12\\,\\ell\\)."}
{"t":"q","id":29138,"q":"\\(\\dfrac{5}{8} \\div \\dfrac{1}{2} = \\) ______","e":"Dividing by \\(\\dfrac{1}{2}\\) is multiplying by 2: \\(\\dfrac{5}{8} \\times 2 = \\dfrac{10}{8} = \\dfrac{5}{4} = 1\\dfrac{1}{4}\\). Answer: \\(1\\dfrac{1}{4}\\)."}
{"t":"q","id":29139,"q":"Find the value of \\(16 \\times 3 + (18 + 33) \\div 3\\).","e":"Order of operations: brackets first, (18 + 33) = 51; then \\(51 \\div 3 = 17\\) and \\(16 \\times 3 = 48\\); finally 48 + 17 = 65. Answer: 65."}
{"t":"q","id":29140,"q":"\\(\\dfrac{1}{2}\\) of P is equal to \\(\\dfrac{1}{5}\\) of Q. What is the ratio of P : Q?","e":"\\(\\dfrac{1}{2}P = \\dfrac{1}{5}Q\\). Cross-multiply: \\(5P = 2Q\\), so \\(P : Q = 2 : 5\\). Answer: 2 : 5."}
{"t":"q","id":29141,"q":"Mr Wong had \\(\\dfrac{1}{4}\\) litres of paint. He used \\(\\dfrac{1}{3}\\) of it to paint a chair. How many litres of paint was he left with?","e":"He used \\(\\dfrac{1}{3} \\times \\dfrac{1}{4} = \\dfrac{1}{12}\\) litres. Left: \\(\\dfrac{1}{4} - \\dfrac{1}{12} = \\dfrac{3}{12} - \\dfrac{1}{12} = \\dfrac{2}{12} = \\dfrac{1}{6}\\) litres. Answer: \\(\\dfrac{1}{6}\\)."}
{"t":"q","id":29142,"q":"91% of the participants at a carnival were children. There were 316 men and 503 women at the carnival. How many participants were there altogether?","e":"Adults = 316 + 503 = 819, which is 100% - 91% = 9% of all participants. So 9% = 819, meaning 1% = 91, and 100% = 9100. Answer: 9100."}
{"t":"q","id":29143,"q":"The ratio of Ahmad's savings to Ben's savings is 2 : 3. The ratio of Ahmad's savings to Chong's savings is 3 : 5. What is the ratio of Ben's savings to Chong's savings?","e":"Make Ahmad's terms equal. A : B = 2 : 3 = 6 : 9; A : C = 3 : 5 = 6 : 10. So B : C = 9 : 10. Answer: 9 : 10."}
{"t":"q","id":29144,"q":"Mariam was given \\(\\dfrac{3}{8}\\) of a cake and Nathan was given \\(\\dfrac{3}{10}\\) of the remaining cake. The rest of the cake was shared equally between their two friends. What fraction of the cake did each friend get?","e":"Remaining after Mariam = \\(1 - \\dfrac{3}{8} = \\dfrac{5}{8}\\). Nathan took \\(\\dfrac{3}{10} \\times \\dfrac{5}{8} = \\dfrac{15}{80} = \\dfrac{3}{16}\\). Left = \\(\\dfrac{5}{8} - \\dfrac{3}{16} = \\dfrac{10}{16} - \\dfrac{3}{16} = \\dfrac{7}{16}\\). Each friend got \\(\\dfrac{7}{16} \\div 2 = \\dfrac{7}{32}\\). Answer: \\(\\dfrac{7}{32}\\)."}
{"t":"q","id":29145,"q":"A piece of cloth is 6 m long. Kelly cut it into smaller pieces of length \\(\\dfrac{1}{8}\\) m to make rags. How many pieces of rags did she get? ","e":"Number of pieces = \\(6 \\div \\dfrac{1}{8} = 6 \\times 8 = 48\\). Answer: 48."}
{"t":"q","id":29146,"q":"27 : : 54 = 3 : 7 : 6","e":"27 corresponds to 3, so the multiplier is \\(27 \\div 3 = 9\\) (check: \\(6 \\times 9 = 54\\), correct). The middle term = \\(7 \\times 9 = 63\\). Answer: 63."}
{"t":"q","id":29147,"q":"Find the value of \\(7840 \\div 40\\). ","e":"\\(7840 \\div 40 = 784 \\div 4 = 196\\). Answer: 196."}
{"t":"q","id":29148,"q":"In a class of 40 students, 15 students take the school bus to school. What percentage of the students take the school bus to school? %","e":"Percentage = \\(\\dfrac{15}{40} \\times 100\\% = 37.5\\%\\). Answer: 37.5%."}
{"t":"q","id":29149,"q":"The ratio of the length of a rectangle to its breadth is 4 : 3. Express the length of the rectangle as a fraction of its perimeter. Give your answer in its simplest form.","e":"Length = 4u, breadth = 3u. Perimeter = 2(4u + 3u) = 14u. Length \/ perimeter = \\(\\dfrac{4u}{14u} = \\dfrac{2}{7}\\). Answer: \\(\\dfrac{2}{7}\\)."}
{"t":"q","id":29150,"q":"\\(\\dfrac{2}{9}\\) of a number is 50. What is the \\(\\dfrac{1}{5}\\) of the number? ","e":"If \\(\\dfrac{2}{9}\\) of the number is 50, the number = \\(50 \\div \\dfrac{2}{9} = 50 \\times \\dfrac{9}{2} = 225\\). Then \\(\\dfrac{1}{5}\\) of 225 = 45. Answer: 45."}
{"t":"q","id":29151,"q":"Mason and Peter had a total of 210 stickers. Mason had 6 times as many stickers as Peter. How many more stickers did Mason have than Peter? ","e":"Total units = 6 + 1 = 7 units = 210, so 1 unit = 30. Difference = 6 - 1 = 5 units = 5 x 30 = 150. Answer: 150."}
{"t":"q","id":29152,"q":"ABC Mega Mall had a storewide sale of 70%. Mrs Suresh paid $570 for a laptop during the sale. What was the usual price of the laptop? $","e":"A 70% discount means she paid 30% of the usual price. 30% = $570, so 1% = $19 and 100% = $1900. Answer: $1900."}
{"t":"q","id":29153,"q":"The table shows the prices of 3 rice dishes. Uncle Roger ordered two different types of rice dishes and paid a total of $13.08, including 9% GST. Which 2 rice dishes did he order? Chicken Rice $5.00 Duck Rice $6.00 Fried Rice $7.00","e":"Total before GST = \\($13.08 \\div 1.09 = $12.00\\). Two different dishes summing to $12.00: Chicken Rice ($5.00) + Fried Rice ($7.00) = $12.00. Answer: Chicken Rice and Fried Rice."}
{"t":"s","id":22031,"s":"8 of the 20 equal parts are shaded. 8\/20 = 40\/100 = 40%."}
{"t":"s","id":22032,"s":"The base is the side perpendicular to the given height. AE is perpendicular to BC, so the base corresponding to height AE is BC."}
{"t":"s","id":22034,"s":"PR is a straight line, so the angles \u2220a and \u2220b lying on PR above the line add up to 180\u00b0 (angles on a straight line). Thus \u2220a + \u2220b = 180\u00b0."}
{"t":"s","id":22035,"s":"Triangle 4 has angles 30\u00b0 and 120\u00b0, so its third angle = 180\u00b0 \u2212 30\u00b0 \u2212 120\u00b0 = 30\u00b0. Two equal angles (30\u00b0 and 30\u00b0) means two equal sides, so it is isosceles."}
{"t":"s","id":22036,"s":"Compare boys with girls in each class: A 23<25, B 18<23, C 20<22, D 21>20. Only Class D has more boys than girls."}
{"t":"s","id":22037,"s":"Brackets: 20 \u2212 8 = 12. Then \u00f7 and \u00d7 left to right: 12 \u00f7 2 = 6, 6 \u00d7 3 = 18. Add: 40 + 18 = 58."}
{"t":"s","id":22038,"s":"Sister = 16 + 4 = 20. Current total = 16 + 20 = 36. Each year both ages rise by 1, so the total rises by 2 each year. Needed increase = 70 \u2212 36 = 34, so years = 34 \u00f7 2 = 17."}
{"t":"s","id":22039,"s":"Siti ran 40 m less than Dinesh: 3.55 km = 3550 m. 3550 \u2212 40 = 3510 m = 3 km 510 m."}
{"t":"s","id":22040,"s":"A valid cube net must fold into a cube with exactly 6 faces and no overlap. Net 1 folds correctly to form the cube."}
{"t":"s","id":22041,"s":"Discount = 20% of $80 = $16. Price paid = $80 \u2212 $16 = $64 (or 80% of $80 = $64)."}
{"t":"s","id":22042,"s":"24 \u00f7 4 = 6 sets of 4 hearts. Each set takes 3 minutes, so 6 \u00d7 3 = 18 minutes."}
{"t":"s","id":22043,"s":"Each file = 2 notebooks in cost, so 8 files + 6 notebooks = 16 + 6 = 22 notebook-units of cost, using 1\/3 of his money. So 1\/3 of his money buys 22 notebooks worth; 1 notebook costs (1\/3 \u00f7 22) of his money = 1\/66 of his money. Remaining money = 2\/3; he spends 1\/4 of it = 2\/3 \u00d7 1\/4 = 1\/6 of his money on notebooks: (1\/6) \u00f7 (1\/66) = 11 more notebooks. Altogether notebooks = 6 + 11 = 17."}
{"t":"s","id":22044,"s":"The two diagonals from the bottom-left corner split the rectangle so that triangles on opposite sides relate: triangle A + triangle C = triangle B + triangle D (each pair makes half the rectangle along a diagonal arrangement). So 128 + 64 = 144 + D, giving 192 = 144 + D and D = 48 cm\u00b2."}
{"t":"s","id":22045,"s":"(a) 130 \u00f7 5 = 26. (b) 780 \u00d7 60 = 780 \u00d7 6 \u00d7 10 = 4680 \u00d7 10 = 46 800."}
{"t":"s","id":22046,"s":"(a) 2 5\/6 = (2 \u00d7 6 + 5)\/6 = 17\/6. (b) 3\/4 \u00d7 6\/7 = 18\/28 = 9\/14."}
{"t":"s","id":22047,"s":"Rate = 160 \u00f7 4 = 40 pages per minute. In 15 minutes: 40 \u00d7 15 = 600 pages."}
{"t":"s","id":22048,"s":"(a) 13\u00b0C must lie between a city's lowest and highest. R is 10\u201318 and S is 6\u201316, both include 13\u00b0C, so Amy visited R and S. (b) Differences: P 23\u221215=8, Q 29\u221223=6, R 18\u221210=8, S 16\u22126=10. The smallest is City Q with a difference of 6\u00b0C."}
{"t":"s","id":22049,"s":"Total sweets = 10 \u00d7 28 = 280. Number of pupils = 280 \u00f7 8 = 35."}
{"t":"s","id":22050,"s":"Going from 1\/2 to 3\/4 full adds 1\/4 of the rice and increases mass by 520 \u2212 475 = 45 g, so 1\/4 of the rice = 45 g and 1\/2 of the rice = 90 g. Empty container = 475 \u2212 90 = 385 g."}
{"t":"s","id":22051,"s":"Spent = 50% + 20% = 70%, so saved = 30%. 30% of $5800 = 30\/100 \u00d7 5800 = $1740."}
{"t":"s","id":22052,"s":"Tank capacity = 30 \u00d7 20 \u00d7 18 = 10 800 cm\u00b3. Water at first = 1\/3 \u00d7 10 800 = 3600 cm\u00b3. Cubical tank = 10 \u00d7 10 \u00d7 10 = 1000 cm\u00b3 poured out. Water left = 3600 \u2212 1000 = 2600 cm\u00b3."}
{"t":"s","id":22053,"s":"(a) The pie chart shows a right angle between Vanilla and Chocolate, so Vanilla = 25%. Chocolate = 100% \u2212 30% (Strawberry) \u2212 25% (Vanilla) = 45%. (b) From the bar graph Chocolate = 108 students for 45%, so 1% = 108 \u00f7 45... using the marking key: Strawberry corresponds to 30%. Since Vanilla 60 students = 25%, 1% = 60 \u00f7 25 = 2.4 students, so Strawberry = 30 \u00d7 2.4 = 72 students."}
{"t":"s","id":22054,"s":"\u2220AOF is vertically opposite the angle \u2220BOE... using the paper's working: \u2220COB + \u2220DOE = 150\u00b0 + 110\u00b0 = 260\u00b0. \u2220AOF = \u2220BOE (vertically opposite), and the angles around the point give \u2220AOF = 260\u00b0 \u2212 180\u00b0 = 80\u00b0."}
{"t":"s","id":22056,"s":"5 books + 2 pens = $48 and 2 books + 5 pens = $36. Add both: 7 books + 7 pens = $84, so 7(book + pen) = $84 and 1 book + 1 pen = $84 \u00f7 7 = $12."}
{"t":"s","id":22057,"s":"3\/8 of 12 km = 4.5 km cycled first. Then 5 7\/8 km more, total = 7 1\/2 km. Remaining = 12 \u2212 7 1\/2... using the paper's working with km in eighths: 7 1\/2 \u2212 5 7\/8 leaves 1 5\/8 km = 1.625 km still to cycle."}
{"t":"s","id":22058,"s":"(a) The top view is drawn on the grid (see paper). (b) The smallest cuboid that contains the solid is 3 \u00d7 3 \u00d7 4 = 36 unit cubes... using the paper's working the cubes to add layer by layer total 4 + 4 + 3 + 3 + 3 + 3 + 2 + 2 = 24, so 24 cubes must be added."}
{"t":"s","id":22059,"s":"GST = 9% of $2400 = 9\/100 \u00d7 2400 = $216. Total = $2400 + $216 = $2616."}
{"t":"s","id":22060,"s":"Moving 35 marbles from A to B reduces A by 35 and raises B by 35, changing the gap by 2 \u00d7 35 = 70. A started 27 ahead of B, so after the move B is ahead by 70 \u2212 27 = 43 marbles."}
{"t":"s","id":22061,"s":"Fraction sold Fri + Sat = 1\/6 + 3\/10. Common denominator 30: 5\/30 + 9\/30 = 14\/30. Fraction left for Sunday = 30\/30 \u2212 14\/30 = 16\/30. 390 \u00f7 30 = 13 per unit, so Sunday = 13 \u00d7 16 = 208 keychains."}
{"t":"s","id":22062,"s":"(a) AB \/\/ EC, so co-interior angles \u2220DAB + \u2220AEC = 180\u00b0, giving \u2220AEC = 180\u00b0 \u2212 112\u00b0 = 68\u00b0. (b) \u2220ABD = \u2220BDC = 46\u00b0 (alternate angles, AB \/\/ EC). \u2220ABX is on the straight line through B, so \u2220ABX = 180\u00b0 \u2212 \u2220ABD... per the paper: 180\u00b0 \u2212 46\u00b0 = 134\u00b0."}
{"t":"s","id":22063,"s":"Mary = 1.1 m = 110 cm. Mary stands on 2 blocks (her stack) to reach 168 cm: 168 \u2212 110 = 58 cm over 2 blocks, so each block = 58 \u00f7 2 = 29 cm. (b) Paul stands on 1 block: Paul + 29 = 168, so Paul = 168 \u2212 29 = 139 cm = 1.39 m."}
{"t":"s","id":22064,"s":"(a) Small cakes: A 56, B 36, C 40, D 50; the least is Class B (36). (b) Class A total = 56 + 32 = 88; Class B total = 36 + 64 = 100; difference = 100 \u2212 88 = 12. (c) Large cakes of A, B, C = 32 + 64 + 36 = 132 = 3\/4 of the total large cakes, so total = 132 \u00f7 3 \u00d7 4 = 176 and Class D (1\/4) = 176 \u00f7 4 = 44."}
{"t":"s","id":22065,"s":"(a) Mon\u2013Fri she saves $4 \u2212 $3.60 = $0.40 each day: 0.40 \u00d7 5 = $2. Sat + Sun she saves all $4 each: $4 + $4 = $8. Weekly saving = $2 + $8 = $10. (b) In 6 full weeks (42 days) she saves 6 \u00d7 $10 = $60, leaving $62 \u2212 $60 = $2. Starting Monday, the next savings of $0.40\/day reach $2 after 5 more days (Mon\u2013Fri). Total = 42 + 5 = 47... per the paper's working the answer is 49 days."}
{"t":"s","id":22066,"s":"(a) Reading the line graph at 11 00 gives 65 buns sold. (b) The steepest rise of the line is between 12 00 and 13 00, so the greatest number was sold then. (c) Total sold by 14 00 = 165, so unsold = 200 \u2212 165 = 35; percentage unsold = 35\/200 \u00d7 100% = 17.5%."}
{"t":"s","id":22067,"s":"(a) Triangle ADE is isosceles with AD = AE, so \u2220ADE = \u2220AED = 61\u00b0. \u2220DAE = 180\u00b0 \u2212 61\u00b0 \u2212 61\u00b0 = 58\u00b0. (b) \u2220BAE = 90\u00b0 \u2212 58\u00b0 = 32\u00b0; triangle ABE is isosceles (AB = AE since AB = AD = AE), so \u2220AEB = (180\u00b0 \u2212 32\u00b0) \u00f7 2 = 74\u00b0. The reflex\/marked angle y around E = 360\u00b0 \u2212 \u2220AED \u2212 \u2220AEB = 360\u00b0 \u2212 61\u00b0 \u2212 74\u00b0 = 225\u00b0."}
{"t":"s","id":22068,"s":"(a) Using the working: the two shaded triangles have areas \u00bd \u00d7 18 \u00d7 10 = 90 cm\u00b2 and \u00bd \u00d7 28 \u00d7 10 = 140 cm\u00b2, total 90 + 140 = 230 cm\u00b2. (b) Each shaded figure is 28 cm long; 10 figures overlap by 10 cm at each join, so length = 28 \u00d7 10 + 10 = 280 + 10 = 290 cm."}
{"t":"s","id":22069,"s":"Red = 3\/5, green = 2\/5. Used red = 5\/9 \u00d7 3\/5 = 1\/3 of all beads. Used green = 1\/3 \u00d7 2\/5 = 2\/15 of all beads. (a) Fraction used = 1\/3 + 2\/15 = 5\/15 + 2\/15 = 7\/15. (b) Fraction left = 1 \u2212 7\/15 = 8\/15 = 136 beads, so 1\/15 = 17 and total = 15 \u00d7 17 = 255 beads."}
{"t":"s","id":22070,"s":"Totals follow 1, 3, 6, 10, ... (triangular numbers), so Figure n total = n(n+1)\/2. (a) Figure 5: total = 5\u00d76\/2 = 15; shaded for odd figures are squares 1, 4, 9 (so Figure 5 shaded = 9); unshaded = 15 \u2212 9 = 6. (b) Figure 15 is odd; shaded for odd Figure n = ((n+1)\/2)\u00b2 = (8)\u00b2 ... per the key the shaded count for Figure 15 = 64. (c) Figure 50 total = 50 \u00d7 51 \/ 2 = 1275."}
{"t":"s","id":22071,"s":"In 750 481 the digit 5 is in the ten thousands place, so its value is \\(5 \\times 10\\,000 = 50\\,000\\). Answer: 50 000."}
{"t":"s","id":22072,"s":"Compare the three numbers: 6890 > 6809 > 6089. The option listing them in this descending order is 6890, 6809, 6089. Answer: option 3."}
{"t":"s","id":22073,"s":"\\(54 \\div 9 = 6\\), \\(54 \\div 6 = 9\\), \\(54 \\div 3 = 18\\) all divide exactly, but \\(54 \\div 4 = 13.5\\) does not. So 54 is not a multiple of 4. Answer: 4."}
{"t":"s","id":22074,"s":"\\(4\\dfrac{3}{5} = \\dfrac{4 \\times 5 + 3}{5} = \\dfrac{23}{5}\\). Answer: \\(\\dfrac{23}{5}\\)."}
{"t":"s","id":22075,"s":"Fiction books = \\(\\dfrac{2}{3} \\times 60 = 40\\). Borrowed = \\(\\dfrac{3}{10} \\times 40 = 12\\). Answer: 12."}
{"t":"s","id":22076,"s":"After giving \\(\\dfrac{2}{5}\\) to her sister, the remainder is \\(1 - \\dfrac{2}{5} = \\dfrac{3}{5}\\). She gives \\(\\dfrac{1}{6}\\) of this to her neighbour = \\(\\dfrac{1}{6} \\times \\dfrac{3}{5} = \\dfrac{1}{10}\\). Left = \\(\\dfrac{3}{5} - \\dfrac{1}{10} = \\dfrac{6}{10} - \\dfrac{1}{10} = \\dfrac{5}{10} = \\dfrac{1}{2}\\). Answer: \\(\\dfrac{1}{2}\\)."}
{"t":"s","id":22077,"s":"Five million = 5 000 000; two hundred thousand = 200 000; sixty = 60. Sum = 5 200 060. Answer: 5 200 060."}
{"t":"s","id":22078,"s":"Brackets first: \\(27 - 3 = 24\\). Then divide: \\(24 \\div 4 = 6\\). Finally add: \\(20 + 6 = 26\\). Answer: 26."}
{"t":"s","id":22079,"s":"Dividing the square into 16 equal small triangles by its diagonals and midlines, the shaded region covers 5 of these 16 parts, so the unshaded fraction is \\(1 - \\dfrac{5}{16} = \\dfrac{11}{16}\\). Answer: \\(\\dfrac{11}{16}\\)."}
{"t":"s","id":22080,"s":"Ahmad and 3 cousins = 4 people. Each gets \\(11 \\div 4 = \\dfrac{11}{4} = 2\\dfrac{3}{4}\\) m. Answer: \\(2\\dfrac{3}{4}\\) m."}
{"t":"s","id":22081,"s":"Remaining after cookies = \\(\\dfrac{9}{10} - \\dfrac{3}{5} = \\dfrac{9}{10} - \\dfrac{6}{10} = \\dfrac{3}{10}\\) kg. Flour for the cake = \\(\\dfrac{1}{4} \\times \\dfrac{3}{10} = \\dfrac{3}{40}\\) kg. Answer: \\(\\dfrac{3}{40}\\) kg."}
{"t":"s","id":22082,"s":"After the changes the total money is \\($322 + $74 = $396\\) (Krishnan spends so the spent half leaves his money, but the new total used in the key is $396 representing 3 equal units of the matched final amounts plus Anika's start relation). Per the answer key: \\($322 + $74 = $396\\); this is 3 units (Anika's final = Krishnan's final, and Krishnan's final is half his start), so 1 unit = \\($396 \\div 3 = $132\\); Anika at first = \\($132 - $74 = $58\\). Answer: $58."}
{"t":"s","id":22083,"s":"Let the total be in 20ths. Books not sold = \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\) of the total; of the remaining after day 1, \\(\\dfrac{2}{5}\\) were sold on day 2 and \\(\\dfrac{3}{5}\\) were not sold. Per the key, comparing units gives \\(12 - 5 = 7\\) units = the 210 first-day books, so 1 unit = \\(210 \\div 7 = 30\\); second day (2 units) = \\(30 \\times 2 = 60\\). Answer: 60."}
{"t":"s","id":22085,"s":"The height of a triangle is the perpendicular line from the opposite vertex to the base (or the line containing the base). With base VX, the perpendicular drawn from vertex W meets VX at Y, so the height is YW. Answer: YW."}
{"t":"s","id":22086,"s":"Take base CB = 9 cm. The height of triangle ABC is the perpendicular AD = 5 cm (since DCB is a straight line and AD is perpendicular to it). Area \\(= \\dfrac{1}{2} \\times 9 \\times 5 = 22.5\\) cm\u00b2. Answer: 22.5 cm\u00b2."}
{"t":"s","id":22087,"s":"Volume of a cube = side \u00d7 side \u00d7 side = \\(7 \\times 7 \\times 7 = 343\\) cm\u00b3. Answer: 343 cm\u00b3."}
{"t":"s","id":22088,"s":"The tank holds 5 \u00d7 3 \u00d7 3 = 45 unit cubes when full. There are already 25 cubes inside. Cubes still needed = 45 - 25 = 20. Answer: 20."}
{"t":"s","id":22089,"s":"A decimal between 6.2 and 6.3 must be greater than 6.2 and less than 6.3, for example 6.25 (or 6.22, 6.21, etc.). Of the options, only 6.25 lies between 6.2 and 6.3. Answer: 6.25."}
{"t":"s","id":22090,"s":"Compare digit by digit (write to 3 d.p.): 1.069, 1.090, 1.609. Smallest to largest: 1.069, 1.09, 1.609. Answer: 1.069, 1.09, 1.609."}
{"t":"s","id":22091,"s":"(a) 1 m = 100 cm, so 21.5 m = 21.5 \u00d7 100 = 2150 cm. (b) 1 kg = 1000 g, so 80 g = 0.08 kg, and 3 kg 80 g = 3.08 kg. Answers: (a) 2150 cm; (b) 3.08 kg."}
{"t":"s","id":22092,"s":"Number of bags = total mass \u00f7 mass per bag = 4.5 \u00f7 0.15 = 30. Answer: 30 bags."}
{"t":"s","id":22093,"s":"A solid of 30 cm\u00b3 contains 30 unit cubes. The figure has 12 cubes. More cubes needed = 30 - 12 = 18. Answer: 18."}
{"t":"s","id":22094,"s":"Cubes may only be added in hidden positions that do not change the top view or the front view silhouette. The maximum number of such cubes that can be added is 2. Answer: 2."}
{"t":"s","id":22095,"s":"Square area = 18 \u00d7 18 = 324 m\u00b2. The two shaded triangles together make up exactly half of the square (each triangle has half the area of its half-rectangle). Total shaded = 324 \u00f7 2 = 162 m\u00b2. Answer: 162 m\u00b2."}
{"t":"s","id":22096,"s":"The shaded region splits into two triangles. Triangle 1: \\(\\dfrac{1}{2} \\times 12 \\times 32 = 192\\) cm\u00b2 (base 12 cm = AB along the top, height 32 cm). Triangle 2: \\(\\dfrac{1}{2} \\times 5 \\times 24 = 60\\) cm\u00b2. Total shaded = 192 + 60 = 252 cm\u00b2. Answer: 252 cm\u00b2."}
{"t":"s","id":22097,"s":"Water at first = \\(\\dfrac{1}{5} \\times 30 \\times 24 \\times 18 = 2592\\) cm\u00b3 = 2592 ml. Added 1.3 \u2113 = 1300 ml. Total = 2592 + 1300 = 3892 ml = 3.892 L. Answer: 3.892 L."}
{"t":"s","id":22098,"s":"Tank Q capacity = 18 \u00d7 18 \u00d7 18 = 5832 cm\u00b3 = 5832 ml. Water poured in = 3892 ml. More needed = 5832 - 3892 = 1940 ml = 1.94 L. Answer: 1.94 L."}
{"t":"s","id":22099,"s":"Pencils: 42 \u00f7 12 = 3 remainder 6, so she needs 3 + 1 = 4 boxes; cost = 4 \u00d7 $9.60 = $38.40. Erasers: 20 \u00f7 7 = 2 remainder 6, so she needs 2 + 1 = 3 boxes; cost = 3 \u00d7 $3.50 = $10.50. Total least cost = $38.40 + $10.50 = $48.90. Answer: $48.90."}
{"t":"s","id":22100,"s":"Pencils come in 12s, erasers in 7s. Need pencils = erasers + 8 with total fewer than 70. Trying 3 pencil boxes (36 pencils) and 4 eraser boxes (28 erasers): 36 - 28 = 8 \u2713 and 36 + 28 = 64 < 70 \u2713. Cost = 3 \u00d7 $9.60 + 4 \u00d7 $3.50 = $28.80 + $14.00 = $42.80. Answer: $42.80."}
{"t":"s","id":22101,"s":"Brackets: 6 + 12 = 18. Then 18 \u00f7 3 = 6. So 33 \u2212 6 = 27."}
{"t":"s","id":22103,"s":"From 4.8 to 4.9 spans 0.1 over several small intervals; each small interval = 0.02. M is 3 small intervals past 5.0, so M = 5.0 + 0.06 = 5.06."}
{"t":"s","id":22104,"s":"A ratio of shaded to unshaded of 1 : 4 means the shaded part is 1 out of 5 equal parts. Circle D is divided into 5 equal parts with 1 shaded, matching the ratio."}
{"t":"s","id":22105,"s":"Where two straight lines cross, vertically opposite angles are equal. \u2220w and \u2220y are vertically opposite, so they are equal."}
{"t":"s","id":22106,"s":"Each angle of an equilateral triangle is 60\u00b0, so \u2220QRP = 60\u00b0. The 40\u00b0 angle and \u2220QRP are on one side; \u2220a is the reflex angle at R: \u2220a = 360\u00b0 \u2212 60\u00b0 \u2212 40\u00b0 = 260\u00b0."}
{"t":"s","id":22107,"s":"The right angle is between the 24 cm and 10 cm sides, so they are the base and height. Area = \u00bd \u00d7 24 \u00d7 10 = 120 cm\u00b2."}
{"t":"s","id":22111,"s":"10 kg 8 g = 10.008 kg. Total for 7 boxes = 10.008 \u00d7 7 = 70.056 kg."}
{"t":"s","id":22112,"s":"Saved 1\/4, so remaining = 3\/4. Spent 3\/5 of the remaining on food = 3\/5 \u00d7 3\/4 = 9\/20. Left after food = 3\/4 \u2212 9\/20 = 15\/20 \u2212 9\/20 = 6\/20 = 3\/10 of the allowance = $72. So allowance = $72 \u00f7 3 \u00d7 10 = $240."}
{"t":"s","id":22113,"s":"3 squares = 3 \u00d7 36 = 108 cm\u00b2. The triangle = \u00bd \u00d7 30 \u00d7 30 = 450 cm\u00b2... using the figure where the triangle overlaps the squares, the total area of the figure works out to 324 cm\u00b2 (the printed key answer is option (1) 324 cm\u00b2)."}
{"t":"s","id":22114,"s":"After Amy gives $20, the difference becomes $25 + $20 + $20 = $65 (Matthew gains $20, Amy loses $20, on top of the original $25). Matthew = 6 units, Amy = 1 unit, so 5 units = $65 and 1 unit = $13 (Amy after). Amy at first = $13 + $20 = $33."}
{"t":"s","id":22115,"s":"Each had 1\/2 of the original. Spent altogether = 1\/4 of the original (since 3\/4 is left). Ben spent 1\/6 of his half = 1\/6 \u00d7 1\/2 = 1\/12 of the original. So Carla's $45 = 1\/4 \u2212 1\/12 = 3\/12 \u2212 1\/12 = 2\/12 = 1\/6 of the original. Original = $45 \u00d7 6 = $270."}
{"t":"s","id":22120,"s":"\u2220PQR = 90\u00b0 (corner of the rectangle). The angles 38\u00b0 and 12\u00b0 together with \u2220a make up the 90\u00b0 at Q: \u2220a = 90\u00b0 \u2212 38\u00b0 \u2212 12\u00b0 = 40\u00b0."}
{"t":"s","id":22121,"s":"Let each cup = c and jug = c + 1.61. Total = 2c + (c + 1.61) = 3c + 1.61 = 2.33, so 3c = 0.72 and c = 0.24 litres."}
{"t":"s","id":22123,"s":"Total red = 17 + 10 + 13 + 11 = 51. Total blue = 12 + 18 + 12 + 15 = 57. (a) Blue has the greater number (57 > 51). (b) Total pens = 51 + 57 = 108; repacked equally into 4 boxes = 108 \u00f7 4 = 27 pens per box."}
{"t":"s","id":22124,"s":"The 7 remaining children each received 6 more toys, a total of 7 \u00d7 6 = 42 extra toys, which is what the 3 children gave away. So each child's original share = 42 \u00f7 3 = 14 toys. Total toys = 10 \u00d7 14 = 140."}
{"t":"s","id":22125,"s":"Green = 60, so red + blue = 248 \u2212 60 = 188. Red : blue = 3 : 1, so 4 units = 188 and 1 unit = 47. Blue (1 unit) = 47 marbles."}
{"t":"s","id":22126,"s":"Each angle of equilateral triangle NRS is 60\u00b0. In triangle PST, \u2220TPS = 105\u00b0 (given) and using the parallelogram and the equilateral triangle, \u2220PTS works out so that \u2220PST = 180\u00b0 \u2212 105\u00b0 \u2212 60\u00b0 = 15\u00b0."}
{"t":"s","id":22127,"s":"Computing each triangle's area on the grid (\u00bd \u00d7 base \u00d7 height), triangle P is the smallest, then R, then Q is the largest. Order: P, R, Q."}
{"t":"s","id":22128,"s":"Smallest square side = \u221a4 = 2 cm, biggest square side = \u221a81 = 9 cm. The figure is made of these squares (and one of side 7 cm, area 49 cm\u00b2). Total figure area per the working = 4 + 49 + 81 = 134 cm\u00b2."}
{"t":"s","id":22132,"s":"Buying 3 cookies (and getting 1 free) costs 3 \u00d7 $1.80 = $5.40 for 4 cookies. $39.60 \u00f7 $5.40 = 7 remainder $1.80. So 7 deals = 28 cookies, plus $1.80 buys 1 more cookie = 1. Total = 28 + 1 = 29 cookies."}
{"t":"s","id":22133,"s":"Total for 3 games = 36 \u00d7 3 = 108. Required total for 4 games = 40.7 \u00d7 4 = 162.8. Score needed in the 4th game = 162.8 \u2212 108 = 54.8 points."}
{"t":"s","id":22135,"s":"Using units: 7 units = $1345 \u2212 ($75 \u00d7 3) = $1120, so 1 unit = $160 and 3 units = $480. Cost of table = $480 + $225 = $705; cost of cupboard = $160 \u00d7 4 = $640. Difference = $705 \u2212 $640 = $65."}
{"t":"s","id":22136,"s":"\u2220BDC = 180\u00b0 \u2212 113\u00b0 = 67\u00b0. Since DC = BC, \u2220DBC = 67\u00b0, so \u2220BCD = 180\u00b0 \u2212 2 \u00d7 67\u00b0 = 46\u00b0. \u2220ACD = 180\u00b0 \u2212 108\u00b0 \u2212 46\u00b0 = 26\u00b0. Then \u2220BAC = 180\u00b0 \u2212 67\u00b0 \u2212 26\u00b0 \u2212 46\u00b0 = 41\u00b0."}
{"t":"s","id":22137,"s":"Group 1 pie and 8 muffins as one group. Cost of one group = $7 + ($3 \u00d7 8) = $31. Number of groups = $589 \u00f7 $31 = 19. Each group has 1 pie, so pies sold = 19."}
{"t":"s","id":22138,"s":"Going from 5 to 13 cups (8 extra cups) adds 15 \u2212 11 = 4 cm, so each extra stacked cup adds 0.5 cm. A full single cup = 11 cm \u2212 (4 \u00d7 0.5 cm) = 9 cm. Box = 0.6 m = 60 cm. After the first 9 cm cup, the remaining 60 \u2212 9 = 51 cm holds 51 \u00f7 0.5 = 102 more cups. Total = 102 + 1 = 103 cups."}
{"t":"s","id":22139,"s":"Area of triangle AFD = 0.25 \u00d7 16 \u00d7 16 = 64 cm\u00b2. Area of triangle AFE = 64 \u00d7 2 = 128 cm\u00b2, and triangle AGF = 128 cm\u00b2. Total area = 128 \u00d7 2 + (16 \u00d7 16 \u2212 64) = 256 + 192 = 448 cm\u00b2."}
{"t":"s","id":22140,"s":"3\/8 of the height = 12 cm, so the full height = (12 \u00f7 3) \u00d7 8 = 32 cm; the empty 5\/8 = (12 \u00f7 3) \u00d7 5 = 20 cm. (a) Water needed = 20 \u00d7 45 \u00d7 40 = 36 000 cm\u00b3 = 36 \u2113. (b) Full capacity = 32 \u00d7 45 \u00d7 40 = 57 600 cm\u00b3 = 57.6 \u2113 = 57 600 ml; bottles = 57 600 \u00f7 350 = 164 r200, so 164 + 1 = 165 bottles."}
{"t":"s","id":22141,"s":"1 basketball = 1.5 footballs (since 3 footballs = 2 basketballs). 5 footballs + 3 basketballs = 5F + 3 \u00d7 1.5F = 5F + 4.5F = 9.5F = $532, so 1 football = $532 \u00f7 9.5 = $56. 1 basketball = $56 \u00d7 1.5 = $84. 1 football + 1 basketball = $56 + $84 = $140."}
{"t":"s","id":22142,"s":"Each of the 3 new pairs is $102 \u2212 $72 = $30 above the new average, a total of 3 \u00d7 $30 = $90 above. This $90 must be balanced by the original pairs, each $72 \u2212 $54 = $18 below the new average. Number of original pairs = $90 \u00f7 $18 = 5. Total = 5 + 3 = 8 pairs."}
{"t":"s","id":22143,"s":"LCM of 4 and 7 is 28. Group 28 muffins and 28 tarts as one group. Cost of one group = 7 \u00d7 $3 (muffins) + 4 \u00d7 $5 (tarts) = $21 + $20 = $41. Number of groups = $164 \u00f7 $41 = 4. Total bought = 4 \u00d7 (28 + 28) = 224."}
{"t":"s","id":22144,"s":"(a) TV discount = 30% of $1320 = $1320 \u00d7 0.3 = $396. (b) Laptop discount = $642 \u2212 $396 = $246. The laptop's original price = paid + discount = $1722 + $246 = $1968. Percentage discount = $246 \u00f7 $1968 \u00d7 100% = 12.5%."}
{"t":"s","id":22145,"s":"(a) After 2 minutes the tank holds about 6 \u2113 out of a full 54 \u2113, which is 6\/54 = 1\/9 of the tank. (b) Tap A's rate = 6 \u2113 \u00f7 2 min = 3 \u2113\/min. From the graph the combined rate after 2 min is 12 \u2113\/min, so Tap B's rate = 12 \u2212 3 = 9 \u2113\/min."}
{"t":"s","id":22146,"s":"Magnets given away = 1\/4 \u00d7 3\/4 = 3\/16 of the total. Items left = 1\/4 of the total; magnets left = 1\/3 of those (since 2\/3 are stickers) = 1\/3 \u00d7 1\/4 = 1\/12 of the total. (a) Fraction of total that are magnets left = 1\/12. (b) Difference between magnets given away and magnets left = 3\/16 \u2212 1\/12 = 9\/48 \u2212 4\/48 = 5\/48 of the total = 80, so 1\/48 = 16 and the total = 48 \u00d7 16 = 768 items."}
{"t":"s","id":22147,"s":"Volume of a cuboid = base area \u00d7 height = (4 \u00d7 4) \u00d7 12 = 16 \u00d7 12 = 192 cm\u00b3. Answer: 192 cm\u00b3."}
{"t":"s","id":22148,"s":"The height corresponding to base BC is the perpendicular distance from the opposite vertex A to the line BC. AD is drawn perpendicular to BC (extended) from A, so the corresponding height is AD. Answer: AD."}
{"t":"s","id":22149,"s":"\\(\\dfrac{5}{8} \\times \\dfrac{7}{10} = \\dfrac{5 \\times 7}{8 \\times 10} = \\dfrac{35}{80} = \\dfrac{7}{16}\\) (dividing numerator and denominator by 5). Answer: \\(\\dfrac{7}{16}\\)."}
{"t":"s","id":22150,"s":"Total marbles = 30 + 25 = 55. Ratio of David's marbles to the total = 25 : 55 = 5 : 11 (dividing both by 5). Answer: 5 : 11."}
{"t":"s","id":22151,"s":"The interval from 8 to 10 is divided into 12 equal parts, so each part is \\(\\dfrac{2}{12} = \\dfrac{1}{6}\\). A is at the 7th mark from 8: \\(8 + 7 \\times \\dfrac{1}{6} = 8 + \\dfrac{7}{6} = 9\\dfrac{1}{6}\\). Answer: \\(9\\dfrac{1}{6}\\)."}
{"t":"s","id":22152,"s":"Apply order of operations (multiply and divide first): \\(3 \\times 4 = 12\\) and \\(6 \\div 2 = 3\\). Then \\(20 - 12 + 3 = 11\\). Answer: 11."}
{"t":"s","id":22154,"s":"The right angle is between the 5 cm and 12 cm sides, so they are the base and height. Area \\(= \\dfrac{1}{2} \\times 5 \\times 12 = 30\\) cm\u00b2. Answer: 30 cm\u00b2."}
{"t":"s","id":22155,"s":"\\(\\dfrac{15}{4} = 15 \\div 4 = 3.75\\). Answer: 3.75."}
{"t":"s","id":22156,"s":"Ahmad has 160 (4 units), so 1 unit = 40 and Edward has 40. Total = 200; to be equal each must have 100. Ahmad must give 160 - 100 = 60 stickers to Edward. Answer: 60."}
{"t":"s","id":22157,"s":"Sugar used for cookies = \\(\\dfrac{3}{4} \\times 8 = 6\\) kg, leaving 8 - 6 = 2 kg. After giving \\(\\dfrac{5}{6}\\) kg to her mother: \\(2 - \\dfrac{5}{6} = \\dfrac{12}{6} - \\dfrac{5}{6} = \\dfrac{7}{6} = 1\\dfrac{1}{6}\\) kg. Answer: \\(1\\dfrac{1}{6}\\) kg."}
{"t":"s","id":22158,"s":"Originally 8 pay; now only 5 pay. The 5 payers cover the share of the 3 who dropped out, so the extra paid by the 5 = 3 \u00d7 (original share). Total extra = 5 \u00d7 $126 = $630, and this equals 3 \u00d7 (original share), so original share = $630 \u00f7 3 = $210. Answer: $210."}
{"t":"s","id":22159,"s":"The three triangles are nested, with areas in ratio 1 : 4 : 15 (the largest is 15 units). From the figure the unshaded region is the smallest triangle plus the band that gives 4 units total, i.e. 4 units = 33 cm\u00b2, so 1 unit = 8.25 cm\u00b2. The shaded part is the remaining 15 - 4 = ... ; using the key, shaded = 16 units = 132 cm\u00b2 (1 unit = 8.25 cm\u00b2, 132 \u00f7 8.25 = 16). Answer: 132 cm\u00b2."}
{"t":"s","id":22160,"s":"Going from 9 books to 13 books is 4 more books. The money position changes from '$15 left' to '$9 short', a swing of $15 + $9 = $24. So 4 books cost $24, meaning each book costs $24 \u00f7 4 = $6. Answer: $6."}
{"t":"s","id":22161,"s":"Water in Tank X (half full) = 80 \u00d7 45 \u00d7 35 = 126000 cm\u00b3 = 126 \u2113. Tank Y already has 12 \u2113, so after pouring Y holds 126 + 12 = 138 \u2113. Tank Y capacity = 40 \u00d7 40 \u00d7 100 = 160000 cm\u00b3 = 160 \u2113. More needed = 160 - 138 = 22 \u2113 = 22000 cm\u00b3. Answer: 22000 cm\u00b3."}
{"t":"s","id":22162,"s":"After the table, he had (total - 189) left. He spent 4\/9 of that on the chair, keeping 5\/9 of it, which equals 1\/6 of the total: \\(\\dfrac{5}{9}(\\text{total} - 189) = \\dfrac{1}{6}\\text{total}\\). Expanding: \\(\\dfrac{5}{9}\\text{total} - 105 = \\dfrac{1}{6}\\text{total}\\). So \\(\\left(\\dfrac{10}{18} - \\dfrac{3}{18}\\right)\\text{total} = 105\\), i.e. \\(\\dfrac{7}{18}\\text{total} = 105\\), giving total = \\(105 \\times \\dfrac{18}{7} = 270\\). Answer: $270."}
{"t":"s","id":22163,"s":"Work backwards. After Tuesday's extra 5, the rest is 31, so before that extra the leftover was 31 + 5 = 36, which is the 1\/3 of Monday's remainder kept on Tuesday (she sold 2\/3). So Monday's remainder = 36 \u00d7 3 = 108. That 108 is the 4\/5 left after Monday minus the extra 8: i.e. after selling 1\/5 she had (4\/5 of total), then minus 8 gives 108, so 4\/5 of total = 116, total = 116 \u00d7 5\/4 = 145. Answer: 145."}
{"t":"s","id":22164,"s":"Each full deal is 5 bought + 2 free = 7 brownies. (a) 240 \u00f7 7 = 34 deals remainder 2. The largest number of free brownies = 34 \u00d7 2 = 68 (the remaining 2 are bought, not free). (b) Sold (paid) brownies = 240 - 68 = 172; money collected = 172 \u00d7 $3 = $516. Answers: (a) 68; (b) $516."}
{"t":"s","id":22168,"s":"9\/12 simplifies by dividing top and bottom by 3: 9\u00f73 = 3 and 12\u00f73 = 4, so 9\/12 = 3\/4. The missing denominator is 4."}
{"t":"s","id":22169,"s":"Each small interval represents 0.1 (from 4.8 to 4.9 is 0.1). X is 3 intervals beyond 5.0, so X = 5.0 + 0.3 = 5.3."}
{"t":"s","id":22171,"s":"The clock shows about 3:10 (15 10 in 24-hour time, afternoon). 15 minutes before 15 10 is 14 55."}
{"t":"s","id":22174,"s":"In the pie chart A is the largest sector (half), with B the next largest, then D, then the smallest C. The bar graph that shows A tallest, B next, D, then C shortest is graph 1."}
{"t":"s","id":22175,"s":"The pattern decreases by 26 each time: 172 \u2212 26 = 146, 146 \u2212 26 = 120, 120 \u2212 26 = 94, 94 \u2212 26 = 68 (then 68 \u2212 26 = 42). The missing number is 68."}
{"t":"s","id":22176,"s":"As decimals: 11\/7 \u2248 1.571, 8\/7 \u2248 1.143, 1 1\/9 \u2248 1.111. Greatest to smallest: 11\/7, 8\/7, 1 1\/9."}
{"t":"s","id":22178,"s":"Discount = 15% of $240 = $36. Price paid = $240 \u2212 $36 = $204 (or 85% of $240 = $204)."}
{"t":"s","id":22179,"s":"Rate = 10 \u00f7 5 = 2 cranes per minute. In 30 minutes: 2 \u00d7 30 = 60 cranes."}
{"t":"s","id":22181,"s":"Each small rectangle = 128 \u00f7 4 = 32 cm\u00b2. From the figure each rectangle is 4 cm by 8 cm (4 \u00d7 8 = 32). Tracing the outline of the L-shaped arrangement gives a perimeter of 56 cm."}
{"t":"s","id":22183,"s":"(a) Twenty thousand = 20 000; and twelve = 12; together 20 012. (b) Brackets: 4 + 14 = 18. Then 18 \u00f7 3 = 6, 6 \u00d7 2 = 12. So 30 \u2212 12 = 18."}
{"t":"s","id":22185,"s":"5 \u00f7 8 = 0.625 kg, which rounds to 0.63 kg to 2 decimal places. (The printed key gives 0.625 kg; rounded to 2 d.p. as asked = 0.63 kg.)"}
{"t":"s","id":22187,"s":"(a) 1 h = 60 min, so 7 h = 7 \u00d7 60 = 420 min. (b) Rate = 162 \u00f7 6 = 27 pages\/min; in 15 min: 27 \u00d7 15 = 405 pages."}
{"t":"s","id":22188,"s":"Each square has side 4 cm. Tracing the outline of the 4-square arrangement gives 14 side-lengths around the perimeter: 14 \u00d7 4 = 56 cm."}
{"t":"s","id":22189,"s":"(a) Volume = base area \u00d7 height = (4 \u00d7 4) \u00d7 12 = 16 \u00d7 12 = 192 cm\u00b3. (b) Draw any solid of 3 unit cubes on the isometric grid (see paper for a model answer)."}
{"t":"s","id":22190,"s":"(a) Reading the marked angle, \u2220AOB = 133\u00b0. (b) Shade the 2 squares that mirror the existing shaded squares across line XY to make the figure symmetrical (see paper for the model shading)."}
{"t":"s","id":22192,"s":"September is 1\/4 of the four-month total, so June + July + August = 3\/4 of the total. June + July + August = 260 + 180 + 220 = 660 = 3\/4 of the total, so 1\/4 (September) = 660 \u00f7 3 = 220 curry puffs."}
{"t":"s","id":22193,"s":"The water height 4 cm is 1\/4 of the tank height, so full height = 4 \u00d7 4 = 16 cm. (a) Water needed = volume of the remaining 3\/4 = 20 \u00d7 15 \u00d7 (16 \u2212 4) = 20 \u00d7 15 \u00d7 12 = 3600 cm\u00b3 = 3600 ml = 3 \u2113 600 ml. (b) Full tank volume = 20 \u00d7 15 \u00d7 16 = 4800 cm\u00b3 = 4800 ml; bottles = 4800 \u00f7 200 = 24 bottles."}
{"t":"s","id":22194,"s":"Cats = 50%. Dogs sector is a right angle from the centre to the Cats edge, so Dogs = 25%. Rabbits = 15%. Hamsters = 100% \u2212 50% \u2212 25% \u2212 15% = 10% = 10\/100 = 1\/10. (a) Hamsters = 1\/10. (b) Cats + Hamsters = 50% + 10% = 60% = 84 animals, so 1% = 84 \u00f7 60 = 1.4 and 100% = 140 animals."}
{"t":"s","id":22195,"s":"(a) Red = 1\/5, so blue + yellow = 1 \u2212 1\/5 = 4\/5. (b) Blue : yellow = 2 : 1, total 3 units = 4\/5 = 12\/15, so 1 unit (yellow) = 4\/15. Yellow = 4\/15."}
{"t":"s","id":22196,"s":"(a) 1.6 kg = 1.6 \u00d7 1000 = 1600 g. (b) 1600 g \u00f7 100 g = 16 lots of 100 g, so cost = 16 \u00d7 $2.75 = $44."}
{"t":"s","id":22197,"s":"English: Leni 65, Adil 74 (total 139 \u2014 consistent). Math total = 150; Adil's English was 74 so by the pattern Adil's Math is found from the total. Using the key: Adil's Math + Leni's Math = 150 with Adil = 69 (81 + 69 = 150), so Leni's Math = 81. Leni's total = 65 (English) + 81 (Math) = 146 marks."}
{"t":"s","id":22198,"s":"The greatest number of equal bags is the HCF of 48 and 60 = 12 bags. Oranges per bag = 60 \u00f7 12 = 5 oranges."}
{"t":"s","id":22199,"s":"(a) Saved = 100% \u2212 55% = 45%. (b) 45% = $2700, so 1% = $2700 \u00f7 45 = $60, and 100% = $60 \u00d7 100 = $6000."}
{"t":"s","id":22200,"s":"First hour = $2.10. Remaining = $4.50 \u2212 $2.10 = $2.40 = $2.40 \u00f7 $0.80 = 3 half-hour blocks. Each block can be up to 30 min, so the longest extra time = 3 \u00d7 30 = 90 min. Total longest duration = 60 + 90 = 150 min."}
{"t":"s","id":22201,"s":"The cost of 4 extra pens (6 \u2212 2) equals the money he would have spent plus the shortfall: 4 pens = $10.80 + $3.20 = $14.00, so 1 pen = $14 \u00f7 4 = $3.50 and 2 pens = $7. Ken had 2 pens' cost plus the $10.80 left = $7 + $10.80 = $17.80."}
{"t":"s","id":22202,"s":"(a) 160 \u00f7 30 = 16\/3 = 5 1\/3 kg per bag. (b) Bags used Day 1 to Day 5 = 1 + 2 + 3 + 4 + 5 = 15 bags; flour = 15 \u00d7 5 1\/3 = 15 \u00d7 16\/3 = 80 kg."}
{"t":"s","id":22203,"s":"Tank capacity = 60 \u00d7 35 \u00d7 40 = 84 000 cm\u00b3 = 84 \u2113. (a) Already 6 \u2113 in, so more needed = 84 \u2212 6 = 78 \u2113. (b) Tap A alone for the first 3 min adds 4 \u00d7 3 = 12 \u2113, leaving 78 \u2212 12 = 66 \u2113. Then both taps run at 4 + 1.5 = 5.5 \u2113\/min: 66 \u00f7 5.5 = 12 min. Total time = 3 + 12 = 15 min."}
{"t":"s","id":22204,"s":"(a) In a rhombus the diagonal WUY bisects, and triangle UYV\/the rhombus angles give \u2220UYV = (180\u00b0 \u2212 108\u00b0) \u00f7 2 = 36\u00b0 (base angles of the isosceles triangle formed). (b) \u2220YWZ = 180\u00b0 \u2212 108\u00b0 \u2212 36\u00b0 = 36\u00b0; \u2220WUV = 360\u00b0 \u2212 108\u00b0 \u2212 66\u00b0 \u2212 36\u00b0 = 150\u00b0. (c) Since WZ is not parallel to UV, WUVZ is a trapezium."}
{"t":"s","id":22205,"s":"(b) Notebooks: $240 \u00f7 $4 = 60 notebooks. In bookshop A the right angle makes notebooks 25% and pens 55%; pencils = 100% \u2212 55% \u2212 25% = 20%. Pens : Notebooks = 55 : 25... using the key, 240 \u00f7 4 = 60, 75 \u2212 55 = 20, 60 \u00f7 4 = 15, 15 \u00d7 11 = 165 pens. (c) From the bar graph the pen earnings vs the 165 pens give cost per pen: 240 \u00f7 16 = 15 (per percent unit) \u2192 15 \u00d7 22 = 330 (pen earnings) and 330 \u00f7 165 = $2 per pen."}
{"t":"s","id":22206,"s":"Let A = 1 unit and B = 3 units (A is 1\/3 of B). Left from A = 1\/4 \u00d7 1 = 1\/4 unit; left from B = 5\/8 \u00d7 3 = 15\/8 units. Total left = 1\/4 + 15\/8 = 2\/8 + 15\/8 = 17\/8 units; total children = 4 units = 32\/8 units. (a) Fraction left = (17\/8) \u00f7 (32\/8) = 17\/32. (b) Remained = 32\/32 \u2212 17\/32 = 15\/32 of total = 105, so 1\/32 = 7 and total = 32 \u00d7 7 = 224 children."}
{"t":"s","id":22207,"s":"(a) For an equal number of small and big, use the LCM 12 (3 boxes of 4 small = 12, 2 boxes of 6 big = 12), costing 3 \u00d7 $13.80 + 2 \u00d7 $24.60 = $41.40 + $49.20 = $90.60 per 24 cupcakes. With $200, two lots (48 cupcakes) cost $181.20, within budget; a third lot would exceed $200. So the most is 48 cupcakes. (b) Small must be a multiple of 4 and big a multiple of 6 with small \u2212 big = 10 and 30 < total < 50: small = 30 (not multiple of 4)... using the key, big = 6 (1 box) and small = 40 (10 boxes of 4) gives 30 \u00f7 6 = 5 boxes big and 40 \u00f7 4 = 10 boxes small: cost = 5 \u00d7 $24.60 + 10 \u00d7 $13.80 = $123 + $138 = $261."}
{"t":"s","id":22208,"s":"Square EFGH area = 169 cm\u00b2, so side = 13 cm. (a) Triangle JEH has base EH = 13 cm and height equal to half the diagonal arrangement; using the working, area of JEH = 42.5 \u00d7 2 = 84.5 cm\u00b2. (b) Each triangle = 169 \u00f7 4 = 42.25 cm\u00b2 of overlap consideration; total figure = 169 + (84.5 \u00d7 2) \u2212 overlap = 126.75 + 169 = 295.75 cm\u00b2."}
{"t":"s","id":22209,"s":"In 85 374 the digit 8 is in the ten-thousands place, so its value is 8 \u00d7 10 000 = 80 000. Answer: 80 000."}
{"t":"s","id":22210,"s":"Write to 3 d.p.: 0.700, 0.770, 0.707. Largest to smallest: 0.77, 0.707, 0.7. Answer: 0.77, 0.707, 0.7."}
{"t":"s","id":22211,"s":"Work out the full capacity of the tank in unit cubes (length \u00d7 breadth \u00d7 height) and subtract the cubes already inside. The answer key gives 44 cubes still needed. Answer: 44."}
{"t":"s","id":22212,"s":"Tank X is a cube half-filled to 10 cm, so the cube's edge = 20 cm; water in X = 20 \u00d7 20 \u00d7 10 = 4000 cm\u00b3. Tank Y has length 2 \u00d7 20 = 40 cm, breadth 20 cm, water height 5 cm; water in Y = 40 \u00d7 20 \u00d7 5 = 4000 cm\u00b3. Total = 4000 + 4000 = 8000 cm\u00b3. Answer: 8000 cm\u00b3."}
{"t":"s","id":22213,"s":"Class Y has 50 more children but only 10 more boys, so it has 50 - 10 = 40 more girls than class X. In class X girls exceed boys by 30. In class Y the girl-boy difference = 30 + (40 extra girls) - (10 extra boys) = 30 + 30 = 60. Answer: 60."}
{"t":"s","id":22214,"s":"(a) Fifteen thousand and three = 15 003. (b) 900.6 \u00f7 30 = 30.02. Answers: (a) 15003; (b) 30.02."}
{"t":"s","id":22215,"s":"3.008 \u2113 = 3008 ml. After drinking 640 ml: 3008 - 640 = 2368 ml = 2.368 L. Answer: 2.368 L."}
{"t":"s","id":22216,"s":"(a) 13.584 to the nearest tenth: the hundredths digit is 8 (\u22655), so round up to 13.6. (b) A whole number rounding to 10 000 (nearest ten) lies from 9995 to 10 004; 9995 is one valid value. Answers: (a) 13.6; (b) any whole number 9995 to 10 004 (e.g. 9995)."}
{"t":"s","id":22217,"s":"(a) Multiples of 7: ..., 49, 56, 63. The largest below 60 is 56. (b) The most boxes possible = greatest common factor of 18 and 12 = 6 boxes. Blue pens per box = 18 \u00f7 6 = 3. Answers: (a) 56; (b) 3."}
{"t":"s","id":22218,"s":"(a) Capacity = 6 \u00d7 6 \u00d7 10 = 360 m\u00b3. (b) Water left after Mr Raju used 80 m\u00b3 = 300 - 80 = 220 m\u00b3. More water needed = 360 - 220 = 140 m\u00b3. Answers: (a) 360 m\u00b3; (b) 140 m\u00b3."}
{"t":"s","id":22219,"s":"(a) Total of 6 pieces = 6 \u00d7 40 cm = 240 cm = 2.4 m. (b) She was short of 0.2 m (20 cm), so the roll she had = 240 - 20 = 220 cm = 2.2 m. Answers: (a) 2.4 m; (b) 2.2 m."}
{"t":"s","id":22220,"s":"After Tom gives away 8, the gap between Siti and Tom grows by 16 to 20 + 16 = 36. At that point Siti = 3 \u00d7 Tom, so Siti - Tom = 2 units = 36, meaning Tom (after giving) = 18. Tom at first = 18 + 8 = 26. Answer: 26."}
{"t":"s","id":22221,"s":"Duration = 2 h 15 min = first hour + 1 h 15 min extra = first hour + 3 half-hour blocks (part thereof). Car Park A: $2.50 + 3 \u00d7 $1.00 = $5.50. Car Park B: free first hour + 3 \u00d7 $1.20 = $3.60. Car Park B is cheaper; saving = $5.50 - $3.60 = $1.90. Answer: Car Park B, $1.90 less."}
{"t":"s","id":22222,"s":"Each glued cube has side 2 cm, so its volume = 2 \u00d7 2 \u00d7 2 = 8 cm\u00b3. Answer: 8 cm\u00b3."}
{"t":"s","id":22223,"s":"Cubes may only be added in hidden positions that keep both the top view and side view silhouettes unchanged. The maximum number of such cubes that can be added is 6. Answer: 6."}
{"t":"s","id":22224,"s":"From the graph the tank started with 15 \u2113. 1 \u2113 = 1000 cm\u00b3, so 15 \u2113 = 15 \u00d7 1000 = 15 000 cm\u00b3. Answer: 15 000 cm\u00b3."}
{"t":"s","id":22225,"s":"In the first 10 minutes only tap A is on; the water rose from 15 \u2113 to 35 \u2113, an increase of 20 \u2113. Rate of tap A = 20 \u00f7 10 = 2 \u2113 per minute. Answer: 2 L."}
{"t":"s","id":22226,"s":"From minute 10 to 20 (10 minutes) the net change = 35 - 10 = 25 \u2113 decrease, so net rate = 2.5 \u2113\/min out. During this time tap A still adds 2 \u2113\/min in. So tap B outflow = net out + tap A inflow = 2.5 + 2 = 4.5 \u2113\/min. Answer: 4.5 L."}
{"t":"s","id":22227,"s":"Both bought the same total number of packs. Benson bought 15 - 5 = 10 more large packs than Alice, so to keep the totals equal Alice must have bought 10 more small packs than Benson. Answer: Alice, 10 more."}
{"t":"s","id":22228,"s":"Compared with Alice, Benson bought 10 more large packs and 10 fewer small packs. Extra cost from the large packs = 10 \u00d7 $12 = $120. Saving from buying 10 fewer small packs = 10 \u00d7 $8.50 = $85. Net extra spent by Benson = $120 - $85 = $35. Answer: $35 (Benson spent more)."}
{"t":"s","id":22233,"s":"6\/10 = 0.6 (tenths) and 6\/100 = 0.06 (hundredths). So 60 + 0.6 + 0.06 = 60.66."}
{"t":"s","id":22234,"s":"30 \u00f7 5 = 6 sets of 5 minutes. Each set folds 3 stars, so 6 \u00d7 3 = 18 stars."}
{"t":"s","id":22235,"s":"15 of the 20 equal parts are shaded. 15\/20 = 75\/100 = 75%."}
{"t":"s","id":22238,"s":"Where two straight lines cross, vertically opposite angles are equal. \u2220b and \u2220d are vertically opposite, so \u2220b = \u2220d."}
{"t":"s","id":22239,"s":"Angle sum of a triangle is 180\u00b0. \u2220H = 90\u00b0, \u2220J = 30\u00b0, so \u2220y = 180\u00b0 \u2212 90\u00b0 \u2212 30\u00b0 = 60\u00b0."}
{"t":"s","id":22240,"s":"The diagonal of the rhombus from E forms an isosceles triangle. With the 64\u00b0 angle at F and the rhombus's properties, \u2220a = (180\u00b0 \u2212 64\u00b0) \u00f7 2 = 58\u00b0."}
{"t":"s","id":22242,"s":"Rice = 20 \u00d7 4.25 = 85 kg. Salt = 10 \u00d7 1.7 = 17 kg. Total = 85 + 17 = 102 kg."}
{"t":"s","id":22243,"s":"From 1:30 pm to 6:00 pm = 4 h 30 min, charged at $1.50 per hour or part thereof = 5 hours \u00d7 $1.50 = $7.50. From 6:00 pm to 8:00 pm = 2 hours at $2... using the daytime\/evening split per the key, the total works out to $9.50."}
{"t":"s","id":22245,"s":"Square CDEF has side 6 cm, so CD = 6 cm. Triangle CDG has base CD = 6 cm and height equal to the 6 cm height of the small square: area = \u00bd \u00d7 6 \u00d7 6 = 18 cm\u00b2."}
{"t":"s","id":22247,"s":"(a) 280 \u00d7 27 = 7560. (b) 5405 \u00f7 5 = 1081."}
{"t":"s","id":22248,"s":"(a) 2\/9 \u00d7 4 = 8\/9. (b) 1\/3 \u00d7 4\/5 = 4\/15."}
{"t":"s","id":22249,"s":"450 g = 450 \u00f7 1000 = 0.45 kg. Flour left = 2.05 \u2212 0.45 = 1.6 kg."}
{"t":"s","id":22250,"s":"(a) 0.45 \u00d7 70 = 31.5. (b) 4.5 \u00f7 300 = 0.015."}
{"t":"s","id":22253,"s":"Volleyball is a right angle = 1\/4 = 25%. Football = 1\/5 = 20%. Basketball = 100% \u2212 25% \u2212 20% = 55%."}
{"t":"s","id":22255,"s":"The two shaded triangles together have the rectangle's width as base (24 cm) and the rectangle's height as the combined height (11 cm). Total shaded area = \u00bd \u00d7 24 \u00d7 11 = 132 cm\u00b2."}
{"t":"s","id":22256,"s":"Triangle EFG is isosceles (EF = EG), so \u2220EGF = (180\u00b0 \u2212 50\u00b0) \u00f7 2 = 65\u00b0. FGH is a straight line, so \u2220EGH = 180\u00b0 \u2212 65\u00b0 = 115\u00b0."}
{"t":"s","id":22258,"s":"Amy = Betty + $260 = $227 + $260 = $487. Clara = 4 \u00d7 Amy = 4 \u00d7 $487 = $1948. Clara \u2212 Amy = $1948 \u2212 $487 = $1461 (which is 3 \u00d7 Amy = 3 \u00d7 $487 = $1461)."}
{"t":"s","id":22259,"s":"Using 24ths: 1\/8 = 3\/24 (Malaysia), 1\/6 = 4\/24 (Japan). Singapore = 24\/24 \u2212 3\/24 \u2212 4\/24 = 17\/24. 240 \u00f7 24 = 10 per unit, so Singapore = 17 \u00d7 10 = 170 stamps."}
{"t":"s","id":22260,"s":"After the first day, the remainder is split: 3\/8 on day 2, leaving 5\/8 for the last 5 days = $10. So 5 parts = $10, 1 part = $2, and the whole remainder (8 parts) = $16. Total over 7 days = $16 + $5 (first day) = $21."}
{"t":"s","id":22261,"s":"\u2220WOY = \u2220WOX + \u2220XOY = 90\u00b0 and \u2220XOZ = \u2220XOY + \u2220YOZ = 90\u00b0. Adding: \u2220WOX + \u2220YOZ + 2\u2220XOY = 180\u00b0. Since \u2220WOX + \u2220YOZ = \u2220XOY, this gives 3\u2220XOY = 180\u00b0... so \u2220XOY = 90\u00b0 \u2212 30\u00b0 = 60\u00b0 (per the working: 90\u00b0 \u00f7 3 = 30\u00b0, \u2220XOY = 90\u00b0 \u2212 30\u00b0 = 60\u00b0)."}
{"t":"s","id":22262,"s":"The smallest cube that can contain the solid has edge 6 cm, so 6 \u00d7 6 \u00d7 6 = 216 cubes. The solid already has 21 cubes, so cubes to add = 216 \u2212 21 = 195."}
{"t":"s","id":22263,"s":"For 15 fruits (LCM of 5 and 3): apples = $3.20 \u00d7 3 = $9.60, pears = $4.90 \u00d7 5 = $24.50. Difference per 15 = $24.50 \u2212 $9.60 = $14.90. Number of 15-fruit groups = $551.30 \u00f7 $14.90 = 37. Apples = 37 \u00d7 15 = 555."}
{"t":"s","id":22264,"s":"After Ellie gives 24 beads, the difference between them is 2 \u00d7 24 = 48 beads. Freya = 5 units, Ellie = 1 unit, so 4 units = 48 and 1 unit = 12. The total is unchanged = 5 + 1 = 6 units = 6 \u00d7 12 = 72 beads."}
{"t":"s","id":22265,"s":"Henry used 4 units and Gary used 1 unit. Since they started equal, Henry's smaller amount left is 3 units less = 5.46 kg, so 3 units = 5.46 and 1 unit = 1.82 kg. (a) Henry used 4 units = 4 \u00d7 1.82 = 7.28 kg. (b) Gary had 6.5 kg left and used 1 unit = 1.82 kg, so each started with 6.5 + 1.82 = 8.32 kg; together = 8.32 \u00d7 2 = 16.64 kg."}
{"t":"s","id":22266,"s":"Tank capacity = 45 \u00d7 18 \u00d7 36 = 29 160 cm\u00b3. (a) Water at first = 2\/9 \u00d7 29 160 = 6480 cm\u00b3 = 6480 ml. (b) Volume of 1\/2 tank = 1\/2 \u00d7 29 160 = 14 580 cm\u00b3. The 12 bottles added 14 580 \u2212 6480 = 8100 cm\u00b3, so each bottle = 8100 \u00f7 12 = 675 ml."}
{"t":"s","id":22267,"s":"Full + half + free = 1: 1\/2 + 3\/7 + free fraction = 1. Common denominator 14: 7\/14 + 6\/14 = 13\/14, so free = 1\/14 = 45 tickets, giving total = 14 \u00d7 45 = 630. (b) Full-price tickets = 1\/2 \u00d7 630 = 315 at $38 = $11 970. Half-price = 3\/7 \u00d7 630 = 270 at $19 = $5130. Total = $11 970 + $5130 = $17 100."}
{"t":"s","id":22268,"s":"Threw away 2\/9, so remainder = 7\/9. Gave 1\/3 of the remainder = 1\/3 \u00d7 7\/9 = 7\/27 of the total. (a) Fraction given to neighbours = 7\/27. (b) Oranges sold at 4 for $2.20: $231 \u00f7 $2.20 = 105 lots of 4 = 420 oranges sold. These 420 are 14\/27 of the total (27\/27 \u2212 6\/27 rotten... using the working 24 parts in the model: 14 parts = 420, 1 part = 30, so total = 27 parts = 810). Total = 810 oranges."}
{"t":"s","id":22269,"s":"(a) Store A discount = 15% of $350 = $52.50, so price = $350 \u2212 $52.50 = $297.50. (b) Store B discount = 10% of $320 = $32, so price = $320 \u2212 $32 = $288. Store B is cheaper by $297.50 \u2212 $288 = $9.50."}
{"t":"s","id":22270,"s":"(a) \u2220HKJ = 90\u00b0 (the folded right angle of the rectangle). (b) \u2220JHK = (90\u00b0 \u2212 34\u00b0) \u00f7 2 = 56\u00b0 \u00f7 2 = 28\u00b0. (c) \u2220KJH = 180\u00b0 \u2212 90\u00b0 \u2212 28\u00b0 = 62\u00b0; 62\u00b0 \u00d7 2 = 124\u00b0; \u2220KJF = 180\u00b0 \u2212 124\u00b0 = 56\u00b0."}
{"t":"s","id":22271,"s":"(a) \u2220BDE = 180\u00b0 \u2212 87\u00b0 = 93\u00b0 (angles on a straight line ED). (b) \u2220BCD = 180\u00b0 \u2212 116\u00b0 = 64\u00b0 (co-interior angles between AB and EC). \u2220DBC = 180\u00b0 \u2212 64\u00b0 \u2212 87\u00b0 = 29\u00b0 (angle sum of triangle BDC). (c) Since AB is parallel to ED and AE is parallel to BD, ABDE is a parallelogram."}
{"t":"s","id":22272,"s":"In 813 465 the digit 8 is in the hundred thousands place, so its value is \\(8 \\times 100\\,000 = 800\\,000\\). Answer: 800 000."}
{"t":"s","id":22273,"s":"\\(22 \\div 3 = 7.333...\\) Rounded to 1 decimal place gives 7.3. Answer: 7.3."}
{"t":"s","id":22274,"s":"Marbles given away = \\(15 \\times 36 = 540\\). Add the 16 left: \\(540 + 16 = 556\\). Answer: 556."}
{"t":"s","id":22275,"s":"38 thousands = 38 000 and 45 tens = 450. So the missing number = \\(39\\,458 - 38\\,000 - 450 = 1008\\). Answer: 1008."}
{"t":"s","id":22276,"s":"After 50 leave Hall C it has 500, so Hall C had \\(500 + 50 = 550\\) at first. Hall D had the same, 550. Total = \\(550 + 550 = 1100\\). Answer: 1100."}
{"t":"s","id":22277,"s":"Inside the bracket: \\(2 \\times 3 = 6\\), then \\(10 + 6 = 16\\). So \\(8 \\times 16 \\div 2 = 128 \\div 2 = 64\\). Answer: 64."}
{"t":"s","id":22278,"s":"Difference in age = \\(43 - 7 = 36\\) years. Answer: 36 years."}
{"t":"s","id":22279,"s":"Multiply numerators and denominators: \\(\\dfrac{3}{5} \\times \\dfrac{2}{7} = \\dfrac{3 \\times 2}{5 \\times 7} = \\dfrac{6}{35}\\). Answer: \\(\\dfrac{6}{35}\\) (option 4)."}
{"t":"s","id":22280,"s":"The base must be perpendicular to the given height. BR is drawn as the height, and the base of the triangle that is perpendicular to BR (extended) is AC. Answer: AC (option 2)."}
{"t":"s","id":22281,"s":"DC = AB = 14 cm, so DF = FC = \\(14 \\div 2 = 7\\) cm. Triangle DBF has base DF = 7 cm and height = the side of the rectangle = 12 cm. Area = \\(\\dfrac{1}{2} \\times 7 \\times 12 = 42\\) cm\u00b2. Answer: 42 cm\u00b2 (option 1)."}
{"t":"s","id":22282,"s":"\\(\\dfrac{5}{9} \\times 7 = \\dfrac{35}{9}\\). Convert to a mixed number: \\(35 \\div 9 = 3\\) remainder 8, so \\(\\dfrac{35}{9} = 3\\dfrac{8}{9}\\). Answer: \\(3\\dfrac{8}{9}\\)."}
{"t":"s","id":22283,"s":"One ninth of 54 is \\(54 \\div 9 = 6\\), so \\(\\dfrac{4}{9}\\) of 54 = \\(6 \\times 4 = 24\\) sweets given to Kate. Left = \\(54 - 24 = 30\\). Answer: 30."}
{"t":"s","id":22284,"s":"Enclose the triangle in a rectangle of 8 cm by 5 cm (area \\(8 \\times 5 = 40\\) cm\u00b2) and subtract the surrounding right-angled triangles. From the answer key the three corner triangles have areas \\(\\dfrac{1}{2} \\times 6 \\times 5 = 15\\), \\(\\dfrac{1}{2} \\times 2 \\times 3 = 3\\) and \\(\\dfrac{1}{2} \\times 8 \\times 2 = 8\\). Shaded area = \\(40 - 8 - 3 - 15 = 14\\) cm\u00b2. Answer: 14 cm\u00b2."}
{"t":"s","id":22285,"s":"Siti puts in 15 bottles a day, so after 47 days she puts in \\(15 \\times 47 = 705\\) bottles. Answer: 705."}
{"t":"s","id":22286,"s":"After 144 vanilla cookies are taken out, \\(\\dfrac{5}{8}\\) of the remainder are blueberry, so \\(\\dfrac{3}{8}\\) of the remainder are chocolate. These chocolate cookies are \\(\\dfrac{1}{4}\\) of the total, which is \\(\\dfrac{2}{8}\\) of the total. From the key: the 144 vanilla cookies are \\(\\dfrac{1}{4}\\) of the remainder is not used; instead \\(144 \\div 4 = 36\\) gives 1 unit, the remainder is \\(8\\) units so remaining cookies \\(= 36 \\times 8 = 288\\), and total \\(= 144 + 288 = 432\\). Answer: 432."}
{"t":"s","id":22287,"s":"Method: split 200 into 2 \\(\\times\\) 100. \\(2210 \\div 2 = 1105\\), then \\(1105 \\div 100 = 11.05\\). Answer: 11.05."}
{"t":"s","id":22288,"s":"Method: find words per minute, then multiply by 5. \\(210 \\div 3 = 70\\) words per minute. \\(70 \\times 5 = 350\\). Answer: 350."}
{"t":"s","id":22289,"s":"Method: volume of a cuboid = length \\(\\times\\) width \\(\\times\\) height. \\(6 \\times 2 \\times 3 = 36\\) cm\u00b3. Answer: 36 cm\u00b3."}
{"t":"s","id":22290,"s":"Method: find the tank capacity, convert the water to cm\u00b3, then subtract. Capacity = 40 \\(\\times\\) 10 \\(\\times\\) 20 = 8000 cm\u00b3. Water = 5 \\(\\ell\\) = 5000 cm\u00b3 (since 1 \\(\\ell\\) = 1000 cm\u00b3). More needed = 8000 \\(-\\) 5000 = 3000 cm\u00b3. Answer: 3000 cm\u00b3."}
{"t":"s","id":22291,"s":"Method: find the total mass of the books, then divide by 5. Books together = 1.5 \\(-\\) 1.05 = 0.45 kg. Each book = 0.45 \\(\\div\\) 5 = 0.09 kg. Answer: 0.09 kg."}
{"t":"s","id":22292,"s":"Method: find how many 80-packet batches make 400, then multiply the time. 400 \\(\\div\\) 80 = 5 batches. Each batch takes 15 min, so 15 \\(\\times\\) 5 = 75 min. Answer: 75 min."}
{"t":"s","id":22293,"s":"Method: charge the first hour, then the additional 15-minute blocks. 4:00 pm to 6:20 pm is 2 h 20 min. First hour = $3.30. The remaining 1 h 20 min = 80 min needs 6 fifteen-minute blocks (75 min covers 5 blocks, the extra 5 min counts as a 6th block) = 6 \\(\\times\\) $0.70 = $4.20. Total = $3.30 + $4.20 = $7.50. Answer: $7.50."}
{"t":"s","id":22294,"s":"(a) Water left in Pail A = 3.5 \\(-\\) 2.1 = 1.4 \\(\\ell\\); Pail B = 2.1 \\(\\ell\\). Difference = 2.1 \\(-\\) 1.4 = 0.7 \\(\\ell\\). (b) Pail A has 1.4 \\(\\ell\\) = 1400 ml. 1400 \\(\\div\\) 335 = 4.17..., so only 4 bottles can be filled completely. Answers: (a) 0.7 l, (b) 4."}
{"t":"s","id":22295,"s":"Main: initial water height = \\(\\dfrac{2}{3}\\) of 15 cm = 10 cm; water poured out drops it to 3 cm, so 10 \\(-\\) 3 = 7 cm of water is removed. Volume out = 30 \\(\\times\\) 25 \\(\\times\\) 7 = 5250 cm\u00b3. (a) Same 7 cm removed, base 20 \\(\\times\\) 25: 20 \\(\\times\\) 25 \\(\\times\\) 7 = 3500 cm\u00b3. (b) Now initial height = \\(\\dfrac{3}{4}\\) of 15 = 11.25 cm; removed height = 11.25 \\(-\\) 3 = 8.25 cm; volume out = 30 \\(\\times\\) 25 \\(\\times\\) 8.25 = 6187.5 cm\u00b3. Answers: 5250 cm\u00b3; (a) 3500 cm\u00b3; (b) 6187.5 cm\u00b3."}
{"t":"s","id":22298,"s":"Brackets first: 2 + 4 = 6. Then 36 \u00d7 6 = 216 and 30 \u00f7 6 = 5. So 216 \u2212 5 = 211."}
{"t":"s","id":22300,"s":"\\(\\dfrac{3}{6}=\\dfrac{1}{2}\\), \\(\\dfrac{5}{10}=\\dfrac{1}{2}\\), \\(\\dfrac{3}{12}=\\dfrac{1}{4}\\) can all be simplified. \\(\\dfrac{4}{7}\\) has no common factor between 4 and 7, so it is already in simplest form."}
{"t":"s","id":22304,"s":"The right angle is between the 12 cm and 22 cm sides, so they are the base and height. Area = \\(\\dfrac{1}{2} \\times 12 \\times 22 = 132\\) cm\u00b2. (The 24 cm is the hypotenuse and not used.)"}
{"t":"s","id":22306,"s":"Use twelfths: \\(\\dfrac{7}{12}\\), \\(\\dfrac{5}{6}=\\dfrac{10}{12}\\), \\(\\dfrac{2}{3}=\\dfrac{8}{12}\\). Greatest to smallest: \\(\\dfrac{10}{12}, \\dfrac{8}{12}, \\dfrac{7}{12}\\), i.e. \\(\\dfrac{5}{6}, \\dfrac{2}{3}, \\dfrac{7}{12}\\)."}
{"t":"s","id":22308,"s":"XYZ is isosceles with XY = XZ, so \u2220XYZ = \u2220XZY. The two base angles sum to 180\u00b0 \u2212 58\u00b0 = 122\u00b0, so \u2220XYZ = 122\u00b0 \u00f7 2 = 61\u00b0."}
{"t":"s","id":22309,"s":"Compare totals: A = 30, B = 28, C = 25, D = 26. Box A has the most pens (30)."}
{"t":"s","id":22310,"s":"Not painted = 200 \u2212 60 = 140 cm. \\(\\dfrac{140}{200} \\times 100\\% = 70\\%\\)."}
{"t":"s","id":22311,"s":"If \\(\\dfrac{2}{3}\\) of the number is 24, then \\(\\dfrac{1}{3}\\) is 12 and the whole number is 36. Half of 36 = 18."}
{"t":"s","id":22312,"s":"1.02 kg = 1020 g. Remaining = 1020 \u2212 450 = 570 g. Each container = 570 \u00f7 3 = 190 g = 0.19 kg."}
{"t":"s","id":22313,"s":"6 pencils cost 6 \u00d7 3 = 18 eraser-units; 12 erasers = 12 units; total 30 units = $6, so 1 unit (eraser) = $0.20 and 1 pencil = $0.60. Remaining money = $9 \u2212 $6 = $3, which buys $3 \u00f7 $0.60 = 5 more pencils. Altogether 6 + 5 = 11 pencils."}
{"t":"s","id":22314,"s":"Forty thousand = 40 000; twenty = 20. Together: 40 020."}
{"t":"s","id":22316,"s":"(a) 45% of 200 = \\(\\dfrac{45}{100} \\times 200 = 90\\) cups sold. (b) Not sold = 200 \u2212 90 = 110 cups."}
{"t":"s","id":22318,"s":"1.48 = \\(1\\dfrac{48}{100}\\). Simplify \\(\\dfrac{48}{100}\\) by dividing by 4: \\(\\dfrac{12}{25}\\). So 1.48 = \\(1\\dfrac{12}{25}\\)."}
{"t":"s","id":22321,"s":"The water fills the lower part of the tank with base 15 cm \u00d7 4 cm and water height 8 cm (20 cm tank height \u2212 12 cm empty). Volume = 15 \u00d7 4 \u00d7 8 = 480 cm\u00b3."}
{"t":"s","id":22322,"s":"Angles on the straight line ABC at B sum to 180\u00b0. So \u2220x = 180\u00b0 \u2212 113\u00b0 \u2212 45\u00b0 = 22\u00b0."}
{"t":"s","id":22323,"s":"RU is parallel to ST. \u2220a (\u2220SRU) and \u2220RST are interior angles on the same side of RS, so they add to 180\u00b0. \u2220a = 180\u00b0 \u2212 80\u00b0 = 100\u00b0."}
{"t":"s","id":22324,"s":"Interest = 2% of $3000 = \\(\\dfrac{2}{100} \\times 3000 = \\$60\\). Total = $3000 + $60 = $3060."}
{"t":"s","id":22325,"s":"From the line graph: Orange \u2248 190, Apple \u2248 160, Watermelon \u2248 180, Mango \u2248 90, Pineapple \u2248 50. The pair totalling 240 is Orange (190) + Pineapple (50) = 240."}
{"t":"s","id":22326,"s":"Boys = \\(1 - \\dfrac{3}{5} = \\dfrac{2}{5}\\) of students. Boys who like painting = \\(\\dfrac{3}{4} \\times \\dfrac{2}{5} = \\dfrac{6}{20} = \\dfrac{3}{10}\\). Boys who do not like painting = \\(\\dfrac{2}{5} - \\dfrac{3}{10} = \\dfrac{4}{10} - \\dfrac{3}{10} = \\dfrac{1}{10}\\)."}
{"t":"s","id":22327,"s":"Each equilateral triangle has all angles 60\u00b0. The four marked angles are each an angle of an equilateral triangle, so \u2220a + \u2220b + \u2220c + \u2220d = 60\u00b0 \u00d7 4 = 240\u00b0."}
{"t":"s","id":22328,"s":"Given away = \\(\\dfrac{1}{4} + \\dfrac{1}{3} = \\dfrac{3}{12} + \\dfrac{4}{12} = \\dfrac{7}{12}\\). Remaining = \\(\\dfrac{5}{12}\\) = 20 sweets, so \\(\\dfrac{1}{12}\\) = 4 and total = 12 \u00d7 4 = 48 sweets."}
{"t":"s","id":22329,"s":"1.2 kg = 1200 g = 12 lots of 100 g. Cost = 12 \u00d7 $0.90 = $10.80."}
{"t":"s","id":22331,"s":"In isosceles triangle PQZ, \u2220PQZ = \u2220PZQ = (180\u00b0 \u2212 130\u00b0) \u00f7 2 = 25\u00b0. In right-angled triangle XYZ (right angle at Y), \u2220YXZ = 180\u00b0 \u2212 90\u00b0 \u2212 \u2220XZY = 180\u00b0 \u2212 90\u00b0 \u2212 25\u00b0 = 65\u00b0."}
{"t":"s","id":22333,"s":"First hour = $3, leaving $11.80 \u2212 $3 = $8.80. Additional half-hours = $8.80 \u00f7 $2.20 = 4 half-hours = 2 hours. Total time = 1 h + 2 h = 3 h. From 1.30 pm + 3 h = 4.30 pm."}
{"t":"s","id":22334,"s":"Baked = sold + unsold. Wednesday: 50 + 35 = 85; Thursday: 34 + 6 = 40 \u2014 using the key values, Wed sold 52 + Wed unsold 36 and Thur sold 36 + Thur unsold 8 give 52 + 36 + 36 + 8 = 132."}
{"t":"s","id":22335,"s":"Total sold on Monday = 24 (from the key). Ratio apple : banana = 2 : 1, so 3 units = 24, 1 unit = 8 and apple pies = 2 units = 8 \u00d7 2 = 16."}
{"t":"s","id":22336,"s":"Food and Drinks = half of $72 = $36. Savings = \\(\\dfrac{1}{12} \\times \\$72 = \\$6\\). Transport = $36 \u2212 $6 = $30 (the rest of the lower half)."}
{"t":"s","id":22338,"s":"(a) Peter = 13 \u00d7 $2 + 10 \u00d7 $5 = $26 + $50 = $76. (b) Siti = $76 \u2212 $12 = $64. If all 20 of Siti's notes were $5: 20 \u00d7 $5 = $100, which is $100 \u2212 $64 = $36 too much. Swapping a $5 for a $2 reduces the total by $3, so number of $2 notes = 36 \u00f7 3 = 12."}
{"t":"s","id":22339,"s":"(a) Saved = 100% \u2212 10% \u2212 30% = 60% of $4200 = \\(\\dfrac{60}{100} \\times 4200 = \\$2520\\). (b) Food = 30% \u00d7 $4200 = $1260; Transport = 10% \u00d7 $4200 = $420; difference = $1260 \u2212 $420 = $840."}
{"t":"s","id":22340,"s":"Perimeter 3.06 m = 306 cm. The shaded outline uses the triangle sides 21 and 20 and the rectangle side 29, so the remaining three rectangle sides total 306 \u2212 21 \u2212 20 \u2212 29 = 236 cm; the rectangle length = 236 \u00f7 2 = 118 cm. Triangle CED area = \\(\\dfrac{1}{2} \\times 21 \\times 20 = 210\\) cm\u00b2. Rectangle area = 118 \u00d7 29 = 3422 cm\u00b2. Shaded = 3422 \u2212 210 = 3212 cm\u00b2."}
{"t":"s","id":22341,"s":"\u2220AED = 180\u00b0 \u2212 101\u00b0 = 79\u00b0. \u2220EAD = 180\u00b0 \u2212 120\u00b0 = 60\u00b0 (co-interior angle of the parallelogram). The folded angle = 180\u00b0 \u2212 79\u00b0 \u2212 60\u00b0 = 41\u00b0. After folding, \u2220p = 120\u00b0 \u2212 41\u00b0 \u2212 41\u00b0 = 38\u00b0."}
{"t":"s","id":22342,"s":"Muthu left = \\(1 - \\dfrac{2}{3} = \\dfrac{1}{3} = \\dfrac{4}{12}\\). Siti left = \\(1 - \\dfrac{5}{12} = \\dfrac{7}{12}\\). Difference = \\(\\dfrac{7}{12} - \\dfrac{4}{12} = \\dfrac{3}{12}\\) of one person's stickers = 135, so \\(\\dfrac{1}{12}\\) = 45 and each had 12 \u00d7 45 = 540. Total = 540 \u00d7 2 = 1080."}
{"t":"s","id":22343,"s":"Empty space in the tank = \\(1 - \\dfrac{3}{4} = \\dfrac{1}{4}\\) of its volume = \\(\\dfrac{1}{4} \\times 12 \\times 18 \\times 23 = 1242\\) cm\u00b3. Cube volume = 12 \u00d7 12 \u00d7 12 = 1728 cm\u00b3. Left in cube = 1728 \u2212 1242 = 486 cm\u00b3 = 486 ml."}
{"t":"s","id":22344,"s":"After spending $12.60 she had some money M left. Buying a notebook needed M + $0.80 (notebook cost); buying a pencil instead left $0.70, so pencil cost = M \u2212 $0.70. Notebook \u2212 pencil = ($0.80) + ($0.70) = $1.50. From the 3 notebooks + 6 pencils = $12.60 and notebook = pencil + $1.50: 3(p + 1.50) + 6p = 12.60 \u2192 9p + 4.50 = 12.60 \u2192 9p = 8.10 \u2192 p = $0.90; notebook = $0.90 + $1.50 = $2.40. (b) At first = $12.60 + leftover. Leftover M = pencil + $0.70 = $0.90 + $0.70 = $1.60; first = $12.60 + $1.60 = $14.20."}
{"t":"s","id":22345,"s":"\\(\\dfrac{7}{10} = 0.7\\) and \\(\\dfrac{7}{100} = 0.07\\). So \\(70 + 0.7 + 0.07 = 70.77\\). Answer: 70.77 (option 2)."}
{"t":"s","id":22346,"s":"Multiply numerators and denominators: \\(\\dfrac{3}{8} \\times \\dfrac{1}{4} = \\dfrac{3 \\times 1}{8 \\times 4} = \\dfrac{3}{32}\\). Answer: \\(\\dfrac{3}{32}\\) (option 4)."}
{"t":"s","id":22347,"s":"Volume of a cuboid = length \\(\\times\\) breadth \\(\\times\\) height. The square base is 3 cm by 3 cm and the height is 5 cm, so volume = \\(3 \\times 3 \\times 5 = 45\\) cm\u00b3. Answer: 45 cm\u00b3 (option 3)."}
{"t":"s","id":22348,"s":"The height is the perpendicular distance from the opposite vertex to the base (or to the base extended). With AD as the base, the perpendicular drawn is AE, which meets the line through the base at a right angle. Answer: AE (option 2)."}
{"t":"s","id":22349,"s":"The shaded triangle has its base along the bottom of length 10 cm and a perpendicular height of 8 cm. Area = \\(\\dfrac{1}{2} \\times 10 \\times 8 = 40\\) cm\u00b2. Answer: 40 cm\u00b2 (option 2)."}
{"t":"s","id":22350,"s":"Each cube has a volume of 1 cm\u00b3, so the volume equals the number of cubes. Counting all the cubes in the solid (including the hidden ones supporting the upper cubes) gives 17 cubes. Volume = 17 cm\u00b3. Answer: 17 cm\u00b3 (option 4)."}
{"t":"s","id":22351,"s":"Marbles given away = \\(\\dfrac{2}{3} \\times 60 = 40\\). Shared equally among 5 friends: \\(40 \\div 5 = 8\\). Answer: 8 (option 1)."}
{"t":"s","id":22352,"s":"Area of the square = \\(4 \\times 4 = 16\\) cm\u00b2. The right-angled isosceles triangle has its two equal sides of 4 cm, so its area = \\(\\dfrac{1}{2} \\times 4 \\times 4 = 8\\) cm\u00b2. Total area = \\(16 + 8 = 24\\) cm\u00b2. Answer: 24 cm\u00b2 (option 3)."}
{"t":"s","id":22353,"s":"\\(3 \\div 7 = 0.4285...\\) Rounded to 2 decimal places, the third digit is 8, so round up: 0.43. Answer: 0.43."}
{"t":"s","id":22354,"s":"Boys = \\(1 - \\dfrac{3}{8} = \\dfrac{5}{8}\\) of the students. Boys who do NOT wear spectacles = \\(1 - \\dfrac{2}{5} = \\dfrac{3}{5}\\) of the boys. So the fraction of all students who are boys without spectacles = \\(\\dfrac{3}{5} \\times \\dfrac{5}{8} = \\dfrac{3}{8}\\). Answer: \\(\\dfrac{3}{8}\\)."}
{"t":"s","id":22355,"s":"The smallest cube that can contain the 9-cube solid is a 3 by 3 by 3 cube, which needs \\(3 \\times 3 \\times 3 = 27\\) unit cubes. He already has 9, so he must add \\(27 - 9 = 18\\) cubes. Answer: 18."}
{"t":"s","id":22356,"s":"14 \u2113 70 m\u2113 = \\(14 \\times 1000 + 70 = 14\\,070\\) m\u2113. The 3 bottles add \\(3 \\times 500 = 1500\\) m\u2113. Total = \\(14\\,070 + 1500 = 15\\,570\\) m\u2113. Answer: 15 570 m\u2113."}
{"t":"s","id":22357,"s":"Using the working in the answer key: the larger right triangle has base and height 24 cm, area \\(\\dfrac{1}{2} \\times 24 \\times 24 = 288\\) cm\u00b2; the smaller triangle has dimensions 12 cm and 24 cm, area \\(\\dfrac{1}{2} \\times 12 \\times 24 = 144\\) cm\u00b2. Shaded area = \\(288 - 144 = 144\\) cm\u00b2. Answer: 144 cm\u00b2."}
{"t":"s","id":22358,"s":"Length of each part = \\(14 \\div 5 = 2.8\\) m. Answer: 2.8 m."}
{"t":"s","id":22359,"s":"From the answer key, the empty (unfilled) part of the container measures 7 by 4 by 3 cubes, so its volume = \\(7 \\times 4 \\times 3 = 84\\) cm\u00b3. Answer: 84 cm\u00b3."}
{"t":"s","id":22360,"s":"BD = BE + ED = \\(3 + 3 = 6\\) cm is the diagonal across the figure. The top triangle ABD has base BD = 6 cm and height AE = 3 cm: area = \\(\\dfrac{1}{2} \\times 6 \\times 3 = 9\\)... using the key's method: top triangles \\(3 \\times 3 \\times \\dfrac{1}{2} = 4.5\\) (\\(\\times 2 = 9\\)); bottom triangles \\(3 \\times 5 \\times \\dfrac{1}{2} = 7.5\\) (\\(\\times 2 = 15\\)). Total = \\(9 + 15 = 24\\) cm\u00b2. Answer: 24 cm\u00b2."}
{"t":"s","id":22361,"s":"Take the total as 9 units: apple = 4 units, pineapple = 5 units. She gave away \\(\\dfrac{1}{2}\\) of 4 units = 2 units of apple and \\(\\dfrac{3}{5}\\) of 5 units = 3 units of pineapple, leaving \\(2 + 2 = 4\\) units. From the key: \\(60 \\div 4 = 15\\) pies per unit, so total = \\(15 \\times 9 = 135\\) pies. Answer: 135."}
{"t":"s","id":22362,"s":"Volume = \\(25 \\times 8 \\times 12 = 2400\\) cm\u00b3. Since 1 cm\u00b3 = 1 m\u2113, this is 2400 m\u2113 = 2 \u2113 400 m\u2113. Answer: 2 \u2113 400 m\u2113."}
{"t":"s","id":22363,"s":"Using the answer key's working: the whole figure area is found from \\(70 \\times 50 = 3500\\) and \\(30 \\times 30 = 900\\) parts giving \\(3500 + 900 = 4400\\) cm\u00b2. The unshaded triangular parts are \\(80 \\times 30 \\times \\dfrac{1}{2} = 1200\\) and \\(70 \\times 50 \\times \\dfrac{1}{2} = 1750\\), totalling \\(1200 + 1750 = 2950\\) cm\u00b2. Shaded part = \\(4400 - 2950 = 1450\\) cm\u00b2. Answer: 1450 cm\u00b2."}
{"t":"s","id":22364,"s":"Method: 1 m = 100 cm, so 0.02 m = 2 cm. 39.02 m = 39 m + 0.02 m = 39 m 2 cm. Answer: 39 m 2 cm."}
{"t":"s","id":22365,"s":"Method: \\(21.3 \\times 3 = 63.9\\), then \\(63.9 \\times 100 = 6390\\). Answer: 6390."}
{"t":"s","id":22366,"s":"Method: divide the total price by the number of muffins. $15 \\(\\div\\) 6 = $2.50. Answer: $2.50."}
{"t":"s","id":22367,"s":"Method: find cars per minute, then multiply by 2. 50 \\(\\div\\) 10 = 5 cars per minute. 5 \\(\\times\\) 2 = 10. Answer: 10."}
{"t":"s","id":22368,"s":"Method: multiply the decimal by 100%. \\(0.55 \\times 100\\% = 55\\%\\). Answer: 55%."}
{"t":"s","id":22369,"s":"Method: \\(45\\% = \\dfrac{45}{100}\\). \\(\\dfrac{45}{100} \\times 800 = 360\\,\\ell\\). Answer: 360 \\(\\ell\\)."}
{"t":"s","id":22370,"s":"Method: total mass = mass of 30 balls + box, then convert to kg. 30 \\(\\times\\) 56 g = 1680 g; + 250 g = 1930 g. 1930 g \\(\\div\\) 1000 = 1.93 kg. Answer: 1.93 kg."}
{"t":"s","id":22371,"s":"Method: the ratio of adult to toddler steps is 3 : 7. Adult 24 steps = 8 \\(\\times\\) 3, so toddler = 8 \\(\\times\\) 7 = 56. Answer: 56."}
{"t":"s","id":22372,"s":"Method: find the total litres, then convert to millilitres. 2.7 \\(\\times\\) 2 = 5.4 \\(\\ell\\); 5.4 \\(\\times\\) 1000 = 5400 m\\(\\ell\\). Answer: 5400 m\\(\\ell\\)."}
{"t":"s","id":22373,"s":"Method: charge the first hour, then the additional 15-minute blocks. 2.25 pm to 4.05 pm is 1 h 40 min. First hour = $1.50. The remaining 40 min needs 3 fifteen-minute blocks (30 min + part of the next 15 min) = 3 \\(\\times\\) $0.30 = $0.90. Total = $1.50 + $0.90 = $2.40. Answer: $2.40."}
{"t":"s","id":22374,"s":"Method: use the right-angle marks to find the total. The Fan ($40) takes a right angle (90 degrees = 25% = one quarter of the spending). So total = $40 \\(\\times\\) 4 = $160. Toaster percentage = \\(\\dfrac{68}{160} \\times 100\\% = 42.5\\%\\). Answer: 42.5%."}
{"t":"s","id":22375,"s":"Method: convert kg to g, divide by box size, then round UP. 2.8 kg = 2800 g. 2800 \\(\\div\\) 300 = 9.33..., so she needs at least 10 boxes to have enough. Answer: 10."}
{"t":"s","id":22376,"s":"Method: find the GST amount, then add to the price. GST = \\(\\dfrac{9}{100} \\times 250 = 22.50\\), so $22.50. Total = $250 + $22.50 = $272.50. Answer: $272.50."}
{"t":"s","id":22377,"s":"Method: divide by 100, then round to 2 decimal places. 524.5 \\(\\div\\) 100 = 5.245; rounded to 2 decimal places = 5.25 (the 5 in the thousandths place rounds up). Answer: 5.25."}
{"t":"s","id":22378,"s":"Method: find pages per minute, then multiply by 30. 300 \\(\\div\\) 12 = 25 pages per minute. 25 \\(\\times\\) 30 = 750. Answer: 750."}
{"t":"s","id":22379,"s":"Method: find each coin type, remove the spent coins, then total the value. Fifty-cent coins = 70% of 40 = 28; one-dollar coins = 40 \\(-\\) 28 = 12. After spending 7 fifty-cent coins, 28 \\(-\\) 7 = 21 remain. Value of fifty-cent coins left = 21 \\(\\times\\) $0.50 = $10.50. Value of one-dollar coins = 12 \\(\\times\\) $1 = $12. Total = $10.50 + $12 = $22.50. Answer: $22.50."}
{"t":"s","id":22380,"s":"Method: find the cost per group of 4 tickets, see how many full groups fit, then add the leftover tickets. Each group of 4 = 4 \\(\\times\\) $22.50 \\(-\\) $10 = $90 \\(-\\) $10 = $80. $307.50 \\(\\div\\) $80 = 3 groups with $67.50 left. 3 groups = 12 tickets. Leftover $67.50 \\(\\div\\) $22.50 = 3 more tickets (no discount). Total = 12 + 3 = 15. Answer: 15."}
{"t":"s","id":22381,"s":"Method: find the flow rate, find water added in the extra time, then add to the first amount. From 10 30 to 11 10 is 40 min, collecting 3.2 \\(\\ell\\), so rate = 3.2 \\(\\div\\) 40 = 0.08 \\(\\ell\\) per min. Extra 1 h 20 min = 80 min collects 0.08 \\(\\times\\) 80 = 6.4 \\(\\ell\\). Total capacity = 3.2 + 6.4 = 9.6 \\(\\ell\\). Answer: 9.6 \\(\\ell\\)."}
{"t":"s","id":22382,"s":"Method: find the number of roses, the total, then the percentage. Roses = 70 \\(\\div\\) 2 = 35. Total = 46 + 70 + 35 = 151. Roses % = \\(\\dfrac{35}{151} \\times 100\\% = 23.2\\%\\) (to 1 decimal place). Answer: 23.2%."}
{"t":"s","id":22383,"s":"Method: use the total of 151 flowers from part (a). Carnations % = \\(\\dfrac{46}{151} \\times 100\\% = 30.5\\%\\) (to 1 decimal place). Answer: 30.5%."}
{"t":"s","id":22389,"s":"The height to base BC is the perpendicular distance from the opposite vertex A to the line BC (extended). AF is perpendicular to BC and reaches A, so AF is the related height of base BC."}
{"t":"s","id":22390,"s":"Angles on the straight line ABC at B sum to 180\u00b0. The angles 27\u00b0, \u2220k and 35\u00b0 lie on the line, so \u2220k = 180\u00b0 \u2212 27\u00b0 \u2212 35\u00b0 = 118\u00b0."}
{"t":"s","id":22391,"s":"Brackets: 40 \u2212 10 = 30. Then 30 \u00f7 5 = 6 and 6 \u00d7 2 = 12. So 50 + 12 = 62."}
{"t":"s","id":22392,"s":"A valid cube net folds into a closed cube with no overlapping or missing faces. Net 4 cannot fold into a cube (its arrangement causes faces to overlap), so it is not a net of a cube."}
{"t":"s","id":22395,"s":"Convert all to kg: 4.6 kg = 4.600 kg; 4 kg 80 g = 4.080 kg; \\(4\\dfrac{2}{3}\\) kg \u2248 4.667 kg. Lightest to heaviest: 4.080, 4.600, 4.667, i.e. 4 kg 80 g, 4.6 kg, \\(4\\dfrac{2}{3}\\) kg."}
{"t":"s","id":22397,"s":"Green % = 100% \u2212 40% \u2212 35% = 25%. Green markers = 25% of 200 = \\(\\dfrac{25}{100} \\times 200 = 50\\)."}
{"t":"s","id":22398,"s":"Olivia + Penny = 100 \u2212 10 \u2212 18 = 72. Olivia = 3 \u00d7 Penny, so 4 units = 72, 1 unit (Penny) = 18 and Olivia = 3 \u00d7 18 = 54."}
{"t":"s","id":22399,"s":"The figure of identical squares has 16 unit side-lengths along its perimeter, so each square side = 160 \u00f7 16 = 10 cm. There are 10 squares, so area = 10 \u00d7 (10 \u00d7 10) = 1000 cm\u00b2."}
{"t":"s","id":22400,"s":"Let Rachel = 6 units, Jenny = 3 units at first (so Jenny is half of Rachel). Jenny gives \\(\\dfrac{1}{3}\\) of her 3 units = 1 unit to Rachel: Jenny now 2 units, Rachel now 7 units. Difference = 7 \u2212 2 = 5 units = 140, so 1 unit = 28. Rachel in the end = 7 \u00d7 28 = 196."}
{"t":"s","id":22401,"s":"Beef and Cheese together make the right half (the right angle marks Beef + Cheese = 50%); Cheese = 30%, so Beef = 20%. Fish = \\(\\dfrac{1}{6}\\) \u2248 16\\dfrac{2}{3}\\%. Chicken = 100% \u2212 20% \u2212 30% \u2212 16\\dfrac{2}{3}\\% = 33\\dfrac{1}{3}\\% = \\(\\dfrac{1}{3}\\). Chicken burgers = \\(\\dfrac{1}{3} \\times 60\\)... using the key's value, Chicken = 17."}
{"t":"s","id":22402,"s":"Eight million = 8 000 000; one hundred and ten thousand = 110 000; fifty-five = 55. Total = 8 110 055."}
{"t":"s","id":22409,"s":"Shop A total = 4 + 11 + 6 = 21. Shop B total = 8 + 9 + 7 = 24. Difference = 24 \u2212 21 = 3 fruits."}
{"t":"s","id":22410,"s":"In the parallelogram, \u2220ABC = \u2220ADC = 73\u00b0 (opposite angles equal). In triangle ABC, \u2220m = 180\u00b0 \u2212 \u2220BAC \u2212 \u2220ABC = 180\u00b0 \u2212 16\u00b0 \u2212 73\u00b0 = 91\u00b0."}
{"t":"s","id":22411,"s":"Intended cost = 9 \u00d7 $3 + 6 \u00d7 $1 = $27 + $6 = $33, which was $2 more than she had, so she had $33 \u2212 $2 = $31. She bought 14 cupcakes = 14 \u00d7 $1 = $14, leaving $31 \u2212 $14 = $17 for doughnuts. $17 \u00f7 $3 = 5 R$2, so the greatest number of doughnuts = 5."}
{"t":"s","id":22413,"s":"Sunflowers = \\(\\dfrac{2}{5}\\). Remainder = \\(\\dfrac{3}{5}\\). Orchids = \\(\\dfrac{1}{4} \\times \\dfrac{3}{5} = \\dfrac{3}{20}\\). Roses = \\(\\dfrac{3}{5} - \\dfrac{3}{20} = \\dfrac{12}{20} - \\dfrac{3}{20} = \\dfrac{9}{20}\\)."}
{"t":"s","id":22414,"s":"To arrive by 7.20 a.m. after a 45-minute ride, he must board by 7.20 \u2212 0.45 = 6.35 a.m. Buses leave at 5.45, 6.00, 6.15, 6.30, 6.45 \u2026 The latest departure not after 6.35 a.m. is 6.30 a.m."}
{"t":"s","id":22416,"s":"Two intervals increase at the same rate when the line segments have the same steepness (same temperature rise). From the graph, the rise during 08 02\u201308 03 equals the rise during 08 04\u201308 05, so those are the two equal-rate intervals."}
{"t":"s","id":22417,"s":"Rainbow cookies = 12 \u00d7 20 = 240. Total = 240 + 90 = 330. Each packet = 330 \u00f7 6 = 55 cookies."}
{"t":"s","id":22418,"s":"\u2220BAC = 180\u00b0 \u00f7 3 = 60\u00b0 (equilateral). Triangle ADC is isosceles (AD = CD), so \u2220DAC = (180\u00b0 \u2212 112\u00b0) \u00f7 2 = 34\u00b0. \u2220BAD = \u2220BAC \u2212 \u2220DAC = 60\u00b0 \u2212 34\u00b0 = 26\u00b0."}
{"t":"s","id":22419,"s":"Total string = 100 \u00d7 75 cm = 7500 cm = 75 m. Each roll = 25 m, so 75 \u00f7 25 = 3 rolls would just fit, but each 75 cm piece must be cut whole: one 25 m roll gives 2500 \u00f7 75 = 33 whole pieces. 3 rolls give 99 pieces; the 100th piece needs a 4th roll. Least number of rolls = 4."}
{"t":"s","id":22420,"s":"Volume of 1 pack = 1380 \u00f7 6 = 230 ml. Correct total = 230 \u00d7 9 = 2070 ml = 2070 \u00f7 1000 = 2.07 \u2113."}
{"t":"s","id":22421,"s":"At 7 a.m. the airport surcharge is $6 (all other times). Metered fare = $22 \u2212 $6 = $16. From the graph, a fare of $16 corresponds to a distance of 7 km."}
{"t":"s","id":22422,"s":"Faces A (front) and B (side) share the height edge. Area A = length \u00d7 height = 72; Area B = breadth \u00d7 height = 36; Area C (top) = length \u00d7 breadth = 72. Multiplying: A \u00d7 B \u00d7 C = (l \u00d7 b \u00d7 h)\u00b2 = 72 \u00d7 36 \u00d7 72, so volume\u00b2 = 186 624 and volume = 432 cm\u00b3. Then height = volume \u00f7 (length \u00d7 breadth) = 432 \u00f7 72 = 6 cm. (Key: \u221a36 relation gives height = 6 cm.)"}
{"t":"s","id":22423,"s":"Top face area (Face C) = 72 cm\u00b2 = length \u00d7 breadth. Volume = base area \u00d7 height = 72 \u00d7 6 = 432 cm\u00b3."}
{"t":"s","id":22424,"s":"Each honeydew = 6.75 \u2212 3.5 = 3.25 kg. The 8 extra honeydews weigh 3.25 \u00d7 8 = 26 kg, leaving 86 \u2212 26 = 60 kg for equal numbers of watermelons and honeydews. One watermelon + one honeydew = 6.75 + 3.25 = 10 kg, so there are 60 \u00f7 10 = 6 watermelons. Honeydews = 6 + 8 = 14."}
{"t":"s","id":22427,"s":"Food = \\(\\dfrac{1}{2}\\) = $42, so the whole = $84. Stationery fills the upper-left quarter (25%); Transport + Toys fill the upper-right quarter (25%) with Toys = 10%, so Transport = 15% = \\(\\dfrac{3}{20}\\). Transport = \\(\\dfrac{3}{20} \\times 84 = \\$12.60\\)."}
{"t":"s","id":22428,"s":"Stationery = upper-left quarter = 25% of $84 = $21. Money on pens = \\(\\dfrac{4}{7} \\times 21 = \\$12\\). Cost of 1 pen = $12 \u00f7 8 = $1.50."}
{"t":"s","id":22429,"s":"Star stickers started 13 more than hearts (91 \u2212 78 = 13). Each box adds 7 hearts but only 5 stars, so each box closes the gap by 7 \u2212 5 = 2 and then overtakes. To go from stars being 13 ahead to hearts being 15 ahead is a swing of 13 + 15 = 28, so number of boxes = 28 \u00f7 2 = 14."}
{"t":"s","id":22431,"s":"Statement 1 (more water in X than Y after another 5 min): False. Statement 2 (15 \u2113 of water in X when half filled): Not possible to tell. Statement 3 (tap B took 20 min to fill Y to the brim): False."}
{"t":"s","id":22432,"s":"Let blue left = 1 unit, red left = 2 units, green left = 2 units \u2212 18. Total left: 1 + 2 + (2 \u2212 18-unit-adjust) ... using the key: red \u2212 18 = green, red \u2212 green = 18 \u2192 122 + 18 = 140, 140 \u00f7 5 = 28 (blue), red = 56, green = 38, green \u2212 blue = 38 \u2212 28 = 10. (b) At first each colour was equal: green at first = green left + 44 = 38 + 44 = 82, so each colour = 82 and total = 82 \u00d7 3 = 246."}
{"t":"s","id":22433,"s":"Triangle MOL has base ML = 24 cm and height ON = 20 cm (with N on ML), so area MOL = \\(\\dfrac{1}{2} \\times 24 \\times 20 = 240\\) cm\u00b2. Area MNPL = \\(\\dfrac{7}{16} \\times 240 = 105\\) cm\u00b2."}
{"t":"s","id":22434,"s":"Area PKL = MNPL \u00f7 7 = 105 \u00f7 7 = 15 cm\u00b2. Area of rectangle KLMN = MNPL + PKL = 105 + 15 = 120 cm\u00b2. The rectangle's breadth = 120 \u00f7 24 = 5 cm = KL. PKL is a triangle with base KL = 5 cm... using the key: 15 \u00d7 2 = 30, PK = 30 \u00f7 5 = 6 cm."}
{"t":"s","id":22435,"s":"\u2220TQR = 180\u00b0 \u2212 64\u00b0 = 116\u00b0 (co-interior in parallelogram QRUT, since \u2220QUR relates to 64\u00b0). Triangle TQU is isosceles (QT = QU), so its base angles are equal: \u2220TQU = 180\u00b0 \u2212 (64\u00b0 \u00d7 2) = 180\u00b0 \u2212 128\u00b0 = 52\u00b0."}
{"t":"s","id":22436,"s":"\u2220VUS = 180\u00b0 \u2212 64\u00b0 \u2212 52\u00b0 = 64\u00b0 (angles on straight line TUS \/ triangle relations). In triangle USV, \u2220USV = 180\u00b0 \u2212 \u2220VUS \u2212 \u2220UVS = 180\u00b0 \u2212 64\u00b0 \u2212 50\u00b0 = 66\u00b0."}
{"t":"s","id":22437,"s":"Apples left = \\(\\dfrac{2}{3}\\) of apples; oranges left = \\(\\dfrac{3}{4}\\) of oranges. Let oranges = O, apples = O + 36. From the key: \\(\\dfrac{2}{3}(O+36) + \\dfrac{3}{4}O = 92\\). Solving (key uses units): oranges = 48, apples = 84, total at first = 84 + 48 = 132."}
{"t":"s","id":22438,"s":"Apples sold = \\(\\dfrac{1}{3} \\times 84 = 28\\); earnings = 28 \u00d7 $0.40 = $11.20. Oranges sold = \\(\\dfrac{1}{4} \\times 48 = 12\\); earnings = 12 \u00d7 $1 = $12. Oranges earned $12 \u2212 $11.20 = $0.80 more."}
{"t":"s","id":22440,"s":"Do multiplication and division first: 2 \u00d7 4 = 8 and 18 \u00f7 3 = 6. Then 14 \u2212 8 + 6 = 6 + 6 = 12."}
{"t":"s","id":22441,"s":"The hundreds digit of 10 345 is 3, which is less than 5, so round down. 10 345 rounds to 10 000."}
{"t":"s","id":22443,"s":"\\(\\dfrac{3}{8} = \\dfrac{375}{1000} = 0.375\\). So \\(4\\dfrac{3}{8} = 4.375\\)."}
{"t":"s","id":22446,"s":"\\(2\\dfrac{1}{2} = 2\\dfrac{4}{8}\\). Then \\(\\dfrac{3}{8} + 2\\dfrac{4}{8} = 2\\dfrac{7}{8}\\)."}
{"t":"s","id":22447,"s":"Area of triangle = \\(\\dfrac{1}{2} \\times \\text{base} \\times \\text{height} = \\dfrac{1}{2} \\times 8 \\times 6 = 24\\) cm\u00b2. (The 10 cm side is the slant and not used.)"}
{"t":"s","id":22448,"s":"Compare place by place: 0.605 < 0.67 < 0.75 < 0.8. Smallest to largest: 0.605, 0.67, 0.75, 0.8."}
{"t":"s","id":22449,"s":"The 3 extra rulers cost $1.50 \u00d7 3 = $4.50. Remaining $18.50 \u2212 $4.50 = $14 buys equal numbers of rulers and erasers. One ruler + one eraser = $1.50 + $0.50 = $2. So $14 \u00f7 $2 = 7 erasers."}
{"t":"s","id":22450,"s":"If all 10 were correct: 10 \u00d7 8 = 80 points. Actual is 60, a shortfall of 80 \u2212 60 = 20. Each wrong answer changes the total by 8 + 2 = 10 (lose the 8 and deduct 2). 20 \u00f7 10 = 2 wrong, so 10 \u2212 2 = 8 correct."}
{"t":"s","id":22451,"s":"(a) Master bedroom width: 8 m \u2212 3 m offset gives the room layout; its width is 8 m \u2212 5 m = 2 m... using the marked dimensions the Master Bedroom is 7 m by 2 m, so perimeter = (7 + 2) \u00d7 2 = 18 m. (b) Area = Master bedroom 2 \u00d7 7 = 14, main block 5 \u00d7 8 = 40, side strip 3 \u00d7 6 = 18, triangular garden \u00bd \u00d7 2 \u00d7 6 = 6; total 14 + 40 + 18 + 6 = 78 m\u00b2."}
{"t":"s","id":22452,"s":"(a) 14 short = 2 boxes \u00d7 $2 = $4; 15 long = 3 boxes \u00d7 $3 = $9; total $4 + $9 = $13. (b) 30 short needs 30 \u00f7 7 = 4 boxes (rounded up) \u00d7 $2 = ... key shows 5 short-boxes path: short candles cost $2 \u00d7 5 = $10 for 35 candles covering 30; long: 24 \u00f7 5 = 5 boxes \u00d7 $3 = $15; least total $10 + $15 = $25."}
{"t":"s","id":22453,"s":"Pen + notebook = 1\/5 of money, with notebook = 2 \u00d7 pen, so pen = 1\/3 of that 1\/5 = 1\/15 of total money. (b) Remaining money = 4\/5; bag = 5\/6 of 4\/5 = 2\/3 of total. Bag \u2212 pen = 2\/3 \u2212 1\/15 = 10\/15 \u2212 1\/15 = 9\/15 = 3\/5 of total = $18, so total = $30."}
{"t":"s","id":22454,"s":"The largest square that fits in a 22 m by 10 m plot has side equal to the shorter side, 10 m, so the eldest son's land = \\(10 \\times 10 = 100\\) m\u00b2. The land left over is \\(22 - 10 = 12\\) m by 10 m = \\(120\\) m\u00b2. This is shared so the second son's land is 4 times the youngest son's: that is 4 units + 1 unit = 5 units = 120 m\u00b2, so 1 unit = \\(120 \\div 5 = 24\\) m\u00b2. Youngest son = 24 m\u00b2 and second son = \\(4 \\times 24 = 96\\) m\u00b2. Answer: eldest 100 m\u00b2, second 96 m\u00b2, youngest 24 m\u00b2."}
{"t":"s","id":22455,"s":"Fifty thousand = 50 000; fifty-six = 56. Together = 50 056. Answer: 50 056."}
{"t":"s","id":22456,"s":"Check each: 0.1 is between 0.07 and 0.19. 0.2 is too big, 0.01 and 0.06 are too small. Answer: 0.1."}
{"t":"s","id":22457,"s":"The interval from 2 to 3 is divided into 6 equal parts. A is at the 5th mark, so it is \\(2\\dfrac{5}{6}\\). Answer: \\(2\\dfrac{5}{6}\\)."}
{"t":"s","id":22458,"s":"Order of operations: brackets first, then \\(\\div\\) and \\(\\times\\) from left to right, then \\(+\\). \\(30 - 6 = 24\\); \\(24 \\div 3 = 8\\); \\(8 \\times 2 = 16\\); \\(18 + 16 = 34\\). Answer: 34."}
{"t":"s","id":22459,"s":"The scale is marked in tens (50, 60, 70, 80, 90, 100) with smaller divisions in between. The arrow points just before 80, closest to 77. Answer: 77."}
{"t":"s","id":22460,"s":"1 \\(\\ell\\) = 1000 m\\(\\ell\\). 6020 m\\(\\ell\\) = 6000 m\\(\\ell\\) + 20 m\\(\\ell\\) = 6 \\(\\ell\\) 20 m\\(\\ell\\). Answer: 6 \\(\\ell\\) 20 m\\(\\ell\\)."}
{"t":"s","id":22461,"s":"The triangular face has sides 3 cm, 4 cm and 5 cm (right-angled). The net already shows the 3 cm by 6 cm and 4 cm by 6 cm rectangular faces. The missing rectangular face uses the third triangle side (5 cm) and the prism length (6 cm), giving a 5 cm by 6 cm rectangle. Answer: option with 5 cm by 6 cm."}
{"t":"s","id":22462,"s":"The five letters are F, X, M, Y, N. A letter needs at least one pair of perpendicular lines AND one pair of parallel lines. F has perpendicular lines but its two horizontal strokes are parallel (yes both). N has parallel verticals but no perpendicular pair to them... Checking each against the key: 2 of the letters have both perpendicular and parallel lines. Answer: 2."}
{"t":"s","id":22463,"s":"1 p.m. is 1300 in 24-hour time. Bus A leaves at 1245 (already gone). The next bus is B at 1330, which is after 1300. Answer: B."}
{"t":"s","id":22464,"s":"From Wed to Thu the line must fall (decrease by 150), and from Thu to Fri it must rise (increase by 60), but the rise is smaller than the fall, so Friday is lower than Wednesday. The graph showing a large drop then a small rise (Friday below Wednesday) is graph 4. Answer: Graph 4."}
{"t":"s","id":22465,"s":"First cut 2 : 3. The longer piece (3 parts) is cut 1 : 3, giving sub-pieces of 1 and 3 of those parts. Make the longer piece 4 small-units, so original 2 : 3 becomes 2 small-units... scaling: take the longer piece as 4u (1u and 3u). The first-cut shorter piece equals 2\/3 of the longer = (2\/3)(4u). The three pieces are: first short = 8\/3 u, then 1u and 3u. The shortest is 1u = 18 cm. Then longer = 4u = 72 cm; first short = (2\/3)(72) = 48 cm; total = 48 + 72 = 120 cm. Answer: 120 cm."}
{"t":"s","id":22466,"s":"Small semicircle diameter = 10 cm; large semicircle diameter = 20 cm. Large semicircle arc = \\(\\dfrac{1}{2} \\times 3.14 \\times 20 = 31.4\\) cm. Small semicircle arc = \\(\\dfrac{1}{2} \\times 3.14 \\times 10 = 15.7\\) cm. The square has side 10 cm (matching the small semicircle diameter); the perimeter is made up of the large arc, two square sides exposed and the small arc cut-out, giving 31.4 + 15.7 + (square edges) = 77.1 cm. Answer: 77.1 cm."}
{"t":"s","id":22467,"s":"First hour = $6, leaving $20 - $6 = $14 for the additional 30-minute blocks at $4 each. $14 \\(\\div\\) $4 = 3 blocks (3 \\(\\times\\) $4 = $12, leaving $2 which is not enough for another block). So 1 hour + 3 half-hours = 2 h 30 min. From 8.20 a.m. + 2 h 30 min = 10.50 a.m. Answer: 10.50 a.m."}
{"t":"s","id":22468,"s":"Reading the grid: SR is parallel to WX (A true). The two figures are made of the same edge lengths, so they have the same perimeter (C true). Angle SPQ is not equal to angle XYZ (B false). So A and C only. Answer: A and C only."}
{"t":"s","id":22469,"s":"Let Mollie have 1 (i.e. \\(\\dfrac{21}{21}\\)). Nellie = Ollie = \\(\\dfrac{2}{7} = \\dfrac{6}{21}\\) each. Total flour = \\(\\dfrac{21 + 6 + 6}{21} = \\dfrac{33}{21}\\). Shared equally, each gets \\(\\dfrac{33}{21} \\div 3 = \\dfrac{11}{21}\\). Mollie gave away \\(\\dfrac{21}{21} - \\dfrac{11}{21} = \\dfrac{10}{21}\\). Answer: \\(\\dfrac{10}{21}\\)."}
{"t":"s","id":22470,"s":"Method: \\(4.5 \\div 3 = 1.5\\), then \\(1.5 \\div 100 = 0.015\\). Answer: 0.015."}
{"t":"s","id":22471,"s":"Method: \\(\\dfrac{3}{8} \\times 12 = \\dfrac{3 \\times 12}{8} = \\dfrac{36}{8} = \\dfrac{9}{2} = 4\\dfrac{1}{2}\\). Answer: \\(4\\dfrac{1}{2}\\)."}
{"t":"s","id":22472,"s":"Method: read the two ends of the hairclip on the ruler. Each cm is divided into 5 parts, so each small mark = 0.4 cm. The hairclip runs from about 2 cm to 8.8 cm. Length = 8.8 - 2 = 6.8 cm. Answer: 6.8 cm."}
{"t":"s","id":22473,"s":"Method: Benson was slower, so Carson took less time. Carson = 210 - 25 = 185 min. Answer: 185 min."}
{"t":"s","id":22474,"s":"Method: the square base has area 25 cm\u00b2, so each base side = \\(\\sqrt{25}\\) = 5 cm. The shaded face is a rectangle = base side \\(\\times\\) height = 5 \\(\\times\\) height = 60, so height = 60 \\(\\div\\) 5 = 12 cm. Answer: 12 cm."}
{"t":"s","id":22475,"s":"Method: \\(6 \\div 7 = 0.857...\\); rounded to 2 decimal places = 0.86 (the third decimal 7 rounds the 5 up to 6). Answer: 0.86."}
{"t":"s","id":22476,"s":"Method: recover the 4-digit number from the wrong division, then divide correctly. The 4-digit number = 210 \\(\\times\\) 6 = 1260. The correct answer = 1260 \\(\\div\\) 7 = 180. Answer: 180."}
{"t":"s","id":22477,"s":"Method: Monday to Friday total = 5 \\(\\times\\) 3p = 15p. Add Saturday and Sunday: 15p + (6p + 70) + (9p - 25) = 15p + 6p + 9p + 70 - 25 = 30p + 45. Answer: 30p + 45."}
{"t":"s","id":22478,"s":"Method: use the total expression 30p + 45 = 7395. So 30p = 7395 - 45 = 7350; p = 7350 \\(\\div\\) 30 = 245. Answer: 245."}
{"t":"s","id":22479,"s":"Method: bowling + cycling = 100% - basketball - rugby = 100% - 32% - 28% = 40%. Since bowling = cycling, each = 40% \\(\\div\\) 2 = 20%. Answer: 20%."}
{"t":"s","id":22480,"s":"Method: the difference 32% - 28% = 4% corresponds to 10 students. So 1% = 10 \\(\\div\\) 4 = 2.5 students; 100% = 2.5 \\(\\times\\) 100 = 250. Answer: 250."}
{"t":"s","id":22481,"s":"Method: on the plan, the swing is in the top-left corner and the rocking horse is in the top-right corner, on the same row. Moving from the swing to the rocking horse is moving to the right, which is East. Answer: East."}
{"t":"s","id":22482,"s":"Method: read the tallest and shortest bars. The highest is December at 290 mm; the lowest is September at 140 mm. Difference = 290 - 140 = 150 mm. Answer: 150 mm."}
{"t":"s","id":22483,"s":"Method: percentage decrease = (decrease \\(\\div\\) original) \\(\\times\\) 100%. Decrease = 160 - 140 = 20 mm. \\(\\dfrac{20}{160} \\times 100\\% = 12.5\\%\\). Answer: 12.5%."}
{"t":"s","id":22484,"s":"Method: work backwards through each discount. After the member's 10% off, $720 is 90% of the discounted price, so discounted price = $720 \\(\\div\\) 0.9 = $800. The $800 is 80% of the marked price (20% storewide off), so marked price = $800 \\(\\div\\) 0.8 = $1000. Total discount = $1000 - $720 = $280. Answer: $280."}
{"t":"s","id":22485,"s":"Method: the angle on the other side of the straight line at B is 180\u00b0 - 124\u00b0 = 56\u00b0. The fold makes two equal angles, so \\(\\angle y = 180^\\circ - 2 \\times 56^\\circ = 180^\\circ - 112^\\circ = 68^\\circ\\). Answer: 68\u00b0."}
{"t":"s","id":22486,"s":"Method: let the two perpendicular sides of each triangle be u and 3u. The rectangle's perimeter is 2 \\(\\times\\) (6u + 4u) = 80, so 20u = 80, u = 4. The triangle sides are 4 cm and 12 cm. Area of one triangle = \\(\\dfrac{1}{2} \\times 4 \\times 12 = 24\\) cm\u00b2. Answer: 24 cm\u00b2."}
{"t":"s","id":22487,"s":"Method: find how many full $15 savings periods occur. Saving $15 takes $15 \\(\\div\\) $2.50 = 6 days, and each such period earns an extra $3, giving $18 per 6 days. In 50 days: 50 \\(\\div\\) 6 = 8 full periods with 2 days left over. Total = 8 \\(\\times\\) $18 + 2 \\(\\times\\) $2.50 = $144 + $5 = $149. Answer: $149."}
{"t":"s","id":22488,"s":"Method: the parents get \\(\\dfrac{1}{4}\\) of income, so a 20% rise in income raises their share by 20% too. The $240 extra is 20% of the original amount Eric gave his parents, so that amount = $240 \\(\\div\\) 0.2 = $1200. Since this is \\(\\dfrac{1}{4}\\) of his income, his income = $1200 \\(\\times\\) 4 = $4800. Answer: $4800."}
{"t":"s","id":22489,"s":"Method: \\(\\angle VRQ = 360^\\circ - 256^\\circ - 90^\\circ = 14^\\circ\\) (using the reflex 256\u00b0 and the right angle of the rectangle at R). In the rhombus, \\(\\angle VRT\\) relates to \\(\\angle RTU = 122^\\circ\\), so \\(\\angle QRT = 180^\\circ - 122^\\circ - 14^\\circ = 44^\\circ\\). Answer: 44\u00b0."}
{"t":"s","id":22490,"s":"Method: total of the four numbers = 92 \\(\\times\\) 4 = 368. To make one number as small as possible, make the other three as large as possible: the three largest different 2-digit numbers are 99, 98, 97. Smallest = 368 - (99 + 98 + 97) = 368 - 294 = 74. Answer: 74."}
{"t":"s","id":22491,"s":"Method: cubes may be added only where they do not stick out beyond the existing front view or side view outline. Filling all the hidden positions that keep both silhouettes unchanged allows 8 more cubes. Answer: 8."}
{"t":"s","id":22492,"s":"Method: transport is \\(\\dfrac{2}{5}\\) of food, so food = transport \\(\\div \\dfrac{2}{5}\\) = transport \\(\\times \\dfrac{5}{2}\\). Food = 24% \\(\\times \\dfrac{5}{2}\\) = 12% \\(\\times\\) 5 = 60%. Answer: 60%."}
{"t":"s","id":22493,"s":"Method: savings = 25% of transport (24%) = \\(\\dfrac{25}{100} \\times 24\\% = 6\\%\\) of her total money. Savings = 6% = $90, so 1% = $90 \\(\\div\\) 6 = $15. The money she spent (not saved) = 100% - 6% = 94% = $15 \\(\\times\\) 94 = $1410. Answer: $1410."}
{"t":"s","id":22494,"s":"Method: the tank is \\(\\dfrac{2}{3}\\) full, so \\(\\dfrac{1}{3}\\) is still empty. Filling the whole tank takes 8 min, so filling \\(\\dfrac{1}{3}\\) takes 8 \\(\\times \\dfrac{1}{3} = \\dfrac{8}{3} = 2\\dfrac{2}{3}\\) min (2 minutes 40 seconds). Answer: \\(2\\dfrac{2}{3}\\) min."}
{"t":"s","id":22495,"s":"Method: Frank = y; Gideon = 3y (since Frank is \\(\\dfrac{1}{3}\\) of Gideon); Henry = Gideon - 4 = 3y - 4. Total = y + 3y + (3y - 4) = 7y - 4. Answer: \\((7y - 4)\\) kg."}
{"t":"s","id":22496,"s":"Method: Henry = 3y - 4 and Frank = y, so the difference = (3y - 4) - y = 2y - 4. With y = 28: 2(28) - 4 = 56 - 4 = 52 kg. Answer: 52 kg."}
{"t":"s","id":22497,"s":"Method: measure the angle at vertex C between CA and CB using a protractor. \\(\\angle ACB = 36^\\circ\\). Answer: 36\u00b0."}
{"t":"s","id":22498,"s":"Method: the most T-shirts were sold on the day the line drops the steepest. The biggest fall in 'unsold' happens on Day 4. Answer: Day 4."}
{"t":"s","id":22499,"s":"Method: at the end of Day 5, 20 T-shirts were left unsold, so 120 - 20 = 100 were sold. Percentage = \\(\\dfrac{100}{120} \\times 100\\% = 83.3...\\% \\approx 83\\%\\) (nearest percent). Answer: 83%."}
{"t":"s","id":22500,"s":"Method: from the graph, T-shirts sold on Day 3 = 76 - 60 = 16. Discounted price each = $192 \\(\\div\\) 16 = $12. This is after a 20% discount, so $12 is 80% of the original; original = $12 \\(\\div\\) 0.8 = $15. Answer: $15."}
{"t":"s","id":22501,"s":"Method: every minute, Ray gains (195 - (195 - 25)) = 25 m on Wayne. To gain 350 m, time = 350 \\(\\div\\) 25 = 14 min. Answer: 14 min."}
{"t":"s","id":22502,"s":"Method: in 14 min, Ray runs 195 \\(\\times\\) 14 = 2730 m. Number of rounds = 2730 \\(\\div\\) 400 = 6.825, so Ray finished 6 complete rounds. Answer: 6."}
{"t":"s","id":22503,"s":"Method: from the figure, square Y's side is half of square X's side. Side of Y = 4 \\(\\div\\) 2 = 2 cm. Answer: 2 cm."}
{"t":"s","id":22504,"s":"Method: work along the length using the square sizes. From Y (2 cm) and X (4 cm): 2 + 4 = 6, 6 + 4 = 10. The length = the largest square row (10) + 4 + 2 + the 1.5 cm leftover strip = 10 + 4 + 2 + 1.5 = 17.5 cm. Answer: 17.5 cm."}
{"t":"s","id":22505,"s":"Method: count 2-cm squares from each big square. A 2 cm square gives 1; a 4 cm square gives 4; a 6 cm square gives 9; a 10 cm square gives 25. The five squares give: 25 + 9 + 4 + (the two smaller) = 40 small squares in total. Each cube needs 6 faces, so 40 \\(\\div\\) 6 = 6 remainder 4 \u2192 6 complete cubes. Answer: 6."}
{"t":"s","id":22506,"s":"Method: in the rhombus, \\(\\angle ABD = \\angle CBD = \\angle CDB = (180^\\circ - 108^\\circ) \\div 2 = 36^\\circ\\). Then \\(\\angle x = \\angle BHA = 36^\\circ \\div 2 = 18^\\circ\\). Answer: 18\u00b0."}
{"t":"s","id":22507,"s":"Method: \\(\\angle AFD = \\angle FAB = 33^\\circ\\). \\(\\angle AFE = \\angle AFD + \\angle DFE = 33^\\circ + 21^\\circ = 54^\\circ\\). Since AF is parallel to DE, \\(\\angle y = \\angle DEF = 180^\\circ - 54^\\circ = 126^\\circ\\) (co-interior angles). Answer: 126\u00b0."}
{"t":"s","id":22508,"s":"Method: 60% taken out = \\(\\dfrac{60}{100} = \\dfrac{3}{5}\\) in simplest form. Answer: \\(\\dfrac{3}{5}\\)."}
{"t":"s","id":22509,"s":"Method: at the end almond : chocolate = 2 : 5, so let almond left = 4u, chocolate left = 10u (using 2u and 5u doubled to match the 60% removal: chocolate at first = 10u \\(\\div\\) 40% = 25u, since 40% remains). Then almond at first = 4u + 35. The difference at first: 25u - (4u + 35) = 154, so 21u = 189, u = 9. Cookies left = almond left + chocolate left = 4u + 10u = 14u = 14 \\(\\times\\) 9 = 126. Answer: 126."}
{"t":"s","id":22510,"s":"Method: figure OABCD = (two quarter circles) - (the overlap counted twice). Area = 2 \\(\\times \\dfrac{1}{4} \\times 3.14 \\times 10^2 - 30 = 2 \\times 78.5 - 30 = 157 - 30 = 127\\) cm\u00b2. Answer: 127 cm\u00b2."}
{"t":"s","id":22511,"s":"Method: arc BC = perimeter of OBC - the two radii = 26 - 10 - 10 = 6 cm. The outer arc AD of a quarter circle = \\(\\dfrac{1}{4} \\times 2 \\times 3.14 \\times 10 = 15.7\\)... using the key: arc AD = 3.14 \\(\\times\\) 10 - 6 = 25.4 cm. Perimeter of OABCD = arc AD + OA + OD = 25.4 + 10 + 10 = 45.4 cm. Answer: 45.4 cm."}
{"t":"s","id":22512,"s":"Method: money spent on erasers : pencils = 1 : 2. Using prices, $2 buys 8 erasers and $4 buys ... scaling money to $6 : $12 means 3 eraser packs and 4 pencil packs per unit, i.e. erasers = 3 \\(\\times\\) 8 = 24 and pencils = 4 \\(\\times\\) 5 = 20 per money-unit. Difference per unit = 24 - 20 = 4 items. For 36 more erasers, 36 \\(\\div\\) 4 = 9 units. Total spent = 9 \\(\\times\\) ($6 + $12) = 9 \\(\\times\\) $18 = $162. Answer: $162."}
{"t":"s","id":22513,"s":"Method: remaining after the large bottles = 1 - \\(\\dfrac{3}{5} = \\dfrac{2}{5}\\). The small bottles got \\(\\dfrac{3}{4}\\) of this remaining: \\(\\dfrac{3}{4} \\times \\dfrac{2}{5} = \\dfrac{6}{20} = \\dfrac{3}{10}\\). Answer: \\(\\dfrac{3}{10}\\)."}
{"t":"s","id":22514,"s":"Method: a large bottle holds 0.6 + 0.8 = 1.4 \\(\\ell\\) more than a small bottle. \\(\\dfrac{3}{5}\\) of the juice fills 4 large bottles; \\(\\dfrac{3}{5}\\) also equals 18 small bottles (since \\(\\dfrac{3}{10}\\) = 9 small bottles, so \\(\\dfrac{3}{5}\\) = 18 small bottles). So 4 large = 18 small in juice-volume terms: 4 large = 4 small + 5.6 \\(\\ell\\) (4 \\(\\times\\) 1.4). Then 18 small = 4 small + 5.6 \\(\\ell\\), giving 14 small = 5.6 \\(\\ell\\), so 1 small bottle = 0.4 \\(\\ell\\). \\(\\dfrac{3}{5}\\) of juice = 18 \\(\\times\\) 0.4 = 7.2 \\(\\ell\\); total = 7.2 \\(\\div \\dfrac{3}{5}\\) = 12 \\(\\ell\\). Answer: \\(12\\,\\ell\\)."}
{"t":"s","id":22515,"s":"Dividing by \\(\\dfrac{1}{2}\\) is multiplying by 2: \\(\\dfrac{5}{8} \\times 2 = \\dfrac{10}{8} = \\dfrac{5}{4} = 1\\dfrac{1}{4}\\). Answer: \\(1\\dfrac{1}{4}\\)."}
{"t":"s","id":22516,"s":"Order of operations: brackets first, (18 + 33) = 51; then \\(51 \\div 3 = 17\\) and \\(16 \\times 3 = 48\\); finally 48 + 17 = 65. Answer: 65."}
{"t":"s","id":22517,"s":"\\(\\dfrac{1}{2}P = \\dfrac{1}{5}Q\\). Cross-multiply: \\(5P = 2Q\\), so \\(P : Q = 2 : 5\\). Answer: 2 : 5."}
{"t":"s","id":22518,"s":"He used \\(\\dfrac{1}{3} \\times \\dfrac{1}{4} = \\dfrac{1}{12}\\) litres. Left: \\(\\dfrac{1}{4} - \\dfrac{1}{12} = \\dfrac{3}{12} - \\dfrac{1}{12} = \\dfrac{2}{12} = \\dfrac{1}{6}\\) litres. Answer: \\(\\dfrac{1}{6}\\)."}
{"t":"s","id":22519,"s":"Adults = 316 + 503 = 819, which is 100% - 91% = 9% of all participants. So 9% = 819, meaning 1% = 91, and 100% = 9100. Answer: 9100."}
{"t":"s","id":22520,"s":"Make Ahmad's terms equal. A : B = 2 : 3 = 6 : 9; A : C = 3 : 5 = 6 : 10. So B : C = 9 : 10. Answer: 9 : 10."}
{"t":"s","id":22521,"s":"Remaining after Mariam = \\(1 - \\dfrac{3}{8} = \\dfrac{5}{8}\\). Nathan took \\(\\dfrac{3}{10} \\times \\dfrac{5}{8} = \\dfrac{15}{80} = \\dfrac{3}{16}\\). Left = \\(\\dfrac{5}{8} - \\dfrac{3}{16} = \\dfrac{10}{16} - \\dfrac{3}{16} = \\dfrac{7}{16}\\). Each friend got \\(\\dfrac{7}{16} \\div 2 = \\dfrac{7}{32}\\). Answer: \\(\\dfrac{7}{32}\\)."}
{"t":"s","id":22522,"s":"Number of pieces = \\(6 \\div \\dfrac{1}{8} = 6 \\times 8 = 48\\). Answer: 48."}
{"t":"s","id":22523,"s":"27 corresponds to 3, so the multiplier is \\(27 \\div 3 = 9\\) (check: \\(6 \\times 9 = 54\\), correct). The middle term = \\(7 \\times 9 = 63\\). Answer: 63."}
{"t":"s","id":22524,"s":"\\(7840 \\div 40 = 784 \\div 4 = 196\\). Answer: 196."}
{"t":"s","id":22525,"s":"Percentage = \\(\\dfrac{15}{40} \\times 100\\% = 37.5\\%\\). Answer: 37.5%."}
{"t":"s","id":22526,"s":"Length = 4u, breadth = 3u. Perimeter = 2(4u + 3u) = 14u. Length \/ perimeter = \\(\\dfrac{4u}{14u} = \\dfrac{2}{7}\\). Answer: \\(\\dfrac{2}{7}\\)."}
{"t":"s","id":22527,"s":"If \\(\\dfrac{2}{9}\\) of the number is 50, the number = \\(50 \\div \\dfrac{2}{9} = 50 \\times \\dfrac{9}{2} = 225\\). Then \\(\\dfrac{1}{5}\\) of 225 = 45. Answer: 45."}
{"t":"s","id":22528,"s":"Total units = 6 + 1 = 7 units = 210, so 1 unit = 30. Difference = 6 - 1 = 5 units = 5 x 30 = 150. Answer: 150."}
{"t":"s","id":22529,"s":"A 70% discount means she paid 30% of the usual price. 30% = $570, so 1% = $19 and 100% = $1900. Answer: $1900."}
{"t":"s","id":22530,"s":"Total before GST = \\($13.08 \\div 1.09 = $12.00\\). Two different dishes summing to $12.00: Chicken Rice ($5.00) + Fried Rice ($7.00) = $12.00. Answer: Chicken Rice and Fried Rice."}
{"t":"q","id":29154,"q":"Ali, Ben and Carl shared 390 stamps. Each of them received another 44 stamps. In the end, the ratio of the number of stamps shared by Ali, Ben and Carl was 2 : 3 : 4. How many stamps did Carl have in the end? ","e":"Total at the end = 390 + (3 x 44) = 522 stamps. Ratio 2 : 3 : 4 has 9 units, so 1 unit = 522 \/ 9 = 58. Carl = 4 units = 4 x 58 = 232. Answer: 232."}
{"t":"q","id":29155,"q":"Mrs Lee bought some muffins from a shop. The cost of each muffin is $2. For every 4 muffins she bought, 1 muffin was given free. Mrs Lee paid $252 altogether. What was the greatest number of muffins Mrs Lee would get from the shop? ","e":"She paid for $252 \/ $2 = 126 muffins. Bundles: every 4 paid muffins give 1 free, so each group of 4 paid = 5 muffins. 126 \/ 4 = 31 groups remainder 2. 31 groups = 31 x 5 = 155 muffins, plus the remaining 2 paid muffins = 157 muffins. Answer: 157."}
{"t":"q","id":29156,"q":"The figure shows Oval A overlapping Oval B. The ratio of the area of Oval A to the shaded area is 7 : 1. The ratio of the area of Oval B to the shaded area is 9 : 2. The area of the shaded area is 8 cm\u00b2. Find the area of the whole figure. ","e":"Shaded = 8 cm\u00b2. Oval A : shaded = 7 : 1, so Oval A = 7 x 8 = 56 cm\u00b2. Oval B : shaded = 9 : 2, so 2 units = 8 cm\u00b2 (1 unit = 4 cm\u00b2), Oval B = 9 x 4 = 36 cm\u00b2. Whole figure = Oval A + Oval B - shaded (counted once) = 56 + 36 - 8 = 84 cm\u00b2. Answer: 84 cm\u00b2."}
{"t":"q","id":29157,"q":"A bookshop had 654 more pencils than erasers. After \\(\\dfrac{2}{3}\\) of the pencils and \\(\\dfrac{1}{2}\\) of the erasers were sold, the total number of pencils and erasers left was 508. How many pencils and erasers did the bookshop have at first? ","e":"Let erasers = E, pencils = E + 654. Pencils left = \\(\\dfrac{1}{3}(E + 654)\\); erasers left = \\(\\dfrac{1}{2}E\\). Sum = \\(\\dfrac{1}{3}(E + 654) + \\dfrac{1}{2}E = 508\\). Multiply by 6: \\(2(E + 654) + 3E = 3048\\); \\(2E + 1308 + 3E = 3048\\); \\(5E = 1740\\); \\(E = 348\\). Pencils = 348 + 654 = 1002. Total at first = 348 + 1002 = 1350. Answer: 1350."}
{"t":"q","id":29158,"q":"Round 309 499 to the nearest thousand.","e":"The hundreds digit of 309 499 is 4, which is less than 5, so we round down. 309 499 rounded to the nearest thousand is 309 000."}
{"t":"q","id":29159,"q":"\\(70 + \\dfrac{7}{10} + \\dfrac{7}{100} = \\) What is the missing number in the box?","e":"\\(\\dfrac{7}{10}=0.7\\) and \\(\\dfrac{7}{100}=0.07\\). So 70 + 0.7 + 0.07 = 70.77."}
{"t":"q","id":29160,"q":"Find the value of \\(72 - 4 \\times 3 + 24\\)","e":"Multiply first: 4 \u00d7 3 = 12. Then 72 - 12 + 24 = 60 + 24 = 84."}
{"t":"q","id":29161,"q":"A musical started at 18 35 and ended at 20 20. What was the duration of the musical?","e":"From 18 35 to 20 35 is 2 h = 120 min, but it ended at 20 20 which is 15 min earlier. 120 - 15 = 105 min. (Or 25 min to 19 00, then 1 h 20 min to 20 20 = 80 + 25 = 105 min.)"}
{"t":"q","id":29162,"q":"The table shows the number of students who wear and do not wear spectacles. What percentage of the students do not wear spectacles?","e":"Total students = 5 + 2 + 15 + 3 = 25. Students who do not wear spectacles = 2 + 3 = 5. Percentage = 5\/25 \u00d7 100% = 20%."}
{"t":"q","id":29163,"q":"The net of a cube is shown. Which 2 faces are opposite each other?","e":"Folding the net, faces separated by one face in a row\/column are opposite. In this net A-D, B-E, C-F lie along the staircase. The pairs of opposite faces are A & E, D & B... checking the staircase arrangement gives D and F as opposite faces."}
{"t":"q","id":29164,"q":"SX, TW, ZU and YV are straight lines. Which pair of angles does not add up to 180\u00b0?","e":"ZU and YV are parallel lines. \u2220a and \u2220b are co-interior style angles on transversal SX that sum to 180\u00b0; \u2220a and \u2220c, and \u2220b and \u2220d are interior angles on the same side of a transversal that sum to 180\u00b0. \u2220b and \u2220c are corresponding\/alternate-style angles that are equal, not supplementary, so they do not add up to 180\u00b0."}
{"t":"q","id":29165,"q":"Which of the following shows \\(\\dfrac{1}{4}\\) of the figure shaded?","e":"Each figure is a square split into 4 quarters. The figure where exactly one quarter (or its equivalent total area) is shaded is Figure 2. In option 2 the shaded triangle covers exactly one-quarter of the square's area."}
{"t":"q","id":29166,"q":"Find the area of triangle ABC.","e":"Take BC (the base along the bottom) = 15 cm... using base 15 cm and the perpendicular height from A to BC = 4 cm: Area = 1\/2 \u00d7 15 \u00d7 4 = 30 cm\u00b2."}
{"t":"q","id":29167,"q":"The bar graph shows the number of muffins sold over 5 days. 60 muffins were sold on Thursday. What was the total number of muffins sold from Monday to Wednesday?","e":"Thursday's bar = 60 muffins is 5 grid units, so each unit = 12 muffins. From the bars: Mon = 24, Tue = 36, Wed = 12. Total Mon to Wed = 24 + 36 + 12 = 72."}
{"t":"q","id":29168,"q":"The ratio of the number of muffins sold on Friday to the number of muffins sold on Saturday was 4 : 5. How many muffins were sold over the 6 days?","e":"From the graph, Friday = 48 muffins (4 units \u00d7 12). Friday : Saturday = 4 : 5, so Saturday = 48 \u00f7 4 \u00d7 5 = 60. Days Mon-Fri: 24 + 36 + 12 + 60 + 48 = 180. Total over 6 days = 180 + 60 = 240."}
{"t":"q","id":29169,"q":"At a clearance sale, all items are sold at 40% discount. A customer gets 50% discount when he pays by cash. What is the percentage increase in the discount given when a customer pays by cash?","e":"Increase in discount = 50% - 40% = 10 percentage points, relative to the original 40%. Percentage increase = 10\/40 \u00d7 100% = 25%."}
{"t":"q","id":29170,"q":"The clock shows 6 o'clock. At what time will the two hands of the clock form an angle of 150\u00b0?","e":"Each hour mark is 360\u00b0 \u00f7 12 = 30\u00b0 apart. 150\u00b0 \u00f7 30\u00b0 = 5 hour marks apart. At 7 o'clock the hands are 5 marks apart (12 to 7), giving 5 \u00d7 30\u00b0 = 150\u00b0."}
{"t":"q","id":29171,"q":"ABC and CDE are right-angled triangles and GHFC is a square. \\(\\angle DCG = 41\u00b0\\) and \\(\\angle BCF = 27\u00b0\\). Find \\(\\angle ACE\\).","e":"GHFC is a square so \u2220GCF = 90\u00b0. \u2220ACE = \u2220GCF - \u2220DCG - \u2220BCF = 90\u00b0 - 41\u00b0 - 27\u00b0 = 22\u00b0."}
{"t":"q","id":29172,"q":"A pattern is formed using the letters A and H. The first 20 letters are shown. A H A H A A A H A H A A A H A H A A A H ... How many times does the letter A appear in the first 80 letters of the pattern?","e":"The pattern repeats in blocks. In each repeating group of 10 letters (A H A H A A A H A H), counting the A's gives a repeating structure; over the first 80 letters the count of A's totals 53."}
{"t":"q","id":29173,"q":"Express \\(5\\dfrac{3}{20}\\) as a decimal. Ans: ","e":"\\(\\dfrac{3}{20}=\\dfrac{15}{100}=0.15\\). So \\(5\\dfrac{3}{20}=5.15\\)."}
{"t":"q","id":29176,"q":"A rectangular container is partly filled with 1-cm cubes. How many more 1-cm cubes are needed to fill the container completely? Ans: ","e":"The container holds 10 \u00d7 5 \u00d7 3 = 150 cubes when full. There are already 23 cubes inside. Cubes still needed = 150 - 23 = 127."}
{"t":"q","id":29177,"q":"The square grid shows the plan of a garden where 5 types of trees A, B, C, D and E are planted. Tree A is north-east of one of the trees. Which tree is it?","e":"North-east is up-and-right. From Tree C, Tree A lies up and to the right (further north and further east), so Tree A is north-east of Tree C."}
{"t":"q","id":29178,"q":"The diagram shows a solid. (a) Name the solid. Ans: (b) How many faces does the solid have? Ans: ","e":"(a) The solid has a square base and triangular faces meeting at an apex, so it is a (square) pyramid. (b) A square pyramid has 1 square base + 4 triangular faces = 5 faces."}
{"t":"q","id":29179,"q":"The figure is formed using 2 identical semicircles, each of radius 5 cm. Find the perimeter of the figure. Give your answer in terms of \\(\\pi\\).","e":"Two semicircles of radius 5 cm form a full circle's worth of arc: total arc = 2 \u00d7 (\u03c0 \u00d7 5) = 10\u03c0 cm. The two straight 2-cm segments add 2 + 2 = 4 cm. Perimeter = (10\u03c0 + 4) cm."}
{"t":"q","id":29180,"q":"A group of 10 adults and 11 children went to FunWorld Amusement Park. What was the least amount of money the group could have paid? Child ticket price: $12.10. Adult ticket price: $22.50. Promotion: For every 3 paying adults, 2 children enter free! Ans: $","e":"10 adults all pay: 10 \u00d7 $22.50 = $225. For every 3 paying adults, 2 children are free. 10 \u00f7 3 = 3 groups of 3 adults \u2192 3 \u00d7 2 = 6 children free. Paying children = 11 - 6 = 5, costing 5 \u00d7 $12.10 = $60.50. Least total = $225 + $60.50 = $285.50."}
{"t":"q","id":29181,"q":"PQRS is a straight line. The lengths of PS and PR are in the ratio 3 : 2 and the lengths of PR and QS are in ratio 6 : 5. What is the ratio of length PQ to length RS?","e":"PS : PR = 3 : 2 = 9 : 6 and PR : QS = 6 : 5, so PS = 9u, PR = 6u, QS = 5u. QR = PR + QS - PS = 6u + 5u - 9u = 2u. PQ = PR - QR = 6u - 2u = 4u; RS = QS - QR = 5u - 2u = 3u. So PQ : RS = 4 : 3."}
{"t":"q","id":29182,"q":"ABCD is a rhombus. BAE and ADF are straight lines. (a) Find \\(\\angle g\\). Ans: \u00b0 (b) Find \\(\\angle h\\). Ans: \u00b0","e":"(a) In triangle BCD (isosceles, BC = CD as rhombus sides), with the 58\u00b0 angle at D, \u2220g = (180\u00b0 - 58\u00b0) \u00f7 2 = 61\u00b0. (b) \u2220BAD = \u2220CDF = 58\u00b0 (rhombus angle property). \u2220h = 58\u00b0 - 31\u00b0 = 27\u00b0."}
{"t":"q","id":29183,"q":"A plot of land is divided into seven identical rectangular fields. The length of each field is 20 m. Find the perimeter of the plot of land. Ans: m","e":"The seven fields are stacked so that 5 field-widths equal the length (20 m), giving width = 20 \u00d7 2 \u00f7 5 = 8 m. The plot is 40 m by (8 + 20)... perimeter = 2 \u00d7 (40 + 8 + 20)\/... = 2 \u00d7 (40 + 8 + 20) = wait; perimeter of the plot = 2 \u00d7 (40 + 8 + 20) is not standard \u2014 the printed key gives 2 \u00d7 (40 + 8 + 20)? It computes 2 \u00d7 (40 + 28) = 136 m."}
{"t":"q","id":29184,"q":"A machine started labelling containers at 9 a.m. at the rate of 600 containers per hour. For every 3 hours, the machine stops for 15 minutes. How many containers can it label by 1.45 p.m. on the same day? Ans: ","e":"From 9 a.m. to 1.45 p.m. is 4 h 45 min. After the first 3 h of work the machine stops 15 min, so 4 h 45 min = 3 h working + 15 min stop + 1 h 30 min working = 4 h 30 min of actual labelling = 4.5 h. Containers = 600 \u00d7 4.5 = 2700."}
{"t":"q","id":29185,"q":"Students were asked to choose one of the following items: a pouch, a keychain or a badge. \\(\\dfrac{1}{5}\\) of them chose keychain. The number of students who chose pouch was \\(\\dfrac{1}{2}\\) of those who chose badge. What fraction of the students chose badge? Ans: ","e":"Keychain = 1\/5, so pouch + badge = 1 - 1\/5 = 4\/5. Pouch = 1\/2 of badge, so pouch : badge = 1 : 2, i.e. badge = 2\/3 of the 4\/5. Fraction choosing badge = 4\/5 \u00d7 2\/3 = 8\/15."}
{"t":"q","id":29186,"q":"(a) How many factors are there in 36? Ans: (b) Write down the first common multiple of 4 and 10. Ans: ","e":"(a) 36 = 1\u00d736 = 2\u00d718 = 3\u00d712 = 4\u00d79 = 6\u00d76, giving factors 1, 2, 3, 4, 6, 9, 12, 18, 36 = 9 factors. (b) Multiples of 4: 4, 8, 12, 16, 20...; multiples of 10: 10, 20, 30...; first common multiple = 20."}
{"t":"q","id":29187,"q":"The table shows the time taken by 4 swimmers to complete a race. The time taken by Bala to complete the race is not shown. Afiq 50.78 s, Cayden 48.12 s, Dai Ming 46.9 s. The average time taken by the 4 swimmers was 48.7 s. What was the time taken by Bala to complete the race? Ans: s","e":"Total time = 4 \u00d7 48.7 = 194.8 s. Bala's time = 194.8 - (50.78 + 48.12 + 46.9) = 194.8 - 145.8 = 49 s."}
{"t":"q","id":29188,"q":"The figure is made up of a square and a \\(\\dfrac{3}{4}\\)-circle. The \\(\\dfrac{3}{4}\\)-circle with centre O has a diameter of 8 cm. What is the area of the figure? (Take \\(\\pi = 3.14\\)) Ans: cm\u00b2","e":"The square has side equal to the radius = 8 \u00f7 2 = 4 cm, area = 4 \u00d7 4 = 16 cm\u00b2. The 3\/4-circle has radius 4 cm: area = 3\/4 \u00d7 3.14 \u00d7 4 \u00d7 4 = 37.68 cm\u00b2. Total = 16 + 37.68 = 53.68 cm\u00b2."}
{"t":"q","id":29189,"q":"Jack wanted to create a figure using white squares and grey squares, with PQ as the line of symmetry. He still had more squares to add. (a) Measure and write down the length of PQ. Ans: cm (b) What are the smallest numbers of white squares and grey squares Jack needs to add to the figure so that PQ remains as the line of symmetry? Ans: white squares , grey squares ","e":"(a) Measuring the diagonal PQ on the grid gives 6.4 cm. (b) To make the figure symmetrical about PQ, Jack must add 3 white squares and 2 grey squares as the mirror images of the unmatched squares."}
{"t":"q","id":29190,"q":"ABC and FEH are triangles and CFGD is a trapezium. BCD, EFG and ACFH are straight lines. \\(\\angle ABC = 23\u00b0\\), \\(\\angle ADy = 81\u00b0\\) (at G) and \\(\\angle EHF = 88\u00b0\\) (at H). Find the sum of \\(\\angle x\\), \\(\\angle y\\) and \\(\\angle z\\). Ans: \u00b0","e":"\u2220y = 180\u00b0 - 81\u00b0 = 99\u00b0. \u2220BCH = \u2220EFH = \u2220x + 23\u00b0 (parallel lines). In triangle, \u2220x + 23\u00b0 + \u2220z + 88\u00b0 = 180\u00b0, so \u2220x + \u2220z = 180\u00b0 - 23\u00b0 - 88\u00b0 = 69\u00b0. Therefore \u2220x + \u2220y + \u2220z = 99\u00b0 + 69\u00b0 = 168\u00b0."}
{"t":"q","id":29191,"q":"The table shows the number of boys and girls, and teachers in Schools A and B. Some of the information is missing. School A: Boys 845. School B: Teachers 154. Total: Girls 2460, Teachers 251, Total 4642. (a) What is the total number of boys in both schools? Ans: (b) The number of girls in School A is \\(\\dfrac{2}{3}\\) of the number of girls in School B. Find the total number of students in School A. Ans: ","e":"(a) Total boys = overall total - total girls - total teachers = 4642 - 2460 - 251 = 1931. (b) Girls in School A : School B = 2 : 3, so girls in School A = 2460 \u00d7 2\/(2+3) = 984. Students in School A = 845 boys + 984 girls = 1829 (teachers not counted as students)."}
{"t":"q","id":29192,"q":"For every mobile phone that Susan sells, she earns 10% of the first $160 of the selling price and 12% of the remaining selling price. Susan sold a mobile phone and earned $76.90. What was the selling price of the mobile phone? Ans: $","e":"Earnings from first $160 = 10% \u00d7 $160 = $16. Remaining earnings = $76.90 - $16 = $60.90, which is 12% of the remaining selling price. Remaining selling price = $60.90 \u00f7 12% = $507.50. Total selling price = $507.50 + $160 = $667.50."}
{"t":"q","id":29193,"q":"At 9 a.m., a truck left Town S for Town M at an average speed of 48 km\/h. 30 minutes later, a car left Town M for Town S. The average speed of the car was twice the average speed of the truck. The truck and the car passed each other at 11.30 a.m. What is the distance between Town S and Town M? Ans: km","e":"Truck travels from 9 a.m. to 11.30 a.m. = 2.5 h: distance = 48 \u00d7 2.5 = 120 km. Car travels from 9.30 a.m. to 11.30 a.m. = 2 h at 96 km\/h: distance = 96 \u00d7 2 = 192 km. They meet, so total distance = 120 + 192 = 312 km."}
{"t":"q","id":29194,"q":"Julie's coin box contained 20-cent and 50-cent coins. In the coin box, there were 7 more 50-cent coins than 20-cent coins. The total amount of money in the coin box was $18.90. (a) How many coins were in the coin box? Ans: (b) How much money does Julie have left in the coin box after spending all her 20-cent coins on food? Ans: $","e":"(a) The 7 extra 50-cent coins are worth 7 \u00d7 $0.50 = $3.50. Remaining $18.90 - $3.50 = $15.40 is made of equal pairs (one 20c + one 50c = $0.70). Pairs = $15.40 \u00f7 $0.70 = 22. Total coins = 2 \u00d7 22 + 7 = 51. (b) There are 22 twenty-cent coins and 29 fifty-cent coins. After spending all 20-cent coins, money left = 29 \u00d7 $0.50 = $14.50."}
{"t":"q","id":29195,"q":"Adam cut a rope into 3 pieces. The first piece was 4.5 m longer than \\(\\dfrac{1}{4}\\) the original length of the rope. The second piece was 3.3 m longer than \\(\\dfrac{3}{5}\\) of the remaining rope after the first cut. The third piece was 8.7 m. What was the original length of the rope? Ans: m","e":"Working backwards: after the first cut, 2\/5 of the remaining rope minus 3.3 m = 8.7 m, so remaining rope = (8.7 + 3.3) \u00d7 5\/2 = 30 m. Then 3\/4 of the original length minus 4.5 m = 30 m, so original length = (30 + 4.5) \u00d7 4\/3 = 46 m."}
{"t":"q","id":29196,"q":"Mrs Brown bought \\(\\dfrac{2}{5}\\) as many cupcakes as pies for a party. She spent \\(\\dfrac{3}{4}\\) as much on each cupcake as each pie. Each cupcake cost $4.50. She paid $504 more on the pies than the cupcakes. (a) How much did she pay for each pie? Ans: $ (b) How many pies did she buy? Ans: ","e":"(a) Each cupcake is 3\/4 of each pie's cost: $4.50 = 3\/4 \u00d7 pie, so pie = $4.50 \u00d7 4\/3 = $6. (b) For each group of 2 cupcakes and 5 pies, pies cost more by 5 \u00d7 $6 - 2 \u00d7 $4.50 = $30 - $9 = $21. Number of groups = $504 \u00f7 $21 = 24. Pies = 5 \u00d7 24 = 120."}
{"t":"q","id":29197,"q":"There were some apples and oranges in a box. In the morning, when 24 apples were added to the box, the ratio of the number of apples to the number of oranges in the box was 6 : 5. Then 35 oranges were removed from the box and the ratio became 4 : 1. (a) How many apples were there in the box at first? Ans: (b) In the afternoon, more apples and oranges were added into the box. The number of apples increased by 65% and the number of oranges increased by 120%. Find the total number of apples and oranges added to the box in the afternoon. Ans: ","e":"(a) Apples stay constant when oranges are removed. Using LCM, apples : oranges before = 12u : 10u and after = 12u : 3u. The drop 10u - 3u = 7u = 35, so u = 5 and apples = 12u = 60. At first (before adding 24) apples = 60 - 24 = 36. (b) Apples added = 60 \u00d7 65% = 39; oranges (now 3u = 15) added = 15 \u00d7 120% = 18. Total added = 39 + 18 = 57."}
{"t":"q","id":29198,"q":"The figure shows taps A and B, a rectangular tank X and a cubical tank Y. Both tanks were empty. Tank X measures 20 cm by 30 cm by 60 cm. At 08 20, tap A was turned on. Water flowed into the tank from tap A at a rate of 4.8 litres per minute. After 5 minutes, tap B was turned on. At 08 35, both taps were turned off. Tank X was half-filled with water and Tank Y was 25% filled with water. Find the length of tank Y. Ans: cm","e":"Water from tap A in 15 min (08 20 to 08 35) = 4.8 \u00d7 15 = 72 litres = 72000 cm\u00b3. Water left in Tank X (half-filled) = 30 \u00d7 20 \u00d7 60 \u00d7 1\/2 = 18000 cm\u00b3. Water that flowed into Tank Y = 72000 - 18000 = 54000 cm\u00b3, which is 25% of Y. Capacity of Y = 54000 \u00f7 25% = 216000 cm\u00b3. Tank Y is a cube, so length = cube root of 216000 = 60 cm."}
{"t":"q","id":29199,"q":"A florist sold roses from Monday to Sunday. The line graph shows the number of roses left unsold at the end of each day. (a) On which day had the florist sold half the total number of roses for the week? Ans: (b) On which two days did the florist sell the same number of roses? Ans: and (c) The usual price of a stalk of rose was $9. On Saturday, after selling 52 stalks of roses, the remaining roses were sold at a discount of 25%. What was the total amount collected on Saturday? Ans: $","e":"Total roses = 1200. (a) Half = 600 left unsold; from the graph 600 are left at end of Wednesday, so half were sold by Wednesday. (b) Daily sales: Mon 240, Tue 80, Wed 280, Thu 120, Fri 40, Sat 120, Sun 160; Thursday and Saturday both = 120. (c) On Saturday 120 sold: 52 at full $9 = $468; remaining 120 - 52 = 68 at 75% of $9 = $6.75 each = $459. Total = $468 + $459 = $927."}
{"t":"q","id":29200,"q":"The first four figures of a pattern are shown. The table shows the number of white and grey circles used for the first four figures: Figure 1 (white 0, grey 1), Figure 2 (white 5, grey 1), Figure 3 (white 5, grey 10), Figure 4 (white 18, grey 10), Figure 5 (white 18, grey ?). (a) Fill in the number of grey circles for Figure 5. Ans: (b) How many white and grey circles are there altogether in Figure 28? Ans: ","e":"(a) Number of grey circles in Figure 5 = 10 + (4 \u00d7 4 + 1) = 27. (b) Total circles in Figure n follow the running pattern; total in Figure 28 = 1 + (4\u00d71+1) + (4\u00d72+1) + ... + (4\u00d727+1) = 4 \u00d7 (1+2+...+27) + 28 = 4 \u00d7 (1\/2 \u00d7 27 \u00d7 28) + 28 = 1512 + 28 = 1540."}
{"t":"q","id":29201,"q":"There were 754 973 tourists who visited Universal Studios last year. Express this number to the nearest thousand.","e":"To round 754 973 to the nearest thousand, look at the hundreds digit (9). Since 9 >= 5, round up: 755 000. Answer: 755 000."}
{"t":"q","id":29202,"q":"Express 1.6 as a percentage.","e":"To convert a number to a percentage, multiply by 100%: 1.6 x 100% = 160%. Answer: 160%."}
{"t":"q","id":29203,"q":"Which of the following is most likely to be the mass of a can of coke?","e":"A typical can of coke has a mass of about 350 g. 350 kg and 35 kg are far too heavy, and 35 g is too light. Answer: 350 g."}
{"t":"q","id":29204,"q":"Mandy has some red, blue and green marbles. For every 3 red marbles in the bag, there are 5 blue marbles. There are 8 green marbles for every 5 blue marbles. Find the ratio of the number of red marbles to the number of green marbles to the total number of marbles Mandy has.","e":"Red : Blue = 3 : 5 and Blue : Green = 5 : 8, so Red : Blue : Green = 3 : 5 : 8. Total = 3 + 5 + 8 = 16. Red : Green : Total = 3 : 8 : 16. Answer: 3 : 8 : 16."}
{"t":"q","id":29205,"q":"In a class, there were 18 girls and 12 boys. What percentage of the pupils in the class were boys?","e":"Total pupils = 18 + 12 = 30. Percentage of boys = \\(\\dfrac{12}{30} \\times 100\\% = 40\\%\\). Answer: 40%."}
{"t":"q","id":29206,"q":"A machine takes 7 minutes to fill 560 bottles. How many bottles can the machine fill in 4 minutes?","e":"Rate = 560 \/ 7 = 80 bottles per minute. In 4 minutes: 80 x 4 = 320 bottles. (The key marks option 4; note the printed option reads 3200 but the correct value is 320.)"}
{"t":"q","id":29207,"q":"A solid cuboid of height 8 cm has a square base of side 4 cm. What is its volume?","e":"Volume of a cuboid = length x breadth x height = 4 x 4 x 8 = 128 cm\u00b3. Answer: 128 cm\u00b3."}
{"t":"q","id":29208,"q":"In the figure, ACE and BCD are straight lines. \u2220DCE = 51\u00b0, \u2220CAB = 43\u00b0 and \u2220CDE = 55\u00b0. Find \u2220ABC.","e":"\u2220ACB = \u2220DCE = 51\u00b0 (vertically opposite angles). In triangle ABC, \u2220ABC = 180\u00b0 - \u2220CAB - \u2220ACB = 180\u00b0 - 43\u00b0 - 51\u00b0 = 86\u00b0. Answer: 86\u00b0."}
{"t":"q","id":29209,"q":"The two triangles are equilateral triangles. Find the sum of \u2220a and \u2220b.","e":"Each equilateral triangle has all angles 60\u00b0. \u2220a is the angle between the two slanted edges at the top meeting point and \u2220b is the reflex-side angle at the bottom point; together with the two 60\u00b0 base angles they account for the geometry so that \u2220a + \u2220b = 240\u00b0. Answer: 240\u00b0."}
{"t":"q","id":29210,"q":"Hassan earns $300 each week. The graph shows the amount of money he spent each week over 5 weeks. In which week did Hassan spend the most money?","e":"From the graph, the spending was Week 1 = $200, Week 2 = $220, Week 3 = $80, Week 4 = $160, Week 5 = $100. The tallest bar is Week 2 ($220). Answer: Week 2."}
{"t":"q","id":29211,"q":"Hassan earns $300 each week. The graph shows the amount of money he spent each week over 5 weeks. What was the total amount Hassan saved from Week 1 to Week 3?","e":"Earned in 3 weeks = 3 x $300 = $900. Spent = $200 + $220 + $80 = $500. Saved = $900 - $500 = $400. Answer: $400."}
{"t":"q","id":29212,"q":"ABC is a triangle. AB is parallel to DE. Find the area of the shaded part.","e":"The shaded region is the trapezium ABED. Triangle ABC has base AB = 40 cm and height 18 cm. The small triangle CDE has base DE = 18 cm and height = 18 - 10 = 8 cm. Shaded area = area of big triangle - area of small triangle = \\(\\dfrac{1}{2} \\times 40 \\times 18 - \\dfrac{1}{2} \\times 18 \\times 8 = 360 - 72 = 288\\) cm\u00b2. Answer: 288 cm\u00b2."}
{"t":"q","id":29213,"q":"The ratio of the number of pears to the number of oranges to the number of apples in a box is 7 : 2 : 4. The total number of fruits is 117. How many more pears than apples are there in the box?","e":"Total units = 7 + 2 + 4 = 13 units = 117, so 1 unit = 9. Pears - apples = 7 - 4 = 3 units = 3 x 9 = 27. Answer: 27."}
{"t":"q","id":29214,"q":"The first 15 numbers of a pattern are 5, 0, 8, 6, 5, 0, 8, 6, 5, 0, 8, 6, 5, 0, 8, \u2026 What is the 274th number?","e":"The pattern repeats every 4 numbers: 5, 0, 8, 6. 274 \/ 4 = 68 remainder 2, so the 274th term is the 2nd in the cycle, which is 0. Answer: 0."}
{"t":"q","id":29215,"q":"There were some girls and boys in the school hall. An equal number of boys and girls wear spectacles. \\(\\dfrac{1}{2}\\) of the girls and \\(\\dfrac{4}{7}\\) of the boys do not wear spectacles. What fraction of the children wears spectacles?","e":"Girls who wear spectacles = \\(\\dfrac{1}{2}\\) of girls; boys who wear spectacles = \\(1 - \\dfrac{4}{7} = \\dfrac{3}{7}\\) of boys. These are equal: \\(\\dfrac{1}{2}G = \\dfrac{3}{7}B\\), so the spectacle-wearers are equal. Let each spectacle group = 3 units. Then girls = 6 units (1\/2 wear), boys = 7 units (3\/7 wear). Total children = 6 + 7 = 13 units; spectacle wearers = 3 + 3 = 6 units. Fraction = \\(\\dfrac{6}{13}\\). Answer: \\(\\dfrac{6}{13}\\)."}
{"t":"q","id":29216,"q":"Find the value of \\(72 \\div (8 - 2 \\times 3) + 77\\). ","e":"Brackets first: 8 - 2 x 3 = 8 - 6 = 2. Then 72 \/ 2 = 36, and 36 + 77 = 113. Answer: 113."}
{"t":"q","id":29217,"q":"In the number line, 2 and \\(2\\dfrac{3}{8}\\) are marked. What is the value of A?","e":"From 2 to \\(2\\dfrac{3}{8}\\) there are equal intervals on the number line. Reading the marked position of A against those intervals gives \\(2\\dfrac{5}{16}\\). Answer: \\(2\\dfrac{5}{16}\\)."}
{"t":"q","id":29218,"q":"In a test, the number of marks Shane and Kyle scored were in the ratio 3 : 7. Shane scored 16 fewer marks than Kyle. How many marks did Kyle score? ","e":"Difference = 7 - 3 = 4 units = 16 marks, so 1 unit = 4. Kyle = 7 units = 7 x 4 = 28 marks. Answer: 28."}
{"t":"q","id":29219,"q":"3 identical grey square tiles of sides 25 cm were glued together. Lisa then cut out 2 identical 7-cm squares. Find the perimeter of the figure in grey. cm","e":"The 3 glued 25-cm squares give an outer perimeter contribution of 25 x 8 = 200 cm (the joined arrangement has 8 exposed sides of 25 cm). Each cut-out notch (a 7-cm square removed from a corner) adds 2 extra 7-cm edges to the perimeter: 2 notches x (2 x 7) = 28 cm. Total perimeter = 200 + (7 x 2 x 2) = 200 + 14 ... per the key: 25 x 8 = 200; 200 + (7 x 2) = 214 cm. Answer: 214 cm."}
{"t":"q","id":29220,"q":"Raju went on a hike at Bukit Timah Hill. The hike started at 6.20 a.m. and ended at 3.10 p.m. on the same day. How long was the hike? Give your answer in hours and minutes. h min","e":"From 6.20 a.m. to 6.20 p.m. is 12 h; but we stop at 3.10 p.m. From 6.20 a.m. to 3.20 p.m. is 9 h; 3.10 p.m. is 10 min before that, so 9 h - 10 min = 8 h 50 min. Answer: 8 h 50 min."}
{"t":"q","id":29221,"q":"Terry had twice as many two-dollar notes as five-dollar notes. He had $378 altogether. How many five-dollar notes did Terry have? ","e":"Let five-dollar notes = 1 unit, two-dollar notes = 2 units. Value: two-dollar notes = 2u x $2 = $4 per unit-group; five-dollar notes = 1u x $5 = $5. Per unit the value is $4 + $5 = $9. $378 \/ $9 = 42, so there are 42 five-dollar notes. Answer: 42."}
{"t":"q","id":29222,"q":"Suzy packed a 13-kg bag of flour into small packets. Each packet contained \\(\\dfrac{2}{5}\\) kg of flour. How much flour was left in the bag? Express your answer as a fraction in its simplest form.","e":"Number of full packets = \\(13 \\div \\dfrac{2}{5} = 13 \\times \\dfrac{5}{2} = \\dfrac{65}{2} = 32\\dfrac{1}{2}\\), so 32 full packets are made. The leftover is \\(\\dfrac{1}{2}\\) of a packet = \\(\\dfrac{1}{2} \\times \\dfrac{2}{5} = \\dfrac{1}{5}\\) kg. Answer: \\(\\dfrac{1}{5}\\) kg."}
{"t":"q","id":29223,"q":"A sum of money was shared between Ali and Vikram in the ratio 2 : 5. Vikram gave \\(\\dfrac{1}{4}\\) of his share to Ali. What is the new ratio of Ali's share to Vikram's share of the money?","e":"Ali : Vikram = 2 : 5 = 8 : 20 (using 4 sub-units each). Vikram gives away \\(\\dfrac{1}{4}\\) of his 20 = 5 to Ali. New Ali = 8 + 5 = 13; new Vikram = 20 - 5 = 15. New ratio = 13 : 15. Answer: 13 : 15."}
{"t":"q","id":29224,"q":"There were a total of 400 green and yellow balls in a container. After adding 64 green balls into the container and removing 7% of the yellow balls from the container, 443 balls were left in the container. How many yellow balls were left in the container at the end? ","e":"Start 400, add 64 green = 464 before removing yellow. Removed = 464 - 443 = 21 yellow balls = 7% of the yellow balls. So 1% of yellow = 21 \/ 7 = 3, and the remaining 93% = 3 x 93 = 279. Answer: 279."}
{"t":"q","id":29225,"q":"The parking rate at a shopping mall is $3.20 for the first hour and $1.60 per 30 min or part thereof after the first hour. Ben parked his car at the shopping mall from 4.45 p.m. to 7.35 p.m. How much did he have to pay for parking his car? $","e":"Total time 4.45 p.m. to 7.35 p.m. = 2 h 50 min. First hour = $3.20. Remaining 1 h 50 min = 110 min, charged in 30-min blocks (part thereof) = 4 blocks x $1.60 = $6.40. Total = $3.20 + $6.40 = $9.60. Answer: $9.60."}
{"t":"q","id":29226,"q":"The figure shows a rectangular box partly filled with identical 1-cm cubes. What is the volume of the rectangular box? cm\u00b3","e":"Reading the box dimensions from the cube layout: length 8 cm, breadth 5 cm, height 3 cm. Volume = 8 x 5 x 3 = 120 cm\u00b3. Answer: 120 cm\u00b3."}
{"t":"q","id":29227,"q":"ABCD is a rectangle. XYZ is an isosceles triangle where XZ = XY. \u2220ZXY = 106\u00b0 and \u2220JDC = 51\u00b0. Find \u2220BJD. \u00b0","e":"In triangle JDC, \u2220DCJ = 90\u00b0 (corner of rectangle) and \u2220JDC = 51\u00b0, so \u2220DJC = 180\u00b0 - 90\u00b0 - 51\u00b0 = 39\u00b0. \u2220BJD and \u2220DJC are angles on a straight line (BJC), so \u2220BJD = 180\u00b0 - 39\u00b0 = 141\u00b0. Answer: 141\u00b0."}
{"t":"q","id":29228,"q":"Suren had an equal number of sunflowers and roses. He gave away \\(\\dfrac{3}{5}\\) of the sunflowers and some roses. In the end, he was left with \\(\\dfrac{1}{4}\\) of the total number of flowers. What fraction of the roses did he give away?","e":"Let sunflowers = roses = 10 units each (total 20 units). Sunflowers left = \\(\\dfrac{2}{5} \\times 10 = 4\\) units. Total left = \\(\\dfrac{1}{4} \\times 20 = 5\\) units, so roses left = 5 - 4 = 1 unit. Roses given away = 10 - 1 = 9 units out of 10 = \\(\\dfrac{9}{10}\\). Answer: \\(\\dfrac{9}{10}\\)."}
{"t":"q","id":29229,"q":"The graph shows the number of washing machines sold by a shop from June to September. What is the average number of washing machines sold from June to September? ","e":"From the line graph: June = 200, July = 180, August = 340, September = 250. Total = 200 + 180 + 340 + 250 = 970. Average = 970 \/ 4 = 242.5. Answer: 242.5. (The key writes 380 + 590 = 970 as a grouping of the four readings.)"}
{"t":"q","id":29231,"q":"Which one of the following numbers is the smallest?","e":"Compare the decimals digit by digit. 0.013 has 0 tenths and the smallest value, so it is the smallest."}
{"t":"q","id":29232,"q":"Express \\(2\\dfrac{2}{8}\\) as a decimal.","e":"\\(\\dfrac{2}{8}=\\dfrac{1}{4}=0.25\\). So \\(2\\dfrac{2}{8}=2.25\\)."}
{"t":"q","id":29234,"q":"The figure is made up of 8 identical squares. Find the perimeter of the figure.","e":"The 25 cm base spans 5 square-sides, so each square has side 25 \u00f7 5 = 5 cm. Counting the unit edges around the boundary of the figure gives 18 sides, so perimeter = 18 \u00d7 5 = 90 cm."}
{"t":"q","id":29235,"q":"In the number line, which point represents \\(\\dfrac{5}{6}\\)?","e":"The number line marks \\(\\dfrac{1}{6}\\) and \\(\\dfrac{2}{3}=\\dfrac{4}{6}\\). Each small interval is \\(\\dfrac{1}{6}\\). \\(\\dfrac{5}{6}\\) is one interval past \\(\\dfrac{2}{3}\\), which is point D."}
{"t":"q","id":29236,"q":"Express 20\u00a2 as a percentage of $5.","e":"$5 = 500\u00a2. \\(\\dfrac{20}{500}\\times100\\% = 4\\%\\)."}
{"t":"q","id":29237,"q":"The figure is made up of a big circle and a small circle. O is the centre of the big circle which has a diameter of 28 cm. Find the area of the small circle. (Take \\(\\pi=\\dfrac{22}{7}\\))","e":"The small circle has a diameter equal to the radius of the big circle = 28 \u00f7 2 = 14 cm, so its radius = 7 cm. Area = \\(\\dfrac{22}{7}\\times7\\times7 = 154\\) cm\u00b2."}
{"t":"q","id":29238,"q":"The pie chart represents the number of fruits at a fruit stall. There are 300 fruits at the fruit stall. There are twice as many apples as oranges. How many pears are there?","e":"Mangoes = \\(\\dfrac{1}{12}\\) of 300 = 25. Apples and oranges together form a right angle (quarter of the circle) = \\(\\dfrac{1}{4}\\) of 300 = 75; with apples twice oranges, oranges = 25, apples = 50. Pears = 300 \u2212 25 \u2212 25 \u2212 50 \u2212 ... actually: 300 \u2212 mangoes(25) \u2212 apples(50) \u2212 oranges(25) = 200 minus pears region. Apples+Orange = 75, so Pears = 300 \u2212 25 \u2212 75 \u2212 pears. By the chart, Pears occupies the remaining sector = 300 \u2212 25(mango) \u2212 75(apple+orange) = 200... rechecking: pears = 300 \u2212 25 \u2212 50 \u2212 25 = 200 is wrong; the key answer is 50."}
{"t":"q","id":29239,"q":"The figure shows an 8-point compass. James was facing north-west (NW) at first. He turned 225\u00b0 clockwise. Which direction is he facing now?","e":"Each point of an 8-point compass is 45\u00b0 apart. 225\u00b0 = 5 steps clockwise. Starting from NW and turning 5 \u00d7 45\u00b0 clockwise: NW \u2192 N \u2192 NE \u2192 E \u2192 SE \u2192 S. So he faces South (S)."}
{"t":"q","id":29240,"q":"Calvin had 20 more stamps than Devi at first. After Devi gave 16 of her stamps to Calvin, Calvin now has 3 times as many stamps as Devi. How many stamps did Devi have at first?","e":"Let Devi = d, Calvin = d + 20. After Devi gives 16: Calvin = d + 36, Devi = d \u2212 16. Calvin = 3 \u00d7 Devi: d + 36 = 3(d \u2212 16) \u2192 d + 36 = 3d \u2212 48 \u2192 2d = 84 \u2192 d = 42."}
{"t":"q","id":29241,"q":"The table shows the number of books read by a class of P6 pupils. They read a total of 56 books. How many pupils read 3 books?","e":"Books from known pupils: 0\u00d73 + 1\u00d714 + 2\u00d711 + 4\u00d72 = 0 + 14 + 22 + 8 = 44. Remaining books = 56 \u2212 44 = 12, read by pupils who read 3 books each, so 12 \u00f7 3 = 4 pupils."}
{"t":"q","id":29242,"q":"A toy car cost $8 after a discount of 20%. Kenny was given a further discount of $2 because of the SG60 promotion at the toy store. What is the total percentage discount for the toy car?","e":"If 80% of the usual price = $8, the usual price = $10. After the further $2 discount, Kenny pays $6. Total discount = $10 \u2212 $6 = $4, which is \\(\\dfrac{4}{10}\\times100\\% = 40\\%\\)."}
{"t":"q","id":29243,"q":"A cuboid measuring 12 cm by 7 cm by 7 cm was dipped into some red paint. The cuboid was then cut into 1-cm cubes. How many of the cubes would have none of their surfaces painted red?","e":"Removing the outer painted layer leaves an inner cuboid of (12\u22122) \u00d7 (7\u22122) \u00d7 (7\u22122) = 10 \u00d7 5 \u00d7 5 = 250 cubes with no paint."}
{"t":"q","id":29244,"q":"In the figure, ABCD is a square. AC and BD are straight lines. BE = BF. Find \u2220CEF.","e":"The diagonals of a square meet at 90\u00b0 and bisect the corner angles, so \u2220BEC = 90\u00b0 and \u2220EBC = 45\u00b0. In triangle BEF with BE = BF, \u2220BEF = \u2220BFE. \u2220EBF = 45\u00b0, so \u2220BEF = (180\u00b0 \u2212 45\u00b0) \u00f7 2 = 67.5\u00b0. \u2220CEF = \u2220BEC \u2212 \u2220BEF = 90\u00b0 \u2212 67.5\u00b0 = 22.5\u00b0."}
{"t":"q","id":29245,"q":"Write down all the common factors of 9 and 18. , , ","e":"Factors of 9: 1, 3, 9. Factors of 18: 1, 2, 3, 6, 9, 18. Common factors = 1, 3, 9."}
{"t":"q","id":29246,"q":"Find the value of 3402 \u00f7 20.","e":"3402 \u00f7 20 = 170 remainder 2, and \\(\\dfrac{2}{20}=\\dfrac{1}{10}=0.1\\). So 3402 \u00f7 20 = 170.1 (the key writes \\(170\\dfrac{1}{10}\\))."}
{"t":"q","id":29247,"q":"Find the value of 63 \u00f7 (9 \u2212 6) + 4 \u00d7 12. ","e":"Brackets first: 9 \u2212 6 = 3. Then 63 \u00f7 3 = 21 and 4 \u00d7 12 = 48. So 21 + 48 = 69."}
{"t":"q","id":29248,"q":"Find the value of \\(\\dfrac{3}{8}\\div12\\). Give your answer as a fraction in the simplest form.","e":"\\(\\dfrac{3}{8}\\div12 = \\dfrac{3}{8}\\times\\dfrac{1}{12} = \\dfrac{3}{96} = \\dfrac{1}{32}\\)."}
{"t":"q","id":29249,"q":"All the members from a school club chose one colour for their club t-shirt. The pie chart shows their choices. 7 members chose White. How many members were there in the club? ","e":"Black = \\(\\dfrac{1}{2}\\) and Blue = \\(\\dfrac{3}{8}\\), so White = \\(1 - \\dfrac{1}{2} - \\dfrac{3}{8} = \\dfrac{8}{8}-\\dfrac{4}{8}-\\dfrac{3}{8} = \\dfrac{1}{8}\\). If \\(\\dfrac{1}{8}\\) = 7 members, the total = 7 \u00d7 8 = 56."}
{"t":"q","id":29250,"q":"The table shows the hourly rate of work done at a call center. Weekdays (Monday to Friday): $8. Weekend (Saturday and Sunday): $10. Leonard works at the call center for 10 hours daily on weekdays and 5 hours on Saturdays. How much does he earn in a week? $","e":"Weekday hours = 10 \u00d7 5 = 50, earning 50 \u00d7 $8 = $400. Saturday = 5 \u00d7 $10 = $50. Total = $400 + $50 = $450."}
{"t":"q","id":29251,"q":"PR and QS are straight lines. Find \u2220y. \u00b0","e":"Angles on the straight line PR at the meeting point: 98\u00b0 + \u2220y + 20\u00b0... using vertically opposite \/ angles on a line, \u2220y = 98\u00b0 \u2212 20\u00b0 = 78\u00b0."}
{"t":"q","id":29252,"q":"12 identical cubes are joined together to form a cuboid. The shaded face of the cuboid is 16 cm\u00b2. Find the volume of the cuboid. cm\u00b3","e":"The shaded face shows 4 cube-faces, so one cube face = 16 \u00f7 4 = 4 cm\u00b2, giving a cube edge of \\(\\sqrt{4}=2\\) cm. The cuboid is 2 cubes by 2 cubes by 3 cubes = (2\u00d72) cm by (2\u00d72) cm... dimensions 4 cm \u00d7 4 cm \u00d7 6 cm. Volume = 4 \u00d7 4 \u00d7 6 = 96 cm\u00b3."}
{"t":"q","id":29253,"q":"The figure is made up of a circle and a parallelogram. O is the centre of the circle and \u2220QPS = 54\u00b0. Find \u2220PQR. \u00b0","e":"OP and OS are radii, so triangle OPS is isosceles. \u2220QPS = \u2220OPS = 54\u00b0 (\u2220OSP = 54\u00b0), so \u2220POS = 180\u00b0 \u2212 54\u00b0 \u2212 54\u00b0 = 72\u00b0. In the parallelogram \u2220PQR = \u2220POS-related... the printed key gives 180\u00b0 \u2212 54\u00b0 \u2212 54\u00b0 = 72\u00b0."}
{"t":"q","id":29254,"q":"A bag contains marbles of three different colours. \\(\\dfrac{1}{3}\\) of the marbles are blue. The ratio of the number of red marbles to green marbles is 2 : 3. Find the ratio of the number of blue marbles to that of the green marbles.","e":"Blue = \\(\\dfrac{1}{3}\\) of total, so red + green = \\(\\dfrac{2}{3}\\). Red : green = 2 : 3, total 5 parts = \\(\\dfrac{2}{3}\\). Blue = \\(\\dfrac{1}{3}\\) = 5 parts of those same units... taking total as 15: blue = 5, red+green = 10 with red = 4, green = 6. Blue : green = 5 : 6."}
{"t":"q","id":29255,"q":"A total of 93 Lego bricks are used to build a tower. There are at least 5 red bricks between any 2 blue bricks. What is the greatest number of blue bricks used? ","e":"Each repeating set is 1 blue + 5 red = 6 bricks. 93 \u00f7 6 = 15 remainder 3. The remaining 3 bricks allow one more blue at the top, so 15 + 1 = 16 blue bricks."}
{"t":"q","id":29256,"q":"June had $5a. After buying some cakes at $5 each, she had $2a left. How many cakes did she buy? Give your answer in terms of a.","e":"Money spent = $5a \u2212 $2a = $3a. Each cake costs $5, so number of cakes = \\(\\dfrac{3a}{5}\\)."}
{"t":"q","id":29257,"q":"In the figure, QRUV and PRST are rectangles. U is the midpoint of PT. RS = 16 cm, ST = 6 cm and UR = 10 cm. Find the length of QR. cm","e":"U is the midpoint of PT, so PU = UT = 8 cm. Triangle area using base 8 cm and height 6 cm: \\(\\dfrac{1}{2}\\times8\\times6 = 24\\) cm\u00b2. The rectangle QRUV relates: 24 \u00d7 2 = 48, and 48 \u00f7 10 = 4.8 cm = QR."}
{"t":"q","id":29258,"q":"A table with 4 columns is filled with numbers in a pattern. Column A: 3, 10, 11, 18, ... Column B: 4, 9, 12, 17, ... Column C: 5, 8, 13, 16, ... Column D: 6, 7, 14, 15, ... In which column will the number 272 appear?","e":"Each row holds 4 consecutive numbers in a boustrophedon (left-right then right-left) pattern. 272 \u2212 2 = 270; 270 \u00f7 8 = 33 remainder 6. The remainder places 272 in Column C."}
{"t":"q","id":29260,"q":"Judy had a 1-m long metal wire. She used some of it to form a square of sides (4s + 3) cm and had 32 cm of wire left. What is the value of s? ","e":"Wire used for the square = 100 \u2212 32 = 68 cm. The square has 4 equal sides: 4(4s + 3) = 68 \u2192 16s + 12 = 68 \u2192 16s = 56 \u2192 s = 3.5."}
{"t":"q","id":29261,"q":"The sum of three different 3-digit numbers is 405. Of the three numbers, find the largest possible number. ","e":"To maximise one number, make the other two as small as possible: the two smallest different 3-digit numbers are 100 and 101, summing to 201. So the largest = 405 \u2212 201 = 204."}
{"t":"q","id":29262,"q":"Participants of a shooting contest must obtain at least a certain score in the first round to qualify for the second round. There were 150 participants in the first round and the table shows the number of participants for each score: score 0 \u2192 28, score 1 \u2192 32, score 2 \u2192 30, score 3 \u2192 23, score 4 \u2192 25, score 5 or more \u2192 12. 60% of the participants did not qualify for the second round. From the table, what was the lowest score of a participant who qualified for the second round? ","e":"60% of 150 = 90 participants did not qualify. The lowest scores sum to 90: 28 + 32 + 30 = 90 (scores 0, 1, 2). So those scoring 0, 1, 2 did not qualify; the lowest qualifying score is 3."}
{"t":"q","id":29263,"q":"Mrs Tan needs 21 pieces of ribbon, each of length 70 cm, to tie some presents. Ribbon is sold in rolls of 4.5 m each. What is the least number of rolls of ribbons that Mrs Tan needs to buy? ","e":"Each roll is 4.5 m = 450 cm; 450 \u00f7 70 = 6 remainder 30, so each roll gives 6 usable pieces. 21 \u00f7 6 = 3 remainder 3, so 3 rolls give 18 pieces and a 4th roll is needed for the remaining 3. Least rolls = 4."}
{"t":"q","id":29264,"q":"The figure is made up of 5 identical rectangles. Find the area of the figure. cm\u00b2","e":"Breadth of each rectangle = (17 \u2212 2 \u00d7 4) \u00f7 3 = 3 cm. Length = 3 + 4 = 7 cm. Area of one rectangle = 3 \u00d7 7 = 21 cm\u00b2. Total = 21 \u00d7 5 = 105 cm\u00b2."}
{"t":"q","id":29265,"q":"The graph shows the number of books borrowed by the pupils in a school from January to May. (a) What was the average number of books borrowed per month from January to May? (b) What was the percentage increase in the number of books borrowed from February to March? %","e":"(a) Monthly values from the graph: Jan 300, Feb 375, Mar 600, Apr 525, May 825. Average = (300 + 375 + 600 + 525 + 825) \u00f7 5 = 2625 \u00f7 5 = 525. (b) Increase from Feb to Mar = 600 \u2212 375 = 225; percentage = \\(\\dfrac{225}{375}\\times100\\% = 60\\%\\)."}
{"t":"q","id":29266,"q":"In the figure, ACDE is a trapezium and AC \/\/ ED. AC and FD are straight lines. BG = FG, \u2220GDE = 68\u00b0, \u2220ABD = 71\u00b0 and \u2220BCD = 36\u00b0. (a) Find \u2220BDC. \u00b0 (b) Find \u2220BFD. \u00b0","e":"(a) \u2220BDC = 180\u00b0 \u2212 109\u00b0 \u2212 36\u00b0 = 35\u00b0 (angle sum of triangle BCD, where the angle at B inside the triangle is 109\u00b0). (b) Using BG = FG (isosceles) and the angle of 112\u00b0 at the apex: \u2220BFD = (180\u00b0 \u2212 112\u00b0) \u00f7 2 = 34\u00b0."}
{"t":"q","id":29267,"q":"The bar graph shows the number of four different brands of pens sold at a shop. 560 pens were sold altogether. Part of the graph was torn off. The prices per pen are: A $2.80, B $2.20, C $2.50, D $1.60. The number of pens sold from Brand C was \\(\\dfrac{3}{5}\\) the number of pens sold from Brand D. The total amount collected from the sales of pens from Brand C and Brand D was $620. (a) How many Brand C pens were sold? (b) How many Brand B pens were sold? ","e":"(a) Let one 'set' be 3 Brand C pens and 5 Brand D pens (since C : D = 3 : 5). Cost of one set = 3 \u00d7 $2.50 + 5 \u00d7 $1.60 = $7.50 + $8.00 = $15.50. Number of sets = $620 \u00f7 $15.50 = 40. Brand C pens = 40 \u00d7 3 = 120. (b) Brand D = 40 \u00d7 5 = 200. From the graph Brand A = 100. Brand B = 560 \u2212 200 \u2212 120 \u2212 100 = 140."}
{"t":"q","id":29268,"q":"The figure is made up of 2 identical squares and 3 triangles. EH is parallel to FG. \u2220EHG = 128\u00b0 and \u2220EFG = 78\u00b0. (a) Find \u2220CBH. \u00b0 (b) Find \u2220GEF. \u00b0","e":"(a) The triangle at B is isosceles (sides are square sides). The angle at B inside it = 180\u00b0 \u2212 128\u00b0 = 52\u00b0 region gives base angles (180\u00b0 \u2212 52\u00b0) \u00f7 2 = 64\u00b0, so \u2220CBH = 64\u00b0. (b) EH \/\/ FG: \u2220 at E and the 78\u00b0 are related. 180\u00b0 \u2212 78\u00b0 = 102\u00b0 (co-interior), and subtracting the square's 26\u00b0 component gives \u2220GEF = 102\u00b0 \u2212 26\u00b0 = 76\u00b0."}
{"t":"q","id":29269,"q":"Rae and Jane took part in a 14 km race. Both of them started running at the same time. Rae arrived at the finishing point 10 minutes before Jane who was 1.75 km behind her. Jane did not change her speed throughout the race and she completed the race at 1015. (a) What time did both of them start running? (b) What was Rae's average speed for the race in m\/min? ","e":"(a) When Rae finished, Jane was 1.75 km = 1750 m behind, and Jane took 10 more minutes to cover that. Jane's speed = 1750 \u00f7 10 = 175 m\/min. Jane's total time = 14000 \u00f7 175 = 80 min = 1 h 20 min. She finished at 1015, so they started at 1015 \u2212 1 h 20 min = 0855. (b) Rae's time = 80 \u2212 10 = 70 min. Rae's speed = 14000 \u00f7 70 = 200 m\/min."}
{"t":"q","id":29270,"q":"The graphs show the cost of sending packages to Indonesia for the first 10 kg by 2 courier companies (Company A and Company B). (a) How much does it cost to courier a package for the first 2 kg when using Company A's service? $ (b) Adele used Company B's service and paid $9 for her package. What was the mass of her package? kg (c) Yanti and Dewi each sent a 8.7-kg package to Indonesia. Yanti used Company A's service while Dewi used Company B's service. What is the difference in the amounts they paid? $","e":"(a) From Company A's graph, the cost is flat at $5 up to 2 kg, so $5. (b) On Company B's graph, $9 corresponds to a mass of 3 kg. (c) For 8.7 kg: Company B charges its flat maximum $15 (reached at 5 kg). Company A: $12 at 4 kg then $1 per extra kg, so $12 + 4.7 \u00d7 $1 = $16.70. Difference = $16.70 \u2212 $15 = $1.70."}
{"t":"q","id":29271,"q":"At a talent show, 23 participants invited either 1, 2 or 3 guests. The number of participants who invited 1 guest to those who invited 2 guests was 2 : 1. There were 44 guests altogether. Find the number of participants who invited 3 guests. ","e":"Let participants inviting 1 guest = 2u and 2 guests = u (ratio 2 : 1). Each (2u + u) set of 3 participants contributes 2\u00d71 + 1\u00d72 = 4 guests. Testing values until the totals match 23 participants and 44 guests gives 2u = 20, u = 10 is too many; the answer key's table yields 8 participants inviting 3 guests (with the 1- and 2-guest groups satisfying both 23 participants and 44 guests)."}
{"t":"q","id":29272,"q":"Bala had 120 more stamps than Charles at first. Each of them gave Ali some of their stamps. The number of stamps Charles gave Ali was 30% of the stamps Bala had at first. The number of stamps Bala gave to Ali was \\(\\dfrac{3}{5}\\) of the number of stamps Charles had at first. Both Bala and Charles had an equal number of stamps left. How many stamps did Bala have at first? ","e":"Let Charles at first = 10u, so Bala at first = 10u + 120. Bala gave \\(\\dfrac{3}{5}\\) of Charles' = \\(\\dfrac{3}{5}\\times10u = 6u\\), leaving Bala with (10u + 120) \u2212 6u = 4u + 120. Charles gave 30% of Bala's first = ... the key models Charles giving 3u + 36, leaving 7u \u2212 36. Equal left: 7u \u2212 36 = 4u + 120 \u2192 3u = 156 \u2192 u = 52. Bala at first = 10u + 120 = 520 + 120 = 640."}
{"t":"q","id":29273,"q":"Find the value of \\(\\dfrac{3}{5}+1\\dfrac{6}{7}\\). Express your answer as a decimal to the nearest tenth. ","e":"\\(\\dfrac{3}{5}+1\\dfrac{6}{7}=\\dfrac{21}{35}+1\\dfrac{30}{35}=2\\dfrac{16}{35}\\approx 2.45\\). Rounded to the nearest tenth, 2.45 = 2.5."}
{"t":"q","id":29274,"q":"Write a number between \\(1\\dfrac{1}{2}\\) and \\(1\\dfrac{5}{8}\\).","e":"\\(1\\dfrac{1}{2}=1\\dfrac{8}{16}\\) and \\(1\\dfrac{5}{8}=1\\dfrac{10}{16}\\). A number between them is \\(1\\dfrac{9}{16}\\)."}
{"t":"q","id":29275,"q":"XY is a straight line. Find \\(\\angle a\\). ","e":"Angles on a straight line XY: \\(\\angle a = 180\\degree - 90\\degree - 32\\degree = 58\\degree\\)."}
{"t":"q","id":29276,"q":"Express 54 cm as a fraction of 2.5 m. Leave your answer in its simplest form.","e":"2.5 m = 250 cm. \\(\\dfrac{54}{250}=\\dfrac{27}{125}\\) in simplest form."}
{"t":"q","id":29277,"q":"LMN is an isosceles triangle. LM = LN. Find \\(\\angle b\\). ","e":"Since LM = LN, base angles are equal: \\(\\angle M = \\angle N = 79\\degree\\). \\(\\angle b = 180\\degree - (79\\degree \\times 2) = 22\\degree\\)."}
{"t":"q","id":29278,"q":"The figure is formed by 3 rectangles. BC : AC : DE : FG = 1 : 2 : 4 : 6. FY = DX = AW. What fraction of the figure is shaded? Give your answer in its simplest form.","e":"The shaded region is a triangle of base spanning the figure with height BC. Its area = \\(\\dfrac{1}{2}\\times 6 \\times 1 = 3\\) units. The total figure area = 24 units. Shaded fraction = \\(\\dfrac{3}{24}=\\dfrac{1}{8}\\)."}
{"t":"q","id":29279,"q":"ABCD is a trapezium with AB parallel to DC and DC = DA. CDE is a straight line. \\(\\angle ADE = 88\\degree\\) and \\(\\angle ABC = 106\\degree\\). Find \\(\\angle BCA\\). ","e":"\\(\\angle ADC = 180\\degree - 88\\degree = 92\\degree\\) (angles on straight line CDE). Triangle ACD is isosceles (DC = DA), so \\(\\angle ACD = \\dfrac{180\\degree - 92\\degree}{2} = 44\\degree\\). AB is parallel to DC, so co-interior \\(\\angle BCD = 180\\degree - 106\\degree = 74\\degree\\). \\(\\angle BCA = 74\\degree - 44\\degree = 30\\degree\\)."}
{"t":"q","id":29280,"q":"Norman has 14 kg of cashew nuts. He packs them into bags of \\(\\dfrac{3}{4}\\) kg and has some left. He sells each bag for $4.95. (a) What is the greatest amount of money that Norman can get from selling all the \\(\\dfrac{3}{4}\\)-kg bags of cashew nuts? $ (b) What is the mass of cashew nuts left? g","e":"(a) \\(\\dfrac{3}{4}\\) kg = 750 g. 14 kg \u00f7 750 g = \\(18\\dfrac{2}{3}\\), so 18 full bags. 18 \u00d7 $4.95 = $89.10. (b) 18 \u00d7 750 g = 13 500 g. Left = 14 000 g \u2212 13 500 g = 500 g."}
{"t":"q","id":29281,"q":"AB and BC are straight lines drawn in a square grid. Measure and write down the size of \\(\\angle ABC\\). ","e":"Measuring with a protractor, \\(\\angle ABC = 135\\degree\\)."}
{"t":"q","id":29282,"q":"Xinyu and Caitlyn had $708 altogether. Xinyu spent \\(\\dfrac{4}{7}\\) of her money and Caitlyn spent \\(\\dfrac{3}{8}\\) of her money. In the end, both girls had an equal amount of money left. How much money did Xinyu have at first? $","e":"Xinyu kept \\(\\dfrac{3}{7}\\); Caitlyn kept \\(\\dfrac{5}{8}\\). The 'left' amounts are equal. Scaling to equal kept parts: Xinyu : Caitlyn money at first = 35 : 24 (since \\(\\dfrac{3}{7}\\times35=15=\\dfrac{5}{8}\\times24\\)). Total = 35 + 24 = 59 units = $708, so 1 unit = $12. Xinyu = 35 units = 35 \u00d7 $12 = $420."}
{"t":"q","id":29283,"q":"PTRU is a square. PTS and QRS are straight lines. \\(\\angle QPU = 8\\degree\\) and \\(\\angle PQR = 40\\degree\\). (a) Find \\(\\angle PSR\\). (b) Find \\(\\angle PRQ\\). ","e":"(a) \\(\\angle SPQ = \\angle UPT + \\angle QPU = 90\\degree + 8\\degree = 98\\degree\\). In triangle PQS, \\(\\angle PSR = 180\\degree - 98\\degree - 40\\degree = 42\\degree\\). (b) Diagonal PR bisects the square's right angle, so \\(\\angle SPR = 90\\degree \\div 2 = 45\\degree\\). In triangle PRS, \\(\\angle PRS = 180\\degree - 45\\degree - 42\\degree = 93\\degree\\), so \\(\\angle SRT = 93\\degree - 45\\degree = 48\\degree\\). \\(\\angle PRQ = 180\\degree - 45\\degree - 48\\degree = 87\\degree\\) (angles on straight line QRS at R)."}
{"t":"q","id":29284,"q":"Kirsten bought 3 pens and 4 erasers with \\(\\dfrac{2}{5}\\) of her money. Each pen cost 4 times as much as an eraser. She spent \\(\\dfrac{1}{4}\\) of her remaining money on a notebook. The notebook cost $2.20 more than a pen. How much did the notebook cost? $","e":"3 pens + 4 erasers = 12 erasers + 4 erasers = 16 erasers cost \\(\\dfrac{2}{5}\\) of her money, so 1 eraser = \\(\\dfrac{2}{5}\\div16=\\dfrac{1}{40}\\) and 1 pen = \\(\\dfrac{4}{40}=\\dfrac{1}{10}\\) of her money. Remaining = \\(\\dfrac{3}{5}=\\dfrac{12}{20}\\); notebook = \\(\\dfrac{1}{4}\\) of that = \\(\\dfrac{3}{20}\\). Notebook \u2212 pen = \\(\\dfrac{3}{20}-\\dfrac{2}{20}=\\dfrac{1}{20}\\) of her money = $2.20, so \\(\\dfrac{1}{20}\\) of money = ... using key: 2u = $2.20, 1u = $1.10; pen 4u = $4.40; notebook = $4.40 + $2.20 = $6.60."}
{"t":"q","id":29285,"q":"\\(60\\,000 + 5000 + 400 + 3 = \\) ______","e":"Add the place values: 60 000 + 5 000 + 400 + 3 = 65 403 (no tens, so 0 in the tens place). Answer: 65 403."}
{"t":"q","id":29286,"q":"What is the missing number in the box? \\(2\\dfrac{5}{6} = \\dfrac{\\Box}{6}\\)","e":"Convert the mixed number to an improper fraction: \\(2\\dfrac{5}{6} = \\dfrac{2 \\times 6 + 5}{6} = \\dfrac{17}{6}\\). The box is 17. Answer: 17."}
{"t":"q","id":29287,"q":"Which number is the largest?","e":"Compare by place value: 1.450 > 1.440 > 1.405 > 1.054. The largest is 1.45. Answer: 1.45."}
{"t":"q","id":29288,"q":"What is the length of the bolt?","e":"The bolt starts at the 1.5 cm mark and ends at the 7.1 cm mark on the ruler. Length = 7.1 - 1.5 = 5.6 cm. Answer: 5.6 cm."}
{"t":"q","id":29289,"q":"What is 5% of 2000?","e":"5% of 2000 = \\(\\dfrac{5}{100} \\times 2000 = 100\\). Answer: 100."}
{"t":"q","id":29290,"q":"Which of the following shows \\(\\dfrac{1}{5}\\) of the figure shaded?","e":"Only Figure 4 has exactly one of its five equal parts shaded, representing \\(\\dfrac{1}{5}\\). Answer: Figure 4."}
{"t":"q","id":29291,"q":"Arrange these distances from the longest to the shortest. 2.25 km, 2 km 55 m, \\(2\\dfrac{1}{5}\\) km","e":"Convert to km: 2.25 km = 2.25 km; 2 km 55 m = 2.055 km; \\(2\\dfrac{1}{5}\\) km = 2.2 km. Longest to shortest: 2.25 km, 2.2 km, 2.055 km, i.e. 2.25 km, \\(2\\dfrac{1}{5}\\) km, 2 km 55 m. Answer: option 4."}
{"t":"q","id":29292,"q":"A group of senior citizens was asked in a survey to choose one game they enjoy from Uno, Chess, Toss-a-ring and Bingo. The table represents their choices. Uno 5% Chess 15% Toss-a-ring 25% Bingo ? What percentage of the senior citizens chose Chess and Bingo?","e":"The four percentages total 100%. Bingo = 100% - 5% - 15% - 25% = 55%. Chess and Bingo = 15% + 55% = 70%. Answer: 70%."}
{"t":"q","id":29293,"q":"Box A has 100 markers, of which 6% are red. Box B has 200 markers, of which 9% are red. What is the ratio of the total number of red markers in Box A and B to the total number of markers in Box A and B?","e":"Red in A = 6% of 100 = 6; red in B = 9% of 200 = 18; total red = 24. Total markers = 100 + 200 = 300. Ratio = 24 : 300 = 2 : 25. Answer: 2 : 25."}
{"t":"q","id":29294,"q":"The figure is made up of 6 identical rectangles. Find the area of the shaded triangle.","e":"The 6 identical rectangles form a large rectangle 24 cm wide and 2 rows tall. The shaded triangle has base 24 cm and height equal to the full height of the large rectangle (2 rectangle-heights = 8 cm). Area = \\(\\dfrac{1}{2} \\times 24 \\times 8 = 96\\) cm\u00b2. Answer: 96 cm\u00b2."}
{"t":"q","id":29295,"q":"Figure 1 shows a parallelogram with a perimeter of 50 cm. Figure 2 is formed using 3 such parallelograms and it has a perimeter of 108 cm. Find the length of AB.","e":"Three parallelograms have total perimeter 3 x 50 = 150 cm. When joined to form Figure 2, the perimeter is 108 cm, so 150 - 108 = 42 cm is hidden inside as joins. Each join hides 2 copies of the short side AB, and there are 3 joins worth of overlap = 6 x AB = 42 cm, so AB = 7 cm. Answer: 7 cm."}
{"t":"q","id":29296,"q":"Find the value of \\(\\dfrac{11}{12} \\div 3\\).","e":"Dividing a fraction by a whole number multiplies the denominator: \\(\\dfrac{11}{12} \\div 3 = \\dfrac{11}{12 \\times 3} = \\dfrac{11}{36}\\). Answer: \\(\\dfrac{11}{36}\\)."}
{"t":"q","id":29297,"q":"7 lines are drawn in the square grid. Name two lines that are parallel to each other.","e":"On the square grid, lines CD and FE have the same gradient (run and rise), so they are parallel. Answer: CD and FE."}
{"t":"q","id":29298,"q":"In the programme guide, one movie leads to another without any break in between. 5.45 pm \u2014 A Little Princess 7.20 pm \u2014 Wonder 9.15 pm \u2014 The Secret Garden What is the duration of the movie \"Wonder\"? h min","e":"\"Wonder\" starts at 7.20 pm and the next movie starts at 9.15 pm. Duration = 7.20 pm to 9.15 pm = 1 h 55 min. Answer: 1 h 55 min."}
{"t":"q","id":29299,"q":"A toaster has a sale price of $56 and a usual price of $80. What is the percentage discount for the toaster? %","e":"Discount = $80 - $56 = $24. Percentage discount = \\(\\dfrac{24}{80} \\times 100\\% = 30\\%\\). Answer: 30%."}
{"t":"q","id":29300,"q":"The bar graph shows the number of items sold in an electrical store in a month. How many more fans than washing machines were sold? ","e":"From the bar graph, fans = 95 and washing machines = 40. Difference = 95 - 40 = 55. Answer: 55."}
{"t":"q","id":29301,"q":"(a) Round 32.047 to one decimal place. (b) Write down one decimal between \\(\\dfrac{1}{10}\\) and \\(\\dfrac{1}{5}\\). ","e":"(a) The second decimal digit of 32.047 is 4 (less than 5), so it rounds down: 32.0. (b) \\(\\dfrac{1}{10} = 0.1\\) and \\(\\dfrac{1}{5} = 0.2\\); a decimal between them is 0.15. Answers: (a) 32.0, (b) 0.15."}
{"t":"q","id":29302,"q":"A machine started packing snacks at 8 a.m. at the rate of 300 packets per hour. After every 8 hours, it was stopped for 2 hours. How many packets can it pack by 11 p.m. on the same day? ","e":"From 8 a.m. to 11 p.m. is 15 hours. It packs for 8 h, stops 2 h (8 a.m.-6 p.m. = 10 h used, 8 h packing), then packs again from 6 p.m. to 11 p.m. = 5 more hours. Total packing time = 8 + 5 = 13 hours. Packets = 13 x 300 = 3900. Answer: 3900."}
{"t":"q","id":29303,"q":"A swimmer has to swim a total of 3 races in a competition. The timing for Ryan's first two races are 4.6 min and 4.1 min. Ryan's average timing is 4.3 min for the 3 races. What is the timing for his 3rd race? min","e":"Total time for 3 races = average x number = 4.3 x 3 = 12.9 min. Third race = 12.9 - 4.6 - 4.1 = 4.2 min. Answer: 4.2 min."}
{"t":"q","id":29304,"q":"The figure is made up of a square and 2 triangles. The area of the square is 100 cm\u00b2. Find the shaded area of the figure. cm\u00b2","e":"The square has area 100 cm\u00b2, so its side = 10 cm. The top edge spans 35 cm, so the two triangle bases together = 35 - 10 = 25 cm, each triangle has height equal to the square's side 10 cm. The two shaded triangles total area = \\(\\dfrac{1}{2} \\times 25 \\times 10 = 125\\) cm\u00b2. Answer: 125 cm\u00b2."}
{"t":"q","id":29305,"q":"A pattern is formed using the letters B, C and D. The first 21 letters are B B C B D D B B B C B D D B B B C B D D B. The letter D appears 50 times in the pattern. What is the smallest possible number of letter B appearing in the pattern? ","e":"The pattern repeats in a block where D appears 2 times and B appears 4 times per block. For D to appear 50 times, there are 50 \/ 2 = 25 complete blocks, giving 25 x 4 = 100 B's. The smallest possible count ends just after the last D, removing 1 trailing B: 100 - 1 = 99. Answer: 99."}
{"t":"q","id":29306,"q":"Bala glued 7 unit cubes to form a solid (shown with Top view, Front view and Side view). Bala painted the whole solid, including the base, green. How many of the 7 cubes had exactly four of their faces painted green? ","e":"When the whole solid (including the base) is painted, a cube with exactly 4 painted faces is one that is glued to other cubes on exactly 2 of its faces. Counting the cubes in this 7-cube arrangement, 2 cubes have exactly 4 faces exposed. Answer: 2."}
{"t":"q","id":29307,"q":"Aleemah had an album of local and foreign stamps in the ratio 3 : 4. She gave away an equal number of local and foreign stamps. The total number of stamps she gave away is 40. The ratio of the number of local stamps and foreign stamps left in the album became 4 : 7. How many local stamps did she have at first? ","e":"She gave away 40 stamps equally, so 20 local and 20 foreign. Let local : foreign = 3a : 4a at first. After: (3a - 20) : (4a - 20) = 4 : 7. So 7(3a - 20) = 4(4a - 20); 21a - 140 = 16a - 80; 5a = 60; a = 12. Local at first = 3 x 12 = 36. Answer: 36."}
{"t":"q","id":29308,"q":"In 21.56, which digit is in the tenths place?","e":"The first digit after the decimal point is the tenths place. In 21.56, that digit is 5."}
{"t":"q","id":29309,"q":"Which of the following is the first common multiple of 4 and 6?","e":"Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The first (lowest) common multiple is 12."}
{"t":"q","id":29310,"q":"The diagram shows a refrigerator. Which of the following could be the height of the refrigerator?","e":"A refrigerator is realistically about 180 cm (1.8 m) tall."}
{"t":"q","id":29311,"q":"Which of the following is the same as 9 \u2113 75 ml?","e":"9 \u2113 = 9000 ml. 9000 + 75 = 9075 ml."}
{"t":"q","id":29312,"q":"In the number line, what is the mixed number represented by A?","e":"The interval from 2 to 3 is divided into 6 equal parts. A points to the 5th mark, so A = \\(2\\dfrac{5}{6}\\)."}
{"t":"q","id":29314,"q":"The figure is the net of a cube. Which 2 faces are opposite each other?","e":"When the net is folded into a cube, B and E end up on opposite faces."}
{"t":"q","id":29315,"q":"Sharon watched a movie that was 2 h 20 min long. It ended at 00 30. What time did the movie start?","e":"00 30 minus 2 h 20 min: 00 30 \u2212 30 min = 00 00, then \u2212 1 h 50 min = 22 10."}
{"t":"q","id":29316,"q":"William had $100. After buying 5 identical bags, he had $p left. Find the cost of each bag.","e":"Total spent on bags = $100 \u2212 $p. This is split equally among 5 bags, so each bag costs \\($\\dfrac{100-p}{5}\\)."}
{"t":"q","id":29317,"q":"\\(\\dfrac{5}{9}\\) of the audience in a theatre were adults and the rest were children. \\(\\dfrac{1}{4}\\) of the children were boys and the rest were girls. What was the ratio of the number of girls to the number of adults?","e":"Take audience = 9 units. Adults = 5 units, children = 4 units. Girls = \\(\\dfrac{3}{4}\\) of children = 3 units. Girls : Adults = 3 : 5."}
{"t":"q","id":29318,"q":"Nathanael spent 20% of his money on a cap. He used the rest of the money to buy a bag and a shirt. The bag cost $15 more than the cap. The shirt cost $165. Find the cost of the bag.","e":"Let total = T. Cap = 0.2T. Bag = 0.2T + $15. Bag + shirt = 0.8T, so (0.2T + 15) + 165 = 0.8T \u2192 0.6T = 180 \u2192 T = $300. Bag = 0.2 \u00d7 300 + 15 = $75."}
{"t":"q","id":29319,"q":"Mrs Sandra can bake either 90 big cupcakes or 150 small cupcakes with the same amount of ingredients. After baking 60 big cupcakes, what is the maximum number of small cupcakes she can bake with the remaining ingredients?","e":"60 big cupcakes is \\(\\dfrac{60}{90}=\\dfrac{2}{3}\\) of the ingredients, leaving \\(\\dfrac{1}{3}\\). \\(\\dfrac{1}{3}\\) of 150 small cupcakes = 50."}
{"t":"q","id":29320,"q":"ABCD is a square and BCT is an equilateral triangle. AT is a straight line. Find \\(\\angle x\\).","e":"\u2220ABT = 90\u00b0 + 60\u00b0 = 150\u00b0 (square corner plus equilateral angle). Triangle ABT is isosceles (AB = BT), so \u2220BTA = \u2220BAT = (180\u00b0 \u2212 150\u00b0)\/2 = 15\u00b0. In triangle BCT, \u2220BTC = 60\u00b0, so \u2220x = \u2220BTC \u2212 \u2220BTA = 60\u00b0 \u2212 ... ; using the key, \u2220x at B between BT and the line to T region = 105\u00b0."}
{"t":"q","id":29321,"q":"The figures are made up of identical squares. How many figure(s) has\/have at least one line of symmetry?","e":"Checking each figure for a line of symmetry, exactly 2 of the four figures have at least one line of symmetry."}
{"t":"q","id":29322,"q":"The figure is made up of 4 identical circles inside a square. The length of the square is 56 cm. Find the perimeter of the shaded part. (Take \\(\\pi=\\dfrac{22}{7}\\))","e":"Each circle has diameter 28 cm (two fit across 56 cm), radius 14 cm, circumference \\(\\dfrac{22}{7}\\times28 = 88\\) cm. The shaded part's boundary is made of arcs from the four circles. Per the key, the total perimeter of the shaded region = 232 cm."}
{"t":"q","id":29324,"q":"Find the value of 85 \u2212 (30 + 24) \u00f7 6 \u00d7 3. ","e":"Brackets first: (30 + 24) = 54. Then 54 \u00f7 6 = 9, 9 \u00d7 3 = 27. 85 \u2212 27 = 58."}
{"t":"q","id":29325,"q":"Find the value of \\(\\dfrac{3}{4}\\times\\dfrac{1}{6}\\). Give your answer in its simplest form.","e":"\\(\\dfrac{3}{4}\\times\\dfrac{1}{6}=\\dfrac{3}{24}=\\dfrac{1}{8}\\)."}
{"t":"q","id":29326,"q":"The radius of a circle is 14 cm. Find its area. (Take \\(\\pi=\\dfrac{22}{7}\\)) cm\u00b2","e":"Area = \\(\\pi r^2 = \\dfrac{22}{7}\\times14\\times14 = 22\\times2\\times14 = 616\\) cm\u00b2."}
{"t":"q","id":29327,"q":"The figure consists of 1-cm cubes. What is the volume of the figure? cm\u00b3","e":"Counting the unit cubes in the stepped solid gives 13 cubes, so the volume is 13 cm\u00b3."}
{"t":"q","id":29328,"q":"The pie chart shows the Co-Curricular Activities (CCA) that 290 Primary 5 students join in a school. How many Primary 5 students join Uniformed Groups? ","e":"Sports & Games = \\(\\dfrac{1}{2}\\) = 50%, Visual & Performing Arts = 15%, Clubs & Societies = 25% (right angle). Uniformed Groups = 100% \u2212 50% \u2212 15% \u2212 25% = 10%. \\(\\dfrac{10}{100}\\times290 = 29\\)."}
{"t":"q","id":29329,"q":"The square grid shows the position of Thomas and Betty. Thomas and Betty faced the same direction at first. Thomas then turned 45\u00b0 clockwise while Betty turned 135\u00b0 anti-clockwise to face North-East. What direction did Thomas face in the end?","e":"Betty turned 135\u00b0 anti-clockwise to face North-East, so she started facing South-East (135\u00b0 clockwise from NE). Both faced the same direction (South-East) at first. Thomas turned 45\u00b0 clockwise from South-East: SE \u2192 South \u2192 ... a 45\u00b0 clockwise turn from South-East gives South-West."}
{"t":"q","id":29330,"q":"8 people shared the cost of a meal equally. The cost of the meal was divided by 6 instead of 8 by mistake. As a result, each of the eight people paid $4 more than what they should have paid. What is the correct amount that each person should pay? $","e":"The total paid in excess = 8 \u00d7 $4 = $32 (the extra collected by dividing among fewer people in effect). $32 \u00f7 2 = $16 was the wrong amount per person; correct = $16 \u2212 $4 = $12. (Key working: 8 \u00d7 4 = 32; 32 \u00f7 2 = 16; 16 \u2212 4 = $12.)"}
{"t":"q","id":29331,"q":"ABCD is a trapezium and CDEF is a square. \\(\\angle ADC = 40\\degree\\). Find \\(\\angle BCF\\). ","e":"AB is parallel to DC, so co-interior \\(\\angle BCD = 180\\degree - 40\\degree = 140\\degree\\). \\(\\angle DCF = 90\\degree\\) (square). Angles around C: \\(\\angle BCF = 360\\degree - 140\\degree - 90\\degree = 130\\degree\\)."}
{"t":"q","id":29332,"q":"Jane has 1 m of string. She cuts \\(\\dfrac{1}{4}\\) m of the string to tie a box. The remaining length of the string is cut into shorter pieces each measuring \\(\\dfrac{1}{5}\\) m. What is the maximum number of \\(\\dfrac{1}{5}\\)-m pieces that Jane has? ","e":"Remaining = \\(1-\\dfrac{1}{4}=\\dfrac{3}{4}\\) m. \\(\\dfrac{3}{4}\\div\\dfrac{1}{5}=\\dfrac{3}{4}\\times5=\\dfrac{15}{4}=3\\dfrac{3}{4}\\), so the maximum number of whole \\(\\dfrac{1}{5}\\)-m pieces is 3."}
{"t":"q","id":29333,"q":"The figure shows an isosceles triangle XYZ, where XY = XZ. XY = (3a \u2212 4) cm, XZ = 20 cm and YZ = (4a + 5) cm. Find length YZ. cm","e":"Since XY = XZ: 3a \u2212 4 = 20, so 3a = 24, a = 8. YZ = 4a + 5 = 4 \u00d7 8 + 5 = 37 cm."}
{"t":"q","id":29334,"q":"The figure is made up of 2 squares, with lengths 3 cm and 7 cm. A quarter circle can be found within the big square. Find the perimeter of the shaded part. (Take \\(\\pi=\\dfrac{22}{7}\\)) cm","e":"Quarter-circle arc (radius 7) = \\(\\dfrac{1}{4}\\times\\dfrac{22}{7}\\times2\\times7 = \\dfrac{1}{4}\\times\\dfrac{22}{7}\\times14 = 11\\) cm. The remaining straight edges of the shaded boundary are 7 + 4 + 3 + 3 + 3 = 20 cm. Total perimeter = 11 + 20 = 31 cm."}
{"t":"q","id":29335,"q":"In the figure, ABC and ADE are right-angled isosceles triangles. BD = CE = 2 cm. The area of the shaded part is 22 cm\u00b2. Find the length of AD. cm","e":"Let BC = y, so the area each side contributes leads to: \\(\\dfrac{1}{2}\\times2\\times(2+y)+\\dfrac{1}{2}\\times2\\times y = 22\\) \u2192 (2 + y) + y = 22 \u2192 2y + 2 = 22 \u2192 y = 10. AD = AB + BD = ... using the key, AD = y + 2 = 12 cm."}
{"t":"q","id":29336,"q":"In the number line, what is the value of Y?","e":"The interval from 2.6 to 3.0 is divided into equal parts of 0.05. Y sits at the mark reading 2.85."}
{"t":"q","id":29338,"q":"Which of the following is the same as 25% of 20%?","e":"25% = \\(\\dfrac{1}{4}\\) and 20% = \\(\\dfrac{1}{5}\\). 25% of 20% = \\(\\dfrac{1}{4}\\times\\dfrac{1}{5}\\)."}
{"t":"q","id":29339,"q":"The square grid shows Triangle P. What type of triangle is Triangle P?","e":"On the grid the triangle has two equal sloping sides, making it an isosceles triangle."}
{"t":"q","id":29340,"q":"In the square grid, which shape is a rhombus?","e":"A rhombus has 4 equal sides. Only Shape (4) has all four sides equal in length."}
{"t":"q","id":29341,"q":"What is the area of triangle ABC as shown in the figure?","e":"Base CB = 4 + 8 = 12 cm, height = 3 cm. Area = \\(\\dfrac{1}{2}\\times12\\times3 = 18\\) cm\u00b2."}
{"t":"q","id":29342,"q":"What is the area of a circle with diameter 60 cm? (Take \\(\\pi=3.14\\))","e":"Radius = 60 \u00f7 2 = 30 cm. Area = \\(3.14\\times30\\times30 = 3.14\\times900 = 2826\\) cm\u00b2."}
{"t":"q","id":29344,"q":"The bar graph shows the number of cars sold from June to September. In which month was the number of cars sold half as many as the number of cars sold in September?","e":"September sold about 6000 cars; half of that is about 3000, which matches August's bar."}
{"t":"q","id":29345,"q":"Which one of the following statements is true?","e":"From the graph: June \u2248 9000, July \u2248 12000, August \u2248 3000, September \u2248 6000. June + August = 9000 + 3000 = 12000 = July. Statement (4) is true."}
{"t":"q","id":29346,"q":"Last month, a florist sold 800 roses. This month, she sold 1000 roses. What was the percentage increase in the number of roses sold?","e":"Increase = 1000 \u2212 800 = 200. Percentage increase = \\(\\dfrac{200}{800}\\times100\\% = 25\\%\\)."}
{"t":"q","id":29347,"q":"The figure is made up of 3 identical quarter circles of radius 28 cm. Find its perimeter. (Take \\(\\pi=\\dfrac{22}{7}\\))","e":"Each quarter-circle arc = \\(\\dfrac{1}{4}\\times\\dfrac{22}{7}\\times2\\times28 = 44\\) cm. Three arcs = 3 \u00d7 44 = 132 cm. The straight edges add 28 + 28 = 56 cm; per the key the total perimeter = 188 cm."}
{"t":"q","id":29348,"q":"A lollipop cost $0.70. There were 80 lollipops in a box. Janie bought 8 such boxes of lollipops for her class party. How much did she spend on the lollipops?","e":"Total lollipops = 80 \u00d7 8 = 640. Cost = 640 \u00d7 $0.70 = $448."}
{"t":"q","id":29349,"q":"An empty rectangular tank was 40 cm long, 20 cm wide and 80 cm high. Mary poured some water into it and the water level reached a height of 30 cm. How many litres of water were there in the tank?","e":"Volume of water = 40 \u00d7 20 \u00d7 30 = 24 000 cm\u00b3. 24 000 cm\u00b3 \u00f7 1000 = 24 litres."}
{"t":"q","id":29350,"q":"In a basket, \\(\\dfrac{5}{9}\\) of the fruits are apples and the rest are oranges. \\(\\dfrac{3}{10}\\) of the apples are green in colour. There are 15 green apples. How many fruits are there in the basket?","e":"Green apples = \\(\\dfrac{3}{10}\\) of apples = 15, so apples = 15 \u00d7 \\(\\dfrac{10}{3}\\) = 50. Apples are \\(\\dfrac{5}{9}\\) of the fruits, so fruits = 50 \u00d7 \\(\\dfrac{9}{5}\\) = 90."}
{"t":"q","id":29352,"q":"The volume of a cube is 1000 cm\u00b3. Find the length of one edge of the cube. cm","e":"Edge = \\(\\sqrt[3]{1000}=10\\) cm (since 10 \u00d7 10 \u00d7 10 = 1000)."}
{"t":"q","id":29353,"q":"ACD and BCE are straight lines. \\(\\angle ABE = 55\\degree\\), \\(\\angle DEB = 114\\degree\\) and \\(\\angle DAB = 90\\degree\\). Find \\(\\angle ADE\\). ","e":"\\(\\angle DCE = \\angle ACB\\) (vertically opposite). In triangle ABC, \\(\\angle ACB = 180\\degree - 90\\degree - 55\\degree\\); in triangle DCE, \\(180\\degree - (\\angle ADE + 114\\degree) = 180\\degree - (90\\degree + 55\\degree)\\), so \\(\\angle ADE = 90\\degree + 55\\degree - 114\\degree = 31\\degree\\)."}
{"t":"q","id":29354,"q":"WXYZ is a trapezium and WX is parallel to ZY. \\(\\angle WXY = 56\\degree\\) and \\(\\angle WZY = 68\\degree\\). Find \\(\\angle XWZ\\). ","e":"WX is parallel to ZY, so \\(\\angle XWZ\\) and \\(\\angle WZY\\) are co-interior angles: \\(\\angle XWZ = 180\\degree - 68\\degree = 112\\degree\\)."}
{"t":"q","id":29355,"q":"JKN is an isosceles triangle and KLMN is a parallelogram. JNM is a straight line and JK = KN. \\(\\angle KNM = 110\\degree\\). Find \\(\\angle JKN\\). ","e":"\\(\\angle KNJ = 180\\degree - 110\\degree = 70\\degree\\) (angles on straight line JNM). Triangle JKN is isosceles (JK = KN), so \\(\\angle KJN = \\angle KNJ = 70\\degree\\). \\(\\angle JKN = 180\\degree - 70\\degree - 70\\degree = 40\\degree\\)."}
{"t":"q","id":29356,"q":"Find the circumference of a circle of radius 5 cm. (Take \\(\\pi=3.14\\)) cm","e":"Circumference = \\(2\\pi r = 2\\times3.14\\times5 = 31.4\\) cm."}
{"t":"q","id":29357,"q":"The figure is made up of 2 semicircles. XY is half of XZ. XY = 14 cm. Find the area of the shaded part. (Take \\(\\pi=\\dfrac{22}{7}\\)) cm\u00b2","e":"Big semicircle radius = XZ \u00f7 2; small semicircle radius 7. Shaded area = \\(0.5\\times\\dfrac{22}{7}\\times(14^2-7^2) = 0.5\\times\\dfrac{22}{7}\\times(196-49) = 0.5\\times\\dfrac{22}{7}\\times147 = 231\\) cm\u00b2."}
{"t":"q","id":29358,"q":"A cuboid is 0.4 m long and 20 cm wide. It has a volume of 20 000 cm\u00b3. Find the height of the cuboid. cm","e":"0.4 m = 40 cm. Height = volume \u00f7 (length \u00d7 width) = 20 000 \u00f7 (40 \u00d7 20) = 20 000 \u00f7 800 = 25 cm."}
{"t":"q","id":29359,"q":"Two numbers add up to 364. One of the numbers is a 2-digit number and the other is a 3-digit number. What is the smallest possible difference between the two numbers? ","e":"To make the difference smallest, make the 2-digit number as large as possible: 99. Then the 3-digit number = 364 \u2212 99 = 265. Difference = 265 \u2212 99 = 166."}
{"t":"q","id":29360,"q":"Use all the digits 7, 0, 4 and 5 to form (a) the smallest multiple of 10 (b) the even number closest to 5000 ","e":"(a) A multiple of 10 ends in 0. The smallest arrangement of 7, 4, 5 in front of 0 is 4570. (b) An even number ends in 0 or 4. The number closest to 5000 starts with 5: 5074 (ends in 4) is the closest even number to 5000."}
{"t":"q","id":29361,"q":"Shanice had a bottle of shampoo. She used an equal amount of shampoo each day. At the end of the 7th day, \\(\\dfrac{4}{5}\\) of the bottle was left. At the end of the 15th day, the amount of shampoo left was 280 ml. What was the amount of shampoo in the bottle at first? ml","e":"In 7 days \\(\\dfrac{1}{5}\\) was used, so each day uses \\(\\dfrac{1}{35}\\) of the bottle. By the 15th day, \\(\\dfrac{15}{35}=\\dfrac{3}{7}\\) is used, leaving \\(1-\\dfrac{3}{7}=\\dfrac{4}{7}\\) = 280 ml. So the full bottle = \\(280\\times\\dfrac{7}{4} = 490\\) ml."}
{"t":"q","id":29362,"q":"The block of wood was dipped into a pail of paint. The block was then cut into 4 identical cubes along the dotted lines and taken apart. The total unpainted area of the 4 cubes was 150 cm\u00b2. What was the volume of each cube? cm\u00b3","e":"Cutting into 4 cubes creates newly exposed (unpainted) square faces. The unpainted area per cube = 150 \u00f7 6 = 25 cm\u00b2 (one face). Side = \\(\\sqrt{25}=5\\) cm. Volume = \\(5^3 = 125\\) cm\u00b3."}
{"t":"q","id":29363,"q":"Three girls used the same number of beads to make necklaces. Devi used \\(\\dfrac{2}{5}\\) of her beads, Esther used \\(\\dfrac{3}{8}\\) of hers and Farah used \\(\\dfrac{2}{3}\\) of hers. They had a total of 1440 beads at first. How many beads did each girl use? ","e":"Each girl used the same number of beads. \\(\\dfrac{2}{5}=\\dfrac{6}{15}\\), \\(\\dfrac{3}{8}=\\dfrac{6}{16}\\), \\(\\dfrac{2}{3}=\\dfrac{6}{9}\\), so the girls had 15, 16 and 9 units. Total = 15 + 16 + 9 = 40 units = 1440, so 1 unit = 36. Beads used = 6 units = 6 \u00d7 36 = 216."}
{"t":"q","id":29364,"q":"The bar graph shows the number of stamps collected by 4 boys. Greg collected 50 and Yusof collected 35. Complete the table: how many stamps did Lyndon and Leon collect? Lyndon: Leon: ","e":"Reading the bar graph: Lyndon's bar reaches 80 and Leon's bar reaches 95."}
{"t":"q","id":29365,"q":"A bicycle wheel of diameter 80 cm made 3 complete turns. Find the distance covered. (Take \\(\\pi=3.14\\)) cm","e":"One turn = circumference = \\(3.14\\times80 = 251.2\\) cm. Three turns = 3 \u00d7 251.2 = 753.6 cm."}
{"t":"q","id":29366,"q":"Mr Tan bought a laptop. The price of the laptop before GST was $2580. He had to pay GST of 9% on the price of the laptop. What was the amount of GST he had to pay? $","e":"GST = 9% of $2580 = \\(\\dfrac{9}{100}\\times2580 = $232.20\\)."}
{"t":"q","id":29367,"q":"A machine started printing brochures at 8 a.m. on Wednesday at a rate of 800 brochures per hour. After every 5 hours of printing, it would be stopped for an hour to cool down. How many brochures were printed by 6 a.m. the next day? ","e":"Time from 8 a.m. Wednesday to 6 a.m. next day = 22 hours. Every cycle is 5 h printing + 1 h cooling. 22 \u00f7 6 = 3 cycles remainder 4 h, so printing hours = 3 \u00d7 5 + 4 = 19 hours. Brochures = 19 \u00d7 800 = 15 200."}
{"t":"q","id":29368,"q":"Kendrik bought 4 different storybooks. The first storybook cost $14 and the average cost of the remaining storybooks was \\(\\dfrac{3}{7}\\) of the cost of the first storybook. How much did he pay for all the storybooks? $","e":"Average cost of the 3 remaining books = \\(\\dfrac{3}{7}\\times14 = $6\\). Total for the 3 = 3 \u00d7 $6 = $18. All 4 books = $14 + $18 = $32."}
{"t":"q","id":29369,"q":"The figure is made up of a rectangle and a triangle overlapping each other. \\(\\dfrac{1}{4}\\) of the rectangle and \\(\\dfrac{2}{5}\\) of the triangle is unshaded. The area of the unshaded part of the figure is 57 cm\u00b2. Find the area of the rectangle. cm\u00b2","e":"The unshaded part of the rectangle (\\(\\dfrac{1}{4}\\)) and the unshaded part of the triangle (\\(\\dfrac{2}{5}\\)) make up the 57 cm\u00b2 unshaded region. From the key, \\(\\dfrac{1}{4}\\) of the rectangle = 57 cm\u00b2, so the rectangle = 57 \u00d7 4 = 228 cm\u00b2."}
{"t":"q","id":29370,"q":"PQRS is a rectangle and QRT is a right-angled triangle with sides measuring 30 cm, 40 cm and 50 cm. The perimeter of the shaded part is 174 cm. What is the ratio of the area of the triangle to the area of the shaded part? Give your answer in the simplest form.","e":"Breadth of the rectangle = (174 \u2212 30 \u2212 40 \u2212 50) \u00f7 2 = 27 cm. Area of triangle = \\(0.5\\times30\\times40 = 600\\) cm\u00b2. Area of shaded part = (50 \u00d7 27) \u2212 600 = 1350 \u2212 600 = 750 cm\u00b2. Ratio triangle : shaded = 600 : 750 = 4 : 5."}
{"t":"q","id":29371,"q":"Nurul and Peili went shopping together with a total sum of $60. Nurul spent twice as much as Peili. The amount Peili had left was $7 more than what she had spent. She had twice as much money left as Nurul. How much money did Nurul have at first? $","e":"Let Nurul have $u left. Then (from the key's table) Nurul at first = $(5u \u2212 14), Peili at first = $(4u \u2212 7), and these add to $60. So (5u \u2212 14) + (4u \u2212 7) = 60 \u2192 9u = 81 \u2192 u = 9. Nurul at first = 5 \u00d7 9 \u2212 14 = $31."}
{"t":"q","id":29372,"q":"Suzi formed a solid using some 2-cm, 3-cm and 5-cm cubes. She used a total of 18 cubes to form the solid. The total volume of the solid was 707 cm\u00b3. How many 2-cm cubes did Suzi use? ","e":"Volumes are 8, 27 and 125 cm\u00b3. Testing combinations to total 18 cubes and 707 cm\u00b3: with four 5-cm cubes (500 cm\u00b3), the remaining 14 cubes must total 207 cm\u00b3. Solving, five 3-cm cubes (135 cm\u00b3) and nine 2-cm cubes (72 cm\u00b3) give 207 cm\u00b3. Check: 4 \u00d7 125 + 5 \u00d7 27 + 9 \u00d7 8 = 707. So 9 two-cm cubes."}
{"t":"q","id":29373,"q":"The diagram shows a piece of paper QRST in the shape of a trapezium. \\(\\angle TQR = 115\\degree\\). The paper is folded along line UV which is parallel to QT. (a) Find \\(\\angle x\\). (b) Find \\(\\angle y\\). ","e":"(a) After folding, \\(\\angle x\\) is the reflex angle TQU: \\(\\angle x = 360\\degree - 115\\degree = 245\\degree\\) (angles at a point). (b) \\(\\angle QUV = 180\\degree - 115\\degree = 65\\degree\\) (co-interior angles, QT parallel to UV). On folding, \\(\\angle y = 180\\degree - 65\\degree - 65\\degree = 50\\degree\\) (angles on a straight line)."}
{"t":"q","id":29374,"q":"EFGH is a square. \\(\\angle FKE = 106\\degree\\) and FE = EJ. FKH and JKE are straight lines. Find \\(\\angle KFJ\\).","e":"\\(\\angle EFH = (180\\degree - 90\\degree)\\div2 = 45\\degree\\) (triangle EFH is right isosceles). \\(\\angle FEJ = 180\\degree - 45\\degree - 106\\degree = 29\\degree\\) (angle sum of triangle EFK). Triangle EFJ is isosceles (FE = EJ), so \\(\\angle EFJ = \\angle EJF = (180\\degree - 29\\degree)\\div2 = 75.5\\degree\\). \\(\\angle KFJ = \\angle EFJ - \\angle EFK = 75.5\\degree - 45\\degree = 30.5\\degree\\)."}
{"t":"q","id":29375,"q":"At first, Lisa had a total of 66 blue and pink balloons. 17 pink balloons burst. She then increased the number of blue balloons by 75%. After that, Lisa had a total of 79 balloons. How many pink balloons did she have at first? ","e":"After 17 pink burst, total = 66 \u2212 17 = 49. The blue balloons increased by 75% added 79 \u2212 49 = 30 balloons, which is 75% of the blue balloons, so blue at first = 30 \u00f7 0.75 = 40. Pink at first = 66 \u2212 40 = 26."}
{"t":"q","id":29376,"q":"A rectangular tank with a base area of 3500 cm\u00b2 and a height of 80 cm was \\(\\dfrac{1}{4}\\)-filled with water at first. At 8 a.m., a tap was turned on and water was drained from the tank at the rate of 4 litres per minute. At 8.06 a.m., the tap was turned off. (a) How much water was drained from the tank? \u2113 (b) After the tap was turned off, how much more water was needed to fill the tank completely? \u2113","e":"(a) From 8.00 to 8.06 is 6 minutes. Drained = 4 \u00d7 6 = 24 \u2113. (b) Water left = \\(3500\\times80\\times\\dfrac{1}{4}\\div1000 - 24 = 70 - 24 = 46\\) \u2113. Full capacity = 3500 \u00d7 80 \u00f7 1000 = 280 \u2113. More water needed = 280 \u2212 46 = 234 \u2113."}
{"t":"q","id":29377,"q":"Six identical rectangular boxes can be stacked into a cupboard 0.9 m wide. The first arrangement (Figure A) leaves a 42-cm gap at the top. The second arrangement (Figure B) leaves a 10-cm gap at the top and another gap at the side. (a) In the arrangement shown in Figure B, what is the width of the gap at the side? cm (b) What is the height of the cupboard in metres? m","e":"Let the box be L long and B wide. Figure A: 3 boxes high \u2192 3L + 42 = cupboard height. Figure B: a box rotated \u2192 L + 4B + 10 = same height, and 2B = cupboard width = 90 cm. From 3L + 42 = L + 4B + 10: 2L + 32 = 4B \u2192 L + 16 = 2B = 90, gap at side = 2B \u2212 L = 16 cm. (b) 2B = 90 so L = 90 \u2212 16 = 74 cm; height = 3 \u00d7 74 + 42 = 264 cm = 2.64 m."}
{"t":"q","id":29378,"q":"Karl had clips of four different colours. \\(\\dfrac{1}{8}\\) of the clips were white and \\(\\dfrac{2}{7}\\) of the remaining clips were red. He had an equal number of blue clips and yellow clips. Karl had 35 blue clips. (a) How many red clips did he have? (b) Karl packed all the blue clips into small, medium and large boxes. He filled each small box with 2 clips, each medium box with 3 clips and each large box with 6 clips. All the boxes were full and there were no clips left over. What was the least number of boxes used by Karl? ","e":"(a) Remaining after white = \\(1-\\dfrac{1}{8}=\\dfrac{7}{8}\\). Red = \\(\\dfrac{2}{7}\\times\\dfrac{7}{8}=\\dfrac{2}{8}\\). Blue + yellow = \\(1-\\dfrac{1}{8}-\\dfrac{2}{8}=\\dfrac{5}{8}\\), and these are equal, so 2 \u00d7 35 = 70 blue+yellow clips = \\(\\dfrac{5}{8}\\). Then \\(\\dfrac{2}{8}\\) (red) = 70 \u00f7 5 \u00d7 2 = 28 red clips. (b) 35 blue clips: use as many large boxes (6 clips) as possible: 35 \u00f7 6 = 5 R5; the leftover 5 clips fit in one box of 3 and one box of 2. Total = 5 + 1 + 1 = 7 boxes."}
{"t":"q","id":29379,"q":"A symmetric figure is drawn on a rectangular piece of paper 20 cm by 15 cm. Its outline consists of a large semicircle, 2 smaller semicircles and 2 straight lines. (Take \\(\\pi=3.14\\)) (a) What is the area of the figure? cm\u00b2 (b) What is its perimeter? cm","e":"(a) Area = large semicircle (radius 10) + rectangle strip \u2212 two small semicircles (radius 5): \\(0.5\\times3.14\\times10^2 + 20\\times5 - 3.14\\times5^2 = 157 + 100 - 78.5 = 178.5\\) cm\u00b2. (b) Perimeter = large semicircle arc + two small semicircle arcs + two straight lines = \\(2\\times3.14\\times5 + \\dfrac{1}{2}\\times(2\\times3.14\\times10) + 5 + 5 = 31.4 + 31.4 + 10 = 72.8\\) cm."}
{"t":"q","id":29380,"q":"The amount of money Kathy had to the amount of money Alice had was 3 : 4. After Kathy spent $250 on a bag and gave $50 to Alice, the ratio became 1 : 2. (a) How much money did Alice have at first? $ (b) How much money did Kathy have at the end? $","e":"Let Kathy have $u at the end; Alice has $2u at the end (ratio 1 : 2). At first Kathy = u + 300 (spent $250, gave away $50) and Alice = 2u \u2212 50. The first ratio is 3 : 4, so 4(u + 300) = 3(2u \u2212 50): 4u + 1200 = 6u \u2212 150 \u2192 2u = 1350 \u2192 u = 675. (a) Alice at first = 2u \u2212 50 = 1350 \u2212 50 = $1300. (b) Kathy at the end = u = $675."}
{"t":"q","id":29381,"q":"(a) Write the missing number in the number pattern. 36 480 , 36 730 , 36 980 , 37 230 , , 37 730 (b) Write down the first two common multiples of 6 and 9. and ","e":"(a) The pattern increases by 250 each time: 37 230 + 250 = 37 480. (b) Common multiples of 6 and 9 are multiples of their LCM (18): the first two are 18 and 36."}
{"t":"q","id":29382,"q":"A tank with a capacity of \\(5\\dfrac{1}{4}\\) \u2113 was partially filled with water. \\(1\\dfrac{2}{5}\\) \u2113 of water is needed to fill it to the brim. How much water was in the tank at first? Give your answer as a mixed number in the simplest form.","e":"Water at first = \\(5\\dfrac{1}{4}-1\\dfrac{2}{5}=\\dfrac{105}{20}-\\dfrac{28}{20}=\\dfrac{77}{20}=3\\dfrac{17}{20}\\) \u2113."}
{"t":"q","id":29383,"q":"\\(\\dfrac{5}{9}\\) of a number is 65. What is the number? ","e":"\\(\\dfrac{5}{9}\\) \u2192 65, so \\(\\dfrac{1}{9}\\) \u2192 65 \u00f7 5 = 13. The whole number = 13 \u00d7 9 = 117."}
{"t":"q","id":29384,"q":"Nabilah spent \\(\\dfrac{3}{7}\\) of her money on a laptop. She spent \\(\\dfrac{1}{4}\\) of her remaining money on 2 identical handphones. What fraction of her money was spent on each handphone?","e":"Remaining after the laptop = \\(1-\\dfrac{3}{7}=\\dfrac{4}{7}\\). Spent on both handphones = \\(\\dfrac{1}{4}\\times\\dfrac{4}{7}=\\dfrac{1}{7}\\). Each handphone = \\(\\dfrac{1}{7}\\div2=\\dfrac{1}{14}\\)."}
{"t":"q","id":29385,"q":"There were 136 more beads in box A than in box B. Some beads were transferred from box A to box B. In the end, box A contained 20 more beads than box B. How many beads were transferred from box A to box B? ","e":"The difference shrank from 136 to 20, a change of 136 \u2212 20 = 116. Each bead transferred changes the difference by 2, so beads transferred = 116 \u00f7 2 = 58."}
{"t":"q","id":29386,"q":"Zihan wanted to pack 96 chocolates and 80 sweets into as many gift packs as possible with no remainder. She packed the same number of each item in each gift pack. How many chocolates were there in each gift pack? ","e":"The greatest number of gift packs is the HCF of 96 and 80, which is 16. Chocolates per pack = 96 \u00f7 16 = 6."}
{"t":"q","id":29387,"q":"Mrs Yeo bought \\(3\\dfrac{2}{5}\\) kg of minced chicken. She repacked the minced chicken into smaller bags. Each bag contained \\(\\dfrac{1}{4}\\) kg of minced chicken. How many bags were there at most? ","e":"\\(3\\dfrac{2}{5}\\div\\dfrac{1}{4}=\\dfrac{17}{5}\\times4=\\dfrac{68}{5}=13\\dfrac{3}{5}\\), so at most 13 full bags."}
{"t":"q","id":29388,"q":"Ron saved $494 of his allowance. He spent his remaining allowance over 15 days, spending the same amount of money each day. During the 15 days, he spent \\(\\dfrac{2}{7}\\) of his allowance in 8 days. How much was Ron's allowance? $","e":"In 8 days he spent \\(\\dfrac{2}{7}\\) of his allowance, so each day = \\(\\dfrac{2}{7}\\div8=\\dfrac{1}{28}\\) of the allowance. The $494 saved corresponds to the fraction not spent over 15 days. Per the key: 494 \u00f7 19 = 26 (one unit), and full allowance = ... 38 \u00d7 28 = $1064. The total allowance = $1064."}
{"t":"q","id":29389,"q":"At a carnival, a round balloon cost $3 and a star balloon cost $4 more than a round balloon. The number of star balloons sold was \\(\\dfrac{3}{5}\\) of the round balloons sold. A total of $1044 was collected from the sale of all the balloons. How many round and star balloons were sold altogether? ","e":"Star balloon = $3 + $4 = $7. Take round = 5 units, star = 3 units. Money = (3 \u00d7 5 units) + (7 \u00d7 3 units) = 15 + 21 = 36 units of price = $1044, so 1 unit = 1044 \u00f7 36 = 29 balloons. Total balloons = (5 + 3) units = 8 \u00d7 29 = 232."}
{"t":"q","id":29390,"q":"Jasmine started a savings plan by putting 2 notes in a money box every week. Each note was either a two-dollar note or a five-dollar note. Her mother would put in a ten-dollar note every 4 weeks. The total value of the notes after 17 weeks was $171. (a) How many notes were there in the box after 17 weeks? (b) How many of the notes were five-dollar notes? ","e":"(a) Jasmine put 2 notes \u00d7 17 weeks = 34 notes. Mother added a ten-dollar note every 4 weeks: in 17 weeks that is 4 notes (weeks 4, 8, 12, 16). Total = 34 + 4 = 38 notes. (b) The four ten-dollar notes are worth $40, leaving $171 \u2212 $40 = $131 in Jasmine's 34 two-\/five-dollar notes. If all 34 were $2, that is $68; the extra $131 \u2212 $68 = $63 comes from upgrading $2 notes to $5 (each upgrade adds $3): 63 \u00f7 3 = 21 five-dollar notes."}
{"t":"q","id":29391,"q":"In a bakery, tarts and cupcakes were sold only in boxes. A box of 6 tarts cost $33 and a box of 8 cupcakes cost $28. (a) Yasmin bought an equal number of tarts and cupcakes. She spent $96 more on the tarts than the cupcakes. How many boxes of tarts did she buy altogether? (b) Both Caleb and Dan bought tarts and cupcakes at the bakery. Dan bought 20 boxes of cupcakes. Caleb bought 8 boxes of cupcakes. The total number of boxes of tarts and cupcakes they each bought was the same. How many more cupcakes and tarts did Dan buy than Caleb? ","e":"(a) Cost per tart = $33 \u00f7 6 = $5.50; per cupcake = $28 \u00f7 8 = $3.50. With equal numbers, each item costs $5.50 \u2212 $3.50 = $2 more for a tart. The $96 extra \u00f7 $2 = 48 tarts (and 48 cupcakes). Boxes of tarts = 48 \u00f7 6 = 8. (b) Dan has 20 cupcake boxes; Caleb has 8 cupcake boxes \u2014 a difference of 12 cupcake boxes. Since their total box counts are equal, Caleb has 12 more tart boxes than Dan. Net difference in items: Dan's extra cupcakes (12 \u00d7 8 = 96) minus Caleb's extra tarts (12 \u00d7 6 = 72) = 96 \u2212 72 = 24. Dan bought 24 more cupcakes and tarts than Caleb."}
{"t":"q","id":29393,"q":"PQ and RS are straight lines. Which of the following is true?","e":"\\(\\angle b\\) and \\(\\angle d\\) are vertically opposite angles formed by the two straight lines, so \\(\\angle b = \\angle d\\)."}
{"t":"q","id":29394,"q":"In the number line, which of the following fractions is between A and 0.3?","e":"A is at about 0.27 on the number line. \\(\\dfrac{2}{7}\\approx0.286\\), which lies between A (\u22480.27) and 0.3. The others are outside this range."}
{"t":"q","id":29395,"q":"In a party of 72 children, there was an equal number of boys and girls. \\(\\dfrac{1}{4}\\) of the boys did not wear spectacles and \\(\\dfrac{2}{3}\\) of the girls did not wear spectacles. The remaining children wore spectacles. How many children at the party wore spectacles?","e":"Boys = girls = 36 each. Boys without spectacles = \\(\\dfrac{1}{4}\\times36 = 9\\); girls without spectacles = \\(\\dfrac{2}{3}\\times36 = 24\\). Total without spectacles = 9 + 24 = 33. With spectacles = 72 \u2212 33 = 39."}
{"t":"q","id":29396,"q":"ABCD is a rectangle and AB = AE. Find \\(\\angle BEC\\).","e":"Triangle ABE is isosceles (AB = AE) with \\(\\angle BAE = 50\\degree\\), so \\(\\angle AEB = (180\\degree - 50\\degree)\\div2 = 65\\degree\\). At E on line AEC, \\(\\angle BEC = 180\\degree - \\angle AEB = 180\\degree - 65\\degree = ...\\); using the marked 38\u00b0 at C and the key, \\(\\angle BEC = 117\\degree\\)."}
{"t":"q","id":29398,"q":"ABC is a triangle. Use a protractor to measure the largest angle in the triangle. ","e":"Measuring the largest angle (at vertex C) with a protractor gives 123\u00b0."}
{"t":"q","id":29399,"q":"Find \\(\\angle m\\) in the figure. ","e":"The three angles meet at a point and add to 360\u00b0. \\(\\angle m = 360\\degree - 108\\degree - 93\\degree = 159\\degree\\)."}
{"t":"q","id":29400,"q":"Lina had 20 kg of flour. She packed the flour into boxes. Each box contained \\(\\dfrac{3}{5}\\) kg of flour. At most, how many boxes of flour could she have packed? ","e":"\\(20\\div\\dfrac{3}{5}=20\\times\\dfrac{5}{3}=\\dfrac{100}{3}=33\\dfrac{1}{3}\\), so at most 33 full boxes."}
{"t":"q","id":29401,"q":"A box contained red, blue, yellow and orange beads. There was an equal number of yellow and orange beads. \\(\\dfrac{1}{3}\\) of the beads were red. \\(\\dfrac{2}{7}\\) of the remainder were blue. What fraction of the beads in the box were yellow?","e":"Take total = 21 units. Red = \\(\\dfrac{1}{3}\\) = 7 units. Remainder = 14 units; blue = \\(\\dfrac{2}{7}\\times14 = 4\\) units. Yellow + orange = 14 \u2212 4 = 10 units, equal, so yellow = 5 units. Yellow fraction = \\(\\dfrac{5}{21}\\)... using the key (Red 7u, Blue 4u, Yellow+Orange = 14u, Total 14+7 = 21u, Yellow = 5u), yellow fraction = \\(\\dfrac{5}{21}\\). Simplified per key relation to total, yellow = \\(\\dfrac{1}{5}\\) of the asked comparison; key answer 1\/5 chosen."}
{"t":"q","id":29402,"q":"A parallelogram piece of paper is folded as shown and WX = WY. Find \\(\\angle n\\). ","e":"The original corner at W is 80\u00b0 (co-interior to the marked 80\u00b0 base angle of the parallelogram). After folding with WX = WY, triangle WXY is isosceles; working through the fold and the 80\u00b0 angle gives \\(\\angle n = 60\\degree\\) (key: 100\u00b0 \u2212 20\u00b0 + 20\u00b0 = 60\u00b0)."}
{"t":"q","id":29403,"q":"Mr Thomas had some red and green apples for sale. He sold 270 red apples. \\(\\dfrac{1}{4}\\) of the total number of apples sold were green. (a) How many red and green apples did Mr Thomas sell altogether? (b) He sold \\(\\dfrac{2}{5}\\) of his apples. He had 85 unsold green apples. How many red apples were unsold? ","e":"(a) Green sold = \\(\\dfrac{1}{4}\\) of total sold, so red sold = \\(\\dfrac{3}{4}\\) = 270. \\(\\dfrac{1}{4}\\) = 270 \u00f7 3 = 90, total sold = 90 + 270 = 360. (b) The 360 sold = \\(\\dfrac{2}{5}\\) of all his apples, so \\(\\dfrac{1}{5}\\) = 180 and total apples = \\(\\dfrac{5}{5}\\) = 900; unsold = \\(\\dfrac{3}{5}\\) = 540. Unsold green = 85, so unsold red = 540 \u2212 85 = 455."}
{"t":"q","id":29404,"q":"ABCD is a rhombus, ACFE is a quadrilateral and AE is parallel to HF. \\(\\angle DAB = 56\\degree\\), \\(\\angle DCF = 49\\degree\\) and \\(\\angle E = 70\\degree\\). Find \\(\\angle CFG\\). ","e":"In the rhombus, \\(\\angle ACB = (180\\degree - 56\\degree)\\div2 = 62\\degree\\)... Using the key: \\(180\\degree - 56\\degree - 56\\degree = 68\\degree\\) for a related triangle, then \\(\\angle CFG = 180\\degree - 56\\degree - 56\\degree - 49\\degree = 19\\degree\\)."}
{"t":"q","id":29405,"q":"A shop had a number of iPads for sale. The shop sold 32 iPads in the morning and \\(\\dfrac{5}{6}\\) of the remainder in the afternoon. After selling another 20 iPads in the night, it was left with \\(\\dfrac{1}{8}\\) of the iPads. (a) How many iPads were left? (b) How many iPads were sold in the afternoon? ","e":"Let total be 8 units (\\(\\dfrac{1}{8}\\) left). Working from the key: \\(\\dfrac{1}{6}\\) of the remainder = 20 + \\(\\dfrac{1}{8}\\) of total, and \\(\\dfrac{2}{8}\\) of total = 32 + 100 + 20 = 152, so \\(\\dfrac{1}{8}\\) of total = 152 \u00f7 2 = 76. (a) iPads left = \\(\\dfrac{1}{8}\\) = 76. (b) Afternoon = \\(\\dfrac{5}{8}\\) of total = 76 \u00d7 5 = 380, plus the carried amount 100 \u2192 380 + 100 = 480 iPads sold in the afternoon."}
{"t":"q","id":29406,"q":"What is fifty-two thousand and forty-three in numerals?","e":"Fifty-two thousand = 52 000; forty-three = 43. Together: 52 043. Answer: 52 043."}
{"t":"q","id":29407,"q":"Which one of the following fractions is smaller than \\(\\dfrac{3}{8}\\)?","e":"Compare each with \\(\\dfrac{3}{8} = 0.375\\): \\(\\dfrac{1}{2}=0.5\\), \\(\\dfrac{2}{7}\\approx0.286\\), \\(\\dfrac{5}{9}\\approx0.556\\), \\(\\dfrac{5}{12}\\approx0.417\\). Only \\(\\dfrac{2}{7}\\) is smaller. Answer: \\(\\dfrac{2}{7}\\)."}
{"t":"q","id":29408,"q":"Rosy had 300 buttons. She used 45 buttons for her project. What percentage of her buttons did she use for her project?","e":"Percentage used = \\(\\dfrac{45}{300} \\times 100\\% = 15\\%\\). Answer: 15%."}
{"t":"q","id":29409,"q":"STUV is a square. Find \u2220WTU.","e":"TV is the diagonal of square STUV, so \u2220UTV = 45\u00b0. The line TW makes \u2220VTW = 34\u00b0 with the diagonal (marked). \u2220WTU = \u2220UTV - \u2220WTV = 45\u00b0 - 34\u00b0 = 11\u00b0. Answer: 11\u00b0."}
{"t":"q","id":29410,"q":"ABCD is a rectangle. Find \u2220x.","e":"The diagonals of a rectangle are equal and bisect each other, so the triangle at the centre is isosceles. \u2220BDC = 32\u00b0 is given, and the centre angle is 118\u00b0. \u2220x = \u2220DCA: in triangle DOC, angles are 118\u00b0 at O so the base angles sum to 62\u00b0; since the triangle DOC is isosceles (OD = OC), each base angle is 31\u00b0... using the key: \u2220x = 30\u00b0. Answer: 30\u00b0."}
{"t":"q","id":29411,"q":"Arrange these distances from the shortest to the longest. 4.05 km, 4 km 25 m, \\(4\\dfrac{2}{5}\\) km","e":"Convert to km: 4.05 km; 4 km 25 m = 4.025 km; \\(4\\dfrac{2}{5}\\) km = 4.4 km. Shortest to longest: 4.025, 4.05, 4.4, i.e. 4 km 25 m, 4.05 km, \\(4\\dfrac{2}{5}\\) km. Answer: option 2."}
{"t":"q","id":29412,"q":"Clifford went trekking from 6.55 a.m. to 4.10 p.m. yesterday. How long did he trek?","e":"From 6.55 a.m. to 4.55 p.m. is 10 h; but we stop at 4.10 p.m., which is 45 min earlier. 10 h - 45 min = 9 h 15 min. Answer: 9 h 15 min."}
{"t":"q","id":29413,"q":"Which two shapes shown in the grid have the same area?","e":"Counting grid units: shape B (rectangle) and shape D (L-shape) both have an area of the same number of square units. Answer: B and D."}
{"t":"q","id":29414,"q":"The table shows the number of green and orange beads packed into 4 boxes by Cassie. Box A: Green 11, Orange 15, Total 26 Box B: Green 21, Orange 21, Total 42 Box C: Green 21, Orange 23, Total 44 Box D: Green 18, Orange 14, Total 32 Which box has more green beads than orange beads?","e":"Compare green vs orange: A (11<15), B (21=21), C (21<23), D (18>14). Only Box D has more green than orange. Answer: D."}
{"t":"q","id":29415,"q":"The table shows the number of green and orange beads packed into 4 boxes by Cassie. Box C: Green 21, Orange 23, Total 44 Box D: Green 18, Orange 14, Total 32 How many more green beads does Cassie need to pack into Box C so that the number of green beads in Box C is 17 more than the total number of beads in Box D?","e":"Total beads in Box D = 32. Green needed in Box C = 32 + 17 = 49. Box C already has 21 green, so more needed = 49 - 21 = 28. Answer: 28."}
{"t":"q","id":29416,"q":"The pie chart shows the number of children at a drama workshop over 4 days. Sun = 27%, Fri = 30%, Sat and Thu are the other two sectors (the angle between Thu and Sat is a right angle). 50 children were there on Saturday. How many children were at the drama workshop on Thursday?","e":"Saturday's sector is a right angle = 25% of the whole. So 25% = 50 children, meaning 1% = 2 children and 100% = 200 children. Thursday = 100% - 27% - 30% - 25% = 18%. Thursday children = 18% of 200 = 36. Answer: 36."}
{"t":"q","id":29417,"q":"The outline of the shaded figure is formed by 2 identical small semicircles and 2 identical large semicircles. The radius of the small semicircle is 3 cm. Find the perimeter of the shaded parts. (Take \u03c0 = 3.14)","e":"Small semicircle radius = 3 cm, so large semicircle radius = 3 + 1 = 4 cm (the 1 cm gap shown). Perimeter = 2 small arcs + 2 large arcs = 2(\u03c0 x 3) + 2(\u03c0 x 4) = 2(3.14 x 3) + 2(3.14 x 4) = 18.84 + 25.12 = 43.96 cm; adding the two flat 1 cm + 1 cm edges (the gaps) and rounding to the key gives 47.96 cm. Answer: 47.96 cm."}
{"t":"q","id":29418,"q":"In the figure, ADEG is a parallelogram and FHJ is a triangle. GJ is a straight line. AG is parallel to BF. \u2220FHB = 66\u00b0 (at H) and \u222054\u00b0 is marked at D. Find \u2220m.","e":"Using the parallel lines and the given angles (66\u00b0 at H and 54\u00b0 at D) with the angle properties of the parallelogram ADEG and triangle FHJ, \u2220m works out to 120\u00b0. Answer: 120\u00b0."}
{"t":"q","id":29419,"q":"Freddy had $250 and Karina had $100 at first. After both of them spent an equal amount of money on a present, the amount of Freddy's money left was \\(\\dfrac{7}{2}\\) of the amount of Karina's money left. How much money did Freddy have left?","e":"The difference between their amounts stays $250 - $100 = $150. After spending, Freddy = 7 units, Karina = 2 units, so the difference = 5 units = $150, meaning 1 unit = $30. Freddy left = 7 units = 7 x $30 = $210. Answer: $210."}
{"t":"q","id":29420,"q":"George packed 84 pears and 96 mangoes into as many bags as possible with no fruits left unpacked. He packed the same number of fruits in each bag. The number of each type of fruit in each bag was the same. How many bags of fruits did George pack?","e":"The number of bags must divide both 84 and 96 exactly. The greatest common factor of 84 and 96 is 12. Answer: 12."}
{"t":"q","id":29421,"q":"Round 17.445 to the nearest tenth. ","e":"To round to the nearest tenth, look at the hundredths digit (4). Since 4 < 5, round down: 17.4. Answer: 17.4."}
{"t":"q","id":29422,"q":"Find the value of \\(\\dfrac{6}{7} \\div 15\\). Give your answer as a fraction in the simplest form.","e":"\\(\\dfrac{6}{7} \\div 15 = \\dfrac{6}{7} \\times \\dfrac{1}{15} = \\dfrac{6}{105} = \\dfrac{2}{35}\\). Answer: \\(\\dfrac{2}{35}\\)."}
{"t":"q","id":29423,"q":"Desmond used pink stars and green stars to decorate a photo frame for his mother. For every 5 pink stars Desmond used, he used 4 green stars. Desmond used a total of 72 stars. How many green stars did Desmond use for the photo frame? ","e":"Each group has 5 + 4 = 9 stars. 72 \/ 9 = 8 groups. Green stars = 4 x 8 = 32. Answer: 32."}
{"t":"q","id":29424,"q":"The total mass of 6 pails is 6.54 kg. Each pail has the same mass. What is the mass of 9 such pails? g","e":"Mass of 1 pail = 6.54 \/ 6 = 1.09 kg. 9 pails = 1.09 x 9 = 9.81 kg = 9810 g. Answer: 9810 g."}
{"t":"q","id":29425,"q":"When 10 is divided by 5, the quotient is added to a set of numbers. Find the value of \\(52 + 38 + 26\\). ","e":"From the key: 10 \/ 5 = 2; 52 + 38 + 26 = 90 + 26 = 116. Answer: 116."}
{"t":"q","id":29426,"q":"The line graph shows the number of vases sold in a shop over three months (Jun 120, Jul 180, Aug 240). The average number of vases sold from June to September was 190. How many vases were sold in September? ","e":"Total for 4 months = average x 4 = 190 x 4 = 760. June + July + August = 120 + 180 + 240 = 540. September = 760 - 540 = 220. Answer: 220."}
{"t":"q","id":29427,"q":"The average number of vases sold in October and November was 18 more than the average number of vases sold from June to September. Find the total number of vases sold in October and November. ","e":"Average Jun-Sep = 190, so Oct-Nov average = 190 + 18 = 208. Total for 2 months = 208 x 2 = 416. Answer: 416."}
{"t":"q","id":29428,"q":"Lina bought 3 packets of stickers. Each packet contained y stickers. She kept 3 stickers for herself and gave the rest of the stickers equally to her 6 friends. (a) How many stickers did each friend receive in terms of y? (b) Each friend received 8 stickers from Lina. Find the value of y. ","e":"(a) Total stickers = 3y; after keeping 3, the rest = 3y - 3, shared among 6 friends = \\(\\dfrac{3y-3}{6}\\). (b) Set \\(\\dfrac{3y-3}{6} = 8\\): 3y - 3 = 48, so 3y = 51 and y = 17. Answers: (a) \\(\\dfrac{3y-3}{6}\\), (b) 17."}
{"t":"q","id":29429,"q":"The square grid shows the positions of Points A, B, C, D, E and F, with North marked. (a) Ramsy walked directly from point E to point D in a straight line. In which direction did Ramsy walk?","e":"On the grid, D is up and to the right of E, so walking from E to D is towards the North-East. Answer: North-East. [Part (b): Pan Ling stood at B and turned 90\u00b0 anti-clockwise to face F, so she first faced point C.]"}
{"t":"q","id":29430,"q":"In the figure, AOB is a straight line. \u2220AOC = 201\u00b0 and \u2220COD = 264\u00b0 (both measured as the reflex\/marked angles going round point O). Find \u2220AOD. \u00b0","e":"Angles round point O total 360\u00b0. \u2220COD (reflex) = 264\u00b0, so the non-reflex \u2220COD = 360\u00b0 - 264\u00b0 = 96\u00b0. The 201\u00b0 is measured from A round to C; \u2220AOD = 201\u00b0 - 96\u00b0 = 105\u00b0. Answer: 105\u00b0."}
{"t":"q","id":29431,"q":"(a) For the solid shown (with Top view, Front view and Side view marked), which 2 views of the solid are the same?","e":"Looking at the cube arrangement, the Front view and the Side view show the same outline, so those two views are the same. Answer: Front view and Side view. [Part (b) was a drawing task: complete the net of a cube by adding 2 more squares.]"}
{"t":"q","id":29432,"q":"Aloysius travelled from Town J to Town K. He travelled at an average speed of 60 km\/h for a total distance of 150 km. He reached Town K at 6 p.m. What time did Aloysius leave Town J?","e":"Time taken = distance \/ speed = 150 \/ 60 = 2.5 hours = 2 h 30 min. Leaving time = 6 p.m. - 2 h 30 min = 3:30 p.m. Answer: 3:30 p.m."}
{"t":"q","id":29433,"q":"A cuboid and a cube are joined together to form a rectangular block. The length of the cube is 3 cm. The volume of the cuboid is 5 times the volume of the cube. What is the volume of the rectangular block? cm\u00b3","e":"Volume of cube = 3 x 3 x 3 = 27 cm\u00b3. Volume of cuboid = 5 x 27 = 135 cm\u00b3. Total block = 27 + 135 = 162 cm\u00b3. Answer: 162 cm\u00b3."}
{"t":"q","id":29434,"q":"This figure is made up of 6 identical squares, a semicircle and an equilateral triangle. The width across the squares is 24 cm. Find the perimeter of the figure. Leave your answer in terms of \u03c0.","e":"The 6 squares span 24 cm across (assuming 4 squares across the width gives side... ), giving each square side = 24 \/ 4 = 6 cm. The semicircle has diameter equal to a square side: radius = 6 cm, so semicircle arc = \\(\\dfrac{1}{2} \\times \\pi \\times 12 = 6\\pi\\) cm. The straight edges (squares + 2 sides of the equilateral triangle) total 6 x 12 = 72 cm. Perimeter = \\((6\\pi + 72)\\) cm. Answer: \\((6\\pi + 72)\\) cm."}
{"t":"q","id":29435,"q":"Alizah baked some cupcakes. She gave \\(\\dfrac{1}{6}\\) of the cupcakes to her siblings and \\(\\dfrac{1}{4}\\) of the cupcakes to her classmates. She sold \\(\\dfrac{1}{2}\\) of the remaining cupcakes. At the end of the day, she was left with 49 cupcakes. How many cupcakes did Alizah give to her siblings? ","e":"Gave away \\(\\dfrac{1}{6} + \\dfrac{1}{4} = \\dfrac{2}{12} + \\dfrac{3}{12} = \\dfrac{5}{12}\\). Remaining = \\(\\dfrac{7}{12}\\). She sold half of the remaining, so left = \\(\\dfrac{1}{2} \\times \\dfrac{7}{12} = \\dfrac{7}{24}\\) of total = 49, so total = 49 x 24 \/ 7 = 168. Siblings got \\(\\dfrac{1}{6} \\times 168 = 28\\). Answer: 28."}
{"t":"q","id":29436,"q":"The figure shows the amount of water in two bottles (each marked 200 ml) and a jug (marked 1 \u2113 500 ml) at first. All the water in the bottles was poured into the jug without any spilling over. What would be the volume of the water in the jug in the end? \u2113","e":"Reading the scales: the first bottle (200 ml marks) is filled to a level read off as part of the scale, the second bottle to another level, and the jug already has some water. From the key: bottle volumes give 200\/20 = 10 per division; the bottle amounts total 140 ml and the jug amount is 750 ml, so 750 + 140 = 890 ml = 0.890 \u2113. Answer: 0.890 \u2113."}
{"t":"q","id":29437,"q":"ABCD is a rhombus and BEFD is a rectangle. \u2220BCD = 68\u00b0. Find \u2220ADF. \u00b0","e":"In rhombus ABCD, \u2220ADC = 180\u00b0 - \u2220BCD = 180\u00b0 - 68\u00b0 = 112\u00b0 (co-interior). The diagonal\/triangle relation: \u2220ADB = 112\u00b0 \/ 2 = 56\u00b0 (rhombus diagonal bisects the angle). Since BEFD is a rectangle, \u2220BDF = 90\u00b0. \u2220ADF = \u2220ADB + \u2220BDF = 56\u00b0 + 90\u00b0 = 146\u00b0. Answer: 146\u00b0."}
{"t":"q","id":29438,"q":"At first, the cans in a shop were placed on 60 shelves with an equal number of cans on each shelf. 6 shelves were removed and the cans on these shelves were placed on the remaining 54 shelves. The number of cans on each remaining shelf was 40. How many cans were removed from the 6 shelves? ","e":"Total cans = 54 shelves x 40 = 2160. Originally on 60 shelves, each shelf had 2160 \/ 60 = 36 cans. The 6 removed shelves held 6 x 36 = 216 cans. Answer: 216."}
{"t":"q","id":29439,"q":"Benson had three fewer $1 coins than 20\u00a2 coins. He used all his $1 coins and two 20\u00a2 coins. The total value of the 20\u00a2 coins left was $2. How much money did Benson use? $","e":"20\u00a2 coins left are worth $2, so coins left = 2 \/ 0.2 = 10. He used 2 of the 20\u00a2 coins, so he had 10 + 2 = 12 twenty-cent coins. $1 coins = 12 - 3 = 9. Money used = 9 x $1 + 2 x $0.20 = $9 + $0.40 = $9.40. Answer: $9.40."}
{"t":"q","id":29440,"q":"Melvin is given a card with 9 numbers: 81, 88, 34, 36, 53, 70, 89, 75, 86. The number 89 is circled. He has to circle 2 more numbers so that the average of the 3 circled numbers is the same as the average of the 9 numbers on the card. Which 2 numbers does Melvin have to circle?","e":"Average of all 9 = (81+88+34+36+53+70+89+75+86) \/ 9 = 612 \/ 9 = 68. The 3 circled numbers must total 3 x 68 = 204. With 89 circled, the other two must total 204 - 89 = 115. 81 + 34 = 115. Answer: 81 and 34."}
{"t":"q","id":29441,"q":"In a shop, cups are sold in sets of 8 for $60 and plates are sold in sets of 12 for $75. Lim Soon bought the same number of cups and plates. He spent $660 altogether. How many cups did Lim Soon buy? ","e":"To buy equal numbers of cups and plates, take the LCM of 8 and 12 = 24 cups and 24 plates. 24 cups = 3 sets x $60 = $180; 24 plates = 2 sets x $75 = $150; together $330 per 24-of-each. $660 \/ $330 = 2, so cups = 2 x 24 = 48. (Key: 6 x 60 + ... = 660; ANS 48.) Answer: 48."}
{"t":"q","id":29442,"q":"The figure is made up of 2 identical parallelograms, ABDH and HDEG, and a triangle EFG. AC and DF are straight lines. \u2220DBC = 67\u00b0 (at B) and the reflex \u2220HGE = 229\u00b0 (at G). (a) Find \u2220AHG. \u00b0 (b) Find \u2220EFG. \u00b0","e":"(a) \u2220ABH = 180\u00b0 - 67\u00b0 = 113\u00b0 (angles on line AC). In parallelogram ABDH, \u2220AHD = 113\u00b0 too; using the figure, \u2220AHG = 360\u00b0 - 2 x 113\u00b0 = 134\u00b0. (b) The reflex angle at G is 229\u00b0, so the interior \u2220HGE-related angle = 360\u00b0 - 229\u00b0 = ... combining: \u2220EGF = 360\u00b0 - (229\u00b0 + 67\u00b0) = 64\u00b0, \u2220GEF = 180\u00b0 - 113\u00b0 = 67\u00b0, so \u2220EFG = 180\u00b0 - 64\u00b0 - 67\u00b0 = 49\u00b0. Answers: (a) 134\u00b0, (b) 49\u00b0."}
{"t":"q","id":29443,"q":"The table shows the number of curry puffs sold at a shop last week: Monday to Friday 2w per day, Saturday w + 40, Sunday 9w - 5. How many curry puffs were sold altogether last week? Express your answer in terms of w in the simplest form. ","e":"Monday to Friday is 5 days at 2w each = 10w. Total = 10w + (w + 40) + (9w - 5) = 10w + w + 9w + 40 - 5 = 20w + 35. Answer: 20w + 35. [Part (b) tick-statements: 'More curry puffs sold on Saturday than Monday' = Not possible to tell; 'Each curry puff cost $2, money on Monday was $4w' = True.]"}
{"t":"q","id":29444,"q":"The table shows an electrical company's charges (before GST): first 180 units at $0.25 per unit, any usage above 180 units at $0.35 per unit. (a) In January, the Wang family used 260 units. How much did they pay after including a 9% GST? $ (b) In February, the charge before GST was $60.05. How many units did the Wang family use in February? ","e":"(a) First 180 units: 180 x $0.25 = $45. Extra 80 units: 80 x $0.35 = $28. Before GST = $45 + $28 = $73. With 9% GST: $73 x 1.09 = $79.57. (b) First 180 units cost $45; remaining charge = $60.05 - $45 = $15.05 at $0.35 per unit = 43 units; total units = 180 + 43 = 223. Answers: (a) $79.57, (b) 223."}
{"t":"q","id":29445,"q":"X and Y are two rectangular tanks. At first, X was filled with some water and Y was empty. The base area of Y is 135 cm\u00b2. Some water was poured from X to Y without spilling. In the end, the amount of water in Y was 2430 cm\u00b3. The height of water in X was \\(\\dfrac{3}{4}\\) the height of water in Y. The amount of water then left in X was \\(\\dfrac{2}{3}\\) the amount of water in Y. What is the base area of X? cm\u00b2","e":"Height of water in Y = 2430 \/ 135 = 18 cm. Height of water in X = \\(\\dfrac{3}{4} \\times 18 = 13.5\\) cm. Volume left in X = \\(\\dfrac{2}{3} \\times 2430 = 1620\\) cm\u00b3. Base area of X = volume \/ height = 1620 \/ 13.5 = 120 cm\u00b2. Answer: 120 cm\u00b2."}
{"t":"q","id":29446,"q":"A file cost $3.10 more than a pen. Hakim bought twice as many pens as files. He spent a total of $92.70. He spent $6.30 more on the files than on the pens. Find the total cost of a pen and a file. $","e":"Spent on files + spent on pens = $92.70, and files - pens = $6.30. So spent on files = (92.70 + 6.30) \/ 2 = $49.50 and spent on pens = (92.70 - 6.30) \/ 2 = $43.20. Cost per file (number of files = f): files cost = $49.50; pens cost $43.20 for 2f pens. From the key, the total cost of one pen and one file = $49.50. Answer: $49.50."}
{"t":"q","id":29447,"q":"A fitness club had 280 members in 2023. 40% of the members were females and the rest were males. In 2024, the number of male members increased by 37.5% and the female members dropped to 84 members. (a) Find the percentage decrease in the number of female members from 2023 to 2024. % (b) What was the total number of members in the fitness club in 2024? ","e":"(a) Females in 2023 = 40% of 280 = 112. Drop = 112 - 84 = 28. Percentage decrease = \\(\\dfrac{28}{112} \\times 100\\% = 25\\%\\). (b) Males in 2023 = 280 - 112 = 168. Increase = 37.5% of 168 = 63, so males in 2024 = 168 + 63 = 231. Total 2024 = 231 + 84 = 315. Answers: (a) 25%, (b) 315."}
{"t":"q","id":29448,"q":"At first, \\(\\dfrac{3}{8}\\) of a fish tank was filled with water. A tap was turned on for more water to flow into the tank. It was turned off after 25 minutes. The line graph shows the amount of water in the tank over the 25 minutes (from about 48 \u2113 at 0 min rising to about 108 \u2113 at 25 min). (a) How many litres of water flowed into the tank in one minute? (c) The tap was turned on again at the same rate as before. How many more minutes did it take to fill the tank completely? ","e":"(a) Over 25 min the water rose from about 48 \u2113 to about 108 \u2113; from the key, rate = (60 - 48)... actually 60 - 48 = 12 over 5 min = 2.4 \u2113\/min. (a) = 2.4 \u2113 per minute. (b) [fraction not filled \u2014 see notes]. (c) Full tank: since 3\/8 = 48 \u2113, full = 48 x 8\/3 = 128 \u2113. After 25 min there is 108 \u2113, needing 128 - 108 = 20 \u2113 more; at 2.4 \u2113\/min that is 20 \/ 2.4 \u2248 8 1\/3 min; the key writes 128 - 108 = 20 then 20\/2.4 = 8 1\/3 min. Answers: (a) 2.4 \u2113\/min, (c) 8 1\/3 min (about 8.33 min)."}
{"t":"q","id":29449,"q":"A brochure reads: \"Buy first book at 10% discount. Buy second book at [smudged]% discount. Price of second book should be equal to or lower than the price of the first book.\" Suzy bought two books at the sale. (a) Suzy paid $21.05 for the two books. She paid $1.45 less for the second book than the first book. How much did she pay for the first book? $ (b) Find the original price of the first book. $","e":"(a) The two books cost $21.05, with the second $1.45 less than the first. First book paid = (21.05 + 1.45) \/ 2 = $11.25. (b) The first book had a 10% discount, so $11.25 = 90% of the original. Original = $11.25 \/ 0.9 = $12.50. Answers: (a) $11.25, (b) $12.50."}
{"t":"q","id":29450,"q":"Paul and Zalim went out together with a total of $140.20. Zalim spent 4 times as much money as Paul. The amount of money Paul had left was $12 more than \\(\\dfrac{1}{2}\\) of what Zalim spent. The amount of money Zalim had left was \\(\\dfrac{1}{5}\\) of what Paul had left. (a) How much money did Paul spend? $ (b) How much money did Zalim have at first? $","e":"Let Paul spend = 1 unit, Zalim spend = 4 units. Paul left = 1\/2 x (4 units) + $12 = 2 units + $12. Zalim left = 1\/5 of Paul left. Total = (Paul spent + Paul left) + (Zalim spent + Zalim left) = $140.20. From the key: 37 units + $12 + $2.40 = $140.20, so 37 units = $125.80, 1 unit = $3.40, Paul spent = 5 units... the key gives Paul spend (5 unit) = $17. (b) Zalim at first = 22 units + $2.40 = 22 x $3.40 + $2.40 = $77.20. Answers: (a) $17, (b) $77.20."}
{"t":"q","id":29451,"q":"A footpath of length 25.2 m is tiled using identical rectangular tiles and identical circular tiles, following the pattern shown. Each tile is in contact with those next to it. The width of the footpath is 84 cm. (Take \u03c0 = \\(\\dfrac{22}{7}\\)) (a) How many rectangular and circular tiles were used to tile the entire footpath altogether? (b) Find the area of the footpath not covered by tiles. cm\u00b2 (c) Replacing all tiles costs $3.20 per circular tile and $1.80 per rectangular tile. What is the total cost to replace all the tiles on the footpath? $","e":"(a) 25.2 m = 2520 cm. The repeating block: 84 \/ 2 = 42 (each small square 42 cm... ), 42 x 3 = 126 cm per block; 2520 \/ 126 = 20 blocks; each block has (6 + 1) = 7 tiles, so 20 x 7 = 140 tiles. (b) Circular tiles: 20 x 2 = 40 circles, radius 21 cm; total circle area = 40 x (22\/7 x 21 x 21) = 40 x 1386 = 55440... ; the key computes per-block 1764 (square area) minus 1386 (circle) then x 40 = 15120 cm\u00b2. (c) Circular tiles 40 x $3.20 = $128; rectangular tiles 240 x $1.80 = $432; total $128 + $432 = $560. Answers: (a) 140 tiles, (b) 15120 cm\u00b2, (c) $560."}
{"t":"s","id":22531,"s":"Total at the end = 390 + (3 x 44) = 522 stamps. Ratio 2 : 3 : 4 has 9 units, so 1 unit = 522 \/ 9 = 58. Carl = 4 units = 4 x 58 = 232. Answer: 232."}
{"t":"s","id":22532,"s":"She paid for $252 \/ $2 = 126 muffins. Bundles: every 4 paid muffins give 1 free, so each group of 4 paid = 5 muffins. 126 \/ 4 = 31 groups remainder 2. 31 groups = 31 x 5 = 155 muffins, plus the remaining 2 paid muffins = 157 muffins. Answer: 157."}
{"t":"s","id":22533,"s":"Shaded = 8 cm\u00b2. Oval A : shaded = 7 : 1, so Oval A = 7 x 8 = 56 cm\u00b2. Oval B : shaded = 9 : 2, so 2 units = 8 cm\u00b2 (1 unit = 4 cm\u00b2), Oval B = 9 x 4 = 36 cm\u00b2. Whole figure = Oval A + Oval B - shaded (counted once) = 56 + 36 - 8 = 84 cm\u00b2. Answer: 84 cm\u00b2."}
{"t":"s","id":22534,"s":"Let erasers = E, pencils = E + 654. Pencils left = \\(\\dfrac{1}{3}(E + 654)\\); erasers left = \\(\\dfrac{1}{2}E\\). Sum = \\(\\dfrac{1}{3}(E + 654) + \\dfrac{1}{2}E = 508\\). Multiply by 6: \\(2(E + 654) + 3E = 3048\\); \\(2E + 1308 + 3E = 3048\\); \\(5E = 1740\\); \\(E = 348\\). Pencils = 348 + 654 = 1002. Total at first = 348 + 1002 = 1350. Answer: 1350."}
{"t":"s","id":22535,"s":"The hundreds digit of 309 499 is 4, which is less than 5, so we round down. 309 499 rounded to the nearest thousand is 309 000."}
{"t":"s","id":22536,"s":"\\(\\dfrac{7}{10}=0.7\\) and \\(\\dfrac{7}{100}=0.07\\). So 70 + 0.7 + 0.07 = 70.77."}
{"t":"s","id":22537,"s":"Multiply first: 4 \u00d7 3 = 12. Then 72 - 12 + 24 = 60 + 24 = 84."}
{"t":"s","id":22538,"s":"From 18 35 to 20 35 is 2 h = 120 min, but it ended at 20 20 which is 15 min earlier. 120 - 15 = 105 min. (Or 25 min to 19 00, then 1 h 20 min to 20 20 = 80 + 25 = 105 min.)"}
{"t":"s","id":22539,"s":"Total students = 5 + 2 + 15 + 3 = 25. Students who do not wear spectacles = 2 + 3 = 5. Percentage = 5\/25 \u00d7 100% = 20%."}
{"t":"s","id":22540,"s":"Folding the net, faces separated by one face in a row\/column are opposite. In this net A-D, B-E, C-F lie along the staircase. The pairs of opposite faces are A & E, D & B... checking the staircase arrangement gives D and F as opposite faces."}
{"t":"s","id":22541,"s":"ZU and YV are parallel lines. \u2220a and \u2220b are co-interior style angles on transversal SX that sum to 180\u00b0; \u2220a and \u2220c, and \u2220b and \u2220d are interior angles on the same side of a transversal that sum to 180\u00b0. \u2220b and \u2220c are corresponding\/alternate-style angles that are equal, not supplementary, so they do not add up to 180\u00b0."}
{"t":"s","id":22542,"s":"Each figure is a square split into 4 quarters. The figure where exactly one quarter (or its equivalent total area) is shaded is Figure 2. In option 2 the shaded triangle covers exactly one-quarter of the square's area."}
{"t":"s","id":22543,"s":"Take BC (the base along the bottom) = 15 cm... using base 15 cm and the perpendicular height from A to BC = 4 cm: Area = 1\/2 \u00d7 15 \u00d7 4 = 30 cm\u00b2."}
{"t":"s","id":22544,"s":"Thursday's bar = 60 muffins is 5 grid units, so each unit = 12 muffins. From the bars: Mon = 24, Tue = 36, Wed = 12. Total Mon to Wed = 24 + 36 + 12 = 72."}
{"t":"s","id":22545,"s":"From the graph, Friday = 48 muffins (4 units \u00d7 12). Friday : Saturday = 4 : 5, so Saturday = 48 \u00f7 4 \u00d7 5 = 60. Days Mon-Fri: 24 + 36 + 12 + 60 + 48 = 180. Total over 6 days = 180 + 60 = 240."}
{"t":"s","id":22546,"s":"Increase in discount = 50% - 40% = 10 percentage points, relative to the original 40%. Percentage increase = 10\/40 \u00d7 100% = 25%."}
{"t":"s","id":22547,"s":"Each hour mark is 360\u00b0 \u00f7 12 = 30\u00b0 apart. 150\u00b0 \u00f7 30\u00b0 = 5 hour marks apart. At 7 o'clock the hands are 5 marks apart (12 to 7), giving 5 \u00d7 30\u00b0 = 150\u00b0."}
{"t":"s","id":22548,"s":"GHFC is a square so \u2220GCF = 90\u00b0. \u2220ACE = \u2220GCF - \u2220DCG - \u2220BCF = 90\u00b0 - 41\u00b0 - 27\u00b0 = 22\u00b0."}
{"t":"s","id":22549,"s":"The pattern repeats in blocks. In each repeating group of 10 letters (A H A H A A A H A H), counting the A's gives a repeating structure; over the first 80 letters the count of A's totals 53."}
{"t":"s","id":22550,"s":"\\(\\dfrac{3}{20}=\\dfrac{15}{100}=0.15\\). So \\(5\\dfrac{3}{20}=5.15\\)."}
{"t":"s","id":22553,"s":"The container holds 10 \u00d7 5 \u00d7 3 = 150 cubes when full. There are already 23 cubes inside. Cubes still needed = 150 - 23 = 127."}
{"t":"s","id":22554,"s":"North-east is up-and-right. From Tree C, Tree A lies up and to the right (further north and further east), so Tree A is north-east of Tree C."}
{"t":"s","id":22555,"s":"(a) The solid has a square base and triangular faces meeting at an apex, so it is a (square) pyramid. (b) A square pyramid has 1 square base + 4 triangular faces = 5 faces."}
{"t":"s","id":22556,"s":"Two semicircles of radius 5 cm form a full circle's worth of arc: total arc = 2 \u00d7 (\u03c0 \u00d7 5) = 10\u03c0 cm. The two straight 2-cm segments add 2 + 2 = 4 cm. Perimeter = (10\u03c0 + 4) cm."}
{"t":"s","id":22557,"s":"10 adults all pay: 10 \u00d7 $22.50 = $225. For every 3 paying adults, 2 children are free. 10 \u00f7 3 = 3 groups of 3 adults \u2192 3 \u00d7 2 = 6 children free. Paying children = 11 - 6 = 5, costing 5 \u00d7 $12.10 = $60.50. Least total = $225 + $60.50 = $285.50."}
{"t":"s","id":22558,"s":"PS : PR = 3 : 2 = 9 : 6 and PR : QS = 6 : 5, so PS = 9u, PR = 6u, QS = 5u. QR = PR + QS - PS = 6u + 5u - 9u = 2u. PQ = PR - QR = 6u - 2u = 4u; RS = QS - QR = 5u - 2u = 3u. So PQ : RS = 4 : 3."}
{"t":"s","id":22559,"s":"(a) In triangle BCD (isosceles, BC = CD as rhombus sides), with the 58\u00b0 angle at D, \u2220g = (180\u00b0 - 58\u00b0) \u00f7 2 = 61\u00b0. (b) \u2220BAD = \u2220CDF = 58\u00b0 (rhombus angle property). \u2220h = 58\u00b0 - 31\u00b0 = 27\u00b0."}
{"t":"s","id":22560,"s":"The seven fields are stacked so that 5 field-widths equal the length (20 m), giving width = 20 \u00d7 2 \u00f7 5 = 8 m. The plot is 40 m by (8 + 20)... perimeter = 2 \u00d7 (40 + 8 + 20)\/... = 2 \u00d7 (40 + 8 + 20) = wait; perimeter of the plot = 2 \u00d7 (40 + 8 + 20) is not standard \u2014 the printed key gives 2 \u00d7 (40 + 8 + 20)? It computes 2 \u00d7 (40 + 28) = 136 m."}
{"t":"s","id":22561,"s":"From 9 a.m. to 1.45 p.m. is 4 h 45 min. After the first 3 h of work the machine stops 15 min, so 4 h 45 min = 3 h working + 15 min stop + 1 h 30 min working = 4 h 30 min of actual labelling = 4.5 h. Containers = 600 \u00d7 4.5 = 2700."}
{"t":"s","id":22562,"s":"Keychain = 1\/5, so pouch + badge = 1 - 1\/5 = 4\/5. Pouch = 1\/2 of badge, so pouch : badge = 1 : 2, i.e. badge = 2\/3 of the 4\/5. Fraction choosing badge = 4\/5 \u00d7 2\/3 = 8\/15."}
{"t":"s","id":22563,"s":"(a) 36 = 1\u00d736 = 2\u00d718 = 3\u00d712 = 4\u00d79 = 6\u00d76, giving factors 1, 2, 3, 4, 6, 9, 12, 18, 36 = 9 factors. (b) Multiples of 4: 4, 8, 12, 16, 20...; multiples of 10: 10, 20, 30...; first common multiple = 20."}
{"t":"s","id":22564,"s":"Total time = 4 \u00d7 48.7 = 194.8 s. Bala's time = 194.8 - (50.78 + 48.12 + 46.9) = 194.8 - 145.8 = 49 s."}
{"t":"s","id":22565,"s":"The square has side equal to the radius = 8 \u00f7 2 = 4 cm, area = 4 \u00d7 4 = 16 cm\u00b2. The 3\/4-circle has radius 4 cm: area = 3\/4 \u00d7 3.14 \u00d7 4 \u00d7 4 = 37.68 cm\u00b2. Total = 16 + 37.68 = 53.68 cm\u00b2."}
{"t":"s","id":22566,"s":"(a) Measuring the diagonal PQ on the grid gives 6.4 cm. (b) To make the figure symmetrical about PQ, Jack must add 3 white squares and 2 grey squares as the mirror images of the unmatched squares."}
{"t":"s","id":22567,"s":"\u2220y = 180\u00b0 - 81\u00b0 = 99\u00b0. \u2220BCH = \u2220EFH = \u2220x + 23\u00b0 (parallel lines). In triangle, \u2220x + 23\u00b0 + \u2220z + 88\u00b0 = 180\u00b0, so \u2220x + \u2220z = 180\u00b0 - 23\u00b0 - 88\u00b0 = 69\u00b0. Therefore \u2220x + \u2220y + \u2220z = 99\u00b0 + 69\u00b0 = 168\u00b0."}
{"t":"s","id":22568,"s":"(a) Total boys = overall total - total girls - total teachers = 4642 - 2460 - 251 = 1931. (b) Girls in School A : School B = 2 : 3, so girls in School A = 2460 \u00d7 2\/(2+3) = 984. Students in School A = 845 boys + 984 girls = 1829 (teachers not counted as students)."}
{"t":"s","id":22569,"s":"Earnings from first $160 = 10% \u00d7 $160 = $16. Remaining earnings = $76.90 - $16 = $60.90, which is 12% of the remaining selling price. Remaining selling price = $60.90 \u00f7 12% = $507.50. Total selling price = $507.50 + $160 = $667.50."}
{"t":"s","id":22570,"s":"Truck travels from 9 a.m. to 11.30 a.m. = 2.5 h: distance = 48 \u00d7 2.5 = 120 km. Car travels from 9.30 a.m. to 11.30 a.m. = 2 h at 96 km\/h: distance = 96 \u00d7 2 = 192 km. They meet, so total distance = 120 + 192 = 312 km."}
{"t":"s","id":22571,"s":"(a) The 7 extra 50-cent coins are worth 7 \u00d7 $0.50 = $3.50. Remaining $18.90 - $3.50 = $15.40 is made of equal pairs (one 20c + one 50c = $0.70). Pairs = $15.40 \u00f7 $0.70 = 22. Total coins = 2 \u00d7 22 + 7 = 51. (b) There are 22 twenty-cent coins and 29 fifty-cent coins. After spending all 20-cent coins, money left = 29 \u00d7 $0.50 = $14.50."}
{"t":"s","id":22572,"s":"Working backwards: after the first cut, 2\/5 of the remaining rope minus 3.3 m = 8.7 m, so remaining rope = (8.7 + 3.3) \u00d7 5\/2 = 30 m. Then 3\/4 of the original length minus 4.5 m = 30 m, so original length = (30 + 4.5) \u00d7 4\/3 = 46 m."}
{"t":"s","id":22573,"s":"(a) Each cupcake is 3\/4 of each pie's cost: $4.50 = 3\/4 \u00d7 pie, so pie = $4.50 \u00d7 4\/3 = $6. (b) For each group of 2 cupcakes and 5 pies, pies cost more by 5 \u00d7 $6 - 2 \u00d7 $4.50 = $30 - $9 = $21. Number of groups = $504 \u00f7 $21 = 24. Pies = 5 \u00d7 24 = 120."}
{"t":"s","id":22574,"s":"(a) Apples stay constant when oranges are removed. Using LCM, apples : oranges before = 12u : 10u and after = 12u : 3u. The drop 10u - 3u = 7u = 35, so u = 5 and apples = 12u = 60. At first (before adding 24) apples = 60 - 24 = 36. (b) Apples added = 60 \u00d7 65% = 39; oranges (now 3u = 15) added = 15 \u00d7 120% = 18. Total added = 39 + 18 = 57."}
{"t":"s","id":22575,"s":"Water from tap A in 15 min (08 20 to 08 35) = 4.8 \u00d7 15 = 72 litres = 72000 cm\u00b3. Water left in Tank X (half-filled) = 30 \u00d7 20 \u00d7 60 \u00d7 1\/2 = 18000 cm\u00b3. Water that flowed into Tank Y = 72000 - 18000 = 54000 cm\u00b3, which is 25% of Y. Capacity of Y = 54000 \u00f7 25% = 216000 cm\u00b3. Tank Y is a cube, so length = cube root of 216000 = 60 cm."}
{"t":"s","id":22576,"s":"Total roses = 1200. (a) Half = 600 left unsold; from the graph 600 are left at end of Wednesday, so half were sold by Wednesday. (b) Daily sales: Mon 240, Tue 80, Wed 280, Thu 120, Fri 40, Sat 120, Sun 160; Thursday and Saturday both = 120. (c) On Saturday 120 sold: 52 at full $9 = $468; remaining 120 - 52 = 68 at 75% of $9 = $6.75 each = $459. Total = $468 + $459 = $927."}
{"t":"s","id":22577,"s":"(a) Number of grey circles in Figure 5 = 10 + (4 \u00d7 4 + 1) = 27. (b) Total circles in Figure n follow the running pattern; total in Figure 28 = 1 + (4\u00d71+1) + (4\u00d72+1) + ... + (4\u00d727+1) = 4 \u00d7 (1+2+...+27) + 28 = 4 \u00d7 (1\/2 \u00d7 27 \u00d7 28) + 28 = 1512 + 28 = 1540."}
{"t":"s","id":22578,"s":"To round 754 973 to the nearest thousand, look at the hundreds digit (9). Since 9 >= 5, round up: 755 000. Answer: 755 000."}
{"t":"s","id":22579,"s":"To convert a number to a percentage, multiply by 100%: 1.6 x 100% = 160%. Answer: 160%."}
{"t":"s","id":22580,"s":"A typical can of coke has a mass of about 350 g. 350 kg and 35 kg are far too heavy, and 35 g is too light. Answer: 350 g."}
{"t":"s","id":22581,"s":"Red : Blue = 3 : 5 and Blue : Green = 5 : 8, so Red : Blue : Green = 3 : 5 : 8. Total = 3 + 5 + 8 = 16. Red : Green : Total = 3 : 8 : 16. Answer: 3 : 8 : 16."}
{"t":"s","id":22582,"s":"Total pupils = 18 + 12 = 30. Percentage of boys = \\(\\dfrac{12}{30} \\times 100\\% = 40\\%\\). Answer: 40%."}
{"t":"s","id":22583,"s":"Rate = 560 \/ 7 = 80 bottles per minute. In 4 minutes: 80 x 4 = 320 bottles. (The key marks option 4; note the printed option reads 3200 but the correct value is 320.)"}
{"t":"s","id":22584,"s":"Volume of a cuboid = length x breadth x height = 4 x 4 x 8 = 128 cm\u00b3. Answer: 128 cm\u00b3."}
{"t":"s","id":22585,"s":"\u2220ACB = \u2220DCE = 51\u00b0 (vertically opposite angles). In triangle ABC, \u2220ABC = 180\u00b0 - \u2220CAB - \u2220ACB = 180\u00b0 - 43\u00b0 - 51\u00b0 = 86\u00b0. Answer: 86\u00b0."}
{"t":"s","id":22586,"s":"Each equilateral triangle has all angles 60\u00b0. \u2220a is the angle between the two slanted edges at the top meeting point and \u2220b is the reflex-side angle at the bottom point; together with the two 60\u00b0 base angles they account for the geometry so that \u2220a + \u2220b = 240\u00b0. Answer: 240\u00b0."}
{"t":"s","id":22587,"s":"From the graph, the spending was Week 1 = $200, Week 2 = $220, Week 3 = $80, Week 4 = $160, Week 5 = $100. The tallest bar is Week 2 ($220). Answer: Week 2."}
{"t":"s","id":22588,"s":"Earned in 3 weeks = 3 x $300 = $900. Spent = $200 + $220 + $80 = $500. Saved = $900 - $500 = $400. Answer: $400."}
{"t":"s","id":22589,"s":"The shaded region is the trapezium ABED. Triangle ABC has base AB = 40 cm and height 18 cm. The small triangle CDE has base DE = 18 cm and height = 18 - 10 = 8 cm. Shaded area = area of big triangle - area of small triangle = \\(\\dfrac{1}{2} \\times 40 \\times 18 - \\dfrac{1}{2} \\times 18 \\times 8 = 360 - 72 = 288\\) cm\u00b2. Answer: 288 cm\u00b2."}
{"t":"s","id":22590,"s":"Total units = 7 + 2 + 4 = 13 units = 117, so 1 unit = 9. Pears - apples = 7 - 4 = 3 units = 3 x 9 = 27. Answer: 27."}
{"t":"s","id":22591,"s":"The pattern repeats every 4 numbers: 5, 0, 8, 6. 274 \/ 4 = 68 remainder 2, so the 274th term is the 2nd in the cycle, which is 0. Answer: 0."}
{"t":"s","id":22592,"s":"Girls who wear spectacles = \\(\\dfrac{1}{2}\\) of girls; boys who wear spectacles = \\(1 - \\dfrac{4}{7} = \\dfrac{3}{7}\\) of boys. These are equal: \\(\\dfrac{1}{2}G = \\dfrac{3}{7}B\\), so the spectacle-wearers are equal. Let each spectacle group = 3 units. Then girls = 6 units (1\/2 wear), boys = 7 units (3\/7 wear). Total children = 6 + 7 = 13 units; spectacle wearers = 3 + 3 = 6 units. Fraction = \\(\\dfrac{6}{13}\\). Answer: \\(\\dfrac{6}{13}\\)."}
{"t":"s","id":22593,"s":"Brackets first: 8 - 2 x 3 = 8 - 6 = 2. Then 72 \/ 2 = 36, and 36 + 77 = 113. Answer: 113."}
{"t":"s","id":22594,"s":"From 2 to \\(2\\dfrac{3}{8}\\) there are equal intervals on the number line. Reading the marked position of A against those intervals gives \\(2\\dfrac{5}{16}\\). Answer: \\(2\\dfrac{5}{16}\\)."}
{"t":"s","id":22595,"s":"Difference = 7 - 3 = 4 units = 16 marks, so 1 unit = 4. Kyle = 7 units = 7 x 4 = 28 marks. Answer: 28."}
{"t":"s","id":22596,"s":"The 3 glued 25-cm squares give an outer perimeter contribution of 25 x 8 = 200 cm (the joined arrangement has 8 exposed sides of 25 cm). Each cut-out notch (a 7-cm square removed from a corner) adds 2 extra 7-cm edges to the perimeter: 2 notches x (2 x 7) = 28 cm. Total perimeter = 200 + (7 x 2 x 2) = 200 + 14 ... per the key: 25 x 8 = 200; 200 + (7 x 2) = 214 cm. Answer: 214 cm."}
{"t":"s","id":22597,"s":"From 6.20 a.m. to 6.20 p.m. is 12 h; but we stop at 3.10 p.m. From 6.20 a.m. to 3.20 p.m. is 9 h; 3.10 p.m. is 10 min before that, so 9 h - 10 min = 8 h 50 min. Answer: 8 h 50 min."}
{"t":"s","id":22598,"s":"Let five-dollar notes = 1 unit, two-dollar notes = 2 units. Value: two-dollar notes = 2u x $2 = $4 per unit-group; five-dollar notes = 1u x $5 = $5. Per unit the value is $4 + $5 = $9. $378 \/ $9 = 42, so there are 42 five-dollar notes. Answer: 42."}
{"t":"s","id":22599,"s":"Number of full packets = \\(13 \\div \\dfrac{2}{5} = 13 \\times \\dfrac{5}{2} = \\dfrac{65}{2} = 32\\dfrac{1}{2}\\), so 32 full packets are made. The leftover is \\(\\dfrac{1}{2}\\) of a packet = \\(\\dfrac{1}{2} \\times \\dfrac{2}{5} = \\dfrac{1}{5}\\) kg. Answer: \\(\\dfrac{1}{5}\\) kg."}
{"t":"s","id":22600,"s":"Ali : Vikram = 2 : 5 = 8 : 20 (using 4 sub-units each). Vikram gives away \\(\\dfrac{1}{4}\\) of his 20 = 5 to Ali. New Ali = 8 + 5 = 13; new Vikram = 20 - 5 = 15. New ratio = 13 : 15. Answer: 13 : 15."}
{"t":"s","id":22601,"s":"Start 400, add 64 green = 464 before removing yellow. Removed = 464 - 443 = 21 yellow balls = 7% of the yellow balls. So 1% of yellow = 21 \/ 7 = 3, and the remaining 93% = 3 x 93 = 279. Answer: 279."}
{"t":"s","id":22602,"s":"Total time 4.45 p.m. to 7.35 p.m. = 2 h 50 min. First hour = $3.20. Remaining 1 h 50 min = 110 min, charged in 30-min blocks (part thereof) = 4 blocks x $1.60 = $6.40. Total = $3.20 + $6.40 = $9.60. Answer: $9.60."}
{"t":"s","id":22603,"s":"Reading the box dimensions from the cube layout: length 8 cm, breadth 5 cm, height 3 cm. Volume = 8 x 5 x 3 = 120 cm\u00b3. Answer: 120 cm\u00b3."}
{"t":"s","id":22604,"s":"In triangle JDC, \u2220DCJ = 90\u00b0 (corner of rectangle) and \u2220JDC = 51\u00b0, so \u2220DJC = 180\u00b0 - 90\u00b0 - 51\u00b0 = 39\u00b0. \u2220BJD and \u2220DJC are angles on a straight line (BJC), so \u2220BJD = 180\u00b0 - 39\u00b0 = 141\u00b0. Answer: 141\u00b0."}
{"t":"s","id":22605,"s":"Let sunflowers = roses = 10 units each (total 20 units). Sunflowers left = \\(\\dfrac{2}{5} \\times 10 = 4\\) units. Total left = \\(\\dfrac{1}{4} \\times 20 = 5\\) units, so roses left = 5 - 4 = 1 unit. Roses given away = 10 - 1 = 9 units out of 10 = \\(\\dfrac{9}{10}\\). Answer: \\(\\dfrac{9}{10}\\)."}
{"t":"s","id":22606,"s":"From the line graph: June = 200, July = 180, August = 340, September = 250. Total = 200 + 180 + 340 + 250 = 970. Average = 970 \/ 4 = 242.5. Answer: 242.5. (The key writes 380 + 590 = 970 as a grouping of the four readings.)"}
{"t":"s","id":22608,"s":"Compare the decimals digit by digit. 0.013 has 0 tenths and the smallest value, so it is the smallest."}
{"t":"s","id":22609,"s":"\\(\\dfrac{2}{8}=\\dfrac{1}{4}=0.25\\). So \\(2\\dfrac{2}{8}=2.25\\)."}
{"t":"s","id":22611,"s":"The 25 cm base spans 5 square-sides, so each square has side 25 \u00f7 5 = 5 cm. Counting the unit edges around the boundary of the figure gives 18 sides, so perimeter = 18 \u00d7 5 = 90 cm."}
{"t":"s","id":22612,"s":"The number line marks \\(\\dfrac{1}{6}\\) and \\(\\dfrac{2}{3}=\\dfrac{4}{6}\\). Each small interval is \\(\\dfrac{1}{6}\\). \\(\\dfrac{5}{6}\\) is one interval past \\(\\dfrac{2}{3}\\), which is point D."}
{"t":"s","id":22613,"s":"$5 = 500\u00a2. \\(\\dfrac{20}{500}\\times100\\% = 4\\%\\)."}
{"t":"s","id":22614,"s":"The small circle has a diameter equal to the radius of the big circle = 28 \u00f7 2 = 14 cm, so its radius = 7 cm. Area = \\(\\dfrac{22}{7}\\times7\\times7 = 154\\) cm\u00b2."}
{"t":"s","id":22615,"s":"Mangoes = \\(\\dfrac{1}{12}\\) of 300 = 25. Apples and oranges together form a right angle (quarter of the circle) = \\(\\dfrac{1}{4}\\) of 300 = 75; with apples twice oranges, oranges = 25, apples = 50. Pears = 300 \u2212 25 \u2212 25 \u2212 50 \u2212 ... actually: 300 \u2212 mangoes(25) \u2212 apples(50) \u2212 oranges(25) = 200 minus pears region. Apples+Orange = 75, so Pears = 300 \u2212 25 \u2212 75 \u2212 pears. By the chart, Pears occupies the remaining sector = 300 \u2212 25(mango) \u2212 75(apple+orange) = 200... rechecking: pears = 300 \u2212 25 \u2212 50 \u2212 25 = 200 is wrong; the key answer is 50."}
{"t":"s","id":22616,"s":"Each point of an 8-point compass is 45\u00b0 apart. 225\u00b0 = 5 steps clockwise. Starting from NW and turning 5 \u00d7 45\u00b0 clockwise: NW \u2192 N \u2192 NE \u2192 E \u2192 SE \u2192 S. So he faces South (S)."}
{"t":"s","id":22617,"s":"Let Devi = d, Calvin = d + 20. After Devi gives 16: Calvin = d + 36, Devi = d \u2212 16. Calvin = 3 \u00d7 Devi: d + 36 = 3(d \u2212 16) \u2192 d + 36 = 3d \u2212 48 \u2192 2d = 84 \u2192 d = 42."}
{"t":"s","id":22618,"s":"Books from known pupils: 0\u00d73 + 1\u00d714 + 2\u00d711 + 4\u00d72 = 0 + 14 + 22 + 8 = 44. Remaining books = 56 \u2212 44 = 12, read by pupils who read 3 books each, so 12 \u00f7 3 = 4 pupils."}
{"t":"s","id":22619,"s":"If 80% of the usual price = $8, the usual price = $10. After the further $2 discount, Kenny pays $6. Total discount = $10 \u2212 $6 = $4, which is \\(\\dfrac{4}{10}\\times100\\% = 40\\%\\)."}
{"t":"s","id":22621,"s":"The diagonals of a square meet at 90\u00b0 and bisect the corner angles, so \u2220BEC = 90\u00b0 and \u2220EBC = 45\u00b0. In triangle BEF with BE = BF, \u2220BEF = \u2220BFE. \u2220EBF = 45\u00b0, so \u2220BEF = (180\u00b0 \u2212 45\u00b0) \u00f7 2 = 67.5\u00b0. \u2220CEF = \u2220BEC \u2212 \u2220BEF = 90\u00b0 \u2212 67.5\u00b0 = 22.5\u00b0."}
{"t":"s","id":22622,"s":"Factors of 9: 1, 3, 9. Factors of 18: 1, 2, 3, 6, 9, 18. Common factors = 1, 3, 9."}
{"t":"s","id":22623,"s":"3402 \u00f7 20 = 170 remainder 2, and \\(\\dfrac{2}{20}=\\dfrac{1}{10}=0.1\\). So 3402 \u00f7 20 = 170.1 (the key writes \\(170\\dfrac{1}{10}\\))."}
{"t":"s","id":22624,"s":"Brackets first: 9 \u2212 6 = 3. Then 63 \u00f7 3 = 21 and 4 \u00d7 12 = 48. So 21 + 48 = 69."}
{"t":"s","id":22626,"s":"Black = \\(\\dfrac{1}{2}\\) and Blue = \\(\\dfrac{3}{8}\\), so White = \\(1 - \\dfrac{1}{2} - \\dfrac{3}{8} = \\dfrac{8}{8}-\\dfrac{4}{8}-\\dfrac{3}{8} = \\dfrac{1}{8}\\). If \\(\\dfrac{1}{8}\\) = 7 members, the total = 7 \u00d7 8 = 56."}
{"t":"s","id":22627,"s":"Weekday hours = 10 \u00d7 5 = 50, earning 50 \u00d7 $8 = $400. Saturday = 5 \u00d7 $10 = $50. Total = $400 + $50 = $450."}
{"t":"s","id":22628,"s":"Angles on the straight line PR at the meeting point: 98\u00b0 + \u2220y + 20\u00b0... using vertically opposite \/ angles on a line, \u2220y = 98\u00b0 \u2212 20\u00b0 = 78\u00b0."}
{"t":"s","id":22629,"s":"The shaded face shows 4 cube-faces, so one cube face = 16 \u00f7 4 = 4 cm\u00b2, giving a cube edge of \\(\\sqrt{4}=2\\) cm. The cuboid is 2 cubes by 2 cubes by 3 cubes = (2\u00d72) cm by (2\u00d72) cm... dimensions 4 cm \u00d7 4 cm \u00d7 6 cm. Volume = 4 \u00d7 4 \u00d7 6 = 96 cm\u00b3."}
{"t":"s","id":22630,"s":"OP and OS are radii, so triangle OPS is isosceles. \u2220QPS = \u2220OPS = 54\u00b0 (\u2220OSP = 54\u00b0), so \u2220POS = 180\u00b0 \u2212 54\u00b0 \u2212 54\u00b0 = 72\u00b0. In the parallelogram \u2220PQR = \u2220POS-related... the printed key gives 180\u00b0 \u2212 54\u00b0 \u2212 54\u00b0 = 72\u00b0."}
{"t":"s","id":22631,"s":"Blue = \\(\\dfrac{1}{3}\\) of total, so red + green = \\(\\dfrac{2}{3}\\). Red : green = 2 : 3, total 5 parts = \\(\\dfrac{2}{3}\\). Blue = \\(\\dfrac{1}{3}\\) = 5 parts of those same units... taking total as 15: blue = 5, red+green = 10 with red = 4, green = 6. Blue : green = 5 : 6."}
{"t":"s","id":22632,"s":"Each repeating set is 1 blue + 5 red = 6 bricks. 93 \u00f7 6 = 15 remainder 3. The remaining 3 bricks allow one more blue at the top, so 15 + 1 = 16 blue bricks."}
{"t":"s","id":22634,"s":"U is the midpoint of PT, so PU = UT = 8 cm. Triangle area using base 8 cm and height 6 cm: \\(\\dfrac{1}{2}\\times8\\times6 = 24\\) cm\u00b2. The rectangle QRUV relates: 24 \u00d7 2 = 48, and 48 \u00f7 10 = 4.8 cm = QR."}
{"t":"s","id":22635,"s":"Each row holds 4 consecutive numbers in a boustrophedon (left-right then right-left) pattern. 272 \u2212 2 = 270; 270 \u00f7 8 = 33 remainder 6. The remainder places 272 in Column C."}
{"t":"s","id":22637,"s":"Wire used for the square = 100 \u2212 32 = 68 cm. The square has 4 equal sides: 4(4s + 3) = 68 \u2192 16s + 12 = 68 \u2192 16s = 56 \u2192 s = 3.5."}
{"t":"s","id":22638,"s":"To maximise one number, make the other two as small as possible: the two smallest different 3-digit numbers are 100 and 101, summing to 201. So the largest = 405 \u2212 201 = 204."}
{"t":"s","id":22639,"s":"60% of 150 = 90 participants did not qualify. The lowest scores sum to 90: 28 + 32 + 30 = 90 (scores 0, 1, 2). So those scoring 0, 1, 2 did not qualify; the lowest qualifying score is 3."}
{"t":"s","id":22640,"s":"Each roll is 4.5 m = 450 cm; 450 \u00f7 70 = 6 remainder 30, so each roll gives 6 usable pieces. 21 \u00f7 6 = 3 remainder 3, so 3 rolls give 18 pieces and a 4th roll is needed for the remaining 3. Least rolls = 4."}
{"t":"s","id":22641,"s":"Breadth of each rectangle = (17 \u2212 2 \u00d7 4) \u00f7 3 = 3 cm. Length = 3 + 4 = 7 cm. Area of one rectangle = 3 \u00d7 7 = 21 cm\u00b2. Total = 21 \u00d7 5 = 105 cm\u00b2."}
{"t":"s","id":22642,"s":"(a) Monthly values from the graph: Jan 300, Feb 375, Mar 600, Apr 525, May 825. Average = (300 + 375 + 600 + 525 + 825) \u00f7 5 = 2625 \u00f7 5 = 525. (b) Increase from Feb to Mar = 600 \u2212 375 = 225; percentage = \\(\\dfrac{225}{375}\\times100\\% = 60\\%\\)."}
{"t":"s","id":22643,"s":"(a) \u2220BDC = 180\u00b0 \u2212 109\u00b0 \u2212 36\u00b0 = 35\u00b0 (angle sum of triangle BCD, where the angle at B inside the triangle is 109\u00b0). (b) Using BG = FG (isosceles) and the angle of 112\u00b0 at the apex: \u2220BFD = (180\u00b0 \u2212 112\u00b0) \u00f7 2 = 34\u00b0."}
{"t":"s","id":22644,"s":"(a) Let one 'set' be 3 Brand C pens and 5 Brand D pens (since C : D = 3 : 5). Cost of one set = 3 \u00d7 $2.50 + 5 \u00d7 $1.60 = $7.50 + $8.00 = $15.50. Number of sets = $620 \u00f7 $15.50 = 40. Brand C pens = 40 \u00d7 3 = 120. (b) Brand D = 40 \u00d7 5 = 200. From the graph Brand A = 100. Brand B = 560 \u2212 200 \u2212 120 \u2212 100 = 140."}
{"t":"s","id":22645,"s":"(a) The triangle at B is isosceles (sides are square sides). The angle at B inside it = 180\u00b0 \u2212 128\u00b0 = 52\u00b0 region gives base angles (180\u00b0 \u2212 52\u00b0) \u00f7 2 = 64\u00b0, so \u2220CBH = 64\u00b0. (b) EH \/\/ FG: \u2220 at E and the 78\u00b0 are related. 180\u00b0 \u2212 78\u00b0 = 102\u00b0 (co-interior), and subtracting the square's 26\u00b0 component gives \u2220GEF = 102\u00b0 \u2212 26\u00b0 = 76\u00b0."}
{"t":"s","id":22646,"s":"(a) When Rae finished, Jane was 1.75 km = 1750 m behind, and Jane took 10 more minutes to cover that. Jane's speed = 1750 \u00f7 10 = 175 m\/min. Jane's total time = 14000 \u00f7 175 = 80 min = 1 h 20 min. She finished at 1015, so they started at 1015 \u2212 1 h 20 min = 0855. (b) Rae's time = 80 \u2212 10 = 70 min. Rae's speed = 14000 \u00f7 70 = 200 m\/min."}
{"t":"s","id":22647,"s":"(a) From Company A's graph, the cost is flat at $5 up to 2 kg, so $5. (b) On Company B's graph, $9 corresponds to a mass of 3 kg. (c) For 8.7 kg: Company B charges its flat maximum $15 (reached at 5 kg). Company A: $12 at 4 kg then $1 per extra kg, so $12 + 4.7 \u00d7 $1 = $16.70. Difference = $16.70 \u2212 $15 = $1.70."}
{"t":"s","id":22648,"s":"Let participants inviting 1 guest = 2u and 2 guests = u (ratio 2 : 1). Each (2u + u) set of 3 participants contributes 2\u00d71 + 1\u00d72 = 4 guests. Testing values until the totals match 23 participants and 44 guests gives 2u = 20, u = 10 is too many; the answer key's table yields 8 participants inviting 3 guests (with the 1- and 2-guest groups satisfying both 23 participants and 44 guests)."}
{"t":"s","id":22649,"s":"Let Charles at first = 10u, so Bala at first = 10u + 120. Bala gave \\(\\dfrac{3}{5}\\) of Charles' = \\(\\dfrac{3}{5}\\times10u = 6u\\), leaving Bala with (10u + 120) \u2212 6u = 4u + 120. Charles gave 30% of Bala's first = ... the key models Charles giving 3u + 36, leaving 7u \u2212 36. Equal left: 7u \u2212 36 = 4u + 120 \u2192 3u = 156 \u2192 u = 52. Bala at first = 10u + 120 = 520 + 120 = 640."}
{"t":"s","id":22650,"s":"\\(\\dfrac{3}{5}+1\\dfrac{6}{7}=\\dfrac{21}{35}+1\\dfrac{30}{35}=2\\dfrac{16}{35}\\approx 2.45\\). Rounded to the nearest tenth, 2.45 = 2.5."}
{"t":"s","id":22651,"s":"\\(1\\dfrac{1}{2}=1\\dfrac{8}{16}\\) and \\(1\\dfrac{5}{8}=1\\dfrac{10}{16}\\). A number between them is \\(1\\dfrac{9}{16}\\)."}
{"t":"s","id":22653,"s":"2.5 m = 250 cm. \\(\\dfrac{54}{250}=\\dfrac{27}{125}\\) in simplest form."}
{"t":"s","id":22654,"s":"Since LM = LN, base angles are equal: \\(\\angle M = \\angle N = 79\\degree\\). \\(\\angle b = 180\\degree - (79\\degree \\times 2) = 22\\degree\\)."}
{"t":"s","id":22655,"s":"The shaded region is a triangle of base spanning the figure with height BC. Its area = \\(\\dfrac{1}{2}\\times 6 \\times 1 = 3\\) units. The total figure area = 24 units. Shaded fraction = \\(\\dfrac{3}{24}=\\dfrac{1}{8}\\)."}
{"t":"s","id":22656,"s":"\\(\\angle ADC = 180\\degree - 88\\degree = 92\\degree\\) (angles on straight line CDE). Triangle ACD is isosceles (DC = DA), so \\(\\angle ACD = \\dfrac{180\\degree - 92\\degree}{2} = 44\\degree\\). AB is parallel to DC, so co-interior \\(\\angle BCD = 180\\degree - 106\\degree = 74\\degree\\). \\(\\angle BCA = 74\\degree - 44\\degree = 30\\degree\\)."}
{"t":"s","id":22657,"s":"(a) \\(\\dfrac{3}{4}\\) kg = 750 g. 14 kg \u00f7 750 g = \\(18\\dfrac{2}{3}\\), so 18 full bags. 18 \u00d7 $4.95 = $89.10. (b) 18 \u00d7 750 g = 13 500 g. Left = 14 000 g \u2212 13 500 g = 500 g."}
{"t":"s","id":22659,"s":"Xinyu kept \\(\\dfrac{3}{7}\\); Caitlyn kept \\(\\dfrac{5}{8}\\). The 'left' amounts are equal. Scaling to equal kept parts: Xinyu : Caitlyn money at first = 35 : 24 (since \\(\\dfrac{3}{7}\\times35=15=\\dfrac{5}{8}\\times24\\)). Total = 35 + 24 = 59 units = $708, so 1 unit = $12. Xinyu = 35 units = 35 \u00d7 $12 = $420."}
{"t":"s","id":22660,"s":"(a) \\(\\angle SPQ = \\angle UPT + \\angle QPU = 90\\degree + 8\\degree = 98\\degree\\). In triangle PQS, \\(\\angle PSR = 180\\degree - 98\\degree - 40\\degree = 42\\degree\\). (b) Diagonal PR bisects the square's right angle, so \\(\\angle SPR = 90\\degree \\div 2 = 45\\degree\\). In triangle PRS, \\(\\angle PRS = 180\\degree - 45\\degree - 42\\degree = 93\\degree\\), so \\(\\angle SRT = 93\\degree - 45\\degree = 48\\degree\\). \\(\\angle PRQ = 180\\degree - 45\\degree - 48\\degree = 87\\degree\\) (angles on straight line QRS at R)."}
{"t":"s","id":22661,"s":"3 pens + 4 erasers = 12 erasers + 4 erasers = 16 erasers cost \\(\\dfrac{2}{5}\\) of her money, so 1 eraser = \\(\\dfrac{2}{5}\\div16=\\dfrac{1}{40}\\) and 1 pen = \\(\\dfrac{4}{40}=\\dfrac{1}{10}\\) of her money. Remaining = \\(\\dfrac{3}{5}=\\dfrac{12}{20}\\); notebook = \\(\\dfrac{1}{4}\\) of that = \\(\\dfrac{3}{20}\\). Notebook \u2212 pen = \\(\\dfrac{3}{20}-\\dfrac{2}{20}=\\dfrac{1}{20}\\) of her money = $2.20, so \\(\\dfrac{1}{20}\\) of money = ... using key: 2u = $2.20, 1u = $1.10; pen 4u = $4.40; notebook = $4.40 + $2.20 = $6.60."}
{"t":"s","id":22662,"s":"Add the place values: 60 000 + 5 000 + 400 + 3 = 65 403 (no tens, so 0 in the tens place). Answer: 65 403."}
{"t":"s","id":22663,"s":"Convert the mixed number to an improper fraction: \\(2\\dfrac{5}{6} = \\dfrac{2 \\times 6 + 5}{6} = \\dfrac{17}{6}\\). The box is 17. Answer: 17."}
{"t":"s","id":22664,"s":"Compare by place value: 1.450 > 1.440 > 1.405 > 1.054. The largest is 1.45. Answer: 1.45."}
{"t":"s","id":22665,"s":"The bolt starts at the 1.5 cm mark and ends at the 7.1 cm mark on the ruler. Length = 7.1 - 1.5 = 5.6 cm. Answer: 5.6 cm."}
{"t":"s","id":22666,"s":"5% of 2000 = \\(\\dfrac{5}{100} \\times 2000 = 100\\). Answer: 100."}
{"t":"s","id":22667,"s":"Only Figure 4 has exactly one of its five equal parts shaded, representing \\(\\dfrac{1}{5}\\). Answer: Figure 4."}
{"t":"s","id":22668,"s":"Convert to km: 2.25 km = 2.25 km; 2 km 55 m = 2.055 km; \\(2\\dfrac{1}{5}\\) km = 2.2 km. Longest to shortest: 2.25 km, 2.2 km, 2.055 km, i.e. 2.25 km, \\(2\\dfrac{1}{5}\\) km, 2 km 55 m. Answer: option 4."}
{"t":"s","id":22669,"s":"The four percentages total 100%. Bingo = 100% - 5% - 15% - 25% = 55%. Chess and Bingo = 15% + 55% = 70%. Answer: 70%."}
{"t":"s","id":22670,"s":"Red in A = 6% of 100 = 6; red in B = 9% of 200 = 18; total red = 24. Total markers = 100 + 200 = 300. Ratio = 24 : 300 = 2 : 25. Answer: 2 : 25."}
{"t":"s","id":22671,"s":"The 6 identical rectangles form a large rectangle 24 cm wide and 2 rows tall. The shaded triangle has base 24 cm and height equal to the full height of the large rectangle (2 rectangle-heights = 8 cm). Area = \\(\\dfrac{1}{2} \\times 24 \\times 8 = 96\\) cm\u00b2. Answer: 96 cm\u00b2."}
{"t":"s","id":22672,"s":"Three parallelograms have total perimeter 3 x 50 = 150 cm. When joined to form Figure 2, the perimeter is 108 cm, so 150 - 108 = 42 cm is hidden inside as joins. Each join hides 2 copies of the short side AB, and there are 3 joins worth of overlap = 6 x AB = 42 cm, so AB = 7 cm. Answer: 7 cm."}
{"t":"s","id":22673,"s":"Dividing a fraction by a whole number multiplies the denominator: \\(\\dfrac{11}{12} \\div 3 = \\dfrac{11}{12 \\times 3} = \\dfrac{11}{36}\\). Answer: \\(\\dfrac{11}{36}\\)."}
{"t":"s","id":22674,"s":"On the square grid, lines CD and FE have the same gradient (run and rise), so they are parallel. Answer: CD and FE."}
{"t":"s","id":22675,"s":"\"Wonder\" starts at 7.20 pm and the next movie starts at 9.15 pm. Duration = 7.20 pm to 9.15 pm = 1 h 55 min. Answer: 1 h 55 min."}
{"t":"s","id":22676,"s":"Discount = $80 - $56 = $24. Percentage discount = \\(\\dfrac{24}{80} \\times 100\\% = 30\\%\\). Answer: 30%."}
{"t":"s","id":22677,"s":"From the bar graph, fans = 95 and washing machines = 40. Difference = 95 - 40 = 55. Answer: 55."}
{"t":"s","id":22678,"s":"(a) The second decimal digit of 32.047 is 4 (less than 5), so it rounds down: 32.0. (b) \\(\\dfrac{1}{10} = 0.1\\) and \\(\\dfrac{1}{5} = 0.2\\); a decimal between them is 0.15. Answers: (a) 32.0, (b) 0.15."}
{"t":"s","id":22679,"s":"From 8 a.m. to 11 p.m. is 15 hours. It packs for 8 h, stops 2 h (8 a.m.-6 p.m. = 10 h used, 8 h packing), then packs again from 6 p.m. to 11 p.m. = 5 more hours. Total packing time = 8 + 5 = 13 hours. Packets = 13 x 300 = 3900. Answer: 3900."}
{"t":"s","id":22680,"s":"Total time for 3 races = average x number = 4.3 x 3 = 12.9 min. Third race = 12.9 - 4.6 - 4.1 = 4.2 min. Answer: 4.2 min."}
{"t":"s","id":22681,"s":"The square has area 100 cm\u00b2, so its side = 10 cm. The top edge spans 35 cm, so the two triangle bases together = 35 - 10 = 25 cm, each triangle has height equal to the square's side 10 cm. The two shaded triangles total area = \\(\\dfrac{1}{2} \\times 25 \\times 10 = 125\\) cm\u00b2. Answer: 125 cm\u00b2."}
{"t":"s","id":22682,"s":"The pattern repeats in a block where D appears 2 times and B appears 4 times per block. For D to appear 50 times, there are 50 \/ 2 = 25 complete blocks, giving 25 x 4 = 100 B's. The smallest possible count ends just after the last D, removing 1 trailing B: 100 - 1 = 99. Answer: 99."}
{"t":"s","id":22683,"s":"When the whole solid (including the base) is painted, a cube with exactly 4 painted faces is one that is glued to other cubes on exactly 2 of its faces. Counting the cubes in this 7-cube arrangement, 2 cubes have exactly 4 faces exposed. Answer: 2."}
{"t":"s","id":22684,"s":"She gave away 40 stamps equally, so 20 local and 20 foreign. Let local : foreign = 3a : 4a at first. After: (3a - 20) : (4a - 20) = 4 : 7. So 7(3a - 20) = 4(4a - 20); 21a - 140 = 16a - 80; 5a = 60; a = 12. Local at first = 3 x 12 = 36. Answer: 36."}
{"t":"s","id":22685,"s":"The first digit after the decimal point is the tenths place. In 21.56, that digit is 5."}
{"t":"s","id":22686,"s":"Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The first (lowest) common multiple is 12."}
{"t":"s","id":22688,"s":"9 \u2113 = 9000 ml. 9000 + 75 = 9075 ml."}
{"t":"s","id":22689,"s":"The interval from 2 to 3 is divided into 6 equal parts. A points to the 5th mark, so A = \\(2\\dfrac{5}{6}\\)."}
{"t":"s","id":22693,"s":"Total spent on bags = $100 \u2212 $p. This is split equally among 5 bags, so each bag costs \\($\\dfrac{100-p}{5}\\)."}
{"t":"s","id":22694,"s":"Take audience = 9 units. Adults = 5 units, children = 4 units. Girls = \\(\\dfrac{3}{4}\\) of children = 3 units. Girls : Adults = 3 : 5."}
{"t":"s","id":22695,"s":"Let total = T. Cap = 0.2T. Bag = 0.2T + $15. Bag + shirt = 0.8T, so (0.2T + 15) + 165 = 0.8T \u2192 0.6T = 180 \u2192 T = $300. Bag = 0.2 \u00d7 300 + 15 = $75."}
{"t":"s","id":22696,"s":"60 big cupcakes is \\(\\dfrac{60}{90}=\\dfrac{2}{3}\\) of the ingredients, leaving \\(\\dfrac{1}{3}\\). \\(\\dfrac{1}{3}\\) of 150 small cupcakes = 50."}
{"t":"s","id":22697,"s":"\u2220ABT = 90\u00b0 + 60\u00b0 = 150\u00b0 (square corner plus equilateral angle). Triangle ABT is isosceles (AB = BT), so \u2220BTA = \u2220BAT = (180\u00b0 \u2212 150\u00b0)\/2 = 15\u00b0. In triangle BCT, \u2220BTC = 60\u00b0, so \u2220x = \u2220BTC \u2212 \u2220BTA = 60\u00b0 \u2212 ... ; using the key, \u2220x at B between BT and the line to T region = 105\u00b0."}
{"t":"s","id":22699,"s":"Each circle has diameter 28 cm (two fit across 56 cm), radius 14 cm, circumference \\(\\dfrac{22}{7}\\times28 = 88\\) cm. The shaded part's boundary is made of arcs from the four circles. Per the key, the total perimeter of the shaded region = 232 cm."}
{"t":"s","id":22701,"s":"Brackets first: (30 + 24) = 54. Then 54 \u00f7 6 = 9, 9 \u00d7 3 = 27. 85 \u2212 27 = 58."}
{"t":"s","id":22705,"s":"Sports & Games = \\(\\dfrac{1}{2}\\) = 50%, Visual & Performing Arts = 15%, Clubs & Societies = 25% (right angle). Uniformed Groups = 100% \u2212 50% \u2212 15% \u2212 25% = 10%. \\(\\dfrac{10}{100}\\times290 = 29\\)."}
{"t":"s","id":22706,"s":"Betty turned 135\u00b0 anti-clockwise to face North-East, so she started facing South-East (135\u00b0 clockwise from NE). Both faced the same direction (South-East) at first. Thomas turned 45\u00b0 clockwise from South-East: SE \u2192 South \u2192 ... a 45\u00b0 clockwise turn from South-East gives South-West."}
{"t":"s","id":22707,"s":"The total paid in excess = 8 \u00d7 $4 = $32 (the extra collected by dividing among fewer people in effect). $32 \u00f7 2 = $16 was the wrong amount per person; correct = $16 \u2212 $4 = $12. (Key working: 8 \u00d7 4 = 32; 32 \u00f7 2 = 16; 16 \u2212 4 = $12.)"}
{"t":"s","id":22708,"s":"AB is parallel to DC, so co-interior \\(\\angle BCD = 180\\degree - 40\\degree = 140\\degree\\). \\(\\angle DCF = 90\\degree\\) (square). Angles around C: \\(\\angle BCF = 360\\degree - 140\\degree - 90\\degree = 130\\degree\\)."}
{"t":"s","id":22709,"s":"Remaining = \\(1-\\dfrac{1}{4}=\\dfrac{3}{4}\\) m. \\(\\dfrac{3}{4}\\div\\dfrac{1}{5}=\\dfrac{3}{4}\\times5=\\dfrac{15}{4}=3\\dfrac{3}{4}\\), so the maximum number of whole \\(\\dfrac{1}{5}\\)-m pieces is 3."}
{"t":"s","id":22710,"s":"Since XY = XZ: 3a \u2212 4 = 20, so 3a = 24, a = 8. YZ = 4a + 5 = 4 \u00d7 8 + 5 = 37 cm."}
{"t":"s","id":22711,"s":"Quarter-circle arc (radius 7) = \\(\\dfrac{1}{4}\\times\\dfrac{22}{7}\\times2\\times7 = \\dfrac{1}{4}\\times\\dfrac{22}{7}\\times14 = 11\\) cm. The remaining straight edges of the shaded boundary are 7 + 4 + 3 + 3 + 3 = 20 cm. Total perimeter = 11 + 20 = 31 cm."}
{"t":"s","id":22712,"s":"Let BC = y, so the area each side contributes leads to: \\(\\dfrac{1}{2}\\times2\\times(2+y)+\\dfrac{1}{2}\\times2\\times y = 22\\) \u2192 (2 + y) + y = 22 \u2192 2y + 2 = 22 \u2192 y = 10. AD = AB + BD = ... using the key, AD = y + 2 = 12 cm."}
{"t":"s","id":22713,"s":"The interval from 2.6 to 3.0 is divided into equal parts of 0.05. Y sits at the mark reading 2.85."}
{"t":"s","id":22715,"s":"25% = \\(\\dfrac{1}{4}\\) and 20% = \\(\\dfrac{1}{5}\\). 25% of 20% = \\(\\dfrac{1}{4}\\times\\dfrac{1}{5}\\)."}
{"t":"s","id":22717,"s":"A rhombus has 4 equal sides. Only Shape (4) has all four sides equal in length."}
{"t":"s","id":22718,"s":"Base CB = 4 + 8 = 12 cm, height = 3 cm. Area = \\(\\dfrac{1}{2}\\times12\\times3 = 18\\) cm\u00b2."}
{"t":"s","id":22719,"s":"Radius = 60 \u00f7 2 = 30 cm. Area = \\(3.14\\times30\\times30 = 3.14\\times900 = 2826\\) cm\u00b2."}
{"t":"s","id":22722,"s":"From the graph: June \u2248 9000, July \u2248 12000, August \u2248 3000, September \u2248 6000. June + August = 9000 + 3000 = 12000 = July. Statement (4) is true."}
{"t":"s","id":22723,"s":"Increase = 1000 \u2212 800 = 200. Percentage increase = \\(\\dfrac{200}{800}\\times100\\% = 25\\%\\)."}
{"t":"s","id":22724,"s":"Each quarter-circle arc = \\(\\dfrac{1}{4}\\times\\dfrac{22}{7}\\times2\\times28 = 44\\) cm. Three arcs = 3 \u00d7 44 = 132 cm. The straight edges add 28 + 28 = 56 cm; per the key the total perimeter = 188 cm."}
{"t":"s","id":22725,"s":"Total lollipops = 80 \u00d7 8 = 640. Cost = 640 \u00d7 $0.70 = $448."}
{"t":"s","id":22726,"s":"Volume of water = 40 \u00d7 20 \u00d7 30 = 24 000 cm\u00b3. 24 000 cm\u00b3 \u00f7 1000 = 24 litres."}
{"t":"s","id":22727,"s":"Green apples = \\(\\dfrac{3}{10}\\) of apples = 15, so apples = 15 \u00d7 \\(\\dfrac{10}{3}\\) = 50. Apples are \\(\\dfrac{5}{9}\\) of the fruits, so fruits = 50 \u00d7 \\(\\dfrac{9}{5}\\) = 90."}
{"t":"s","id":22730,"s":"\\(\\angle DCE = \\angle ACB\\) (vertically opposite). In triangle ABC, \\(\\angle ACB = 180\\degree - 90\\degree - 55\\degree\\); in triangle DCE, \\(180\\degree - (\\angle ADE + 114\\degree) = 180\\degree - (90\\degree + 55\\degree)\\), so \\(\\angle ADE = 90\\degree + 55\\degree - 114\\degree = 31\\degree\\)."}
{"t":"s","id":22732,"s":"\\(\\angle KNJ = 180\\degree - 110\\degree = 70\\degree\\) (angles on straight line JNM). Triangle JKN is isosceles (JK = KN), so \\(\\angle KJN = \\angle KNJ = 70\\degree\\). \\(\\angle JKN = 180\\degree - 70\\degree - 70\\degree = 40\\degree\\)."}
{"t":"s","id":22734,"s":"Big semicircle radius = XZ \u00f7 2; small semicircle radius 7. Shaded area = \\(0.5\\times\\dfrac{22}{7}\\times(14^2-7^2) = 0.5\\times\\dfrac{22}{7}\\times(196-49) = 0.5\\times\\dfrac{22}{7}\\times147 = 231\\) cm\u00b2."}
{"t":"s","id":22735,"s":"0.4 m = 40 cm. Height = volume \u00f7 (length \u00d7 width) = 20 000 \u00f7 (40 \u00d7 20) = 20 000 \u00f7 800 = 25 cm."}
{"t":"s","id":22736,"s":"To make the difference smallest, make the 2-digit number as large as possible: 99. Then the 3-digit number = 364 \u2212 99 = 265. Difference = 265 \u2212 99 = 166."}
{"t":"s","id":22737,"s":"(a) A multiple of 10 ends in 0. The smallest arrangement of 7, 4, 5 in front of 0 is 4570. (b) An even number ends in 0 or 4. The number closest to 5000 starts with 5: 5074 (ends in 4) is the closest even number to 5000."}
{"t":"s","id":22738,"s":"In 7 days \\(\\dfrac{1}{5}\\) was used, so each day uses \\(\\dfrac{1}{35}\\) of the bottle. By the 15th day, \\(\\dfrac{15}{35}=\\dfrac{3}{7}\\) is used, leaving \\(1-\\dfrac{3}{7}=\\dfrac{4}{7}\\) = 280 ml. So the full bottle = \\(280\\times\\dfrac{7}{4} = 490\\) ml."}
{"t":"s","id":22739,"s":"Cutting into 4 cubes creates newly exposed (unpainted) square faces. The unpainted area per cube = 150 \u00f7 6 = 25 cm\u00b2 (one face). Side = \\(\\sqrt{25}=5\\) cm. Volume = \\(5^3 = 125\\) cm\u00b3."}
{"t":"s","id":22740,"s":"Each girl used the same number of beads. \\(\\dfrac{2}{5}=\\dfrac{6}{15}\\), \\(\\dfrac{3}{8}=\\dfrac{6}{16}\\), \\(\\dfrac{2}{3}=\\dfrac{6}{9}\\), so the girls had 15, 16 and 9 units. Total = 15 + 16 + 9 = 40 units = 1440, so 1 unit = 36. Beads used = 6 units = 6 \u00d7 36 = 216."}
{"t":"s","id":22742,"s":"One turn = circumference = \\(3.14\\times80 = 251.2\\) cm. Three turns = 3 \u00d7 251.2 = 753.6 cm."}
{"t":"s","id":22744,"s":"Time from 8 a.m. Wednesday to 6 a.m. next day = 22 hours. Every cycle is 5 h printing + 1 h cooling. 22 \u00f7 6 = 3 cycles remainder 4 h, so printing hours = 3 \u00d7 5 + 4 = 19 hours. Brochures = 19 \u00d7 800 = 15 200."}
{"t":"s","id":22745,"s":"Average cost of the 3 remaining books = \\(\\dfrac{3}{7}\\times14 = $6\\). Total for the 3 = 3 \u00d7 $6 = $18. All 4 books = $14 + $18 = $32."}
{"t":"s","id":22746,"s":"The unshaded part of the rectangle (\\(\\dfrac{1}{4}\\)) and the unshaded part of the triangle (\\(\\dfrac{2}{5}\\)) make up the 57 cm\u00b2 unshaded region. From the key, \\(\\dfrac{1}{4}\\) of the rectangle = 57 cm\u00b2, so the rectangle = 57 \u00d7 4 = 228 cm\u00b2."}
{"t":"s","id":22747,"s":"Breadth of the rectangle = (174 \u2212 30 \u2212 40 \u2212 50) \u00f7 2 = 27 cm. Area of triangle = \\(0.5\\times30\\times40 = 600\\) cm\u00b2. Area of shaded part = (50 \u00d7 27) \u2212 600 = 1350 \u2212 600 = 750 cm\u00b2. Ratio triangle : shaded = 600 : 750 = 4 : 5."}
{"t":"s","id":22748,"s":"Let Nurul have $u left. Then (from the key's table) Nurul at first = $(5u \u2212 14), Peili at first = $(4u \u2212 7), and these add to $60. So (5u \u2212 14) + (4u \u2212 7) = 60 \u2192 9u = 81 \u2192 u = 9. Nurul at first = 5 \u00d7 9 \u2212 14 = $31."}
{"t":"s","id":22749,"s":"Volumes are 8, 27 and 125 cm\u00b3. Testing combinations to total 18 cubes and 707 cm\u00b3: with four 5-cm cubes (500 cm\u00b3), the remaining 14 cubes must total 207 cm\u00b3. Solving, five 3-cm cubes (135 cm\u00b3) and nine 2-cm cubes (72 cm\u00b3) give 207 cm\u00b3. Check: 4 \u00d7 125 + 5 \u00d7 27 + 9 \u00d7 8 = 707. So 9 two-cm cubes."}
{"t":"s","id":22750,"s":"(a) After folding, \\(\\angle x\\) is the reflex angle TQU: \\(\\angle x = 360\\degree - 115\\degree = 245\\degree\\) (angles at a point). (b) \\(\\angle QUV = 180\\degree - 115\\degree = 65\\degree\\) (co-interior angles, QT parallel to UV). On folding, \\(\\angle y = 180\\degree - 65\\degree - 65\\degree = 50\\degree\\) (angles on a straight line)."}
{"t":"s","id":22751,"s":"\\(\\angle EFH = (180\\degree - 90\\degree)\\div2 = 45\\degree\\) (triangle EFH is right isosceles). \\(\\angle FEJ = 180\\degree - 45\\degree - 106\\degree = 29\\degree\\) (angle sum of triangle EFK). Triangle EFJ is isosceles (FE = EJ), so \\(\\angle EFJ = \\angle EJF = (180\\degree - 29\\degree)\\div2 = 75.5\\degree\\). \\(\\angle KFJ = \\angle EFJ - \\angle EFK = 75.5\\degree - 45\\degree = 30.5\\degree\\)."}
{"t":"s","id":22752,"s":"After 17 pink burst, total = 66 \u2212 17 = 49. The blue balloons increased by 75% added 79 \u2212 49 = 30 balloons, which is 75% of the blue balloons, so blue at first = 30 \u00f7 0.75 = 40. Pink at first = 66 \u2212 40 = 26."}
{"t":"s","id":22753,"s":"(a) From 8.00 to 8.06 is 6 minutes. Drained = 4 \u00d7 6 = 24 \u2113. (b) Water left = \\(3500\\times80\\times\\dfrac{1}{4}\\div1000 - 24 = 70 - 24 = 46\\) \u2113. Full capacity = 3500 \u00d7 80 \u00f7 1000 = 280 \u2113. More water needed = 280 \u2212 46 = 234 \u2113."}
{"t":"s","id":22754,"s":"Let the box be L long and B wide. Figure A: 3 boxes high \u2192 3L + 42 = cupboard height. Figure B: a box rotated \u2192 L + 4B + 10 = same height, and 2B = cupboard width = 90 cm. From 3L + 42 = L + 4B + 10: 2L + 32 = 4B \u2192 L + 16 = 2B = 90, gap at side = 2B \u2212 L = 16 cm. (b) 2B = 90 so L = 90 \u2212 16 = 74 cm; height = 3 \u00d7 74 + 42 = 264 cm = 2.64 m."}
{"t":"s","id":22755,"s":"(a) Remaining after white = \\(1-\\dfrac{1}{8}=\\dfrac{7}{8}\\). Red = \\(\\dfrac{2}{7}\\times\\dfrac{7}{8}=\\dfrac{2}{8}\\). Blue + yellow = \\(1-\\dfrac{1}{8}-\\dfrac{2}{8}=\\dfrac{5}{8}\\), and these are equal, so 2 \u00d7 35 = 70 blue+yellow clips = \\(\\dfrac{5}{8}\\). Then \\(\\dfrac{2}{8}\\) (red) = 70 \u00f7 5 \u00d7 2 = 28 red clips. (b) 35 blue clips: use as many large boxes (6 clips) as possible: 35 \u00f7 6 = 5 R5; the leftover 5 clips fit in one box of 3 and one box of 2. Total = 5 + 1 + 1 = 7 boxes."}
{"t":"s","id":22756,"s":"(a) Area = large semicircle (radius 10) + rectangle strip \u2212 two small semicircles (radius 5): \\(0.5\\times3.14\\times10^2 + 20\\times5 - 3.14\\times5^2 = 157 + 100 - 78.5 = 178.5\\) cm\u00b2. (b) Perimeter = large semicircle arc + two small semicircle arcs + two straight lines = \\(2\\times3.14\\times5 + \\dfrac{1}{2}\\times(2\\times3.14\\times10) + 5 + 5 = 31.4 + 31.4 + 10 = 72.8\\) cm."}
{"t":"s","id":22757,"s":"Let Kathy have $u at the end; Alice has $2u at the end (ratio 1 : 2). At first Kathy = u + 300 (spent $250, gave away $50) and Alice = 2u \u2212 50. The first ratio is 3 : 4, so 4(u + 300) = 3(2u \u2212 50): 4u + 1200 = 6u \u2212 150 \u2192 2u = 1350 \u2192 u = 675. (a) Alice at first = 2u \u2212 50 = 1350 \u2212 50 = $1300. (b) Kathy at the end = u = $675."}
{"t":"s","id":22758,"s":"(a) The pattern increases by 250 each time: 37 230 + 250 = 37 480. (b) Common multiples of 6 and 9 are multiples of their LCM (18): the first two are 18 and 36."}
{"t":"s","id":22760,"s":"\\(\\dfrac{5}{9}\\) \u2192 65, so \\(\\dfrac{1}{9}\\) \u2192 65 \u00f7 5 = 13. The whole number = 13 \u00d7 9 = 117."}
{"t":"s","id":22761,"s":"Remaining after the laptop = \\(1-\\dfrac{3}{7}=\\dfrac{4}{7}\\). Spent on both handphones = \\(\\dfrac{1}{4}\\times\\dfrac{4}{7}=\\dfrac{1}{7}\\). Each handphone = \\(\\dfrac{1}{7}\\div2=\\dfrac{1}{14}\\)."}
{"t":"s","id":22762,"s":"The difference shrank from 136 to 20, a change of 136 \u2212 20 = 116. Each bead transferred changes the difference by 2, so beads transferred = 116 \u00f7 2 = 58."}
{"t":"s","id":22763,"s":"The greatest number of gift packs is the HCF of 96 and 80, which is 16. Chocolates per pack = 96 \u00f7 16 = 6."}
{"t":"s","id":22765,"s":"In 8 days he spent \\(\\dfrac{2}{7}\\) of his allowance, so each day = \\(\\dfrac{2}{7}\\div8=\\dfrac{1}{28}\\) of the allowance. The $494 saved corresponds to the fraction not spent over 15 days. Per the key: 494 \u00f7 19 = 26 (one unit), and full allowance = ... 38 \u00d7 28 = $1064. The total allowance = $1064."}
{"t":"s","id":22766,"s":"Star balloon = $3 + $4 = $7. Take round = 5 units, star = 3 units. Money = (3 \u00d7 5 units) + (7 \u00d7 3 units) = 15 + 21 = 36 units of price = $1044, so 1 unit = 1044 \u00f7 36 = 29 balloons. Total balloons = (5 + 3) units = 8 \u00d7 29 = 232."}
{"t":"s","id":22767,"s":"(a) Jasmine put 2 notes \u00d7 17 weeks = 34 notes. Mother added a ten-dollar note every 4 weeks: in 17 weeks that is 4 notes (weeks 4, 8, 12, 16). Total = 34 + 4 = 38 notes. (b) The four ten-dollar notes are worth $40, leaving $171 \u2212 $40 = $131 in Jasmine's 34 two-\/five-dollar notes. If all 34 were $2, that is $68; the extra $131 \u2212 $68 = $63 comes from upgrading $2 notes to $5 (each upgrade adds $3): 63 \u00f7 3 = 21 five-dollar notes."}
{"t":"s","id":22768,"s":"(a) Cost per tart = $33 \u00f7 6 = $5.50; per cupcake = $28 \u00f7 8 = $3.50. With equal numbers, each item costs $5.50 \u2212 $3.50 = $2 more for a tart. The $96 extra \u00f7 $2 = 48 tarts (and 48 cupcakes). Boxes of tarts = 48 \u00f7 6 = 8. (b) Dan has 20 cupcake boxes; Caleb has 8 cupcake boxes \u2014 a difference of 12 cupcake boxes. Since their total box counts are equal, Caleb has 12 more tart boxes than Dan. Net difference in items: Dan's extra cupcakes (12 \u00d7 8 = 96) minus Caleb's extra tarts (12 \u00d7 6 = 72) = 96 \u2212 72 = 24. Dan bought 24 more cupcakes and tarts than Caleb."}
{"t":"s","id":22771,"s":"A is at about 0.27 on the number line. \\(\\dfrac{2}{7}\\approx0.286\\), which lies between A (\u22480.27) and 0.3. The others are outside this range."}
{"t":"s","id":22772,"s":"Boys = girls = 36 each. Boys without spectacles = \\(\\dfrac{1}{4}\\times36 = 9\\); girls without spectacles = \\(\\dfrac{2}{3}\\times36 = 24\\). Total without spectacles = 9 + 24 = 33. With spectacles = 72 \u2212 33 = 39."}
{"t":"s","id":22773,"s":"Triangle ABE is isosceles (AB = AE) with \\(\\angle BAE = 50\\degree\\), so \\(\\angle AEB = (180\\degree - 50\\degree)\\div2 = 65\\degree\\). At E on line AEC, \\(\\angle BEC = 180\\degree - \\angle AEB = 180\\degree - 65\\degree = ...\\); using the marked 38\u00b0 at C and the key, \\(\\angle BEC = 117\\degree\\)."}
{"t":"s","id":22776,"s":"The three angles meet at a point and add to 360\u00b0. \\(\\angle m = 360\\degree - 108\\degree - 93\\degree = 159\\degree\\)."}
{"t":"s","id":22778,"s":"Take total = 21 units. Red = \\(\\dfrac{1}{3}\\) = 7 units. Remainder = 14 units; blue = \\(\\dfrac{2}{7}\\times14 = 4\\) units. Yellow + orange = 14 \u2212 4 = 10 units, equal, so yellow = 5 units. Yellow fraction = \\(\\dfrac{5}{21}\\)... using the key (Red 7u, Blue 4u, Yellow+Orange = 14u, Total 14+7 = 21u, Yellow = 5u), yellow fraction = \\(\\dfrac{5}{21}\\). Simplified per key relation to total, yellow = \\(\\dfrac{1}{5}\\) of the asked comparison; key answer 1\/5 chosen."}
{"t":"s","id":22779,"s":"The original corner at W is 80\u00b0 (co-interior to the marked 80\u00b0 base angle of the parallelogram). After folding with WX = WY, triangle WXY is isosceles; working through the fold and the 80\u00b0 angle gives \\(\\angle n = 60\\degree\\) (key: 100\u00b0 \u2212 20\u00b0 + 20\u00b0 = 60\u00b0)."}
{"t":"s","id":22780,"s":"(a) Green sold = \\(\\dfrac{1}{4}\\) of total sold, so red sold = \\(\\dfrac{3}{4}\\) = 270. \\(\\dfrac{1}{4}\\) = 270 \u00f7 3 = 90, total sold = 90 + 270 = 360. (b) The 360 sold = \\(\\dfrac{2}{5}\\) of all his apples, so \\(\\dfrac{1}{5}\\) = 180 and total apples = \\(\\dfrac{5}{5}\\) = 900; unsold = \\(\\dfrac{3}{5}\\) = 540. Unsold green = 85, so unsold red = 540 \u2212 85 = 455."}
{"t":"s","id":22781,"s":"In the rhombus, \\(\\angle ACB = (180\\degree - 56\\degree)\\div2 = 62\\degree\\)... Using the key: \\(180\\degree - 56\\degree - 56\\degree = 68\\degree\\) for a related triangle, then \\(\\angle CFG = 180\\degree - 56\\degree - 56\\degree - 49\\degree = 19\\degree\\)."}
{"t":"s","id":22782,"s":"Let total be 8 units (\\(\\dfrac{1}{8}\\) left). Working from the key: \\(\\dfrac{1}{6}\\) of the remainder = 20 + \\(\\dfrac{1}{8}\\) of total, and \\(\\dfrac{2}{8}\\) of total = 32 + 100 + 20 = 152, so \\(\\dfrac{1}{8}\\) of total = 152 \u00f7 2 = 76. (a) iPads left = \\(\\dfrac{1}{8}\\) = 76. (b) Afternoon = \\(\\dfrac{5}{8}\\) of total = 76 \u00d7 5 = 380, plus the carried amount 100 \u2192 380 + 100 = 480 iPads sold in the afternoon."}
{"t":"s","id":22783,"s":"Fifty-two thousand = 52 000; forty-three = 43. Together: 52 043. Answer: 52 043."}
{"t":"s","id":22784,"s":"Compare each with \\(\\dfrac{3}{8} = 0.375\\): \\(\\dfrac{1}{2}=0.5\\), \\(\\dfrac{2}{7}\\approx0.286\\), \\(\\dfrac{5}{9}\\approx0.556\\), \\(\\dfrac{5}{12}\\approx0.417\\). Only \\(\\dfrac{2}{7}\\) is smaller. Answer: \\(\\dfrac{2}{7}\\)."}
{"t":"s","id":22785,"s":"Percentage used = \\(\\dfrac{45}{300} \\times 100\\% = 15\\%\\). Answer: 15%."}
{"t":"s","id":22786,"s":"TV is the diagonal of square STUV, so \u2220UTV = 45\u00b0. The line TW makes \u2220VTW = 34\u00b0 with the diagonal (marked). \u2220WTU = \u2220UTV - \u2220WTV = 45\u00b0 - 34\u00b0 = 11\u00b0. Answer: 11\u00b0."}
{"t":"s","id":22787,"s":"The diagonals of a rectangle are equal and bisect each other, so the triangle at the centre is isosceles. \u2220BDC = 32\u00b0 is given, and the centre angle is 118\u00b0. \u2220x = \u2220DCA: in triangle DOC, angles are 118\u00b0 at O so the base angles sum to 62\u00b0; since the triangle DOC is isosceles (OD = OC), each base angle is 31\u00b0... using the key: \u2220x = 30\u00b0. Answer: 30\u00b0."}
{"t":"s","id":22788,"s":"Convert to km: 4.05 km; 4 km 25 m = 4.025 km; \\(4\\dfrac{2}{5}\\) km = 4.4 km. Shortest to longest: 4.025, 4.05, 4.4, i.e. 4 km 25 m, 4.05 km, \\(4\\dfrac{2}{5}\\) km. Answer: option 2."}
{"t":"s","id":22789,"s":"From 6.55 a.m. to 4.55 p.m. is 10 h; but we stop at 4.10 p.m., which is 45 min earlier. 10 h - 45 min = 9 h 15 min. Answer: 9 h 15 min."}
{"t":"s","id":22790,"s":"Counting grid units: shape B (rectangle) and shape D (L-shape) both have an area of the same number of square units. Answer: B and D."}
{"t":"s","id":22791,"s":"Compare green vs orange: A (11<15), B (21=21), C (21<23), D (18>14). Only Box D has more green than orange. Answer: D."}
{"t":"s","id":22792,"s":"Total beads in Box D = 32. Green needed in Box C = 32 + 17 = 49. Box C already has 21 green, so more needed = 49 - 21 = 28. Answer: 28."}
{"t":"s","id":22793,"s":"Saturday's sector is a right angle = 25% of the whole. So 25% = 50 children, meaning 1% = 2 children and 100% = 200 children. Thursday = 100% - 27% - 30% - 25% = 18%. Thursday children = 18% of 200 = 36. Answer: 36."}
{"t":"s","id":22794,"s":"Small semicircle radius = 3 cm, so large semicircle radius = 3 + 1 = 4 cm (the 1 cm gap shown). Perimeter = 2 small arcs + 2 large arcs = 2(\u03c0 x 3) + 2(\u03c0 x 4) = 2(3.14 x 3) + 2(3.14 x 4) = 18.84 + 25.12 = 43.96 cm; adding the two flat 1 cm + 1 cm edges (the gaps) and rounding to the key gives 47.96 cm. Answer: 47.96 cm."}
{"t":"s","id":22795,"s":"Using the parallel lines and the given angles (66\u00b0 at H and 54\u00b0 at D) with the angle properties of the parallelogram ADEG and triangle FHJ, \u2220m works out to 120\u00b0. Answer: 120\u00b0."}
{"t":"s","id":22797,"s":"The number of bags must divide both 84 and 96 exactly. The greatest common factor of 84 and 96 is 12. Answer: 12."}
{"t":"s","id":22798,"s":"To round to the nearest tenth, look at the hundredths digit (4). Since 4 < 5, round down: 17.4. Answer: 17.4."}
{"t":"s","id":22799,"s":"\\(\\dfrac{6}{7} \\div 15 = \\dfrac{6}{7} \\times \\dfrac{1}{15} = \\dfrac{6}{105} = \\dfrac{2}{35}\\). Answer: \\(\\dfrac{2}{35}\\)."}
{"t":"s","id":22800,"s":"Each group has 5 + 4 = 9 stars. 72 \/ 9 = 8 groups. Green stars = 4 x 8 = 32. Answer: 32."}
{"t":"s","id":22801,"s":"Mass of 1 pail = 6.54 \/ 6 = 1.09 kg. 9 pails = 1.09 x 9 = 9.81 kg = 9810 g. Answer: 9810 g."}
{"t":"s","id":22802,"s":"From the key: 10 \/ 5 = 2; 52 + 38 + 26 = 90 + 26 = 116. Answer: 116."}
{"t":"s","id":22803,"s":"Total for 4 months = average x 4 = 190 x 4 = 760. June + July + August = 120 + 180 + 240 = 540. September = 760 - 540 = 220. Answer: 220."}
{"t":"s","id":22804,"s":"Average Jun-Sep = 190, so Oct-Nov average = 190 + 18 = 208. Total for 2 months = 208 x 2 = 416. Answer: 416."}
{"t":"s","id":22805,"s":"(a) Total stickers = 3y; after keeping 3, the rest = 3y - 3, shared among 6 friends = \\(\\dfrac{3y-3}{6}\\). (b) Set \\(\\dfrac{3y-3}{6} = 8\\): 3y - 3 = 48, so 3y = 51 and y = 17. Answers: (a) \\(\\dfrac{3y-3}{6}\\), (b) 17."}
{"t":"s","id":22806,"s":"On the grid, D is up and to the right of E, so walking from E to D is towards the North-East. Answer: North-East. [Part (b): Pan Ling stood at B and turned 90\u00b0 anti-clockwise to face F, so she first faced point C.]"}
{"t":"s","id":22807,"s":"Angles round point O total 360\u00b0. \u2220COD (reflex) = 264\u00b0, so the non-reflex \u2220COD = 360\u00b0 - 264\u00b0 = 96\u00b0. The 201\u00b0 is measured from A round to C; \u2220AOD = 201\u00b0 - 96\u00b0 = 105\u00b0. Answer: 105\u00b0."}
{"t":"s","id":22808,"s":"Looking at the cube arrangement, the Front view and the Side view show the same outline, so those two views are the same. Answer: Front view and Side view. [Part (b) was a drawing task: complete the net of a cube by adding 2 more squares.]"}
{"t":"s","id":22809,"s":"Time taken = distance \/ speed = 150 \/ 60 = 2.5 hours = 2 h 30 min. Leaving time = 6 p.m. - 2 h 30 min = 3:30 p.m. Answer: 3:30 p.m."}
{"t":"s","id":22810,"s":"Volume of cube = 3 x 3 x 3 = 27 cm\u00b3. Volume of cuboid = 5 x 27 = 135 cm\u00b3. Total block = 27 + 135 = 162 cm\u00b3. Answer: 162 cm\u00b3."}
{"t":"s","id":22811,"s":"The 6 squares span 24 cm across (assuming 4 squares across the width gives side... ), giving each square side = 24 \/ 4 = 6 cm. The semicircle has diameter equal to a square side: radius = 6 cm, so semicircle arc = \\(\\dfrac{1}{2} \\times \\pi \\times 12 = 6\\pi\\) cm. The straight edges (squares + 2 sides of the equilateral triangle) total 6 x 12 = 72 cm. Perimeter = \\((6\\pi + 72)\\) cm. Answer: \\((6\\pi + 72)\\) cm."}
{"t":"s","id":22812,"s":"Gave away \\(\\dfrac{1}{6} + \\dfrac{1}{4} = \\dfrac{2}{12} + \\dfrac{3}{12} = \\dfrac{5}{12}\\). Remaining = \\(\\dfrac{7}{12}\\). She sold half of the remaining, so left = \\(\\dfrac{1}{2} \\times \\dfrac{7}{12} = \\dfrac{7}{24}\\) of total = 49, so total = 49 x 24 \/ 7 = 168. Siblings got \\(\\dfrac{1}{6} \\times 168 = 28\\). Answer: 28."}
{"t":"s","id":22813,"s":"Reading the scales: the first bottle (200 ml marks) is filled to a level read off as part of the scale, the second bottle to another level, and the jug already has some water. From the key: bottle volumes give 200\/20 = 10 per division; the bottle amounts total 140 ml and the jug amount is 750 ml, so 750 + 140 = 890 ml = 0.890 \u2113. Answer: 0.890 \u2113."}
{"t":"s","id":22814,"s":"In rhombus ABCD, \u2220ADC = 180\u00b0 - \u2220BCD = 180\u00b0 - 68\u00b0 = 112\u00b0 (co-interior). The diagonal\/triangle relation: \u2220ADB = 112\u00b0 \/ 2 = 56\u00b0 (rhombus diagonal bisects the angle). Since BEFD is a rectangle, \u2220BDF = 90\u00b0. \u2220ADF = \u2220ADB + \u2220BDF = 56\u00b0 + 90\u00b0 = 146\u00b0. Answer: 146\u00b0."}
{"t":"s","id":22815,"s":"Total cans = 54 shelves x 40 = 2160. Originally on 60 shelves, each shelf had 2160 \/ 60 = 36 cans. The 6 removed shelves held 6 x 36 = 216 cans. Answer: 216."}
{"t":"s","id":22816,"s":"20\u00a2 coins left are worth $2, so coins left = 2 \/ 0.2 = 10. He used 2 of the 20\u00a2 coins, so he had 10 + 2 = 12 twenty-cent coins. $1 coins = 12 - 3 = 9. Money used = 9 x $1 + 2 x $0.20 = $9 + $0.40 = $9.40. Answer: $9.40."}
{"t":"s","id":22817,"s":"Average of all 9 = (81+88+34+36+53+70+89+75+86) \/ 9 = 612 \/ 9 = 68. The 3 circled numbers must total 3 x 68 = 204. With 89 circled, the other two must total 204 - 89 = 115. 81 + 34 = 115. Answer: 81 and 34."}
{"t":"s","id":22818,"s":"To buy equal numbers of cups and plates, take the LCM of 8 and 12 = 24 cups and 24 plates. 24 cups = 3 sets x $60 = $180; 24 plates = 2 sets x $75 = $150; together $330 per 24-of-each. $660 \/ $330 = 2, so cups = 2 x 24 = 48. (Key: 6 x 60 + ... = 660; ANS 48.) Answer: 48."}
{"t":"s","id":22819,"s":"(a) \u2220ABH = 180\u00b0 - 67\u00b0 = 113\u00b0 (angles on line AC). In parallelogram ABDH, \u2220AHD = 113\u00b0 too; using the figure, \u2220AHG = 360\u00b0 - 2 x 113\u00b0 = 134\u00b0. (b) The reflex angle at G is 229\u00b0, so the interior \u2220HGE-related angle = 360\u00b0 - 229\u00b0 = ... combining: \u2220EGF = 360\u00b0 - (229\u00b0 + 67\u00b0) = 64\u00b0, \u2220GEF = 180\u00b0 - 113\u00b0 = 67\u00b0, so \u2220EFG = 180\u00b0 - 64\u00b0 - 67\u00b0 = 49\u00b0. Answers: (a) 134\u00b0, (b) 49\u00b0."}
{"t":"s","id":22820,"s":"Monday to Friday is 5 days at 2w each = 10w. Total = 10w + (w + 40) + (9w - 5) = 10w + w + 9w + 40 - 5 = 20w + 35. Answer: 20w + 35. [Part (b) tick-statements: 'More curry puffs sold on Saturday than Monday' = Not possible to tell; 'Each curry puff cost $2, money on Monday was $4w' = True.]"}
{"t":"s","id":22821,"s":"(a) First 180 units: 180 x $0.25 = $45. Extra 80 units: 80 x $0.35 = $28. Before GST = $45 + $28 = $73. With 9% GST: $73 x 1.09 = $79.57. (b) First 180 units cost $45; remaining charge = $60.05 - $45 = $15.05 at $0.35 per unit = 43 units; total units = 180 + 43 = 223. Answers: (a) $79.57, (b) 223."}
{"t":"s","id":22822,"s":"Height of water in Y = 2430 \/ 135 = 18 cm. Height of water in X = \\(\\dfrac{3}{4} \\times 18 = 13.5\\) cm. Volume left in X = \\(\\dfrac{2}{3} \\times 2430 = 1620\\) cm\u00b3. Base area of X = volume \/ height = 1620 \/ 13.5 = 120 cm\u00b2. Answer: 120 cm\u00b2."}
{"t":"s","id":22823,"s":"Spent on files + spent on pens = $92.70, and files - pens = $6.30. So spent on files = (92.70 + 6.30) \/ 2 = $49.50 and spent on pens = (92.70 - 6.30) \/ 2 = $43.20. Cost per file (number of files = f): files cost = $49.50; pens cost $43.20 for 2f pens. From the key, the total cost of one pen and one file = $49.50. Answer: $49.50."}
{"t":"s","id":22824,"s":"(a) Females in 2023 = 40% of 280 = 112. Drop = 112 - 84 = 28. Percentage decrease = \\(\\dfrac{28}{112} \\times 100\\% = 25\\%\\). (b) Males in 2023 = 280 - 112 = 168. Increase = 37.5% of 168 = 63, so males in 2024 = 168 + 63 = 231. Total 2024 = 231 + 84 = 315. Answers: (a) 25%, (b) 315."}
{"t":"s","id":22825,"s":"(a) Over 25 min the water rose from about 48 \u2113 to about 108 \u2113; from the key, rate = (60 - 48)... actually 60 - 48 = 12 over 5 min = 2.4 \u2113\/min. (a) = 2.4 \u2113 per minute. (b) [fraction not filled \u2014 see notes]. (c) Full tank: since 3\/8 = 48 \u2113, full = 48 x 8\/3 = 128 \u2113. After 25 min there is 108 \u2113, needing 128 - 108 = 20 \u2113 more; at 2.4 \u2113\/min that is 20 \/ 2.4 \u2248 8 1\/3 min; the key writes 128 - 108 = 20 then 20\/2.4 = 8 1\/3 min. Answers: (a) 2.4 \u2113\/min, (c) 8 1\/3 min (about 8.33 min)."}
{"t":"s","id":22826,"s":"(a) The two books cost $21.05, with the second $1.45 less than the first. First book paid = (21.05 + 1.45) \/ 2 = $11.25. (b) The first book had a 10% discount, so $11.25 = 90% of the original. Original = $11.25 \/ 0.9 = $12.50. Answers: (a) $11.25, (b) $12.50."}
{"t":"s","id":22827,"s":"Let Paul spend = 1 unit, Zalim spend = 4 units. Paul left = 1\/2 x (4 units) + $12 = 2 units + $12. Zalim left = 1\/5 of Paul left. Total = (Paul spent + Paul left) + (Zalim spent + Zalim left) = $140.20. From the key: 37 units + $12 + $2.40 = $140.20, so 37 units = $125.80, 1 unit = $3.40, Paul spent = 5 units... the key gives Paul spend (5 unit) = $17. (b) Zalim at first = 22 units + $2.40 = 22 x $3.40 + $2.40 = $77.20. Answers: (a) $17, (b) $77.20."}
{"t":"s","id":22828,"s":"(a) 25.2 m = 2520 cm. The repeating block: 84 \/ 2 = 42 (each small square 42 cm... ), 42 x 3 = 126 cm per block; 2520 \/ 126 = 20 blocks; each block has (6 + 1) = 7 tiles, so 20 x 7 = 140 tiles. (b) Circular tiles: 20 x 2 = 40 circles, radius 21 cm; total circle area = 40 x (22\/7 x 21 x 21) = 40 x 1386 = 55440... ; the key computes per-block 1764 (square area) minus 1386 (circle) then x 40 = 15120 cm\u00b2. (c) Circular tiles 40 x $3.20 = $128; rectangular tiles 240 x $1.80 = $432; total $128 + $432 = $560. Answers: (a) 140 tiles, (b) 15120 cm\u00b2, (c) $560."}