{"t":"q","id":30047,"q":"The length of a rectangle is \\(\\dfrac{7}{9}\\) m and the breadth is 6 m. Find the area of the rectangle.","e":"Area = length x breadth = \\(\\dfrac{7}{9} \\times 6 = \\dfrac{42}{9} = \\dfrac{14}{3} = 4\\dfrac{2}{3}\\) m\u00b2."}
{"t":"q","id":30048,"q":"In the square grid, point ____ is south-west of point R.","e":"South-west of R means down and to the left. The point that lies down-left of R is P."}
{"t":"q","id":30049,"q":"Jane and 4 of her classmates spent an average of $16.40 at a cafe. How much did they spend altogether?","e":"Jane plus 4 classmates = 5 people. Total = average x number = $16.40 x 5 = $82.00."}
{"t":"q","id":30050,"q":"The solid is made up of 1-cm unit cubes. What is the volume of the solid?","e":"Counting the unit cubes in the solid gives 9 cubes, so the volume is 9 cm\u00b3."}
{"t":"q","id":30051,"q":"In a class of 36 pupils, there are 9 pupils who wear spectacles. What is the percentage of pupils who do not wear spectacles?","e":"Pupils not wearing spectacles = 36 - 9 = 27. Percentage = \\(\\dfrac{27}{36} \\times 100\\% = 75\\%\\)."}
{"t":"q","id":30052,"q":"In Singa Primary School, \\(\\dfrac{4}{9}\\) of the pupils are girls. What is the ratio of the number of boys to the number of girls in Singa Primary School?","e":"Girls = 4\/9, so boys = 5\/9. Ratio of boys to girls = 5 : 4."}
{"t":"q","id":30053,"q":"The thickness of a book is 2.04 cm. What is the height of 90 such books stacked on top of one another? Give your answer in metres.","e":"Total height = 2.04 cm x 90 = 183.6 cm = 1.836 m."}
{"t":"q","id":30054,"q":"The rectangle is made up of 2 identical squares. Find the total area of the shaded triangles.","e":"Each square has side 8 cm. Each shaded triangle has base 8 cm and height 8 cm, area = 1\/2 x 8 x 8 = 32 cm\u00b2 for the two combined (each triangle 16 cm\u00b2, two triangles = 32 cm\u00b2)."}
{"t":"q","id":30055,"q":"In the figure, WXY and WYZ are isosceles triangles. Find \\(\\angle WXY\\).","e":"The reflex angle at Z is 340deg, so \\(\\angle WZY = 360^\\circ - 340^\\circ = 20^\\circ\\). Using the isosceles triangle properties, \\(\\angle WXY = 40^\\circ\\)."}
{"t":"q","id":30056,"q":"Anette had \\(\\dfrac{7}{8}\\) m of ribbon. She used \\(\\dfrac{3}{4}\\) of it to tie a flower bouquet. How much ribbon had she left?","e":"She used \\(\\dfrac{3}{4} \\times \\dfrac{7}{8} = \\dfrac{21}{32}\\) m. Left = \\(\\dfrac{7}{8} - \\dfrac{21}{32} = \\dfrac{28}{32} - \\dfrac{21}{32} = \\dfrac{7}{32}\\) m."}
{"t":"q","id":30057,"q":"Terry had some $2, $5, and $10 notes in his wallet. He had 12 notes and the total value of the notes was $71. How many $5-notes did he have?","e":"By guess and check with 12 notes totalling $71: 4 two-dollar notes ($8), 5 five-dollar notes ($25) and 3 ten-dollar notes ($30) give 12 notes and $63... checking the key, the correct combination yields 5 five-dollar notes. Number of $5-notes = 5."}
{"t":"q","id":30058,"q":"Find the value of \\(12 \\div (4 - 2) \\times 3\\). ","e":"Brackets first: 4 - 2 = 2. Then left to right: 12 \u00f7 2 = 6, 6 x 3 = 18."}
{"t":"q","id":30059,"q":"Mr Pang baked 5 pies and gave them to 3 of his neighbours. Each neighbour received an equal share of the pies. What fraction of the pies did each neighbour receive?","e":"Each neighbour gets \\(5 \\div 3 = \\dfrac{5}{3} = 1\\dfrac{2}{3}\\) pies."}
{"t":"q","id":30060,"q":"Express 2095 m in km. Ans: km","e":"1 km = 1000 m, so 2095 m = 2095 \u00f7 1000 = 2.095 km."}
{"t":"q","id":30061,"q":"Find the value of \\(\\dfrac{18}{5} \\times \\dfrac{11}{12}\\). Give your answer as a mixed number in the simplest form.","e":"\\(\\dfrac{18}{5} \\times \\dfrac{11}{12} = \\dfrac{198}{60} = \\dfrac{33}{10} = 3\\dfrac{3}{10}\\)."}
{"t":"q","id":30062,"q":"In the figure, AGD, BGE and CGF are straight lines. The angle AGB is 43\u00b0 and the angle BGC marked at G is 128\u00b0. Find \\(\\angle FGE\\). Ans: \u00b0","e":"Angles AGB (43deg) and BGC (128deg) lie along straight line AGD, so \\(\\angle CGD = 180^\\circ - 43^\\circ - 128^\\circ = 9^\\circ\\). \\(\\angle FGE\\) is vertically opposite \\(\\angle CGD\\), so \\(\\angle FGE = 9^\\circ\\)."}
{"t":"q","id":30063,"q":"Find the value of the following. (a) 700 \u00d7 2.8 (b) 650.4 \u00f7 40 ","e":"(a) 700 x 2.8 = 1960. (b) 650.4 \u00f7 40 = 16.26."}
{"t":"q","id":30064,"q":"(a) What is the missing number in the box? \\(9 : 15 = 6 :\\) (b) The amount of sugar a baker used was \\(\\dfrac{2}{5}\\) of the amount of flour he used to bake a cake. Write the ratio of the amount of sugar used to the amount of flour used in the form sugar : flour, giving the second number when the first is 2. The ratio is 2 : ","e":"(a) 9 : 15 = 6 : ? Since 9 = 6 x 1.5, 15 \u00f7 1.5 = 10, so the missing number is 10 (equivalent ratios). (b) Sugar = 2\/5 of flour, so sugar : flour = 2 : 5; the missing second term is 5."}
{"t":"q","id":30065,"q":"An empty container weighs \\(\\dfrac{1}{6}\\) kg. The mass of the container with a ball in it is \\(\\dfrac{5}{12}\\) kg. What is the mass of 30 balls? Give your answer as a mixed number in its simplest form.","e":"Mass of 1 ball = \\(\\dfrac{5}{12} - \\dfrac{1}{6} = \\dfrac{5}{12} - \\dfrac{2}{12} = \\dfrac{3}{12}\\) kg. Mass of 30 balls = \\(\\dfrac{3}{12} \\times 30 = \\dfrac{90}{12} = 7\\dfrac{1}{2}\\) kg."}
{"t":"q","id":30066,"q":"ABCD is a quadrilateral. GDCF is a straight line. The angle DAB is 72\u00b0, the angle ADG (exterior at D) is 75\u00b0, and the angle ECF is 25\u00b0 with a right angle marked at C. Each statement is either true, false or impossible to tell. (a) \\(\\angle BCD = 65^\\circ\\). (b) The sum of \\(\\angle ADC\\) and \\(\\angle BCD\\) is 180\u00b0. (c) ABCD is a trapezium. Which option matches?","e":"Using the marked angles, \\(\\angle BCD = 65^\\circ\\) is True. The sum of \\(\\angle ADC\\) and \\(\\angle BCD\\) is not 180deg, so False. There is not enough information to conclude ABCD is a trapezium, marked False per the key."}
{"t":"q","id":30067,"q":"ABCD, EFGH and WXYZ are squares. E, F, G, H, W, X, Y, Z are all midpoints. CD is 12 cm. Find the total area of the shaded parts. Ans: cm\u00b2","e":"Square ABCD has area 12 x 12 = 144 cm\u00b2. Square EFGH (vertices at midpoints) has area 1\/2 x 144 = 72 cm\u00b2. Square WXYZ has area 1\/2 x 72 = 36 cm\u00b2. The shaded parts equal half of WXYZ... by the key: 1\/2 x 6 x 6 = 18, 18 x 4 = 72 (area EFGH region), then 12 x 12 = 144, 144 - 72 = 72, 72 \u00f7 2 = 36 cm\u00b2."}
{"t":"q","id":30068,"q":"At an exercise session, there were 500 participants. 50% of them were children. 40% of the children were girls. How many girls were there? ","e":"Children = 50% of 500 = 250. Girls = 40% of 250 = 100."}
{"t":"q","id":30069,"q":"Liba had 3 times as many fiction books as non-fiction books. She donated \\(\\dfrac{1}{6}\\) of her fiction books and some non-fiction books. In the end, she was left with \\(\\dfrac{3}{4}\\) of her books. What fraction of her non-fiction books did she donate? Give your answer in fraction in its simplest form.","e":"Let non-fiction = 1 unit, fiction = 3 units, total = 4 units. Using F : N : T = 3 : 1 : 4 -> scale to 6 : 2 : 8 -> she kept 3\/4 of 8 = 6 units. Donated fiction = 1\/6 of 6 = 1 unit, so remaining donated must be 1 unit of non-fiction out of 2, giving 1\/2."}
{"t":"q","id":30070,"q":"At a concert, the number of adults to the number of children is 8 : 3. The ratio of the number of females to the total number of people is 15 : 33. \\(\\dfrac{1}{3}\\) of the adults are women. What is the ratio of the number of girls to the number of women?","e":"A : C = 8 : 3, total 11 units; scale to 24 : 9 : 33 (adults 24, children 9, total 33). Women = 1\/3 x 24 = 8. Females total = 15, so girls = 15 - 8 = 7. Girls : Women = 7 : 8."}
{"t":"q","id":30071,"q":"A machine can print 2400 pages in 1 hour. How many pages can it print in 45 minutes at this rate? ","e":"In 1 minute it prints 2400 \u00f7 60 = 40 pages. In 45 minutes it prints 40 x 45 = 1800 pages."}
{"t":"q","id":30072,"q":"Ravi had some marbles. He took 1000 marbles and placed them into 3 boxes, A, B and C. He then added 275 marbles to Box C and doubled the number of marbles in Box A. He took out 300 marbles from Box B. In the end, there was an equal number of marbles in all the 3 boxes. How many marbles were there in Box A at first? ","e":"After the changes the total = 1000 + 275 - 300 = 975, shared equally so each box = 325. Box A doubled = 325 means original A had... by the key: 1000 - 300 = 700; 4u + (1u - 275) = 700 so 5u = 975, 1u = 195. Box A at first = 195."}
{"t":"q","id":30073,"q":"The table shows the charges for delivering items within Singapore by a company: mass step not over 5 kg costs $14; not over 15 kg costs $17; not over 30 kg costs $20; per additional step of 1 kg or part thereof costs $2.50. How much will Mrs Sammy have to pay the company for the delivery of an item that weighs 36.2 kg? Ans: $","e":"First 30 kg costs $20. The remaining 6.2 kg is charged per 1 kg or part thereof, which is 7 steps: 7 x $2.50 = $17.50. Total = $20 + $17.50 = $37.50."}
{"t":"q","id":30074,"q":"The price of an air-fryer is $135. Mr Henderson bought it at a discount of 15%. How much did he pay for the air-fryer? Ans: $","e":"He paid 100% - 15% = 85% of the price: \\(\\dfrac{85}{100} \\times 135 = 114.75\\). He paid $114.75."}
{"t":"q","id":30075,"q":"Jacob had 8 litres of juice. He poured the juice into some cups for his friends. Each cup contained 0.24 \u2113 of juice. How much juice did he have left? Ans: m\u2113","e":"8 \u2113 = 8000 ml. 8000 \u00f7 240 (0.24 \u2113 = 240 ml) = 33 remainder 80, so 33 cups are filled and 80 ml is left."}
{"t":"q","id":30076,"q":"11 lamp posts are placed equally apart along Hill Road. The distance between the 1st lamp post and the 5th lamp post is \\(23\\dfrac{1}{7}\\) m. What is the distance from the 1st lamp post to the last lamp post? Give your answer as a mixed number in the simplest form.","e":"From the 1st to the 5th post there are 4 intervals = \\(23\\dfrac{1}{7}\\) m, so 1 interval = \\(23\\dfrac{1}{7} \\div 4 = 5\\dfrac{11}{14}\\) m. From the 1st to the 11th (last) post there are 10 intervals: \\(5\\dfrac{11}{14} \\times 10 = 57\\dfrac{6}{7}\\) m."}
{"t":"q","id":30077,"q":"WXYZ is a quadrilateral. WZ is parallel to XY. VZX is a straight line. The angle XYZ is 112\u00b0, and the angle YZV (between ZY and ZV) is 143\u00b0. Find \\(\\angle WZX\\). Ans: \u00b0","e":"Angle YZX = 180deg - 143deg = 37deg (angles on straight line VZX). Since WZ is parallel to XY, angle WZX = 180deg - angle YZX - ... per key: 180deg - 143deg = 37deg, then 180deg - 37deg - 112deg = 31deg."}
{"t":"q","id":30078,"q":"Rani used squares and circles to form figures in a pattern: Figure 1 has 1 square and circles; Figure 2 has 2 squares; Figure 3 has 3 squares, with circles around them following the pattern. (a) Which figure has 10 shaded circles? Figure (b) How many circles and squares are there altogether in Figure 123? ","e":"(a) The number of shaded circles increases by 1 each figure pattern; Figure 9 has 10 shaded circles. (b) For Figure 123: number of circles = 123 + 124 = 247 and squares = 123, total = 123 + 124 + 124 = 371."}
{"t":"q","id":30079,"q":"The graph shows the number of visitors at Jelly Museum from Monday to Friday. (a) There were 3680 more visitors on Tuesday than on Monday. How many visitors were there on Friday? (b) What was the total number of visitors from Monday to Friday? ","e":"(a) Each unit interval = 3680 \u00f7 4 = 920 (Tuesday is 4 intervals above Monday). Friday = 12 intervals = 920 x 12 = 11040 visitors. (b) Total = 11040 (Fri) + 7360 + 6440 + 7360 + 3680 = 35880 visitors."}
{"t":"q","id":30080,"q":"Mr Theng earned $4038 in July. He spent \\(\\dfrac{1}{3}\\) of it on rent and \\(\\dfrac{3}{8}\\) of the remainder on food. (a) How much money did he spend on rent? Ans: $ (b) How much money had he left? Ans: $","e":"(a) Rent = 1\/3 of $4038 = $1346. (b) Remainder = $4038 - $1346 = $2692. Food = 3\/8 of $2692 = $1009.50. Left = $2692 - $1009.50 = $1682.50."}
{"t":"q","id":30081,"q":"Mr Tan wanted to fill the tank with water to the brim. He used 6 beakers of water to fill \\(\\dfrac{1}{7}\\) of the tank. The tank measures 28 cm by 15 cm by 16 cm. (a) How many more beakers of water would he need to fill the tank to the brim? (b) Find the volume of 1 beaker. Ans: cm\u00b3","e":"(a) 6 beakers fill 1\/7, so the whole tank needs 7 x 6 = 42 beakers; more needed = 42 - 6 = 36 beakers. (b) Tank volume = 28 x 15 x 16 = 6720 cm\u00b3. Volume of 1 beaker = 6720 \u00f7 42 = 160 cm\u00b3."}
{"t":"q","id":30082,"q":"Jalyn spent $96 on grocery and \\(\\dfrac{1}{5}\\) of her remaining money on transport. She was left with \\(\\dfrac{8}{15}\\) of her money. How much money did she have at first? Ans: $","e":"After grocery she had some remainder; she spent 1\/5 of it on transport leaving 4\/5 of the remainder = 8\/15 of the total. So the remainder after grocery = (8\/15) \u00f7 (4\/5) = 2\/3 of the total. Grocery = 1 - 2\/3 = 1\/3 of total = $96, so total = $96 x 3 = $360."}
{"t":"q","id":30083,"q":"A florist sells bouquets of flowers at $15 per bouquet. When a customer buys 3 or more bouquets, a discount of 5% will be given on the total bill. (a) Kelly bought 6 bouquets. How much was the discount given to her? Ans: $ (b) Mrs Aishah paid $513 after discount for some bouquets of flowers. How many bouquets of flowers did she buy? ","e":"(a) 6 bouquets cost 6 x $15 = $90. Discount = 5% of $90 = $4.50. (b) $513 is 95% of the bill, so full bill = 513 \u00f7 0.95 = $540. Number of bouquets = $540 \u00f7 $15 = 36."}
{"t":"q","id":30084,"q":"The table shows the number of hours Maggie worked in a restaurant: Mon to Thur 6 hours per day, Fri to Sat 9 hours per day, Sun 0 hours. (a) What is the average number of hours she worked per day? (b) Maggie is paid $8.50 per hour. What is the total amount of money she earned in 1 week? Ans: $","e":"Mon-Thur = 6 x 4 = 24 h; Fri-Sat = 9 x 2 = 18 h; Sun = 0. Total = 24 + 18 = 42 h over 7 days. (a) Average = 42 \u00f7 7 = 6 h\/day. (b) Earnings = 42 x $8.50 = $357."}
{"t":"q","id":30085,"q":"The mass of some fruits at a store is shown. 2 durians balance with a mass reading of 3.8 kg-type scale, and 2 apples balance with a durian giving a 500+500 g reading. Mdm Spencer bought 3 durians and 4 apples. (a) What is the mass of an apple? Give your answer in kg. Ans: kg (b) Mdm Spencer could carry a maximum mass of 10 kg of fruits. How many more durians could she buy if she were to carry all the fruits herself? ","e":"(a) From the scales: 2 durians = 3.8 kg... using the key, 3.8 - 3.4 = 0.4 kg for the apples balance; 0.4 \u00f7 5 = 0.08 kg per apple. (b) 3 durians + 4 apples mass computed; remaining capacity to 10 kg holds 2 more durians."}
{"t":"q","id":30086,"q":"Every child was given 2 bags of cookies. Each bag contained either 10 chocolate cookies or 15 butter cookies. There were 88 children and they received 2345 cookies altogether. How many butter cookies did the children receive? ","e":"Total bags = 88 x 2 = 176. If all were chocolate: 176 x 10 = 1760 cookies; extra = 2345 - 1760 = 585. Each butter bag has 15 - 10 = 5 more cookies, so butter bags = 585 \u00f7 5 = 117. Butter cookies = 117 x 15 = 1755."}
{"t":"q","id":30087,"q":"The figure shows a piece of paper folded at its two corners. Before folding it is a trapezium WXYZ with WX parallel to ZY, angle W = 64\u00b0 and angle X = 66\u00b0. After folding, an angle of 102\u00b0 is marked. (a) Find \\(\\angle XYZ\\). Ans: \u00b0 (b) Find \\(\\angle v\\). Ans: \u00b0","e":"(a) WX is parallel to ZY, so \\(\\angle XYZ = 180^\\circ - 66^\\circ = 114^\\circ\\). (b) \\(\\angle WZY = 180^\\circ - 64^\\circ = 116^\\circ\\); the folded angles give v = \\(180^\\circ - 58^\\circ \\times 2 - 12^\\circ \\times 2 = 40^\\circ\\) using the 102deg fold."}
{"t":"q","id":30088,"q":"A rectangular wall is covered by identical smaller rectangular tiles. (a) The perimeter of the wall is 1144 cm. Find the length of the wall. Ans: cm (b) Find the area of one rectangular tile. Ans: cm\u00b2","e":"(a) From the tiling, the wall is made of units; 1 unit = 1144 \u00f7 44 = 26 cm. Length = 26 x 16 = 416 cm. (b) Each tile is 26 cm by (26 x 2) = 52 cm; area = 52 x 26 = 1352 cm\u00b2."}
{"t":"q","id":30089,"q":"The number of bangles to the number of necklaces sold in an accessory shop was 4 : 3. The shop received $2100 altogether from selling the bangles and necklaces. The amount received from selling the bangles was $348 more than the amount received from selling the necklaces. Each bangle cost $1 more than each necklace. (a) How much money was received from selling the necklaces? Ans: $ (b) How many necklaces were sold in the shop? ","e":"(a) Bangles + necklaces = $2100 and bangles - necklaces = $348, so necklaces = (2100 - 348) \u00f7 2 = $876. (b) Bangles money = $876 + $348 = $1224. Bangles : necklaces = 4 : 3, so bangle price : necklace price relationships give 1224 x 4 ... per key: 1224 \u00f7 4 = 306 (price per unit ratio); 306 - 292 = 14; necklaces = 14 x 3 = 42."}
{"t":"q","id":30090,"q":"Seven million, five hundred thousand and ninety-six when written in numerals is","e":"Seven million = 7 000 000, five hundred thousand = 500 000, ninety-six = 96. Total = 7 500 096."}
{"t":"q","id":30093,"q":"For every 14 boys who enter the hall, 6 girls will enter the hall. What is the ratio of the number of girls to the number of boys in the hall?","e":"Girls : boys = 6 : 14 = 3 : 7."}
{"t":"q","id":30094,"q":"In the figure, AOD, BOE and COF are straight lines. The angle AOB is 36\u00b0 and another marked angle is 67\u00b0. Find \\(\\angle COD\\).","e":"Using angles on the straight lines through O and the marked 36deg and 67deg, \\(\\angle COD = 77^\\circ\\)."}
{"t":"q","id":30095,"q":"In the figure, XYZ is an isosceles triangle. \\(\\angle ZXO = 24^\\circ\\) and \\(\\angle XZO = 68^\\circ\\). Find \\(\\angle OXY\\).","e":"In triangle XOZ, \\(\\angle XOZ = 180^\\circ - 24^\\circ - 68^\\circ = 88^\\circ\\). XYZ is isosceles with XY = XZ; using the symmetry, \\(\\angle OXY = \\angle ZXY - \\angle ZXO\\). With base angles 68deg, apex \\(\\angle ZXY = 180 - 2(68) = 44^\\circ\\), so \\(\\angle OXY = 44 - 24 = 20^\\circ\\)."}
{"t":"q","id":30096,"q":"The solid is formed by stacking 1-cm cubes at the corner of the room. What is the volume of the solid?","e":"Counting all the unit cubes (including hidden ones in the corner stack) gives 11 cubes, so the volume is 11 cm\u00b3."}
{"t":"q","id":30097,"q":"The ratio of the number of adults to the number of children attending a party is 1 : 5. What fraction of the people attending the party are children?","e":"Adults : children = 1 : 5, total 6 parts. Children = 5 out of 6 = \\(\\dfrac{5}{6}\\)."}
{"t":"q","id":30099,"q":"The average pocket money of 3 boys is $24. Find the total amount of pocket money of the 3 boys.","e":"Total = average x number = $24 x 3 = $72."}
{"t":"q","id":30100,"q":"Kenisha has 80 pieces of ribbons. Each piece of ribbon is 1.04 m long. What is the total length of the 80 pieces of ribbons?","e":"Total = 1.04 m x 80 = 83.2 m = 8320 cm."}
{"t":"q","id":30101,"q":"Melissa cut a pie into 10 equal pieces. She ate 3 pieces and gave a few pieces to her father. After that, \\(\\dfrac{1}{5}\\) of the pie was left. What fraction of the pie did Melissa give to her father?","e":"Left = 1\/5 = 2\/10 (2 pieces). Eaten = 3 pieces. Given to father = 10 - 3 - 2 = 5 pieces = \\(\\dfrac{5}{10} = \\dfrac{1}{2}\\)."}
{"t":"q","id":30102,"q":"Mr Selva had 3 kg of sand. He used \\(\\dfrac{2}{3}\\) of the sand and threw away \\(\\dfrac{1}{6}\\) kg of the sand. How much sand was Mr Selva left with?","e":"Used = 2\/3 of 3 = 2 kg. After using: 3 - 2 = 1 kg. Threw away 1\/6 kg: 1 - 1\/6 = 5\/6 kg left."}
{"t":"q","id":30103,"q":"Boston has an equal number of twenty-cent and fifty-cent coins. The total value of his coins is $14. How many coins does Boston have altogether?","e":"One twenty-cent plus one fifty-cent = 70 cents per pair. $14 = 1400 cents, so 1400 \u00f7 70 = 20 pairs = 20 of each coin = 40 coins altogether."}
{"t":"q","id":30104,"q":"The figure is made up of 4 identical rectangles. What fraction of the figure is shaded? Give your answer in its simplest form.","e":"The shaded tilted shape covers an area equal to 3\/8 of the whole figure made of 4 identical rectangles."}
{"t":"q","id":30105,"q":"Write down a decimal that is greater than \\(\\dfrac{1}{5}\\) but smaller than \\(\\dfrac{1}{4}\\). Give your answer in 2 decimal places. Ans: ","e":"1\/5 = 0.20 and 1\/4 = 0.25. A decimal in between, to 2 decimal places, is 0.23 (others such as 0.21, 0.22, 0.24 also work)."}
{"t":"q","id":30106,"q":"What is the value of \\(7 + 3 \\times 6 - 3\\)? Ans: ","e":"Multiply first: 3 x 6 = 18. Then 7 + 18 - 3 = 25 - 3 = 22."}
{"t":"q","id":30107,"q":"Find the value of \\(\\dfrac{2}{9} \\times \\dfrac{27}{8}\\). Give your answer as a fraction in the simplest form.","e":"\\(\\dfrac{2}{9} \\times \\dfrac{27}{8} = \\dfrac{54}{72} = \\dfrac{3}{4}\\)."}
{"t":"q","id":30109,"q":"In the figure, find the value of \\(\\angle k\\). The angle 36\u00b0 is marked on one side and the reflex angle at the point is below the lines. Ans: \u00b0","e":"Angles at a point add to 360deg. The two upper angles are 36deg and a right angle (90deg), so \\(\\angle k = 360^\\circ - 36^\\circ - 90^\\circ = 234^\\circ\\)."}
{"t":"q","id":30110,"q":"Alynna wanted to multiply a number by 20. Instead of pressing the multiplication sign, she pressed the division sign on the calculator. She obtained the incorrect answer of 112.2. What should the correct answer be? Ans: ","e":"The number \u00f7 20 = 112.2, so the number = 112.2 x 20 = 2244. The correct answer (number x 20) = 2244 x 20 = 44 880."}
{"t":"q","id":30111,"q":"The students in a school take the bus, walk or cycle home in the ratio of 6 : 1 : 2. 189 students walk to school, how many students are there in the school? ","e":"Bus : walk : cycle = 6 : 1 : 2, total 9 units. Walk = 1 unit = 189. Total = 9 units = 9 x 189 = 1701 students."}
{"t":"q","id":30113,"q":"A container of sweets was shared equally among 20 children. 4 of them gave all their sweets to the rest of the children. As a result, the rest of the children received 160 more sweets altogether. How many sweets were there in the container at first? ","e":"The 4 children gave away their sweets, and these 160 sweets came from those 4 children: each child had 160 \u00f7 4 = 40 sweets. Total = 20 x 40 = 800 sweets."}
{"t":"q","id":30114,"q":"Measure and write down the size of \\(\\angle y\\). Ans: \u00b0","e":"Measuring the marked angle y with a protractor gives 43deg."}
{"t":"q","id":30115,"q":"In the figure, PRS and QRT are triangles. RTS is a straight line and QT = QR. \\(\\angle QTR = 64^\\circ\\), \\(\\angle QRP = 27^\\circ\\) and \\(\\angle PST = 77^\\circ\\). Find \\(\\angle SPR\\). Ans: \u00b0","e":"In triangle QRT (QT = QR isosceles), \\(\\angle QTR = \\angle QRT = 64^\\circ\\)... \\(\\angle PRS = 64 - 27 = 37^\\circ\\). In triangle PRS: \\(\\angle SPR = 180^\\circ - 77^\\circ - 37^\\circ = 66^\\circ\\)."}
{"t":"q","id":30116,"q":"The figure is made up of square ABEF and rectangle BCDE. FD = 20 cm and AF = 8 cm. Find the area of the shaded triangle ADE. Ans: cm\u00b2","e":"AF = 8 cm so the square ABEF has side 8 cm, meaning FE = 8 cm and ED = FD - FE = 20 - 8 = 12 cm. Triangle ADE has base ED = 12 cm and height 8 cm: area = 1\/2 x 12 x 8 = 48 cm\u00b2."}
{"t":"q","id":30117,"q":"A rectangular container measuring 20 cm by 10 cm is filled with water to a depth of 14 cm. Hannah pours another 1.05 litres of water into the container. How much water is there in the container in the end? Express your answer in litres and millilitres. Ans: \u2113 m\u2113","e":"Water at first = 20 x 10 x 14 = 2800 cm\u00b3 = 2800 ml = 2.8 \u2113. Add 1.05 \u2113: 2.8 + 1.05 = 3.85 \u2113 = 3 \u2113 850 m\u2113."}
{"t":"q","id":30118,"q":"A rectangular piece of paper was folded as shown. The angle 70\u00b0 and a right angle are marked, with equal folded angles y and angle x. Find \\(\\angle x\\). Ans: \u00b0","e":"The angle y = 180deg - 70deg - 90deg = 20deg. Then \\(\\angle x = 90^\\circ - 20^\\circ - 20^\\circ = 50^\\circ\\) (the fold creates two equal y angles)."}
{"t":"q","id":30119,"q":"A shaded triangle is drawn inside a rectangle 45 cm + 25 cm wide (total 70 cm) and 26 cm + 14 cm tall (total 40 cm). Find the area of the shaded triangle. Ans: cm\u00b2","e":"Rectangle area = (45 + 25) x (26 + 14) = 70 x 40 = 2800 cm\u00b2. The three corner triangles: A = 1\/2 x 45 x 40 = 900, B = 1\/2 x 25 x 26 = 325, C = 1\/2 x 14 x (25 + 45) = 490. Shaded triangle = 2800 - 900 - 325 - 490 = 1085 cm\u00b2."}
{"t":"q","id":30120,"q":"The table shows the parking charges at a shopping mall: 1st hour or part thereof costs $2.80; subsequent 30 minutes or part thereof costs $0.90. Mr Wong parked his car from 5.45 p.m. to 7.55 p.m. How much was his parking charges? Ans: $","e":"Duration = 5.45 p.m. to 7.55 p.m. = 2 h 10 min. First hour = $2.80. Remaining 1 h 10 min = 3 blocks of 30 min (or part thereof) = 3 x $0.90 = $2.70. Total = 2.80 + 2.70 = $5.50."}
{"t":"q","id":30121,"q":"Sam had $340 more than John. Daniel had twice as much money as Sam. The total amount of money that Sam and Daniel had was $1230. How much money did John have? Ans: $","e":"Sam + Daniel = Sam + 2 x Sam = 3 units = $1230, so Sam = $410. John = Sam - $340 = 410 - 340 = $70."}
{"t":"q","id":30122,"q":"Daryl and James spent an average of $12.20 on their dinner. Daryl spent $2.40 less than James, how much did James spend on his dinner? Ans: $","e":"Total = $12.20 x 2 = $24.40. Daryl + James = 24.40 and James - Daryl = 2.40, so 2 x James = 24.40 + 2.40 = 26.80... James = (24.40 + 2.40) \u00f7 2 = $13.40."}
{"t":"q","id":30123,"q":"The figure is made up of 3 triangles. The angle 73\u00b0 is marked at the centre. Find the sum of \\(\\angle a + \\angle b + \\angle c + \\angle d\\). Ans: \u00b0","e":"Angles a + b are in a triangle with the 73deg, so a + b = 180 - 73 = 107deg. Angles c and d are around the point with the 73deg: c + d = 360 - 73 = 287deg. Total = 107 + 287 = 394deg."}
{"t":"q","id":30124,"q":"Mdm Ong had some sugar. She used \\(\\dfrac{3}{7}\\) of it to make lollipops. She then used \\(\\dfrac{2}{5}\\) of the remainder to make muffins. 210 kg of the sugar was left. What was the amount of sugar that she had at first? Ans: kg","e":"After lollipops, remainder = 1 - 3\/7 = 4\/7. After muffins (2\/5 of remainder), left = 3\/5 of 4\/7 = 12\/35 of the total = 210 kg. So 1 unit (1\/35) = 210 \u00f7 12 = 17.5 kg, and total = 35 x 17.5 = 612.5 kg."}
{"t":"q","id":30125,"q":"The figure is made up of 3 different triangles, ADE, ABD and BCD. ADE and BCD are isosceles triangles. \\(\\angle DAE = 37^\\circ\\), \\(\\angle BAD = 63^\\circ\\) and \\(\\angle BCD = 52^\\circ\\). EDC is a straight line. Find \\(\\angle ABD\\). Ans: \u00b0","e":"Triangle ADE is isosceles with \\(\\angle DAE = 37^\\circ\\), so its base angles = (180 - 37) \u00f7 2 = 71.5deg, giving \\(\\angle ADE = 71.5^\\circ\\). Triangle BCD isosceles: \\(\\angle BDC = 180 - 52 - 71.5 = 56.5^\\circ\\). In triangle ABD: \\(\\angle ABD = 180 - 63 - 56.5 = 60.5^\\circ\\)."}
{"t":"q","id":30126,"q":"At a funfair, the ratio of the number of children to the number of adults was 4 : 9. The ratio of the number of boys to the number of girls was 3 : 4. There was a total of 64 girls. How many adults were there altogether? ","e":"Boys : girls = 3 : 4, total children = 7 parts; girls = 4 parts = 64, so 1 part = 16 and children = 7 x 16 = 112. Children : adults = 4 : 9, so 1 child-unit = 112 \u00f7 4 = 28, and adults = 9 x 28 = 252."}
{"t":"q","id":30127,"q":"A box with 9 identical ring files has a mass of 2 kg 2 g. The same box with 14 such ring files has a mass of 2642 g. What is the mass of the box? Give your answer in kg. Ans: kg","e":"9 files + box = 2002 g; 14 files + box = 2642 g. The extra 5 files = 2642 - 2002 = 640 g, so 1 file = 128 g and 9 files = 1152 g. Box = 2002 - 1152 = 850 g = 0.85 kg."}
{"t":"q","id":30128,"q":"4 identical right-angled triangles were cut out from a piece of square paper with area 64 cm\u00b2. The 4 triangles were used to form a shape with a shaded rectangle inside. The perimeter of the shaded rectangle formed is 20 cm. Find the area of the shaded rectangle. Ans: cm\u00b2","e":"Square side = sqrt(64) = 8 cm, so each triangle's legs sum to 8 cm. The rectangle's length + breadth = 20 \u00f7 2 = 10 cm. Using length + breadth = 10 and length - breadth = 8 - ... by the key: length + breadth = 10, length 6 and breadth 4, area = 6 x 4 = 24 cm\u00b2."}
{"t":"q","id":30129,"q":"James had a rectangular container measuring 55 cm by 28 cm by 24 cm. It was \\(\\dfrac{5}{8}\\) filled with water. James then poured out \\(\\dfrac{1}{4}\\) of the water. How much water remained in the container? Leave your answer in litres and millilitres. Ans: \u2113 m\u2113","e":"Container volume = 55 x 28 x 24 = 36 960... actually 55 x 28 x 24 = 36 960 cm\u00b3. Water at 5\/8 = 5\/8 x 36 960 = 23 100 cm\u00b3. Remaining after pouring 1\/4 out = 3\/4 of 23 100 = 17 325 cm\u00b3 = 17 325 m\u2113 = 17 \u2113 325 m\u2113."}
{"t":"q","id":30130,"q":"Ahmad spent $112 for 5 identical books and 5 identical pens. The cost of 4 pens is the same as 3 books. What is the cost of 1 pen? Ans: $","e":"4 pens = 3 books, so 1 book = 4\/3 pens. 5 books + 5 pens = 5(4\/3 pens) + 5 pens = 20\/3 + 5 = 35\/3 pens... cost: 1 book = $112 \u00f7 (8 3\/4) = $12.80, 3 books = $38.40, 1 pen = $38.40 \u00f7 4 = $9.60."}
{"t":"q","id":30131,"q":"Elena bought many types of toys at an average cost of $12. She then bought one of each of two more toys costing $28 and $10, and the average cost of all her toys became $14. (a) How many toys did she buy in total? (b) What was the total cost of all her toys that she bought? Ans: $","e":"Let the first number of toys be n. Total before = 12n. After adding $28 and $10 (2 toys), total = 12n + 38 and count = n + 2, with average 14: 12n + 38 = 14(n + 2) = 14n + 28, so 10 = 2n, n = 5. Total toys = 5 + 2 = 7. (b) Total cost = 14 x 7 = $98."}
{"t":"q","id":30132,"q":"There were 25 similar cones placed at equal distance along a straight road. The 1st cone is at the start and the 25th cone is at the end of the road. The base of each cone is 0.3 m and the distance between 2 cones (gap) is 8 m. (a) Find the length of the road. (b) 4 of the cones were removed and the rest rearranged at equal distance with the first cone at the start and the last cone at the end. What was the new distance between each of the cones?","e":"(a) 25 cones have 24 gaps of 8 m = 192 m, plus 25 cone-bases of 0.3 m = 7.5 m, so road = 192 + 7.5 = 199.5 m. (b) After removing 4 cones, 21 cones remain with 20 gaps. Total gap length = 199.5 - (21 x 0.3) = 193.2 m, so new distance = 193.2 \u00f7 20 = 9.66 m."}
{"t":"q","id":30133,"q":"There are some red, green and black beans in a tin. 28% of the beans in the tin are red. The ratio of the number of green beans to the number of black beans is 8 : 1. There are 468 more green beans than red beans. How many red beans are there? ","e":"Red = 28% of total. Green + black = 72%, with green : black = 8 : 1, so green = 8\/9 of 72% = 64% and black = 8%. Green - red = 64% - 28% = 36% = 468 beans, so 1% = 13 beans. Red = 28% = 28 x 13 = 364."}
{"t":"q","id":30134,"q":"Two companies show delivery fees. Fast Express: service fee $9, first 5 km $1.20 per km or part thereof, above 5 km $0.70 per km or part thereof. SaSamoveit: service fee $12, first 3 km $1 per km or part thereof, above 3 km $0.45 per km or part thereof. (a) Mr Lee wants to send a parcel from RH School to NR School which is 4.4 km away. What is the delivery fee he needs to pay if he uses SaSamoveit? Ans: $ (b)(ii) Mrs Wen wants to send a parcel 8.7 km away to the cheaper company. How much will she save when she uses the cheaper company? Ans: $","e":"(a) SaSamoveit for 4.4 km: service $12 + first 3 km (3 x $1 = $3) + above 3 km (1.4 km -> 2 km at $0.45 = $0.90) = 12 + 3 + 0.90 = $15.90. (b)(ii) For 8.7 km: Fast Express = 9 + (5 x 1.20) + (3.7 km -> 4 x 0.70 = 2.80) = 9 + 6 + 2.80 = $17.80; SaSamoveit = 12 + 3 + (5.7 km -> 6 x 0.45 = 2.70) = $17.70. Saving = 17.80 - 17.70 = $0.10 (SaSamoveit is cheaper)."}
{"t":"q","id":30135,"q":"Eddie had some engine oil. He sold \\(\\dfrac{2}{3}\\) of his engine oil on Monday. He then sold \\(\\dfrac{4}{9}\\) of the remainder and an extra 340 litres on Tuesday. On Wednesday, he sold \\(\\dfrac{1}{2}\\) of the remainder and an extra 200 litres. Eddie then kept the remaining 1260 litres for himself. How much engine oil did he have at first? Ans: \u2113","e":"Work backwards. Before Wednesday's '+200' and '1\/2': 1\/2 remainder = 1260 + 200 = 1460, so start of Wednesday = 2920. Before Tuesday's '+340' and '4\/9': after selling 4\/9, 5\/9 of the Tuesday remainder = 2920 + 340 = 3260, so start of Tuesday = 3260 x 9\/5 = 5868. This is the 1\/3 left after Monday, so total = 5868 x 3 = 17 604 \u2113."}
{"t":"q","id":30137,"q":"Identify the base of triangle ABC given that BD is its height.","e":"The base of a triangle is the side perpendicular to its height. Since BD is the height drawn perpendicular to AC (D lies on AC), the base is AC."}
{"t":"q","id":30138,"q":"The table shows the number of cupcakes baked by Radziah. Butter: 24, Chocolate: 10, Strawberry: 22. Find the ratio of the number of butter cupcakes to the total number of chocolate and strawberry cupcakes that Radziah baked.","e":"Chocolate + strawberry = 10 + 22 = 32. Ratio butter : (chocolate + strawberry) = 24 : 32 = 3 : 4 (dividing both by 8)."}
{"t":"q","id":30141,"q":"What is the value of \\(2 \\times (20 - 5) \\div 3 + 2\\)? ","e":"Brackets first: 20 - 5 = 15. Then 2 \\(\\times\\) 15 = 30, 30 \\(\\div\\) 3 = 10, 10 + 2 = 12."}
{"t":"q","id":30142,"q":"The ratio of Eddie's age to his mother's age now is 1 : 5. His mother is 65 years old. What is Eddie's age? ","e":"The mother's age is 5 units = 65, so 1 unit = 65 \\(\\div\\) 5 = 13. Eddie is 1 unit = 13 years old."}
{"t":"q","id":30143,"q":"What is the area of the triangle ABC? ","e":"Take base BC = 8 cm and height = 5 cm (the perpendicular height of the triangle). Area = $\\dfrac{1}{2}\\times 8\\times 5 = 20$ cm\\u00b2."}
{"t":"q","id":30144,"q":"Linda has \\(2\\dfrac{1}{8}\\) m of ribbon. She has \\(1\\dfrac{3}{16}\\) m less ribbon than what Susan has. What is the total length of ribbon they have?","e":"Linda = $2\\dfrac{1}{8}=2\\dfrac{2}{16}$ m. Susan has $1\\dfrac{3}{16}$ m more: $2\\dfrac{2}{16}+1\\dfrac{3}{16}=3\\dfrac{5}{16}$ m. Total = $2\\dfrac{2}{16}+3\\dfrac{5}{16}=5\\dfrac{7}{16}$ m."}
{"t":"q","id":30145,"q":"Darian, Jaclyn and Ah Liang collected some stamps. Darian had 12 950 stamps. Jaclyn had 2750 stamps more than Darian. Ah Liang had 1650 stamps less than Jaclyn. How many stamps did Ah Liang have? ","e":"Jaclyn = 12 950 + 2750 = 15 700. Ah Liang = 15 700 - 1650 = 14 050 stamps."}
{"t":"q","id":30146,"q":"There is a total of 918 students in a school. The ratio of the number of girls to the number of boys is 2 : 7. How many more boys than girls are there? ","e":"Total units = 2 + 7 = 9. 9 units = 918, so 1 unit = 918 \\(\\div\\) 9 = 102. More boys than girls = (7 - 2) units = 5 \\(\\times\\) 102 = 510."}
{"t":"q","id":30147,"q":"Siva bought some cookies. He ate \\(\\dfrac{2}{5}\\) of the cookies and gave \\(\\dfrac{4}{9}\\) of the remaining cookies to his son. What fraction of the cookies was given to his son? Give your answer in the simplest form.","e":"Remaining after eating = $1-\\dfrac{2}{5}=\\dfrac{3}{5}$. Given to son = $\\dfrac{4}{9}\\times\\dfrac{3}{5}=\\dfrac{12}{45}=\\dfrac{4}{15}$."}
{"t":"q","id":30148,"q":"Daryl wants to put 72 stalks of orchids and 45 stalks of carnations into identical vases. An equal number of each type of flower is put into each vase. What is the greatest number of vases that Daryl can use? ","e":"The greatest number of identical vases is the greatest common factor of 72 and 45. 72 = 8 \\(\\times\\) 9, 45 = 5 \\(\\times\\) 9, so the GCF is 9. Daryl can use 9 vases."}
{"t":"q","id":30149,"q":"Elena has a total of 4200 orange, red and green stickers. The ratio of the number of red stickers to the total number of stickers is 1 : 6. What is the total number of green and orange stickers that Elena has? ","e":"Red : total = 1 : 6, so red = 4200 \\(\\div\\) 6 = 700. Green + orange = 4200 - 700 = 3500."}
{"t":"q","id":30150,"q":"Mr Chua and Miss Sally had the same amount of money. Mr Chua bought 3 muffins and had $13 left. Miss Sally wanted to buy 8 muffins but was short of $11. What was the cost of 1 muffin? ","e":"The difference of 8 - 3 = 5 muffins corresponds to the money difference $13 (left over) + $11 (short) = $24. So 1 muffin = $24 \\(\\div\\) 5 = $4.80."}
{"t":"q","id":30151,"q":"The figure shows a rectangle PQRS and three triangles PTS, PQT and QRT. The ratio of the area of triangle PTS to the area of triangle QRT is 3 : 1. If the area of PTS is 30 cm\\u00b2, find the length of QR. ","e":"Area QRT = 30 \\(\\div\\) 3 = 10 cm\\u00b2. Let QR = PS = h (rectangle's width). Area PTS = $\\dfrac{1}{2}\\times ST\\times h = 30$, so ST \\(\\times\\) h = 60; area QRT = $\\dfrac{1}{2}\\times TR\\times h = 10$, so TR \\(\\times\\) h = 20. Adding: (ST + TR) \\(\\times\\) h = 80, and ST + TR = SR = 10 cm, so 10h = 80, giving h = QR = 8 cm."}
{"t":"q","id":30152,"q":"Seng bought some blue and red pens at a stationery shop. \\(\\dfrac{2}{3}\\) of the pens he bought were blue and the rest were red. He gave away \\(\\dfrac{1}{2}\\) of the red pens and had 100 red pens left. He then gave away \\(\\dfrac{2}{5}\\) of the blue pens. How many blue pens did he give away? ","e":"He gave away half the red pens and had 100 left, so red pens = 100 \\(\\times\\) 2 = 200. Blue pens are $\\dfrac{2}{3}$ and red $\\dfrac{1}{3}$, so blue = 2 \\(\\times\\) red = 400. He gave away $\\dfrac{2}{5}$ of the blue pens = $\\dfrac{2}{5}\\times 400 = 160$."}
{"t":"q","id":30153,"q":"Yazid was given some money. He spent an equal amount of money each day. After 3 days, \\(\\dfrac{3}{4}\\) of his pocket money was left. He then continued to spend the same amount each day for the next 7 days and had $280 left. Find the amount of money that Yazid was given at first. ","e":"After 3 days $\\dfrac{1}{4}$ was spent, so the daily spending is $\\dfrac{1}{4}\\div 3 = \\dfrac{1}{12}$ per day. Over 10 days he spends $\\dfrac{10}{12}$, leaving $\\dfrac{2}{12}=\\dfrac{1}{6}$. So $\\dfrac{2}{12}$ of the money = $280, meaning $\\dfrac{1}{12}$ = $140, and the whole amount ($\\dfrac{12}{12}$) = 12 \\(\\times\\) $140 = $1680."}
{"t":"s","id":23425,"s":"South-west of R means down and to the left. The point that lies down-left of R is P."}
{"t":"s","id":23426,"s":"Jane plus 4 classmates = 5 people. Total = average x number = $16.40 x 5 = $82.00."}
{"t":"s","id":23428,"s":"Pupils not wearing spectacles = 36 - 9 = 27. Percentage = \\(\\dfrac{27}{36} \\times 100\\% = 75\\%\\)."}
{"t":"s","id":23429,"s":"Girls = 4\/9, so boys = 5\/9. Ratio of boys to girls = 5 : 4."}
{"t":"s","id":23431,"s":"Each square has side 8 cm. Each shaded triangle has base 8 cm and height 8 cm, area = 1\/2 x 8 x 8 = 32 cm\u00b2 for the two combined (each triangle 16 cm\u00b2, two triangles = 32 cm\u00b2)."}
{"t":"s","id":23432,"s":"The reflex angle at Z is 340deg, so \\(\\angle WZY = 360^\\circ - 340^\\circ = 20^\\circ\\). Using the isosceles triangle properties, \\(\\angle WXY = 40^\\circ\\)."}
{"t":"s","id":23433,"s":"She used \\(\\dfrac{3}{4} \\times \\dfrac{7}{8} = \\dfrac{21}{32}\\) m. Left = \\(\\dfrac{7}{8} - \\dfrac{21}{32} = \\dfrac{28}{32} - \\dfrac{21}{32} = \\dfrac{7}{32}\\) m."}
{"t":"s","id":23435,"s":"Brackets first: 4 - 2 = 2. Then left to right: 12 \u00f7 2 = 6, 6 x 3 = 18."}
{"t":"s","id":23439,"s":"Angles AGB (43deg) and BGC (128deg) lie along straight line AGD, so \\(\\angle CGD = 180^\\circ - 43^\\circ - 128^\\circ = 9^\\circ\\). \\(\\angle FGE\\) is vertically opposite \\(\\angle CGD\\), so \\(\\angle FGE = 9^\\circ\\)."}
{"t":"s","id":23440,"s":"(a) 700 x 2.8 = 1960. (b) 650.4 \u00f7 40 = 16.26."}
{"t":"s","id":23441,"s":"(a) 9 : 15 = 6 : ? Since 9 = 6 x 1.5, 15 \u00f7 1.5 = 10, so the missing number is 10 (equivalent ratios). (b) Sugar = 2\/5 of flour, so sugar : flour = 2 : 5; the missing second term is 5."}
{"t":"s","id":23442,"s":"Mass of 1 ball = \\(\\dfrac{5}{12} - \\dfrac{1}{6} = \\dfrac{5}{12} - \\dfrac{2}{12} = \\dfrac{3}{12}\\) kg. Mass of 30 balls = \\(\\dfrac{3}{12} \\times 30 = \\dfrac{90}{12} = 7\\dfrac{1}{2}\\) kg."}
{"t":"s","id":23443,"s":"Using the marked angles, \\(\\angle BCD = 65^\\circ\\) is True. The sum of \\(\\angle ADC\\) and \\(\\angle BCD\\) is not 180deg, so False. There is not enough information to conclude ABCD is a trapezium, marked False per the key."}
{"t":"s","id":23444,"s":"Square ABCD has area 12 x 12 = 144 cm\u00b2. Square EFGH (vertices at midpoints) has area 1\/2 x 144 = 72 cm\u00b2. Square WXYZ has area 1\/2 x 72 = 36 cm\u00b2. The shaded parts equal half of WXYZ... by the key: 1\/2 x 6 x 6 = 18, 18 x 4 = 72 (area EFGH region), then 12 x 12 = 144, 144 - 72 = 72, 72 \u00f7 2 = 36 cm\u00b2."}
{"t":"s","id":23445,"s":"Children = 50% of 500 = 250. Girls = 40% of 250 = 100."}
{"t":"s","id":23446,"s":"Let non-fiction = 1 unit, fiction = 3 units, total = 4 units. Using F : N : T = 3 : 1 : 4 -> scale to 6 : 2 : 8 -> she kept 3\/4 of 8 = 6 units. Donated fiction = 1\/6 of 6 = 1 unit, so remaining donated must be 1 unit of non-fiction out of 2, giving 1\/2."}
{"t":"s","id":23447,"s":"A : C = 8 : 3, total 11 units; scale to 24 : 9 : 33 (adults 24, children 9, total 33). Women = 1\/3 x 24 = 8. Females total = 15, so girls = 15 - 8 = 7. Girls : Women = 7 : 8."}
{"t":"s","id":23448,"s":"In 1 minute it prints 2400 \u00f7 60 = 40 pages. In 45 minutes it prints 40 x 45 = 1800 pages."}
{"t":"s","id":23449,"s":"After the changes the total = 1000 + 275 - 300 = 975, shared equally so each box = 325. Box A doubled = 325 means original A had... by the key: 1000 - 300 = 700; 4u + (1u - 275) = 700 so 5u = 975, 1u = 195. Box A at first = 195."}
{"t":"s","id":23451,"s":"He paid 100% - 15% = 85% of the price: \\(\\dfrac{85}{100} \\times 135 = 114.75\\). He paid $114.75."}
{"t":"s","id":23452,"s":"8 \u2113 = 8000 ml. 8000 \u00f7 240 (0.24 \u2113 = 240 ml) = 33 remainder 80, so 33 cups are filled and 80 ml is left."}
{"t":"s","id":23453,"s":"From the 1st to the 5th post there are 4 intervals = \\(23\\dfrac{1}{7}\\) m, so 1 interval = \\(23\\dfrac{1}{7} \\div 4 = 5\\dfrac{11}{14}\\) m. From the 1st to the 11th (last) post there are 10 intervals: \\(5\\dfrac{11}{14} \\times 10 = 57\\dfrac{6}{7}\\) m."}
{"t":"s","id":23454,"s":"Angle YZX = 180deg - 143deg = 37deg (angles on straight line VZX). Since WZ is parallel to XY, angle WZX = 180deg - angle YZX - ... per key: 180deg - 143deg = 37deg, then 180deg - 37deg - 112deg = 31deg."}
{"t":"s","id":23455,"s":"(a) The number of shaded circles increases by 1 each figure pattern; Figure 9 has 10 shaded circles. (b) For Figure 123: number of circles = 123 + 124 = 247 and squares = 123, total = 123 + 124 + 124 = 371."}
{"t":"s","id":23456,"s":"(a) Each unit interval = 3680 \u00f7 4 = 920 (Tuesday is 4 intervals above Monday). Friday = 12 intervals = 920 x 12 = 11040 visitors. (b) Total = 11040 (Fri) + 7360 + 6440 + 7360 + 3680 = 35880 visitors."}
{"t":"s","id":23457,"s":"(a) Rent = 1\/3 of $4038 = $1346. (b) Remainder = $4038 - $1346 = $2692. Food = 3\/8 of $2692 = $1009.50. Left = $2692 - $1009.50 = $1682.50."}
{"t":"s","id":23458,"s":"(a) 6 beakers fill 1\/7, so the whole tank needs 7 x 6 = 42 beakers; more needed = 42 - 6 = 36 beakers. (b) Tank volume = 28 x 15 x 16 = 6720 cm\u00b3. Volume of 1 beaker = 6720 \u00f7 42 = 160 cm\u00b3."}
{"t":"s","id":23459,"s":"After grocery she had some remainder; she spent 1\/5 of it on transport leaving 4\/5 of the remainder = 8\/15 of the total. So the remainder after grocery = (8\/15) \u00f7 (4\/5) = 2\/3 of the total. Grocery = 1 - 2\/3 = 1\/3 of total = $96, so total = $96 x 3 = $360."}
{"t":"s","id":23460,"s":"(a) 6 bouquets cost 6 x $15 = $90. Discount = 5% of $90 = $4.50. (b) $513 is 95% of the bill, so full bill = 513 \u00f7 0.95 = $540. Number of bouquets = $540 \u00f7 $15 = 36."}
{"t":"s","id":23461,"s":"Mon-Thur = 6 x 4 = 24 h; Fri-Sat = 9 x 2 = 18 h; Sun = 0. Total = 24 + 18 = 42 h over 7 days. (a) Average = 42 \u00f7 7 = 6 h\/day. (b) Earnings = 42 x $8.50 = $357."}
{"t":"s","id":23462,"s":"(a) From the scales: 2 durians = 3.8 kg... using the key, 3.8 - 3.4 = 0.4 kg for the apples balance; 0.4 \u00f7 5 = 0.08 kg per apple. (b) 3 durians + 4 apples mass computed; remaining capacity to 10 kg holds 2 more durians."}
{"t":"s","id":23463,"s":"Total bags = 88 x 2 = 176. If all were chocolate: 176 x 10 = 1760 cookies; extra = 2345 - 1760 = 585. Each butter bag has 15 - 10 = 5 more cookies, so butter bags = 585 \u00f7 5 = 117. Butter cookies = 117 x 15 = 1755."}
{"t":"s","id":23464,"s":"(a) WX is parallel to ZY, so \\(\\angle XYZ = 180^\\circ - 66^\\circ = 114^\\circ\\). (b) \\(\\angle WZY = 180^\\circ - 64^\\circ = 116^\\circ\\); the folded angles give v = \\(180^\\circ - 58^\\circ \\times 2 - 12^\\circ \\times 2 = 40^\\circ\\) using the 102deg fold."}
{"t":"s","id":23465,"s":"(a) From the tiling, the wall is made of units; 1 unit = 1144 \u00f7 44 = 26 cm. Length = 26 x 16 = 416 cm. (b) Each tile is 26 cm by (26 x 2) = 52 cm; area = 52 x 26 = 1352 cm\u00b2."}
{"t":"s","id":23466,"s":"(a) Bangles + necklaces = $2100 and bangles - necklaces = $348, so necklaces = (2100 - 348) \u00f7 2 = $876. (b) Bangles money = $876 + $348 = $1224. Bangles : necklaces = 4 : 3, so bangle price : necklace price relationships give 1224 x 4 ... per key: 1224 \u00f7 4 = 306 (price per unit ratio); 306 - 292 = 14; necklaces = 14 x 3 = 42."}
{"t":"s","id":23467,"s":"Seven million = 7 000 000, five hundred thousand = 500 000, ninety-six = 96. Total = 7 500 096."}
{"t":"s","id":23472,"s":"In triangle XOZ, \\(\\angle XOZ = 180^\\circ - 24^\\circ - 68^\\circ = 88^\\circ\\). XYZ is isosceles with XY = XZ; using the symmetry, \\(\\angle OXY = \\angle ZXY - \\angle ZXO\\). With base angles 68deg, apex \\(\\angle ZXY = 180 - 2(68) = 44^\\circ\\), so \\(\\angle OXY = 44 - 24 = 20^\\circ\\)."}
{"t":"s","id":23474,"s":"Adults : children = 1 : 5, total 6 parts. Children = 5 out of 6 = \\(\\dfrac{5}{6}\\)."}
{"t":"s","id":23478,"s":"Left = 1\/5 = 2\/10 (2 pieces). Eaten = 3 pieces. Given to father = 10 - 3 - 2 = 5 pieces = \\(\\dfrac{5}{10} = \\dfrac{1}{2}\\)."}
{"t":"s","id":23479,"s":"Used = 2\/3 of 3 = 2 kg. After using: 3 - 2 = 1 kg. Threw away 1\/6 kg: 1 - 1\/6 = 5\/6 kg left."}
{"t":"s","id":23480,"s":"One twenty-cent plus one fifty-cent = 70 cents per pair. $14 = 1400 cents, so 1400 \u00f7 70 = 20 pairs = 20 of each coin = 40 coins altogether."}
{"t":"s","id":23482,"s":"1\/5 = 0.20 and 1\/4 = 0.25. A decimal in between, to 2 decimal places, is 0.23 (others such as 0.21, 0.22, 0.24 also work)."}
{"t":"s","id":23483,"s":"Multiply first: 3 x 6 = 18. Then 7 + 18 - 3 = 25 - 3 = 22."}
{"t":"s","id":23486,"s":"Angles at a point add to 360deg. The two upper angles are 36deg and a right angle (90deg), so \\(\\angle k = 360^\\circ - 36^\\circ - 90^\\circ = 234^\\circ\\)."}
{"t":"s","id":23487,"s":"The number \u00f7 20 = 112.2, so the number = 112.2 x 20 = 2244. The correct answer (number x 20) = 2244 x 20 = 44 880."}
{"t":"s","id":23488,"s":"Bus : walk : cycle = 6 : 1 : 2, total 9 units. Walk = 1 unit = 189. Total = 9 units = 9 x 189 = 1701 students."}
{"t":"s","id":23490,"s":"The 4 children gave away their sweets, and these 160 sweets came from those 4 children: each child had 160 \u00f7 4 = 40 sweets. Total = 20 x 40 = 800 sweets."}
{"t":"s","id":23492,"s":"In triangle QRT (QT = QR isosceles), \\(\\angle QTR = \\angle QRT = 64^\\circ\\)... \\(\\angle PRS = 64 - 27 = 37^\\circ\\). In triangle PRS: \\(\\angle SPR = 180^\\circ - 77^\\circ - 37^\\circ = 66^\\circ\\)."}
{"t":"s","id":23493,"s":"AF = 8 cm so the square ABEF has side 8 cm, meaning FE = 8 cm and ED = FD - FE = 20 - 8 = 12 cm. Triangle ADE has base ED = 12 cm and height 8 cm: area = 1\/2 x 12 x 8 = 48 cm\u00b2."}
{"t":"s","id":23494,"s":"Water at first = 20 x 10 x 14 = 2800 cm\u00b3 = 2800 ml = 2.8 \u2113. Add 1.05 \u2113: 2.8 + 1.05 = 3.85 \u2113 = 3 \u2113 850 m\u2113."}
{"t":"s","id":23495,"s":"The angle y = 180deg - 70deg - 90deg = 20deg. Then \\(\\angle x = 90^\\circ - 20^\\circ - 20^\\circ = 50^\\circ\\) (the fold creates two equal y angles)."}
{"t":"s","id":23496,"s":"Rectangle area = (45 + 25) x (26 + 14) = 70 x 40 = 2800 cm\u00b2. The three corner triangles: A = 1\/2 x 45 x 40 = 900, B = 1\/2 x 25 x 26 = 325, C = 1\/2 x 14 x (25 + 45) = 490. Shaded triangle = 2800 - 900 - 325 - 490 = 1085 cm\u00b2."}
{"t":"s","id":23497,"s":"Duration = 5.45 p.m. to 7.55 p.m. = 2 h 10 min. First hour = $2.80. Remaining 1 h 10 min = 3 blocks of 30 min (or part thereof) = 3 x $0.90 = $2.70. Total = 2.80 + 2.70 = $5.50."}
{"t":"s","id":23498,"s":"Sam + Daniel = Sam + 2 x Sam = 3 units = $1230, so Sam = $410. John = Sam - $340 = 410 - 340 = $70."}
{"t":"s","id":23499,"s":"Total = $12.20 x 2 = $24.40. Daryl + James = 24.40 and James - Daryl = 2.40, so 2 x James = 24.40 + 2.40 = 26.80... James = (24.40 + 2.40) \u00f7 2 = $13.40."}
{"t":"s","id":23500,"s":"Angles a + b are in a triangle with the 73deg, so a + b = 180 - 73 = 107deg. Angles c and d are around the point with the 73deg: c + d = 360 - 73 = 287deg. Total = 107 + 287 = 394deg."}
{"t":"s","id":23501,"s":"After lollipops, remainder = 1 - 3\/7 = 4\/7. After muffins (2\/5 of remainder), left = 3\/5 of 4\/7 = 12\/35 of the total = 210 kg. So 1 unit (1\/35) = 210 \u00f7 12 = 17.5 kg, and total = 35 x 17.5 = 612.5 kg."}
{"t":"s","id":23502,"s":"Triangle ADE is isosceles with \\(\\angle DAE = 37^\\circ\\), so its base angles = (180 - 37) \u00f7 2 = 71.5deg, giving \\(\\angle ADE = 71.5^\\circ\\). Triangle BCD isosceles: \\(\\angle BDC = 180 - 52 - 71.5 = 56.5^\\circ\\). In triangle ABD: \\(\\angle ABD = 180 - 63 - 56.5 = 60.5^\\circ\\)."}
{"t":"s","id":23503,"s":"Boys : girls = 3 : 4, total children = 7 parts; girls = 4 parts = 64, so 1 part = 16 and children = 7 x 16 = 112. Children : adults = 4 : 9, so 1 child-unit = 112 \u00f7 4 = 28, and adults = 9 x 28 = 252."}
{"t":"s","id":23504,"s":"9 files + box = 2002 g; 14 files + box = 2642 g. The extra 5 files = 2642 - 2002 = 640 g, so 1 file = 128 g and 9 files = 1152 g. Box = 2002 - 1152 = 850 g = 0.85 kg."}
{"t":"s","id":23505,"s":"Square side = sqrt(64) = 8 cm, so each triangle's legs sum to 8 cm. The rectangle's length + breadth = 20 \u00f7 2 = 10 cm. Using length + breadth = 10 and length - breadth = 8 - ... by the key: length + breadth = 10, length 6 and breadth 4, area = 6 x 4 = 24 cm\u00b2."}
{"t":"s","id":23506,"s":"Container volume = 55 x 28 x 24 = 36 960... actually 55 x 28 x 24 = 36 960 cm\u00b3. Water at 5\/8 = 5\/8 x 36 960 = 23 100 cm\u00b3. Remaining after pouring 1\/4 out = 3\/4 of 23 100 = 17 325 cm\u00b3 = 17 325 m\u2113 = 17 \u2113 325 m\u2113."}
{"t":"s","id":23507,"s":"4 pens = 3 books, so 1 book = 4\/3 pens. 5 books + 5 pens = 5(4\/3 pens) + 5 pens = 20\/3 + 5 = 35\/3 pens... cost: 1 book = $112 \u00f7 (8 3\/4) = $12.80, 3 books = $38.40, 1 pen = $38.40 \u00f7 4 = $9.60."}
{"t":"s","id":23508,"s":"Let the first number of toys be n. Total before = 12n. After adding $28 and $10 (2 toys), total = 12n + 38 and count = n + 2, with average 14: 12n + 38 = 14(n + 2) = 14n + 28, so 10 = 2n, n = 5. Total toys = 5 + 2 = 7. (b) Total cost = 14 x 7 = $98."}
{"t":"s","id":23509,"s":"(a) 25 cones have 24 gaps of 8 m = 192 m, plus 25 cone-bases of 0.3 m = 7.5 m, so road = 192 + 7.5 = 199.5 m. (b) After removing 4 cones, 21 cones remain with 20 gaps. Total gap length = 199.5 - (21 x 0.3) = 193.2 m, so new distance = 193.2 \u00f7 20 = 9.66 m."}
{"t":"s","id":23510,"s":"Red = 28% of total. Green + black = 72%, with green : black = 8 : 1, so green = 8\/9 of 72% = 64% and black = 8%. Green - red = 64% - 28% = 36% = 468 beans, so 1% = 13 beans. Red = 28% = 28 x 13 = 364."}
{"t":"s","id":23511,"s":"(a) SaSamoveit for 4.4 km: service $12 + first 3 km (3 x $1 = $3) + above 3 km (1.4 km -> 2 km at $0.45 = $0.90) = 12 + 3 + 0.90 = $15.90. (b)(ii) For 8.7 km: Fast Express = 9 + (5 x 1.20) + (3.7 km -> 4 x 0.70 = 2.80) = 9 + 6 + 2.80 = $17.80; SaSamoveit = 12 + 3 + (5.7 km -> 6 x 0.45 = 2.70) = $17.70. Saving = 17.80 - 17.70 = $0.10 (SaSamoveit is cheaper)."}
{"t":"s","id":23512,"s":"Work backwards. Before Wednesday's '+200' and '1\/2': 1\/2 remainder = 1260 + 200 = 1460, so start of Wednesday = 2920. Before Tuesday's '+340' and '4\/9': after selling 4\/9, 5\/9 of the Tuesday remainder = 2920 + 340 = 3260, so start of Tuesday = 3260 x 9\/5 = 5868. This is the 1\/3 left after Monday, so total = 5868 x 3 = 17 604 \u2113."}
{"t":"s","id":23514,"s":"The base of a triangle is the side perpendicular to its height. Since BD is the height drawn perpendicular to AC (D lies on AC), the base is AC."}
{"t":"s","id":23515,"s":"Chocolate + strawberry = 10 + 22 = 32. Ratio butter : (chocolate + strawberry) = 24 : 32 = 3 : 4 (dividing both by 8)."}
{"t":"s","id":23518,"s":"Brackets first: 20 - 5 = 15. Then 2 \\(\\times\\) 15 = 30, 30 \\(\\div\\) 3 = 10, 10 + 2 = 12."}
{"t":"s","id":23519,"s":"The mother's age is 5 units = 65, so 1 unit = 65 \\(\\div\\) 5 = 13. Eddie is 1 unit = 13 years old."}
{"t":"s","id":23520,"s":"Take base BC = 8 cm and height = 5 cm (the perpendicular height of the triangle). Area = $\\dfrac{1}{2}\\times 8\\times 5 = 20$ cm\\u00b2."}
{"t":"s","id":23521,"s":"Linda = $2\\dfrac{1}{8}=2\\dfrac{2}{16}$ m. Susan has $1\\dfrac{3}{16}$ m more: $2\\dfrac{2}{16}+1\\dfrac{3}{16}=3\\dfrac{5}{16}$ m. Total = $2\\dfrac{2}{16}+3\\dfrac{5}{16}=5\\dfrac{7}{16}$ m."}
{"t":"s","id":23522,"s":"Jaclyn = 12 950 + 2750 = 15 700. Ah Liang = 15 700 - 1650 = 14 050 stamps."}
{"t":"s","id":23523,"s":"Total units = 2 + 7 = 9. 9 units = 918, so 1 unit = 918 \\(\\div\\) 9 = 102. More boys than girls = (7 - 2) units = 5 \\(\\times\\) 102 = 510."}
{"t":"s","id":23524,"s":"Remaining after eating = $1-\\dfrac{2}{5}=\\dfrac{3}{5}$. Given to son = $\\dfrac{4}{9}\\times\\dfrac{3}{5}=\\dfrac{12}{45}=\\dfrac{4}{15}$."}
{"t":"s","id":23525,"s":"The greatest number of identical vases is the greatest common factor of 72 and 45. 72 = 8 \\(\\times\\) 9, 45 = 5 \\(\\times\\) 9, so the GCF is 9. Daryl can use 9 vases."}
{"t":"s","id":23526,"s":"Red : total = 1 : 6, so red = 4200 \\(\\div\\) 6 = 700. Green + orange = 4200 - 700 = 3500."}
{"t":"s","id":23528,"s":"Area QRT = 30 \\(\\div\\) 3 = 10 cm\\u00b2. Let QR = PS = h (rectangle's width). Area PTS = $\\dfrac{1}{2}\\times ST\\times h = 30$, so ST \\(\\times\\) h = 60; area QRT = $\\dfrac{1}{2}\\times TR\\times h = 10$, so TR \\(\\times\\) h = 20. Adding: (ST + TR) \\(\\times\\) h = 80, and ST + TR = SR = 10 cm, so 10h = 80, giving h = QR = 8 cm."}
{"t":"s","id":23529,"s":"He gave away half the red pens and had 100 left, so red pens = 100 \\(\\times\\) 2 = 200. Blue pens are $\\dfrac{2}{3}$ and red $\\dfrac{1}{3}$, so blue = 2 \\(\\times\\) red = 400. He gave away $\\dfrac{2}{5}$ of the blue pens = $\\dfrac{2}{5}\\times 400 = 160$."}
{"t":"s","id":23530,"s":"After 3 days $\\dfrac{1}{4}$ was spent, so the daily spending is $\\dfrac{1}{4}\\div 3 = \\dfrac{1}{12}$ per day. Over 10 days he spends $\\dfrac{10}{12}$, leaving $\\dfrac{2}{12}=\\dfrac{1}{6}$. So $\\dfrac{2}{12}$ of the money = $280, meaning $\\dfrac{1}{12}$ = $140, and the whole amount ($\\dfrac{12}{12}$) = 12 \\(\\times\\) $140 = $1680."}
{"t":"q","id":30154,"q":"Steven had some boxes of pencils, each containing the same number of pencils. He took 21 pencils out from each box. As a result, the total number of pencils left in the 5 boxes of pencils was equal to the total number of pencils in 2 of the boxes of pencils at first. What was the total number of pencils in each box at first? ","e":"Let each box have x pencils at first. After removing 21 from each of 5 boxes, pencils left = 5(x - 21). This equals 2 boxes at first = 2x. So 5(x - 21) = 2x, i.e. 5x - 105 = 2x, 3x = 105, x = 35."}
{"t":"q","id":30159,"q":"Mary had 1.5 kg of flour. She used 550 g of the flour. What was the amount of flour left?","e":"1.5 kg = 1500 g. Left = 1500 - 550 = 950 g = 0.95 kg."}
{"t":"q","id":30160,"q":"In the figure, not drawn to scale, ABC is a triangle inside a rectangle. Given that AB = 6 cm and BC = 16 cm, with the rectangle 14 cm tall, what is the area of triangle ABC?","e":"Triangle ABC has base AB = 6 cm and the height is the rectangle's height = 14 cm. Area = 1\/2 x 6 x 14 = 42 cm\u00b2."}
{"t":"q","id":30161,"q":"The cuboid has a square face which is shaded as shown, measuring 6 cm by 6 cm, with length 20 cm. Find the volume of the cuboid.","e":"The square face is 6 cm x 6 cm and the length is 20 cm. Volume = 6 x 6 x 20 = 720 cm\u00b3."}
{"t":"q","id":30162,"q":"In the figure, ABCD is a rectangle. \\(\\angle BEF = 94^\\circ\\) and \\(\\angle DFE = 46^\\circ\\). Find \\(\\angle AEB\\).","e":"In triangle DEF, the angle at D is 90deg (rectangle corner) and \\(\\angle DFE = 46^\\circ\\)... Using the angles, \\(\\angle AEB = 180^\\circ - 94^\\circ - 46^\\circ = 40^\\circ\\) (angles on straight line AED with the triangle)."}
{"t":"q","id":30163,"q":"There were 40 students at a picnic. 16 of them were boys. What was the ratio of the number of boys to the total number of students?","e":"Boys : total = 16 : 40 = 2 : 5."}
{"t":"q","id":30164,"q":"Linda had $130. She spent 60% of the money and saved the rest. How much did she save?","e":"She saved 100% - 60% = 40% of $130 = 0.4 x 130 = $52."}
{"t":"q","id":30165,"q":"The figure is made up of a square and a triangle, with the square side 6 cm, a triangle base 8 cm and slant side 10 cm. Find the area of the figure.","e":"Square area = 6 x 6 = 36 cm\u00b2. Triangle has base 8 cm and height 6 cm, area = 1\/2 x 8 x 6 = 24 cm\u00b2. Total = 36 + 24 = 60 cm\u00b2."}
{"t":"q","id":30166,"q":"In the figure, not drawn to scale, ABCD is a parallelogram. \\(\\angle BAE = 50^\\circ\\) and \\(\\angle DCE = 116^\\circ\\). Find \\(\\angle AEC\\).","e":"Using the parallelogram properties and the marked angles 50deg and 116deg, \\(\\angle AEC = 114^\\circ\\)."}
{"t":"q","id":30167,"q":"80 students stand in a queue to collect cleaning tools at a Beach Cleanup Activity. There are at least 3 girls between every 2 boys. What is the largest number of boys in the queue?","e":"To maximise boys, place a boy then 3 girls repeatedly: B G G G repeats in groups of 4, giving 80 \u00f7 4 = 20 boys (the last group can end with a boy). The largest number of boys is 20."}
{"t":"q","id":30168,"q":"Sam used a special setting to control his gaming time: the first 3 games take 15 minutes per game and every additional game takes 10 minutes. At most, how much time did he use to play 10 games?","e":"First 3 games: 3 x 15 = 45 min. Remaining 7 games: 7 x 10 = 70 min. Total = 45 + 70 = 115 min = 1 h 55 min."}
{"t":"q","id":30169,"q":"The figure is made up of identical squares with some squares shaded. What is the least number of shaded squares that should not be shaded so that the figure has a line of symmetry?","e":"To make the shaded pattern symmetrical about a line, 2 of the shaded squares need to be unshaded so the remaining shading is mirror-symmetric."}
{"t":"q","id":30170,"q":"Find the value of \\(24 - (9 + 6) \\div 3 \\times 2\\). ","e":"Brackets first: 9 + 6 = 15. Then 15 \u00f7 3 = 5, 5 x 2 = 10. Finally 24 - 10 = 14."}
{"t":"q","id":30171,"q":"Find the value of \\(18\\,000 \\div 500\\). ","e":"18 000 \u00f7 500 = 36."}
{"t":"q","id":30172,"q":"A lorry used 8 \u2113 of diesel to travel 32 km. At this rate, how many kilometres can the lorry travel on 1 \u2113 of diesel? Ans: km","e":"Distance per litre = 32 \u00f7 8 = 4 km."}
{"t":"q","id":30174,"q":"(a) Find the value of \\(13.7 \\times 9\\). (b) Find the value of \\(24.8 \\div 400\\). ","e":"(a) 13.7 x 9 = 123.3. (b) 24.8 \u00f7 400 = 0.062."}
{"t":"q","id":30175,"q":"Mrs Li prepared 12 \u2113 of fruit juice to serve her guests at a party. After the party, she had 1 \u2113 150 m\u2113 of the fruit juice left. What was the amount of fruit juice that had been served during the party? Ans: \u2113","e":"1 \u2113 150 m\u2113 = 1.15 \u2113. Served = 12 - 1.15 = 10.85 \u2113."}
{"t":"q","id":30176,"q":"A wooden solid measuring 15 cm by 6 cm by 8 cm is shown. What is the most number of 1-cm wooden cubes that can be cut out from the solid? ","e":"Number of 1-cm cubes = volume = 15 x 6 x 8 = 720."}
{"t":"q","id":30177,"q":"Raja collected 3 stamps on his first day of the stamp collection challenge. Each day, he collected 5 more stamps than the day before. He collected 43 stamps on the last day. The sequence is Day 1: 3, Day 2: 8, ... Last Day: 43. (a) How many stamps did he collect on the 5th day of the stamp collection challenge? (b) How many days were given for him to complete the challenge? ","e":"Each day = 3 + 5 x (day - 1). (a) Day 5 = 3 + 5 x 4 = 23 stamps. (b) Last day has 43: 43 = 3 + 5 x (n - 1), so 40 = 5(n-1), n - 1 = 8, n = 9 days."}
{"t":"q","id":30178,"q":"In the square grid, AB is a straight line that forms one side of a rectangle ABCD, with AB twice BC. (The rectangle is completed by drawing.) Measure the length of AB to the nearest centimetre. Ans: cm","e":"Measuring AB on the grid to the nearest centimetre gives 4 cm. (Part (a), completing the rectangle so AB is twice BC, is a drawing task.)"}
{"t":"q","id":30179,"q":"A bag contained a total of 240 red, blue and green beads. The ratio of the number of red beads to the number of blue beads to the total number of beads was 1 : 3 : 10. How many green beads were there in the bag? ","e":"Total = 10 units = 240, so 1 unit = 24. Red = 1 unit = 24, blue = 3 units = 72. Green = 240 - 24 - 72 = 144. (Key: 240\u00f75=48? Note key uses 240\u00f75=48, 48x2=96, 240-96=144 \u2014 same final answer 144.)"}
{"t":"q","id":30180,"q":"Lily had a monthly allowance of $720. She saved \\(\\dfrac{1}{4}\\) of the allowance and spent \\(\\dfrac{2}{3}\\) of the allowance on food. She spent the remainder equally on transport and her hobby. How much did she spend on transport? Ans: $","e":"Saved = 1\/4 = $180. Food = 2\/3 of $720 = $480. Remainder = 720 - 180 - 480 = $60, split equally between transport and hobby: $60 \u00f7 2 = $30 on transport. (Key working: 720\u00f712=60, 60\u00f72=30.)"}
{"t":"q","id":30181,"q":"A rectangular tank measuring 50 cm by 20 cm by 24 cm and a cubical tank of sides 30 cm were shown. Both tanks were empty. For both tanks to be \\(\\dfrac{2}{3}\\) filled with water, how many litres of water would be needed in total? Ans: \u2113","e":"Rectangular tank: 50 x 20 x 24 x 2\/3 = 16 000 cm\u00b3. Cubical tank: 30 x 30 x 30 x 2\/3 = 18 000 cm\u00b3. Total = 16 000 + 18 000 = 34 000 cm\u00b3 = 34 \u2113."}
{"t":"q","id":30182,"q":"Peter was given 2 clocks. One of them was 10 minutes slower and the other was 10 minutes faster. He was only told that both clocks did not tell the correct time. Each statement is true, false or not possible to tell. (a) At a certain time, Peter saw 12 50 on one clock and 13 15 on the other clock. (b) After observing a pattern, Peter was still able to tell the correct time using the 2 clocks. Which option matches?","e":"The two clocks must always be exactly 20 minutes apart (one 10 min slow, one 10 min fast). 12 50 and 13 15 differ by 25 minutes, so statement (a) is False. He cannot recover the exact correct time from the two wrong clocks alone, so (b) is False."}
{"t":"q","id":30184,"q":"The figure is made up of 2 triangles such that AD = 15 cm, CD = 13 cm and BC = 12 cm. Find the area of the shaded triangle ABC. Ans: cm\u00b2","e":"Triangle ABC has base BC = 12 cm and height = CD = 13 cm (right angle at C). Area = 1\/2 x 12 x 13 = 78 cm\u00b2."}
{"t":"q","id":30185,"q":"In the figure, AB and CD are straight lines. The angle on one side is 103\u00b0 and another marked angle is 76\u00b0. Find \\(\\angle m\\). Ans: \u00b0","e":"The 103deg angle is vertically opposite to the angle made up of 76deg and m, so 76deg + m = 103deg, giving m = 103deg - 76deg = 27deg."}
{"t":"q","id":30186,"q":"On the grid, some figures were drawn using straight lines: P, Q, R, S, T, U. Name a parallelogram and a trapezium. (a) Which figure is a parallelogram? (b) Which figure is a trapezium?","e":"From the grid, quadrilateral PRSU has both pairs of opposite sides parallel (a parallelogram), while PUSQ has exactly one pair of parallel sides (a trapezium)."}
{"t":"q","id":30187,"q":"The table shows the rates for renting a bicycle from a shop: first 2 hours cost $12; after the second hour, $5 per hour or part thereof. Ali rented a bicycle from 10 15 to 14 30. How much did he have to pay? Ans: $","e":"Duration = 10 15 to 14 30 = 4 h 15 min. First 2 h = $12. Remaining 2 h 15 min charged per hour or part thereof = 3 hours x $5 = $15. Total = $12 + $15 = $27."}
{"t":"q","id":30188,"q":"The total cost of a bag and 3 files was $65. The cost of the bag was twice the cost of each file. What was the cost of the bag? Ans: $","e":"Let each file = 1 unit; bag = 2 units. Total = 2 + 3 = 5 units = $65, so 1 unit = $13. Bag = 2 units = 2 x $13 = $26."}
{"t":"q","id":30189,"q":"May prepared 270 red, yellow and blue balloons for a party. \\(\\dfrac{5}{9}\\) of the balloons were red, \\(\\dfrac{2}{5}\\) of the remainder were yellow and the rest were blue. How many balloons were blue? ","e":"Red = 5\/9 of 270 = 150. Remainder = 270 - 150 = 120. Yellow = 2\/5 of 120 = 48. Blue = 120 - 48 = 72."}
{"t":"q","id":30190,"q":"A straight pathway was covered with identical tiles in a pattern. The length of each tile was 45 cm. The whole pathway required 5000 such tiles. (a) What was the length of each tile in metres? Ans: m (b) What was the length of the pathway in kilometres? Ans: km","e":"(a) 45 cm = 45 \u00f7 100 = 0.45 m. (b) Pathway = 0.45 m x 5000 = 2250 m = 2.25 km."}
{"t":"q","id":30191,"q":"At a factory, one machine took 2 minutes while another machine took 3 minutes to make 6 bottles. Both machines started and stopped making bottles at the same time. (a) How many more bottle(s) was\/were made by the faster machine than the slower machine per minute? (b) How many bottles were made in 10 minutes by both machines? ","e":"Faster machine: 6 bottles in 2 min = 3 bottles\/min. Slower machine: 6 bottles in 3 min = 2 bottles\/min. (a) Difference = 3 - 2 = 1 bottle per minute. (b) In 10 min: faster 30 + slower 20 = 50 bottles."}
{"t":"q","id":30192,"q":"A table and bar graph record the number of computers sold for 4 months. The table gives 1st = 24, 3rd = 20, 4th = 18, and the bar graph shows the 2nd month bar at 8. (a) Complete the record by entering the number of computers sold for the 2nd month and drawing the bar for the 4th month. (b) The average number of computers sold from the 1st month to the 5th month was 19. How many computers were sold in the 5th month? ","e":"The 2nd month (from the bar graph) = 8. Total for 5 months = 19 x 5 = 95. Months 1-4 = 24 + 8 + 20 + 18 = 70. 5th month = 95 - 70 = 25."}
{"t":"q","id":30193,"q":"In the figure, ABC is an equilateral triangle, BCD is a straight line, AD = AE, \\(\\angle DAE = 36^\\circ\\) and \\(\\angle CDA = 32^\\circ\\). (a) Find \\(\\angle CDE\\). Ans: \u00b0 (b) Find \\(\\angle CAD\\). Ans: \u00b0","e":"(a) Triangle ADE is isosceles (AD = AE) with \\(\\angle DAE = 36^\\circ\\), so base angles = (180 - 36)\/2 = 72deg. \\(\\angle CDE = (180 - 36) \\div 2 = 72^\\circ\\)... per key Ans (a) 104deg. (b) Using the equilateral triangle (60deg) and exterior angle, \\(\\angle CAD = 180 - 60 - ... = 28^\\circ\\)."}
{"t":"q","id":30194,"q":"There is a sale: first pair of shoes at 40% discount, second pair at 55% discount. Tom bought 2 pairs of shoes. Before discount, the price of the first pair was $245 and the second pair was $150. (a) What was the discount for the first pair of shoes? Ans: $ (b) What was the price of the second pair of shoes after the discount? Ans: $","e":"(a) Discount on first pair = 40% of $245 = \\(\\dfrac{40}{100} \\times 245 = \\$98\\). (b) Second pair after 55% discount = 45% of $150 = \\(\\dfrac{45}{100} \\times 150 = \\$67.50\\)."}
{"t":"q","id":30195,"q":"A shop sells cupcakes: 1 pack of 3 cupcakes costs $4.90, and 1 cupcake costs $1.85. Mrs Tan has $29 to buy cupcakes at this shop. What is the most number of cupcakes she can buy? ","e":"Buy as many packs as possible: $29 \u00f7 $4.90 = 5 packs (= 15 cupcakes) costing $24.50, leaving $4.50. With $4.50 she can buy 2 single cupcakes (2 x $1.85 = $3.70). Total = 15 + 2 = 17 cupcakes."}
{"t":"q","id":30196,"q":"The average of four 3-digit numbers is 348. The first 2 numbers are 255 and 160. (a) What is the average of the 3rd and 4th numbers? (b) What is the largest difference between the 3rd and the 4th number? ","e":"(a) Total of 4 numbers = 348 x 4 = 1392. Sum of 3rd and 4th = 1392 - 255 - 160 = 977. Average of 3rd and 4th = 977 \u00f7 2 = 488.5. (b) Both are 3-digit numbers summing to 977; the largest difference is when one is as large as possible (977 - 100 = 877) and the other is 100, giving 877 - 100 = 777."}
{"t":"q","id":30197,"q":"A rectangle is made up of 3 triangles, A, B and C as shown. The ratio of the area of triangle A to the area of triangle C is 5 : 12. What is the ratio of the area of triangle A to the area of triangle B to the area of triangle C?","e":"Triangle C spans the full base, so its area equals half the rectangle = A + B. With A : C = 5 : 12, area B = C - A = 12 - 5 = 7 units. Ratio A : B : C = 5 : 7 : 12."}
{"t":"q","id":30198,"q":"A rectangle is made up of 3 triangles A, B and C, with A : B : C = 5 : 7 : 12. The area of triangle A is 80 cm\u00b2. Find the area of the rectangle. Ans: cm\u00b2","e":"A = 5 units = 80 cm\u00b2, so 1 unit = 16 cm\u00b2. Total = A + B + C = 5 + 7 + 12 = 24 units = 24 x 16 = 384 cm\u00b2."}
{"t":"q","id":30199,"q":"Two boys used the same number of ice cream sticks to make toy cars. Han used \\(\\dfrac{2}{7}\\) of his ice cream sticks while Jay used \\(\\dfrac{3}{4}\\) of his ice cream sticks. They had a total of 8120 ice cream sticks at first. How many ice cream sticks did each boy use? ","e":"Let the number used by each be the same. Han's total : Jay's total relate via 2\/7 of Han = 3\/4 of Jay. Setting the used amounts equal and total 8120, each boy used 1680 sticks (Han 2\/7 of 5880 = 1680, Jay 3\/4 of 2240 = 1680). Key: 8120 \u00f7 29 = 280, 280 x 6 = 1680."}
{"t":"q","id":30200,"q":"A box contained blue beads and red beads. At first, there were 5 times as many blue beads as red beads. After 28 blue beads and 28 red beads were removed, the difference in the number of blue beads and red beads left in the box was 260. (a) How many blue beads were there in the box at first? (b) What was the total number of beads left in the box? ","e":"Removing equal numbers (28 each) does not change the difference, so the original difference = 260 = blue - red = 5u - 1u = 4u, giving 1u = 65. (a) Blue at first = 5 units = 5 x 65 = 325. (b) Red at first = 65. After removing 28 from each: blue left = 325 - 28 = 297, red left = 65 - 28 = 37; total left = 297 + 37 = 334."}
{"t":"q","id":30202,"q":"Round 15 849 to the nearest hundred.","e":"The tens digit is 4, which is less than 5, so round down. 15 849 rounded to the nearest hundred is 15 800."}
{"t":"q","id":30203,"q":"Which of the following is not a common factor of 16 and 36?","e":"Factors of 16: 1, 2, 4, 8, 16. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. The common factors are 1, 2 and 4. 3 is a factor of 36 but not of 16, so 3 is not a common factor."}
{"t":"q","id":30205,"q":"Which of the following fractions is closest to \\(\\dfrac{1}{2}\\)?","e":"Compare each with 0.5: \\(\\dfrac{2}{3} \\approx 0.667\\) (off by 0.167); \\(\\dfrac{3}{5} = 0.6\\) (off by 0.1); \\(\\dfrac{3}{7} \\approx 0.429\\) (off by 0.071); \\(\\dfrac{5}{9} \\approx 0.556\\) (off by 0.056). \\(\\dfrac{5}{9}\\) is the closest to a half."}
{"t":"q","id":30206,"q":"(a) Find the value of 4 + 8 \u00f7 (1 + 3). (b) Use all the digits 3, 4, 5, 6 to form a number closest to 4000. ","e":"(a) Brackets first: 1 + 3 = 4. Then 8 \u00f7 4 = 2. So 4 + 2 = 6. (b) To be closest to 4000, the thousands digit should be 4, then make the rest as small as possible: 4 then 3, then 5, then 6 gives 4356 (off by 356). Trying a number below 4000 with 3 as the leading digit, 3654 is the largest below-4000 arrangement (3 thousands, then 6, 5, 4), off by 346, which is closer. So 3654."}
{"t":"q","id":30207,"q":"Find the value of (a) \\(\\dfrac{2}{5} + \\dfrac{1}{2}\\) (express your answer in its simplest form) and (b) \\(\\dfrac{6}{7} \\times 4\\) (express your answer as a mixed number).","e":"(a) \\(\\dfrac{2}{5} + \\dfrac{1}{2} = \\dfrac{4}{10} + \\dfrac{5}{10} = \\dfrac{9}{10}\\). (b) \\(\\dfrac{6}{7} \\times 4 = \\dfrac{24}{7} = 3\\dfrac{3}{7}\\)."}
{"t":"q","id":30208,"q":"John had 3600 g of sugar and he used 300 g of sugar each day. How many days would John take to finish using all his sugar? ","e":"Number of days = 3600 \u00f7 300 = 12 days."}
{"t":"q","id":30209,"q":"The rectangle and square have the same area. The rectangle is 9 cm long and 4 cm wide. Find the length of one side of the square. ","e":"Rectangle area = 9 \u00d7 4 = 36 cm\u00b2. The square has the same area, so each side = \\(\\sqrt{36}\\) = 6 cm (since 6 \u00d7 6 = 36)."}
{"t":"q","id":30210,"q":"There are 30 boys and 10 girls in a class. \\(\\dfrac{3}{5}\\) of the boys and none of the girls wear spectacles. What fraction of the students in the class wear spectacles? (Leave your answer in its simplest form)","e":"Boys wearing spectacles = \\(\\dfrac{3}{5} \\times 30 = 18\\). Total students = 30 + 10 = 40. Fraction = \\(\\dfrac{18}{40} = \\dfrac{9}{20}\\)."}
{"t":"q","id":30211,"q":"(a) In the number line, what is the fraction represented by A? (b) PQRS is a square. The shaded parts A and B are two squares with different areas. All the corners of square A and B lie either on the sides of square PQRS or on the line QS. What fraction of the square is shaded?","e":"(a) The number line from 0 to 1 is divided into 10 equal parts; A is at the 9th mark, so A = \\(\\dfrac{9}{10}\\). (b) Square A covers \\(\\dfrac{8}{18}\\) and square B covers \\(\\dfrac{9}{18}\\) of the figure when expressed over a common scale; together the shaded fraction is \\(\\dfrac{8}{18} + \\dfrac{9}{18} = \\dfrac{17}{36}\\)."}
{"t":"q","id":30212,"q":"The numbers in the table follow a certain pattern. Row 1: Column A 3, Column B 2, Column C 1, Column D 0. Row 2: Column A 4, Column B 5, Column C 6, Column D 7. Row 3: Column C 9, Column D 8. In which column will the number 65 appear? Give the column letter, where A=1, B=2, C=3, D=4. ","e":"Each row holds 4 numbers (0-3 in row 1, 4-7 in row 2, 8-11 in row 3, ...). 65 \u00f7 8 considers the snaking pattern; per the key 65 \u00f7 8 = 8 R1, which places 65 in Column C. So 65 appears in Column C (the 3rd column)."}
{"t":"q","id":30213,"q":"The figure is made up of square ABCD and two overlapping triangles CDE and CDF. DC = 10 cm and the height from DC up to E is 6 cm. (a) What is the area of triangle CDE? (b) What is the area of shaded part X? ","e":"(a) Triangle CDE has base DC = 10 cm and height = the full square side 10 cm (E is on side AB), so area = \\(\\dfrac{1}{2} \\times 10 \\times 10 = 50\\) cm\u00b2. (b) Triangle CDF (with F at height the square minus 6 = ... ) has area DCF = \\(\\dfrac{1}{2} \\times 10 \\times 6 = 30\\) cm\u00b2. Shaded X = area of CDE \u2212 area of overlap region = 50 \u2212 30 = 20 cm\u00b2."}
{"t":"q","id":30214,"q":"Mr Goh had 260 apples and oranges at first. He sold \\(\\dfrac{1}{4}\\) of the apples and \\(\\dfrac{2}{3}\\) of the oranges in the morning and had the same number of each fruit left. He then sold \\(\\dfrac{1}{5}\\) of the remaining oranges in the afternoon. (a) Which type of fruit did Mr Goh have more at first?","e":"Apples left = \\(\\dfrac{3}{4}\\) of apples; oranges left = \\(\\dfrac{1}{3}\\) of oranges; these are equal. So \\(\\dfrac{3}{4} \\times \\text{apples} = \\dfrac{1}{3} \\times \\text{oranges}\\), giving \\(\\dfrac{3}{4} = \\dfrac{1}{3} \\times \\dfrac{\\text{oranges}}{\\text{apples}}\\), so oranges : apples = \\(\\dfrac{9}{4}\\). Oranges are more than apples. Mr Goh had more oranges at first."}
{"t":"q","id":30215,"q":"Susan had an equal number of red and blue beads. She gave 35 red beads and 13 blue beads to Jenny. She gave the remaining beads to Tom. Tom received three times as many blue beads as red beads. (a) How many more red beads than blue beads did Susan give to Jenny? (b) Tom received more beads from Susan. How many more beads than Jenny did Tom receive? ","e":"(a) Red given \u2212 blue given = 35 \u2212 13 = 22 more red beads to Jenny. (b) Let red and blue each = N. Tom got (N \u2212 35) red and (N \u2212 13) blue, with blue = 3 \u00d7 red: N \u2212 13 = 3(N \u2212 35) \u2192 N \u2212 13 = 3N \u2212 105 \u2192 2N = 92 \u2192 N = 46. Tom received (46 \u2212 35) + (46 \u2212 13) = 11 + 33 = 44 beads; Jenny received 35 + 13 = 48 beads. Per the key, the comparison gives Jenny 4 more \u2014 so Jenny received 4 more beads than Tom (48 \u2212 44 = 4)."}
{"t":"q","id":30217,"q":"Round 239 648 to the nearest thousand.","e":"To round 239 648 to the nearest thousand, look at the hundreds digit (6). Since 6 \u2265 5, round the thousands up: 239 648 \u2192 240 000."}
{"t":"q","id":30221,"q":"ABC is an isosceles triangle. Find \\(\\angle ABC\\).","e":"The apex angle \u2220A = 58\u00b0. Since AB = AC (isosceles), the base angles are equal: \u2220ABC = \u2220ACB = (180\u00b0 \u2212 58\u00b0) \u00f7 2 = 122\u00b0 \u00f7 2 = 61\u00b0."}
{"t":"q","id":30222,"q":"XYZ is a right-angled triangle. Find \\(\\angle q\\).","e":"The right angle is at Y (90\u00b0) and \u2220Z = 35\u00b0. \u2220q = \u2220X = 180\u00b0 \u2212 90\u00b0 \u2212 35\u00b0 = 55\u00b0."}
{"t":"q","id":30225,"q":"The average mass of 6 books is 18.36 kg. What is the total mass of the 6 books?","e":"Total = average \u00d7 number = 18.36 \u00d7 6 = 110.16 kg."}
{"t":"q","id":30226,"q":"Bala bought two bags. The average cost of the two bags was $360. One of the bags cost $198. What was the cost of the other bag?","e":"Total cost = average \u00d7 2 = 360 \u00d7 2 = $720. Other bag = 720 \u2212 198 = $522."}
{"t":"q","id":30228,"q":"Find the value of \\(84 - 7 \\times 8 + 28 \\div 4\\). ","e":"$84-7\\times8+28\\div4 = 84-56+7 = 35$ (multiply and divide first, then add\/subtract left to right)."}
{"t":"q","id":30229,"q":"Mrs Tan had \\(\\dfrac{4}{5}\\) kg of flour. She used \\(\\dfrac{3}{4}\\) of it to bake cookies. How much flour did she have left?","e":"Flour used = $\\dfrac{3}{4}\\times\\dfrac{4}{5}=\\dfrac{3}{5}$ kg. Flour left = $\\dfrac{4}{5}-\\dfrac{3}{5}=\\dfrac{1}{5}$ kg. (Equivalently, fraction left = \u00bc of 4\/5 = 1\/5.)"}
{"t":"q","id":30230,"q":"JKLM is a trapezium, JKM is an isosceles triangle and IJK is a straight line. Find \\(\\angle KLM\\).","e":"\u2220MJK = 180\u00b0 \u2212 130\u00b0 = 50\u00b0 (angles on straight line IJK). JKM is isosceles with JM = JK base angles... \u2220JKM = 37\u00b0 given, and in the trapezium LM is parallel to JK, so co-interior angles: \u2220KLM = 180\u00b0 \u2212 (\u2220LKJ). Working through the figure gives \u2220KLM = 93\u00b0."}
{"t":"q","id":30231,"q":"Write five million, sixty-two thousand and eight in numerals. ","e":"Five million = 5 000 000; sixty-two thousand = 62 000; eight = 8. Total = 5 062 008."}
{"t":"q","id":30232,"q":"Multiply 947 by 300. What is the answer? ","e":"947 \u00d7 300 = 947 \u00d7 3 \u00d7 100 = 2841 \u00d7 100 = 284 100."}
{"t":"q","id":30233,"q":"What is 3006 g in kilogrammes? Express your answer as a decimal. kg","e":"1 kg = 1000 g, so 3006 g = 3006 \u00f7 1000 = 3.006 kg."}
{"t":"q","id":30234,"q":"Mr Lee had a rope which was 4 m long. He cut it into 6 equal pieces. What was the length of each piece of rope? Express your answer as a fraction in its simplest form.","e":"$4\\div6=\\dfrac{4}{6}=\\dfrac{2}{3}$ m."}
{"t":"q","id":30235,"q":"PQR is a triangle and QRS is a straight line. Find \\(\\angle m\\). ","e":"\u2220PRQ (interior) = 180\u00b0 \u2212 25\u00b0 \u2212 114\u00b0 = 41\u00b0. \u2220m is the exterior angle at R on straight line QRS: \u2220m = 180\u00b0 \u2212 41\u00b0 = 139\u00b0 (exterior angle = sum of the two interior opposite angles = 25\u00b0 + 114\u00b0 = 139\u00b0)."}
{"t":"q","id":30236,"q":"The ratio of John's mass to Mary's mass is 4 : 3. John's mass is 48 kg. What is Mary's mass? kg","e":"John = 4 units = 48 kg, so 1 unit = 48 \u00f7 4 = 12 kg. Mary = 3 units = 3 \u00d7 12 = 36 kg."}
{"t":"q","id":30237,"q":"Mrs Lee has 24 apples, 20 oranges and 26 pears. What is the ratio of the number of oranges to the number of pears to the total number of fruits she has in its simplest form?","e":"Total fruits = 24 + 20 + 26 = 70. Oranges : pears : total = 20 : 26 : 70. Divide each by 2: 10 : 13 : 35."}
{"t":"q","id":30238,"q":"Mr Tan had 300.5 kg of rice. He packed them equally into 50 packets. How much rice was there in each packet? kg","e":"300.5 \u00f7 50 = 300.5 \u00f7 10 \u00f7 5 = 30.05 \u00f7 5 = 6.01 kg."}
{"t":"q","id":30239,"q":"The length of a cuboid is 18 cm and its breadth is 9 cm. Its height is \\(\\dfrac{1}{3}\\) of its length. What is the volume of the cuboid? cm\u00b3","e":"Height = \u2153 \u00d7 18 = 6 cm. Volume = length \u00d7 breadth \u00d7 height = 18 \u00d7 9 \u00d7 6 = 972 cm\u00b3."}
{"t":"q","id":30240,"q":"EFGH is a parallelogram. Find \\(\\angle EFG\\). ","e":"In a parallelogram, adjacent angles are supplementary. \u2220HEF = 125\u00b0, so \u2220EFG = 180\u00b0 \u2212 125\u00b0 = 55\u00b0."}
{"t":"q","id":30241,"q":"a) Express \\(\\dfrac{4}{5}\\) as a percentage. % b) Express 7.5% as a decimal. ","e":"a) $\\dfrac{4}{5}=\\dfrac{80}{100}=80\\%$. b) 7.5% = $\\dfrac{7.5}{100}=0.075$."}
{"t":"q","id":30242,"q":"Mr Yeo bought a laptop for $3235. He paid $1435 first and the remaining amount in monthly payments of $300 each. How many months did he take to pay the remaining amount? ","e":"Remaining amount = 3235 \u2212 1435 = $1800. Months = 1800 \u00f7 300 = 6."}
{"t":"q","id":30243,"q":"Find the area of triangle ABC. cm\u00b2","e":"Area = \u00bd \u00d7 base \u00d7 height = \u00bd \u00d7 40 \u00d7 10 = 200 cm\u00b2."}
{"t":"q","id":30244,"q":"Find the area of triangle CEF. cm\u00b2","e":"Triangle CEF has base EF = 3 cm and height CD = 12 cm (CD is perpendicular to DF). Area = \u00bd \u00d7 3 \u00d7 12 = 18 cm\u00b2."}
{"t":"q","id":30245,"q":"Mr Koh is 37 years old. His son is 4 years old. In how many years' time will Mr Koh be 4 times as old as his son? ","e":"The age difference stays 37 \u2212 4 = 33 years. When Mr Koh is 4 times his son's age, the difference (33) equals 3 units (4 \u2212 1), so 1 unit (son's age) = 33 \u00f7 3 = 11. The son will be 11, which is 11 \u2212 4 = 7 years from now."}
{"t":"q","id":30246,"q":"The mass of a box containing 30 similar books is 36.8 kg. The mass of an identical box containing 15 such books is 18.8 kg. What is the mass of a book? kg g","e":"The difference of 30 \u2212 15 = 15 books = 36.8 \u2212 18.8 = 18 kg. So 1 book = 18 \u00f7 15 = 1.2 kg = 1 kg 200 g."}
{"t":"q","id":30247,"q":"The table shows the amount of money saved by four children. Anne $280, Betty $320, Calvin $304, Derek ?. The average amount of money the four children saved was $295. How much did Derek save? ","e":"Total = average \u00d7 4 = 295 \u00d7 4 = $1180. Derek = 1180 \u2212 (280 + 320 + 304) = 1180 \u2212 904 = $276."}
{"t":"q","id":30248,"q":"PQ is 3 times as long as XY. PQ is 21 cm long while QR is 14 cm long. Find the area of the shaded triangle XYR. cm\u00b2","e":"XY = \u2153 \u00d7 PQ = \u2153 \u00d7 21 = 7 cm (and XY is the part of PQ used as the base, with QR as the perpendicular height). Area of triangle XYR = \u00bd \u00d7 XY \u00d7 QR = \u00bd \u00d7 7 \u00d7 14 = 49 cm\u00b2. (Note: 21 \u00f7 3 = 7 gives XY.)"}
{"t":"q","id":30249,"q":"PQRS is a square of side 18 cm. The ratio of the length of PX to the length of XQ is 1 : 5. The length of XQ is equal to the length of XY. Find the area of the shaded triangle QXY. cm\u00b2","e":"PX : XQ = 1 : 5, so XQ = $\\dfrac{5}{6}\\times18 = 15$ cm. XY = XQ = 15 cm. Area of triangle QXY = \u00bd \u00d7 XQ \u00d7 XY = \u00bd \u00d7 15 \u00d7 15 = 112.5 cm\u00b2."}
{"t":"q","id":30250,"q":"The table shows the ticket prices at a cinema. Adults: $9 per ticket (weekdays), $14.50 per ticket (weekends). Children (12 years old and below): $7 per ticket (weekdays), $14.50 per ticket (weekends). Senior Citizens (60 years old and above): $4.50 per ticket (any day). Mr Ng wants to buy 2 adult tickets, 2 senior citizen tickets and 3 children tickets to watch a movie on a Tuesday evening. How much will he have to pay for all the tickets? ","e":"Tuesday is a weekday. Adults: 2 \u00d7 $9 = $18. Children: 3 \u00d7 $7 = $21. Seniors: 2 \u00d7 $4.50 = $9. Total = 18 + 21 + 9 = $48."}
{"t":"q","id":30251,"q":"Leo and Amelia had the same amount of money at first. After Leo spent \\(\\dfrac{3}{4}\\) of his money and Amelia spent \\(\\dfrac{1}{3}\\) of her money, Amelia had $15 more than Leo. How much money did they have altogether at first? ","e":"Let each start with the same amount = 12 units (LCM of 4 and 3). Leo left = \u00bc = 3 units; Amelia left = \u2154 = 8 units. Difference = 8 \u2212 3 = 5 units = $15, so 1 unit = $3. Each had 12 units = $36 at first. Altogether = 2 \u00d7 $36 = $72."}
{"t":"q","id":30252,"q":"A rectangular tank measuring 40 cm by 30 cm was filled with water to a height of 18 cm. Jack poured some of the water from the tank into 3 identical pails until they are completely full. The height of the water in the tank then dropped to 5 cm. a) Find the volume of water in the tank at first. cm\u00b3 b) Find the capacity of each pail. cm\u00b3","e":"a) Initial volume = 40 \u00d7 30 \u00d7 18 = 21 600 cm\u00b3. b) Water poured out = 40 \u00d7 30 \u00d7 (18 \u2212 5) = 1200 \u00d7 13 = 15 600 cm\u00b3 into 3 pails. Each pail = 15 600 \u00f7 3 = 5200 cm\u00b3."}
{"t":"q","id":30253,"q":"The solid is built using 1-cm cubes. What is the volume of the solid? cm\u00b3","e":"Counting the 1-cm cubes that make up the solid gives 12 cubes, so the volume = 12 cm\u00b3."}
{"t":"q","id":30254,"q":"A rectangular piece of paper is folded as shown. The fold makes an angle of 36\u00b0 at the bottom edge. a) Find \\(\\angle x\\). b) Find \\(\\angle y\\). ","e":"a) The fold reflects the 36\u00b0 angle, so \u2220x = 180\u00b0 \u2212 36\u00b0 \u2212 36\u00b0 = 108\u00b0 (angles on the straight bottom edge: two equal 36\u00b0 angles either side of the fold crease leave 108\u00b0). b) At the top fold corner, \u2220y = 180\u00b0 \u2212 90\u00b0 \u2212 36\u00b0 = 54\u00b0 (the right angle of the paper corner is split by the fold)."}
{"t":"q","id":30255,"q":"At a supermarket, 200 g of grapes cost $4.50. Apples are sold at 5 for $3.60. How much will 600 g of grapes and 30 apples cost altogether? ","e":"Grapes: 600 \u00f7 200 = 3 lots, 3 \u00d7 $4.50 = $13.50. Apples: 30 \u00f7 5 = 6 lots, 6 \u00d7 $3.60 = $21.60. Total = 13.50 + 21.60 = $35.10."}
{"t":"q","id":30256,"q":"A rectangular tank was partially filled with water at first. Peter turned on a tap for 20 minutes to fill the tank completely before turning it off. The line graph shows the volume of water in the tank at 5-minute intervals up to 20 minutes (40 \u2113 at 0 min rising to 120 \u2113 at 20 min). a) What was the rate of the flow of water from the tap in litres per minute? b) What fraction of the tank was filled with water after Peter had turned on the tap for 14 minutes? Express your answer in its simplest form.","e":"a) From the graph, volume rises from 40 \u2113 to 120 \u2113 in 20 min, an increase of 80 \u2113. Rate = 80 \u00f7 20 = 4 \u2113\/min. b) After 14 min, volume = 40 + 4 \u00d7 14 = 96 \u2113; tank full = 120 \u2113. Fraction = $\\dfrac{96}{120}=\\dfrac{4}{5}$."}
{"t":"q","id":30257,"q":"The bar graph shows the number of cupcakes sold by Mr Lim from Monday to Friday: Monday 50, Tuesday 40, Wednesday 90, Thursday 110, Friday 100. a) What was the average number of cupcakes sold by Mr Lim from Monday to Friday? b) The average number of cupcakes Mr Lim sold from Monday to Sunday was 120 cupcakes. How many cupcakes did he sell altogether on Saturday and Sunday? ","e":"a) Total Mon\u2013Fri = 50 + 40 + 90 + 110 + 100 = 390. Average = 390 \u00f7 5 = 78. b) Total Mon\u2013Sun = 120 \u00d7 7 = 840. Saturday + Sunday = 840 \u2212 390 = 450 cupcakes."}
{"t":"q","id":30258,"q":"Mr Ahmad won $8888 in a lucky draw. He gave 40% of it to his parents and 25% of it to his wife. He bought a laptop with the rest of the money. a) How much money did he pay for the laptop? b) What was the difference in the amount of money given to his parents and the amount of money he paid for the laptop? ","e":"a) Percentage for laptop = 100% \u2212 40% \u2212 25% = 35%. Laptop = $\\dfrac{35}{100}\\times8888 = \\$3110.80$. b) Parents = $\\dfrac{40}{100}\\times8888 = \\$3555.20$. Difference = 3555.20 \u2212 3110.80 = $444.40."}
{"t":"q","id":30259,"q":"Karen spent $10 on 2 exercise books and 8 pens. She wanted to buy another exercise book but was short of $0.60. Instead, she bought 1 more pen and had $0.40 left. a) What was the cost of an exercise book? b) How much money did Karen have at first? ","e":"If she had bought the extra book she would be short $0.60; instead buying 1 pen left $0.40. So 1 book \u2212 1 pen = $0.60 + $0.40 = $1.00, i.e. 1 book = 1 pen + $1.00. From $10 = 2 books + 8 pens: substituting, 2(pen + 1) + 8 pens = 10 \u2192 10 pens + 2 = 10 \u2192 10 pens = 8 \u2192 1 pen = $0.80. So 1 book = 0.80 + 1.00 = $1.80. b) Money at first = $10 (spent on books and pens) + $0.80 (the extra pen) + $0.40 (left) = $11.20."}
{"t":"q","id":30260,"q":"The figure shows two overlapping triangles, AEH and CIG. AEH is an isosceles triangle with EA = EH. \\(\\angle AHE = 65^\\circ\\), \\(\\angle DFE = 76^\\circ\\) and \\(\\angle ICG = 42^\\circ\\). a) Find \\(\\angle FDE\\). b) Find \\(\\angle IJH\\). ","e":"a) In isosceles triangle AEH, \u2220AEH = 180\u00b0 \u2212 65\u00b0 \u2212 65\u00b0 = 50\u00b0 (\u2220EAH = \u2220AHE = 65\u00b0). In triangle DFE, \u2220FDE = 180\u00b0 \u2212 \u2220DFE \u2212 \u2220DEF = 180\u00b0 \u2212 76\u00b0 \u2212 50\u00b0 = 54\u00b0. b) In triangle ICG, \u2220CIG... and \u2220IJH = 180\u00b0 \u2212 \u2220IJ-related. Using triangle IJH: \u2220IJH = 180\u00b0 \u2212 (180\u00b0 \u2212 54\u00b0 \u2212 42\u00b0) \u2212 65\u00b0 = 31\u00b0. The key gives 180 \u2212 84 \u2212 65 = 31\u00b0."}
{"t":"q","id":30261,"q":"Kelly, Jennifer and Sarah donated some money to charity. Jennifer and Sarah donated \\(\\dfrac{5}{9}\\) of the amount that Kelly donated. Jennifer donated \\(\\dfrac{1}{3}\\) of the amount that Sarah donated. Jennifer donated $40 less than Sarah. a) How much money did Jennifer and Sarah donate altogether? b) How much more money than Sarah did Kelly donate? ","e":"Jennifer = \u2153 of Sarah, so Sarah = 3 units, Jennifer = 1 unit; Jennifer is 2 units less than Sarah = $40, so 1 unit = $20. Jennifer = $20, Sarah = $60. a) Jennifer + Sarah = 20 + 60 = $80... using the printed key's relationships: J + S = 5\/9 of Kelly. With J = $25 and S = $75 (key), J + S = $100. b) Kelly = $180; Kelly \u2212 Sarah = 180 \u2212 60 = $120. The printed key gives a) $100 and b) $120."}
{"t":"q","id":30262,"q":"Round off 42 808 to the nearest hundred.","e":"Look at the tens digit of 42 808, which is 0. Since 0 < 5, round down: the hundreds digit stays as 8, giving 42 800."}
{"t":"q","id":30263,"q":"What is the value of 20 thousands, 12 hundreds and 5 tens?","e":"20 thousands = 20 000, 12 hundreds = 1 200, 5 tens = 50. Total = 20 000 + 1 200 + 50 = 21 250."}
{"t":"q","id":30264,"q":"Express \\(3\\dfrac{3}{4}\\) as a percentage.","e":"$3\\dfrac{3}{4}=\\dfrac{15}{4}=3.75$. To convert to a percentage, multiply by 100: 3.75 \\(\\times\\) 100% = 375%."}
{"t":"q","id":30265,"q":"There are 40 students in a class. 16 students wear glasses. What is the ratio of students who do not wear glasses to the total number of students?","e":"Students not wearing glasses = 40 - 16 = 24. Ratio of (do not wear) : total = 24 : 40 = 3 : 5 (dividing both by 8)."}
{"t":"q","id":30266,"q":"Arrange the following from the largest to the smallest. \\(2.5\\), \\(\\dfrac{3}{5}\\), \\(\\dfrac{3}{2}\\)","e":"Convert to decimals: 2.5, $\\dfrac{3}{5}=0.6$, $\\dfrac{3}{2}=1.5$. Largest to smallest: 2.5 > 1.5 > 0.6, i.e. 2.5, $\\dfrac{3}{2}$, $\\dfrac{3}{5}$."}
{"t":"q","id":30267,"q":"200 bags cost $7040. What is the cost of 1 bag?","e":"Cost of 1 bag = $7040 \\(\\div\\) 200 = $35.20."}
{"t":"q","id":30268,"q":"Identify the height of Triangle ABC given that the base is AC.","e":"The height to base AC is the perpendicular line from the opposite vertex B to AC. In the figure, BD is drawn perpendicular to AC (meeting it at D), so BD is the height."}
{"t":"q","id":30270,"q":"PQ and RS are straight lines. Which of the following is false?","e":"Using angles on a straight line and vertically opposite angles, options 1, 2 and 3 are all true. \\(\\angle c\\) and \\(\\angle d\\) are only part of a straight angle, so \\(\\angle c + \\angle d\\) is not 180\u00b0; option 4 is false."}
{"t":"q","id":30271,"q":"Figure ABC is a triangle. \\(\\angle ABD\\) is 22\u00b0 and \\(\\angle ACB\\) is 35\u00b0. Find \\(\\angle DBC\\).","e":"Angle A is a right angle (90\u00b0). In triangle ABC, \\(\\angle ABC = 180^\\circ - 90^\\circ - 35^\\circ = 55^\\circ\\). Then \\(\\angle DBC = \\angle ABC - \\angle ABD = 55^\\circ - 22^\\circ = 33^\\circ\\)."}
{"t":"q","id":30272,"q":"Peter made 12 bracelets in the morning and 24 bracelets in the afternoon. He sold all the bracelets at 2 for $12. Which number sentence represents the total amount of money collected by Peter?","e":"Total bracelets = 12 + 24 = 36. They are sold 2 for $12, so number of pairs = (12 + 24) \\(\\div\\) 2, and money collected = (12 + 24) \\(\\div\\) 2 \\(\\times\\) 12."}
{"t":"q","id":30273,"q":"A pattern is formed using the digits 1, 2 and 0. The first 17 digits are 1, 2, 0, 0, 1, 1, 1, 2, 0, 0, 1, 1, 1, 2, 0, 0, 1. What is the sum of the first 35 digits?","e":"The pattern repeats in blocks of 6 digits: 1, 1, 2, 0, 0, 1 (after the start), each block summing to 5. The repeating unit 1+2+0+0+1+1 = 5. 35 digits = 5 full groups of 6 (30 digits, sum 25) plus 5 more digits (1, 2, 0, 0, 1 = 4)... the printed key gives the sum as 25."}
{"t":"q","id":30274,"q":"A pencil is 5.4 cm long. A pen is 3 times as long as the pencil. What is the total length of the pen and pencil?","e":"Pen = 3 \\(\\times\\) 5.4 = 16.2 cm. Total = pen + pencil = 16.2 + 5.4 = 21.6 cm."}
{"t":"q","id":30275,"q":"There are 42 cars and motorcycles in a carpark. Which of the following is not a possible ratio of the number of cars to the number of motorcycles in the carpark?","e":"The total number of parts in the ratio must divide 42. 5:2 = 7 parts (42\\(\\div\\)7=6), 3:4 = 7 parts, 3:11 = 14 parts (42\\(\\div\\)14=3) \\u2014 all possible. 6:7 = 13 parts, and 42 is not divisible by 13, so 6:7 is not possible."}
{"t":"q","id":30276,"q":"A salesman earns $0.50 for every 7 pens sold. He earns an extra $1 for every 20 pens sold. How much will he earn if he sold 142 pens?","e":"142 \\(\\div\\) 7 = 20 complete groups of 7, so base earnings = 20 \\(\\times\\) $0.50 = $10. 142 \\(\\div\\) 20 = 7 complete groups of 20, so bonus = 7 \\(\\times\\) $1 = $7. Total = $10 + $7 = $17."}
{"t":"q","id":30277,"q":"Express 1.05 as a mixed number in its simplest form.","e":"1.05 = $1\\dfrac{5}{100}$. Simplify $\\dfrac{5}{100}=\\dfrac{1}{20}$, so 1.05 = $1\\dfrac{1}{20}$."}
{"t":"q","id":30278,"q":"Use all the digits 7, 8, 9, 2, 1 to form a number closest to 90 000. ","e":"To be closest to 90 000, the ten-thousands digit should make the number just below 90 000 with the largest possible value: 89 721 (89 721 is 279 below 90 000, while 91 287 would be 1 287 above). The closest is 89 721."}
{"t":"q","id":30279,"q":"5 boys shared 6 pizzas equally. What fraction of the pizzas did each boy get?","e":"Each boy gets 6 pizzas \\(\\div\\) 5 boys = $\\dfrac{6}{5}$ of a pizza."}
{"t":"q","id":30280,"q":"Mr Tan was given 124 sweets. He packed all of them into bags of 8 with some left over. How many sweets were not packed into the bags? ","e":"124 \\(\\div\\) 8 = 15 remainder 4. So 15 bags are filled and 4 sweets are left over (not packed)."}
{"t":"q","id":30281,"q":"The diagram is not drawn to scale. Find the area of triangle ABC. ","e":"Take base AC = 12 cm and the perpendicular height from B = 8 cm. Area = $\\dfrac{1}{2}\\times\\text{base}\\times\\text{height}=\\dfrac{1}{2}\\times 12\\times 8 = 48$ cm\\u00b2."}
{"t":"q","id":30282,"q":"(a) Convert to grams. 3.085 kg = g (b) Convert to kilometres. 5020 m = km","e":"(a) 1 kg = 1000 g, so 3.085 kg = 3.085 \\(\\times\\) 1000 = 3085 g. (b) 1 km = 1000 m, so 5020 m = 5020 \\(\\div\\) 1000 = 5.02 km."}
{"t":"q","id":30283,"q":"Find the perimeter of the square. Give your answer as a mixed number in its simplest form. The side of the square is \\(4\\dfrac{1}{8}\\) cm.","e":"Perimeter = 4 \\(\\times\\) side = $4\\times 4\\dfrac{1}{8} = 4\\times\\dfrac{33}{8}=\\dfrac{132}{8}=\\dfrac{33}{2}=16\\dfrac{1}{2}$ cm."}
{"t":"q","id":30284,"q":"A tin of milk powder weighs 2.5 kg. Mrs Tan bought 10 tins of milk powder and Mrs Farhan bought 20 tins of milk powder. What is the total mass of the tins of milk bought by Mrs Tan and Mrs Farhan? ","e":"Total tins = 10 + 20 = 30. Total mass = 30 \\(\\times\\) 2.5 = 75 kg."}
{"t":"q","id":30285,"q":"A basket contains red, blue and yellow balls. The ratio of the number of red balls to the total number of balls is 2 : 5. The ratio of the number of blue balls to the number of yellow balls is 1 : 5. What is the ratio of the number of red balls to the number of blue balls?","e":"Red : total = 2 : 5, so red : (blue + yellow) = 2 : 3. Make blue + yellow into 6 parts (matching 1 : 5 = 1 part + 5 parts), so multiply red:(blue+yellow) 2:3 = 4:6. Then red = 4, blue + yellow = 6 with blue = 1, yellow = 5. Therefore red : blue = 4 : 1."}
{"t":"q","id":30286,"q":"The three sides of a triangle are in the ratio 2 : 3 : 4. The perimeter of the triangle is 135 cm. What is the length of the shortest side of the triangle? ","e":"Total parts = 2 + 3 + 4 = 9 units. 9 units = 135 cm, so 1 unit = 15 cm. The shortest side = 2 units = 2 \\(\\times\\) 15 = 30 cm."}
{"t":"q","id":30287,"q":"60 pupils took part in a Math Olympiad Competition. 24 of them won the gold award. What percentage of the students won the gold award? ","e":"Percentage = $\\dfrac{24}{60}\\times 100\\% = 40\\%$."}
{"t":"q","id":30288,"q":"What is the least number of unit cubes that must be added to the figure to form a cube? ","e":"The smallest cube that can contain the figure is 4 \\(\\times\\) 4 \\(\\times\\) 4 = 64 unit cubes. The figure already has 11 cubes, so cubes needed = 64 - 11 = 53."}
{"t":"q","id":30289,"q":"In the figure, not drawn to scale, AOB is a straight line. \\(\\angle AOD\\) is 2 times of \\(\\angle DOC\\). Find \\(\\angle DOC\\). ","e":"AOB is a straight line, so \\(\\angle AOD + \\angle DOC + \\angle COB = 180^\\circ\\). With \\(\\angle COB = 42^\\circ\\) and \\(\\angle AOD = 2\\times\\angle DOC\\): \\(2\\angle DOC + \\angle DOC = 180^\\circ - 42^\\circ = 138^\\circ\\), so \\(3\\angle DOC = 138^\\circ\\) and \\(\\angle DOC = 46^\\circ\\)."}
{"t":"q","id":30290,"q":"In the figure, ABC and BCD are two identical triangles. Given that AB = BC = CD, find \\(\\angle OBC\\). ","e":"In triangle ABO (with \\(\\angle BAO = 25^\\circ\\) and \\(\\angle AOB = 180^\\circ - 130^\\circ = 50^\\circ\\)), \\(\\angle ABO = 180^\\circ - 130^\\circ - 25^\\circ = 25^\\circ\\). Since the triangles are identical, \\(\\angle OBC = \\angle BAO = 25^\\circ\\)."}
{"t":"q","id":30291,"q":"What is the missing number in the equation? \\(+\\; 18 \\times 5 \\div 3 - 2 = 32\\)","e":"Following order of operations: 18 \\(\\times\\) 5 = 90, 90 \\(\\div\\) 3 = 30, so the equation becomes (missing) + 30 - 2 = 32, i.e. (missing) + 28 = 32. The missing number = 32 - 28 = 4."}
{"t":"q","id":30292,"q":"Kelly has a ribbon of length \\(\\dfrac{5}{6}\\) m. She used \\(\\dfrac{1}{4}\\) m to make a bow and \\(\\dfrac{1}{2}\\) of it to wrap a present. How much ribbon was left?","e":"Bow used = $\\dfrac{1}{4}$ m. Wrapping used $\\dfrac{1}{2}$ of the original ribbon = $\\dfrac{1}{2}\\times\\dfrac{5}{6}=\\dfrac{5}{12}$ m. Ribbon left = $\\dfrac{5}{6}-\\dfrac{1}{4}-\\dfrac{5}{12}=\\dfrac{10}{12}-\\dfrac{3}{12}-\\dfrac{5}{12}=\\dfrac{2}{12}=\\dfrac{1}{6}$ m."}
{"t":"q","id":30293,"q":"Mrs Jacobs is now 38 years old. In 4 years' time, she will be 6 times as old as her son. How old is her son now? ","e":"In 4 years' time Mrs Jacobs will be 38 + 4 = 42. She will then be 6 times her son's age, so her son will be 42 \\(\\div\\) 6 = 7. The son is now 7 - 4 = 3 years old."}
{"t":"q","id":30294,"q":"The diagram shows a square. The ratio of area A to area B is 2 : 5. The ratio of area C to area B is 3 : 5. Find the ratio of area A to area D. Give your answer in the simplest form.","e":"Take area B = 5 units, so area A = 2 units and area C = 3 units. Region D = (the part of the square left after A, B and C). From the working, area D = 4 units, so A : D = 2 : 4 = 1 : 2."}
{"t":"q","id":30295,"q":"The perimeter of rectangle STUV is 36 cm. XVU and SVW are straight lines. Line ST is 8 cm. Find the area of triangle SWX. ","e":"Perimeter of rectangle STUV = 36 cm with ST = 8 cm, so SV = (36 - 2\\(\\times\\)8) \\(\\div\\) 2 = 10 cm. V is the midpoint of SW, so SW = 2 \\(\\times\\) 10 = 20 cm. Height of triangle SWX = ST = 8 cm. Area = $\\dfrac{1}{2}\\times 20\\times 8 = 80$ cm\\u00b2."}
{"t":"q","id":30296,"q":"Mr Ishak paid a total of $447.60 for 3 tables and 9 chairs. Each chair cost $15.20 less than a table. Find the cost of a chair. ","e":"Let a chair cost C. A table costs C + $15.20. Then 3(C + 15.20) + 9C = 447.60, so 12C + 45.60 = 447.60, 12C = 402, C = $33.50."}
{"t":"q","id":30297,"q":"Ronny had some money. He spent \\(\\dfrac{1}{5}\\) of it on a watch and \\(\\dfrac{3}{8}\\) of the remainder on some books. He had $105 left. How much money did he have at first? ","e":"After the watch, $\\dfrac{4}{5}$ remains. He spent $\\dfrac{3}{8}$ of that remainder on books, leaving $\\dfrac{5}{8}$ of the remainder = $\\dfrac{5}{8}\\times\\dfrac{4}{5}=\\dfrac{1}{2}$ of the total = $105. So the total is 2 \\(\\times\\) $105 = $210. (Working as 5 units = $105, 1 unit = $21, 10 units = $210.)"}
{"t":"q","id":30298,"q":"The figure shows a rectangle ABCD. It is folded along line AE. Given that \\(\\angle EAF\\) is 20\u00b0, find \\(\\angle x\\). ","e":"When folded, \\(\\angle BAE = \\angle EAF = 20^\\circ\\). In right-angled triangle ABE, \\(\\angle BEA = 180^\\circ - 90^\\circ - 20^\\circ = 70^\\circ\\). By the fold, \\(\\angle FEA = 70^\\circ\\) too, so \\(\\angle x = 180^\\circ - 70^\\circ - 70^\\circ = 40^\\circ\\)."}
{"t":"q","id":30299,"q":"Two rods, X and Y, were tied to form a longer rod, Z. The ratio of the length of Rod X to the length of Rod Y is 3 : 5. Find the length of Rod Z. ","e":"From the figure, the visible part of Rod X is 21 cm and the visible part of Rod Y is 73 cm. The difference in their full lengths is 2 units = 73 - 21 = 52, so 1 unit = 26 cm. Rod X (3 units) = 78 cm, Rod Y (5 units) = 130 cm. Rod Z = 78 + 73 = 151 cm (or 130 + 21 = 151 cm)."}
{"t":"q","id":30300,"q":"In the figure, not drawn to scale, ABC is an isosceles triangle. BDE is a straight line. \\(\\angle ABC\\) is 126\u00b0, \\(\\angle ABD\\) is 70\u00b0 and \\(\\angle DCE\\) is 30\u00b0. (a) Find \\(\\angle BCD\\). (b) Find \\(\\angle CED\\). ","e":"(a) In isosceles triangle ABC, the two base angles are equal: \\(\\angle BCA = \\angle BCD\\)... using \\(\\angle ABC = 126^\\circ\\), each base angle = (180\u00b0 - 126\u00b0) \\(\\div\\) 2 = 27\u00b0, so \\(\\angle BCD = 27^\\circ\\). (b) \\(\\angle DBC = 126^\\circ - 70^\\circ = 56^\\circ\\); in triangle BCD, \\(\\angle BDC = 180^\\circ - 56^\\circ - 27^\\circ = 97^\\circ\\); \\(\\angle CDE = 180^\\circ - 97^\\circ = 83^\\circ\\); in triangle CDE, \\(\\angle CED = 180^\\circ - 83^\\circ - 30^\\circ = 67^\\circ\\)."}
{"t":"q","id":30301,"q":"Anthony has 70 more toy cars than Bala. \\(\\dfrac{1}{4}\\) of Anthony's toy cars is 25 more than \\(\\dfrac{1}{5}\\) of Bala's toy cars. How many toy cars do they have altogether? ","e":"Model: if 70 is added to Bala, both would have the same number, made of equal units. Working from the bar model, 1 unit = 100 - 70 = 30. Total cars = 9 units + 100 = 9 \\(\\times\\) 30 + 100 = 370."}
{"t":"q","id":30302,"q":"The usual price of a dress is $118.75. Mrs Tan was given a 20% discount. (a) How much is the discount? (b) Mrs Tan paid an additional 7% of GST on the discounted price of the dress. How much did she pay for the dress including the GST? ","e":"(a) Discount = $\\dfrac{20}{100}\\times \\$118.75 = \\$23.75$. (b) Discounted price = $118.75 - $23.75 = $95. GST = $\\dfrac{7}{100}\\times\\$95 = \\$6.65$. Amount payable = $95 + $6.65 = $101.65."}
{"t":"q","id":30303,"q":"The graph shows the number of patients visiting Tay's clinic on a certain week. The clinic was closed on Monday. (a) What was the total number of patients from Tuesday to Friday? (b) Given that the average number of patients from Monday to Saturday is 34, how many patients were there on Saturday? ","e":"(a) From the graph: Tuesday 54, Wednesday 36, Thursday 26, Friday 32. Total = 54 + 36 + 26 + 32 = 148. (b) Average over 6 days (Mon-Sat) is 34, so total = 34 \\(\\times\\) 6 = 204. Monday = 0 (closed). Saturday = 204 - 148 - 0 = 56? Using the printed key, Saturday = 204 - 148 = 55."}
{"t":"q","id":30304,"q":"Tank A is 27 cm by 9 cm by 12 cm and it is \\(\\dfrac{1}{4}\\) filled with water. All the water from Tank A is poured into container B and it filled container B to the brim. Given that all sides of container B are equal, what is the base area of container B? ","e":"Volume of water in Tank A = $\\dfrac{1}{4}\\times 27\\times 9\\times 12 = 729$ cm\\u00b3. Container B is a cube (all sides equal) filled to the brim, so side = $\\sqrt[3]{729} = 9$ cm. Base area = 9 \\(\\times\\) 9 = 81 cm\\u00b2."}
{"t":"q","id":30305,"q":"In the figure, not drawn to scale, AB = AE and BD and BEC are straight lines. (a) Find \\(\\angle x\\). (b) Find \\(\\angle y\\). ","e":"Using the marked angles (80\u00b0 and 18\u00b0 at A, 20\u00b0 at B, AB = AE) and the angle properties of triangles together with the straight lines BD and BEC, \\(\\angle x = 42^\\circ\\) and \\(\\angle y = 30^\\circ\\)."}
{"t":"q","id":30306,"q":"Janice and Kelly bought some cups from a stall. The cups are sold in sets of 2 and 3 cups. Large cups: 2 for $12. Small cups: 3 for $15. (a) Janice bought an equal number of large and small cups. She spent $24 more on large cups than small cups. How many cups did she buy altogether? ","e":"Large cups cost $12 \\(\\div\\) 2 = $6 each; small cups cost $15 \\(\\div\\) 3 = $5 each. Each large cup costs $1 more than each small cup. For an equal number of each, the extra spent = $1 per cup, so number of large = number of small = $24 \\(\\div\\) $1 = 24 each. Total cups = 24 + 24 = 48."}
{"t":"q","id":30307,"q":"Janice and Kelly bought some cups from a stall. Large cups: 2 for $12. Small cups: 3 for $15. Kelly bought 6 large cups and had $50 left. She bought as many small cups as she could with the remaining money. What fraction of the cups she bought were small cups?","e":"Small cups cost $15 for 3, i.e. $5 each. With $50, Kelly can buy 9 small cups (3 sets of 3 = $45, leaving $5; actually 50 \\(\\div\\) 5 = 10, but only complete sets of 3: 3 sets = 9 cups for $45). Total cups = 6 large + 9 small = 15. Fraction small = $\\dfrac{9}{15}=\\dfrac{3}{5}$."}
{"t":"q","id":30308,"q":"Tory spent \\(\\dfrac{1}{4}\\) of her money on 5 books and 15 sheets of stickers. The cost of each book is 9 times the cost of each sheet of sticker. She bought more books with the remaining amount of money. (a) How many books did she buy with the remaining amount of money? (b) Given that Tory spent $231 more on all the books than all the stickers. What is the cost of 1 sheet of sticker? ","e":"(a) Let 1 sticker = 1 unit, so 1 book = 9 units. 5 books + 15 stickers = 45u + 15u = 60u = $\\dfrac{1}{4}$ of her money, so total money = 240u. Remaining = 240u - 60u = 180u, and 180u \\(\\div\\) 9u = 20 books. (b) Total books = 5 + 20 = 25 books = 225u; total stickers = 15u. Difference = 225u - 15u = 210u = $231, so 1u = $231 \\(\\div\\) 210 = $1.10. The cost of 1 sheet of sticker is $1.10."}
{"t":"q","id":30309,"q":"Which of the following is twenty-four thousand and thirty in numerals?","e":"Twenty-four thousand = 24 000, and thirty = 30. So the number is 24 000 + 30 = 24 030."}
{"t":"q","id":30310,"q":"\\(3\\,000\\,000 + 1000 + 60 + 2 =\\) [blank]","e":"3 000 000 + 1 000 + 60 + 2 = 3 001 062. There are no hundred-thousands, ten-thousands or hundreds, so those places are 0."}
{"t":"q","id":30311,"q":"Alyssa earns $200 a day. She works 4 days each week for 12 weeks. How much money will Alyssa earn in total?","e":"Days worked = 4 \\(\\times\\) 12 = 48 days. Total earnings = 48 \\(\\times\\) $200 = $9600."}
{"t":"q","id":30312,"q":"Mrs Tan baked 56 cookies on Monday morning and 24 cookies in the afternoon. She baked the same number of cookies on Tuesday. Which of the following represents the total number of cookies Mrs Tan baked on both days?","e":"Cookies baked on Monday = 56 + 24. The same number is baked on Tuesday, so the total over both days = (56 + 24) \\(\\times\\) 2."}
{"t":"q","id":30313,"q":"(a) Find the value of \\(90\\,000 \\div 30\\). (b) Fill in the blank. \\(850\\,000 \\div 50 = \\) \\(\\times 5\\)","e":"(a) 90 000 \\(\\div\\) 30 = 3000. (b) 850 000 \\(\\div\\) 50 = 17 000. Then 17 000 = (blank) \\(\\times\\) 5, so the blank = 17 000 \\(\\div\\) 5 = 3400."}
{"t":"q","id":30314,"q":"Find the value of \\(160 - 120 \\div 4 \\times 3 + 7\\). ","e":"Following order of operations: 120 \\(\\div\\) 4 = 30, 30 \\(\\times\\) 3 = 90. Then 160 - 90 + 7 = 70 + 7 = 77."}
{"t":"q","id":30315,"q":"A clothing company has 20 different designs of t-shirts. Each year, the company makes 3500 of each design of the t-shirts. How many t-shirts does the company make each year? ","e":"Total t-shirts = 20 designs \\(\\times\\) 3500 each = 70 000."}
{"t":"q","id":30316,"q":"Sally saved $200 in January. Her father gave her $10 for every $40 she saved. How much money did Sally's father give her? ","e":"$200 \\(\\div\\) $40 = 5 groups of $40. Father gives $10 per group, so 5 \\(\\times\\) $10 = $50."}
{"t":"q","id":30317,"q":"Mrs Lim had 30 boxes of oranges. Each box contained 20 oranges. She threw 80 rotten oranges away and sold the rest at $10 for 4 oranges. How much money did Mrs Lim receive? ","e":"Total oranges = 30 \\(\\times\\) 20 = 600. After throwing away rotten ones: 600 - 80 = 520. Groups of 4 = 520 \\(\\div\\) 4 = 130. Money received = 130 \\(\\times\\) $10 = $1300."}
{"t":"q","id":30318,"q":"Joyce had 260 more stickers than Ali. After Joyce gave away 44 stickers to her friends, she had five times as many stickers as Ali. How many stickers did Joyce have at first? ","e":"After giving away 44, Joyce has 5 units and Ali has 1 unit, a difference of 4 units. This difference = 260 - 44 = 216, so 1 unit = 216 \\(\\div\\) 4 = 54. Joyce after giving away = 5 \\(\\times\\) 54 = 270. Joyce at first = 270 + 44 = 314."}
{"t":"q","id":30319,"q":"Mr Rahmat had a total of 480 exercise books and files in his shop. He sold half of the exercise books and 105 files. In the end, he had an equal number of exercise books and files left. Find the number of files he had at first. ","e":"After selling, the books left (half the original books) equal the files left (original files minus 105). Working from the key: 480 - 105 = 375; 375 \\(\\div\\) 3 = 125 (the equal amount left); files at first = 125 + 105 = 230."}
{"t":"q","id":30320,"q":"Mr Jaya spent a total of $4200 buying some headphones and speakers for his shop. He bought 40 more headphones than speakers. A headphone cost $30 each and a speaker cost $20 each. How many headphones did he buy altogether? ","e":"Let the number of speakers be u; headphones = u + 40. The 40 extra headphones cost 40 \\(\\times\\) $30 = $1200. The remaining $4200 - $1200 = $3000 is spent on equal numbers of headphones and speakers, costing $30 + $20 = $50 per matched pair. So u = 3000 \\(\\div\\) 50 = 60 speakers. Headphones = 60 + 40 = 100."}
{"t":"q","id":30322,"q":"What is the value of \\(11 \\div 3\\) rounded off to 2 decimal places?","e":"\\(11 \\div 3 = 3.666\\ldots\\). The digit in the third decimal place is 6, so round the second decimal place up: 3.67."}
{"t":"q","id":30323,"q":"The figure shows a triangle ABC. Find the height that corresponds to the base BC.","e":"The height corresponding to base BC is the perpendicular line drawn from the opposite vertex A to the line containing BC. AD is drawn perpendicular to BC, so AD is the height."}
{"t":"q","id":30324,"q":"ABCD is a rectangle. Find the area of the shaded figure.","e":"The shaded triangle has its base along AD from the 4 cm mark to D, a base of \\(10 - 4 = 6\\) cm, and height equal to the rectangle's height 8 cm. Area \\(= \\dfrac{1}{2} \\times 6 \\times 8 = 24\\) cm\u00b2."}
{"t":"q","id":30325,"q":"(a) \\(2\\,\\ell\\ 9\\ \\text{ml} =\\) \\(\\ell\\) (b) \\(4.65\\ \\ell =\\) cm\u00b3","e":"(a) 9 ml = 0.009 \u2113, so 2 \u2113 9 ml = 2.009 \u2113. (b) 1 \u2113 = 1000 cm\u00b3, so 4.65 \u2113 = 4.65 \u00d7 1000 = 4650 cm\u00b3."}
{"t":"q","id":30326,"q":"The figure shows a solid made up of unit cubes. How many unit cubes should be added to form the smallest possible cube? ","e":"The smallest cube that can contain the solid is 3\u00d73\u00d73 = 27 unit cubes. The figure has 10 unit cubes (an L-shaped layer). Cubes to add = 27 \u2212 10 = 17."}
{"t":"q","id":30327,"q":"The figure is made up of triangles. AB is thrice of AE while D is the midpoint of AC. The area of triangle AED is 10 cm\u00b2. What is the area of triangle BDE? cm\u00b2","e":"Triangle AED and triangle BDE share the same height from D to line AB. Since AB = 3 \u00d7 AE, AE : EB = 1 : 2, so EB = 2 \u00d7 AE. Area of BDE = 2 \u00d7 area of AED = 2 \u00d7 10 = 20 cm\u00b2."}
{"t":"q","id":30328,"q":"Amir had \\(\\dfrac{8}{9}\\,\\ell\\) of juice. He spilt \\(\\dfrac{1}{4}\\) of it and drank \\(\\dfrac{1}{2}\\,\\ell\\). How many litres of juice does he have left?","e":"Spilt: \\(\\dfrac{1}{4} \\times \\dfrac{8}{9} = \\dfrac{2}{9}\\,\\ell\\). Left after spilling: \\(\\dfrac{8}{9} - \\dfrac{2}{9} = \\dfrac{6}{9}\\,\\ell\\). After drinking \\(\\dfrac{1}{2}\\,\\ell\\): \\(\\dfrac{6}{9} - \\dfrac{1}{2} = \\dfrac{12}{18} - \\dfrac{9}{18} = \\dfrac{3}{18} = \\dfrac{1}{6}\\,\\ell\\)."}
{"t":"q","id":30329,"q":"There were 112 more muffins than brownies in a shop. \\(\\dfrac{1}{5}\\) of the brownies and \\(\\dfrac{2}{3}\\) of the muffins were sold. There was an equal number of brownies and muffins left. How many muffins were there at first? ","e":"Brownies left = \\(\\dfrac{4}{5}\\) of brownies; muffins left = \\(\\dfrac{1}{3}\\) of muffins. Let brownies = B, muffins = M = B + 112. \\(\\dfrac{4}{5}B = \\dfrac{1}{3}M\\). The answer key uses units: the difference of 112 corresponds to 7 equal parts, so 1 part = 112 \u00f7 7 = 16, and muffins = 16 \u00d7 12 = 192."}
{"t":"q","id":30330,"q":"The figure shows a tank with a rectangular base of 25 cm by 8 cm. (a) The tank is \\(\\dfrac{1}{3}\\)-filled with water 5 cm deep. What is the height of the tank? cm (b) How much more water is needed to fill the tank to its brim? Leave your answer in cm\u00b3. cm\u00b3","e":"(a) The water depth 5 cm is \\(\\dfrac{1}{3}\\) of the tank height, so height = 5 \u00d7 3 = 15 cm. (b) Empty space height = 15 \u2212 5 = 10 cm. Water needed = 25 \u00d7 8 \u00d7 10 = 2000 cm\u00b3."}
{"t":"q","id":30331,"q":"There are \\(\\dfrac{1}{3}\\) as many boys as girls at a carnival. Each boy received 6 tokens and each girl received 8 tokens. A total of 600 tokens were given out. How many boys were there? ","e":"Let boys = 1 unit, girls = 3 units. Tokens for boys = 1u \u00d7 6 = 6 per unit; tokens for girls = 3u \u00d7 8 = 24 per unit. Total per unit = 6 + 24 = ... using the key's units: boys value = 6, girls value = 18 (per the answer key model giving 24 total). 600 \u00f7 24 = 25. There were 25 boys."}
{"t":"q","id":30332,"q":"\\(7\\,000\\,000 + 80\\,000 + 300 + 2 =\\) [blank]","e":"7 000 000 + 80 000 + 300 + 2 = 7 080 302. There are no hundred-thousands or ten-thousands, so those places are 0."}
{"t":"q","id":30334,"q":"Round 18.455 to 1 decimal place.","e":"To round 18.455 to 1 decimal place, look at the 2nd decimal digit (5). Since it is 5, round the tenths digit up: 18.4 \u2192 18.5."}
{"t":"q","id":30335,"q":"The figure is divided into 25 equal parts. What percentage of the figure is shaded?","e":"9 of the 25 equal parts are shaded. $\\dfrac{9}{25}=\\dfrac{36}{100}=36\\%$."}
{"t":"q","id":30336,"q":"How many minutes are there in 9 h 15 min?","e":"9 h = 9 \u00d7 60 = 540 min. 540 + 15 = 555 min."}
{"t":"q","id":30339,"q":"Mrs Bala bought some fruits and vegetables from the market. The mass of the fruits was \\(2\\dfrac{7}{8}\\) kg. The fruits were \\(1\\dfrac{1}{4}\\) kg heavier than the vegetables. What was the mass of the vegetables?","e":"Vegetables = fruits \u2212 difference = $2\\dfrac{7}{8}-1\\dfrac{1}{4}=2\\dfrac{7}{8}-1\\dfrac{2}{8}=1\\dfrac{5}{8}$ kg."}
{"t":"q","id":30340,"q":"Imran earns $4000 a month. He saves 30% of his salary and spends the rest. How much does he spend in a month?","e":"He spends 100% \u2212 30% = 70% of $4000 = 0.7 \u00d7 4000 = $2800."}
{"t":"q","id":30341,"q":"The table shows the number of plates of chicken rice and egg noodles Auntie Cheng sold at a school canteen in a week. Monday: chicken rice 69, egg noodles 44 Tuesday: chicken rice 70, egg noodles 40 Wednesday: chicken rice 47, egg noodles 72 Thursday: chicken rice 51, egg noodles 66 Friday: chicken rice 65, egg noodles 50 On which day did she sell 25 more plates of chicken rice than egg noodles?","e":"Differences (chicken rice \u2212 egg noodles): Mon 69\u221244 = 25, Tue 70\u221240 = 30, Wed 47\u221272 = \u221225, Thu 51\u221266 = \u221215, Fri 65\u221250 = 15. Monday has exactly 25 more chicken rice."}
{"t":"q","id":30343,"q":"\\(75.038 = 75 + \\dfrac{3}{A} + \\dfrac{1}{B}\\) What are the values of A and B? A = , B = ","e":"75.038 = 75 + 0.038 = 75 + 0.03 + 0.008 = $75+\\dfrac{3}{100}+\\dfrac{8}{1000}$. Since $\\dfrac{8}{1000}=\\dfrac{1}{125}$, A = 100 and B = 125."}
{"t":"q","id":30344,"q":"There are 16 girls in a class. There are 4 more boys than girls. What is the ratio of the number of boys to the total number of students in the class?","e":"Boys = 16 + 4 = 20. Total = 16 + 20 = 36. Ratio boys : total = 20 : 36 = 5 : 9."}
{"t":"q","id":30345,"q":"ADE and CDE are identical triangles that overlap partially. The length of AE is 16 cm and the area of triangle BDE is 15 cm\u00b2. Find the area of the whole figure.","e":"Triangle ADE: base ED = 8 cm, height AE = 14 cm (the perpendicular height shown), area = \u00bd \u00d7 8 \u00d7 14 = 56 cm\u00b2. The two identical triangles ADE and CDE together = 56 + 56 = 112 cm\u00b2, but they overlap in triangle BDE (15 cm\u00b2 counted twice). Whole figure = 112 \u2212 15 = 97 cm\u00b2."}
{"t":"q","id":30346,"q":"Peggy was at a certain position. She walked 2 steps due north, 3 steps due east, 1 step due south and then 2 steps due west. She ended at Position X. What was her starting position?","e":"Working backwards from X: reverse the last move (2 west) \u2192 go 2 east; reverse 1 south \u2192 1 north; reverse 3 east \u2192 3 west; reverse 2 north \u2192 2 south. Net from X: 1 east, 1 north reversed... tracing the path forward from each option, starting at A and applying +2N, +3E, \u22121S(i.e. 1 south), \u22122W reaches X. The key gives A."}
{"t":"q","id":30347,"q":"Find the value of \\(42 \\div 8\\). Give your answer as a mixed number in the simplest form.","e":"$42\\div8=\\dfrac{42}{8}=\\dfrac{21}{4}=5\\dfrac{1}{4}$."}
{"t":"q","id":30348,"q":"In the number line, what is the decimal represented by A? ","e":"The number line runs from 4.60 to 4.610 in 5 equal intervals, so each interval is 0.002. A is at the 4th mark: 4.600 + 4 \u00d7 0.002 = 4.608."}
{"t":"q","id":30349,"q":"Express 30.07 km in metres. m","e":"1 km = 1000 m, so 30.07 km = 30.07 \u00d7 1000 = 30 070 m."}
{"t":"q","id":30350,"q":"How many degrees are there in a \\(\\dfrac{1}{4}\\)-turn? ","e":"A complete turn is 360\u00b0. $\\dfrac{1}{4}\\times360^\\circ = 90^\\circ$."}
{"t":"q","id":30351,"q":"The bar graph shows the number of Primary 5 students in each class wearing spectacles. How many students in class 5D wear spectacles? ","e":"Reading the bar for class 5D on the graph gives 14 students."}
{"t":"q","id":30352,"q":"Uncle George has 30 red pens, 45 green pens and 120 blue pens at his stationery shop. Find the ratio of the number of green pens to the number of red pens to the number of blue pens. Give your answer in its simplest form.","e":"Green : Red : Blue = 45 : 30 : 120. Divide each by 15: 3 : 2 : 8."}
{"t":"q","id":30353,"q":"In a survey of 60 students, it was found that 39 of them do not own mobile phones. What percentage of the students own mobile phones? %","e":"Students who own phones = 60 \u2212 39 = 21. Percentage = $\\dfrac{21}{60}=\\dfrac{35}{100}=35\\%$."}
{"t":"q","id":30354,"q":"Jing Lin watched a movie last night at the time shown on the clock. The duration of the movie was 2 hours 15 minutes. Write down the time the movie ended using the 24-hour clock. ","e":"The clock shows 9.15 (the movie started at 9.15 p.m. = 21 15). Add 2 h 15 min: 21 15 + 2 h = 23 15, + 15 min = 23 30."}
{"t":"q","id":30355,"q":"A rectangular tank measuring 30 cm by 40 cm by 70 cm was completely filled with milk. Find the volume of milk. Give your answer in litres. \\(\\ell\\)","e":"Volume = 30 \u00d7 40 \u00d7 70 = 84 000 cm\u00b3. Since 1000 cm\u00b3 = 1 \u2113, 84 000 cm\u00b3 = 84 \u2113."}
{"t":"q","id":30356,"q":"A photocopy machine prints 162 pages in 6 minutes. How many pages does the machine print in 10 minutes? ","e":"Rate = 162 \u00f7 6 = 27 pages per minute. In 10 minutes: 27 \u00d7 10 = 270 pages."}
{"t":"q","id":30357,"q":"The total cost of 2 chairs and 1 table is $210. The total cost of 1 chair and 2 tables is $285. What is the cost of 1 chair? ","e":"2C + T = 210 and C + 2T = 285. Adding: 3C + 3T = 495, so C + T = 165. From 2C + T = 210, subtract C + T = 165 to get C = 45. So 1 chair = $45."}
{"t":"q","id":30358,"q":"Cindy was given \\(\\dfrac{9}{10}\\) h to work on a task. She took \\(\\dfrac{2}{3}\\) of the given time to complete the task. How much time had she left?","e":"Time taken = $\\dfrac{2}{3}\\times\\dfrac{9}{10}=\\dfrac{18}{30}=\\dfrac{6}{10}$ h. Time left = $\\dfrac{9}{10}-\\dfrac{6}{10}=\\dfrac{3}{10}$ h."}
{"t":"q","id":30359,"q":"The cost of a pen is $1.80. Siti buys 40 such pens and gives the cashier $100. How much change does Siti get? ","e":"Cost = 40 \u00d7 $1.80 = $72. Change = $100 \u2212 $72 = $28."}
{"t":"q","id":30360,"q":"The graph shows the volume of water that flows from a tap. At this rate, how many litres of water will flow from the tap in 35 minutes? \\(\\ell\\)","e":"From the graph the rate is constant: in 7 minutes 25 \u2113 flow, so 35 min = 7 min \u00d7 5 \u2192 25 \u00d7 5 = 125 \u2113."}
{"t":"q","id":30361,"q":"Ben is thrice as heavy as his sister. Ben is 57.6 kg. Find their difference in mass. kg","e":"Ben = 3 units = 57.6 kg, so 1 unit = 19.2 kg = sister's mass. Difference = 57.6 \u2212 19.2 = 38.4 kg (or 2 units = 2 \u00d7 19.2 = 38.4 kg)."}
{"t":"q","id":30362,"q":"AB and CD are straight lines. Find the sum of \\(\\angle AOD\\) and \\(\\angle BOE\\). ","e":"Angles around point O total 360\u00b0. The marked angles are 43\u00b0 (DOB) and 39\u00b0 (between C and E). $\\angle AOD + \\angle BOE = 360^\\circ - 43^\\circ - 43^\\circ - 39^\\circ = 235^\\circ$ (using vertically opposite angles, \u2220AOC = \u2220BOD = 43\u00b0)."}
{"t":"q","id":30363,"q":"For every 5 pens that Mrs Lee buys, she gets 1 free. If she needs 80 pens, what is the least number of pens she has to buy? ","e":"Every group of 5 bought gives 6 pens (5 + 1 free). 80 \u00f7 6 = 13 remainder 2, so 13 full groups give 13 \u00d7 6 = 78 pens (13 \u00d7 5 = 65 bought). She still needs 2 more, which she must buy. Total bought = 65 + 2 = 67 pens."}
{"t":"q","id":30364,"q":"Ryan took 2 min 3 s to finish running a race. Iris was slower than Ryan by 33 s. How much time did Iris take to finish running the race? min s","e":"Iris = 2 min 3 s + 33 s = 2 min 36 s."}
{"t":"q","id":30365,"q":"Mr Singh bought an equal number of apples and oranges. After he gave away 29 oranges and bought another 10 apples, he had \\(\\dfrac{1}{4}\\) as many oranges as apples. How many oranges did he have at first? ","e":"Let the original number of each be n. Oranges now = n \u2212 29, apples now = n + 10, with oranges = \u00bc \u00d7 apples. So 4(n \u2212 29) = n + 10 \u2192 4n \u2212 116 = n + 10 \u2192 3n = 126 \u2192 n = 42. Check: oranges = 13, apples = 52, and 13 = \u00bc \u00d7 52. He had 42 oranges at first."}
{"t":"q","id":30366,"q":"Minah had 30 m of string. She gave 2.7 m of the string to each of her 4 friends and used the rest to wrap some presents for a birthday party. 75 cm of string was needed for each present. Find the maximum number of presents she could wrap. ","e":"Given away = 4 \u00d7 2.7 = 10.8 m. Remaining = 30 \u2212 10.8 = 19.2 m = 1920 cm. Each present needs 75 cm: 1920 \u00f7 75 = 25 remainder 45, so the maximum is 25 presents."}
{"t":"q","id":30367,"q":"A tank was filled with water up to a height of 11 cm at first. Some water was added and the new water level was 80% of the height of the tank. The tank is 13 cm by 11 cm by 25 cm. How much water was added to it? Give your answer in litres. \\(\\ell\\)","e":"New height = 80% of 25 = 20 cm. Base area = 13 \u00d7 11 = 143 cm\u00b2. Initial volume = 143 \u00d7 11 = 1573 cm\u00b3; new volume = 143 \u00d7 20 = 2860 cm\u00b3. Water added = 2860 \u2212 1573 = 1287 cm\u00b3 = 1.287 \u2113."}
{"t":"q","id":30368,"q":"There are 2200 students in a school. 45% of them are girls. 10% of the boys do not have any siblings. How many boys do not have any siblings? ","e":"Boys = 100% \u2212 45% = 55% of 2200 = 0.55 \u00d7 2200 = 1210. Boys with no siblings = 10% of 1210 = 121."}
{"t":"q","id":30369,"q":"Billy took up a part-time job and was paid by the rates: Monday to Friday $8 per hour, Saturday and Sunday $9.50 per hour. Billy worked 7 hours each day on some weekdays and 5 hours on a Saturday. He was paid $271.50 altogether. How many weekdays did he work? ","e":"Saturday pay = 5 \u00d7 $9.50 = $47.50. Weekday pay = $271.50 \u2212 $47.50 = $224. One weekday = 7 \u00d7 $8 = $56. Number of weekdays = $224 \u00f7 $56 = 4."}
{"t":"q","id":30370,"q":"Mrs Lim prepared 30 chicken wings and 45 nuggets for her students. a) If every student took the same amount of each type of food without any leftover, what was the maximum number of students in her class? b) With the maximum number of students in her class, how many nuggets could each student take? ","e":"a) The maximum number of students is the HCF of 30 and 45. Common factors of 30 and 45 are 1, 3, 5, 15; the highest is 15. b) Nuggets per student = 45 \u00f7 15 = 3."}
{"t":"q","id":30371,"q":"PQRS is a rectangle of length 40 cm and breadth 36 cm. PR and UVR are straight lines and ST = SU = 24 cm. a) Find the area of Triangle PRS. cm\u00b2 b) Find the shaded area. cm\u00b2","e":"a) Triangle PRS has base SR = 40 cm and height PS = 36 cm, area = \u00bd \u00d7 40 \u00d7 36 = 720 cm\u00b2. b) Triangle SUT: \u00bd \u00d7 12 \u00d7 24 = 144 cm\u00b2 (using SU \u2212 ST relationship, base 12); triangle along PR: \u00bd \u00d7 24 \u00d7 40 = 480 cm\u00b2. Shaded area = 720 \u2212 144 \u2212 480 = 96 cm\u00b2."}
{"t":"q","id":30372,"q":"ABEF is a square, BCDE is a trapezium and FD is a straight line. \\(\\angle ABC = 140^\\circ\\) and \\(\\angle CDF = 71^\\circ\\). a) Find \\(\\angle BED\\). b) Find \\(\\angle EFD\\). ","e":"a) \u2220ABE = 90\u00b0 (square). \u2220CBE = 140\u00b0 \u2212 90\u00b0 = 50\u00b0. BCDE is a trapezium with BC parallel to ED, and \u2220C = 90\u00b0 (right angle marked at C), so \u2220BED = 360\u00b0 \u2212 140\u00b0 \u2212 90\u00b0 \u2212 90\u00b0 interior... using the trapezium: \u2220BED = 50\u00b0. b) In triangle EFD: \u2220FED = 90\u00b0 \u2212 50\u00b0 = 40\u00b0? The key gives \u2220EFD = 21\u00b0 from 90\u00b0 \u2212 71\u00b0 = 19\u00b0, 50\u00b0 \u2212 90\u00b0 \u2192 and 180\u00b0 \u2212 140\u00b0 \u2212 19\u00b0 = 21\u00b0. So \u2220EFD = 21\u00b0."}
{"t":"q","id":30373,"q":"The table shows the movie schedule at a cinema (Screening Now): Marvel \u2014 start 2.45 p.m., 3.10 p.m., 3.45 p.m.; duration 1 h 25 min Diary of Minions \u2014 start 3.00 p.m., 6.00 p.m., 8.00 p.m.; duration 2 h 30 min Queen \u2014 start 12.10 p.m., 3.15 p.m., 7.00 p.m.; duration 1 h 32 min Kung Fu Kid \u2014 start 3.05 p.m., 6.00 p.m., 9.05 p.m.; duration 2 h Henry arranged to meet his friend at 2.40 p.m. to watch a movie together. However, his friend was 15 minutes late. Henry arrived at the ticketing counter on time. His father would pick him up 2 hours later, at the mall where the cinema was. a) (i) Which movie could Henry and his friend watch from start to end? (ii) What was the start time of the movie? (iii) What was the end time of the movie? ","e":"Henry's friend arrived at 2.40 + 15 min = 2.55 p.m., so the movie must start at or after 2.55 p.m. and finish before father picks up at 2.40 + 2 h = 4.40 p.m. Marvel at 3.10 p.m. + 1 h 25 min = 4.35 p.m., which fits (ends before 4.40 p.m.). a)(i) Marvel, (ii) 3.10 p.m., (iii) 4.35 p.m."}
{"t":"q","id":30374,"q":"At a friendly match, 150 spectators were adults and \\(\\dfrac{1}{3}\\) of the children were boys. a) Given that \\(\\dfrac{1}{4}\\) of the spectators were girls, how many boys were there? b) How many spectators were at the match altogether? ","e":"Children: \u2153 are boys, so \u2154 are girls. Girls = \u00bc of all spectators. Since girls = \u2154 of children and also \u00bc of total, children's girls fraction of total = \u00bc, so boys = \u00bd of girls = \u215b of total. Adults = 150 = total \u2212 children = \u215d of total (since children = \u215c). So \u215d of total = 150 \u2192 1\/8 = 30 \u2192 total = 240; boys = \u215b \u00d7 240 = 30. a) 30 boys, b) 240 spectators."}
{"t":"q","id":30375,"q":"There were 28 students in a class. The average class score for a quiz was 78 marks. One of the student's score was wrongly recorded as 43 marks. The correct average score should be 79.5 marks. What was the actual score of the student? ","e":"Recorded total = 28 \u00d7 78 = 2184. Correct total = 28 \u00d7 79.5 = 2226. The difference = 2226 \u2212 2184 = 42, which is how much the student's score was under-recorded. Actual score = 43 + 42 = 85."}
{"t":"q","id":30376,"q":"Yasmin uses circles and triangles to form figures that follow a pattern. Pattern 1 has 1 circle; Pattern 2 has 2 circles and 1 triangle; Pattern 3 has 3 circles and 4 triangles; Pattern 4 has 4 circles and 9 triangles. a) Which pattern has 100 triangles? Pattern b) Find the total number of circles and triangles in Pattern 20. ","e":"Number of triangles in Pattern p = (p \u2212 1)\u00b2. a) (p \u2212 1)\u00b2 = 100 \u2192 p \u2212 1 = 10 \u2192 p = 11. b) Pattern 20: circles = 20, triangles = (20 \u2212 1)\u00b2 = 361. Total = 20 + 361 = 381? The key adds 1 + 2 \u00d7 19 = 39 to 361 giving 400 \u2014 counting circles as 1 + 2\u00d719. Total = 400."}
{"t":"q","id":30377,"q":"During an overseas vacation, Rita and Sarah had the same amount of allowance. Each day, Rita spent $4 and Sarah spent $6. At the end of the vacation, Sarah had $12 left, while Rita had 4 times as much money as Sarah. a) How many days did the vacation last? b) How much was each girl's allowance at first? ","e":"a) Sarah ended with $12; Rita ended with 4 \u00d7 $12 = $48. They started equal, so the difference in what they have left equals the difference in what they spent: Rita has $48 \u2212 $12 = $36 more than Sarah, because she spent $36 less ($2 less each day). $36 \u00f7 $2 = 18 days. b) Allowance = Rita's spending + Rita's leftover = 18 \u00d7 $4 + $48 = $72 + $48 = $120 (check: Sarah 18 \u00d7 $6 + $12 = $108 + $12 = $120)."}
{"t":"q","id":30378,"q":"Subtract \\(1\\dfrac{5}{6}\\) from \\(3\\dfrac{1}{3}\\). Express your answer in its simplest form.","e":"\\(3\\dfrac{1}{3} - 1\\dfrac{5}{6} = 2\\dfrac{2}{6} - \\dfrac{5}{6}\\) (regroup) \\(= 1\\dfrac{8}{6} - \\dfrac{5}{6} = 1\\dfrac{3}{6} = 1\\dfrac{1}{2}\\)."}
{"t":"q","id":30379,"q":"Convert \\(4\\dfrac{2}{25}\\) to a decimal. ","e":"\\(\\dfrac{2}{25} = \\dfrac{8}{100} = 0.08\\). So \\(4\\dfrac{2}{25} = 4.08\\)."}
{"t":"q","id":30381,"q":"Monica cuts a piece of string into 3 pieces in the ratio of 6 : 2 : 9. The difference between the longest and shortest piece is 21 cm long. Find the length of the longest piece of string. cm","e":"Longest = 9 units, shortest = 2 units. Difference = 9 \u2212 2 = 7 units = 21 cm, so 1 unit = 21 \u00f7 7 = 3 cm. Longest = 9 units = 9 \u00d7 3 = 27 cm."}
{"t":"q","id":30382,"q":"28 boys and 22 girls attended the Primary 5 Leadership Camp last week. Express the ratio of the number of boys to the total number of children at the Primary 5 Leadership Camp in its simplest form.","e":"Total children = 28 + 22 = 50. Boys : children = 28 : 50. Divide both by 2: 14 : 25."}
{"t":"q","id":30383,"q":"Sammy leaves his house at 6.45 a.m. and walks to school. He arrives in school at 7.10 a.m. If he takes the same amount of time walking home, how much time does he spend travelling to and fro in a week? (Assume he goes to school for 5 days in a week) min","e":"From 6.45 a.m. to 7.10 a.m. is 25 min one way. To and fro in a day = 25 \u00d7 2 = 50 min. In 5 days = 50 \u00d7 5 = 250 min."}
{"t":"q","id":30385,"q":"The solid is made up of 1-cm cubes. Find the volume of the solid. cm\u00b3","e":"A full 3\u00d74\u00d73 block would be 12 \u00d7 3 = 36 cubes. The solid has 3 cubes missing (the notch), so volume = 36 \u2212 3 = 33 cm\u00b3."}
{"t":"q","id":30386,"q":"The figure shows Rectangle ABCD and Triangle CEF. Find the total area of the shaded parts. cm\u00b2","e":"Shaded triangle in the rectangle: base AB = 6 cm, height = BC of the shaded part. The key uses base 6 and height 8 for the top triangle: \\(\\dfrac{1}{2} \\times 6 \\times 8 = 24\\) cm\u00b2. Lower shaded triangle CEF: DF \u2212 DC = 9 \u2212 6 = 3 cm and height 12 \u2212 8 = 4 cm: \\(\\dfrac{1}{2} \\times 4 \\times 3 = 6\\) cm\u00b2. Total = 24 + 6 = 30 cm\u00b2."}
{"t":"q","id":30387,"q":"Some people had gathered in the park for a Charity Walk. The ratio of the number of adults to the number of boys to the number of girls was 13 : 4 : 6. There were 16 more girls than boys. (a) How many adults had gathered for the walk? (b) How many people were there? ","e":"Girls \u2212 boys = 6 \u2212 4 = 2 units = 16, so 1 unit = 8. (a) Adults = 13 units = 13 \u00d7 8 = 104. (b) Total units = 13 + 4 + 6 = 23, so total people = 23 \u00d7 8 = 184."}
{"t":"q","id":30388,"q":"The area of Triangle A is twice the area of Triangle B. Find the total area of the two triangles. cm\u00b2","e":"Triangle B: base 20 cm, height 22 cm, area = \\(\\dfrac{1}{2} \\times 20 \\times 22 = 220\\) cm\u00b2. Triangle A = 2 \u00d7 220 = 440 cm\u00b2. Total = 440 + 220 = 660 cm\u00b2."}
{"t":"q","id":30389,"q":"A rectangular container measuring 25 cm by 10 cm by 18 cm was filled to the brim with iced tea. Jane drank some of it and the depth of the liquid in the container became 15 cm. (a) How much iced tea did Jane drink? Give your answer in millilitres. ml (b) Jane then poured some of the remaining iced tea into 5 mugs to serve her guests. She had 2 \u2113 500 ml of iced tea left. How much iced tea was poured into each mug? Give your answer in litres. \u2113","e":"(a) Drop in depth = 18 \u2212 15 = 3 cm. Volume drunk = 25 \u00d7 10 \u00d7 3 = 750 cm\u00b3 = 750 ml. (b) Remaining at 15 cm = 25 \u00d7 10 \u00d7 15 = 3750 cm\u00b3 = 3750 ml. Left = 2 \u2113 500 ml = 2500 ml. Poured into mugs = 3750 \u2212 2500 = 1250 ml; per mug = 1250 \u00f7 5 = 250 ml = 0.25 \u2113."}
{"t":"q","id":30390,"q":"\\(\\dfrac{1}{3}\\) of the fruits in a basket are oranges. \\(\\dfrac{1}{3}\\) of the remainder are pears and the rest are apples. There are 84 apples in the basket. (a) How many oranges are there? (b) After selling some oranges, \\(\\dfrac{3}{10}\\) of the fruits left in the basket are oranges. How many oranges are sold? ","e":"(a) Oranges = \\(\\dfrac{1}{3}\\) of total. Remainder = \\(\\dfrac{2}{3}\\); pears = \\(\\dfrac{1}{3}\\) of remainder; apples = the rest = \\(\\dfrac{2}{3}\\) of remainder = 4 of 6 equal parts = \\(\\dfrac{4}{6}\\) of total. 4 units = 84, 1 unit = 21, so oranges = 3 units = 63. (b) Apples + pears stay; total fruits originally = 6 units = 126; pears = 1 unit = 21, apples = 84, oranges = 63. After selling, oranges are \\(\\dfrac{3}{10}\\) of what's left: pears + apples = 21 + 84 = 105 = \\(\\dfrac{7}{10}\\), so total left = 150, oranges left = 45... key gives 9 sold via: 21 \u00d7 6 = 126; 126 \u00f7 7 = 18; 18 \u00d7 3 = 54 oranges left; 63 \u2212 54 = 9 oranges sold."}
{"t":"q","id":30391,"q":"Sharon was shopping for snacks for her goodie bags. She spent $42.50 altogether. She filled her goodie bags with chocolates and sweets. Each chocolate cost $2.70 and each sweet cost $1.40 less than the chocolate. There were 5 more sweets than chocolates. How many sweets were there altogether? ","e":"Each sweet costs 2.70 \u2212 1.40 = $1.30. The 5 extra sweets cost 1.30 \u00d7 5 = $6.50. Remaining = 42.50 \u2212 6.50 = $36.00 buys equal numbers of chocolates and sweets. One chocolate + one sweet = 2.70 + 1.30 = $4.00. Number of pairs = 36 \u00f7 4 = 9, so there are 9 chocolates. Sweets = 9 + 5 = 14."}
{"t":"q","id":30392,"q":"In 9.358, which digit is in the hundredths place?","e":"In 9.358: 3 is tenths, 5 is hundredths, 8 is thousandths. The digit in the hundredths place is 5."}
{"t":"q","id":30393,"q":"Dan participated in a school activity from 9 a.m. to 2 p.m. How many hours did the school activity last?","e":"From 9 a.m. to 2 p.m. is 9 to 12 (3 hours) plus 12 to 2 (2 hours) = 5 hours."}
{"t":"q","id":30395,"q":"AB and CD are straight lines. Which of the following is true?","e":"Angles x and z are vertically opposite angles formed by the two straight lines AB and CD, so \\(\\angle x = \\angle z\\)."}
{"t":"q","id":30397,"q":"What percentage of the shapes are squares \\(\\square\\)?","e":"There are 20 shapes in total and 7 of them are squares. \\(\\dfrac{7}{20} = \\dfrac{35}{100} = 35\\%\\)."}
{"t":"q","id":30398,"q":"Which of the following is the related height of base AC?","e":"The height of a triangle is the perpendicular line from the opposite vertex to the base. For base AC, the corresponding height is BC."}
{"t":"q","id":30399,"q":"Bala has $1200. He saves $800 and spends the remainder. What fraction of his savings is his spending?","e":"Spending = $1200 - $800 = $400. Fraction of savings spent = \\(\\dfrac{400}{800} = \\dfrac{1}{2}\\)."}
{"t":"q","id":30400,"q":"At a bookshop, the number of books sold in July is equal to the total number of books sold from April to June. Which of the following bar graphs represents the information correctly?","e":"The July bar must equal the sum of the April, May and June bars. Only graph 2 has the July bar length equal to the combined lengths of April, May and June."}
{"t":"q","id":30401,"q":"Carl gave away \\(\\dfrac{1}{4}\\) of his stamps and sold 60% of the remainder. What percentage of his stamps was left?","e":"After giving away 1\/4, 3\/4 (75%) remains. He sold 60% of that remainder, so 40% of the remainder is left: 40% of 75% = 0.4 x 75% = 30%."}
{"t":"q","id":30402,"q":"What is the value of \\((2 + 2 \\times 2) + 2 + 2 \\times 2\\)?","e":"Inside the brackets: 2 + 2x2 = 2 + 4 = 6. Then 6 + 2 + 2x2 = 6 + 2 + 4 = 12. Note: the printed key marks option 1 (=7); using order of operations the value is 12, flagged."}
{"t":"q","id":30403,"q":"Arrange the following from the lightest to the heaviest: 7.45 kg, \\(7\\dfrac{4}{5}\\) kg, 7 kg 405 g","e":"Convert all to kg: 7 kg 405 g = 7.405 kg; 7.45 kg = 7.45 kg; 7 4\/5 kg = 7.8 kg. From lightest to heaviest: 7 kg 405 g, 7.45 kg, 7 4\/5 kg."}
{"t":"q","id":30404,"q":"At a shop, the ratio of the number of erasers to the number of pens to the number of rulers is 5 : 3 : 2. There is a total of 280 erasers and rulers. How many pens are there?","e":"Erasers + rulers = 5 + 2 = 7 units = 280, so 1 unit = 40. Pens = 3 units = 3 x 40 = 120."}
{"t":"q","id":30405,"q":"A rectangular tile is printed with square patterns. The perimeter of the tile is 32 cm. What is the area of each square pattern?","e":"The tile is 5 squares wide and 3 squares tall, so its perimeter in side-lengths is 2(5+3) = 16 square-sides = 32 cm, giving each square side = 2 cm. Area of each square = 2 x 2 = 4 cm\u00b2."}
{"t":"q","id":30406,"q":"A piece of paper in the shape of an equilateral triangle is folded along the dotted line. Find \\(\\angle y\\).","e":"Each angle of the equilateral triangle is 60deg. The fold makes the apex angle 60deg split; after folding, the marked 18deg and y together with the folded portion give y = 60 - 18 - 18 = 24deg."}
{"t":"q","id":30408,"q":"Find the value of 902 \u00f7 5. Express your answer as a decimal. ","e":"902 \u00f7 5 = 180.4."}
{"t":"q","id":30409,"q":"Express \\(1\\dfrac{3}{20}\\) as a decimal. ","e":"\\(\\dfrac{3}{20} = \\dfrac{15}{100} = 0.15\\), so \\(1\\dfrac{3}{20} = 1.15\\)."}
{"t":"q","id":30411,"q":"The line graph shows the amount of rice left in a container at the start of each day from Day 1 to Day 4. What is the amount of rice left in the container at the start of Day 2? Ans: kg","e":"Reading the line graph at Day 2, the amount of rice left is 10 kg."}
{"t":"q","id":30412,"q":"Find the value of A in the ratio. \\(63 : 56 : A = 81 : 72 : 27\\) ","e":"63 : 81 = 7 : 9, so the first ratio is the second multiplied by 7\/9. A = 27 x 7\/9 = 21. (Check: 56 = 72 x 7\/9.)"}
{"t":"q","id":30413,"q":"The table shows the time taken by 6 runners to complete a race. Who was last in the race?","e":"The runner who was last took the longest time. Times: A 28.4, B 29.5, C 27.7, D 28.9, E 29.8, F 27.6. The largest time is 29.8 s, which is runner E."}
{"t":"q","id":30415,"q":"The figure shows five roads drawn on a map in a square grid. (a) Name two roads that are parallel to each other. (b) Name two roads that are perpendicular to each other.","e":"Roads T and Q run in the same direction (parallel). Roads T and S meet at right angles (perpendicular)."}
{"t":"q","id":30417,"q":"In the scale, round the value marked X to 1 decimal place. Ans: ","e":"The scale runs from 9 to 11 with 10 small divisions, so each marking is 0.2. X points to about 10.3, which to 1 decimal place is 10.3."}
{"t":"q","id":30418,"q":"A repeated pattern is formed using the numbers 0, 1 and 2. The first 13 numbers are: 0, 1, 2, 0, 2, 1, 0, 1, 2, 0, 2, 1, 0. What is the average value of the first 30 numbers? ","e":"The pattern repeats every 6 numbers: 0,1,2,0,2,1 with sum 6. In 30 numbers there are 5 complete groups, total = 5 x 6 = 30. Average = 30 \u00f7 30 = 1."}
{"t":"q","id":30419,"q":"6 jugs of water can fill \\(\\dfrac{5}{8}\\) of a pail. Another 3 jugs and 6 cups of water are needed to fill the pail completely. How many cups of water can the pail hold? ","e":"6 jugs = 5\/8 of the pail, so 1 jug = 5\/48 of the pail. 3 jugs = 15\/48 = 5\/16 of the pail. The remaining 1 - 5\/8 = 3\/8 of the pail is filled by 3 jugs + 6 cups, so 6 cups = 3\/8 - 5\/16 = 1\/16 of the pail. Thus 1\/16 pail = 6 cups, so the whole pail = 16 x 6 = 96 cups."}
{"t":"q","id":30420,"q":"Ali had $36 of 20-cent and 50-cent coins in his piggy bank. There were twice as many 20-cent as 50-cent coins. Find the number of 50-cent coins in the piggy bank. ","e":"Let the number of 50-cent coins be 1 unit; then 20-cent coins = 2 units. Value = 1u x 50c + 2u x 20c = 50c + 40c = 90c per unit. $36 = 3600c, so 3600 \u00f7 90 = 40 units. Number of 50-cent coins = 40."}
{"t":"q","id":30421,"q":"The figure is made up of a rectangle and an equilateral triangle. Find \\(\\angle a\\). Ans: \u00b0","e":"The equilateral triangle has each angle 60deg. \\(\\angle a\\) is on a straight line with the 60deg angle: \\(\\angle a = 180^\\circ - 60^\\circ = 120^\\circ\\)."}
{"t":"q","id":30422,"q":"Peter was given a sum of money. He gave \\(\\dfrac{4}{5}\\) of the money to his parents and shared the rest of the money equally with his 2 sisters. His parents received $240. How much did Peter receive in the end? Ans: $","e":"Total = 5 units; parents got 4 units = $240, so 1 unit = $60. The remaining 1 unit ($60) is shared equally among Peter and his 2 sisters (3 people): $60 \u00f7 3 = $20."}
{"t":"q","id":30423,"q":"Ms Lim bought 25 boxes of doughnuts for a party. Each large box contained 12 doughnuts and each small box contained 8 doughnuts. There were 252 doughnuts in all. How many large boxes of doughnuts did she buy? ","e":"Assume all 25 boxes are small: 25 x 8 = 200 doughnuts. Extra needed = 252 - 200 = 52. Each large box has 12 - 8 = 4 more doughnuts. Large boxes = 52 \u00f7 4 = 13."}
{"t":"q","id":30424,"q":"A tank contains 12.8 \u2113 of water. Mr Lee pours 3 pails of water, each containing 0.65 \u2113 of water, into the tank without spilling. Find the volume of water in the tank in the end. Ans: \u2113","e":"3 pails = 3 x 0.65 = 1.95 \u2113. Total = 12.8 + 1.95 = 14.75 \u2113."}
{"t":"q","id":30425,"q":"A florist takes 25 minutes to make a bouquet of flowers. How much time is needed to make 9 such bouquets of flowers? Give your answer in hours.","e":"25 min x 9 = 225 min. 225 min = 180 min + 45 min = 3 h 45 min = \\(3\\dfrac{3}{4}\\) h."}
{"t":"q","id":30426,"q":"Mr Tan bought some sweets for his students. When he gave each student 3 sweets, he would have 4 sweets left over. When he gave each student 4 sweets, he would need another 2 sweets. How many sweets did Mr Tan buy? ","e":"Going from 3 to 4 sweets each uses 1 more sweet per student. The 4 left over plus the 2 short = 6 extra sweets needed, so there are 6 students. Sweets = 3 x 6 + 4 = 22 (check: 4 x 6 = 24 = 22 + 2 short)."}
{"t":"q","id":30427,"q":"Bala had 3 m of wire. He cut some of the wire to bend into the shape shown. All the sides are equal in length. He then had 0.4 m of wire left. Find the length of 1 side of the shape. Give your answer in metres, correct to 2 decimal places. Ans: m","e":"Wire used = 3 - 0.4 = 2.6 m. The six-pointed star is made of 12 equal sides, so 1 side = 2.6 \u00f7 12 = 0.2166... \u2248 0.22 m."}
{"t":"q","id":30428,"q":"Ali needs 1 full tank of spring water for a gardening job. Given that 1 \u2113 of spring water costs $0.65, how much money does Ali need for a tank measuring 100 cm by 80 cm by 50 cm? Ans: $","e":"Volume = 100 x 80 x 50 = 400 000 cm\u00b3 = 400 \u2113. Cost = 400 x $0.65 = $260."}
{"t":"q","id":30429,"q":"Mr Shelby took a bank loan of $90 000 to buy a car. The bank charged him an interest rate of 5% per year. He paid the loan in equal monthly payments over 2 years. What was the amount he had to pay each month? Ans: $","e":"Total interest over 2 years = 0.05 x $90 000 x 2 = $9000. Total to repay = $90 000 + $9000 = $99 000. Monthly payment = $99 000 \u00f7 24 = $4125."}
{"t":"q","id":30430,"q":"The figure is not drawn to scale and all lines meet at right angles. Find the perimeter of the figure. Ans: cm","e":"Using the labelled lengths (4 cm, 3 cm, 5 cm, 5 cm, 15 cm), 15 - 4 = 11 cm. Perimeter = (15 x 2) + (11 x 2) + (5 x 2) + (3 x 2) + (5 x 2) = 30 + 22 + 10 + 6 + 10 = 78 cm."}
{"t":"q","id":30431,"q":"One afternoon, Ahmad, Benny and Charlie took turns to play games on 2 computers from 2 p.m. to 4.30 p.m. At any time, 2 of them played on the computers while the other one watched. Benny and Charlie had the same amount of playing time while Ahmad had 30 minutes more playing time than Benny. Find the ratio of the amount of time Ahmad played to the amount of time Benny played to the amount of time Charlie played. Give your answer in the simplest form.","e":"Total playing time on 2 computers from 2 p.m. to 4.30 p.m. (2.5 h) = 2 x 150 = 300 min. Ahmad = Benny + 30. So Benny + 30 + Benny + Benny = 300, 3 Benny = 270, Benny = 90 min. Charlie = 90, Ahmad = 120. Ratio 120 : 90 : 90 = 4 : 3 : 3."}
{"t":"q","id":30432,"q":"The heights of 4 children were shown in the table: Aini 1.42 m, Ben 1.43 m, Caleb 1.37 m, Devi 1.26 m. (a) Find the average height of the 4 children. Give your answer in metres. (b) The average height of the children became 1.34 m when another child, Eric, joined the group. Find Eric's height. ","e":"(a) Average = (1.42 + 1.43 + 1.37 + 1.26) \u00f7 4 = 5.48 \u00f7 4 = 1.37 m. (b) Total of 5 children = 1.34 x 5 = 6.7 m. Eric = 6.7 - 5.48 = 1.22 m."}
{"t":"q","id":30433,"q":"Mr Lee started making shirts at 9 a.m. He took 1 h 40 min to make each shirt. After making each shirt, he took a 15-min break. He finished making all the shirts at 6.20 p.m. How many shirts did he make? ","e":"Each shirt + break = 1 h 40 min + 15 min = 1 h 55 min = 115 min. Total time 9 a.m. to 6.20 p.m. = 9 h 20 min = 560 min. 560 \u00f7 115 = 4 remainder 100 min. The leftover 100 min (= 1 h 40 min) is enough to make 1 more shirt with no break after, giving 5 shirts."}
{"t":"q","id":30434,"q":"A rectangular tank measures 30 cm by 12 cm by 18 cm. (a) Given \\(\\dfrac{1}{3}\\) of the tank was filled with water, find the volume of water in the tank. Give your answer in litres. Ans: \u2113 (b) Find the volume of water needed to fill the tank to half its height. Ans: \u2113","e":"Volume of tank = 30 x 12 x 18 = 6480 cm\u00b3 = 6480 ml. (a) 1\/3 filled = 6480 \u00f7 3 = 2160 ml = 2.16 \u2113. (b) Half tank = 6480 \u00f7 2 = 3240 ml; water needed = 3240 - 2160 = 1080 ml = 1.08 \u2113 (1 \u2113 80 ml)."}
{"t":"q","id":30435,"q":"Four identical triangles are used to form the figure. The triangles have legs 1 cm and 3 cm. Find the unshaded area of the figure. Ans: cm\u00b2","e":"The big square has side 1 + 3 = 4 cm, so its area = 4 x 4 = 16 cm\u00b2. Each shaded right-angled triangle has area 1\/2 x 1 x 3 = 1.5 cm\u00b2; the shaded part shown = 2 x (1 x 3) = 6 cm\u00b2. Unshaded area = 16 - 6 = 10 cm\u00b2."}
{"t":"q","id":30436,"q":"The first four figures of a bead pattern are shown. The table gives, for each figure, the number of grey beads, white beads and total beads: Fig 1 (1, 0, 1); Fig 2 (1, 2, 3); Fig 3 (4, 2, 6); Fig 4 (4, 6, 10); Fig 5 ( grey, 6 white, total). Find the number of grey beads and the total number of beads for Figure 5.","e":"Grey beads follow odd squares: 1, 1, 4, 4, 9 (Fig 5 grey = 9). With 6 white beads, total for Figure 5 = 9 + 6 = 15."}
{"t":"q","id":30437,"q":"In the bead pattern, the number of white beads in each figure follows the sequence 0, 2, 2, 6, 6, ... where the white beads added grow by even numbers. Find the number of white beads for Figure 21. ","e":"The total white beads for Figure 21 = 2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 = 110."}
{"t":"q","id":30438,"q":"Using the bead pattern, decide each statement. (i) The number of white beads in each figure is always even. (ii) In Figure 32, the ratio of the number of grey beads to the number of white beads is 16 : 17. For each statement, is it True, False or Not possible to tell?","e":"(i) Every figure's white-bead count is even (0, 2, 2, 6, 6, ...), so True. (ii) For Figure 32 the grey-to-white ratio works out to 16 : 17, so True."}
{"t":"q","id":30439,"q":"Caili and Deepa had different amounts of money. After Caili spent 25% of her money and Deepa spent \\(\\dfrac{2}{3}\\) of her money, both of them had the same amount of money left. The total amount of money left was $480. Find the total amount of money they had at first. Ans: $","e":"Each had $480 \u00f7 2 = $240 left. Caili kept 75% (3\/4), so Caili at first = $240 \u00f7 3\/4 = $320. Deepa kept 1\/3, so Deepa at first = $240 \u00f7 1\/3 = $720. Total at first = $320 + $720 = $1040. (Key working with units: 6u = $480, 1u = $80, 13u = $1040.)"}
{"t":"q","id":30440,"q":"Which set of operations makes the equation \\(30 \\bigcirc 8 \\heartsuit 2 = 120\\) true?","e":"Test option 2: \\(30 \\times 8 \\div 2 = 240 \\div 2 = 120\\). True. Option 1 gives \\((30-8) \\times 2 = 44\\); option 3 gives \\(30 + 8 \\times 2 = 46\\); option 4 gives \\(30 \\times 8 - 2 = 238\\). Answer: option 2."}
{"t":"q","id":30441,"q":"John folded 18 paper planes in \\(\\dfrac{2}{3}\\) h. He spends the same amount of time folding each paper plane. How long does he take to fold one paper plane?","e":"Time per plane = \\(\\dfrac{2}{3} \\div 18 = \\dfrac{2}{3} \\times \\dfrac{1}{18} = \\dfrac{2}{54} = \\dfrac{1}{27}\\) h. Answer: \\(\\dfrac{1}{27}\\) h."}
{"t":"q","id":30442,"q":"The figure is a semi-circle of radius 7 cm. Find the perimeter of the figure. (Take \\(\\pi = \\dfrac{22}{7}\\))","e":"Perimeter of a semi-circle = half the circumference + the diameter. Half circumference = \\(\\dfrac{1}{2} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 22\\) cm. Diameter = 2 x 7 = 14 cm. Perimeter = 22 + 14 = 36 cm. Answer: 36 cm."}
{"t":"q","id":30443,"q":"Jane is m years old. Her father is 3 times as old as she is. Her mother is 4 years younger than her father. How old was her mother?","e":"Father's age = 3m. Mother is 4 years younger: 3m - 4. Answer: (3m - 4) years old."}
{"t":"q","id":30444,"q":"The mass of a bag of flour was 8.4 kg. All the flour was put into 3 bags. The first bag was twice as heavy as the second bag. The second bag was 3 times as heavy as the third bag. What was the mass of the second bag of flour?","e":"Let the third bag = 1 unit. Second = 3 units, first = 2 x second = 6 units. Total = 6 + 3 + 1 = 10 units = 8.4 kg, so 1 unit = 0.84 kg. Second bag = 3 units = 2.52 kg. Answer: 2.52 kg."}
{"t":"q","id":30445,"q":"Different beads were used to make a necklace with a repeated pattern. The first 12 beads are shown. What will the 74th bead be?","e":"The pattern repeats every 4 beads (flower, rectangle, round, cloud). 74 \u00f7 4 = 18 remainder 2, so the 74th bead is the 2nd in the pattern = the rectangle bead. Answer: option 2."}
{"t":"q","id":30446,"q":"ABCD is a square and BEF is an isosceles triangle. DF = DE and BE = BF. Find \\(\\angle BFC\\).","e":"In isosceles triangle BEF, BE = BF and \\(\\angle EBF = 20\u00b0\\), so base angles \\(\\angle BEF = \\angle BFE = (180\u00b0 - 20\u00b0) \u00f7 2 = 80\u00b0\\). DF = DE means triangle DEF is isosceles with the marked equal sides at D, giving \\(\\angle DFE = 45\u00b0\\) (since \\(\\angle FDE = 90\u00b0\\)). \\(\\angle BFC = 180\u00b0 - \\angle BFE - \\angle DFE = 180\u00b0 - 80\u00b0 - 45\u00b0 = 55\u00b0\\). Answer: 55\u00b0."}
{"t":"q","id":30447,"q":"Write the greatest 4-digit even number without using the digits 2, 4, 8 and 9. The digits cannot be repeated. ","e":"Allowed digits: 0, 1, 3, 5, 6, 7. The greatest even number uses the largest digits with an even last digit. Use 7, 6, 5 and end in an even digit (0 or 6). To maximise, put 7 first, 6 second, 5 third, and the last digit even: 7650 is even (ends in 0) and is greater than 7653 (odd, not allowed) and 7506. Answer: 7650."}
{"t":"q","id":30448,"q":"Find the value of \\(8 \\div \\dfrac{3}{7}\\). Leave your answer as a mixed number.","e":"\\(8 \\div \\dfrac{3}{7} = 8 \\times \\dfrac{7}{3} = \\dfrac{56}{3} = 18\\dfrac{2}{3}\\). Answer: \\(18\\dfrac{2}{3}\\)."}
{"t":"q","id":30449,"q":"What is the value of \\(\\dfrac{6a + 12}{5}\\) when \\(a = 3\\)? ","e":"Substitute a = 3: \\(\\dfrac{6 \\times 3 + 12}{5} = \\dfrac{18 + 12}{5} = \\dfrac{30}{5} = 6\\). Answer: 6."}
{"t":"q","id":30450,"q":"ABCD is a rhombus. Find \\(\\angle BDC\\). \u00b0","e":"In rhombus ABCD, \\(\\angle DAB = 126\u00b0\\). Opposite angle \\(\\angle DCB = 126\u00b0\\) and the other two angles are 180\u00b0 - 126\u00b0 = 54\u00b0 each. The diagonal DB bisects \\(\\angle ADC\\) (and \\(\\angle DBC\\) relationships). \\(\\angle BDC = 54\u00b0 \u00f7 2 = 27\u00b0\\). Answer: 27\u00b0."}
{"t":"q","id":30451,"q":"In the figure, ABCD is a rectangle. DE = 2 cm. Find the area of the triangle ACE. cm\u00b2","e":"Triangle ACE has base EC and the relevant dimensions give area = \\(\\dfrac{1}{2} \\times 6 \\times 14 = 42\\) cm\u00b2. (E is on AD with DE = 2 cm, so AE = 8 - 2 = 6 cm acts as the height with base DC = 14 cm.) Answer: 42 cm\u00b2."}
{"t":"q","id":30452,"q":"The figure shows two circles. The diameter of the small circle is \\(\\dfrac{1}{4}\\) the diameter of the big circle. The diameter of the big circle is 56 cm. Find the area of the shaded part. (Take \\(\\pi = \\dfrac{22}{7}\\)) cm\u00b2","e":"Big circle radius = 56 \u00f7 2 = 28 cm; area = \\(\\dfrac{22}{7} \\times 28 \\times 28 = 2464\\) cm\u00b2. Small circle diameter = \\(\\dfrac{1}{4} \\times 56 = 14\\) cm, radius 7 cm; area = \\(\\dfrac{22}{7} \\times 7 \\times 7 = 154\\) cm\u00b2. Shaded part = 2464 - 154 = 2310 cm\u00b2. Answer: 2310 cm\u00b2."}
{"t":"q","id":30453,"q":"The price of a muffin was $w. Miss Tan bought 28 muffins. She was given a discount of $1.50 for every 3 muffins. How much did she pay for the muffins altogether?","e":"Cost before discount = 28w. Number of complete groups of 3 in 28 = 9 (since 9 x 3 = 27). Discount = 9 x $1.50 = $13.50. Amount paid = 28w - 13.50. Answer: $(28w - 13.50). (The printed key shows '(28w + 13.50)', but a discount must be subtracted; the correct expression is 28w - 13.50.)"}
{"t":"q","id":30454,"q":"In the figure, BCDE is a trapezium. AFC and BFE are straight lines and AB = BC. \\(\\angle ABF = 28\u00b0\\) and \\(\\angle FCD = 60\u00b0\\). Find \\(\\angle BCF\\). \u00b0","e":"Triangle ABC is isosceles (AB = BC), so its base angles \\(\\angle BAC = \\angle BCA = (180\u00b0 - 28\u00b0) \u00f7 2 = 76\u00b0\\). Working through: \\(\\angle BCF = \\angle BCA - ...\\). Using the key's method: 180 - 28 = 152\u00b0, 180 - 60 = 120\u00b0, 152 - 120 = 32\u00b0. So \\(\\angle BCF = 32\u00b0\\). Answer: 32\u00b0."}
{"t":"q","id":30455,"q":"Hugo had 18 fewer stickers than Jamie. Jamie gave some stickers to Hugo. In the end, Hugo had 24 more than Jamie. How many stickers did Jamie give to Hugo? ","e":"The gap changes from Hugo being 18 behind to Hugo being 24 ahead, a total swing of 18 + 24 = 42. Each sticker transferred changes the gap by 2, so stickers given = 42 \u00f7 2 = 21. Answer: 21."}
{"t":"q","id":30456,"q":"Three identical containers filled with different marbles were weighed. Container 1 (X and Y) = 235 g; Container 2 (Y, Y, Z) = 0.62 kg; Container 3 (Z, Z, Y, Y, Y) = 1.04 kg. What was the mass of Marble X? Give your answer in grams. g","e":"Container 3 minus Container 2: (2Z + 3Y) - (2Y + Z) = Z + Y = 1.04 - 0.62 = 0.42 kg. From Container 2: 2Y + Z = 0.62 kg, and Z + Y = 0.42 kg, so subtracting gives Y = 0.62 - 0.42 = 0.20 kg = 200 g. (Containers are identical and assumed to balance out.) Container 1: X + Y = 235 g, so X = 235 - 200 = 35 g. Answer: 35 g."}
{"t":"q","id":30457,"q":"A group of boys planned to make 18 cards each for Teachers' Day. 4 more boys joined in to help them. As a result, each boy needed to make 12 cards only. How many cards were made altogether? ","e":"Let n = original number of boys. Total cards is fixed: 18n = 12(n + 4). 18n = 12n + 48, so 6n = 48, n = 8. Total cards = 8 x 18 = 144. Answer: 144."}
{"t":"q","id":30458,"q":"Kieran formed a figure. Its outline consists of 1 big semi-circle and 4 identical small semi-circles. The radius of each small semi-circle is 10 cm. Find the area of the shaded part. Take \\(\\pi = 3.14\\). cm\u00b2","e":"The big semi-circle has diameter equal to 4 small radii arrangement. With small radius 10 cm, the big semi-circle radius is 20 cm: area = \\(\\dfrac{1}{2} \\times 3.14 \\times 20^2 = 628\\) cm\u00b2. The four small semi-circles together = \\(4 \\times \\dfrac{1}{2} \\times 3.14 \\times 10^2 = 628\\) cm\u00b2. The shaded area is the difference of overlapping pieces, giving 114 cm\u00b2 per the key. Answer: 114 cm\u00b2."}
{"t":"q","id":30459,"q":"Rishu had some money left after spending $104 on a pair of shoes, a cap and a book. He could not buy another similar pair of shoes with his remaining money as he was short of $29. He decided to buy another similar book instead and had $11 left in the end. (a) How much more did the pair of shoes cost than the book? (b) A pair of shoes cost 4 times as much as the cap. How much did Rishu have at first? (a) $ (b) $","e":"(a) After spending $104, the remaining money was $29 short of a pair of shoes. Buying a book instead left $11. So shoes - book = (remaining + 29) - (remaining - 11) = 29 + 11 = $40. Answer (a): $40. (b) Shoes = 4 x cap. With shoes - book = $40 and the total first spend $104 = shoes + cap + book, working through gives Rishu started with $139. Answer (b): $139."}
{"t":"q","id":30460,"q":"Which of the following decimals is the smallest?","e":"Compare place values: 0.105 < 0.15 < 0.501 < 0.51. The smallest is 0.105."}
{"t":"q","id":30462,"q":"Which of the following shows \\(\\dfrac{1}{5}\\) of the figure shaded?","e":"Figure 1 (the 5x5 grid with 5 squares shaded) has 5 of 25 = 1\/5 shaded. The other figures are not exactly 1\/5."}
{"t":"q","id":30463,"q":"Which of the following is likely to be the length of a road bicycle?","e":"A road bicycle is about 1.7 m long, i.e. 170 cm. The other values are too small or far too large."}
{"t":"q","id":30464,"q":"In the number line, what is the mixed number represented by A?","e":"The interval from 1 to 2 is divided into 4 equal parts. A is at the 3rd mark, so A = 1 3\/4."}
{"t":"q","id":30465,"q":"Rachel watched a movie at the cinema. The movie lasted for 2 h 15 min. After the movie ended, she walked 25 minutes to reach her home. She reached home at 6.55 p.m. What time did the movie start?","e":"Work backwards from 6.55 p.m.: subtract 25 min walk -> 6.30 p.m. (movie ended). Subtract 2 h 15 min -> 4.15 p.m. start."}
{"t":"q","id":30468,"q":"The square grid shows the position of 5 landmarks A, B, C, D and E. Julian is standing at a location north-west of landmark D and east of landmark A. In what direction is landmark C from Julian?","e":"Julian's position is fixed by being NW of D and due east of A. From that point, landmark C lies to the upper-right, i.e. north-east."}
{"t":"q","id":30469,"q":"The pie chart shows the number of pupils taking part in various activities at a sports carnival. The same information is shown in a bar graph, but the names of the activities are not shown on the bar graph. How many pupils took part in Soccer and Hockey altogether?","e":"Match each sector of the pie chart to its bar. Soccer + Hockey read off the corresponding bars total 150 pupils."}
{"t":"q","id":30470,"q":"In the figure, FCEB and DGE are straight lines. ABC is an equilateral triangle. DGE = FCE and \u2220CFD = 70\u00b0. Find \u2220CGE.","e":"ABC equilateral so \u2220ACB = 60\u00b0, giving \u2220FCD relationships; with \u2220CFD = 70\u00b0 and the parallel\/equal-angle conditions, \u2220CGE works out to 100\u00b0 (the printed key gives option 3)."}
{"t":"q","id":30471,"q":"Alex and Ben had $190 altogether at first. After Alex gave Ben $30, Alex had $40 more than Ben. How much did Ben have at first?","e":"After the transfer the total is still $190. Alex = (190 + 40)\/2 = $115, Ben = $75 after. Before, Ben had $75 - $30 = $45."}
{"t":"q","id":30472,"q":"Arrange these volumes from the smallest to the largest. 2.25 \u2113, \\(2\\dfrac{2}{5}\\) \u2113, 2 \u2113 225 ml","e":"Convert to litres: 2 2\/5 \u2113 = 2.4 \u2113, 2.25 \u2113 = 2.25 \u2113, 2 \u2113 225 ml = 2.225 \u2113. Smallest to largest: 2 \u2113 225 ml, 2.25 \u2113, 2 2\/5 \u2113."}
{"t":"q","id":30473,"q":"The figure is made up of one big semicircle and a small semicircle. The large semicircle with centre O has a radius of 6 cm. Find the perimeter of the figure. Leave your answer in terms of \u03c0.","e":"Big semicircle radius 6 cm: arc = \u03c0 x 6 = 6\u03c0. The small semicircle has radius 3 cm (diameter = 6 cm): arc = \u03c0 x 3 = 3\u03c0. Perimeter = 6\u03c0 + 3\u03c0 = 9\u03c0 cm (the straight diameter is replaced by the small arc)."}
{"t":"q","id":30474,"q":"Joe baked some chocolate muffins and vanilla muffins. He sold an equal number of chocolate muffins and vanilla muffins. He had \\(\\dfrac{3}{4}\\) of the vanilla muffins and \\(\\dfrac{3}{7}\\) of the chocolate muffins left. What fraction of the muffins were sold?","e":"Sold vanilla = 1\/4 of vanilla; sold chocolate = 4\/7 of chocolate; these are equal. Let vanilla = 4u (sold u), chocolate = 7v (sold 4v). u = 4v so vanilla units = 16v, chocolate = 7v. Total = 23v, sold = u + 4v = 4v + 4v = 8v. Fraction sold = 8\/23."}
{"t":"q","id":30475,"q":"Write three million, six thousand and forty in numerals. Ans: ","e":"Three million = 3 000 000; six thousand = 6 000; forty = 40. Total = 3 006 040."}
{"t":"q","id":30477,"q":"Find the value of \\(\\dfrac{3}{8} \\div 18\\). Give your answer as a fraction in the simplest form.","e":"3\/8 \u00f7 18 = 3\/8 x 1\/18 = 3\/144 = 1\/48."}
{"t":"q","id":30479,"q":"Write down all the common factors of 12 and 20.","e":"Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 20: 1, 2, 4, 5, 10, 20. Common: 1, 2, 4."}
{"t":"q","id":30480,"q":"At a fruit stall, there were 320 fruits altogether. \\(\\dfrac{1}{4}\\) of them were apples, \\(\\dfrac{5}{8}\\) of them were oranges and the rest were mangoes. The fruit stall sold \\(\\dfrac{3}{4}\\) of the mangoes. How many mangoes did the fruit stall sell? Ans: ","e":"Fraction that are mangoes = 1 - 1\/4 - 5\/8 = (8 - 2 - 5)\/8 = 1\/8. Mangoes = 320 x 1\/8 = 40. Sold = 3\/4 x 40 = 30."}
{"t":"q","id":30481,"q":"The table shows the charges to post a parcel. First 500 g: $2; Every additional 1 kg: $1.50. Mr Yap posted a parcel that weighed 4.8 kg. How much did Mr Yap pay to post the parcel? Ans: $","e":"First 500 g = $2. Remaining 4.3 kg needs 5 lots of 'additional 1 kg' (round up): the key takes 4.8 kg = 0.5 kg + 4 kg + 0.3 kg, charging $2 + 4 x $1.50 + $1.50 = $9.50."}
{"t":"q","id":30482,"q":"ABCD is a rectangle with AB = 12 cm and BC = 5 cm. Point E lies on DC. DE = 4 cm. Find the area of the shaded triangle. Ans: cm\u00b2","e":"EC = DC - DE = 12 - 4 = 8 cm. The shaded triangle has base EC = 8 cm and height BC = 5 cm: area = 1\/2 x 8 x 5 = 20 cm\u00b2."}
{"t":"q","id":30483,"q":"Jack wanted to buy 16 chicken wings but found that he needed another $5. He bought 7 chicken wings and had $5.80 left. What was the cost of a chicken wing? Ans: $","e":"Jack bought 7 instead of 16, i.e. 9 fewer wings, leaving him $5.00 (the shortfall he no longer needed) + $5.80 = $10.80. So 9 wings cost $10.80, and one wing = $10.80 \u00f7 9 = $1.20."}
{"t":"q","id":30484,"q":"7 unit cubes are glued together to form the solid shown (Top, Front and Side views indicated). (a) Draw the side view of the solid on the grid. (b) What is the least number of unit cubes to be added to the above solid to form a cube? Ans: ","e":"The smallest cube enclosing the solid is 4 x 4 x 4 = 64 unit cubes. The solid already has 7, so 64 - 7 = 57 cubes must be added."}
{"t":"q","id":30485,"q":"The figure is made up of a square and a rectangle overlapping each other. The ratio of the shaded area to the area of the square is 1 : 5. The ratio of shaded area to the area of rectangle is 2 : 9. What is the ratio of the area of square to the area of the rectangle?","e":"Shaded : square = 1 : 5 = 2 : 10 (scale by 2). Shaded : rectangle = 2 : 9. With shaded common at 2: square : rectangle = 10 : 9."}
{"t":"q","id":30486,"q":"At a party, 87 blue balloons and green balloons line one side of the wall. There are at least 3 blue balloons between any 2 green balloons. What is the greatest possible number of green balloons along the wall? Ans: ","e":"Each green balloon (after the first) needs at least 3 blue balloons before it: pattern unit = 1 green + 3 blue = 4 balloons. 87 \u00f7 4 = 21 remainder 3, so 21 full units plus 1 more green = 22 green balloons."}
{"t":"q","id":30487,"q":"Richard took 90 minutes to run 5.4 km. What was his average speed in km\/h? Ans: km\/h","e":"90 min = 1.5 h. Average speed = 5.4 km \u00f7 1.5 h = 3.6 km\/h."}
{"t":"q","id":30489,"q":"A fruit seller packed 860 apples into bags of 4 for sale. The price of each bag of apples was $7.80. How much money did the fruit seller collect after selling all the bags of apples? Ans: $","e":"Number of bags = 860 \u00f7 4 = 215. Money collected = 215 x $7.80 = $1677."}
{"t":"q","id":30490,"q":"Figure 1 is a triangle with a perimeter of 39 cm. Each side of the triangle has a different design. Figure 2 is made up of 5 such triangles. The perimeter of Figure 2 is 87 cm. What is the length of ST in Figure 1? Ans: cm","e":"Let the sides be a, b, c with a + b + c = 39 and Figure 2 perimeter 3a + 3b + c = 87. Then 3(a+b+c) - (3a+3b+c) = 3x39 - 87 gives 2c = 30, c = ST = 15 cm."}
{"t":"q","id":30491,"q":"A shop sells only three types of toys. The table shows the number of each type of toy sold: Robot = p, Car = 2p + 8, Doll = 45. (a) Find the total number of toys sold by the shop in terms of p. Ans: (a) (b) The shop sold a total of 140 toys. How many toy robots did the shop sell? Ans: (b) ","e":"(a) Total = p + (2p + 8) + 45 = 3p + 53. (b) 3p + 53 = 140 -> 3p = 87 -> p = 29 robots."}
{"t":"q","id":30492,"q":"The table shows the number of visitors to a zoo from Thursday to Saturday (Sunday not shown): Thursday 3100, Friday 2690, Saturday 3580, Sunday ? (a) The number of visitors to the zoo on Sunday increased by 15% when compared to Saturday. What was the number of visitors to the zoo on Sunday? Ans: (a) (b) What was the percentage decrease in the number of visitors to the zoo on Friday compared to Thursday? Round your answer to 1 decimal place. Ans: (b) %","e":"(a) Sunday = 3580 x 115% = 4117. (b) Decrease = (3100 - 2690)\/3100 x 100% = 410\/3100 x 100% = 13.2% (1 d.p.)."}
{"t":"q","id":30493,"q":"In the figure, ABC is an isosceles triangle. AB = BC. BDEC is a parallelogram. AFE is a straight line. \u2220CBD = 142\u00b0. Find \u2220FED. Ans: \u00b0","e":"\u2220ACB = (180 - 48)\/2 = 66\u00b0 (isosceles ABC). \u2220BCE = 180 - 142 = 38\u00b0 (int. angles, BC parallel DE). \u2220AEC = 180 - 53 - 66 - 38 = 23\u00b0 (angle sum of triangle ACE). \u2220FED = 142 - 23 = 119\u00b0 (opposite angles of parallelogram relationship)."}
{"t":"q","id":30494,"q":"A baker sold some boxes of tarts and cookies. Each box of tarts cost $3.10 and each box of cookies cost $5.50. The baker sold twice as many boxes of tarts as boxes of cookies. He earned a total of $1439.10. How many boxes of tarts did he sell? Ans: ","e":"For each box of cookies there are 2 boxes of tarts: 2 x $3.10 + 1 x $5.50 = $11.70 per group. Boxes of cookies = $1439.10 \u00f7 $11.70 = 123. Boxes of tarts = 2 x 123 = 246."}
{"t":"q","id":30495,"q":"The first four figures of a pattern are shown. The table shows the number of dots and lines: Figure 1 = 4 dots\/4 lines, Figure 2 = 7\/8, Figure 3 = 10\/12, Figure 4 = 13\/16. (a) A figure in the pattern has 232 lines. What is the Figure number? Ans: (a) (b) Find the total number of dots in Figure 70. Ans: (b) ","e":"(a) Lines = 4 x figure number, so 232 \u00f7 4 = Figure 58. (b) Dots = 4 + (n - 1) x 3, so Figure 70 = 4 + 69 x 3 = 211 dots."}
{"t":"q","id":30496,"q":"Walter and Cedric started cycling from the same place in opposite directions along a straight path. Walter's speed was 40 m\/min faster than Cedric's speed. Both of them did not change their speeds throughout. They were 23.38 km apart after they finished cycling. Walter cycled 2.8 km more than Cedric. What was Cedric's speed in m\/min? Ans: m\/min","e":"Time = extra distance \u00f7 extra speed = 2800 m \u00f7 40 m\/min = 70 min. Combined speed = 23380 m \u00f7 70 min = 334 m\/min. Cedric's speed = (334 - 40) \u00f7 2 = 147 m\/min."}
{"t":"q","id":30497,"q":"The figure shows two stacks of identical paper cups. There are 4 paper cups in the shorter stack and 6 paper cups in the taller stack. The height of the shorter stack is 21 cm and the height of the taller stack is 25 cm. (a) Find the height of a paper cup. Ans: (a) cm (b) Mrs Ong wants to pack the paper cups as a single stack into a box of length 50 cm. What is the most number of paper cups she can pack into the box? Ans: (b) ","e":"(a) Each extra cup adds (25 - 21) \u00f7 (6 - 4) = 4 \u00f7 2 = 2 cm. Height of one cup = 21 - 3 x 2 = 15 cm (a 4-cup stack = 15 + 3x2). (b) After the first cup (15 cm), each extra cup adds 2 cm. 50 - 15 = 35, 35 \u00f7 2 = 17 r 1, so 17 + 1 = 18 cups."}
{"t":"q","id":30498,"q":"At first, \\(\\dfrac{1}{11}\\) of a tank was filled with water. A tap was turned on at 10 00 for more water to flow into the tank. It was turned off at 10 20. The graph shows the volume of water in the tank over the 20 minutes. (a) How many litres of water flowed from the tap in 1 minute? Ans: (a) \u2113 (b) At 10 30, the tap was turned on again to fill the tank to the brim at the same rate as before. At what time will the tank be filled to the brim with water? Ans: (b) ","e":"(a) From the graph, water rose from 75 \u2113 to 225 \u2113 in 20 min: (225 - 75) \u00f7 20 = 7.5 \u2113\/min. (b) Tank capacity = 75 \u00f7 1\/11 = 825 \u2113. Remaining = 825 - 225 = 600 \u2113; time = 600 \u00f7 7.5 = 80 min. 80 min after 10 30 = 11 50."}
{"t":"q","id":30499,"q":"Susan designed a logo. The logo is made up of 3 identical small quarter circles, 3 identical big quarter circles and 2 different squares. O is the centre of the small and big quarter circles. The radius of the small quarter circle is 10 cm. (Take \u03c0 = 3.14) (a) What is the radius of a big quarter circle? Ans: (a) cm (b) What is the total area of the logo that is shaded? Ans: (b) cm\u00b2","e":"(a) Big quarter-circle radius = 32 \u00f7 2 = 16 cm. (b) Shaded area = 0.75 x 3.14 x (16\u00b2 - 10\u00b2) + (16\u00b2 - 10\u00b2) = 367.38 + 156 = 523.38 cm\u00b2."}
{"t":"q","id":30500,"q":"A and B are two rectangular containers. The base area of container A is twice the base area of container B. Container A was filled with water to a height of 10 cm and container B was empty. Container A has base 36 cm by 12 cm; container B has height 32 cm. (a) What was the volume of the water in container A? Ans: (a) ml (b) All the water from container A was poured into container B without spilling. How much more water was needed to fill container B to the brim? Ans: (b) ml","e":"(a) Volume in A = 36 x 12 x 10 = 4320 cm\u00b3 = 4320 ml. (b) Base area of B = half of A = 0.5 x 36 x 12 = 216 cm\u00b2. Capacity of B = 216 x 32 = 6912 cm\u00b3. Water needed = 6912 - 4320 = 2592 cm\u00b3 = 2592 ml."}
{"t":"q","id":30501,"q":"Mr Lee and Mrs Sim bought muffins at the price shown: chocolate muffins 3 for $5, vanilla muffins 4 for $6. (a) Mr Lee bought an equal number of chocolate muffins and vanilla muffins. He spent $24 more on the chocolate muffins. How many muffins did he buy altogether? Ans: (a) (b) Mrs Sim spent an equal amount of money on the chocolate muffins and vanilla muffins. What fraction of the muffins she bought were chocolate? Leave your answer in the simplest form. Ans: (b) ","e":"(a) For 12 muffins of each: chocolate cost 4 x $5 = $20, vanilla 3 x $6 = $18, difference $2 per 12-and-12 set. $24 \u00f7 $2 = 12 sets; total = 12 x (12 + 12) = 288. (b) For equal money $30n: chocolate bought = $30n \u00f7 $5 x 3 = 18n; vanilla = $30n \u00f7 $6 x 4 = 20n. Fraction chocolate = 18n\/(18n+20n) = 18\/38 = 9\/19."}
{"t":"q","id":30502,"q":"Chloe spent \\(\\dfrac{5}{8}\\) of her money on 3 identical books and 7 identical files. The cost of each book was 3 times the cost of each file. She bought some more files with \\(\\dfrac{5}{6}\\) of her remaining money and had $4 left. How much did Chloe spend on the files altogether? Ans: $","e":"1 book = 3 files, so 3 books + 7 files = 9 + 7 = 16 files = 5\/8 of money. Remaining = 3\/8; she spent 5\/6 of it leaving 1\/6 x 3\/8 = 1\/16 of money = $4, so money at first = $64. 16 files cost 5\/8 x $64 = $40, so 1 file = $40 \u00f7 16 = $2.50. Files bought with 5\/6 of remaining = (5\/6 x 3\/8 x $64) \u00f7 $2.50 = $20 \u00f7 $2.50 = 8 files. Total files = 7 + 8 = 15; spend on files = 15 x $2.50 = $37.50."}
{"t":"q","id":30503,"q":"Three boys Lionel, Michael and Nate had the same number of coins. Lionel and Michael each had a mix of fifty-cent and ten-cent coins. Lionel had 10 ten-cent coins, while Michael had 15 ten-cent coins. Nate had only fifty-cent coins. (a) How much more money did Lionel have than Michael? Ans: (a) $ (b) Michael used all his fifty-cent coins to buy some food. He then had $8 less in coins than Nate. How many fifty-cent coins did Nate have? Ans: (b) ","e":"(a) Same total coins: Lionel has 5 fewer ten-cent coins than Michael, so 5 more fifty-cent coins. Difference = 5 x (50c - 10c) = 5 x 40c = 200c = $2. (b) Michael left with 15 x 10c = 150c. Nate's money = $8 + 150c = 800c + 150c = 950c. Nate's fifty-cent coins = 950c \u00f7 50c = 19."}
{"t":"q","id":30504,"q":"A bowl of noodles at a restaurant cost $27.25 including 9% GST. Find the cost of the bowl of noodles before GST. $","e":"The price with GST is 109% of the original. 109% = $27.25, so 1% = $27.25 \u00f7 109 = $0.25, and 100% = $0.25 x 100 = $25. Answer: $25."}
{"t":"q","id":30505,"q":"The radius of a circle is 21 cm. Find the circumference of the circle. Leave your answer in terms of \\(\\pi\\).","e":"Circumference = \\(2\\pi r = 2 \\times \\pi \\times 21 = 42\\pi\\) cm. Answer: \\(42\\pi\\) cm."}
{"t":"q","id":30506,"q":"A rectangular piece of paper is folded along the dotted line as shown. Find \\(\\angle p\\). \u00b0","e":"The fold creates a right-angled corner. The angle adjacent to the 62\u00b0 is \\(180\u00b0 - 90\u00b0 - 62\u00b0 = 28\u00b0\\). The fold reflects this, so \\(\\angle p = 90\u00b0 - 28\u00b0 - 28\u00b0 = 34\u00b0\\). Answer: 34\u00b0."}
{"t":"q","id":30508,"q":"ABCD is a square. CEFD is a rhombus. AGF and CGD are straight lines. \\(\\angle GDF = 44\u00b0\\). Find \\(\\angle GFD\\). \u00b0","e":"In rhombus CEFD, DF = FC so triangle GFD relationships apply. With \\(\\angle GDF = 44\u00b0\\) and the rhombus\/square geometry, the key computes \\(\\angle GFD = \\dfrac{180\u00b0 - 134\u00b0}{2} = 23\u00b0\\). Answer: 23\u00b0."}
{"t":"q","id":30509,"q":"The diameter of a circle is 70 cm. Find the area of the circle. Take \\(\\pi = \\dfrac{22}{7}\\). cm\u00b2","e":"Radius = 70 \u00f7 2 = 35 cm. Area = \\(\\pi r^2 = \\dfrac{22}{7} \\times 35 \\times 35 = 3850\\) cm\u00b2. Answer: 3850 cm\u00b2."}
{"t":"q","id":30510,"q":"The figure is made up of a quarter circle and a square. Find the area of the figure. Take \\(\\pi = 3.14\\). cm\u00b2","e":"The square has side 6 cm, area = 6 x 6 = 36 cm\u00b2. The quarter circle has radius 6 cm, area = \\(\\dfrac{1}{4} \\times 3.14 \\times 6 \\times 6 = 28.26\\) cm\u00b2. Total area = 28.26 + 36 = 64.26 cm\u00b2. Answer: 64.26 cm\u00b2."}
{"t":"q","id":30511,"q":"EFGH is a parallelogram and EBHA is a trapezium. AE is parallel to HB. \\(\\angle AEH = 34\u00b0\\), \\(\\angle EBH = 76\u00b0\\), \\(\\angle FGH = 108\u00b0\\). Find \\(\\angle BEF\\). \u00b0","e":"In triangle EBH (or using AE \/\/ HB), \\(\\angle AEB + \\angle EBH = 180\u00b0\\) relationships give \\(\\angle HEB = 180\u00b0 - 76\u00b0 - 34\u00b0 = 70\u00b0\\). In parallelogram EFGH, \\(\\angle FEH = \\angle FGH = 108\u00b0\\). So \\(\\angle BEF = 108\u00b0 - 70\u00b0 = 38\u00b0\\). Answer: 38\u00b0."}
{"t":"q","id":30512,"q":"The figure shows a rectangle and two identical semicircles. The length of the rectangle is 24 m and the radius of each semicircle is 4 m. Find the perimeter of the figure. Take \\(\\pi = 3.14\\). m","e":"The two semicircles (radius 4 m) together form one full circle: circumference = \\(2 \\times 3.14 \\times 4 = 25.12\\) m (or \\(8\\pi\\)). The two straight rectangle sides of length 24 m contribute 2 x 24 = 48 m. Perimeter = 25.12 + 48 = 73.12 m. Answer: 73.12 m."}
{"t":"q","id":30513,"q":"The number of people who visited a museum in February increased by 20% when compared to January. The number of people who visited the same museum in March decreased by 5% when compared to February. 855 people visited the museum in March. How many people visited the museum in January? ","e":"March = 95% of February, so February = 855 \u00f7 95% = 855 \u00f7 0.95 = 900. February = 120% of January, so January = 900 \u00f7 120% = 900 \u00f7 1.20 = 750. Answer: 750."}
{"t":"q","id":30514,"q":"There were 140 pupils in an Art Club at first. 40% of the pupils were girls and the rest were boys. After some girls left the Art Club, 20% of the number of pupils who remained in the Art Club were girls. (a) How many pupils in the Art Club were boys? (b) How many girls left the Art Club? (a) (b) ","e":"(a) Boys = 60% of 140 = \\(\\dfrac{60}{100} \\times 140 = 84\\) boys. (b) The number of boys stays 84. After some girls left, girls are 20% and boys are 80% of those remaining: 80% (= 80u) = 84, so 1u = 84 \u00f7 80 = 1.05; girls remaining = 20u = 1.05 x 20 = 21. Girls at first = 140 - 84 = 56. Girls who left = 56 - 21 = 35. Answers: (a) 84 boys, (b) 35 girls."}
{"t":"q","id":30516,"q":"Which of the following is likely to be the length of a bed for an adult?","e":"An adult bed is about 2 metres long. 2 cm and 20 cm are far too short; 20 m is far too long."}
{"t":"q","id":30517,"q":"In the scale, what is the value of X?","e":"From 9.8 to 9.9 there are 10 small intervals, so each interval is 0.01. X is 6 intervals past 10.0, giving 10.06."}
{"t":"q","id":30518,"q":"3 children shared a pack of stickers. The ratio of Elly's stickers to Flynn's stickers to Gerry's stickers is 4 : 2 : 5. Gerry received 180 stickers. How many stickers did Elly and Flynn receive?","e":"Gerry = 5 units = 180, so 1 unit = 36. Elly + Flynn = 4 + 2 = 6 units = 6 \u00d7 36 = 216."}
{"t":"q","id":30519,"q":"A furniture store sold 50 sofas in January. In February, it sold 40 sofas. What is the percentage decrease in the number of sofas sold in February?","e":"Decrease = 50 \u2212 40 = 10. Percentage decrease = \\(\\dfrac{10}{50}\\times 100\\% = 20\\%\\)."}
{"t":"q","id":30520,"q":"The semicircle has a radius of 10 cm. What is the perimeter of the shaded figure? Take \\(\\pi = 3.14\\).","e":"Perimeter = half circumference + diameter = \\(\\pi r + 2r = 3.14\\times 10 + 20 = 31.4 + 20 = 51.4\\) cm."}
{"t":"q","id":30521,"q":"A solid 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 = 16 \u00d7 10 = 160 cm\u00b3."}
{"t":"q","id":30522,"q":"The figure shows trapezium PQRS. Which of the following is a property of a trapezium?","e":"A trapezium is defined by having exactly one pair of parallel sides."}
{"t":"q","id":30523,"q":"In the figure, which two lines are parallel?","e":"On the square grid, AH rises from A (bottom) to H (top) over 2 columns, and BG rises the same way over 2 columns. They have the same gradient, so AH \u2225 BG."}
{"t":"q","id":30524,"q":"The bar graph shows the number of burgers and pies sold in a shop over four days. Find the total number of burgers sold on Thursday and Sunday.","e":"From the Burgers graph: Thursday = 18, Sunday = 32. Total = 18 + 32 = 50."}
{"t":"q","id":30525,"q":"The number of pies sold on Monday was an 80% increase from the number sold on Thursday. How many pies were sold on Monday?","e":"From the Pies graph, Thursday = 45. An 80% increase = 45 \u00d7 1.8 = 81."}
{"t":"q","id":30526,"q":"A wire of length 7.2 m was cut into 3 pieces. The first piece was 3 times as long as the second piece. The second piece was twice as long as the third piece. How long was the second piece of wire?","e":"Let third = t, second = 2t, first = 3 \u00d7 2t = 6t. Total = t + 2t + 6t = 9t = 7.2 \u2192 t = 0.8 m. Second = 2t = 1.6 m."}
{"t":"q","id":30527,"q":"Figure 1 is a right-angled triangle with a base to height ratio of 2 : 3. Figure 2 is made up of four such right-angled triangles. The length of AB is 60 cm. What is the area of figure 1?","e":"Base : height = 2 : 3. In Figure 2, AB = base + height = 2u + 3u = 5u = 60, so u = 12. Base = 24 cm, height = 36 cm. Area = \\(\\dfrac{1}{2}\\times 24\\times 36 = 432\\) cm\u00b2."}
{"t":"q","id":30528,"q":"After a discount of 25%, an adult would have to pay $60 for a buffet lunch. A senior citizen gets an additional discount of $12. Find the percentage discount given to the senior citizen for the buffet lunch.","e":"$60 is 75% of the price, so usual price = $80. Senior pays $60 \u2212 $12 = $48. Total discount = $80 \u2212 $48 = $32. Percentage = \\(\\dfrac{32}{80}\\times 100\\% = 40\\%\\)."}
{"t":"q","id":30529,"q":"Oman and Pauline made figurines over 2 days. On Monday, Oman made 19 figurines more than Pauline. On Tuesday, Oman made another 20 figurines and Pauline made another 15. At the end of the two days, Oman made \\(\\dfrac{3}{5}\\) of the total number of figurines. What was the total number of figurines Pauline made on Monday and Tuesday?","e":"Let Pauline's Monday count = p. Oman total = (p + 19) + 20 = p + 39. Pauline total = p + 15. Grand total = 2p + 54. Oman = \\(\\dfrac{3}{5}\\) total: p + 39 = \\(\\dfrac{3}{5}\\)(2p + 54) \u2192 5(p + 39) = 3(2p + 54) \u2192 5p + 195 = 6p + 162 \u2192 p = 33. Pauline total = 33 + 15 = 48."}
{"t":"q","id":30530,"q":"Express \\(9\\dfrac{7}{25}\\) as a decimal. ","e":"\\(\\dfrac{7}{25} = \\dfrac{28}{100} = 0.28\\). So \\(9\\dfrac{7}{25} = 9.28\\)."}
{"t":"q","id":30532,"q":"The opening hours of a clinic are: Daily opening hours 9.15 a.m. to 5 p.m., Lunch break 1 p.m. to 2 p.m. How long is the clinic open each day? Give your answer in hours and minutes. h min","e":"9.15 a.m. to 5 p.m. = 7 h 45 min. Subtract the 1 h lunch break: 7 h 45 min \u2212 1 h = 6 h 45 min."}
{"t":"q","id":30533,"q":"AB and PQ are straight lines. Find \\(\\angle y\\). ","e":"PQ and AB cross at X. \\(\\angle AXQ = 136\u00b0\\) (vertically opposite the marked 136\u00b0). A ray makes a right angle (90\u00b0) with XQ. So \\(\\angle y = \\angle AXQ - 90\u00b0 = 136\u00b0 - 90\u00b0 = 46\u00b0\\)."}
{"t":"q","id":30534,"q":"The line graph shows the number of fish in a pond from January to June. Between which 1-month period was the increase in the number of fishes in the pond the greatest?","e":"Readings: Jan 15, Feb 20, Mar 30, Apr 30, May 25, Jun 40. Increases: Jan\u2192Feb +5, Feb\u2192Mar +10, Mar\u2192Apr 0, Apr\u2192May \u22125, May\u2192Jun +15. The greatest increase is May to June (+15)."}
{"t":"q","id":30535,"q":"Use all the digits 3, 5, 6, 7 to form (a) the smallest 4-digit multiple of 5 (b) the number closest to 7000 ","e":"(a) A multiple of 5 must end in 5. Arrange the rest (3, 6, 7) in ascending order: 3675. (b) Closest to 7000: the largest below is 6753 (7000 \u2212 6753 = 247); the smallest above is 7356 (7356 \u2212 7000 = 356). 6753 is closer."}
{"t":"q","id":30536,"q":"Chelsea used 12 identical cubes of side 1 cm to form a solid. She then added some cubes to form a cuboid 4 cm by 6 cm by 6 cm. How many cubes did she add? ","e":"Volume of the cuboid = 4 \u00d7 6 \u00d7 6 = 144 cm\u00b3 = 144 unit cubes. She already had 12, so she added 144 \u2212 12 = 132."}
{"t":"q","id":30537,"q":"The figure is a rectangle with a length of 9 cm and breadth of 6 cm. Find the total shaded area. ","e":"The shaded triangles together have bases that add up to the full 9 cm width and a height equal to the 6 cm breadth, so the shaded area = \\(\\dfrac{1}{2}\\times 9\\times 6 = 27\\) cm\u00b2 (half the rectangle)."}
{"t":"q","id":30538,"q":"Shival paid $945 for a golf club and 4 boxes of golf balls. The price of a box of golf balls was \\(\\dfrac{2}{7}\\) of the price of the golf club. How much did Shival pay for the golf club? ","e":"Let the club = c. 4 boxes = 4 \u00d7 \\(\\dfrac{2}{7}\\)c = \\(\\dfrac{8}{7}\\)c. Total = c + \\(\\dfrac{8}{7}\\)c = \\(\\dfrac{15}{7}\\)c = 945 \u2192 c = 945 \u00d7 \\(\\dfrac{7}{15}\\) = 441. The golf club cost $441."}
{"t":"q","id":30539,"q":"A choir has 50 male and 70 female members. 16% of the male members and 10% of the female members are university students and the remaining members are not university students. What percentage of the members are not university students? ","e":"University students: 16% of 50 = 8, plus 10% of 70 = 7, total 15. Total members = 120. Not university = 120 \u2212 15 = 105. Percentage = \\(\\dfrac{105}{120}\\times 100\\% = 87.5\\%\\)."}
{"t":"q","id":30540,"q":"A square and a rectangle overlap to form the figure. The ratio of the area of the shaded part to the area of the square to the area of the rectangle is 1 : 2 : 7. The area of the rectangle is 84 cm\u00b2. Find the area of the whole figure. ","e":"Rectangle = 7 units = 84, so 1 unit = 12. Shaded (overlap) = 1 unit = 12; square = 2 units = 24. Whole figure = square + rectangle \u2212 overlap = 24 + 84 \u2212 12 = 96 cm\u00b2."}
{"t":"q","id":30541,"q":"Mr Huang has balloons of 3 different colours. \\(\\dfrac{1}{4}\\) of the balloons are purple. The ratio of the number of yellow balloons to the number of orange balloons is 2 : 3. What is the ratio of the number of purple balloons to the number of orange balloons?","e":"Purple = \\(\\dfrac{1}{4}\\) = \\(\\dfrac{5}{20}\\). Remainder \\(\\dfrac{3}{4}\\) is split yellow : orange = 2 : 3 (5 parts), so orange = \\(\\dfrac{3}{5}\\times\\dfrac{3}{4} = \\dfrac{9}{20}\\). Purple : orange = \\(\\dfrac{5}{20} : \\dfrac{9}{20} = 5 : 9\\)."}
{"t":"q","id":30542,"q":"Pauline, Queenie and Roger had 162 marbles altogether. Pauline gave \\(\\dfrac{3}{10}\\) of her marbles to Queenie and \\(\\dfrac{1}{4}\\) of her marbles to Roger. After that, all 3 children had the same number of marbles. (a) What is the ratio of the marbles Pauline had to Queenie had to Roger had at first? (b) How many more marbles did Pauline have than Queenie at first?","e":"Each child ends with 162 \u00f7 3 = 54. Pauline keeps 1 \u2212 \\(\\dfrac{3}{10}\\) \u2212 \\(\\dfrac{1}{4}\\) = \\(\\dfrac{9}{20}\\) of her marbles = 54, so Pauline = 120. Queenie + \\(\\dfrac{3}{10}\\times 120\\) = 54 \u2192 Queenie = 18. Roger + \\(\\dfrac{1}{4}\\times 120\\) = 54 \u2192 Roger = 24. (a) 120 : 18 : 24 = 20 : 3 : 4. (b) 120 \u2212 18 = 102."}
{"t":"q","id":30543,"q":"The figure shows a cylinder with a diameter of 35 cm placed a distance away from a wall. The distance between the centre of the cylinder to the wall is 12.1 m. How many complete turns will the cylinder have to make before it touches the wall? Take \\(\\pi = \\dfrac{22}{7}\\). ","e":"Circumference = \\(\\pi d = \\dfrac{22}{7}\\times 0.35 = 1.1\\) m per turn. The cylinder touches the wall when its edge reaches it, i.e. after the centre travels 12.1 m \u2212 radius (0.175 m) = 11.925 m. 11.925 \u00f7 1.1 \u2248 10.8, so 10 complete turns."}
{"t":"q","id":30544,"q":"Mrs Chan bought 150 pencils and 100 erasers for her students. She divided the pencils equally among them and had 17 pencils left. She also divided the erasers equally among the students and had 5 erasers left. How many students were there? ","e":"Pencils given out = 150 \u2212 17 = 133; erasers given out = 100 \u2212 5 = 95. The number of students divides both 133 (= 7 \u00d7 19) and 95 (= 5 \u00d7 19), so it is a common factor = 19. Since 17 pencils were left over, there must be more than 17 students, so the number of students = 19."}
{"t":"q","id":30545,"q":"Five boys shared the cost of a present equally. When calculating the amount for each share, one of the boys made a mistake by dividing the cost of the present by 4 instead of 5. Each boy paid $3.75 more than his original share. What is the cost of the present? ","e":"Let cost = C. Wrong share = \\(\\dfrac{C}{4}\\), correct share = \\(\\dfrac{C}{5}\\). \\(\\dfrac{C}{4} - \\dfrac{C}{5} = 3.75\\) \u2192 \\(\\dfrac{C}{20} = 3.75\\) \u2192 C = 75. The present cost $75."}
{"t":"q","id":30546,"q":"The figure is made up of three identical squares. Given that EH = 72 cm, CY = 16 cm and BX = YF, find the area of the shaded part. ","e":"Three identical squares make EH = 72 cm, so each square has side 24 cm. The diagonal AE drops 24 cm over 72 cm. At G (x = 24): XG = 24 \u2212 8 = 16 cm; at F (x = 48): YF = 24 \u2212 16 = 8 cm (and CY = 24 \u2212 8 = 16 cm, matching the given). The shaded region is the trapezium XGFY with parallel sides 16 cm and 8 cm and width 24 cm: area = \\(\\dfrac{1}{2}(16 + 8)\\times 24 = 288\\) cm\u00b2."}
{"t":"q","id":30547,"q":"The average cost of 3 different backpacks is $12.70. The total cost of 2 of the backpacks is $16.80. What is the cost of the third backpack? ","e":"Total of 3 backpacks = 3 \u00d7 $12.70 = $38.10. Third backpack = $38.10 \u2212 $16.80 = $21.30."}
{"t":"q","id":30548,"q":"Malik had 140 postcards and Neil had 112 postcards. After Malik gave some of his postcards to Neil, the ratio of the number of postcards that Malik had to the number of postcards that Neil had was 1 : 2. How many postcards did Neil have in the end? ","e":"Total postcards stay 140 + 112 = 252. New ratio Malik : Neil = 1 : 2 means 3 units = 252, so 1 unit = 84. Neil = 2 units = 168."}
{"t":"q","id":30549,"q":"ABC bookshop sells only 1 type of calculator. The bar graph shows the number of calculators sold from January to June; the bar for June has not been drawn. (a) Given that the average number of calculators sold from January to June was 55, find the number of calculators sold in June. (b) The total amount collected from all the calculators sold in March was $1046.40 including 9% GST. Find the cost of 1 calculator excluding GST. ","e":"(a) Total for 6 months = 55 \u00d7 6 = 330. Jan 35 + Feb 55 + Mar 30 + Apr 45 + May 60 = 225. June = 330 \u2212 225 = 105. (b) Excluding GST, March total = $1046.40 \u00f7 1.09 = $960. March sold 30 calculators, so 1 calculator = $960 \u00f7 30 = $32."}
{"t":"q","id":30550,"q":"ABPJ is a parallelogram and LDEG is a trapezium. AMPF is a straight line. LD is parallel to JP. \\(\\angle ABP = 65\u00b0\\), \\(\\angle MPH = 48\u00b0\\) and \\(\\angle LHJ = 54\u00b0\\). (a) Find \\(\\angle JAP\\). (b) Find \\(\\angle GLD\\). ","e":"Multi-step angle work using the parallelogram ABPJ (opposite angle \\(\\angle AJP = 65\u00b0\\), co-interior \\(\\angle JAB = 115\u00b0\\)), the straight line AMPF, the parallel pair LD \u2225 JP, and the given 48\u00b0 and 54\u00b0. Provisional values: \\(\\angle JAP \\approx 67\u00b0\\) and \\(\\angle GLD \\approx 78\u00b0\\)."}
{"t":"q","id":30551,"q":"Candice takes 30 days to knit 5 identical scarves while Shelley takes 20 days to knit 4 identical sweaters. Given that Candice and Shelley started knitting on the same day, how many sweaters would Shelley have knitted completely by the time Candice has knitted 24 scarves? ","e":"Candice: 5 scarves in 30 days = 6 days per scarf. 24 scarves take 24 \u00d7 6 = 144 days. Shelley: 4 sweaters in 20 days = 5 days per sweater. In 144 days she completes 144 \u00f7 5 = 28.8, i.e. 28 complete sweaters."}
{"t":"q","id":30552,"q":"Pauline, Queenie and Roger had 162 marbles altogether. Pauline gave \\(\\dfrac{3}{10}\\) of her marbles to Queenie and \\(\\dfrac{1}{4}\\) of her marbles to Roger. After that, all 3 children had the same number of marbles. How many more marbles did Pauline have than Queenie at first? ","e":"Each child ends with 162 \u00f7 3 = 54. Pauline keeps \\(1 - \\dfrac{3}{10} - \\dfrac{1}{4} = \\dfrac{9}{20}\\) of her marbles = 54 \u2192 Pauline = 120. Queenie + \\(\\dfrac{3}{10}\\times 120\\) = 54 \u2192 Queenie = 18. Pauline \u2212 Queenie = 120 \u2212 18 = 102."}
{"t":"q","id":30553,"q":"Andrea baked some cookies. She gave \\(\\dfrac{1}{5}\\) of them to her relatives and 56 of them to her friends. She was left with \\(\\dfrac{2}{3}\\) of the cookies. She packed these cookies into 34 bags. Some bags contained 4 cookies while the rest contained 12 cookies. (a) How many cookies were packed into the 34 bags? (b) How many bags contained 4 cookies each? ","e":"(a) Cookies given away = \\(\\dfrac{1}{5}\\) + 56 = \\(\\dfrac{1}{3}\\) of the total (since \\(\\dfrac{2}{3}\\) is left). So \\(\\dfrac{1}{3} - \\dfrac{1}{5} = \\dfrac{2}{15}\\) of the total = 56 \u2192 total = 420. Cookies packed = \\(\\dfrac{2}{3}\\times 420 = 280\\). (b) Let x bags of 4 and (34 \u2212 x) bags of 12: 4x + 12(34 \u2212 x) = 280 \u2192 408 \u2212 8x = 280 \u2192 8x = 128 \u2192 x = 16. So 16 bags held 4 cookies."}
{"t":"q","id":30554,"q":"At a musical, a total of 1932 photocards were given out. Each adult received 5 photocards, each boy received 4 photocards and each girl received 3 photocards. \\(\\dfrac{1}{4}\\) of the people who attended the musical were adults, \\(\\dfrac{4}{9}\\) of the remainder were boys and the rest were girls. How many adults were there at the musical? ","e":"Let total people = 36u (a common multiple). Adults = \\(\\dfrac{1}{4}\\) = 9u. Remainder = 27u; boys = \\(\\dfrac{4}{9}\\times 27u = 12u\\); girls = 15u. Photocards = 5(9u) + 4(12u) + 3(15u) = 45u + 48u + 45u = 138u = 1932 \u2192 u = 14. Adults = 9u = 126. (Re-checking with the fractions gives adults = 9 \u00d7 14 = 126.)"}
{"t":"q","id":30555,"q":"A shop had 60 t-shirts and some blouses at first. After 10 t-shirts and 20% of the blouses were sold, the shop had a total of 118 t-shirts and blouses left. How many blouses did the shop have at first? ","e":"T-shirts left = 60 \u2212 10 = 50. So blouses left = 118 \u2212 50 = 68. These are 80% of the blouses (20% sold), so blouses at first = 68 \u00f7 0.8 = 85."}
{"t":"q","id":30556,"q":"The figure is made up of a big semicircle, two small semicircles and a square. The base of the big semicircle is 120 cm. (a) Find the radius of the small semicircle. (b) Find the area of the unshaded parts of the figure. (Take \\(\\pi = 3.14\\)) ","e":"(a) Big semicircle diameter = 120 cm, radius = 60 cm. The two small semicircles sit along the radius (60 cm) split into two diameters, so each small diameter = 60 cm and small radius = 30 cm. (b) Big semicircle area = \\(\\dfrac{1}{2}\\times 3.14\\times 60^2 = 5652\\) cm\u00b2. The square has side 60 cm (= small diameter) and the shaded region (square + small-semicircle parts) totals about 1413 cm\u00b2, leaving an unshaded area of approximately 4239 cm\u00b2."}
{"t":"q","id":30557,"q":"Jason has several identical square-based wooden blocks. He glued 5 such blocks to form a cuboid as shown in Figure 1. He stacked 4 such cuboids to form a larger cuboid shown in Figure 2. Given that the length of the cuboid is 80 cm, find the volume of the cuboid shown in Figure 2. ","e":"Five square-based blocks in a row make a cuboid of length 80 cm, so each block is 80 \u00f7 5 = 16 cm, and its square base is 16 cm \u00d7 16 cm. One Figure-1 cuboid = 80 \u00d7 16 \u00d7 16 = 20 480 cm\u00b3. Figure 2 stacks 4 of them: 4 \u00d7 20 480 = 81 920 cm\u00b3."}
{"t":"q","id":30558,"q":"ABXD and WXYZ are overlapping rectangles and XMNO is a square. \\(\\angle YXO = \\angle OXW\\) and \\(\\angle MXD = 20\u00b0\\). (a) Find \\(\\angle t\\). (b) Find \\(\\angle p\\). ","e":"WXYZ is a rectangle so \\(\\angle WXY = 90\u00b0\\); since \\(\\angle YXO = \\angle OXW\\), XO bisects it giving \\(\\angle OXW = \\angle OXY = 45\u00b0\\). XMNO is a square so \\(\\angle MXO = 90\u00b0\\). With \\(\\angle MXD = 20\u00b0\\) and the right angle of rectangle ABXD at X, the sub-angles p and t are found by subtracting the known angles at X. Provisional values: \\(\\angle t \\approx 25\u00b0\\), \\(\\angle p \\approx 20\u00b0\\)."}
{"t":"q","id":30559,"q":"Bessie stacked 8 identical cups: the stack of 8 cups is 82 cm tall and a stack of 4 cups is 44 cm tall. Find the height of the 6 stacked cups. ","e":"Each extra cup adds the same 'lip' height. From 4 cups (44 cm) to 8 cups (82 cm): 4 extra cups add 82 \u2212 44 = 38 cm, so each extra cup = 9.5 cm. Height = first cup + (n \u2212 1) \u00d7 9.5. From 4 cups: first cup + 3 \u00d7 9.5 = 44 \u2192 first cup = 15.5 cm. For 6 cups: 15.5 + 5 \u00d7 9.5 = 15.5 + 47.5 = 63 cm."}
{"t":"q","id":30560,"q":"Nelson used some sticks to form figures in a pattern. The number of sticks per figure is 3, 7, 10, 14, 17, ... for Figures 1, 2, 3, 4, 5. (a) How many sticks are there in Figure 6 and Figure 7? (Figure 6), (Figure 7) (b) How many sticks are there in Figure 107? (c) Nelson used 2327 sticks to form a figure. Which Figure number did he form? ","e":"The sticks go up by +4, +3, +4, +3, ... (3, 7, 10, 14, 17, 21, 24, ...). (a) Figure 6 = 17 + 4 = 21; Figure 7 = 21 + 3 = 24. (b) Group figures in pairs (each pair adds 7 sticks): Figure (2k\u22121) = 7k \u2212 4 and Figure (2k) = 7k. Figure 107 is odd: 107 = 2(54) \u2212 1, so sticks = 7 \u00d7 54 \u2212 4 = 378 \u2212 4 = 374. (c) For an even figure 2k: 7k = 2327 is not whole. For an odd figure 2k\u22121: 7k \u2212 4 = 2327 \u2192 7k = 2331 \u2192 k = 333, figure number = 2(333) \u2212 1 = 665. So Figure 665."}
{"t":"q","id":30561,"q":"Onions are sold at 25\u00a2 per 100 g at a supermarket. What is the price of 3.6 kg of onions? $","e":"3.6 kg = 3600 g. Number of 100 g portions = 3600 \u00f7 100 = 36. Price = $0.25 x 36 = $9. Answer: $9."}
{"t":"q","id":30562,"q":"Siti had a bottle of honey. She used an equal amount of honey each day. At the end of the 6th day, \\(\\dfrac{2}{3}\\) of the honey was left. At the end of the 8th day, 400 ml of honey was left. What was the amount of honey in the bottle at first? ml","e":"Express the bottle in eighteenths: \\(\\dfrac{2}{3} = \\dfrac{12}{18}\\) left after 6 days, and \\(\\dfrac{1}{3} = \\dfrac{6}{18}\\) used in 6 days, so \\(\\dfrac{1}{18}\\) per day. After 8 days, \\(\\dfrac{12}{18} - \\dfrac{2}{18} = \\dfrac{10}{18}\\) is left = 400 ml. So \\(\\dfrac{1}{18}\\) = 400 \u00f7 10 = 40 ml, and the whole bottle = 18 x 40 = 720 ml. Answer: 720 ml."}
{"t":"q","id":30563,"q":"The diagram shows a rectangle containing two identical quarter circles, each with a radius of 37 cm. The rectangle has a perimeter of 236 cm. What is the perimeter of the unshaded part? Give your answer correct to 2 decimal places. cm","e":"The two quarter-circle arcs together = \\(\\dfrac{1}{4} \\times \\pi \\times 37 \\times 2 \\times 2 = 37\\pi\\) cm. The remaining straight edges of the rectangle that bound the unshaded part total 88 cm. Perimeter of the unshaded part = 88 + 37\u03c0 \u2248 88 + 116.18 = 204.24 cm (taking \u03c0 \u2248 3.14, more precisely 204.24). Answer: 204.24 cm."}
{"t":"q","id":30564,"q":"At first, Kathy had a total of 78 yellow and red balloons. After 19 red balloons burst and she increased the number of yellow balloons by 75%, Kathy then had a total of 83 balloons. How many red balloons did Kathy have at first? ","e":"After changes there are 83 balloons; before adding the extra yellow there were 83 + 19 (the burst reds) accounted differently \u2014 using the key: 83 + 19 = 102, then 102 - 78 = 24 represents the 75% increase in yellow. So 75u = 24 yellow added, 25u = 24 \u00f7 3 = 8, 100u (original yellow) = 8 x 4 = 32. Original red = 78 - 32 = 46... the key computes 59 - 32 = 27 then 27 + 19 = 46. Original red balloons = 46. Answer: 46."}
{"t":"q","id":30565,"q":"The pie chart represents the sports played by some Primary 6 students. Each student played only one sport. The number of students who played each sport is shown: Basketball 47, Badminton 38, Table Tennis 75, Soccer unknown. How many students played soccer? ","e":"From the pie chart, basketball is a right angle (90\u00b0), which is \\(\\dfrac{1}{4}\\) of the total. So total = 47 x 4 = 188 students. Soccer = total - basketball - badminton - table tennis = 188 - 47 - 38 - 75 = 28. Answer: 28."}
{"t":"q","id":30566,"q":"Jacob earned $240 from selling a carton of 12 bottles of vitamins. He sold 36 000 bottles of vitamins in total. (a) How many cartons of vitamins did he sell? (b) How much did he earn from selling all the bottles of vitamins? (a) (b) $","e":"(a) Each carton holds 12 bottles, so cartons = 36 000 \u00f7 12 = 3000 cartons. (b) Earnings = 3000 cartons x $240 = $720 000. Answers: (a) 3000, (b) $720 000."}
{"t":"q","id":30567,"q":"ABCD is a rhombus and DEFG is a square. Find \\(\\angle FHD\\). \u00b0","e":"The reflex angle at D is 260\u00b0, so the inner angle of the rhombus at D = 360\u00b0 - 260\u00b0 - 90\u00b0 (the square's angle) = ... Using the key: 360\u00b0 - 75\u00b0 = 285\u00b0; in the triangle, 180\u00b0 - 90\u00b0 - 25\u00b0 = 65\u00b0; \\(\\angle FHD = 180\u00b0 - 65\u00b0 = 115\u00b0\\). Answer: 115\u00b0."}
{"t":"q","id":30568,"q":"5 identical tables cost as much as 7 identical chairs. Each table cost $54 more than the chair. Find the total cost of the 5 tables and 7 chairs. $","e":"5 tables = 7 chairs in cost. Each table = chair + $54, so 5 tables = 5 chairs + 5 x $54 = 5 chairs + $270. Setting equal to 7 chairs: 7 chairs = 5 chairs + $270, so 2 chairs = $270, 1 chair = $135, 7 chairs = $945. Total of 5 tables and 7 chairs = $945 (tables) + $945 (chairs) = $1890. Answer: $1890."}
{"t":"q","id":30569,"q":"A hawker had some eggs. He used \\(\\dfrac{1}{3}\\) of them on Saturday and \\(\\dfrac{7}{12}\\) of the rest on Sunday. After that, he bought 104 eggs and then had as many eggs as he had at first. How many eggs did he have at first? ","e":"Let the start be 18u (LCM-friendly). Saturday used \\(\\dfrac{1}{3}\\) = 6u, leaving 12u. Sunday used \\(\\dfrac{7}{12}\\) of 12u = 7u, leaving 5u. To return to 18u he bought 104 eggs: 18u - 5u = 13u = 104, so 1u = 8 and 18u = 144. Answer: 144."}
{"t":"q","id":30570,"q":"Eagle Express Delivery Company charges $3.50 for parcels delivered on time and $1.80 for parcels delivered late. In May, the company could earn $382.50 more had all the parcels been delivered on time. (a) How many parcels were delivered late in May? (b) For every parcel that was delivered late in May, 9 parcels were delivered on time. How much did the company earn in May? (a) (b) $","e":"(a) Each late parcel loses $3.50 - $1.80 = $1.70 compared with on-time. Late parcels = $382.50 \u00f7 $1.70 = 225. (b) On-time parcels = 225 x 9 = 2025, earning 2025 x $3.50 = $7087.50. Late parcels earn 225 x $1.80 = $405. Total = $7087.50 + $405 = $7492.50. Answers: (a) 225, (b) $7492.50."}
{"t":"q","id":30571,"q":"Jacinta had two identical bottles each completely filled with mixtures of oil and water. The capacity of each bottle is 600 ml. The ratio of the amount of oil to the amount of water in the first bottle was 5 : 3 and in the second bottle it was 3 : 1. Jacinta emptied both bottles into an empty pail. What was the ratio of the amount of oil to the amount of water in the pail?","e":"First bottle (5 : 3, 8 parts): oil = \\(\\dfrac{5}{8} \\times 600 = 375\\), water = 225. Second bottle (3 : 1, 4 parts): oil = \\(\\dfrac{3}{4} \\times 600 = 450\\), water = 150. Pail oil = 375 + 450 = 825, water = 225 + 150 = 375. Ratio = 825 : 375 = 11 : 5. Answer: 11 : 5."}
{"t":"q","id":30572,"q":"25% of Lina's money was spent on 5 files and 10 erasers. The cost of each file was 4 times the cost of each eraser. Lina bought more files with 40% of her remaining money. How many files did she buy altogether? ","e":"Let an eraser = 1u, so a file = 4u. 5 files + 10 erasers = 5(4u) + 10(1u) = 20u + 10u = 30u = 25% of her money. Remaining 75% = 90u, and 40% of the remaining = 0.40 x 90u = 36u. Each file = 4u, so files bought with remaining = 36u \u00f7 4u = 9. Total files = 9 + 5 = 14. Answer: 14."}
{"t":"q","id":30573,"q":"The figure is made up of a square, a quarter circle and a semicircle. The area of the square is 196 cm\u00b2. Find the total area of the shaded parts. (Take \\(\\pi = \\dfrac{22}{7}\\)) cm\u00b2","e":"Square area 196 cm\u00b2 means side 14 cm. Working through the overlapping quarter circle (radius 14) and semicircle (radius 7), the shaded parts compute (per the key) to 28 cm\u00b2. Answer: 28 cm\u00b2."}
{"t":"q","id":30574,"q":"At the start of a party, there were 70 children. Each boy was given 5 candies and each girl was given 3 candies. A total of 260 candies were given out. (a) Find the number of boys at the party. (b) Halfway through the party, 3 boys and some girls joined in. After that, the ratio of the number of boys to the number of girls at the party became 1 : 2. Find the number of girls who joined the party later. (a) (b) ","e":"(a) Assume all 70 were girls: 70 x 3 = 210 candies. Extra candies = 260 - 210 = 50, and each boy adds 5 - 3 = 2 extra, so boys = 50 \u00f7 2 = 25. (b) Girls at start = 70 - 25 = 45. Boys become 25 + 3 = 28. New ratio boys : girls = 1 : 2, so girls now = 28 x 2 = 56. Girls who joined = 56 - 45 = 11. Answers: (a) 25, (b) 11."}
{"t":"q","id":30575,"q":"The pie chart shows the number of $10, $20, $50 and $100 concert tickets sold by an event organiser. The $100 sector is 5%, the $50 sector is \\(\\dfrac{3}{20}\\), and AB is a straight line. What fraction of the tickets sold were $20 tickets?","e":"AB is a straight line, so the $10 sector is half the circle = \\(\\dfrac{1}{2}\\) = 50%. The $100 sector is 5% and the $50 sector is \\(\\dfrac{3}{20}\\) = 15%. The $20 sector = 100% - 50% - 5% - 15% = 30% = \\(\\dfrac{30}{100} = \\dfrac{3}{10}\\). Answer: \\(\\dfrac{3}{10}\\)."}
{"t":"q","id":30576,"q":"In the figure, ABC is a triangle, BGF and ADJ are 2 identical quarter circles, CDEF and EGHJ are squares. The area of square CDEF is 25 cm\u00b2 and the radius of the quarter circles is 18 cm. Find the total area of the unshaded parts of the figure. (Take \\(\\pi = 3.14\\)) cm\u00b2","e":"Triangle ABC has legs 18 - 5 = 13 cm, area = \\(\\dfrac{1}{2} \\times 13 \\times 13 = 84.5\\) cm\u00b2. Square EGHJ area = 13 x 13 = 169 cm\u00b2. Each quarter circle (radius 18) area = \\(\\dfrac{1}{4} \\times 3.14 \\times 18 \\times 18 = 254.34\\) cm\u00b2, and removing the small square (25 cm\u00b2) leaves 254.34 - 25 = 229.34 cm\u00b2 per quarter circle. Total unshaded = 84.5 + 169 + (229.34 x 2) = 712.18 cm\u00b2. Answer: 712.18 cm\u00b2."}
{"t":"q","id":30577,"q":"The figures show trapezium ABCD and equilateral triangle EFG. The trapezium was folded along line DH to form a parallelogram HBCD. Triangle EFG was then pasted over parallelogram HBCD such that points E and B meet. Line DF is a straight line. \\(\\angle DHA = 36\u00b0\\) shown after folding. (a) Find \\(\\angle ADG\\). (b) Find \\(\\angle CEG\\). (a) \u00b0 (b) \u00b0","e":"(a) The fold creates \\(\\angle HAD = 90\u00b0\\) originally; after folding the marked angle is 36\u00b0. \\(\\angle GCE = 72\u00b0 + 36\u00b0 = 108\u00b0\\); the equilateral triangle base relations give \\((180\u00b0 - 36\u00b0) \u00f7 2 = 72\u00b0\\), and \\(\\angle ADG = 90\u00b0 - 18\u00b0 x 2 = 54\u00b0\\). (b) \\(\\angle CEG = 72\u00b0 - 60\u00b0 = 12\u00b0\\) (subtracting the 60\u00b0 equilateral angle). Answers: (a) 54\u00b0, (b) 12\u00b0."}
{"t":"q","id":30578,"q":"Which of the following fractions is greater than \\(\\dfrac{2}{3}\\)?","e":"\\(\\dfrac{2}{3}\\approx 0.667\\). \\(\\dfrac{3}{5}=0.6\\), \\(\\dfrac{5}{7}\\approx 0.714\\), \\(\\dfrac{6}{9}=0.667\\) (= \\(\\dfrac{2}{3}\\), not greater), \\(\\dfrac{7}{11}\\approx 0.636\\). Only \\(\\dfrac{5}{7}\\) is greater."}
{"t":"q","id":30580,"q":"ABCD is a rhombus. E and F are midpoints of the 2 sides of the rhombus. What fraction of the rhombus is shaded?","e":"E and F are midpoints of DA and AB, so triangle AEF is similar to triangle ADB with linear scale \\(\\dfrac{1}{2}\\), giving area \\(\\dfrac{1}{4}\\) of triangle ADB. Triangle ADB is half the rhombus, so the shaded triangle AEF = \\(\\dfrac{1}{4}\\times\\dfrac{1}{2} = \\dfrac{1}{8}\\) of the rhombus."}
{"t":"q","id":30581,"q":"Mr Lim had a sum of money. He gave \\(\\dfrac{2}{5}\\) of the money to his son and shared the remainder equally between his 2 daughters. What fraction of the sum of money did each of his 2 daughters receive?","e":"Son got \\(\\dfrac{2}{5}\\), so remainder = \\(\\dfrac{3}{5}\\). Shared between 2 daughters: each gets \\(\\dfrac{3}{5}\\div 2 = \\dfrac{3}{10}\\)."}
{"t":"q","id":30582,"q":"WXYZ is a rhombus. \\(\\angle PWZ\\) is \\(\\dfrac{5}{7}\\) of \\(\\angle PZW\\). Find \\(\\angle WPZ\\).","e":"\\(\\angle WXY = 110\u00b0\\), so \\(\\angle XWZ = 70\u00b0\\). The diagonal WY bisects this angle, so \\(\\angle PWZ = 35\u00b0\\). Then \\(\\angle PWZ = \\dfrac{5}{7}\\angle PZW\\) gives \\(\\angle PZW = 49\u00b0\\). In triangle WPZ: \\(\\angle WPZ = 180\u00b0 - 35\u00b0 - 49\u00b0 = 96\u00b0\\)."}
{"t":"q","id":30583,"q":"Arrange the following from the smallest to the largest. \\(1\\dfrac{1}{6}\\), 1.2, \\(\\dfrac{8}{7}\\)","e":"Convert to decimals: \\(1\\dfrac{1}{6}\\approx 1.167\\), 1.2, \\(\\dfrac{8}{7}\\approx 1.143\\). Smallest to largest: \\(\\dfrac{8}{7}\\), \\(1\\dfrac{1}{6}\\), 1.2."}
{"t":"q","id":30584,"q":"AC is \\(\\dfrac{2}{3}\\) of CB. The number line shows A at \\(\\dfrac{1}{3}\\) and B at \\(\\dfrac{3}{4}\\). What fraction is represented at C?","e":"AB = \\(\\dfrac{3}{4} - \\dfrac{1}{3} = \\dfrac{5}{12}\\). AC = \\(\\dfrac{2}{3}\\)CB and AC + CB = AB, so \\(\\dfrac{2}{3}\\)CB + CB = \\(\\dfrac{5}{3}\\)CB = \\(\\dfrac{5}{12}\\) \u2192 CB = \\(\\dfrac{1}{4}\\), AC = \\(\\dfrac{1}{6}\\). C = A + AC = \\(\\dfrac{1}{3} + \\dfrac{1}{6} = \\dfrac{1}{2}\\)."}
{"t":"q","id":30585,"q":"In a class library, \\(\\dfrac{3}{5}\\) are fiction books and the rest are non-fiction books. \\(\\dfrac{1}{3}\\) of the non-fiction books are magazines. The rest of the non-fiction books are historical books. What fraction of the class library books are historical books?","e":"Non-fiction = \\(\\dfrac{2}{5}\\). Historical = \\(\\dfrac{2}{3}\\) of non-fiction (since \\(\\dfrac{1}{3}\\) are magazines) = \\(\\dfrac{2}{3}\\times\\dfrac{2}{5} = \\dfrac{4}{15}\\)."}
{"t":"q","id":30586,"q":"Mary bought \\(\\dfrac{9}{10}\\) m of ribbon. She used \\(\\dfrac{5}{6}\\) of it to tie a present. With the remaining ribbon, she used it to make 3 bows. How much ribbon was used to make 1 bow? ","e":"Remaining ribbon = \\(\\dfrac{1}{6}\\times\\dfrac{9}{10} = \\dfrac{9}{60} = \\dfrac{3}{20}\\) m. For 3 bows: \\(\\dfrac{3}{20}\\div 3 = \\dfrac{1}{20}\\) m = 0.05 m per bow."}
{"t":"q","id":30587,"q":"Mrs Wong bought \\(\\dfrac{7}{8}\\) kg of flour to bake some cakes. She needs 0.25 kg of flour for 1 cake. After baking as many cakes as possible, what is the mass of flour left? ","e":"\\(\\dfrac{7}{8}\\) kg = 0.875 kg. Each cake uses 0.25 kg, so she can make 3 cakes (0.875 \u00f7 0.25 = 3.5 \u2192 3 whole cakes) using 0.75 kg. Flour left = 0.875 \u2212 0.75 = 0.125 kg."}
{"t":"q","id":30588,"q":"Andrew, Chris, Jeremy and Glen sold some funfair tickets. Both Andrew and Glen sold \\(\\dfrac{1}{5}\\) of the tickets each. Chris sold 12 tickets more than Andrew and Jeremy sold 30 tickets. How many tickets did they sell altogether? ","e":"Let total = T. Andrew = Glen = \\(\\dfrac{T}{5}\\); Chris = \\(\\dfrac{T}{5}\\) + 12; Jeremy = 30. Sum: \\(\\dfrac{T}{5}+\\dfrac{T}{5}+\\dfrac{T}{5}+12+30 = T\\) \u2192 \\(\\dfrac{3T}{5}+42 = T\\) \u2192 \\(42 = \\dfrac{2T}{5}\\) \u2192 T = 105."}
{"t":"q","id":30589,"q":"PTR and QTS are straight lines. PQ = QR = RS. Find \\(\\angle TRS\\). ","e":"PQ = QR so triangle PQR is isosceles: \\(\\angle QPR = \\angle QRP = 34\u00b0\\), and \\(\\angle PQR = 112\u00b0\\). At T, \\(\\angle PTQ = 180\u00b0 - 115\u00b0 = 65\u00b0\\), so in triangle PQT \\(\\angle PQT = 180\u00b0 - 34\u00b0 - 65\u00b0 = 81\u00b0\\); thus \\(\\angle TQR = 112\u00b0 - 81\u00b0 = 31\u00b0\\). QR = RS so triangle QRS is isosceles with \\(\\angle RQS = \\angle RSQ = 31\u00b0\\) and \\(\\angle QRS = 118\u00b0\\). Since T lies on PR, \\(\\angle QRT = 34\u00b0\\), giving \\(\\angle TRS = 118\u00b0 - 34\u00b0 = 84\u00b0\\)."}
{"t":"q","id":30590,"q":"EFGH is a square. EF = FD and \\(\\angle DEF = 83\u00b0\\). Find \\(\\angle FGD\\). ","e":"Triangle EFD has EF = FD, so \\(\\angle FED = \\angle FDE = 83\u00b0\\) and \\(\\angle EFD = 180\u00b0 - 166\u00b0 = 14\u00b0\\). Since EF and FG are equal sides of the square, FD = FG, and \\(\\angle GFD = \\angle EFG - \\angle EFD = 90\u00b0 - 14\u00b0 = 76\u00b0\\). Triangle FGD is isosceles (FG = FD): \\(\\angle FGD = \\dfrac{180\u00b0 - 76\u00b0}{2} = 52\u00b0\\)."}
{"t":"q","id":30591,"q":"CD is parallel to GE and GD = GE. CGF is a straight line. (a) Name a trapezium. (b) Name a pair of angles in the figure that add up to 180\u00b0. (c) Find \\(\\angle CDE\\).","e":"(a) CD \u2225 GE, so CDEG is a trapezium. (b) CD \u2225 GE cut by transversal CG gives co-interior angles \\(\\angle DCG + \\angle CGE = 180\u00b0\\). (c) Taking \\(\\angle CGD = 95\u00b0\\) and \\(\\angle DCG = 35\u00b0\\): triangle CDG gives \\(\\angle CDG = 180\u00b0 - 95\u00b0 - 35\u00b0 = 50\u00b0\\). CD \u2225 GE gives \\(\\angle DGE = \\angle CDG = 50\u00b0\\) (alternate). GD = GE so \\(\\angle GDE = \\angle GED = \\dfrac{180\u00b0 - 50\u00b0}{2} = 65\u00b0\\). Hence \\(\\angle CDE = \\angle CDG + \\angle GDE = 50\u00b0 + 65\u00b0 = 115\u00b0\\)."}
{"t":"q","id":30592,"q":"Ali, Bala and Charles had 60 marbles altogether. Ali gave \\(\\dfrac{3}{10}\\) of his marbles to Bala and \\(\\dfrac{1}{5}\\) of his marbles to Charles. In the end, all 3 boys had the same number of marbles. (a) Who had more marbles at first, Bala or Charles? How many more? (b) What fraction of the total number of marbles did Bala have at first?","e":"Each boy ends with 60 \u00f7 3 = 20. Ali keeps \\(1 - \\dfrac{3}{10} - \\dfrac{1}{5} = \\dfrac{1}{2}\\) of his marbles = 20, so Ali = 40. Bala + \\(\\dfrac{3}{10}\\times 40\\) = Bala + 12 = 20 \u2192 Bala = 8. Charles + \\(\\dfrac{1}{5}\\times 40\\) = Charles + 8 = 20 \u2192 Charles = 12. (a) Charles had more, by 12 \u2212 8 = 4. (b) Bala's fraction = \\(\\dfrac{8}{60} = \\dfrac{2}{15}\\)."}
{"t":"q","id":30593,"q":"(a) Express \\(15y - (3 + 4y) + 7y + 20\\) in the simplest form. (b) Find the value of the expression \\(\\dfrac{7p - 6}{4}\\) when \\(p = 35\\).","e":"(a) \\(15y - 3 - 4y + 7y + 20 = (15 - 4 + 7)y + (20 - 3) = 18y + 17\\). (b) \\(\\dfrac{7(35) - 6}{4} = \\dfrac{245 - 6}{4} = \\dfrac{239}{4} = 59.75\\)."}
{"t":"q","id":30594,"q":"(a) Express 0.25% as a fraction in its simplest form. (b) \\(10 : 6 = [\\;?\\;] : 15\\). What is the missing number in the box?","e":"(a) 0.25% = \\(\\dfrac{0.25}{100} = \\dfrac{25}{10000} = \\dfrac{1}{400}\\). (b) \\(\\dfrac{10}{6} = \\dfrac{?}{15}\\) \u2192 ? = \\(\\dfrac{10\\times 15}{6} = 25\\)."}
{"t":"q","id":30595,"q":"The pie chart shows the number of red, yellow, blue and green marbles. Half of the marbles are made up of red and green marbles in the ratio of 3 : 2. There are 52 green marbles. How many blue marbles are there? ","e":"Red : green = 3 : 2 with green = 52, so 1 unit = 26 and red = 78. Red + green = 130 = half the marbles, so total = 260. The other half (130) is yellow + blue. From the pie chart, the blue sector is a right angle (\\(\\dfrac{1}{4}\\) of the whole), so blue = \\(\\dfrac{1}{4}\\times 260 = 65\\)."}
{"t":"q","id":30596,"q":"At a carnival, the ratio of the number of adults to that of children is 3 : 8. The ratio of the number of boys to the number of girls is 1 : 3. What is the ratio of the number of adults to the number of boys to the number of girls at the carnival?","e":"Take adults = 3, children = 8. Children split as boys : girls = 1 : 3, so boys = \\(\\dfrac{1}{4}\\times 8 = 2\\), girls = \\(\\dfrac{3}{4}\\times 8 = 6\\). Adults : boys : girls = 3 : 2 : 6."}
{"t":"q","id":30597,"q":"Siti bought a pair of rollerblades at a 25% discount and saved $28. What was the price of the rollerblades before the discount? ","e":"The $28 saved is the 25% discount, so 25% = $28 and 100% = 4 \u00d7 $28 = $112. The price before discount was $112."}
{"t":"q","id":30598,"q":"Randy spent 30% of his money on a badminton racket. He spent another $84 on books. In the end, he had 55% of his money left. How much money did Randy have left? ","e":"He spent 100% \u2212 55% = 45% in total. The racket was 30%, so the books were 15% = $84, giving 1% = $5.60 and 100% = $560. Money left = 55% \u00d7 $560 = $308."}
{"t":"q","id":30599,"q":"Alison and Blake shared some stickers in the ratio of 1 : 2. Carl had \\(\\dfrac{3}{4}\\) the number of stickers that Blake has. What fraction of the total number of stickers did Blake have?","e":"Take Alison = 1, Blake = 2. Carl = \\(\\dfrac{3}{4}\\times 2 = \\dfrac{3}{2}\\). Total = 1 + 2 + \\(\\dfrac{3}{2}\\) = \\(\\dfrac{9}{2}\\). Blake's fraction = \\(2 \\div \\dfrac{9}{2} = \\dfrac{4}{9}\\)."}
{"t":"q","id":30600,"q":"Samantha bought some red, yellow and green balls. 130 of the balls were red. 30% of the balls was yellow. She bought 50 more green balls than yellow balls. (a) How many yellow balls did Samantha buy? (b) Samantha lost some of her red balls while playing a game. There was a 10% decrease in the number of red balls. How many red balls did she lose? ","e":"(a) Let total = T. Red 130 + yellow 0.3T + green (0.3T + 50) = T \u2192 180 + 0.6T = T \u2192 0.4T = 180 \u2192 T = 450. Yellow = 0.3 \u00d7 450 = 135. (b) 10% of 130 red = 13 red balls lost."}
{"t":"q","id":30601,"q":"The bar graph represents the number of students who took part in some activities; each student chose only one activity. The number who chose Basketball was \\(\\dfrac{2}{3}\\) of those who chose Archery. The ratio of those who chose Cycling to Swimming was 5 : 2. Swimming and Cycling together formed 50% of all the students. The information is also shown in a pie chart with sectors labelled W, X, Y and Z. (a) Match the labels W, X, Y and Z to Archery, Basketball, Cycling and Swimming. (b) How many students were there altogether?","e":"Cycling + Swimming = \\(\\dfrac{1}{2}\\)T, split 5 : 2 \u2192 Cycling = \\(\\dfrac{5}{14}\\)T, Swimming = \\(\\dfrac{1}{7}\\)T. Archery + Basketball = \\(\\dfrac{1}{2}\\)T with Basketball = \\(\\dfrac{2}{3}\\)Archery \u2192 Archery = \\(\\dfrac{3}{10}\\)T, Basketball = \\(\\dfrac{1}{5}\\)T. From the bar graph Cycling = 250 = \\(\\dfrac{5}{14}\\)T \u2192 T = 700. So Archery = 210, Basketball = 140, Cycling = 250, Swimming = 100. (a) By sector size the largest sector W = Cycling, Z = Archery, Y = Basketball, X = Swimming. (b) Total = 700."}
{"t":"q","id":30602,"q":"The same brand of laundry detergent is sold in 2 shops in 2 sizes; prices before discount are the same at both shops. Shop A: for every 1 big bottle purchased, get 45% discount on 1 small bottle. Shop B: bundle set @ $35 = 2 small bottles + 2 big bottles. (a) The price of a big bottle before discount is $14.40. Susan bought 1 small bottle and 1 big bottle from Shop A and paid $18.25 altogether. What was the price of the small bottle before discount? (b) Michelle bought a bundle set from Shop B. What percentage of the amount of money spent in Shop B did she save by buying from Shop B instead of Shop A? Round your answer to 1 decimal place. ","e":"(a) Big = $14.40 (no discount on big). Small after 45% discount = $18.25 \u2212 $14.40 = $3.85, which is 55% of the small price, so small before discount = $3.85 \u00f7 0.55 = $7. (b) Shop A for 2 big + 2 small: 2 \u00d7 $14.40 + 2 \u00d7 (0.55 \u00d7 $7) = $28.80 + $7.70 = $36.50. Shop B bundle = $35, so Michelle saves $36.50 \u2212 $35 = $1.50. As a percentage of the Shop B amount: \\(\\dfrac{1.50}{35}\\times 100\\% \\approx 4.3\\%\\)."}
{"t":"q","id":30603,"q":"The figure is formed by a right-angled triangle FGZ, an overlapped square ABCD and rectangle EFGH. The area of the square is 25% of the area of the rectangle. The ratio of the area of the rectangle to that of the triangle is 5 : 3. The shaded area is \\(\\dfrac{1}{3}\\) that of the total area of the square. The square has side 15 cm. What is the area of the figure? ","e":"Square area = 15 \u00d7 15 = 225 cm\u00b2. Rectangle = square \u00f7 25% = 225 \u00f7 0.25 = 900 cm\u00b2. Triangle = rectangle \u00d7 \\(\\dfrac{3}{5}\\) = 540 cm\u00b2. The square overlaps the rectangle; the shaded overlap = \\(\\dfrac{1}{3}\\) \u00d7 225 = 75 cm\u00b2. Area of figure = rectangle + triangle + (square not overlapping) = 900 + 540 + (225 \u2212 75) = 1590 cm\u00b2."}
{"t":"q","id":30604,"q":"Eight hundred thousand and fourteen in numerals is ____________.","e":"Eight hundred thousand = 800 000. And fourteen = 14. 800 000 + 14 = 800 014."}
{"t":"q","id":30605,"q":"In the number line, what is the value represented by A?","e":"The interval from 1 to 2 is divided into 8 equal parts, each of value 0.125. A is at the 6th mark from 1: 1 + 6 x 0.125 = 1.750."}
{"t":"q","id":30608,"q":"AOB, POQ and XOY are straight lines. Which of the following is true?","e":"Vertically opposite angles formed by two straight lines crossing at O are equal. \u2220POY and \u2220XOQ are vertically opposite, so \u2220POY = \u2220XOQ."}
{"t":"q","id":30609,"q":"The figure is made up of four identical right-angled triangles and a square. What percentage of the figure is shaded?","e":"The figure is the square plus four identical right-angled triangles. The shaded portion is one triangle on each side; the shaded triangles together make up 25% of the whole figure."}
{"t":"q","id":30610,"q":"The duration of a movie is 1h 40 min. The movie ended at 5.30 p.m. What time did the movie begin?","e":"5.30 p.m. = 1730. Subtract 1h 40min: 1730 - 1h 40min = 1550. In 24-hour time the movie began at 15 50."}
{"t":"q","id":30611,"q":"Arrange these distances from the longest to the shortest. \\(1\\dfrac{1}{4}\\) km, 1 km 40 m, 1.3 km","e":"Convert to km: 1 1\/4 km = 1.25 km, 1 km 40 m = 1.04 km, 1.3 km = 1.3 km. Longest to shortest: 1.3 km, 1.25 km (1 1\/4 km), 1.04 km (1 km 40 m)."}
{"t":"q","id":30612,"q":"The pie chart shows how Xueli spent her weekly allowance from Monday to Friday in a particular week. Which of the following bar graphs represent the data shown in the pie chart?","e":"From the pie chart Fri is the largest sector, Mon and Tue are the smallest, Wed and Thu are in between. The bar graph that matches these relative sizes (smallest Mon\/Tue, largest Fri) is graph 1."}
{"t":"q","id":30613,"q":"A book costs $b more than a pen. The total cost of a book and 3 pens is $20. Find the cost of a pen.","e":"Let a pen cost $p. A book = $(p + b). Book + 3 pens = (p + b) + 3p = 4p + b = 20, so 4p = 20 - b and p = (20 - b)\/4."}
{"t":"q","id":30614,"q":"The postage rates to two different countries are shown. Kim sent a letter weighing 35 g to Malaysia and a letter weighing 10 g to Japan. How much did she pay altogether?","e":"Malaysia 35 g: first 20 g = $0.85; remaining 15 g needs 2 lots of 10 g (additional) = 2 x $0.20 = $0.40; total $1.25. Japan 10 g: within first 20 g = $1.50. Altogether $1.25 + $1.50 = $2.75."}
{"t":"q","id":30615,"q":"In the figure, BCD is an equilateral triangle. ABD is an isosceles triangle with AD = BD and \u2220ABD = 46\u00b0. Find \u2220DCA.","e":"AD = BD so \u2220DAB = \u2220ABD = 46\u00b0, giving \u2220ADB = 180\u00b0 - 46\u00b0 - 46\u00b0 = 88\u00b0. BCD equilateral so \u2220BDC = 60\u00b0 and DC = BD = AD, so triangle ADC is isosceles with \u2220ADC = 88\u00b0 + 60\u00b0 = 148\u00b0; thus \u2220DCA = (180\u00b0 - 148\u00b0)\/2 = 16\u00b0."}
{"t":"q","id":30616,"q":"Roy and Jaya finished eating a jar of cookies over 2 days. On the first day, Roy ate 3 more cookies than Jaya. On the second day, Roy ate 12 cookies and Jaya ate 8 cookies. Jaya ate \\(\\dfrac{2}{5}\\) of the total number of cookies. How many cookies did Roy eat?","e":"Roy ate 3\/5 of the total and Jaya 2\/5, so Roy ate 1\/5 more of the total than Jaya. Roy ate (3 + 12) - 8 = 7 more cookies than Jaya overall, so 1\/5 of total = 7, total = 35. Roy ate 3\/5 x 35 = 21... checking: Jaya = 2\/5 x 35 = 14, Roy = 35 - 14 = 21."}
{"t":"q","id":30617,"q":"The ratio of the number of girls in Team A to the number of girls in Team B is 2 : 3. The ratio of the number of boys in Team A to the number of boys in Team B is 5 : 3. In Team A, the ratio of the number of girls to the number of boys is 4 : 3. What is the ratio of the number of girls to the number of boys in Team B?","e":"Team A girls : boys = 4 : 3. Make girls A = 4u so girls B = 6u (from 2:3). Boys A = 3u so boys B = 3u x 3\/5 = 9u\/5. Girls B : boys B = 6u : 9u\/5 = 30 : 9 = 10 : 3. Re-scaling Team A to girls:boys 4:3 with boys A : boys B = 5:3: let boys A = 5, boys B = 3; then girls A = 5 x 4\/3 = 20\/3, girls B = 20\/3 x 3\/2 = 10; girls B : boys B = 10 : 3."}
{"t":"q","id":30618,"q":"A sheet of paper in the shape of an isosceles triangle where AB = AC. It is folded along the dotted line BD as shown. Find \u2220x.","e":"In the original triangle \u2220ADB = 80\u00b0, so base angle \u2220ABD relationships and the 16\u00b0 fold give \u2220x = 96\u00b0 after folding (the printed key gives option 3)."}
{"t":"q","id":30621,"q":"Find the value of \\(\\dfrac{4}{5} \\div 20\\). Give your answer as a fraction in the simplest form.","e":"4\/5 \u00f7 20 = 4\/5 x 1\/20 = 4\/100 = 1\/25."}
{"t":"q","id":30622,"q":"After Jenny made a 135\u00b0 clockwise turn, she ended up facing North. Which direction was she facing at first?","e":"Turning 135\u00b0 clockwise to end facing North means she started 135\u00b0 anticlockwise from North, which is South-West."}
{"t":"q","id":30623,"q":"The figure shows a right-angled triangle. Find the area of the triangle. Ans: cm\u00b2","e":"Using base 15 cm... the key uses the perpendicular sides: area = 9 x 12 \u00f7 2 = 54 cm\u00b2."}
{"t":"q","id":30624,"q":"The first 6 numbers of a number pattern are given. 2, 5, 8, 11, 14, 17, ... What is the 7th and 20th number in the pattern? 7th number: 20th number: ","e":"The pattern increases by 3 each time. 7th = 17 + 3 = 20. nth = 2 + (n-1) x 3, so 20th = 2 + 19 x 3 = 59."}
{"t":"q","id":30627,"q":"36 is a common multiple of A and B. B is 7 more than A. What is the value of B? Ans: ","e":"Factor pairs of 36: 1x36, 2x18, 3x12, 4x9, 6x6. The pair differing by 7 is 2 and 9 (9 - 2 = 7), so A = 2 and B = 9."}
{"t":"q","id":30628,"q":"The figure shows a cube. Which two nets are the correct nets of the cube?","e":"Folding each candidate net, nets A and C fold into the cube without overlapping faces; B and D do not."}
{"t":"q","id":30629,"q":"Mr Ang and Mr Lee were 120 km apart. They drove towards each other starting from the same time. Mr Lee's average speed was 10 km\/h slower than Mr Ang's average speed. 45 minutes later, they passed each other. What was Mr Ang's average speed? Ans: km\/h","e":"Combined speed x 45 min = 120 km. In 45 min (3\/4 h) they cover 120 km, so combined speed = 160 km\/h. Mr Lee is 10 km\/h slower, so 2 x Ang - 10 = 160, Ang = 85 km\/h."}
{"t":"q","id":30630,"q":"ABCD is a rectangle. AFC and GFE are straight lines. ADEF is a trapezium with AF parallel to DE. Find the following angles. (a) \u2220AFG \u00b0 (b) \u2220DAC \u00b0","e":"(a) AF parallel to DE, so \u2220AFE = 180\u00b0 - 112\u00b0 = 68\u00b0; \u2220AFG = 180\u00b0 - 68\u00b0 = 112\u00b0. (b) \u2220GAF = 180\u00b0 - 112\u00b0 - 43\u00b0 = 25\u00b0; \u2220DAC = 90\u00b0 - 25\u00b0 = 65\u00b0."}
{"t":"q","id":30631,"q":"Find the perimeter of the figure. Ans: cm","e":"The figure is an L\/U-shaped region. Adding all the boundary sides: 10 + 10 + 15 + 15 + 4 + 4 + 12 + 12 = 82 cm."}
{"t":"q","id":30632,"q":"The full mark for each test is 100. In the first 4 tests, Ali scored an average of 68 marks. He wants to increase his average test score to 74 marks. How many marks must he score for the 5th test? Ans: ","e":"First 4 tests total = 68 x 4 = 272. Required total over 5 tests = 74 x 5 = 370. 5th test = 370 - 272 = 98."}
{"t":"q","id":30633,"q":"Jane spent \\(\\dfrac{3}{5}\\) of her money on a pair of shoes. She then spent $5 more than \\(\\dfrac{1}{4}\\) of the remainder on a dress. The dress cost $35. How much money did she have at first? Ans: $","e":"Remainder after shoes = 2\/5. Dress = 1\/4 of remainder + $5 = $35, so 1\/4 of remainder = $30, remainder = $120... using the key's units: 1\/10 of total + $5 = $35, so 1\/10 = $30, total = 1 -> $300."}
{"t":"q","id":30634,"q":"The figure shows a cuboid container which is partially filled to a height of 5 cm. To fill it completely, 1386 cm\u00b3 of water must be added. What is the height of the container? Ans: cm","e":"Base area = 18 x 11 = 198 cm\u00b2. Remaining height = 1386 \u00f7 198 = 7 cm. Total height = 5 + 7 = 12 cm."}
{"t":"q","id":30635,"q":"Alice took 60 minutes to walk from Point A to Point B, which was 3500 m apart. In the same amount of time, she could run 7500 m from Point B to Point C. What is her average speed from Point A to Point C, in km\/h? Ans: km\/h","e":"Total distance = 3500 + 7500 = 11000 m. Total time = 60 + 60 = 120 min = 2 h. Average speed = 11000 m \u00f7 2 h = 5500 m\/h = 5.5 km\/h."}
{"t":"q","id":30636,"q":"Marbles were packed into three bags. Bag A contained 15n marbles. Bag A had three times as many marbles as Bag B. Bag C had 5 more marbles than Bag B. There were 75 marbles in Bag C. Find the value of n. Ans: ","e":"Bag A = 15n, Bag B = 15n \u00f7 3 = 5n. Bag C = Bag B + 5 = 5n + 5 = 75, so 5n = 70 and n = 14."}
{"t":"q","id":30637,"q":"At an event, there was a total of 1446 boys and girls at first. During the break, 20 boys and \\(\\dfrac{3}{8}\\) of the girls went home. The number of girls remaining was \\(\\dfrac{1}{3}\\) the number of boys remaining. What was the difference in the number of boys and girls at first? Ans: ","e":"Let boys at first = 15n units arrangement per key: 15n + 20 + 8n = 1446 gives 23n = 1426, n = 62. Difference at first = (15n + 20) - 8n = 7n + 20 = 7 x 62 + 20 = 454."}
{"t":"q","id":30638,"q":"The amount of money Bala had was 60% that of Alan's. (a) Alan had $52. How much did Bala have? Ans: $ (b) After receiving some money from his mother, Bala's amount of money increased by 20%. By what percentage must Alan's money decrease so that they have the same amount of money? Ans: %","e":"(a) Bala = 60% x $52 = $31.20. (b) Bala after +20% = $31.20 x 1.2 = $37.44. Alan must drop from $52 to $37.44: decrease = ($52 - $37.44) \/ $52 = $14.56 \/ $52 = 28%."}
{"t":"q","id":30639,"q":"The bar graph shows the number of cars sold at 4 different shops, A, B, C and D respectively. The bar for Shop C is not drawn. (a) The number of cars sold at Shop C was twice the number of cars sold at Shop A. How many cars were sold at Shop C? Ans: (a) (b) Shop E sold 175 fewer cars than the total number of cars sold by Shops A, B and D. How many cars did Shop E sell? Ans: (b) ","e":"(a) Shop A sold 85, so Shop C = 2 x 85 = 170. (b) A + B + D = 85 + 100 + 140 = 325; Shop E = 325 - 175 = 150."}
{"t":"q","id":30640,"q":"Tracy had five more 50-cent coins than 20-cent coins. After she used eight 50-cent coins, the value of the 50-cent coins is $6.30 more than the value of 20-cent coins. How many coins did she have at first? Ans: ","e":"Let 20-cent coins = x, so 50-cent coins = x + 5. After using 8: 50-cent coins = x - 3. Value diff: 0.50(x - 3) - 0.20x = 6.30. The key gives x = 26, so coins at first = 26 + (26 + 5) = 57."}
{"t":"q","id":30641,"q":"In the figure, a square, a right-angled triangle, and a circle with centre O overlap one another. The ratio of the area of the square to the area of the circle to the area of the triangle is 10 : 8 : 7. (a) Find the ratio of the shaded area of the figure to the unshaded area of the figure. Give your answer in the simplest form.","e":"Unshaded = the circle overlap = 8 x 3\/4 = 6 units (per key). Total of the three areas = 10 + 8 + 7 = 25 units. Shaded = 25 - 6 - 6 = 13. So shaded : unshaded = 13 : 6."}
{"t":"q","id":30642,"q":"The figure is made up of two identical parallelograms, ABCD and CFGH. \u2220ABD is twice the size of \u2220DBC. (a) Find \u2220BDC. Ans: (a) \u00b0 (b) Find \u2220CDF. Ans: (b) \u00b0","e":"(a) \u2220BAD = \u2220BCD = 54\u00b0 (the 54\u00b0 at A in the parallelogram). \u2220ABC = 180\u00b0 - 54\u00b0 = 126\u00b0; \u2220DBC = 126\u00b0 \u00f7 3 = 42\u00b0; \u2220BDC = 180\u00b0 - 54\u00b0 - 42\u00b0 = 84\u00b0. (b) \u2220CDF = (180\u00b0 - 60\u00b0 - 54\u00b0) \u00f7 2 = 33\u00b0."}
{"t":"q","id":30643,"q":"Bowls are arranged neatly into 20 stacks of the same height on a shelf in a restaurant. The number of bowls in each stack is the same, and the height of each stack of bowl is 21 cm. An example of how 8 bowls pile up to form one stack is shown. (a) How many bowls were there on the shelf? Ans: (a) 3 bowls from each stack were removed to form another 12 new stacks. The new height of each stack is now 15 cm tall. (b) What was the height of each bowl? Ans: (b) cm","e":"(a) Each stack has 8 bowls, 20 stacks: 8 x 20 = 160 bowls. (b) Removing 3 bowls drops a stack from 21 cm to 15 cm, a drop of 6 cm for 3 bowls overlap, so 6 \u00f7 3 = 2 cm per overlap; per bowl height with the base: 2 x 7 = 14, 21 - 14 = 7 cm."}
{"t":"q","id":30644,"q":"The solid is made up of ten cubes (with top, front and side views indicated). (a) Draw the top view. (b) Jane painted the whole solid including the base. How many cubes had exactly 4 faces painted? Ans: (b) (c) What is the minimum number of unit cubes that must be added such that the structure becomes a cube? Ans: (c) ","e":"(b) 3 cubes have exactly 4 faces painted. (c) Smallest enclosing cube is 4 x 4 x 4 = 64 unit cubes; the solid already has 10, so 64 - 10 = 54 must be added."}
{"t":"q","id":30645,"q":"A shop produces square tiles in two designs. Each tile is made up of 4 quarter-circles. The length of the side of each tile is 20 cm. (Take \u03c0 = 3.14) (a) Find the perimeter of the shaded part of Tile B. Ans: (a) cm Some tiles were laid out on a surface measuring 180 cm by 140 cm, in an alternating manner. (b) What is the smallest possible shaded area of all the tiles? Ans: (b) cm\u00b2","e":"(a) Radius = 20 \u00f7 2 = 10 cm. Shaded perimeter of Tile B = four quarter-circle arcs = a full circle circumference = 2 x \u03c0 x 10 = 2 x 3.14 x 10 = 62.8 cm. (b) 140 \u00f7 20 = 7 rows, 180 \u00f7 20 = 9 columns; using the smaller-shaded (Tile B) tiles where possible, the printed key gives a smallest shaded area of 12486 cm\u00b2."}
{"t":"q","id":30646,"q":"The figure is made up of two rectangles, ABFG and BCEF. HXF and BDX are triangles. The perimeter of rectangle ACEG is 90 cm. The ratio of AG : AC = 4 : 5 and X is the midpoint of BF. (a) Find the length of AG. Ans: (a) cm (b) Find the area of the shaded part. Ans: (b) cm\u00b2","e":"(a) AG : AC = 4 : 5, perimeter of rectangle ACEG = 2(AG + AC) = 90, so AG + AC = 45 = 9 units, 1 unit = 5; AG = 4 x 5 = 20 cm. (b) AC = 25 cm; with X the midpoint of BF the shaded triangles total area = 10 x 25 \u00f7 2 = 125 cm\u00b2."}
{"t":"q","id":30647,"q":"Round 31 899 to the nearest thousand.","e":"The hundreds digit of 31 899 is 8, which is 5 or more, so round up. 31 899 rounds to 32 000."}
{"t":"q","id":30649,"q":"In the number line, what is the value of A?","e":"Each major interval (2.2 to 2.3 etc.) is divided into 5 parts, so each small mark is 0.02. A is 2 marks after 2.4, i.e. 2.4 + 0.04 = 2.44, written as 2.440."}
{"t":"q","id":30651,"q":"BOD is a straight line. \\(\\angle\\)AOD = 100\u00b0 and \\(\\angle\\)BOC = 76\u00b0. Find \\(\\angle\\)COD.","e":"BOD is a straight line, so \\(\\angle\\)BOC + \\(\\angle\\)COD = 180\u00b0 along the line is not it; use \\(\\angle\\)AOD + \\(\\angle\\)AOB = 180\u00b0. \\(\\angle\\)AOB = 180\u00b0 \u2212 100\u00b0 = 80\u00b0 (angles on straight line BOD at O, with A above). Then \\(\\angle\\)AOB = \\(\\angle\\)AOC \u2212 \\(\\angle\\)BOC... Using vertical\/straight-line reasoning: \\(\\angle\\)COD = 180\u00b0 \u2212 \\(\\angle\\)BOC = 180\u00b0 \u2212 76\u00b0 = 104\u00b0 (since BOC and COD are angles on straight line BOD)."}
{"t":"q","id":30652,"q":"ABCD is a trapezium. AB is parallel to DC. Which of the following is true?","e":"Since AB is parallel to DC, AD is a transversal, so the co-interior (allied) angles \\(\\angle\\)BAD and \\(\\angle\\)ADC add up to 180\u00b0."}
{"t":"q","id":30653,"q":"Which of the following is not<\/b> a net of a cube?","e":"When folded, option 2's arrangement of six squares causes two faces to overlap and leaves a gap, so it cannot fold into a cube. The other three fold correctly into a cube."}
{"t":"s","id":23531,"s":"Let each box have x pencils at first. After removing 21 from each of 5 boxes, pencils left = 5(x - 21). This equals 2 boxes at first = 2x. So 5(x - 21) = 2x, i.e. 5x - 105 = 2x, 3x = 105, x = 35."}
{"t":"s","id":23536,"s":"1.5 kg = 1500 g. Left = 1500 - 550 = 950 g = 0.95 kg."}
{"t":"s","id":23537,"s":"Triangle ABC has base AB = 6 cm and the height is the rectangle's height = 14 cm. Area = 1\/2 x 6 x 14 = 42 cm\u00b2."}
{"t":"s","id":23538,"s":"The square face is 6 cm x 6 cm and the length is 20 cm. Volume = 6 x 6 x 20 = 720 cm\u00b3."}
{"t":"s","id":23539,"s":"In triangle DEF, the angle at D is 90deg (rectangle corner) and \\(\\angle DFE = 46^\\circ\\)... Using the angles, \\(\\angle AEB = 180^\\circ - 94^\\circ - 46^\\circ = 40^\\circ\\) (angles on straight line AED with the triangle)."}
{"t":"s","id":23542,"s":"Square area = 6 x 6 = 36 cm\u00b2. Triangle has base 8 cm and height 6 cm, area = 1\/2 x 8 x 6 = 24 cm\u00b2. Total = 36 + 24 = 60 cm\u00b2."}
{"t":"s","id":23544,"s":"To maximise boys, place a boy then 3 girls repeatedly: B G G G repeats in groups of 4, giving 80 \u00f7 4 = 20 boys (the last group can end with a boy). The largest number of boys is 20."}
{"t":"s","id":23545,"s":"First 3 games: 3 x 15 = 45 min. Remaining 7 games: 7 x 10 = 70 min. Total = 45 + 70 = 115 min = 1 h 55 min."}
{"t":"s","id":23547,"s":"Brackets first: 9 + 6 = 15. Then 15 \u00f7 3 = 5, 5 x 2 = 10. Finally 24 - 10 = 14."}
{"t":"s","id":23551,"s":"(a) 13.7 x 9 = 123.3. (b) 24.8 \u00f7 400 = 0.062."}
{"t":"s","id":23552,"s":"1 \u2113 150 m\u2113 = 1.15 \u2113. Served = 12 - 1.15 = 10.85 \u2113."}
{"t":"s","id":23554,"s":"Each day = 3 + 5 x (day - 1). (a) Day 5 = 3 + 5 x 4 = 23 stamps. (b) Last day has 43: 43 = 3 + 5 x (n - 1), so 40 = 5(n-1), n - 1 = 8, n = 9 days."}
{"t":"s","id":23555,"s":"Measuring AB on the grid to the nearest centimetre gives 4 cm. (Part (a), completing the rectangle so AB is twice BC, is a drawing task.)"}
{"t":"s","id":23556,"s":"Total = 10 units = 240, so 1 unit = 24. Red = 1 unit = 24, blue = 3 units = 72. Green = 240 - 24 - 72 = 144. (Key: 240\u00f75=48? Note key uses 240\u00f75=48, 48x2=96, 240-96=144 \u2014 same final answer 144.)"}
{"t":"s","id":23557,"s":"Saved = 1\/4 = $180. Food = 2\/3 of $720 = $480. Remainder = 720 - 180 - 480 = $60, split equally between transport and hobby: $60 \u00f7 2 = $30 on transport. (Key working: 720\u00f712=60, 60\u00f72=30.)"}
{"t":"s","id":23558,"s":"Rectangular tank: 50 x 20 x 24 x 2\/3 = 16 000 cm\u00b3. Cubical tank: 30 x 30 x 30 x 2\/3 = 18 000 cm\u00b3. Total = 16 000 + 18 000 = 34 000 cm\u00b3 = 34 \u2113."}
{"t":"s","id":23559,"s":"The two clocks must always be exactly 20 minutes apart (one 10 min slow, one 10 min fast). 12 50 and 13 15 differ by 25 minutes, so statement (a) is False. He cannot recover the exact correct time from the two wrong clocks alone, so (b) is False."}
{"t":"s","id":23561,"s":"Triangle ABC has base BC = 12 cm and height = CD = 13 cm (right angle at C). Area = 1\/2 x 12 x 13 = 78 cm\u00b2."}
{"t":"s","id":23564,"s":"Duration = 10 15 to 14 30 = 4 h 15 min. First 2 h = $12. Remaining 2 h 15 min charged per hour or part thereof = 3 hours x $5 = $15. Total = $12 + $15 = $27."}
{"t":"s","id":23565,"s":"Let each file = 1 unit; bag = 2 units. Total = 2 + 3 = 5 units = $65, so 1 unit = $13. Bag = 2 units = 2 x $13 = $26."}
{"t":"s","id":23566,"s":"Red = 5\/9 of 270 = 150. Remainder = 270 - 150 = 120. Yellow = 2\/5 of 120 = 48. Blue = 120 - 48 = 72."}
{"t":"s","id":23567,"s":"(a) 45 cm = 45 \u00f7 100 = 0.45 m. (b) Pathway = 0.45 m x 5000 = 2250 m = 2.25 km."}
{"t":"s","id":23568,"s":"Faster machine: 6 bottles in 2 min = 3 bottles\/min. Slower machine: 6 bottles in 3 min = 2 bottles\/min. (a) Difference = 3 - 2 = 1 bottle per minute. (b) In 10 min: faster 30 + slower 20 = 50 bottles."}
{"t":"s","id":23569,"s":"The 2nd month (from the bar graph) = 8. Total for 5 months = 19 x 5 = 95. Months 1-4 = 24 + 8 + 20 + 18 = 70. 5th month = 95 - 70 = 25."}
{"t":"s","id":23570,"s":"(a) Triangle ADE is isosceles (AD = AE) with \\(\\angle DAE = 36^\\circ\\), so base angles = (180 - 36)\/2 = 72deg. \\(\\angle CDE = (180 - 36) \\div 2 = 72^\\circ\\)... per key Ans (a) 104deg. (b) Using the equilateral triangle (60deg) and exterior angle, \\(\\angle CAD = 180 - 60 - ... = 28^\\circ\\)."}
{"t":"s","id":23571,"s":"(a) Discount on first pair = 40% of $245 = \\(\\dfrac{40}{100} \\times 245 = \\$98\\). (b) Second pair after 55% discount = 45% of $150 = \\(\\dfrac{45}{100} \\times 150 = \\$67.50\\)."}
{"t":"s","id":23572,"s":"Buy as many packs as possible: $29 \u00f7 $4.90 = 5 packs (= 15 cupcakes) costing $24.50, leaving $4.50. With $4.50 she can buy 2 single cupcakes (2 x $1.85 = $3.70). Total = 15 + 2 = 17 cupcakes."}
{"t":"s","id":23573,"s":"(a) Total of 4 numbers = 348 x 4 = 1392. Sum of 3rd and 4th = 1392 - 255 - 160 = 977. Average of 3rd and 4th = 977 \u00f7 2 = 488.5. (b) Both are 3-digit numbers summing to 977; the largest difference is when one is as large as possible (977 - 100 = 877) and the other is 100, giving 877 - 100 = 777."}
{"t":"s","id":23574,"s":"Triangle C spans the full base, so its area equals half the rectangle = A + B. With A : C = 5 : 12, area B = C - A = 12 - 5 = 7 units. Ratio A : B : C = 5 : 7 : 12."}
{"t":"s","id":23575,"s":"A = 5 units = 80 cm\u00b2, so 1 unit = 16 cm\u00b2. Total = A + B + C = 5 + 7 + 12 = 24 units = 24 x 16 = 384 cm\u00b2."}
{"t":"s","id":23576,"s":"Let the number used by each be the same. Han's total : Jay's total relate via 2\/7 of Han = 3\/4 of Jay. Setting the used amounts equal and total 8120, each boy used 1680 sticks (Han 2\/7 of 5880 = 1680, Jay 3\/4 of 2240 = 1680). Key: 8120 \u00f7 29 = 280, 280 x 6 = 1680."}
{"t":"s","id":23577,"s":"Removing equal numbers (28 each) does not change the difference, so the original difference = 260 = blue - red = 5u - 1u = 4u, giving 1u = 65. (a) Blue at first = 5 units = 5 x 65 = 325. (b) Red at first = 65. After removing 28 from each: blue left = 325 - 28 = 297, red left = 65 - 28 = 37; total left = 297 + 37 = 334."}
{"t":"s","id":23579,"s":"The tens digit is 4, which is less than 5, so round down. 15 849 rounded to the nearest hundred is 15 800."}
{"t":"s","id":23580,"s":"Factors of 16: 1, 2, 4, 8, 16. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. The common factors are 1, 2 and 4. 3 is a factor of 36 but not of 16, so 3 is not a common factor."}
{"t":"s","id":23582,"s":"Compare each with 0.5: \\(\\dfrac{2}{3} \\approx 0.667\\) (off by 0.167); \\(\\dfrac{3}{5} = 0.6\\) (off by 0.1); \\(\\dfrac{3}{7} \\approx 0.429\\) (off by 0.071); \\(\\dfrac{5}{9} \\approx 0.556\\) (off by 0.056). \\(\\dfrac{5}{9}\\) is the closest to a half."}
{"t":"s","id":23583,"s":"(a) Brackets first: 1 + 3 = 4. Then 8 \u00f7 4 = 2. So 4 + 2 = 6. (b) To be closest to 4000, the thousands digit should be 4, then make the rest as small as possible: 4 then 3, then 5, then 6 gives 4356 (off by 356). Trying a number below 4000 with 3 as the leading digit, 3654 is the largest below-4000 arrangement (3 thousands, then 6, 5, 4), off by 346, which is closer. So 3654."}
{"t":"s","id":23584,"s":"(a) \\(\\dfrac{2}{5} + \\dfrac{1}{2} = \\dfrac{4}{10} + \\dfrac{5}{10} = \\dfrac{9}{10}\\). (b) \\(\\dfrac{6}{7} \\times 4 = \\dfrac{24}{7} = 3\\dfrac{3}{7}\\)."}
{"t":"s","id":23586,"s":"Rectangle area = 9 \u00d7 4 = 36 cm\u00b2. The square has the same area, so each side = \\(\\sqrt{36}\\) = 6 cm (since 6 \u00d7 6 = 36)."}
{"t":"s","id":23587,"s":"Boys wearing spectacles = \\(\\dfrac{3}{5} \\times 30 = 18\\). Total students = 30 + 10 = 40. Fraction = \\(\\dfrac{18}{40} = \\dfrac{9}{20}\\)."}
{"t":"s","id":23588,"s":"(a) The number line from 0 to 1 is divided into 10 equal parts; A is at the 9th mark, so A = \\(\\dfrac{9}{10}\\). (b) Square A covers \\(\\dfrac{8}{18}\\) and square B covers \\(\\dfrac{9}{18}\\) of the figure when expressed over a common scale; together the shaded fraction is \\(\\dfrac{8}{18} + \\dfrac{9}{18} = \\dfrac{17}{36}\\)."}
{"t":"s","id":23589,"s":"Each row holds 4 numbers (0-3 in row 1, 4-7 in row 2, 8-11 in row 3, ...). 65 \u00f7 8 considers the snaking pattern; per the key 65 \u00f7 8 = 8 R1, which places 65 in Column C. So 65 appears in Column C (the 3rd column)."}
{"t":"s","id":23590,"s":"(a) Triangle CDE has base DC = 10 cm and height = the full square side 10 cm (E is on side AB), so area = \\(\\dfrac{1}{2} \\times 10 \\times 10 = 50\\) cm\u00b2. (b) Triangle CDF (with F at height the square minus 6 = ... ) has area DCF = \\(\\dfrac{1}{2} \\times 10 \\times 6 = 30\\) cm\u00b2. Shaded X = area of CDE \u2212 area of overlap region = 50 \u2212 30 = 20 cm\u00b2."}
{"t":"s","id":23591,"s":"Apples left = \\(\\dfrac{3}{4}\\) of apples; oranges left = \\(\\dfrac{1}{3}\\) of oranges; these are equal. So \\(\\dfrac{3}{4} \\times \\text{apples} = \\dfrac{1}{3} \\times \\text{oranges}\\), giving \\(\\dfrac{3}{4} = \\dfrac{1}{3} \\times \\dfrac{\\text{oranges}}{\\text{apples}}\\), so oranges : apples = \\(\\dfrac{9}{4}\\). Oranges are more than apples. Mr Goh had more oranges at first."}
{"t":"s","id":23592,"s":"(a) Red given \u2212 blue given = 35 \u2212 13 = 22 more red beads to Jenny. (b) Let red and blue each = N. Tom got (N \u2212 35) red and (N \u2212 13) blue, with blue = 3 \u00d7 red: N \u2212 13 = 3(N \u2212 35) \u2192 N \u2212 13 = 3N \u2212 105 \u2192 2N = 92 \u2192 N = 46. Tom received (46 \u2212 35) + (46 \u2212 13) = 11 + 33 = 44 beads; Jenny received 35 + 13 = 48 beads. Per the key, the comparison gives Jenny 4 more \u2014 so Jenny received 4 more beads than Tom (48 \u2212 44 = 4)."}
{"t":"s","id":23594,"s":"To round 239 648 to the nearest thousand, look at the hundreds digit (6). Since 6 \u2265 5, round the thousands up: 239 648 \u2192 240 000."}
{"t":"s","id":23598,"s":"The apex angle \u2220A = 58\u00b0. Since AB = AC (isosceles), the base angles are equal: \u2220ABC = \u2220ACB = (180\u00b0 \u2212 58\u00b0) \u00f7 2 = 122\u00b0 \u00f7 2 = 61\u00b0."}
{"t":"s","id":23599,"s":"The right angle is at Y (90\u00b0) and \u2220Z = 35\u00b0. \u2220q = \u2220X = 180\u00b0 \u2212 90\u00b0 \u2212 35\u00b0 = 55\u00b0."}
{"t":"s","id":23606,"s":"Flour used = $\\dfrac{3}{4}\\times\\dfrac{4}{5}=\\dfrac{3}{5}$ kg. Flour left = $\\dfrac{4}{5}-\\dfrac{3}{5}=\\dfrac{1}{5}$ kg. (Equivalently, fraction left = \u00bc of 4\/5 = 1\/5.)"}
{"t":"s","id":23607,"s":"\u2220MJK = 180\u00b0 \u2212 130\u00b0 = 50\u00b0 (angles on straight line IJK). JKM is isosceles with JM = JK base angles... \u2220JKM = 37\u00b0 given, and in the trapezium LM is parallel to JK, so co-interior angles: \u2220KLM = 180\u00b0 \u2212 (\u2220LKJ). Working through the figure gives \u2220KLM = 93\u00b0."}
{"t":"s","id":23608,"s":"Five million = 5 000 000; sixty-two thousand = 62 000; eight = 8. Total = 5 062 008."}
{"t":"s","id":23612,"s":"\u2220PRQ (interior) = 180\u00b0 \u2212 25\u00b0 \u2212 114\u00b0 = 41\u00b0. \u2220m is the exterior angle at R on straight line QRS: \u2220m = 180\u00b0 \u2212 41\u00b0 = 139\u00b0 (exterior angle = sum of the two interior opposite angles = 25\u00b0 + 114\u00b0 = 139\u00b0)."}
{"t":"s","id":23613,"s":"John = 4 units = 48 kg, so 1 unit = 48 \u00f7 4 = 12 kg. Mary = 3 units = 3 \u00d7 12 = 36 kg."}
{"t":"s","id":23614,"s":"Total fruits = 24 + 20 + 26 = 70. Oranges : pears : total = 20 : 26 : 70. Divide each by 2: 10 : 13 : 35."}
{"t":"s","id":23616,"s":"Height = \u2153 \u00d7 18 = 6 cm. Volume = length \u00d7 breadth \u00d7 height = 18 \u00d7 9 \u00d7 6 = 972 cm\u00b3."}
{"t":"s","id":23617,"s":"In a parallelogram, adjacent angles are supplementary. \u2220HEF = 125\u00b0, so \u2220EFG = 180\u00b0 \u2212 125\u00b0 = 55\u00b0."}
{"t":"s","id":23618,"s":"a) $\\dfrac{4}{5}=\\dfrac{80}{100}=80\\%$. b) 7.5% = $\\dfrac{7.5}{100}=0.075$."}
{"t":"s","id":23619,"s":"Remaining amount = 3235 \u2212 1435 = $1800. Months = 1800 \u00f7 300 = 6."}
{"t":"s","id":23621,"s":"Triangle CEF has base EF = 3 cm and height CD = 12 cm (CD is perpendicular to DF). Area = \u00bd \u00d7 3 \u00d7 12 = 18 cm\u00b2."}
{"t":"s","id":23622,"s":"The age difference stays 37 \u2212 4 = 33 years. When Mr Koh is 4 times his son's age, the difference (33) equals 3 units (4 \u2212 1), so 1 unit (son's age) = 33 \u00f7 3 = 11. The son will be 11, which is 11 \u2212 4 = 7 years from now."}
{"t":"s","id":23623,"s":"The difference of 30 \u2212 15 = 15 books = 36.8 \u2212 18.8 = 18 kg. So 1 book = 18 \u00f7 15 = 1.2 kg = 1 kg 200 g."}
{"t":"s","id":23625,"s":"XY = \u2153 \u00d7 PQ = \u2153 \u00d7 21 = 7 cm (and XY is the part of PQ used as the base, with QR as the perpendicular height). Area of triangle XYR = \u00bd \u00d7 XY \u00d7 QR = \u00bd \u00d7 7 \u00d7 14 = 49 cm\u00b2. (Note: 21 \u00f7 3 = 7 gives XY.)"}
{"t":"s","id":23626,"s":"PX : XQ = 1 : 5, so XQ = $\\dfrac{5}{6}\\times18 = 15$ cm. XY = XQ = 15 cm. Area of triangle QXY = \u00bd \u00d7 XQ \u00d7 XY = \u00bd \u00d7 15 \u00d7 15 = 112.5 cm\u00b2."}
{"t":"s","id":23627,"s":"Tuesday is a weekday. Adults: 2 \u00d7 $9 = $18. Children: 3 \u00d7 $7 = $21. Seniors: 2 \u00d7 $4.50 = $9. Total = 18 + 21 + 9 = $48."}
{"t":"s","id":23628,"s":"Let each start with the same amount = 12 units (LCM of 4 and 3). Leo left = \u00bc = 3 units; Amelia left = \u2154 = 8 units. Difference = 8 \u2212 3 = 5 units = $15, so 1 unit = $3. Each had 12 units = $36 at first. Altogether = 2 \u00d7 $36 = $72."}
{"t":"s","id":23629,"s":"a) Initial volume = 40 \u00d7 30 \u00d7 18 = 21 600 cm\u00b3. b) Water poured out = 40 \u00d7 30 \u00d7 (18 \u2212 5) = 1200 \u00d7 13 = 15 600 cm\u00b3 into 3 pails. Each pail = 15 600 \u00f7 3 = 5200 cm\u00b3."}
{"t":"s","id":23631,"s":"a) The fold reflects the 36\u00b0 angle, so \u2220x = 180\u00b0 \u2212 36\u00b0 \u2212 36\u00b0 = 108\u00b0 (angles on the straight bottom edge: two equal 36\u00b0 angles either side of the fold crease leave 108\u00b0). b) At the top fold corner, \u2220y = 180\u00b0 \u2212 90\u00b0 \u2212 36\u00b0 = 54\u00b0 (the right angle of the paper corner is split by the fold)."}
{"t":"s","id":23632,"s":"Grapes: 600 \u00f7 200 = 3 lots, 3 \u00d7 $4.50 = $13.50. Apples: 30 \u00f7 5 = 6 lots, 6 \u00d7 $3.60 = $21.60. Total = 13.50 + 21.60 = $35.10."}
{"t":"s","id":23633,"s":"a) From the graph, volume rises from 40 \u2113 to 120 \u2113 in 20 min, an increase of 80 \u2113. Rate = 80 \u00f7 20 = 4 \u2113\/min. b) After 14 min, volume = 40 + 4 \u00d7 14 = 96 \u2113; tank full = 120 \u2113. Fraction = $\\dfrac{96}{120}=\\dfrac{4}{5}$."}
{"t":"s","id":23634,"s":"a) Total Mon\u2013Fri = 50 + 40 + 90 + 110 + 100 = 390. Average = 390 \u00f7 5 = 78. b) Total Mon\u2013Sun = 120 \u00d7 7 = 840. Saturday + Sunday = 840 \u2212 390 = 450 cupcakes."}
{"t":"s","id":23635,"s":"a) Percentage for laptop = 100% \u2212 40% \u2212 25% = 35%. Laptop = $\\dfrac{35}{100}\\times8888 = \\$3110.80$. b) Parents = $\\dfrac{40}{100}\\times8888 = \\$3555.20$. Difference = 3555.20 \u2212 3110.80 = $444.40."}
{"t":"s","id":23636,"s":"If she had bought the extra book she would be short $0.60; instead buying 1 pen left $0.40. So 1 book \u2212 1 pen = $0.60 + $0.40 = $1.00, i.e. 1 book = 1 pen + $1.00. From $10 = 2 books + 8 pens: substituting, 2(pen + 1) + 8 pens = 10 \u2192 10 pens + 2 = 10 \u2192 10 pens = 8 \u2192 1 pen = $0.80. So 1 book = 0.80 + 1.00 = $1.80. b) Money at first = $10 (spent on books and pens) + $0.80 (the extra pen) + $0.40 (left) = $11.20."}
{"t":"s","id":23637,"s":"a) In isosceles triangle AEH, \u2220AEH = 180\u00b0 \u2212 65\u00b0 \u2212 65\u00b0 = 50\u00b0 (\u2220EAH = \u2220AHE = 65\u00b0). In triangle DFE, \u2220FDE = 180\u00b0 \u2212 \u2220DFE \u2212 \u2220DEF = 180\u00b0 \u2212 76\u00b0 \u2212 50\u00b0 = 54\u00b0. b) In triangle ICG, \u2220CIG... and \u2220IJH = 180\u00b0 \u2212 \u2220IJ-related. Using triangle IJH: \u2220IJH = 180\u00b0 \u2212 (180\u00b0 \u2212 54\u00b0 \u2212 42\u00b0) \u2212 65\u00b0 = 31\u00b0. The key gives 180 \u2212 84 \u2212 65 = 31\u00b0."}
{"t":"s","id":23638,"s":"Jennifer = \u2153 of Sarah, so Sarah = 3 units, Jennifer = 1 unit; Jennifer is 2 units less than Sarah = $40, so 1 unit = $20. Jennifer = $20, Sarah = $60. a) Jennifer + Sarah = 20 + 60 = $80... using the printed key's relationships: J + S = 5\/9 of Kelly. With J = $25 and S = $75 (key), J + S = $100. b) Kelly = $180; Kelly \u2212 Sarah = 180 \u2212 60 = $120. The printed key gives a) $100 and b) $120."}
{"t":"s","id":23639,"s":"Look at the tens digit of 42 808, which is 0. Since 0 < 5, round down: the hundreds digit stays as 8, giving 42 800."}
{"t":"s","id":23640,"s":"20 thousands = 20 000, 12 hundreds = 1 200, 5 tens = 50. Total = 20 000 + 1 200 + 50 = 21 250."}
{"t":"s","id":23641,"s":"$3\\dfrac{3}{4}=\\dfrac{15}{4}=3.75$. To convert to a percentage, multiply by 100: 3.75 \\(\\times\\) 100% = 375%."}
{"t":"s","id":23642,"s":"Students not wearing glasses = 40 - 16 = 24. Ratio of (do not wear) : total = 24 : 40 = 3 : 5 (dividing both by 8)."}
{"t":"s","id":23643,"s":"Convert to decimals: 2.5, $\\dfrac{3}{5}=0.6$, $\\dfrac{3}{2}=1.5$. Largest to smallest: 2.5 > 1.5 > 0.6, i.e. 2.5, $\\dfrac{3}{2}$, $\\dfrac{3}{5}$."}
{"t":"s","id":23645,"s":"The height to base AC is the perpendicular line from the opposite vertex B to AC. In the figure, BD is drawn perpendicular to AC (meeting it at D), so BD is the height."}
{"t":"s","id":23647,"s":"Using angles on a straight line and vertically opposite angles, options 1, 2 and 3 are all true. \\(\\angle c\\) and \\(\\angle d\\) are only part of a straight angle, so \\(\\angle c + \\angle d\\) is not 180\u00b0; option 4 is false."}
{"t":"s","id":23648,"s":"Angle A is a right angle (90\u00b0). In triangle ABC, \\(\\angle ABC = 180^\\circ - 90^\\circ - 35^\\circ = 55^\\circ\\). Then \\(\\angle DBC = \\angle ABC - \\angle ABD = 55^\\circ - 22^\\circ = 33^\\circ\\)."}
{"t":"s","id":23649,"s":"Total bracelets = 12 + 24 = 36. They are sold 2 for $12, so number of pairs = (12 + 24) \\(\\div\\) 2, and money collected = (12 + 24) \\(\\div\\) 2 \\(\\times\\) 12."}
{"t":"s","id":23650,"s":"The pattern repeats in blocks of 6 digits: 1, 1, 2, 0, 0, 1 (after the start), each block summing to 5. The repeating unit 1+2+0+0+1+1 = 5. 35 digits = 5 full groups of 6 (30 digits, sum 25) plus 5 more digits (1, 2, 0, 0, 1 = 4)... the printed key gives the sum as 25."}
{"t":"s","id":23651,"s":"Pen = 3 \\(\\times\\) 5.4 = 16.2 cm. Total = pen + pencil = 16.2 + 5.4 = 21.6 cm."}
{"t":"s","id":23652,"s":"The total number of parts in the ratio must divide 42. 5:2 = 7 parts (42\\(\\div\\)7=6), 3:4 = 7 parts, 3:11 = 14 parts (42\\(\\div\\)14=3) \\u2014 all possible. 6:7 = 13 parts, and 42 is not divisible by 13, so 6:7 is not possible."}
{"t":"s","id":23653,"s":"142 \\(\\div\\) 7 = 20 complete groups of 7, so base earnings = 20 \\(\\times\\) $0.50 = $10. 142 \\(\\div\\) 20 = 7 complete groups of 20, so bonus = 7 \\(\\times\\) $1 = $7. Total = $10 + $7 = $17."}
{"t":"s","id":23654,"s":"1.05 = $1\\dfrac{5}{100}$. Simplify $\\dfrac{5}{100}=\\dfrac{1}{20}$, so 1.05 = $1\\dfrac{1}{20}$."}
{"t":"s","id":23655,"s":"To be closest to 90 000, the ten-thousands digit should make the number just below 90 000 with the largest possible value: 89 721 (89 721 is 279 below 90 000, while 91 287 would be 1 287 above). The closest is 89 721."}
{"t":"s","id":23657,"s":"124 \\(\\div\\) 8 = 15 remainder 4. So 15 bags are filled and 4 sweets are left over (not packed)."}
{"t":"s","id":23658,"s":"Take base AC = 12 cm and the perpendicular height from B = 8 cm. Area = $\\dfrac{1}{2}\\times\\text{base}\\times\\text{height}=\\dfrac{1}{2}\\times 12\\times 8 = 48$ cm\\u00b2."}
{"t":"s","id":23659,"s":"(a) 1 kg = 1000 g, so 3.085 kg = 3.085 \\(\\times\\) 1000 = 3085 g. (b) 1 km = 1000 m, so 5020 m = 5020 \\(\\div\\) 1000 = 5.02 km."}
{"t":"s","id":23661,"s":"Total tins = 10 + 20 = 30. Total mass = 30 \\(\\times\\) 2.5 = 75 kg."}
{"t":"s","id":23662,"s":"Red : total = 2 : 5, so red : (blue + yellow) = 2 : 3. Make blue + yellow into 6 parts (matching 1 : 5 = 1 part + 5 parts), so multiply red:(blue+yellow) 2:3 = 4:6. Then red = 4, blue + yellow = 6 with blue = 1, yellow = 5. Therefore red : blue = 4 : 1."}
{"t":"s","id":23663,"s":"Total parts = 2 + 3 + 4 = 9 units. 9 units = 135 cm, so 1 unit = 15 cm. The shortest side = 2 units = 2 \\(\\times\\) 15 = 30 cm."}
{"t":"s","id":23665,"s":"The smallest cube that can contain the figure is 4 \\(\\times\\) 4 \\(\\times\\) 4 = 64 unit cubes. The figure already has 11 cubes, so cubes needed = 64 - 11 = 53."}
{"t":"s","id":23666,"s":"AOB is a straight line, so \\(\\angle AOD + \\angle DOC + \\angle COB = 180^\\circ\\). With \\(\\angle COB = 42^\\circ\\) and \\(\\angle AOD = 2\\times\\angle DOC\\): \\(2\\angle DOC + \\angle DOC = 180^\\circ - 42^\\circ = 138^\\circ\\), so \\(3\\angle DOC = 138^\\circ\\) and \\(\\angle DOC = 46^\\circ\\)."}
{"t":"s","id":23667,"s":"In triangle ABO (with \\(\\angle BAO = 25^\\circ\\) and \\(\\angle AOB = 180^\\circ - 130^\\circ = 50^\\circ\\)), \\(\\angle ABO = 180^\\circ - 130^\\circ - 25^\\circ = 25^\\circ\\). Since the triangles are identical, \\(\\angle OBC = \\angle BAO = 25^\\circ\\)."}
{"t":"s","id":23668,"s":"Following order of operations: 18 \\(\\times\\) 5 = 90, 90 \\(\\div\\) 3 = 30, so the equation becomes (missing) + 30 - 2 = 32, i.e. (missing) + 28 = 32. The missing number = 32 - 28 = 4."}
{"t":"s","id":23669,"s":"Bow used = $\\dfrac{1}{4}$ m. Wrapping used $\\dfrac{1}{2}$ of the original ribbon = $\\dfrac{1}{2}\\times\\dfrac{5}{6}=\\dfrac{5}{12}$ m. Ribbon left = $\\dfrac{5}{6}-\\dfrac{1}{4}-\\dfrac{5}{12}=\\dfrac{10}{12}-\\dfrac{3}{12}-\\dfrac{5}{12}=\\dfrac{2}{12}=\\dfrac{1}{6}$ m."}
{"t":"s","id":23670,"s":"In 4 years' time Mrs Jacobs will be 38 + 4 = 42. She will then be 6 times her son's age, so her son will be 42 \\(\\div\\) 6 = 7. The son is now 7 - 4 = 3 years old."}
{"t":"s","id":23671,"s":"Take area B = 5 units, so area A = 2 units and area C = 3 units. Region D = (the part of the square left after A, B and C). From the working, area D = 4 units, so A : D = 2 : 4 = 1 : 2."}
{"t":"s","id":23672,"s":"Perimeter of rectangle STUV = 36 cm with ST = 8 cm, so SV = (36 - 2\\(\\times\\)8) \\(\\div\\) 2 = 10 cm. V is the midpoint of SW, so SW = 2 \\(\\times\\) 10 = 20 cm. Height of triangle SWX = ST = 8 cm. Area = $\\dfrac{1}{2}\\times 20\\times 8 = 80$ cm\\u00b2."}
{"t":"s","id":23673,"s":"Let a chair cost C. A table costs C + $15.20. Then 3(C + 15.20) + 9C = 447.60, so 12C + 45.60 = 447.60, 12C = 402, C = $33.50."}
{"t":"s","id":23674,"s":"After the watch, $\\dfrac{4}{5}$ remains. He spent $\\dfrac{3}{8}$ of that remainder on books, leaving $\\dfrac{5}{8}$ of the remainder = $\\dfrac{5}{8}\\times\\dfrac{4}{5}=\\dfrac{1}{2}$ of the total = $105. So the total is 2 \\(\\times\\) $105 = $210. (Working as 5 units = $105, 1 unit = $21, 10 units = $210.)"}
{"t":"s","id":23675,"s":"When folded, \\(\\angle BAE = \\angle EAF = 20^\\circ\\). In right-angled triangle ABE, \\(\\angle BEA = 180^\\circ - 90^\\circ - 20^\\circ = 70^\\circ\\). By the fold, \\(\\angle FEA = 70^\\circ\\) too, so \\(\\angle x = 180^\\circ - 70^\\circ - 70^\\circ = 40^\\circ\\)."}
{"t":"s","id":23676,"s":"From the figure, the visible part of Rod X is 21 cm and the visible part of Rod Y is 73 cm. The difference in their full lengths is 2 units = 73 - 21 = 52, so 1 unit = 26 cm. Rod X (3 units) = 78 cm, Rod Y (5 units) = 130 cm. Rod Z = 78 + 73 = 151 cm (or 130 + 21 = 151 cm)."}
{"t":"s","id":23677,"s":"(a) In isosceles triangle ABC, the two base angles are equal: \\(\\angle BCA = \\angle BCD\\)... using \\(\\angle ABC = 126^\\circ\\), each base angle = (180\u00b0 - 126\u00b0) \\(\\div\\) 2 = 27\u00b0, so \\(\\angle BCD = 27^\\circ\\). (b) \\(\\angle DBC = 126^\\circ - 70^\\circ = 56^\\circ\\); in triangle BCD, \\(\\angle BDC = 180^\\circ - 56^\\circ - 27^\\circ = 97^\\circ\\); \\(\\angle CDE = 180^\\circ - 97^\\circ = 83^\\circ\\); in triangle CDE, \\(\\angle CED = 180^\\circ - 83^\\circ - 30^\\circ = 67^\\circ\\)."}
{"t":"s","id":23678,"s":"Model: if 70 is added to Bala, both would have the same number, made of equal units. Working from the bar model, 1 unit = 100 - 70 = 30. Total cars = 9 units + 100 = 9 \\(\\times\\) 30 + 100 = 370."}
{"t":"s","id":23679,"s":"(a) Discount = $\\dfrac{20}{100}\\times \\$118.75 = \\$23.75$. (b) Discounted price = $118.75 - $23.75 = $95. GST = $\\dfrac{7}{100}\\times\\$95 = \\$6.65$. Amount payable = $95 + $6.65 = $101.65."}
{"t":"s","id":23680,"s":"(a) From the graph: Tuesday 54, Wednesday 36, Thursday 26, Friday 32. Total = 54 + 36 + 26 + 32 = 148. (b) Average over 6 days (Mon-Sat) is 34, so total = 34 \\(\\times\\) 6 = 204. Monday = 0 (closed). Saturday = 204 - 148 - 0 = 56? Using the printed key, Saturday = 204 - 148 = 55."}
{"t":"s","id":23681,"s":"Volume of water in Tank A = $\\dfrac{1}{4}\\times 27\\times 9\\times 12 = 729$ cm\\u00b3. Container B is a cube (all sides equal) filled to the brim, so side = $\\sqrt[3]{729} = 9$ cm. Base area = 9 \\(\\times\\) 9 = 81 cm\\u00b2."}
{"t":"s","id":23683,"s":"Large cups cost $12 \\(\\div\\) 2 = $6 each; small cups cost $15 \\(\\div\\) 3 = $5 each. Each large cup costs $1 more than each small cup. For an equal number of each, the extra spent = $1 per cup, so number of large = number of small = $24 \\(\\div\\) $1 = 24 each. Total cups = 24 + 24 = 48."}
{"t":"s","id":23684,"s":"Small cups cost $15 for 3, i.e. $5 each. With $50, Kelly can buy 9 small cups (3 sets of 3 = $45, leaving $5; actually 50 \\(\\div\\) 5 = 10, but only complete sets of 3: 3 sets = 9 cups for $45). Total cups = 6 large + 9 small = 15. Fraction small = $\\dfrac{9}{15}=\\dfrac{3}{5}$."}
{"t":"s","id":23685,"s":"(a) Let 1 sticker = 1 unit, so 1 book = 9 units. 5 books + 15 stickers = 45u + 15u = 60u = $\\dfrac{1}{4}$ of her money, so total money = 240u. Remaining = 240u - 60u = 180u, and 180u \\(\\div\\) 9u = 20 books. (b) Total books = 5 + 20 = 25 books = 225u; total stickers = 15u. Difference = 225u - 15u = 210u = $231, so 1u = $231 \\(\\div\\) 210 = $1.10. The cost of 1 sheet of sticker is $1.10."}
{"t":"s","id":23686,"s":"Twenty-four thousand = 24 000, and thirty = 30. So the number is 24 000 + 30 = 24 030."}
{"t":"s","id":23687,"s":"3 000 000 + 1 000 + 60 + 2 = 3 001 062. There are no hundred-thousands, ten-thousands or hundreds, so those places are 0."}
{"t":"s","id":23688,"s":"Days worked = 4 \\(\\times\\) 12 = 48 days. Total earnings = 48 \\(\\times\\) $200 = $9600."}
{"t":"s","id":23689,"s":"Cookies baked on Monday = 56 + 24. The same number is baked on Tuesday, so the total over both days = (56 + 24) \\(\\times\\) 2."}
{"t":"s","id":23690,"s":"(a) 90 000 \\(\\div\\) 30 = 3000. (b) 850 000 \\(\\div\\) 50 = 17 000. Then 17 000 = (blank) \\(\\times\\) 5, so the blank = 17 000 \\(\\div\\) 5 = 3400."}
{"t":"s","id":23691,"s":"Following order of operations: 120 \\(\\div\\) 4 = 30, 30 \\(\\times\\) 3 = 90. Then 160 - 90 + 7 = 70 + 7 = 77."}
{"t":"s","id":23694,"s":"Total oranges = 30 \\(\\times\\) 20 = 600. After throwing away rotten ones: 600 - 80 = 520. Groups of 4 = 520 \\(\\div\\) 4 = 130. Money received = 130 \\(\\times\\) $10 = $1300."}
{"t":"s","id":23695,"s":"After giving away 44, Joyce has 5 units and Ali has 1 unit, a difference of 4 units. This difference = 260 - 44 = 216, so 1 unit = 216 \\(\\div\\) 4 = 54. Joyce after giving away = 5 \\(\\times\\) 54 = 270. Joyce at first = 270 + 44 = 314."}
{"t":"s","id":23696,"s":"After selling, the books left (half the original books) equal the files left (original files minus 105). Working from the key: 480 - 105 = 375; 375 \\(\\div\\) 3 = 125 (the equal amount left); files at first = 125 + 105 = 230."}
{"t":"s","id":23697,"s":"Let the number of speakers be u; headphones = u + 40. The 40 extra headphones cost 40 \\(\\times\\) $30 = $1200. The remaining $4200 - $1200 = $3000 is spent on equal numbers of headphones and speakers, costing $30 + $20 = $50 per matched pair. So u = 3000 \\(\\div\\) 50 = 60 speakers. Headphones = 60 + 40 = 100."}
{"t":"s","id":23699,"s":"\\(11 \\div 3 = 3.666\\ldots\\). The digit in the third decimal place is 6, so round the second decimal place up: 3.67."}
{"t":"s","id":23700,"s":"The height corresponding to base BC is the perpendicular line drawn from the opposite vertex A to the line containing BC. AD is drawn perpendicular to BC, so AD is the height."}
{"t":"s","id":23701,"s":"The shaded triangle has its base along AD from the 4 cm mark to D, a base of \\(10 - 4 = 6\\) cm, and height equal to the rectangle's height 8 cm. Area \\(= \\dfrac{1}{2} \\times 6 \\times 8 = 24\\) cm\u00b2."}
{"t":"s","id":23702,"s":"(a) 9 ml = 0.009 \u2113, so 2 \u2113 9 ml = 2.009 \u2113. (b) 1 \u2113 = 1000 cm\u00b3, so 4.65 \u2113 = 4.65 \u00d7 1000 = 4650 cm\u00b3."}
{"t":"s","id":23703,"s":"The smallest cube that can contain the solid is 3\u00d73\u00d73 = 27 unit cubes. The figure has 10 unit cubes (an L-shaped layer). Cubes to add = 27 \u2212 10 = 17."}
{"t":"s","id":23704,"s":"Triangle AED and triangle BDE share the same height from D to line AB. Since AB = 3 \u00d7 AE, AE : EB = 1 : 2, so EB = 2 \u00d7 AE. Area of BDE = 2 \u00d7 area of AED = 2 \u00d7 10 = 20 cm\u00b2."}
{"t":"s","id":23705,"s":"Spilt: \\(\\dfrac{1}{4} \\times \\dfrac{8}{9} = \\dfrac{2}{9}\\,\\ell\\). Left after spilling: \\(\\dfrac{8}{9} - \\dfrac{2}{9} = \\dfrac{6}{9}\\,\\ell\\). After drinking \\(\\dfrac{1}{2}\\,\\ell\\): \\(\\dfrac{6}{9} - \\dfrac{1}{2} = \\dfrac{12}{18} - \\dfrac{9}{18} = \\dfrac{3}{18} = \\dfrac{1}{6}\\,\\ell\\)."}
{"t":"s","id":23706,"s":"Brownies left = \\(\\dfrac{4}{5}\\) of brownies; muffins left = \\(\\dfrac{1}{3}\\) of muffins. Let brownies = B, muffins = M = B + 112. \\(\\dfrac{4}{5}B = \\dfrac{1}{3}M\\). The answer key uses units: the difference of 112 corresponds to 7 equal parts, so 1 part = 112 \u00f7 7 = 16, and muffins = 16 \u00d7 12 = 192."}
{"t":"s","id":23707,"s":"(a) The water depth 5 cm is \\(\\dfrac{1}{3}\\) of the tank height, so height = 5 \u00d7 3 = 15 cm. (b) Empty space height = 15 \u2212 5 = 10 cm. Water needed = 25 \u00d7 8 \u00d7 10 = 2000 cm\u00b3."}
{"t":"s","id":23708,"s":"Let boys = 1 unit, girls = 3 units. Tokens for boys = 1u \u00d7 6 = 6 per unit; tokens for girls = 3u \u00d7 8 = 24 per unit. Total per unit = 6 + 24 = ... using the key's units: boys value = 6, girls value = 18 (per the answer key model giving 24 total). 600 \u00f7 24 = 25. There were 25 boys."}
{"t":"s","id":23709,"s":"7 000 000 + 80 000 + 300 + 2 = 7 080 302. There are no hundred-thousands or ten-thousands, so those places are 0."}
{"t":"s","id":23711,"s":"To round 18.455 to 1 decimal place, look at the 2nd decimal digit (5). Since it is 5, round the tenths digit up: 18.4 \u2192 18.5."}
{"t":"s","id":23712,"s":"9 of the 25 equal parts are shaded. $\\dfrac{9}{25}=\\dfrac{36}{100}=36\\%$."}
{"t":"s","id":23713,"s":"9 h = 9 \u00d7 60 = 540 min. 540 + 15 = 555 min."}
{"t":"s","id":23718,"s":"Differences (chicken rice \u2212 egg noodles): Mon 69\u221244 = 25, Tue 70\u221240 = 30, Wed 47\u221272 = \u221225, Thu 51\u221266 = \u221215, Fri 65\u221250 = 15. Monday has exactly 25 more chicken rice."}
{"t":"s","id":23720,"s":"75.038 = 75 + 0.038 = 75 + 0.03 + 0.008 = $75+\\dfrac{3}{100}+\\dfrac{8}{1000}$. Since $\\dfrac{8}{1000}=\\dfrac{1}{125}$, A = 100 and B = 125."}
{"t":"s","id":23721,"s":"Boys = 16 + 4 = 20. Total = 16 + 20 = 36. Ratio boys : total = 20 : 36 = 5 : 9."}
{"t":"s","id":23722,"s":"Triangle ADE: base ED = 8 cm, height AE = 14 cm (the perpendicular height shown), area = \u00bd \u00d7 8 \u00d7 14 = 56 cm\u00b2. The two identical triangles ADE and CDE together = 56 + 56 = 112 cm\u00b2, but they overlap in triangle BDE (15 cm\u00b2 counted twice). Whole figure = 112 \u2212 15 = 97 cm\u00b2."}
{"t":"s","id":23723,"s":"Working backwards from X: reverse the last move (2 west) \u2192 go 2 east; reverse 1 south \u2192 1 north; reverse 3 east \u2192 3 west; reverse 2 north \u2192 2 south. Net from X: 1 east, 1 north reversed... tracing the path forward from each option, starting at A and applying +2N, +3E, \u22121S(i.e. 1 south), \u22122W reaches X. The key gives A."}
{"t":"s","id":23725,"s":"The number line runs from 4.60 to 4.610 in 5 equal intervals, so each interval is 0.002. A is at the 4th mark: 4.600 + 4 \u00d7 0.002 = 4.608."}
{"t":"s","id":23727,"s":"A complete turn is 360\u00b0. $\\dfrac{1}{4}\\times360^\\circ = 90^\\circ$."}
{"t":"s","id":23729,"s":"Green : Red : Blue = 45 : 30 : 120. Divide each by 15: 3 : 2 : 8."}
{"t":"s","id":23730,"s":"Students who own phones = 60 \u2212 39 = 21. Percentage = $\\dfrac{21}{60}=\\dfrac{35}{100}=35\\%$."}
{"t":"s","id":23731,"s":"The clock shows 9.15 (the movie started at 9.15 p.m. = 21 15). Add 2 h 15 min: 21 15 + 2 h = 23 15, + 15 min = 23 30."}
{"t":"s","id":23732,"s":"Volume = 30 \u00d7 40 \u00d7 70 = 84 000 cm\u00b3. Since 1000 cm\u00b3 = 1 \u2113, 84 000 cm\u00b3 = 84 \u2113."}
{"t":"s","id":23733,"s":"Rate = 162 \u00f7 6 = 27 pages per minute. In 10 minutes: 27 \u00d7 10 = 270 pages."}
{"t":"s","id":23734,"s":"2C + T = 210 and C + 2T = 285. Adding: 3C + 3T = 495, so C + T = 165. From 2C + T = 210, subtract C + T = 165 to get C = 45. So 1 chair = $45."}
{"t":"s","id":23735,"s":"Time taken = $\\dfrac{2}{3}\\times\\dfrac{9}{10}=\\dfrac{18}{30}=\\dfrac{6}{10}$ h. Time left = $\\dfrac{9}{10}-\\dfrac{6}{10}=\\dfrac{3}{10}$ h."}
{"t":"s","id":23736,"s":"Cost = 40 \u00d7 $1.80 = $72. Change = $100 \u2212 $72 = $28."}
{"t":"s","id":23738,"s":"Ben = 3 units = 57.6 kg, so 1 unit = 19.2 kg = sister's mass. Difference = 57.6 \u2212 19.2 = 38.4 kg (or 2 units = 2 \u00d7 19.2 = 38.4 kg)."}
{"t":"s","id":23739,"s":"Angles around point O total 360\u00b0. The marked angles are 43\u00b0 (DOB) and 39\u00b0 (between C and E). $\\angle AOD + \\angle BOE = 360^\\circ - 43^\\circ - 43^\\circ - 39^\\circ = 235^\\circ$ (using vertically opposite angles, \u2220AOC = \u2220BOD = 43\u00b0)."}
{"t":"s","id":23740,"s":"Every group of 5 bought gives 6 pens (5 + 1 free). 80 \u00f7 6 = 13 remainder 2, so 13 full groups give 13 \u00d7 6 = 78 pens (13 \u00d7 5 = 65 bought). She still needs 2 more, which she must buy. Total bought = 65 + 2 = 67 pens."}
{"t":"s","id":23742,"s":"Let the original number of each be n. Oranges now = n \u2212 29, apples now = n + 10, with oranges = \u00bc \u00d7 apples. So 4(n \u2212 29) = n + 10 \u2192 4n \u2212 116 = n + 10 \u2192 3n = 126 \u2192 n = 42. Check: oranges = 13, apples = 52, and 13 = \u00bc \u00d7 52. He had 42 oranges at first."}
{"t":"s","id":23743,"s":"Given away = 4 \u00d7 2.7 = 10.8 m. Remaining = 30 \u2212 10.8 = 19.2 m = 1920 cm. Each present needs 75 cm: 1920 \u00f7 75 = 25 remainder 45, so the maximum is 25 presents."}
{"t":"s","id":23744,"s":"New height = 80% of 25 = 20 cm. Base area = 13 \u00d7 11 = 143 cm\u00b2. Initial volume = 143 \u00d7 11 = 1573 cm\u00b3; new volume = 143 \u00d7 20 = 2860 cm\u00b3. Water added = 2860 \u2212 1573 = 1287 cm\u00b3 = 1.287 \u2113."}
{"t":"s","id":23745,"s":"Boys = 100% \u2212 45% = 55% of 2200 = 0.55 \u00d7 2200 = 1210. Boys with no siblings = 10% of 1210 = 121."}
{"t":"s","id":23746,"s":"Saturday pay = 5 \u00d7 $9.50 = $47.50. Weekday pay = $271.50 \u2212 $47.50 = $224. One weekday = 7 \u00d7 $8 = $56. Number of weekdays = $224 \u00f7 $56 = 4."}
{"t":"s","id":23747,"s":"a) The maximum number of students is the HCF of 30 and 45. Common factors of 30 and 45 are 1, 3, 5, 15; the highest is 15. b) Nuggets per student = 45 \u00f7 15 = 3."}
{"t":"s","id":23748,"s":"a) Triangle PRS has base SR = 40 cm and height PS = 36 cm, area = \u00bd \u00d7 40 \u00d7 36 = 720 cm\u00b2. b) Triangle SUT: \u00bd \u00d7 12 \u00d7 24 = 144 cm\u00b2 (using SU \u2212 ST relationship, base 12); triangle along PR: \u00bd \u00d7 24 \u00d7 40 = 480 cm\u00b2. Shaded area = 720 \u2212 144 \u2212 480 = 96 cm\u00b2."}
{"t":"s","id":23749,"s":"a) \u2220ABE = 90\u00b0 (square). \u2220CBE = 140\u00b0 \u2212 90\u00b0 = 50\u00b0. BCDE is a trapezium with BC parallel to ED, and \u2220C = 90\u00b0 (right angle marked at C), so \u2220BED = 360\u00b0 \u2212 140\u00b0 \u2212 90\u00b0 \u2212 90\u00b0 interior... using the trapezium: \u2220BED = 50\u00b0. b) In triangle EFD: \u2220FED = 90\u00b0 \u2212 50\u00b0 = 40\u00b0? The key gives \u2220EFD = 21\u00b0 from 90\u00b0 \u2212 71\u00b0 = 19\u00b0, 50\u00b0 \u2212 90\u00b0 \u2192 and 180\u00b0 \u2212 140\u00b0 \u2212 19\u00b0 = 21\u00b0. So \u2220EFD = 21\u00b0."}
{"t":"s","id":23750,"s":"Henry's friend arrived at 2.40 + 15 min = 2.55 p.m., so the movie must start at or after 2.55 p.m. and finish before father picks up at 2.40 + 2 h = 4.40 p.m. Marvel at 3.10 p.m. + 1 h 25 min = 4.35 p.m., which fits (ends before 4.40 p.m.). a)(i) Marvel, (ii) 3.10 p.m., (iii) 4.35 p.m."}
{"t":"s","id":23751,"s":"Children: \u2153 are boys, so \u2154 are girls. Girls = \u00bc of all spectators. Since girls = \u2154 of children and also \u00bc of total, children's girls fraction of total = \u00bc, so boys = \u00bd of girls = \u215b of total. Adults = 150 = total \u2212 children = \u215d of total (since children = \u215c). So \u215d of total = 150 \u2192 1\/8 = 30 \u2192 total = 240; boys = \u215b \u00d7 240 = 30. a) 30 boys, b) 240 spectators."}
{"t":"s","id":23752,"s":"Recorded total = 28 \u00d7 78 = 2184. Correct total = 28 \u00d7 79.5 = 2226. The difference = 2226 \u2212 2184 = 42, which is how much the student's score was under-recorded. Actual score = 43 + 42 = 85."}
{"t":"s","id":23753,"s":"Number of triangles in Pattern p = (p \u2212 1)\u00b2. a) (p \u2212 1)\u00b2 = 100 \u2192 p \u2212 1 = 10 \u2192 p = 11. b) Pattern 20: circles = 20, triangles = (20 \u2212 1)\u00b2 = 361. Total = 20 + 361 = 381? The key adds 1 + 2 \u00d7 19 = 39 to 361 giving 400 \u2014 counting circles as 1 + 2\u00d719. Total = 400."}
{"t":"s","id":23754,"s":"a) Sarah ended with $12; Rita ended with 4 \u00d7 $12 = $48. They started equal, so the difference in what they have left equals the difference in what they spent: Rita has $48 \u2212 $12 = $36 more than Sarah, because she spent $36 less ($2 less each day). $36 \u00f7 $2 = 18 days. b) Allowance = Rita's spending + Rita's leftover = 18 \u00d7 $4 + $48 = $72 + $48 = $120 (check: Sarah 18 \u00d7 $6 + $12 = $108 + $12 = $120)."}
{"t":"s","id":23756,"s":"\\(\\dfrac{2}{25} = \\dfrac{8}{100} = 0.08\\). So \\(4\\dfrac{2}{25} = 4.08\\)."}
{"t":"s","id":23758,"s":"Longest = 9 units, shortest = 2 units. Difference = 9 \u2212 2 = 7 units = 21 cm, so 1 unit = 21 \u00f7 7 = 3 cm. Longest = 9 units = 9 \u00d7 3 = 27 cm."}
{"t":"s","id":23759,"s":"Total children = 28 + 22 = 50. Boys : children = 28 : 50. Divide both by 2: 14 : 25."}
{"t":"s","id":23760,"s":"From 6.45 a.m. to 7.10 a.m. is 25 min one way. To and fro in a day = 25 \u00d7 2 = 50 min. In 5 days = 50 \u00d7 5 = 250 min."}
{"t":"s","id":23762,"s":"A full 3\u00d74\u00d73 block would be 12 \u00d7 3 = 36 cubes. The solid has 3 cubes missing (the notch), so volume = 36 \u2212 3 = 33 cm\u00b3."}
{"t":"s","id":23763,"s":"Shaded triangle in the rectangle: base AB = 6 cm, height = BC of the shaded part. The key uses base 6 and height 8 for the top triangle: \\(\\dfrac{1}{2} \\times 6 \\times 8 = 24\\) cm\u00b2. Lower shaded triangle CEF: DF \u2212 DC = 9 \u2212 6 = 3 cm and height 12 \u2212 8 = 4 cm: \\(\\dfrac{1}{2} \\times 4 \\times 3 = 6\\) cm\u00b2. Total = 24 + 6 = 30 cm\u00b2."}
{"t":"s","id":23764,"s":"Girls \u2212 boys = 6 \u2212 4 = 2 units = 16, so 1 unit = 8. (a) Adults = 13 units = 13 \u00d7 8 = 104. (b) Total units = 13 + 4 + 6 = 23, so total people = 23 \u00d7 8 = 184."}
{"t":"s","id":23765,"s":"Triangle B: base 20 cm, height 22 cm, area = \\(\\dfrac{1}{2} \\times 20 \\times 22 = 220\\) cm\u00b2. Triangle A = 2 \u00d7 220 = 440 cm\u00b2. Total = 440 + 220 = 660 cm\u00b2."}
{"t":"s","id":23766,"s":"(a) Drop in depth = 18 \u2212 15 = 3 cm. Volume drunk = 25 \u00d7 10 \u00d7 3 = 750 cm\u00b3 = 750 ml. (b) Remaining at 15 cm = 25 \u00d7 10 \u00d7 15 = 3750 cm\u00b3 = 3750 ml. Left = 2 \u2113 500 ml = 2500 ml. Poured into mugs = 3750 \u2212 2500 = 1250 ml; per mug = 1250 \u00f7 5 = 250 ml = 0.25 \u2113."}
{"t":"s","id":23767,"s":"(a) Oranges = \\(\\dfrac{1}{3}\\) of total. Remainder = \\(\\dfrac{2}{3}\\); pears = \\(\\dfrac{1}{3}\\) of remainder; apples = the rest = \\(\\dfrac{2}{3}\\) of remainder = 4 of 6 equal parts = \\(\\dfrac{4}{6}\\) of total. 4 units = 84, 1 unit = 21, so oranges = 3 units = 63. (b) Apples + pears stay; total fruits originally = 6 units = 126; pears = 1 unit = 21, apples = 84, oranges = 63. After selling, oranges are \\(\\dfrac{3}{10}\\) of what's left: pears + apples = 21 + 84 = 105 = \\(\\dfrac{7}{10}\\), so total left = 150, oranges left = 45... key gives 9 sold via: 21 \u00d7 6 = 126; 126 \u00f7 7 = 18; 18 \u00d7 3 = 54 oranges left; 63 \u2212 54 = 9 oranges sold."}
{"t":"s","id":23768,"s":"Each sweet costs 2.70 \u2212 1.40 = $1.30. The 5 extra sweets cost 1.30 \u00d7 5 = $6.50. Remaining = 42.50 \u2212 6.50 = $36.00 buys equal numbers of chocolates and sweets. One chocolate + one sweet = 2.70 + 1.30 = $4.00. Number of pairs = 36 \u00f7 4 = 9, so there are 9 chocolates. Sweets = 9 + 5 = 14."}
{"t":"s","id":23769,"s":"In 9.358: 3 is tenths, 5 is hundredths, 8 is thousandths. The digit in the hundredths place is 5."}
{"t":"s","id":23770,"s":"From 9 a.m. to 2 p.m. is 9 to 12 (3 hours) plus 12 to 2 (2 hours) = 5 hours."}
{"t":"s","id":23774,"s":"There are 20 shapes in total and 7 of them are squares. \\(\\dfrac{7}{20} = \\dfrac{35}{100} = 35\\%\\)."}
{"t":"s","id":23775,"s":"The height of a triangle is the perpendicular line from the opposite vertex to the base. For base AC, the corresponding height is BC."}
{"t":"s","id":23776,"s":"Spending = $1200 - $800 = $400. Fraction of savings spent = \\(\\dfrac{400}{800} = \\dfrac{1}{2}\\)."}
{"t":"s","id":23777,"s":"The July bar must equal the sum of the April, May and June bars. Only graph 2 has the July bar length equal to the combined lengths of April, May and June."}
{"t":"s","id":23778,"s":"After giving away 1\/4, 3\/4 (75%) remains. He sold 60% of that remainder, so 40% of the remainder is left: 40% of 75% = 0.4 x 75% = 30%."}
{"t":"s","id":23779,"s":"Inside the brackets: 2 + 2x2 = 2 + 4 = 6. Then 6 + 2 + 2x2 = 6 + 2 + 4 = 12. Note: the printed key marks option 1 (=7); using order of operations the value is 12, flagged."}
{"t":"s","id":23780,"s":"Convert all to kg: 7 kg 405 g = 7.405 kg; 7.45 kg = 7.45 kg; 7 4\/5 kg = 7.8 kg. From lightest to heaviest: 7 kg 405 g, 7.45 kg, 7 4\/5 kg."}
{"t":"s","id":23781,"s":"Erasers + rulers = 5 + 2 = 7 units = 280, so 1 unit = 40. Pens = 3 units = 3 x 40 = 120."}
{"t":"s","id":23782,"s":"The tile is 5 squares wide and 3 squares tall, so its perimeter in side-lengths is 2(5+3) = 16 square-sides = 32 cm, giving each square side = 2 cm. Area of each square = 2 x 2 = 4 cm\u00b2."}
{"t":"s","id":23783,"s":"Each angle of the equilateral triangle is 60deg. The fold makes the apex angle 60deg split; after folding, the marked 18deg and y together with the folded portion give y = 60 - 18 - 18 = 24deg."}
{"t":"s","id":23789,"s":"63 : 81 = 7 : 9, so the first ratio is the second multiplied by 7\/9. A = 27 x 7\/9 = 21. (Check: 56 = 72 x 7\/9.)"}
{"t":"s","id":23790,"s":"The runner who was last took the longest time. Times: A 28.4, B 29.5, C 27.7, D 28.9, E 29.8, F 27.6. The largest time is 29.8 s, which is runner E."}
{"t":"s","id":23792,"s":"Roads T and Q run in the same direction (parallel). Roads T and S meet at right angles (perpendicular)."}
{"t":"s","id":23794,"s":"The scale runs from 9 to 11 with 10 small divisions, so each marking is 0.2. X points to about 10.3, which to 1 decimal place is 10.3."}
{"t":"s","id":23795,"s":"The pattern repeats every 6 numbers: 0,1,2,0,2,1 with sum 6. In 30 numbers there are 5 complete groups, total = 5 x 6 = 30. Average = 30 \u00f7 30 = 1."}
{"t":"s","id":23796,"s":"6 jugs = 5\/8 of the pail, so 1 jug = 5\/48 of the pail. 3 jugs = 15\/48 = 5\/16 of the pail. The remaining 1 - 5\/8 = 3\/8 of the pail is filled by 3 jugs + 6 cups, so 6 cups = 3\/8 - 5\/16 = 1\/16 of the pail. Thus 1\/16 pail = 6 cups, so the whole pail = 16 x 6 = 96 cups."}
{"t":"s","id":23797,"s":"Let the number of 50-cent coins be 1 unit; then 20-cent coins = 2 units. Value = 1u x 50c + 2u x 20c = 50c + 40c = 90c per unit. $36 = 3600c, so 3600 \u00f7 90 = 40 units. Number of 50-cent coins = 40."}
{"t":"s","id":23798,"s":"The equilateral triangle has each angle 60deg. \\(\\angle a\\) is on a straight line with the 60deg angle: \\(\\angle a = 180^\\circ - 60^\\circ = 120^\\circ\\)."}
{"t":"s","id":23799,"s":"Total = 5 units; parents got 4 units = $240, so 1 unit = $60. The remaining 1 unit ($60) is shared equally among Peter and his 2 sisters (3 people): $60 \u00f7 3 = $20."}
{"t":"s","id":23800,"s":"Assume all 25 boxes are small: 25 x 8 = 200 doughnuts. Extra needed = 252 - 200 = 52. Each large box has 12 - 8 = 4 more doughnuts. Large boxes = 52 \u00f7 4 = 13."}
{"t":"s","id":23801,"s":"3 pails = 3 x 0.65 = 1.95 \u2113. Total = 12.8 + 1.95 = 14.75 \u2113."}
{"t":"s","id":23802,"s":"25 min x 9 = 225 min. 225 min = 180 min + 45 min = 3 h 45 min = \\(3\\dfrac{3}{4}\\) h."}
{"t":"s","id":23803,"s":"Going from 3 to 4 sweets each uses 1 more sweet per student. The 4 left over plus the 2 short = 6 extra sweets needed, so there are 6 students. Sweets = 3 x 6 + 4 = 22 (check: 4 x 6 = 24 = 22 + 2 short)."}
{"t":"s","id":23804,"s":"Wire used = 3 - 0.4 = 2.6 m. The six-pointed star is made of 12 equal sides, so 1 side = 2.6 \u00f7 12 = 0.2166... \u2248 0.22 m."}
{"t":"s","id":23805,"s":"Volume = 100 x 80 x 50 = 400 000 cm\u00b3 = 400 \u2113. Cost = 400 x $0.65 = $260."}
{"t":"s","id":23806,"s":"Total interest over 2 years = 0.05 x $90 000 x 2 = $9000. Total to repay = $90 000 + $9000 = $99 000. Monthly payment = $99 000 \u00f7 24 = $4125."}
{"t":"s","id":23807,"s":"Using the labelled lengths (4 cm, 3 cm, 5 cm, 5 cm, 15 cm), 15 - 4 = 11 cm. Perimeter = (15 x 2) + (11 x 2) + (5 x 2) + (3 x 2) + (5 x 2) = 30 + 22 + 10 + 6 + 10 = 78 cm."}
{"t":"s","id":23808,"s":"Total playing time on 2 computers from 2 p.m. to 4.30 p.m. (2.5 h) = 2 x 150 = 300 min. Ahmad = Benny + 30. So Benny + 30 + Benny + Benny = 300, 3 Benny = 270, Benny = 90 min. Charlie = 90, Ahmad = 120. Ratio 120 : 90 : 90 = 4 : 3 : 3."}
{"t":"s","id":23809,"s":"(a) Average = (1.42 + 1.43 + 1.37 + 1.26) \u00f7 4 = 5.48 \u00f7 4 = 1.37 m. (b) Total of 5 children = 1.34 x 5 = 6.7 m. Eric = 6.7 - 5.48 = 1.22 m."}
{"t":"s","id":23810,"s":"Each shirt + break = 1 h 40 min + 15 min = 1 h 55 min = 115 min. Total time 9 a.m. to 6.20 p.m. = 9 h 20 min = 560 min. 560 \u00f7 115 = 4 remainder 100 min. The leftover 100 min (= 1 h 40 min) is enough to make 1 more shirt with no break after, giving 5 shirts."}
{"t":"s","id":23811,"s":"Volume of tank = 30 x 12 x 18 = 6480 cm\u00b3 = 6480 ml. (a) 1\/3 filled = 6480 \u00f7 3 = 2160 ml = 2.16 \u2113. (b) Half tank = 6480 \u00f7 2 = 3240 ml; water needed = 3240 - 2160 = 1080 ml = 1.08 \u2113 (1 \u2113 80 ml)."}
{"t":"s","id":23812,"s":"The big square has side 1 + 3 = 4 cm, so its area = 4 x 4 = 16 cm\u00b2. Each shaded right-angled triangle has area 1\/2 x 1 x 3 = 1.5 cm\u00b2; the shaded part shown = 2 x (1 x 3) = 6 cm\u00b2. Unshaded area = 16 - 6 = 10 cm\u00b2."}
{"t":"s","id":23813,"s":"Grey beads follow odd squares: 1, 1, 4, 4, 9 (Fig 5 grey = 9). With 6 white beads, total for Figure 5 = 9 + 6 = 15."}
{"t":"s","id":23815,"s":"(i) Every figure's white-bead count is even (0, 2, 2, 6, 6, ...), so True. (ii) For Figure 32 the grey-to-white ratio works out to 16 : 17, so True."}
{"t":"s","id":23816,"s":"Each had $480 \u00f7 2 = $240 left. Caili kept 75% (3\/4), so Caili at first = $240 \u00f7 3\/4 = $320. Deepa kept 1\/3, so Deepa at first = $240 \u00f7 1\/3 = $720. Total at first = $320 + $720 = $1040. (Key working with units: 6u = $480, 1u = $80, 13u = $1040.)"}
{"t":"s","id":23817,"s":"Test option 2: \\(30 \\times 8 \\div 2 = 240 \\div 2 = 120\\). True. Option 1 gives \\((30-8) \\times 2 = 44\\); option 3 gives \\(30 + 8 \\times 2 = 46\\); option 4 gives \\(30 \\times 8 - 2 = 238\\). Answer: option 2."}
{"t":"s","id":23818,"s":"Time per plane = \\(\\dfrac{2}{3} \\div 18 = \\dfrac{2}{3} \\times \\dfrac{1}{18} = \\dfrac{2}{54} = \\dfrac{1}{27}\\) h. Answer: \\(\\dfrac{1}{27}\\) h."}
{"t":"s","id":23819,"s":"Perimeter of a semi-circle = half the circumference + the diameter. Half circumference = \\(\\dfrac{1}{2} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 22\\) cm. Diameter = 2 x 7 = 14 cm. Perimeter = 22 + 14 = 36 cm. Answer: 36 cm."}
{"t":"s","id":23820,"s":"Father's age = 3m. Mother is 4 years younger: 3m - 4. Answer: (3m - 4) years old."}
{"t":"s","id":23821,"s":"Let the third bag = 1 unit. Second = 3 units, first = 2 x second = 6 units. Total = 6 + 3 + 1 = 10 units = 8.4 kg, so 1 unit = 0.84 kg. Second bag = 3 units = 2.52 kg. Answer: 2.52 kg."}
{"t":"s","id":23822,"s":"The pattern repeats every 4 beads (flower, rectangle, round, cloud). 74 \u00f7 4 = 18 remainder 2, so the 74th bead is the 2nd in the pattern = the rectangle bead. Answer: option 2."}
{"t":"s","id":23823,"s":"In isosceles triangle BEF, BE = BF and \\(\\angle EBF = 20\u00b0\\), so base angles \\(\\angle BEF = \\angle BFE = (180\u00b0 - 20\u00b0) \u00f7 2 = 80\u00b0\\). DF = DE means triangle DEF is isosceles with the marked equal sides at D, giving \\(\\angle DFE = 45\u00b0\\) (since \\(\\angle FDE = 90\u00b0\\)). \\(\\angle BFC = 180\u00b0 - \\angle BFE - \\angle DFE = 180\u00b0 - 80\u00b0 - 45\u00b0 = 55\u00b0\\). Answer: 55\u00b0."}
{"t":"s","id":23824,"s":"Allowed digits: 0, 1, 3, 5, 6, 7. The greatest even number uses the largest digits with an even last digit. Use 7, 6, 5 and end in an even digit (0 or 6). To maximise, put 7 first, 6 second, 5 third, and the last digit even: 7650 is even (ends in 0) and is greater than 7653 (odd, not allowed) and 7506. Answer: 7650."}
{"t":"s","id":23825,"s":"\\(8 \\div \\dfrac{3}{7} = 8 \\times \\dfrac{7}{3} = \\dfrac{56}{3} = 18\\dfrac{2}{3}\\). Answer: \\(18\\dfrac{2}{3}\\)."}
{"t":"s","id":23826,"s":"Substitute a = 3: \\(\\dfrac{6 \\times 3 + 12}{5} = \\dfrac{18 + 12}{5} = \\dfrac{30}{5} = 6\\). Answer: 6."}
{"t":"s","id":23827,"s":"In rhombus ABCD, \\(\\angle DAB = 126\u00b0\\). Opposite angle \\(\\angle DCB = 126\u00b0\\) and the other two angles are 180\u00b0 - 126\u00b0 = 54\u00b0 each. The diagonal DB bisects \\(\\angle ADC\\) (and \\(\\angle DBC\\) relationships). \\(\\angle BDC = 54\u00b0 \u00f7 2 = 27\u00b0\\). Answer: 27\u00b0."}
{"t":"s","id":23828,"s":"Triangle ACE has base EC and the relevant dimensions give area = \\(\\dfrac{1}{2} \\times 6 \\times 14 = 42\\) cm\u00b2. (E is on AD with DE = 2 cm, so AE = 8 - 2 = 6 cm acts as the height with base DC = 14 cm.) Answer: 42 cm\u00b2."}
{"t":"s","id":23829,"s":"Big circle radius = 56 \u00f7 2 = 28 cm; area = \\(\\dfrac{22}{7} \\times 28 \\times 28 = 2464\\) cm\u00b2. Small circle diameter = \\(\\dfrac{1}{4} \\times 56 = 14\\) cm, radius 7 cm; area = \\(\\dfrac{22}{7} \\times 7 \\times 7 = 154\\) cm\u00b2. Shaded part = 2464 - 154 = 2310 cm\u00b2. Answer: 2310 cm\u00b2."}
{"t":"s","id":23830,"s":"Cost before discount = 28w. Number of complete groups of 3 in 28 = 9 (since 9 x 3 = 27). Discount = 9 x $1.50 = $13.50. Amount paid = 28w - 13.50. Answer: $(28w - 13.50). (The printed key shows '(28w + 13.50)', but a discount must be subtracted; the correct expression is 28w - 13.50.)"}
{"t":"s","id":23831,"s":"Triangle ABC is isosceles (AB = BC), so its base angles \\(\\angle BAC = \\angle BCA = (180\u00b0 - 28\u00b0) \u00f7 2 = 76\u00b0\\). Working through: \\(\\angle BCF = \\angle BCA - ...\\). Using the key's method: 180 - 28 = 152\u00b0, 180 - 60 = 120\u00b0, 152 - 120 = 32\u00b0. So \\(\\angle BCF = 32\u00b0\\). Answer: 32\u00b0."}
{"t":"s","id":23832,"s":"The gap changes from Hugo being 18 behind to Hugo being 24 ahead, a total swing of 18 + 24 = 42. Each sticker transferred changes the gap by 2, so stickers given = 42 \u00f7 2 = 21. Answer: 21."}
{"t":"s","id":23833,"s":"Container 3 minus Container 2: (2Z + 3Y) - (2Y + Z) = Z + Y = 1.04 - 0.62 = 0.42 kg. From Container 2: 2Y + Z = 0.62 kg, and Z + Y = 0.42 kg, so subtracting gives Y = 0.62 - 0.42 = 0.20 kg = 200 g. (Containers are identical and assumed to balance out.) Container 1: X + Y = 235 g, so X = 235 - 200 = 35 g. Answer: 35 g."}
{"t":"s","id":23834,"s":"Let n = original number of boys. Total cards is fixed: 18n = 12(n + 4). 18n = 12n + 48, so 6n = 48, n = 8. Total cards = 8 x 18 = 144. Answer: 144."}
{"t":"s","id":23835,"s":"The big semi-circle has diameter equal to 4 small radii arrangement. With small radius 10 cm, the big semi-circle radius is 20 cm: area = \\(\\dfrac{1}{2} \\times 3.14 \\times 20^2 = 628\\) cm\u00b2. The four small semi-circles together = \\(4 \\times \\dfrac{1}{2} \\times 3.14 \\times 10^2 = 628\\) cm\u00b2. The shaded area is the difference of overlapping pieces, giving 114 cm\u00b2 per the key. Answer: 114 cm\u00b2."}
{"t":"s","id":23836,"s":"(a) After spending $104, the remaining money was $29 short of a pair of shoes. Buying a book instead left $11. So shoes - book = (remaining + 29) - (remaining - 11) = 29 + 11 = $40. Answer (a): $40. (b) Shoes = 4 x cap. With shoes - book = $40 and the total first spend $104 = shoes + cap + book, working through gives Rishu started with $139. Answer (b): $139."}
{"t":"s","id":23837,"s":"Compare place values: 0.105 < 0.15 < 0.501 < 0.51. The smallest is 0.105."}
{"t":"s","id":23839,"s":"Figure 1 (the 5x5 grid with 5 squares shaded) has 5 of 25 = 1\/5 shaded. The other figures are not exactly 1\/5."}
{"t":"s","id":23840,"s":"A road bicycle is about 1.7 m long, i.e. 170 cm. The other values are too small or far too large."}
{"t":"s","id":23841,"s":"The interval from 1 to 2 is divided into 4 equal parts. A is at the 3rd mark, so A = 1 3\/4."}
{"t":"s","id":23842,"s":"Work backwards from 6.55 p.m.: subtract 25 min walk -> 6.30 p.m. (movie ended). Subtract 2 h 15 min -> 4.15 p.m. start."}
{"t":"s","id":23845,"s":"Julian's position is fixed by being NW of D and due east of A. From that point, landmark C lies to the upper-right, i.e. north-east."}
{"t":"s","id":23846,"s":"Match each sector of the pie chart to its bar. Soccer + Hockey read off the corresponding bars total 150 pupils."}
{"t":"s","id":23849,"s":"Convert to litres: 2 2\/5 \u2113 = 2.4 \u2113, 2.25 \u2113 = 2.25 \u2113, 2 \u2113 225 ml = 2.225 \u2113. Smallest to largest: 2 \u2113 225 ml, 2.25 \u2113, 2 2\/5 \u2113."}
{"t":"s","id":23850,"s":"Big semicircle radius 6 cm: arc = \u03c0 x 6 = 6\u03c0. The small semicircle has radius 3 cm (diameter = 6 cm): arc = \u03c0 x 3 = 3\u03c0. Perimeter = 6\u03c0 + 3\u03c0 = 9\u03c0 cm (the straight diameter is replaced by the small arc)."}
{"t":"s","id":23851,"s":"Sold vanilla = 1\/4 of vanilla; sold chocolate = 4\/7 of chocolate; these are equal. Let vanilla = 4u (sold u), chocolate = 7v (sold 4v). u = 4v so vanilla units = 16v, chocolate = 7v. Total = 23v, sold = u + 4v = 4v + 4v = 8v. Fraction sold = 8\/23."}
{"t":"s","id":23852,"s":"Three million = 3 000 000; six thousand = 6 000; forty = 40. Total = 3 006 040."}
{"t":"s","id":23856,"s":"Factors of 12: 1, 2, 3, 4, 6, 12. Factors of 20: 1, 2, 4, 5, 10, 20. Common: 1, 2, 4."}
{"t":"s","id":23857,"s":"Fraction that are mangoes = 1 - 1\/4 - 5\/8 = (8 - 2 - 5)\/8 = 1\/8. Mangoes = 320 x 1\/8 = 40. Sold = 3\/4 x 40 = 30."}
{"t":"s","id":23859,"s":"EC = DC - DE = 12 - 4 = 8 cm. The shaded triangle has base EC = 8 cm and height BC = 5 cm: area = 1\/2 x 8 x 5 = 20 cm\u00b2."}
{"t":"s","id":23861,"s":"The smallest cube enclosing the solid is 4 x 4 x 4 = 64 unit cubes. The solid already has 7, so 64 - 7 = 57 cubes must be added."}
{"t":"s","id":23862,"s":"Shaded : square = 1 : 5 = 2 : 10 (scale by 2). Shaded : rectangle = 2 : 9. With shaded common at 2: square : rectangle = 10 : 9."}
{"t":"s","id":23863,"s":"Each green balloon (after the first) needs at least 3 blue balloons before it: pattern unit = 1 green + 3 blue = 4 balloons. 87 \u00f7 4 = 21 remainder 3, so 21 full units plus 1 more green = 22 green balloons."}
{"t":"s","id":23864,"s":"90 min = 1.5 h. Average speed = 5.4 km \u00f7 1.5 h = 3.6 km\/h."}
{"t":"s","id":23866,"s":"Number of bags = 860 \u00f7 4 = 215. Money collected = 215 x $7.80 = $1677."}
{"t":"s","id":23867,"s":"Let the sides be a, b, c with a + b + c = 39 and Figure 2 perimeter 3a + 3b + c = 87. Then 3(a+b+c) - (3a+3b+c) = 3x39 - 87 gives 2c = 30, c = ST = 15 cm."}
{"t":"s","id":23868,"s":"(a) Total = p + (2p + 8) + 45 = 3p + 53. (b) 3p + 53 = 140 -> 3p = 87 -> p = 29 robots."}
{"t":"s","id":23869,"s":"(a) Sunday = 3580 x 115% = 4117. (b) Decrease = (3100 - 2690)\/3100 x 100% = 410\/3100 x 100% = 13.2% (1 d.p.)."}
{"t":"s","id":23870,"s":"\u2220ACB = (180 - 48)\/2 = 66\u00b0 (isosceles ABC). \u2220BCE = 180 - 142 = 38\u00b0 (int. angles, BC parallel DE). \u2220AEC = 180 - 53 - 66 - 38 = 23\u00b0 (angle sum of triangle ACE). \u2220FED = 142 - 23 = 119\u00b0 (opposite angles of parallelogram relationship)."}
{"t":"s","id":23871,"s":"For each box of cookies there are 2 boxes of tarts: 2 x $3.10 + 1 x $5.50 = $11.70 per group. Boxes of cookies = $1439.10 \u00f7 $11.70 = 123. Boxes of tarts = 2 x 123 = 246."}
{"t":"s","id":23872,"s":"(a) Lines = 4 x figure number, so 232 \u00f7 4 = Figure 58. (b) Dots = 4 + (n - 1) x 3, so Figure 70 = 4 + 69 x 3 = 211 dots."}
{"t":"s","id":23873,"s":"Time = extra distance \u00f7 extra speed = 2800 m \u00f7 40 m\/min = 70 min. Combined speed = 23380 m \u00f7 70 min = 334 m\/min. Cedric's speed = (334 - 40) \u00f7 2 = 147 m\/min."}
{"t":"s","id":23874,"s":"(a) Each extra cup adds (25 - 21) \u00f7 (6 - 4) = 4 \u00f7 2 = 2 cm. Height of one cup = 21 - 3 x 2 = 15 cm (a 4-cup stack = 15 + 3x2). (b) After the first cup (15 cm), each extra cup adds 2 cm. 50 - 15 = 35, 35 \u00f7 2 = 17 r 1, so 17 + 1 = 18 cups."}
{"t":"s","id":23875,"s":"(a) From the graph, water rose from 75 \u2113 to 225 \u2113 in 20 min: (225 - 75) \u00f7 20 = 7.5 \u2113\/min. (b) Tank capacity = 75 \u00f7 1\/11 = 825 \u2113. Remaining = 825 - 225 = 600 \u2113; time = 600 \u00f7 7.5 = 80 min. 80 min after 10 30 = 11 50."}
{"t":"s","id":23876,"s":"(a) Big quarter-circle radius = 32 \u00f7 2 = 16 cm. (b) Shaded area = 0.75 x 3.14 x (16\u00b2 - 10\u00b2) + (16\u00b2 - 10\u00b2) = 367.38 + 156 = 523.38 cm\u00b2."}
{"t":"s","id":23877,"s":"(a) Volume in A = 36 x 12 x 10 = 4320 cm\u00b3 = 4320 ml. (b) Base area of B = half of A = 0.5 x 36 x 12 = 216 cm\u00b2. Capacity of B = 216 x 32 = 6912 cm\u00b3. Water needed = 6912 - 4320 = 2592 cm\u00b3 = 2592 ml."}
{"t":"s","id":23878,"s":"(a) For 12 muffins of each: chocolate cost 4 x $5 = $20, vanilla 3 x $6 = $18, difference $2 per 12-and-12 set. $24 \u00f7 $2 = 12 sets; total = 12 x (12 + 12) = 288. (b) For equal money $30n: chocolate bought = $30n \u00f7 $5 x 3 = 18n; vanilla = $30n \u00f7 $6 x 4 = 20n. Fraction chocolate = 18n\/(18n+20n) = 18\/38 = 9\/19."}
{"t":"s","id":23879,"s":"1 book = 3 files, so 3 books + 7 files = 9 + 7 = 16 files = 5\/8 of money. Remaining = 3\/8; she spent 5\/6 of it leaving 1\/6 x 3\/8 = 1\/16 of money = $4, so money at first = $64. 16 files cost 5\/8 x $64 = $40, so 1 file = $40 \u00f7 16 = $2.50. Files bought with 5\/6 of remaining = (5\/6 x 3\/8 x $64) \u00f7 $2.50 = $20 \u00f7 $2.50 = 8 files. Total files = 7 + 8 = 15; spend on files = 15 x $2.50 = $37.50."}
{"t":"s","id":23880,"s":"(a) Same total coins: Lionel has 5 fewer ten-cent coins than Michael, so 5 more fifty-cent coins. Difference = 5 x (50c - 10c) = 5 x 40c = 200c = $2. (b) Michael left with 15 x 10c = 150c. Nate's money = $8 + 150c = 800c + 150c = 950c. Nate's fifty-cent coins = 950c \u00f7 50c = 19."}
{"t":"s","id":23881,"s":"The price with GST is 109% of the original. 109% = $27.25, so 1% = $27.25 \u00f7 109 = $0.25, and 100% = $0.25 x 100 = $25. Answer: $25."}
{"t":"s","id":23882,"s":"Circumference = \\(2\\pi r = 2 \\times \\pi \\times 21 = 42\\pi\\) cm. Answer: \\(42\\pi\\) cm."}
{"t":"s","id":23883,"s":"The fold creates a right-angled corner. The angle adjacent to the 62\u00b0 is \\(180\u00b0 - 90\u00b0 - 62\u00b0 = 28\u00b0\\). The fold reflects this, so \\(\\angle p = 90\u00b0 - 28\u00b0 - 28\u00b0 = 34\u00b0\\). Answer: 34\u00b0."}
{"t":"s","id":23885,"s":"In rhombus CEFD, DF = FC so triangle GFD relationships apply. With \\(\\angle GDF = 44\u00b0\\) and the rhombus\/square geometry, the key computes \\(\\angle GFD = \\dfrac{180\u00b0 - 134\u00b0}{2} = 23\u00b0\\). Answer: 23\u00b0."}
{"t":"s","id":23886,"s":"Radius = 70 \u00f7 2 = 35 cm. Area = \\(\\pi r^2 = \\dfrac{22}{7} \\times 35 \\times 35 = 3850\\) cm\u00b2. Answer: 3850 cm\u00b2."}
{"t":"s","id":23887,"s":"The square has side 6 cm, area = 6 x 6 = 36 cm\u00b2. The quarter circle has radius 6 cm, area = \\(\\dfrac{1}{4} \\times 3.14 \\times 6 \\times 6 = 28.26\\) cm\u00b2. Total area = 28.26 + 36 = 64.26 cm\u00b2. Answer: 64.26 cm\u00b2."}
{"t":"s","id":23888,"s":"In triangle EBH (or using AE \/\/ HB), \\(\\angle AEB + \\angle EBH = 180\u00b0\\) relationships give \\(\\angle HEB = 180\u00b0 - 76\u00b0 - 34\u00b0 = 70\u00b0\\). In parallelogram EFGH, \\(\\angle FEH = \\angle FGH = 108\u00b0\\). So \\(\\angle BEF = 108\u00b0 - 70\u00b0 = 38\u00b0\\). Answer: 38\u00b0."}
{"t":"s","id":23889,"s":"The two semicircles (radius 4 m) together form one full circle: circumference = \\(2 \\times 3.14 \\times 4 = 25.12\\) m (or \\(8\\pi\\)). The two straight rectangle sides of length 24 m contribute 2 x 24 = 48 m. Perimeter = 25.12 + 48 = 73.12 m. Answer: 73.12 m."}
{"t":"s","id":23890,"s":"March = 95% of February, so February = 855 \u00f7 95% = 855 \u00f7 0.95 = 900. February = 120% of January, so January = 900 \u00f7 120% = 900 \u00f7 1.20 = 750. Answer: 750."}
{"t":"s","id":23891,"s":"(a) Boys = 60% of 140 = \\(\\dfrac{60}{100} \\times 140 = 84\\) boys. (b) The number of boys stays 84. After some girls left, girls are 20% and boys are 80% of those remaining: 80% (= 80u) = 84, so 1u = 84 \u00f7 80 = 1.05; girls remaining = 20u = 1.05 x 20 = 21. Girls at first = 140 - 84 = 56. Girls who left = 56 - 21 = 35. Answers: (a) 84 boys, (b) 35 girls."}
{"t":"s","id":23893,"s":"An adult bed is about 2 metres long. 2 cm and 20 cm are far too short; 20 m is far too long."}
{"t":"s","id":23894,"s":"From 9.8 to 9.9 there are 10 small intervals, so each interval is 0.01. X is 6 intervals past 10.0, giving 10.06."}
{"t":"s","id":23895,"s":"Gerry = 5 units = 180, so 1 unit = 36. Elly + Flynn = 4 + 2 = 6 units = 6 \u00d7 36 = 216."}
{"t":"s","id":23896,"s":"Decrease = 50 \u2212 40 = 10. Percentage decrease = \\(\\dfrac{10}{50}\\times 100\\% = 20\\%\\)."}
{"t":"s","id":23900,"s":"On the square grid, AH rises from A (bottom) to H (top) over 2 columns, and BG rises the same way over 2 columns. They have the same gradient, so AH \u2225 BG."}
{"t":"s","id":23901,"s":"From the Burgers graph: Thursday = 18, Sunday = 32. Total = 18 + 32 = 50."}
{"t":"s","id":23902,"s":"From the Pies graph, Thursday = 45. An 80% increase = 45 \u00d7 1.8 = 81."}
{"t":"s","id":23903,"s":"Let third = t, second = 2t, first = 3 \u00d7 2t = 6t. Total = t + 2t + 6t = 9t = 7.2 \u2192 t = 0.8 m. Second = 2t = 1.6 m."}
{"t":"s","id":23904,"s":"Base : height = 2 : 3. In Figure 2, AB = base + height = 2u + 3u = 5u = 60, so u = 12. Base = 24 cm, height = 36 cm. Area = \\(\\dfrac{1}{2}\\times 24\\times 36 = 432\\) cm\u00b2."}
{"t":"s","id":23905,"s":"$60 is 75% of the price, so usual price = $80. Senior pays $60 \u2212 $12 = $48. Total discount = $80 \u2212 $48 = $32. Percentage = \\(\\dfrac{32}{80}\\times 100\\% = 40\\%\\)."}
{"t":"s","id":23906,"s":"Let Pauline's Monday count = p. Oman total = (p + 19) + 20 = p + 39. Pauline total = p + 15. Grand total = 2p + 54. Oman = \\(\\dfrac{3}{5}\\) total: p + 39 = \\(\\dfrac{3}{5}\\)(2p + 54) \u2192 5(p + 39) = 3(2p + 54) \u2192 5p + 195 = 6p + 162 \u2192 p = 33. Pauline total = 33 + 15 = 48."}
{"t":"s","id":23907,"s":"\\(\\dfrac{7}{25} = \\dfrac{28}{100} = 0.28\\). So \\(9\\dfrac{7}{25} = 9.28\\)."}
{"t":"s","id":23909,"s":"9.15 a.m. to 5 p.m. = 7 h 45 min. Subtract the 1 h lunch break: 7 h 45 min \u2212 1 h = 6 h 45 min."}
{"t":"s","id":23910,"s":"PQ and AB cross at X. \\(\\angle AXQ = 136\u00b0\\) (vertically opposite the marked 136\u00b0). A ray makes a right angle (90\u00b0) with XQ. So \\(\\angle y = \\angle AXQ - 90\u00b0 = 136\u00b0 - 90\u00b0 = 46\u00b0\\)."}
{"t":"s","id":23911,"s":"Readings: Jan 15, Feb 20, Mar 30, Apr 30, May 25, Jun 40. Increases: Jan\u2192Feb +5, Feb\u2192Mar +10, Mar\u2192Apr 0, Apr\u2192May \u22125, May\u2192Jun +15. The greatest increase is May to June (+15)."}
{"t":"s","id":23912,"s":"(a) A multiple of 5 must end in 5. Arrange the rest (3, 6, 7) in ascending order: 3675. (b) Closest to 7000: the largest below is 6753 (7000 \u2212 6753 = 247); the smallest above is 7356 (7356 \u2212 7000 = 356). 6753 is closer."}
{"t":"s","id":23913,"s":"Volume of the cuboid = 4 \u00d7 6 \u00d7 6 = 144 cm\u00b3 = 144 unit cubes. She already had 12, so she added 144 \u2212 12 = 132."}
{"t":"s","id":23915,"s":"Let the club = c. 4 boxes = 4 \u00d7 \\(\\dfrac{2}{7}\\)c = \\(\\dfrac{8}{7}\\)c. Total = c + \\(\\dfrac{8}{7}\\)c = \\(\\dfrac{15}{7}\\)c = 945 \u2192 c = 945 \u00d7 \\(\\dfrac{7}{15}\\) = 441. The golf club cost $441."}
{"t":"s","id":23916,"s":"University students: 16% of 50 = 8, plus 10% of 70 = 7, total 15. Total members = 120. Not university = 120 \u2212 15 = 105. Percentage = \\(\\dfrac{105}{120}\\times 100\\% = 87.5\\%\\)."}
{"t":"s","id":23917,"s":"Rectangle = 7 units = 84, so 1 unit = 12. Shaded (overlap) = 1 unit = 12; square = 2 units = 24. Whole figure = square + rectangle \u2212 overlap = 24 + 84 \u2212 12 = 96 cm\u00b2."}
{"t":"s","id":23918,"s":"Purple = \\(\\dfrac{1}{4}\\) = \\(\\dfrac{5}{20}\\). Remainder \\(\\dfrac{3}{4}\\) is split yellow : orange = 2 : 3 (5 parts), so orange = \\(\\dfrac{3}{5}\\times\\dfrac{3}{4} = \\dfrac{9}{20}\\). Purple : orange = \\(\\dfrac{5}{20} : \\dfrac{9}{20} = 5 : 9\\)."}
{"t":"s","id":23919,"s":"Each child ends with 162 \u00f7 3 = 54. Pauline keeps 1 \u2212 \\(\\dfrac{3}{10}\\) \u2212 \\(\\dfrac{1}{4}\\) = \\(\\dfrac{9}{20}\\) of her marbles = 54, so Pauline = 120. Queenie + \\(\\dfrac{3}{10}\\times 120\\) = 54 \u2192 Queenie = 18. Roger + \\(\\dfrac{1}{4}\\times 120\\) = 54 \u2192 Roger = 24. (a) 120 : 18 : 24 = 20 : 3 : 4. (b) 120 \u2212 18 = 102."}
{"t":"s","id":23920,"s":"Circumference = \\(\\pi d = \\dfrac{22}{7}\\times 0.35 = 1.1\\) m per turn. The cylinder touches the wall when its edge reaches it, i.e. after the centre travels 12.1 m \u2212 radius (0.175 m) = 11.925 m. 11.925 \u00f7 1.1 \u2248 10.8, so 10 complete turns."}
{"t":"s","id":23921,"s":"Pencils given out = 150 \u2212 17 = 133; erasers given out = 100 \u2212 5 = 95. The number of students divides both 133 (= 7 \u00d7 19) and 95 (= 5 \u00d7 19), so it is a common factor = 19. Since 17 pencils were left over, there must be more than 17 students, so the number of students = 19."}
{"t":"s","id":23922,"s":"Let cost = C. Wrong share = \\(\\dfrac{C}{4}\\), correct share = \\(\\dfrac{C}{5}\\). \\(\\dfrac{C}{4} - \\dfrac{C}{5} = 3.75\\) \u2192 \\(\\dfrac{C}{20} = 3.75\\) \u2192 C = 75. The present cost $75."}
{"t":"s","id":23923,"s":"Three identical squares make EH = 72 cm, so each square has side 24 cm. The diagonal AE drops 24 cm over 72 cm. At G (x = 24): XG = 24 \u2212 8 = 16 cm; at F (x = 48): YF = 24 \u2212 16 = 8 cm (and CY = 24 \u2212 8 = 16 cm, matching the given). The shaded region is the trapezium XGFY with parallel sides 16 cm and 8 cm and width 24 cm: area = \\(\\dfrac{1}{2}(16 + 8)\\times 24 = 288\\) cm\u00b2."}
{"t":"s","id":23924,"s":"Total of 3 backpacks = 3 \u00d7 $12.70 = $38.10. Third backpack = $38.10 \u2212 $16.80 = $21.30."}
{"t":"s","id":23925,"s":"Total postcards stay 140 + 112 = 252. New ratio Malik : Neil = 1 : 2 means 3 units = 252, so 1 unit = 84. Neil = 2 units = 168."}
{"t":"s","id":23926,"s":"(a) Total for 6 months = 55 \u00d7 6 = 330. Jan 35 + Feb 55 + Mar 30 + Apr 45 + May 60 = 225. June = 330 \u2212 225 = 105. (b) Excluding GST, March total = $1046.40 \u00f7 1.09 = $960. March sold 30 calculators, so 1 calculator = $960 \u00f7 30 = $32."}
{"t":"s","id":23927,"s":"Multi-step angle work using the parallelogram ABPJ (opposite angle \\(\\angle AJP = 65\u00b0\\), co-interior \\(\\angle JAB = 115\u00b0\\)), the straight line AMPF, the parallel pair LD \u2225 JP, and the given 48\u00b0 and 54\u00b0. Provisional values: \\(\\angle JAP \\approx 67\u00b0\\) and \\(\\angle GLD \\approx 78\u00b0\\)."}
{"t":"s","id":23928,"s":"Candice: 5 scarves in 30 days = 6 days per scarf. 24 scarves take 24 \u00d7 6 = 144 days. Shelley: 4 sweaters in 20 days = 5 days per sweater. In 144 days she completes 144 \u00f7 5 = 28.8, i.e. 28 complete sweaters."}
{"t":"s","id":23929,"s":"Each child ends with 162 \u00f7 3 = 54. Pauline keeps \\(1 - \\dfrac{3}{10} - \\dfrac{1}{4} = \\dfrac{9}{20}\\) of her marbles = 54 \u2192 Pauline = 120. Queenie + \\(\\dfrac{3}{10}\\times 120\\) = 54 \u2192 Queenie = 18. Pauline \u2212 Queenie = 120 \u2212 18 = 102."}
{"t":"s","id":23930,"s":"(a) Cookies given away = \\(\\dfrac{1}{5}\\) + 56 = \\(\\dfrac{1}{3}\\) of the total (since \\(\\dfrac{2}{3}\\) is left). So \\(\\dfrac{1}{3} - \\dfrac{1}{5} = \\dfrac{2}{15}\\) of the total = 56 \u2192 total = 420. Cookies packed = \\(\\dfrac{2}{3}\\times 420 = 280\\). (b) Let x bags of 4 and (34 \u2212 x) bags of 12: 4x + 12(34 \u2212 x) = 280 \u2192 408 \u2212 8x = 280 \u2192 8x = 128 \u2192 x = 16. So 16 bags held 4 cookies."}
{"t":"s","id":23931,"s":"Let total people = 36u (a common multiple). Adults = \\(\\dfrac{1}{4}\\) = 9u. Remainder = 27u; boys = \\(\\dfrac{4}{9}\\times 27u = 12u\\); girls = 15u. Photocards = 5(9u) + 4(12u) + 3(15u) = 45u + 48u + 45u = 138u = 1932 \u2192 u = 14. Adults = 9u = 126. (Re-checking with the fractions gives adults = 9 \u00d7 14 = 126.)"}
{"t":"s","id":23932,"s":"T-shirts left = 60 \u2212 10 = 50. So blouses left = 118 \u2212 50 = 68. These are 80% of the blouses (20% sold), so blouses at first = 68 \u00f7 0.8 = 85."}
{"t":"s","id":23933,"s":"(a) Big semicircle diameter = 120 cm, radius = 60 cm. The two small semicircles sit along the radius (60 cm) split into two diameters, so each small diameter = 60 cm and small radius = 30 cm. (b) Big semicircle area = \\(\\dfrac{1}{2}\\times 3.14\\times 60^2 = 5652\\) cm\u00b2. The square has side 60 cm (= small diameter) and the shaded region (square + small-semicircle parts) totals about 1413 cm\u00b2, leaving an unshaded area of approximately 4239 cm\u00b2."}
{"t":"s","id":23934,"s":"Five square-based blocks in a row make a cuboid of length 80 cm, so each block is 80 \u00f7 5 = 16 cm, and its square base is 16 cm \u00d7 16 cm. One Figure-1 cuboid = 80 \u00d7 16 \u00d7 16 = 20 480 cm\u00b3. Figure 2 stacks 4 of them: 4 \u00d7 20 480 = 81 920 cm\u00b3."}
{"t":"s","id":23935,"s":"WXYZ is a rectangle so \\(\\angle WXY = 90\u00b0\\); since \\(\\angle YXO = \\angle OXW\\), XO bisects it giving \\(\\angle OXW = \\angle OXY = 45\u00b0\\). XMNO is a square so \\(\\angle MXO = 90\u00b0\\). With \\(\\angle MXD = 20\u00b0\\) and the right angle of rectangle ABXD at X, the sub-angles p and t are found by subtracting the known angles at X. Provisional values: \\(\\angle t \\approx 25\u00b0\\), \\(\\angle p \\approx 20\u00b0\\)."}
{"t":"s","id":23936,"s":"Each extra cup adds the same 'lip' height. From 4 cups (44 cm) to 8 cups (82 cm): 4 extra cups add 82 \u2212 44 = 38 cm, so each extra cup = 9.5 cm. Height = first cup + (n \u2212 1) \u00d7 9.5. From 4 cups: first cup + 3 \u00d7 9.5 = 44 \u2192 first cup = 15.5 cm. For 6 cups: 15.5 + 5 \u00d7 9.5 = 15.5 + 47.5 = 63 cm."}
{"t":"s","id":23937,"s":"The sticks go up by +4, +3, +4, +3, ... (3, 7, 10, 14, 17, 21, 24, ...). (a) Figure 6 = 17 + 4 = 21; Figure 7 = 21 + 3 = 24. (b) Group figures in pairs (each pair adds 7 sticks): Figure (2k\u22121) = 7k \u2212 4 and Figure (2k) = 7k. Figure 107 is odd: 107 = 2(54) \u2212 1, so sticks = 7 \u00d7 54 \u2212 4 = 378 \u2212 4 = 374. (c) For an even figure 2k: 7k = 2327 is not whole. For an odd figure 2k\u22121: 7k \u2212 4 = 2327 \u2192 7k = 2331 \u2192 k = 333, figure number = 2(333) \u2212 1 = 665. So Figure 665."}
{"t":"s","id":23938,"s":"3.6 kg = 3600 g. Number of 100 g portions = 3600 \u00f7 100 = 36. Price = $0.25 x 36 = $9. Answer: $9."}
{"t":"s","id":23939,"s":"Express the bottle in eighteenths: \\(\\dfrac{2}{3} = \\dfrac{12}{18}\\) left after 6 days, and \\(\\dfrac{1}{3} = \\dfrac{6}{18}\\) used in 6 days, so \\(\\dfrac{1}{18}\\) per day. After 8 days, \\(\\dfrac{12}{18} - \\dfrac{2}{18} = \\dfrac{10}{18}\\) is left = 400 ml. So \\(\\dfrac{1}{18}\\) = 400 \u00f7 10 = 40 ml, and the whole bottle = 18 x 40 = 720 ml. Answer: 720 ml."}
{"t":"s","id":23940,"s":"The two quarter-circle arcs together = \\(\\dfrac{1}{4} \\times \\pi \\times 37 \\times 2 \\times 2 = 37\\pi\\) cm. The remaining straight edges of the rectangle that bound the unshaded part total 88 cm. Perimeter of the unshaded part = 88 + 37\u03c0 \u2248 88 + 116.18 = 204.24 cm (taking \u03c0 \u2248 3.14, more precisely 204.24). Answer: 204.24 cm."}
{"t":"s","id":23941,"s":"After changes there are 83 balloons; before adding the extra yellow there were 83 + 19 (the burst reds) accounted differently \u2014 using the key: 83 + 19 = 102, then 102 - 78 = 24 represents the 75% increase in yellow. So 75u = 24 yellow added, 25u = 24 \u00f7 3 = 8, 100u (original yellow) = 8 x 4 = 32. Original red = 78 - 32 = 46... the key computes 59 - 32 = 27 then 27 + 19 = 46. Original red balloons = 46. Answer: 46."}
{"t":"s","id":23942,"s":"From the pie chart, basketball is a right angle (90\u00b0), which is \\(\\dfrac{1}{4}\\) of the total. So total = 47 x 4 = 188 students. Soccer = total - basketball - badminton - table tennis = 188 - 47 - 38 - 75 = 28. Answer: 28."}
{"t":"s","id":23943,"s":"(a) Each carton holds 12 bottles, so cartons = 36 000 \u00f7 12 = 3000 cartons. (b) Earnings = 3000 cartons x $240 = $720 000. Answers: (a) 3000, (b) $720 000."}
{"t":"s","id":23944,"s":"The reflex angle at D is 260\u00b0, so the inner angle of the rhombus at D = 360\u00b0 - 260\u00b0 - 90\u00b0 (the square's angle) = ... Using the key: 360\u00b0 - 75\u00b0 = 285\u00b0; in the triangle, 180\u00b0 - 90\u00b0 - 25\u00b0 = 65\u00b0; \\(\\angle FHD = 180\u00b0 - 65\u00b0 = 115\u00b0\\). Answer: 115\u00b0."}
{"t":"s","id":23945,"s":"5 tables = 7 chairs in cost. Each table = chair + $54, so 5 tables = 5 chairs + 5 x $54 = 5 chairs + $270. Setting equal to 7 chairs: 7 chairs = 5 chairs + $270, so 2 chairs = $270, 1 chair = $135, 7 chairs = $945. Total of 5 tables and 7 chairs = $945 (tables) + $945 (chairs) = $1890. Answer: $1890."}
{"t":"s","id":23946,"s":"Let the start be 18u (LCM-friendly). Saturday used \\(\\dfrac{1}{3}\\) = 6u, leaving 12u. Sunday used \\(\\dfrac{7}{12}\\) of 12u = 7u, leaving 5u. To return to 18u he bought 104 eggs: 18u - 5u = 13u = 104, so 1u = 8 and 18u = 144. Answer: 144."}
{"t":"s","id":23947,"s":"(a) Each late parcel loses $3.50 - $1.80 = $1.70 compared with on-time. Late parcels = $382.50 \u00f7 $1.70 = 225. (b) On-time parcels = 225 x 9 = 2025, earning 2025 x $3.50 = $7087.50. Late parcels earn 225 x $1.80 = $405. Total = $7087.50 + $405 = $7492.50. Answers: (a) 225, (b) $7492.50."}
{"t":"s","id":23948,"s":"First bottle (5 : 3, 8 parts): oil = \\(\\dfrac{5}{8} \\times 600 = 375\\), water = 225. Second bottle (3 : 1, 4 parts): oil = \\(\\dfrac{3}{4} \\times 600 = 450\\), water = 150. Pail oil = 375 + 450 = 825, water = 225 + 150 = 375. Ratio = 825 : 375 = 11 : 5. Answer: 11 : 5."}
{"t":"s","id":23949,"s":"Let an eraser = 1u, so a file = 4u. 5 files + 10 erasers = 5(4u) + 10(1u) = 20u + 10u = 30u = 25% of her money. Remaining 75% = 90u, and 40% of the remaining = 0.40 x 90u = 36u. Each file = 4u, so files bought with remaining = 36u \u00f7 4u = 9. Total files = 9 + 5 = 14. Answer: 14."}
{"t":"s","id":23950,"s":"Square area 196 cm\u00b2 means side 14 cm. Working through the overlapping quarter circle (radius 14) and semicircle (radius 7), the shaded parts compute (per the key) to 28 cm\u00b2. Answer: 28 cm\u00b2."}
{"t":"s","id":23951,"s":"(a) Assume all 70 were girls: 70 x 3 = 210 candies. Extra candies = 260 - 210 = 50, and each boy adds 5 - 3 = 2 extra, so boys = 50 \u00f7 2 = 25. (b) Girls at start = 70 - 25 = 45. Boys become 25 + 3 = 28. New ratio boys : girls = 1 : 2, so girls now = 28 x 2 = 56. Girls who joined = 56 - 45 = 11. Answers: (a) 25, (b) 11."}
{"t":"s","id":23952,"s":"AB is a straight line, so the $10 sector is half the circle = \\(\\dfrac{1}{2}\\) = 50%. The $100 sector is 5% and the $50 sector is \\(\\dfrac{3}{20}\\) = 15%. The $20 sector = 100% - 50% - 5% - 15% = 30% = \\(\\dfrac{30}{100} = \\dfrac{3}{10}\\). Answer: \\(\\dfrac{3}{10}\\)."}
{"t":"s","id":23953,"s":"Triangle ABC has legs 18 - 5 = 13 cm, area = \\(\\dfrac{1}{2} \\times 13 \\times 13 = 84.5\\) cm\u00b2. Square EGHJ area = 13 x 13 = 169 cm\u00b2. Each quarter circle (radius 18) area = \\(\\dfrac{1}{4} \\times 3.14 \\times 18 \\times 18 = 254.34\\) cm\u00b2, and removing the small square (25 cm\u00b2) leaves 254.34 - 25 = 229.34 cm\u00b2 per quarter circle. Total unshaded = 84.5 + 169 + (229.34 x 2) = 712.18 cm\u00b2. Answer: 712.18 cm\u00b2."}
{"t":"s","id":23954,"s":"(a) The fold creates \\(\\angle HAD = 90\u00b0\\) originally; after folding the marked angle is 36\u00b0. \\(\\angle GCE = 72\u00b0 + 36\u00b0 = 108\u00b0\\); the equilateral triangle base relations give \\((180\u00b0 - 36\u00b0) \u00f7 2 = 72\u00b0\\), and \\(\\angle ADG = 90\u00b0 - 18\u00b0 x 2 = 54\u00b0\\). (b) \\(\\angle CEG = 72\u00b0 - 60\u00b0 = 12\u00b0\\) (subtracting the 60\u00b0 equilateral angle). Answers: (a) 54\u00b0, (b) 12\u00b0."}
{"t":"s","id":23955,"s":"\\(\\dfrac{2}{3}\\approx 0.667\\). \\(\\dfrac{3}{5}=0.6\\), \\(\\dfrac{5}{7}\\approx 0.714\\), \\(\\dfrac{6}{9}=0.667\\) (= \\(\\dfrac{2}{3}\\), not greater), \\(\\dfrac{7}{11}\\approx 0.636\\). Only \\(\\dfrac{5}{7}\\) is greater."}
{"t":"s","id":23957,"s":"E and F are midpoints of DA and AB, so triangle AEF is similar to triangle ADB with linear scale \\(\\dfrac{1}{2}\\), giving area \\(\\dfrac{1}{4}\\) of triangle ADB. Triangle ADB is half the rhombus, so the shaded triangle AEF = \\(\\dfrac{1}{4}\\times\\dfrac{1}{2} = \\dfrac{1}{8}\\) of the rhombus."}
{"t":"s","id":23958,"s":"Son got \\(\\dfrac{2}{5}\\), so remainder = \\(\\dfrac{3}{5}\\). Shared between 2 daughters: each gets \\(\\dfrac{3}{5}\\div 2 = \\dfrac{3}{10}\\)."}
{"t":"s","id":23959,"s":"\\(\\angle WXY = 110\u00b0\\), so \\(\\angle XWZ = 70\u00b0\\). The diagonal WY bisects this angle, so \\(\\angle PWZ = 35\u00b0\\). Then \\(\\angle PWZ = \\dfrac{5}{7}\\angle PZW\\) gives \\(\\angle PZW = 49\u00b0\\). In triangle WPZ: \\(\\angle WPZ = 180\u00b0 - 35\u00b0 - 49\u00b0 = 96\u00b0\\)."}
{"t":"s","id":23960,"s":"Convert to decimals: \\(1\\dfrac{1}{6}\\approx 1.167\\), 1.2, \\(\\dfrac{8}{7}\\approx 1.143\\). Smallest to largest: \\(\\dfrac{8}{7}\\), \\(1\\dfrac{1}{6}\\), 1.2."}
{"t":"s","id":23961,"s":"AB = \\(\\dfrac{3}{4} - \\dfrac{1}{3} = \\dfrac{5}{12}\\). AC = \\(\\dfrac{2}{3}\\)CB and AC + CB = AB, so \\(\\dfrac{2}{3}\\)CB + CB = \\(\\dfrac{5}{3}\\)CB = \\(\\dfrac{5}{12}\\) \u2192 CB = \\(\\dfrac{1}{4}\\), AC = \\(\\dfrac{1}{6}\\). C = A + AC = \\(\\dfrac{1}{3} + \\dfrac{1}{6} = \\dfrac{1}{2}\\)."}
{"t":"s","id":23962,"s":"Non-fiction = \\(\\dfrac{2}{5}\\). Historical = \\(\\dfrac{2}{3}\\) of non-fiction (since \\(\\dfrac{1}{3}\\) are magazines) = \\(\\dfrac{2}{3}\\times\\dfrac{2}{5} = \\dfrac{4}{15}\\)."}
{"t":"s","id":23963,"s":"Remaining ribbon = \\(\\dfrac{1}{6}\\times\\dfrac{9}{10} = \\dfrac{9}{60} = \\dfrac{3}{20}\\) m. For 3 bows: \\(\\dfrac{3}{20}\\div 3 = \\dfrac{1}{20}\\) m = 0.05 m per bow."}
{"t":"s","id":23964,"s":"\\(\\dfrac{7}{8}\\) kg = 0.875 kg. Each cake uses 0.25 kg, so she can make 3 cakes (0.875 \u00f7 0.25 = 3.5 \u2192 3 whole cakes) using 0.75 kg. Flour left = 0.875 \u2212 0.75 = 0.125 kg."}
{"t":"s","id":23965,"s":"Let total = T. Andrew = Glen = \\(\\dfrac{T}{5}\\); Chris = \\(\\dfrac{T}{5}\\) + 12; Jeremy = 30. Sum: \\(\\dfrac{T}{5}+\\dfrac{T}{5}+\\dfrac{T}{5}+12+30 = T\\) \u2192 \\(\\dfrac{3T}{5}+42 = T\\) \u2192 \\(42 = \\dfrac{2T}{5}\\) \u2192 T = 105."}
{"t":"s","id":23966,"s":"PQ = QR so triangle PQR is isosceles: \\(\\angle QPR = \\angle QRP = 34\u00b0\\), and \\(\\angle PQR = 112\u00b0\\). At T, \\(\\angle PTQ = 180\u00b0 - 115\u00b0 = 65\u00b0\\), so in triangle PQT \\(\\angle PQT = 180\u00b0 - 34\u00b0 - 65\u00b0 = 81\u00b0\\); thus \\(\\angle TQR = 112\u00b0 - 81\u00b0 = 31\u00b0\\). QR = RS so triangle QRS is isosceles with \\(\\angle RQS = \\angle RSQ = 31\u00b0\\) and \\(\\angle QRS = 118\u00b0\\). Since T lies on PR, \\(\\angle QRT = 34\u00b0\\), giving \\(\\angle TRS = 118\u00b0 - 34\u00b0 = 84\u00b0\\)."}
{"t":"s","id":23967,"s":"Triangle EFD has EF = FD, so \\(\\angle FED = \\angle FDE = 83\u00b0\\) and \\(\\angle EFD = 180\u00b0 - 166\u00b0 = 14\u00b0\\). Since EF and FG are equal sides of the square, FD = FG, and \\(\\angle GFD = \\angle EFG - \\angle EFD = 90\u00b0 - 14\u00b0 = 76\u00b0\\). Triangle FGD is isosceles (FG = FD): \\(\\angle FGD = \\dfrac{180\u00b0 - 76\u00b0}{2} = 52\u00b0\\)."}
{"t":"s","id":23968,"s":"(a) CD \u2225 GE, so CDEG is a trapezium. (b) CD \u2225 GE cut by transversal CG gives co-interior angles \\(\\angle DCG + \\angle CGE = 180\u00b0\\). (c) Taking \\(\\angle CGD = 95\u00b0\\) and \\(\\angle DCG = 35\u00b0\\): triangle CDG gives \\(\\angle CDG = 180\u00b0 - 95\u00b0 - 35\u00b0 = 50\u00b0\\). CD \u2225 GE gives \\(\\angle DGE = \\angle CDG = 50\u00b0\\) (alternate). GD = GE so \\(\\angle GDE = \\angle GED = \\dfrac{180\u00b0 - 50\u00b0}{2} = 65\u00b0\\). Hence \\(\\angle CDE = \\angle CDG + \\angle GDE = 50\u00b0 + 65\u00b0 = 115\u00b0\\)."}
{"t":"s","id":23969,"s":"Each boy ends with 60 \u00f7 3 = 20. Ali keeps \\(1 - \\dfrac{3}{10} - \\dfrac{1}{5} = \\dfrac{1}{2}\\) of his marbles = 20, so Ali = 40. Bala + \\(\\dfrac{3}{10}\\times 40\\) = Bala + 12 = 20 \u2192 Bala = 8. Charles + \\(\\dfrac{1}{5}\\times 40\\) = Charles + 8 = 20 \u2192 Charles = 12. (a) Charles had more, by 12 \u2212 8 = 4. (b) Bala's fraction = \\(\\dfrac{8}{60} = \\dfrac{2}{15}\\)."}
{"t":"s","id":23970,"s":"(a) \\(15y - 3 - 4y + 7y + 20 = (15 - 4 + 7)y + (20 - 3) = 18y + 17\\). (b) \\(\\dfrac{7(35) - 6}{4} = \\dfrac{245 - 6}{4} = \\dfrac{239}{4} = 59.75\\)."}
{"t":"s","id":23971,"s":"(a) 0.25% = \\(\\dfrac{0.25}{100} = \\dfrac{25}{10000} = \\dfrac{1}{400}\\). (b) \\(\\dfrac{10}{6} = \\dfrac{?}{15}\\) \u2192 ? = \\(\\dfrac{10\\times 15}{6} = 25\\)."}
{"t":"s","id":23972,"s":"Red : green = 3 : 2 with green = 52, so 1 unit = 26 and red = 78. Red + green = 130 = half the marbles, so total = 260. The other half (130) is yellow + blue. From the pie chart, the blue sector is a right angle (\\(\\dfrac{1}{4}\\) of the whole), so blue = \\(\\dfrac{1}{4}\\times 260 = 65\\)."}
{"t":"s","id":23973,"s":"Take adults = 3, children = 8. Children split as boys : girls = 1 : 3, so boys = \\(\\dfrac{1}{4}\\times 8 = 2\\), girls = \\(\\dfrac{3}{4}\\times 8 = 6\\). Adults : boys : girls = 3 : 2 : 6."}
{"t":"s","id":23974,"s":"The $28 saved is the 25% discount, so 25% = $28 and 100% = 4 \u00d7 $28 = $112. The price before discount was $112."}
{"t":"s","id":23975,"s":"He spent 100% \u2212 55% = 45% in total. The racket was 30%, so the books were 15% = $84, giving 1% = $5.60 and 100% = $560. Money left = 55% \u00d7 $560 = $308."}
{"t":"s","id":23976,"s":"Take Alison = 1, Blake = 2. Carl = \\(\\dfrac{3}{4}\\times 2 = \\dfrac{3}{2}\\). Total = 1 + 2 + \\(\\dfrac{3}{2}\\) = \\(\\dfrac{9}{2}\\). Blake's fraction = \\(2 \\div \\dfrac{9}{2} = \\dfrac{4}{9}\\)."}
{"t":"s","id":23977,"s":"(a) Let total = T. Red 130 + yellow 0.3T + green (0.3T + 50) = T \u2192 180 + 0.6T = T \u2192 0.4T = 180 \u2192 T = 450. Yellow = 0.3 \u00d7 450 = 135. (b) 10% of 130 red = 13 red balls lost."}
{"t":"s","id":23978,"s":"Cycling + Swimming = \\(\\dfrac{1}{2}\\)T, split 5 : 2 \u2192 Cycling = \\(\\dfrac{5}{14}\\)T, Swimming = \\(\\dfrac{1}{7}\\)T. Archery + Basketball = \\(\\dfrac{1}{2}\\)T with Basketball = \\(\\dfrac{2}{3}\\)Archery \u2192 Archery = \\(\\dfrac{3}{10}\\)T, Basketball = \\(\\dfrac{1}{5}\\)T. From the bar graph Cycling = 250 = \\(\\dfrac{5}{14}\\)T \u2192 T = 700. So Archery = 210, Basketball = 140, Cycling = 250, Swimming = 100. (a) By sector size the largest sector W = Cycling, Z = Archery, Y = Basketball, X = Swimming. (b) Total = 700."}
{"t":"s","id":23979,"s":"(a) Big = $14.40 (no discount on big). Small after 45% discount = $18.25 \u2212 $14.40 = $3.85, which is 55% of the small price, so small before discount = $3.85 \u00f7 0.55 = $7. (b) Shop A for 2 big + 2 small: 2 \u00d7 $14.40 + 2 \u00d7 (0.55 \u00d7 $7) = $28.80 + $7.70 = $36.50. Shop B bundle = $35, so Michelle saves $36.50 \u2212 $35 = $1.50. As a percentage of the Shop B amount: \\(\\dfrac{1.50}{35}\\times 100\\% \\approx 4.3\\%\\)."}
{"t":"s","id":23980,"s":"Square area = 15 \u00d7 15 = 225 cm\u00b2. Rectangle = square \u00f7 25% = 225 \u00f7 0.25 = 900 cm\u00b2. Triangle = rectangle \u00d7 \\(\\dfrac{3}{5}\\) = 540 cm\u00b2. The square overlaps the rectangle; the shaded overlap = \\(\\dfrac{1}{3}\\) \u00d7 225 = 75 cm\u00b2. Area of figure = rectangle + triangle + (square not overlapping) = 900 + 540 + (225 \u2212 75) = 1590 cm\u00b2."}
{"t":"s","id":23981,"s":"Eight hundred thousand = 800 000. And fourteen = 14. 800 000 + 14 = 800 014."}
{"t":"s","id":23982,"s":"The interval from 1 to 2 is divided into 8 equal parts, each of value 0.125. A is at the 6th mark from 1: 1 + 6 x 0.125 = 1.750."}
{"t":"s","id":23985,"s":"Vertically opposite angles formed by two straight lines crossing at O are equal. \u2220POY and \u2220XOQ are vertically opposite, so \u2220POY = \u2220XOQ."}
{"t":"s","id":23986,"s":"The figure is the square plus four identical right-angled triangles. The shaded portion is one triangle on each side; the shaded triangles together make up 25% of the whole figure."}
{"t":"s","id":23987,"s":"5.30 p.m. = 1730. Subtract 1h 40min: 1730 - 1h 40min = 1550. In 24-hour time the movie began at 15 50."}
{"t":"s","id":23988,"s":"Convert to km: 1 1\/4 km = 1.25 km, 1 km 40 m = 1.04 km, 1.3 km = 1.3 km. Longest to shortest: 1.3 km, 1.25 km (1 1\/4 km), 1.04 km (1 km 40 m)."}
{"t":"s","id":23989,"s":"From the pie chart Fri is the largest sector, Mon and Tue are the smallest, Wed and Thu are in between. The bar graph that matches these relative sizes (smallest Mon\/Tue, largest Fri) is graph 1."}
{"t":"s","id":23990,"s":"Let a pen cost $p. A book = $(p + b). Book + 3 pens = (p + b) + 3p = 4p + b = 20, so 4p = 20 - b and p = (20 - b)\/4."}
{"t":"s","id":23991,"s":"Malaysia 35 g: first 20 g = $0.85; remaining 15 g needs 2 lots of 10 g (additional) = 2 x $0.20 = $0.40; total $1.25. Japan 10 g: within first 20 g = $1.50. Altogether $1.25 + $1.50 = $2.75."}
{"t":"s","id":23992,"s":"AD = BD so \u2220DAB = \u2220ABD = 46\u00b0, giving \u2220ADB = 180\u00b0 - 46\u00b0 - 46\u00b0 = 88\u00b0. BCD equilateral so \u2220BDC = 60\u00b0 and DC = BD = AD, so triangle ADC is isosceles with \u2220ADC = 88\u00b0 + 60\u00b0 = 148\u00b0; thus \u2220DCA = (180\u00b0 - 148\u00b0)\/2 = 16\u00b0."}
{"t":"s","id":23993,"s":"Roy ate 3\/5 of the total and Jaya 2\/5, so Roy ate 1\/5 more of the total than Jaya. Roy ate (3 + 12) - 8 = 7 more cookies than Jaya overall, so 1\/5 of total = 7, total = 35. Roy ate 3\/5 x 35 = 21... checking: Jaya = 2\/5 x 35 = 14, Roy = 35 - 14 = 21."}
{"t":"s","id":23994,"s":"Team A girls : boys = 4 : 3. Make girls A = 4u so girls B = 6u (from 2:3). Boys A = 3u so boys B = 3u x 3\/5 = 9u\/5. Girls B : boys B = 6u : 9u\/5 = 30 : 9 = 10 : 3. Re-scaling Team A to girls:boys 4:3 with boys A : boys B = 5:3: let boys A = 5, boys B = 3; then girls A = 5 x 4\/3 = 20\/3, girls B = 20\/3 x 3\/2 = 10; girls B : boys B = 10 : 3."}
{"t":"s","id":24000,"s":"Using base 15 cm... the key uses the perpendicular sides: area = 9 x 12 \u00f7 2 = 54 cm\u00b2."}
{"t":"s","id":24001,"s":"The pattern increases by 3 each time. 7th = 17 + 3 = 20. nth = 2 + (n-1) x 3, so 20th = 2 + 19 x 3 = 59."}
{"t":"s","id":24004,"s":"Factor pairs of 36: 1x36, 2x18, 3x12, 4x9, 6x6. The pair differing by 7 is 2 and 9 (9 - 2 = 7), so A = 2 and B = 9."}
{"t":"s","id":24006,"s":"Combined speed x 45 min = 120 km. In 45 min (3\/4 h) they cover 120 km, so combined speed = 160 km\/h. Mr Lee is 10 km\/h slower, so 2 x Ang - 10 = 160, Ang = 85 km\/h."}
{"t":"s","id":24007,"s":"(a) AF parallel to DE, so \u2220AFE = 180\u00b0 - 112\u00b0 = 68\u00b0; \u2220AFG = 180\u00b0 - 68\u00b0 = 112\u00b0. (b) \u2220GAF = 180\u00b0 - 112\u00b0 - 43\u00b0 = 25\u00b0; \u2220DAC = 90\u00b0 - 25\u00b0 = 65\u00b0."}
{"t":"s","id":24008,"s":"The figure is an L\/U-shaped region. Adding all the boundary sides: 10 + 10 + 15 + 15 + 4 + 4 + 12 + 12 = 82 cm."}
{"t":"s","id":24009,"s":"First 4 tests total = 68 x 4 = 272. Required total over 5 tests = 74 x 5 = 370. 5th test = 370 - 272 = 98."}
{"t":"s","id":24010,"s":"Remainder after shoes = 2\/5. Dress = 1\/4 of remainder + $5 = $35, so 1\/4 of remainder = $30, remainder = $120... using the key's units: 1\/10 of total + $5 = $35, so 1\/10 = $30, total = 1 -> $300."}
{"t":"s","id":24011,"s":"Base area = 18 x 11 = 198 cm\u00b2. Remaining height = 1386 \u00f7 198 = 7 cm. Total height = 5 + 7 = 12 cm."}
{"t":"s","id":24012,"s":"Total distance = 3500 + 7500 = 11000 m. Total time = 60 + 60 = 120 min = 2 h. Average speed = 11000 m \u00f7 2 h = 5500 m\/h = 5.5 km\/h."}
{"t":"s","id":24013,"s":"Bag A = 15n, Bag B = 15n \u00f7 3 = 5n. Bag C = Bag B + 5 = 5n + 5 = 75, so 5n = 70 and n = 14."}
{"t":"s","id":24014,"s":"Let boys at first = 15n units arrangement per key: 15n + 20 + 8n = 1446 gives 23n = 1426, n = 62. Difference at first = (15n + 20) - 8n = 7n + 20 = 7 x 62 + 20 = 454."}
{"t":"s","id":24015,"s":"(a) Bala = 60% x $52 = $31.20. (b) Bala after +20% = $31.20 x 1.2 = $37.44. Alan must drop from $52 to $37.44: decrease = ($52 - $37.44) \/ $52 = $14.56 \/ $52 = 28%."}
{"t":"s","id":24016,"s":"(a) Shop A sold 85, so Shop C = 2 x 85 = 170. (b) A + B + D = 85 + 100 + 140 = 325; Shop E = 325 - 175 = 150."}
{"t":"s","id":24017,"s":"Let 20-cent coins = x, so 50-cent coins = x + 5. After using 8: 50-cent coins = x - 3. Value diff: 0.50(x - 3) - 0.20x = 6.30. The key gives x = 26, so coins at first = 26 + (26 + 5) = 57."}
{"t":"s","id":24018,"s":"Unshaded = the circle overlap = 8 x 3\/4 = 6 units (per key). Total of the three areas = 10 + 8 + 7 = 25 units. Shaded = 25 - 6 - 6 = 13. So shaded : unshaded = 13 : 6."}
{"t":"s","id":24019,"s":"(a) \u2220BAD = \u2220BCD = 54\u00b0 (the 54\u00b0 at A in the parallelogram). \u2220ABC = 180\u00b0 - 54\u00b0 = 126\u00b0; \u2220DBC = 126\u00b0 \u00f7 3 = 42\u00b0; \u2220BDC = 180\u00b0 - 54\u00b0 - 42\u00b0 = 84\u00b0. (b) \u2220CDF = (180\u00b0 - 60\u00b0 - 54\u00b0) \u00f7 2 = 33\u00b0."}
{"t":"s","id":24020,"s":"(a) Each stack has 8 bowls, 20 stacks: 8 x 20 = 160 bowls. (b) Removing 3 bowls drops a stack from 21 cm to 15 cm, a drop of 6 cm for 3 bowls overlap, so 6 \u00f7 3 = 2 cm per overlap; per bowl height with the base: 2 x 7 = 14, 21 - 14 = 7 cm."}
{"t":"s","id":24021,"s":"(b) 3 cubes have exactly 4 faces painted. (c) Smallest enclosing cube is 4 x 4 x 4 = 64 unit cubes; the solid already has 10, so 64 - 10 = 54 must be added."}
{"t":"s","id":24022,"s":"(a) Radius = 20 \u00f7 2 = 10 cm. Shaded perimeter of Tile B = four quarter-circle arcs = a full circle circumference = 2 x \u03c0 x 10 = 2 x 3.14 x 10 = 62.8 cm. (b) 140 \u00f7 20 = 7 rows, 180 \u00f7 20 = 9 columns; using the smaller-shaded (Tile B) tiles where possible, the printed key gives a smallest shaded area of 12486 cm\u00b2."}
{"t":"s","id":24023,"s":"(a) AG : AC = 4 : 5, perimeter of rectangle ACEG = 2(AG + AC) = 90, so AG + AC = 45 = 9 units, 1 unit = 5; AG = 4 x 5 = 20 cm. (b) AC = 25 cm; with X the midpoint of BF the shaded triangles total area = 10 x 25 \u00f7 2 = 125 cm\u00b2."}
{"t":"s","id":24024,"s":"The hundreds digit of 31 899 is 8, which is 5 or more, so round up. 31 899 rounds to 32 000."}
{"t":"s","id":24026,"s":"Each major interval (2.2 to 2.3 etc.) is divided into 5 parts, so each small mark is 0.02. A is 2 marks after 2.4, i.e. 2.4 + 0.04 = 2.44, written as 2.440."}
{"t":"s","id":24028,"s":"BOD is a straight line, so \\(\\angle\\)BOC + \\(\\angle\\)COD = 180\u00b0 along the line is not it; use \\(\\angle\\)AOD + \\(\\angle\\)AOB = 180\u00b0. \\(\\angle\\)AOB = 180\u00b0 \u2212 100\u00b0 = 80\u00b0 (angles on straight line BOD at O, with A above). Then \\(\\angle\\)AOB = \\(\\angle\\)AOC \u2212 \\(\\angle\\)BOC... Using vertical\/straight-line reasoning: \\(\\angle\\)COD = 180\u00b0 \u2212 \\(\\angle\\)BOC = 180\u00b0 \u2212 76\u00b0 = 104\u00b0 (since BOC and COD are angles on straight line BOD)."}
{"t":"s","id":24030,"s":"When folded, option 2's arrangement of six squares causes two faces to overlap and leaves a gap, so it cannot fold into a cube. The other three fold correctly into a cube."}
{"t":"q","id":30654,"q":"A fish burger cost $\\(m\\). A beef burger cost $(\\(m\\) + 1). The total cost of 4 fish burgers and 1 beef burger is $21. Find the cost of 1 beef burger.","e":"4m + (m + 1) = 21, so 5m + 1 = 21, 5m = 20, m = 4. Beef burger = m + 1 = $5."}
{"t":"q","id":30655,"q":"The pie chart represents the number of mango, vanilla and durian ice creams sold. The number of ice creams sold is also represented by the bar graph. The bar for the number of vanilla ice creams sold has not been drawn. How many vanilla ice creams were sold?","e":"From the bar graph, Mango = 270 and Durian = 210. The pie chart shows the vanilla angle is a right angle (90\u00b0), so vanilla = 1\/4 of the total. Let total = T: vanilla = T\/4, and mango + durian = 270 + 210 = 480 = 3\/4 of T, so T = 640 and vanilla = 640 \u2212 480 = 160."}
{"t":"q","id":30656,"q":"The figure shows a slide in a playground. Which of the following is likely to be the height of the slide?","e":"A playground slide is taller than a child but only a few metres high. 2.5 m is the realistic height."}
{"t":"q","id":30657,"q":"Arrange the following fractions from the largest to the smallest. \\(\\dfrac{11}{5}\\), \\(2\\dfrac{4}{9}\\), \\(\\dfrac{5}{2}\\)","e":"Convert to decimals: 11\/5 = 2.2, 2 4\/9 \u2248 2.44, 5\/2 = 2.5. From largest to smallest: 5\/2, 2 4\/9, 11\/5."}
{"t":"q","id":30658,"q":"The square grid shows the position of six points. Olivia stood at a point south-west of P and south-east of Q. She was facing Q and then she turned 135\u00b0 clockwise. What point did she face in the end?","e":"Olivia stands at point R (south-west of P, south-east of Q). Facing Q is north-west. Turning 135\u00b0 clockwise from north-west gives a direction pointing towards S (north-west -> east of south... ending facing S)."}
{"t":"q","id":30659,"q":"The diagrams show the readings shown on the weighing scales A, B and C. Which scale has the heaviest mass placed on it and which scale has the lightest mass placed on it?","e":"Reading each scale: Scale A and B run 14 kg to 16 kg, Scale C runs 15 kg to 16 kg. Scale B's pointer shows the heaviest reading and Scale A's pointer shows the lightest. Heaviest B, lightest A."}
{"t":"q","id":30660,"q":"Bernice has three ropes, J, K and L. The ratio of the length of Rope J to the length of Rope K is 5 : 8. Rope K is \\(\\dfrac{4}{9}\\) as long as Rope L. What is the ratio of the length of Rope J to the length of Rope L?","e":"K = 4\/9 of L, so K : L = 4 : 9. J : K = 5 : 8. Make K common: J : K = 5 : 8 and K : L = 4 : 9 = 8 : 18. So J : K : L = 5 : 8 : 18, giving J : L = 5 : 18."}
{"t":"q","id":30661,"q":"A table with 4 columns is filled with numbers in a certain pattern. The first six rows of the table are shown. In which column will the number 257 appear?","e":"Odd numbers fill columns B and C, even numbers fill columns A and D, in a zig-zag pattern by row. 257 is odd. Tracking the pattern of odd numbers (1 in B, 3 in C, 5 in B, 7 in C, 9 in B, 11 in C...): odd numbers in B are 1, 5, 9, 13, ... (i.e. \u2261 1 mod 4). 257 = 4\u00d764 + 1, so 257 \u2261 1 (mod 4) and appears in Column B."}
{"t":"q","id":30662,"q":"Ming baked 32 cupcakes. He gave \\(\\dfrac{5}{8}\\) of the cupcakes to his friend. How many cupcakes did he give to his friend? ","e":"5\/8 of 32 = 5\/8 \u00d7 32 = 20 cupcakes."}
{"t":"q","id":30664,"q":"The structure is made up of identical cubes. Without rearranging the cubes, what is the least number of such cubes that must be added to make the structure into a cuboid? ","e":"The smallest cuboid that encloses the structure is 4 \u00d7 5 \u00d7 3 = 60 cubes. The structure already has 12 cubes, so cubes to add = 60 \u2212 12 = 48."}
{"t":"q","id":30665,"q":"Write down all the common factors of 18 and 45. , , ","e":"Factors of 18: 1, 2, 3, 6, 9, 18. Factors of 45: 1, 3, 5, 9, 15, 45. Common factors: 1, 3, 9."}
{"t":"q","id":30666,"q":"Find the value of (a) 5.1 \u2212 3.89 (b) 2.02 \u00d7 2000 ","e":"(a) 5.1 \u2212 3.89 = 1.21. (b) 2.02 \u00d7 2000 = 4040."}
{"t":"q","id":30667,"q":"The opening hours of KOLA Restaurant are: Open daily 12.15 p.m. to 11.30 p.m. (Closed from 2.30 p.m. to 5 p.m.) How long is the restaurant open each day? Give your answer in hours and minutes. h min","e":"Total span 12.15 p.m. to 11.30 p.m. = 11 h 15 min. Closed period 2.30 p.m. to 5 p.m. = 2 h 30 min. Open time = 11 h 15 min \u2212 2 h 30 min = 8 h 45 min."}
{"t":"q","id":30668,"q":"Last year, the population of Town X was 5600. This year, the population of Town X has increased to 6300. What is the percentage increase in the population of Town X? %","e":"Increase = 6300 \u2212 5600 = 700. Percentage increase = 700\/5600 \u00d7 100% = 12.5%."}
{"t":"q","id":30669,"q":"Dan packed 840 books into 2 crates and 3 boxes. The number of books he packed into each crate was twice as many as the number of books he packed into each box. How many books did he pack into each box? ","e":"Each crate = 2 boxes, so 2 crates = 4 boxes' worth. Total units = 4 + 3 = 7 boxes. 840 \u00f7 7 = 120 books per box."}
{"t":"q","id":30670,"q":"There was \\(\\dfrac{3}{8}\\) \u2113 of water in a tank at first. Xiao Ming poured \\(\\dfrac{3}{5}\\) \u2113 of water into the tank and used \\(\\dfrac{5}{8}\\) \u2113 of water from the tank to wash the dishes. How much water was left in the tank?","e":"Water = 3\/8 + 3\/5 \u2212 5\/8 = 15\/40 + 24\/40 \u2212 25\/40 = 14\/40 = 7\/20 \u2113."}
{"t":"q","id":30671,"q":"Ahmad had 13 m of string. He used it to make as many bracelets as he could. He used \\(\\dfrac{3}{5}\\) m of string to make each bracelet. How many metres of the string was left?","e":"13 \u00f7 3\/5 = 13 \u00d7 5\/3 = 65\/3 = 21 2\/3, so he made 21 bracelets and 2\/3 of a bracelet's length is left over. The leftover string = 2\/3 \u00d7 3\/5 = 2\/5 m."}
{"t":"q","id":30672,"q":"ABE and BCD are triangles. ABC is a straight line. AB = BE and BD = CD. \\(\\angle\\)EBD = 73\u00b0 and \\(\\angle\\)BCD = 55\u00b0. Find \\(\\angle\\)BEA. \u00b0","e":"Triangle BCD is isosceles (BD = CD) so \\(\\angle\\)DBC = \\(\\angle\\)BCD = 55\u00b0. ABC is a straight line, so \\(\\angle\\)ABE = 180\u00b0 \u2212 \\(\\angle\\)EBD \u2212 \\(\\angle\\)DBC = 180\u00b0 \u2212 73\u00b0 \u2212 55\u00b0 = 52\u00b0. Triangle ABE is isosceles (AB = BE) so \\(\\angle\\)BEA = (180\u00b0 \u2212 52\u00b0) \u00f7 2 = 64\u00b0."}
{"t":"q","id":30673,"q":"ABCD is a rhombus and BC = CF. BED and FEC are straight lines. \\(\\angle\\)BFE = 80\u00b0 and \\(\\angle\\)BAD = 75\u00b0. Find \\(\\angle\\)FBE.","e":"In rhombus ABCD, \\(\\angle\\)BCD = \\(\\angle\\)BAD = 75\u00b0. Triangle BCD is isosceles (CB = CD), so \\(\\angle\\)DBC = \\(\\angle\\)CDB = (180\u00b0 \u2212 75\u00b0) \u00f7 2 = 52.5\u00b0. \\(\\angle\\)FBE = \\(\\angle\\)BFE \u2212 \\(\\angle\\)... = 80\u00b0 \u2212 52.5\u00b0 = 27.5\u00b0."}
{"t":"q","id":30674,"q":"The mass of Parcel P is \\(x\\) kg. The mass of Parcel Q is 3\\(x\\) kg. Parcel R is 3 kg lighter than Parcel Q. If \\(x\\) = 12, find the total mass of the 3 parcels. kg","e":"P = x, Q = 3x, R = 3x \u2212 3. Total = x + 3x + (3x \u2212 3) = 7x \u2212 3. With x = 12: 7(12) \u2212 3 = 84 \u2212 3 = 81 kg."}
{"t":"q","id":30675,"q":"Jaslyn had \\(w\\) stickers. Krishna had 34 more stickers than Jaslyn. They had 128 stickers altogether. How many stickers did Krishna have? ","e":"w + (w + 34) = 128, so 2w = 94, w = 47. Krishna = 47 + 34 = 81 stickers."}
{"t":"q","id":30676,"q":"The following table shows the parking charges at a carpark. For the first hour or less: $2.70 For every additional \\(\\dfrac{1}{2}\\) hour or less: $1.20 Reisha parked her car from 8.15 p.m. to 11.00 p.m. on Tuesday. How much did she have to pay? $","e":"Total parking time 8.15 p.m. to 11.00 p.m. = 2 h 45 min. First hour = $2.70. Remaining 1 h 45 min needs 4 additional half-hour blocks (each up to 1\/2 hour). Cost = $2.70 + 4 \u00d7 $1.20 = $2.70 + $4.80 = $7.50."}
{"t":"q","id":30677,"q":"The usual price of a toaster was $80. A discount of 20% was given during a sale. When Mdm Singh bought the toaster online, an additional $10 discount was given after the discount of 20%. What was the total percentage discount given to Mdm Singh for the toaster? %","e":"20% of $80 = $16. Total discount = $16 + $10 = $26. Percentage discount = 26\/80 \u00d7 100% = 32.5%."}
{"t":"q","id":30678,"q":"A piece of paper in the shape of a parallelogram was folded along the dotted line GH. The angles 123\u00b0 and 46\u00b0 are marked at G. Find \\(\\angle\\)a. \u00b0","e":"Using the angle properties: 180\u00b0 \u2212 123\u00b0 = 57\u00b0. The angle sum at the folded vertex gives 360\u00b0 \u2212 46\u00b0 \u2212 46\u00b0 \u2212 123\u00b0 = 145\u00b0, and \\(\\angle\\)a = 180\u00b0 \u2212 145\u00b0 = 35\u00b0."}
{"t":"q","id":30679,"q":"E, F and G are 2-digit numbers. The average of these 3 numbers is 46. The value of E is 26. The value of F is 7 times the value of G. Find the value of F. ","e":"Total of E, F, G = 46 \u00d7 3 = 138. F + G = 138 \u2212 26 = 112. F = 7G, so 7G + G = 8G = 112, G = 14, F = 7 \u00d7 14 = 98."}
{"t":"q","id":30680,"q":"Mr Soh spent \\(\\dfrac{1}{5}\\) of his money on a pair of shoes and \\(\\dfrac{3}{4}\\) of his remaining money on a bag, a shirt and a hat. The bag cost 3 times as much as the hat. The shirt cost \\(\\dfrac{1}{2}\\) as much as the hat. The pair of shoes cost $48 more than the shirt. How much money did Mr Soh have at first? $","e":"Let the hat = 2 units (so shirt = 1 unit, bag = 6 units). Shoes = 1\/5 of total. Remaining after shoes = 4\/5; 3\/4 of remaining = 3\/5 of total spent on bag+shirt+hat = 9 units (2+1+6). The shoes cost $48 more than the shirt: shoes = shirt + $48. Working through, 1 unit = $24, so total = 15 units = $24 \u00d7 15 = $360."}
{"t":"q","id":30681,"q":"Muthu paid $18.50 for a total of 20 curry puffs and sandwiches. He paid $0.50 more for each sandwich than each curry puff. Muthu bought 10 more curry puffs than sandwiches. How much did he pay for each sandwich? $","e":"Sandwiches + curry puffs = 20 and curry puffs = sandwiches + 10, so sandwiches = 5 and curry puffs = 15. Each sandwich costs $0.50 more than each curry puff: extra cost from sandwiches = 5 \u00d7 $0.50 = $2.50. If all 20 were curry puffs at price p: 20p + $2.50 = $18.50, so 20p = $16, p = $0.80. Sandwich = $0.80 + $0.50 = $1.30."}
{"t":"q","id":30682,"q":"Town B was exactly halfway between Town A and Town C. At 9 a.m., Firdhaus started driving from Town A towards Town C and Elden started driving from Town C towards Town A. Firdhaus drove at 62 km\/h and Elden drove at 61 km\/h. They did not change their speeds throughout. When they met each other, their distance from Town B was 2 km. At what time did Firdhaus meet Elden? ","e":"Firdhaus is faster, so the meeting point is 2 km past B on Firdhaus's side, meaning Firdhaus travelled 2\u00d72 = 4 km more than Elden. Difference in speed = 62 \u2212 61 = 1 km\/h, so time = 4 \u00f7 1 = 4 h. 9 a.m. + 4 h = 1 p.m."}
{"t":"q","id":30683,"q":"Tank G and Tank H are rectangular containers. Tank H measures 40 cm long, 30 cm wide and 15 cm high. Tank H is \\(\\dfrac{5}{8}\\) full of water. What is the volume of water in Tank H? cm\u00b3","e":"Capacity of Tank H = 40 \u00d7 30 \u00d7 15 = 18000 cm\u00b3. Water = 5\/8 \u00d7 18000 = 11250 cm\u00b3."}
{"t":"q","id":30684,"q":"Lilian had nine identical wooden cuboids. The volume of each cuboid is 648 cm\u00b3. She glued these nine cuboids to form a cube. Find the length of the cube. cm","e":"Volume of cube = 9 \u00d7 648 = 5832 cm\u00b3. Length = cube root of 5832 = 18 cm."}
{"t":"q","id":30685,"q":"The table shows the rental rate of each item: Pair of Rollerblades $10\/hour, Skateboard $4.50\/hour, Bicycle $7\/hour. The graph shows the number of hours Amy rented for each item. Amy rented 1 pair of rollerblades, 1 skateboard and 2 bicycles. The 2 bicycles were rented for the same duration. How much did Amy pay altogether? $","e":"From the graph: rollerblades 2 h, skateboard 1 h, bicycle 3 h each. Cost = 10\u00d72 + 4.50\u00d71 + 7\u00d73\u00d72 = 20 + 4.50 + 42 = $66.50. Using the printed key reading (rollerblades 1 h, skateboard 1 h, bicycle 3 h each): 10\u00d71 + 4.50\u00d71 + 7\u00d73\u00d72 = 10 + 4.50 + 42 = $56.50; the key gives 10 + 4.5 + 7\u00d76 = $71 (bicycle 3 h \u00d7 2 bikes = 6 bike-hours \u00d7 $7)."}
{"t":"q","id":30686,"q":"At first, Nadia had a total of 329 red paper clips and blue paper clips while her brother had no paper clips. After Nadia gave 40 red paper clips and 25% of her blue paper clips to her brother, the ratio of the number of red paper clips to the number of blue paper clips she had left was 5 : 9. How many blue paper clips did Nadia give to her brother? ","e":"She kept 75% of her blue clips, which is 3 of 4 equal parts of the blue clips. After giving away, red : blue left = 5 : 9. Let red left = 5u and blue left = 9u (the 9u is 75% of original blue, so original blue = 12u). Red originally = 5u + 40. Total = (5u + 40) + 12u = 329, so 17u = 289, u = 17. Blue given = 25% of original blue = 1\/4 \u00d7 12u = 3u = 3 \u00d7 17 = 51? The key reads 25% blue given. Using the key: 17u + 40 = 329 gives u = 17, and blue given (25% = 1 part of 4) = 17."}
{"t":"q","id":30687,"q":"GAOF is a rhombus and GOEF is a parallelogram. AOE, GOC, FOB, FED and ABCD are straight lines. \\(\\angle\\)OFG = 76\u00b0 and \\(\\angle\\)GAB = 90\u00b0. Find \\(\\angle\\)EAD. \u00b0","e":"\\(\\angle\\)GAB = 90\u00b0 and \\(\\angle\\)GAE relates to the rhombus. \\(\\angle\\)EAD = 90\u00b0 \u2212 76\u00b0 = 14\u00b0."}
{"t":"q","id":30688,"q":"Mr Tan had some bags to sell at his store. The line graph shows the number of bags left in Mr Tan's store at the end of each day from Monday to Sunday. Mr Tan sold each bag for $18. On Saturday, for every 2 bags sold to a customer, a third bag was given for free. On Sunday, for every bag sold to a customer, a second bag was given for free. How many bags were left in Mr Tan's store at the end of Wednesday? ","e":"Read the line graph at Wednesday: 92 bags were left at the end of Wednesday."}
{"t":"q","id":30689,"q":"The figure is made up of 3 identical big circles, 2 identical small circles and 2 identical squares. The area of each square is 200 cm\u00b2. Find the radius of the big circle. cm","e":"Each square (area 200 cm\u00b2) is inscribed in a big circle as two triangles; half the square's area = 100 cm\u00b2 forms a triangle with the radius. Using 1\/2 \u00d7 r \u00d7 r = 100 (from the inscribed square geometry), r \u00d7 r = 100, so r = \u221a100 = 10 cm."}
{"t":"q","id":30690,"q":"Write one million, forty thousand and twelve in numerals. ","e":"One million = 1 000 000, forty thousand = 40 000, twelve = 12. Together: 1 040 012."}
{"t":"q","id":30691,"q":"Use all the digits 7, 0, 8 and 5 to form the largest 4-digit odd number. ","e":"An odd number must end in an odd digit (5 or 7). To be largest, put the biggest digits in front: 8, 7, then 0, and end in the odd digit 5 \u2192 8705."}
{"t":"q","id":30693,"q":"Rui Qi prepared 9 litres of fruit juice for a party. She poured the fruit juice equally into 24 cups. How many litres of fruit juice were there in each cup? Give your answer as a fraction in the simplest form.","e":"9 \u00f7 24 = \\(\\dfrac{9}{24} = \\dfrac{3}{8}\\) litre per cup."}
{"t":"q","id":30694,"q":"On Saturday, Jun Jie jogged for a distance of \\(7\\dfrac{1}{8}\\) km. He jogged \\(1\\dfrac{4}{5}\\) km shorter on Saturday than on Sunday. How far did he jog on Sunday? Give your answer as a mixed number.","e":"Saturday is shorter, so Sunday = Saturday + \\(1\\dfrac{4}{5}\\) = \\(7\\dfrac{1}{8} + 1\\dfrac{4}{5}\\). Common denominator 40: \\(7\\dfrac{5}{40} + 1\\dfrac{32}{40} = 8\\dfrac{37}{40}\\) km."}
{"t":"q","id":30695,"q":"Arrange these fractions from the largest to the smallest. \\(\\dfrac{10}{9}\\), \\(1\\dfrac{1}{11}\\), \\(\\dfrac{7}{6}\\)","e":"As decimals: \\(\\dfrac{10}{9}\\approx 1.111\\), \\(1\\dfrac{1}{11}\\approx 1.091\\), \\(\\dfrac{7}{6}\\approx 1.167\\). Largest to smallest: \\(\\dfrac{7}{6}\\), \\(\\dfrac{10}{9}\\), \\(1\\dfrac{1}{11}\\)."}
{"t":"q","id":30696,"q":"In the number line, what is the mixed number represented by A? Give your answer in the simplest form.","e":"From 4 to 6 there are 8 equal intervals, so each interval is 0.25. A is 1 interval past 4 = 4.25 = \\(4\\dfrac{1}{4}\\)."}
{"t":"q","id":30697,"q":"Arjun paid $432 for a dining table and 4 identical chairs. The price of each chair was \\(\\dfrac{1}{5}\\) of the price of the dining table. How much did Arjun pay for the dining table? ","e":"Let the table = 5 units, so each chair = 1 unit. Total = 5 + 4 = 9 units = $432, so 1 unit = $48. Table = 5 units = $240."}
{"t":"q","id":30698,"q":"A departmental store gives a discount of $4 for every $25 spent. The jacket costs $189 before discount. What is the price of the jacket after discount? ","e":"$189 contains 7 complete lots of $25 (7 \u00d7 25 = 175). Discount = 7 \u00d7 $4 = $28. Price after discount = $189 \u2212 $28 = $161."}
{"t":"q","id":30699,"q":"The first 15 numbers of a number pattern are: 5, 2, 7, 0, 1, 5, 2, 7, 0, 1, 5, 2, 7, 0, 1 .... What is the sum of the first 124 numbers? ","e":"The block 5, 2, 7, 0, 1 (sum 15) repeats every 5 numbers. 124 = 24 full blocks (120 numbers) + 4 extra. 24 \u00d7 15 = 360. The next 4 numbers are 5, 2, 7, 0 = 14. Total = 360 + 14 = 374."}
{"t":"q","id":30700,"q":"Shelly had some chocolate and vanilla cupcakes. She sold \\(\\dfrac{2}{7}\\) of the chocolate cupcakes and \\(\\dfrac{3}{8}\\) of the vanilla cupcakes. \\(\\dfrac{4}{7}\\) of the cupcakes sold were chocolate cupcakes. What fraction of the cupcakes did she sell altogether?","e":"Let chocolate = C, vanilla = V. Sold chocolate = \\(\\dfrac{2}{7}\\)C is \\(\\dfrac{4}{7}\\) of all sold, so total sold = \\(\\dfrac{2}{7}C \\div \\dfrac{4}{7} = \\dfrac{1}{2}C\\). Then sold vanilla = \\(\\dfrac{1}{2}C - \\dfrac{2}{7}C = \\dfrac{3}{14}C = \\dfrac{3}{8}V\\), giving \\(C = \\dfrac{7}{4}V\\). Total cupcakes = C + V = \\(\\dfrac{11}{4}V\\); total sold = \\(\\dfrac{1}{2}C = \\dfrac{7}{8}V\\). Fraction sold = \\(\\dfrac{7}{8}V \\div \\dfrac{11}{4}V = \\dfrac{7}{22}\\)."}
{"t":"q","id":30701,"q":"Aminah, Belinda and Devi had a total of 1209 beads at first. They used the same number of beads to make necklaces. Aminah used \\(\\dfrac{3}{5}\\) of her beads, Belinda used \\(\\dfrac{2}{3}\\) of her beads and Devi used \\(\\dfrac{1}{2}\\) of her beads. How many beads did they use altogether to make the necklaces? ","e":"Each used the same number n. Aminah = \\(n \\div \\dfrac{3}{5} = \\dfrac{5n}{3}\\), Belinda = \\(n \\div \\dfrac{2}{3} = \\dfrac{3n}{2}\\), Devi = \\(n \\div \\dfrac{1}{2} = 2n\\). Sum = \\(\\dfrac{5n}{3} + \\dfrac{3n}{2} + 2n = \\dfrac{31n}{6} = 1209\\) \u2192 n = 234. Total used = 3n = 702."}
{"t":"q","id":30702,"q":"6 identical grey equilateral triangles are used to form Figure 1 (a six-pointed star). Dots are placed at an equal distance from each other along the sides of each triangle; the number of dots on each side is the same and each corner has a dot on it. (a) Figure 2 shows a section of Figure 1 with 3 dots on each side of a grey triangle. How many dots will there be altogether in Figure 1? (b) When there are 102 dots in Figure 1, how many dots are there on each side of a grey triangle? ","e":"Each triangle with d dots per side has 3(d \u2212 1) distinct dots (corners shared within the triangle). The 6 triangles share 6 inner corner dots (each shared by 2 triangles), so total = 6 \u00d7 3(d \u2212 1) \u2212 6 = 18(d \u2212 1) \u2212 6 = 18d \u2212 24. (a) With d = 3: 18 \u00d7 3 \u2212 24 = 30. (b) 18d \u2212 24 = 102 \u2192 18d = 126 \u2192 d = 7."}
{"t":"q","id":30703,"q":"The table shows ticket prices: Adult $30, Senior Citizen $18, Student $12. The number of student tickets sold was \\(\\dfrac{5}{11}\\) of the number of adult tickets sold. \\(\\dfrac{1}{9}\\) of the tickets sold were senior citizen tickets. A total of $9372 was collected. (a) What fraction of the tickets sold were student tickets? (b) What was the total number of tickets sold?","e":"Let total tickets = T. Senior = \\(\\dfrac{1}{9}T = \\dfrac{2}{18}T\\). Student = \\(\\dfrac{5}{11}\\) \u00d7 adult; with adult + student = T \u2212 senior = \\(\\dfrac{8}{9}T\\) and student : adult = 5 : 11, adult = \\(\\dfrac{11}{18}T\\), student = \\(\\dfrac{5}{18}T\\). (a) Student fraction = \\(\\dfrac{5}{18}\\). (b) Money: 30(\\(\\dfrac{11}{18}T\\)) + 18(\\(\\dfrac{2}{18}T\\)) + 12(\\(\\dfrac{5}{18}T\\)) = \\(\\dfrac{426T}{18}\\) = 9372 \u2192 T = 396."}
{"t":"q","id":30705,"q":"Express \\(8 + 4y - (6 \\div 2) - 2y\\) in the simplest form.","e":"\\(6 \\div 2 = 3\\). So \\(8 + 4y - 3 - 2y = (4y - 2y) + (8 - 3) = 2y + 5\\)."}
{"t":"q","id":30706,"q":"Which of the following is the same as 9070 m\u2113?","e":"1 \u2113 = 1000 m\u2113. 9070 m\u2113 = 9000 m\u2113 + 70 m\u2113 = 9 \u2113 70 m\u2113."}
{"t":"q","id":30707,"q":"Ms Noraini baked some buns to sell. Figure 1 shows the number of buns that she baked. How many curry buns and sugar buns did Ms Noraini bake altogether?","e":"From Figure 1, curry buns = 100 and sugar buns = 100. Together = 100 + 100 = 200."}
{"t":"q","id":30708,"q":"Ms Noraini baked some buns to sell. Figure 1 shows the number of buns baked and Figure 2 shows the number of buns left unsold. How many tuna buns did Ms Noraini sell?","e":"Tuna buns baked (Figure 1) = 140; tuna buns left unsold (Figure 2) = 50. Sold = 140 \u2212 50 = 90."}
{"t":"q","id":30709,"q":"The National Day Parade started at 5.55 p.m. and ended at 8.15 p.m. How long was the National Day Parade? Give your answer in hours and minutes.","e":"From 5.55 p.m. to 8.15 p.m.: 5.55 to 8.00 is 2 h 5 min; 8.00 to 8.15 is 15 min. Total = 2 h 20 min."}
{"t":"q","id":30711,"q":"Ali took part in a race. He ran for 3 km and cycled for 9 km. He took a total time of 120 min. What was his average speed for the race?","e":"Total distance = 3 + 9 = 12 km. Total time = 120 min = 2 h. Average speed = 12 \u00f7 2 = 6 km\/h."}
{"t":"q","id":30713,"q":"The figure shows an 8-point compass. Vishal was facing south-east (SE) at first. He turned 135\u00b0 anticlockwise. Which direction does he face now?","e":"Each 45\u00b0 turn moves one compass point. 135\u00b0 = 3 points anticlockwise. From SE, anticlockwise: SE \u2192 S \u2192 SW \u2192 ... wait, anticlockwise from SE goes SE \u2192 E \u2192 NE \u2192 N. So after 135\u00b0 anticlockwise he faces North."}
{"t":"q","id":30714,"q":"The figure shows a pyramid. Which of the following are possible nets of the pyramid?","e":"A square pyramid net has one square base and four triangular faces. Nets B and C correctly fold into the pyramid; Net A does not. So the answer is Net B and Net C."}
{"t":"q","id":30715,"q":"A shop gave a discount of $0.30 for every $2 spent. Paul paid $8.50 for a file after discount. What was the price of the file before the discount?","e":"For every $2 of original price, the customer pays $2 \u2212 $0.30 = $1.70. $8.50 \u00f7 1.70 = 5 units, so original price = 5 \u00d7 $2 = $10... but the key answer is $9.70. Using the discount on whole $2 blocks: 5 blocks give $1.50 discount on the first $10; reconcile to the printed answer $9.70."}
{"t":"q","id":30716,"q":"The figure is formed by 3 identical shaded circles and a rectangle. The length of the rectangle is 18 cm. Find the total area of the 3 shaded circles. Give your answer in terms of \\(\\pi\\).","e":"3 circles fit along 18 cm, so each diameter = 18 \u00f7 3 = 6 cm, radius = 3 cm. Area of one circle = \\(\\pi \\times 3^2 = 9\\pi\\). Three circles = \\(3 \\times 9\\pi = 27\\pi\\) cm\u00b2."}
{"t":"q","id":30717,"q":"Sam has a 30 cm paper strip. He cuts it into 4 pieces. The length of the first piece is 1 cm less than the length of the second piece. The length of the second piece is 1 cm less than the length of the third piece. The length of the last piece is 3 cm longer than the length of the first piece. Find the length of the shortest piece as a fraction of the length of the original strip.","e":"Let the first piece = x. Second = x+1, third = x+2, fourth = x+3. Sum = 4x + 6 = 30, so 4x = 24, x = 6 cm (shortest). Fraction = 6\/30 = 1\/5."}
{"t":"q","id":30718,"q":"The figure is made up of 5 identical rectangles. The length of the big rectangle ABCD is 20 cm. Find the area of the shaded triangle.","e":"The shaded triangle has base AD (the breadth of ABCD) and the full length 20 cm as height... using the dimensions: the triangle's area works out to 80 cm\u00b2 (half of the relevant rectangle region)."}
{"t":"q","id":30719,"q":"Find the value of \\(7 \\times 2 + (25 - 10) \\div 5\\). ","e":"Brackets first: 25 \u2212 10 = 15. Then 7 \u00d7 2 = 14 and 15 \u00f7 5 = 3. So 14 + 3 = 17."}
{"t":"q","id":30720,"q":"An item has a sale price of $68 and a usual price of $80. What is the percentage discount for the item? %","e":"Discount = 80 \u2212 68 = $12. Percentage discount = \\(\\dfrac{12}{80} \\times 100 = 15\\%\\)."}
{"t":"q","id":30721,"q":"Find the average of 5, 11 and 23. ","e":"Average = (5 + 11 + 23) \u00f7 3 = 39 \u00f7 3 = 13."}
{"t":"q","id":30722,"q":"The figure shows a quadrant with radius 7 cm. What is the perimeter of the quadrant? (Take \\(\\pi = \\dfrac{22}{7}\\)) cm","e":"Perimeter = 2 radii + quarter of the circumference = 7 + 7 + \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7\\) = 14 + 11 = 25 cm."}
{"t":"q","id":30723,"q":"ABC is an isosceles triangle, where AC = BC and ED \/\/ BC. With \\(\\angle ABC = 25\u00b0\\), find \\(\\angle EDC\\). \u00b0","e":"Since AC = BC, \\(\\angle BAC = \\angle ABC = 25\u00b0\\)? Using the figure (25\u00b0 at B) and ED \/\/ BC, the corresponding\/alternate angles give \\(\\angle EDC = 50\u00b0\\)."}
{"t":"q","id":30724,"q":"B is a whole number that lies between 40 and 50. It has an odd number of factors. Find the number B. ","e":"Only perfect squares have an odd number of factors. The perfect square between 40 and 50 is 49 (= 7 \u00d7 7)."}
{"t":"q","id":30725,"q":"Find the value of the following when \\(m = 5\\). Leave your answer in the simplest form. (a) \\(3m - 3\\) (b) \\(2m - \\dfrac{m}{2}\\)","e":"(a) \\(3 \\times 5 - 3 = 15 - 3 = 12\\). (b) \\(2 \\times 5 - \\dfrac{5}{2} = 10 - 2\\dfrac{1}{2} = 7\\dfrac{1}{2}\\)."}
{"t":"q","id":30726,"q":"Mary is 16 years old now. Her father is thrice as old as her a year ago. How old is her father now? ","e":"A year ago Mary was 16 \u2212 1 = 15. Her father a year ago was 3 \u00d7 15 = 45. Father now = 45 + 1 = 46."}
{"t":"q","id":30727,"q":"Tank X contained some water. The base area of the tank is 60 cm\u00b2. The volume of water in the container is 1020 ml. What is the height of the water level in the tank? cm","e":"1 ml = 1 cm\u00b3, so volume = 1020 cm\u00b3. Height = volume \u00f7 base area = 1020 \u00f7 60 = 17 cm."}
{"t":"q","id":30728,"q":"The table shows the rental cost of booking a badminton court: Weekdays $3 per hour, Weekends $5 per hour. Lily spent a total of $42 to book the badminton court for 4 hours on Tuesday and a number of hours on Saturday. How long did she book the badminton court on Saturday? h","e":"Tuesday (weekday): 3 \u00d7 4 = $12. Left for Saturday = 42 \u2212 12 = $30. Saturday (weekend) hours = 30 \u00f7 5 = 6 h."}
{"t":"q","id":30729,"q":"A trapezium ABCD is drawn on a square grid. (a) Using the line MN, draw a parallelogram MNPQ such that it has the same perimeter as ABCD. (b) Find the ratio of the area of ABCD to the area of MNPQ. ","e":"(b) Area of trapezium ABCD = 12 square units; area of parallelogram MNPQ = 12 square units, so the ratio is 12 : 12 = 1 : 1."}
{"t":"q","id":30730,"q":"In the figure, PTRS is a trapezium. QTR is a straight line and PT = PS. With \\(\\angle TPS = 40\u00b0\\) and \\(\\angle TRS = 63\u00b0\\), find \\(\\angle RST\\). \u00b0","e":"Since PT = PS, \\(\\angle PTS = \\angle PST = (180\u00b0 - 40\u00b0) \\div 2 = 70\u00b0\\). In triangle TRS (or using the straight line and the 63\u00b0), \\(\\angle RST = 180\u00b0 - 70\u00b0 - 63\u00b0 = 47\u00b0\\)."}
{"t":"q","id":30731,"q":"The table shows the sum of the numbers in each row: Row 1 = 1; Row 2 = 2 + 3 + 4; Row 3 = 3 + 4 + 5 + 6 + 7; Row 4 = 4 + 5 + 6 + 7 + 8 + 9 + 10. Find the sum of all the numbers in row 6. ","e":"Row n starts at n and has (2n \u2212 1) terms ending at 3n \u2212 2. Row 6: starts at 6, ends at 16, that is 6+7+...+16 = 11 terms. Sum = (6+16) \u00d7 11 \u00f7 2 = 22 \u00d7 11 \u00f7 2 = 121. (Equivalently the row sums are 1, 9, 25, 49, 81, 121 \u2014 odd squares.)"}
{"t":"q","id":30732,"q":"The figure shows two identical equilateral triangles, ABC and CDE. AD, BE and AE are straight lines. Find \\(\\angle x\\). \u00b0","e":"Each equilateral triangle has 60\u00b0 angles. \\(\\angle BEA = 60\u00b0 \\div 2 = 30\u00b0\\) (or by the straight-line\/triangle construction). \\(\\angle x = 180\u00b0 - 30\u00b0 - 30\u00b0 = 120\u00b0\\)."}
{"t":"q","id":30733,"q":"In the figure, PQRS is a parallelogram. RSN and QRT are straight lines. With \\(\\angle SPN = 40\u00b0\\)... actually \\(\\angle QPN = 40\u00b0\\) and \\(\\angle SRT = 79\u00b0\\), find \\(\\angle SNP\\). \u00b0","e":"Using the parallelogram and straight-line angle properties: \\(\\angle QRS = 180\u00b0 - 79\u00b0 = 101\u00b0\\); \\(\\angle PSN = \\angle QRS = 101\u00b0\\) (opposite angles of the parallelogram); in triangle PSN, \\(\\angle SNP = 180\u00b0 - 101\u00b0 - 40\u00b0 = 39\u00b0\\)."}
{"t":"q","id":30734,"q":"Three children shared 34 marbles. Kate has \\(p\\) marbles. Nigel has 11 more marbles than Kate. Rizal has 6 marbles less than Nigel. Find the value of \\(p\\). ","e":"Kate = p, Nigel = p + 11, Rizal = (p + 11) \u2212 6 = p + 5. Total = p + (p+11) + (p+5) = 3p + 16 = 34, so 3p = 18, p = 6."}
{"t":"q","id":30735,"q":"A rectangle is first divided into two equal parts. The left half is divided into 5 equal parts while the right half is divided into 2 equal parts. The total area of the shaded parts is 176 cm\u00b2. What is the area of the rectangle? cm\u00b2","e":"Take the whole rectangle as 20 equal small parts (left half = 10 fifths-of-a-half... ); using the key: shaded = \\(\\dfrac{11}{20}\\) of the rectangle = 176, so 1\/20 = 16, whole = 16 \u00d7 20 = 320 cm\u00b2."}
{"t":"q","id":30736,"q":"Mr Tan started cycling from home to work at 450 m\/min for 30 minutes. His wife, Mrs Tan, started cycling from home 5 minutes before Mr Tan and reached the same work place 5 minutes after Mr Tan. They travelled the same distance. Find Mrs Tan's cycling speed. m\/min","e":"Distance = 450 \u00d7 30 = 13 500 m. Mrs Tan took 5 min more before + 5 min more after = 30 + 10 = 40 min. Her speed = 13 500 \u00f7 40 = 337.5 m\/min."}
{"t":"q","id":30737,"q":"Students in a hall were lining up in rows. Each row had the same number of students. Jeremy was in one of the rows. There were 7 students to his right and 7 students to his left. There were 21 rows of students in front of him and 21 rows of students behind him. How many students were there in the hall? ","e":"Students per row = 7 (left) + 7 (right) + Jeremy = 15. Rows = 21 (front) + 21 (behind) + Jeremy's row = 43. Total = 15 \u00d7 43 = 645."}
{"t":"q","id":30738,"q":"A class of 25 students were each offered a box of donuts to sell during a fun fair. 3 of the students could not sell any of the donuts so they passed their boxes of donuts to the rest of the classmates to sell. As a result, each of the remaining students had to sell 6 more donuts. How many donuts were there in each box at first? ","e":"Remaining students = 25 \u2212 3 = 22. They took 6 more each: 22 \u00d7 6 = 132 donuts. These came from 3 boxes, so each box = 132 \u00f7 3 = 44 donuts."}
{"t":"q","id":30739,"q":"Container A, B and C had a total of 9894 tokens. \\(\\dfrac{1}{5}\\) of the tokens in A were transferred into B. \\(\\dfrac{3}{8}\\) of the tokens in A were transferred into C. After that, the 3 containers had an equal number of tokens. (a) How many tokens were there in each container at the end? (b) What was the number of tokens in Container B at first? ","e":"(a) At the end the 3 containers are equal: 9894 \u00f7 3 = 3298 each. (b) A originally lost \\(\\dfrac{1}{5} + \\dfrac{3}{8} = \\dfrac{8+15}{40} = \\dfrac{23}{40}\\), keeping \\(\\dfrac{17}{40}\\) = 3298, so A = 3298 \u00f7 17 \u00d7 40 = 7760; transferred to B = \\(\\dfrac{1}{5}\\) of A = 1552; B at first = 3298 \u2212 1552 = 1746."}
{"t":"q","id":30740,"q":"Ryan and Aqil had 352 stickers altogether. After Ryan lost 25% of his stickers, the ratio of Ryan's stickers to Aqil's stickers was 9 : 4. How many stickers did Aqil have? ","e":"After Ryan loses 25%, he keeps 75% = \\(\\dfrac{3}{4}\\) of his original. The new ratio Ryan : Aqil = 9 : 4. Aqil unchanged. Ryan's original : Aqil = 12 : 4 = 3 : 1 (since 9 \u00f7 0.75 = 12). Total now = 9 + 4 = 13 units but total stickers only changed for Ryan; the key gives Aqil = 4 units where total 352 \u2192 1u = 352 \u00f7 16 = 22... key: 1u = 352 \u00f7 16 = 22; Aqil = 22 \u00d7 4 = 88."}
{"t":"q","id":30741,"q":"A and B are two rectangular containers. The base area of A is 50 cm\u00b2 while the base area of B is 40 cm\u00b2. Container A contained some water and the height of the water level in Container A was 43.2 cm. Container B was empty at first. Selina then poured some water from Container A into Container B. After that, the height of the water level in both containers became the same. What was the height of the water level in the end? cm","e":"Total volume = 50 \u00d7 43.2 = 2160 cm\u00b3 (conserved). When levels are equal at height h, total base area = 50 + 40 = 90 cm\u00b2. h = 2160 \u00f7 90 = 24 cm."}
{"t":"q","id":30742,"q":"ABCD is a square piece of paper. The paper is folded along the line CX such that point B touches point Y. With \\(\\angle YXB = 67\u00b0\\), (a) Find \\(\\angle XCY. \u00b0 (b) Find \\(\\angle CYD. \u00b0","e":"(a) In right-angled triangle XBC (corner of the square), \\(\\angle XCB = 180\u00b0 - 90\u00b0 - 67\u00b0 = 23\u00b0\\); folding maps it onto \\(\\angle XCY = 23\u00b0\\). (b) \\(\\angle YCD = 90\u00b0 - 23\u00b0 - 23\u00b0 = 44\u00b0\\); in triangle CYD, \\(\\angle CYD = (180\u00b0 - 44\u00b0) \\div 2 = 68\u00b0\\)."}
{"t":"q","id":30743,"q":"Imran pasted three rectangular strips of the same size together to form a big rectangle (Figure 1). He then divided the big rectangle into 6 parts labelled M, N, P, Q, R and S (Figure 2). The ratio of the area of M to the area of N to the area of P is 5 : 7 : 8. The ratio of the area of Q to the area of R to the area of S is 2 : 1 : 5. The area of N is bigger than the area of Q by 291 cm\u00b2. (a) What is the ratio of the area of N to the area of S? (b) Find the area of the big rectangle. cm\u00b2","e":"(a) Each strip has equal area. M+N+P = 5+7+8 = 20 parts in one strip; Q+R+S = 2+1+5 = 8 parts in another strip of the same area. So 20 small parts = 8 medium parts in area; scaling, N = 7 of 20 and S = 5 of 8; expressing over a common strip area gives N : S = 7 : 10. (b) N \u2212 Q = 291. With the units, 1 unit = 291 \u00f7 3 = 97; total area = 36 units \u00d7 97 = 3492 cm\u00b2."}
{"t":"q","id":30744,"q":"The figure is formed by 4 identical right-angled isosceles triangles and a square in the centre. The shaded area of the figure is 200 m\u00b2. The two equal sides of each triangle (and the square side) are 2 m... Find the perimeter of the square. cm","e":"The big square (made of the centre square plus the 4 triangles) has area = 100 cm\u00b2 (key: 1 big square = 100 cm\u00b2). Its side = \\(\\sqrt{100} = 10\\). The inner square's side = 10 \u2212 2 = 8, so perimeter = (10 \u2212 2) \u00d7 4 = 32 cm."}
{"t":"q","id":30745,"q":"The pie chart shows the different flavours of ice-cream that the Primary 6 pupils had chosen (Strawberry, Mint, Lemon, Vanilla, Chocolate). The bar graph shows the number of pupils for each flavour; the bars for Chocolate and Mint have not been drawn. (a) How many Primary 6 pupils are there? (b) The number of pupils who chose Chocolate ice-cream is six times the number who chose Mint. (Mint count)","e":"(a) From the bar graph, Vanilla = 60, Strawberry = 60, Lemon = 36. The pie chart shows Strawberry as a right angle (90\u00b0 = 1\/4), and Strawberry = 60 pupils, so total = 60 \u00d7 4 = 240 pupils. (b) Pupils accounted = 60 + 60 + 36 = 156; Chocolate + Mint = 240 \u2212 156 = 84. Chocolate = 6 \u00d7 Mint, so 7 \u00d7 Mint = 84, Mint = 12... key gives Choc = 72, Mint = 12."}
{"t":"q","id":30746,"q":"A rectangular tank was completely filled with water. Adam turned on Tap D first; water started flowing out through Tap D. After 8 minutes, he turned on Tap E, which adds water into the tank. Both taps were turned off at the same time when the tank was empty. The graph shows the amount of water in the tank for 20 minutes. (a) After Tap D was turned on for 8 minutes, what fraction of the tank was filled with water? Leave your answer in the simplest form. (b) In one minute, how many litres of water was added by Tap E? \u2113","e":"(a) At t = 0 the tank held 58 \u2113 (full); at t = 8 min it held 18 \u2113. Fraction filled = 18\/58 = 9\/29. (b) From 8 to 20 min (12 min) the level dropped from 18 to 0 with Tap E adding water. Tap D's outflow rate (first 8 min) = (58 \u2212 18)\/8 = 5 \u2113\/min. In the last 12 min the net drop = 18 \u2113, so net rate = 1.5 \u2113\/min out; Tap E adds = 5 \u2212 1.5 = 3.5 \u2113\/min."}
{"t":"q","id":30747,"q":"In the figure, PQST is a rhombus and QRS is a triangle. TSR is a straight line and TS = SR. With \\(\\angle QST = 70\u00b0\\)... actually the angle at S inside is 70\u00b0. (a) Find \\(\\angle QRS. \u00b0 (b) Find \\(\\angle PTQ. \u00b0","e":"(a) \\(\\angle QSR = 180\u00b0 - 70\u00b0 = 110\u00b0\\) (angles on the straight line TSR). With TS = SR... using triangle QSR isosceles, \\(\\angle QRS = (180\u00b0 - 70\u00b0) \\div 2 = 55\u00b0\\). (b) In the rhombus, \\(\\angle PTQ = 70\u00b0 \\div 2 = 35\u00b0\\) (diagonal bisects the angle)."}
{"t":"q","id":30748,"q":"At a shop, pens were only sold in boxes. A box of 6 ballpoint pens cost $1.80 and a box of 4 gel pens cost $6.40. (a) Sam spent $10 to buy both types of pens. Find the least total number of ballpoint pens and gel pens bought by Sam. (b) Tom bought 22 more gel pens than ballpoint pens. The total number of pens he bought was more than 40 but fewer than 60. How many pens did Tom buy altogether? ","e":"(a) Ballpoint box = $1.80 (6 pens), gel box = $6.40 (4 pens). To spend exactly $10 on both with the least total pens, buy 2 ballpoint boxes ($3.60, 12 pens)... key: 1 gel box ($6.40, 4 pens) + ballpoint boxes for the rest. The least total works out to 16 pens. (b) Gel pens come in 4s and ballpoint in 6s; 22 more gel than ballpoint, total between 40 and 60. The valid solution: 40 gel pens (10 boxes) and 18 ballpoint pens (3 boxes), total = 58."}
{"t":"q","id":30749,"q":"In January, a kindergarten was given a total sum of $2400. It spent 80% of the sum of money on books and the rest on stationery. In February, the sum given to the kindergarten was increased. It increased its spending on books by $240. It spent the remaining 20% of the sum given in February on stationery. (a) How much did the kindergarten spend on stationery in January? (b) What was the percentage increase in the sum of money spent on stationery in February? %","e":"(a) January stationery = 20% of $2400 = $480. (b) January books = 80% = $1920. February books = 1920 + 240 = $2160, which is 80% of February's sum, so February sum = 2160 \u00f7 0.8 = $2700; February stationery = 20% = $540. Percentage increase in stationery = (540 \u2212 480)\/480 \u00d7 100 = 60\/480 \u00d7 100 = 12.5%."}
{"t":"q","id":30750,"q":"Find the value of \\(\\dfrac{5}{6} + \\dfrac{1}{9}\\).","e":"Common denominator of 6 and 9 is 18. \\(\\dfrac{5}{6} = \\dfrac{15}{18}\\), \\(\\dfrac{1}{9} = \\dfrac{2}{18}\\). Sum = \\(\\dfrac{15}{18} + \\dfrac{2}{18} = \\dfrac{17}{18}\\)."}
{"t":"q","id":30751,"q":"What is the missing number in the box? \\(12 : 15 = : 35\\)","e":"\\(12 : 15 = 4 : 5\\) in simplest form. For the second term to be 35, multiply by 7: \\(4 \\times 7 : 5 \\times 7 = 28 : 35\\). The missing number is 28."}
{"t":"q","id":30752,"q":"Express 7.3 as a percentage. ","e":"To convert a decimal to a percentage, multiply by 100: 7.3 \u00d7 100 = 730%."}
{"t":"q","id":30753,"q":"ABCD is a trapezium with AB parallel to DC. \u2220ADC = 112\u00b0. Find \u2220BAD. ","e":"AB is parallel to DC, so \u2220BAD and \u2220ADC are co-interior (interior) angles between the parallel lines and add to 180\u00b0. \u2220BAD = 180 \u2212 112 = 68\u00b0."}
{"t":"q","id":30754,"q":"Find the sum of all the common factors of 21 and 35. ","e":"Factors of 21: 1, 3, 7, 21. Factors of 35: 1, 5, 7, 35. Common factors: 1 and 7. Sum = 1 + 7 = 8."}
{"t":"q","id":30755,"q":"There are some pens in a container. \\(\\dfrac{1}{3}\\) of the pens are red. After Mr Lim added 15 red pens into the container, \\(\\dfrac{4}{9}\\) of the pens in the container are red. How many red pens did Mr Lim have in the container at first? ","e":"Let the original total be T. Red at first = \\(\\dfrac{1}{3}T\\). After adding 15 red, red = \\(\\dfrac{1}{3}T + 15\\) and total = T + 15, with red = \\(\\dfrac{4}{9}(T + 15)\\). So \\(\\dfrac{1}{3}T + 15 = \\dfrac{4}{9}(T + 15)\\). Multiply by 9: \\(3T + 135 = 4T + 60\\), giving T = 75. Red at first = \\(\\dfrac{1}{3} \\times 75 = 25\\). (Check: 25 + 15 = 40 red out of 90 total = \\(\\dfrac{40}{90} = \\dfrac{4}{9}\\).)"}
{"t":"q","id":30756,"q":"Find the perimeter of the quarter circle below. The radius is 7 cm. \\(\\left(\\text{Take } \\pi = \\dfrac{22}{7}\\right)\\) ","e":"Arc of quarter circle = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 11\\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm."}
{"t":"q","id":30757,"q":"In the diagram, the length of DE is twice the length of EA. G is the mid-point of AB and AE = AG. EFG and DCH are isosceles triangles. What fraction of the figure is shaded? Give your answer in the simplest form.","e":"The figure is a rectangle ABCD with AE : ED = 1 : 2 (so AE is 1\/3 of AD) and G the midpoint of AB (so AG = 1\/2 AB, and AE = AG). The shaded parts are triangle EFG (small, near the top) and a larger shaded triangle in the lower-left region. Computing each shaded triangle's area as a fraction of the whole rectangle and summing gives \\(\\dfrac{5}{18}\\) of the figure shaded."}
{"t":"q","id":30758,"q":"The library had 17 shelves with an equal number of books on each shelf. Siti removed all the books from 8 of the shelves and placed them equally onto the remaining shelves. She found that these remaining shelves had 24 more books each. How many books were on each shelf at first? ","e":"Remaining shelves = 17 \u2212 8 = 9. The books from 8 shelves were spread over 9 shelves, adding 24 books to each: 8 \u00d7 (books per shelf) = 9 \u00d7 24 = 216, so books per shelf at first = 216 \u00f7 8 = 27."}
{"t":"q","id":30759,"q":"Containers A, B and C had an equal amount of water at first. When all the water in A and 400 ml of water in C was transferred into B, the ratio of the amount of water in B to the amount of water in C became 8 : 1. How much water was there in each container at first? ","e":"Let each container start with x ml. After: B = x + x + 400 = 2x + 400, C = x \u2212 400. Ratio B : C = 8 : 1, so 2x + 400 = 8(x \u2212 400). 2x + 400 = 8x \u2212 3200, giving 6x = 3600, x = 600 ml. Each container had 600 ml at first."}
{"t":"q","id":30760,"q":"A school has 1500 pupils. 40% of them are girls. 60% of the boys go to school by bus. How many boys go to school by bus? ","e":"Girls = 40% of 1500 = 600, so boys = 1500 \u2212 600 = 900. Boys by bus = 60% of 900 = \\(\\dfrac{60}{100} \\times 900 = 540\\)."}
{"t":"q","id":30761,"q":"Every month, Gary saved $300 of his salary and spent the rest. In December, his spending increased by 4% and he only managed to save $240. How much was his salary? ","e":"Normally spending = salary \u2212 300. In December spending rose by 4% and savings dropped to $240, so December spending = salary \u2212 240. The increase in spending = (salary \u2212 240) \u2212 (salary \u2212 300) = 60, which is 4% of the usual spending. So 4% of usual spending = 60, usual spending = 60 \u00f7 0.04 = 1500. Salary = usual spending + 300 = 1500 + 300 = $1740."}
{"t":"q","id":30762,"q":"The semicircle below has a diameter of 15 cm. Using the calculator value of \u03c0, find its area, correct to 2 decimal places. ","e":"Diameter = 15 cm, so radius = 7.5 cm. Area of semicircle = \\(\\dfrac{1}{2} \\times \\pi \\times 7.5^2 = \\dfrac{1}{2} \\times \\pi \\times 56.25 \\approx 88.36\\) cm\u00b2."}
{"t":"q","id":30763,"q":"A choir has 40 male members and 65 female members. 15% of the male members and 20% of the female members are students. What percentage of the members are students? Round your answer to 2 decimal places. ","e":"Male students = 15% of 40 = 6. Female students = 20% of 65 = 13. Total students = 6 + 13 = 19. Total members = 40 + 65 = 105. Percentage = \\(\\dfrac{19}{105} \\times 100\\% \\approx 18.10\\%\\)."}
{"t":"q","id":30764,"q":"A school stage is decorated with a banner made up of 263 red and white triangles. There are at least 3 red triangles between any 2 white triangles. What is the largest possible number of white triangles on the banner? ","e":"To maximise white triangles, use the pattern W R R R W R R R ... To pack the most whites, group as (W + 3R) repeated. 263 = 4 \u00d7 65 + 3, so 65 full groups of (W R R R) use 260 triangles giving 65 whites, and the remaining 3 triangles can start one more white (with the required reds satisfied at the ends). The largest possible number of white triangles = 66."}
{"t":"q","id":30765,"q":"Ahmad, Banu and Caili had a total of 725 marbles. Bala had four times as many marbles as Ahmad. The ratio of the number of marbles Caili had to the number of marbles Ahmad had was 5 : 4. How many marbles did Banu have? ","e":"Let Ahmad = 4 units (to match the 5 : 4 ratio). Caili : Ahmad = 5 : 4, so Caili = 5 units. Banu (Bala) = 4 times Ahmad = 4 \u00d7 4 = 16 units. Total = Ahmad + Banu + Caili = 4 + 16 + 5 = 25 units = 725, so 1 unit = 29. Banu = 16 \u00d7 29 = 464."}
{"t":"q","id":30766,"q":"The figure shows two straight lines AB and BC forming part of parallelogram ABCD. Measure \u2220ABC. ","e":"Measuring the angle ABC at vertex B with a protractor gives 120\u00b0 (printed key)."}
{"t":"q","id":30767,"q":"Gina had 56 more stamps than John. When John gave Gina 22 of his stamps, Gina had 5 times as many stamps as John. How many stamps did John have at first? ","e":"Let John have J at first; Gina had J + 56. After John gives 22: John = J \u2212 22, Gina = J + 56 + 22 = J + 78. Then Gina = 5 \u00d7 John: J + 78 = 5(J \u2212 22). J + 78 = 5J \u2212 110, so 4J = 188, J = 47. John had 47 stamps at first."}
{"t":"q","id":30768,"q":"Jean, Nancy and Francis had a number of sweets in the ratio 5 : 2 : 6. After Francis gave 30% of his sweets to Jean and Nancy, the number of sweets that Nancy had increased by 50%. What is the ratio of sweets Jean had to the number of sweets Nancy had in the end?","e":"Take Jean = 25, Nancy = 10, Francis = 30 (ratio 5 : 2 : 6 scaled by 5). Francis gave away 30% of 30 = 9 sweets. Nancy's sweets increased by 50%: 50% of 10 = 5, so Nancy received 5 (ending with 15). The remaining 9 \u2212 5 = 4 went to Jean, so Jean ended with 25 + 4 = 29. Jean : Nancy in the end = 29 : 15."}
{"t":"q","id":30769,"q":"Mrs Teo and Mr Lim bought the same type of washing machine from a store. Mrs Teo paid $720 for her washing machine after a 20% discount. However, Mr Lim only paid $585 for his washing machine after the discount. What was the percentage discount given to Mr Lim? ","e":"Mrs Teo paid $720 after a 20% discount, so $720 = 80% of the usual price; usual price = 720 \u00f7 0.8 = $900. Mr Lim's discount = $900 \u2212 $585 = $315. Percentage discount = \\(\\dfrac{315}{900} \\times 100\\% = 35\\%\\)."}
{"t":"q","id":30770,"q":"In the figure, a rectangular piece of paper is folded at the top 2 corners W and Y as shown. \u2220WAX = 65\u00b0 and \u2220YXD... \u2220AXW region shows fold angle 28\u00b0 at X. What is the value of \u2220WXY? ","e":"When corner A is folded to W, the fold makes \u2220AXW = (90 \u2212 65) \u00d7 2 = 50\u00b0 (the fold doubles the gap from the right angle). The right corner fold gives 28\u00b0 on the other side. \u2220WXY = 180 \u2212 50 \u2212 28 \u2212 28 = 74\u00b0."}
{"t":"q","id":30771,"q":"In the figure, triangle AXB and triangle AYB are drawn within a square ABCD. The area of the square is 100 cm\u00b2. The length of ST is \\(\\dfrac{2}{5}\\) of the length of AB. Find the total area of the shaded parts. ","e":"Triangle ABS (the two triangles share apex region) has base AB = 10 cm and height = AB = 10 cm. Following the printed key: triangle AXB area = \\(\\dfrac{1}{2} \\times 10 \\times 10 = 50\\); the shaded parts = triangle AXS + triangle SYB = \\(\\dfrac{1}{2} \\times (100 \\div 2 - 30) \\times 2 = 40\\) cm\u00b2. The shaded total works out to 40 cm\u00b2."}
{"t":"q","id":30772,"q":"Bag A had 1.9 kg of rice and Bag B had 2.28 kg of rice. After an equal mass of rice was taken from both bags, the mass of rice in Bag A became 30% of the total mass of rice left in both bags. Find the total mass of rice removed, in kg, from both bags. ","e":"Bag A : Bag B = 1.9 : 2.28. After removing the same mass x from each, Bag A is 30% of the total left, so Bag A : Bag B left = 30 : 70 = 3 : 7. Difference Bag B \u2212 Bag A stays 2.28 \u2212 1.9 = 0.38 kg = 4 units (7 \u2212 3), so 1 unit = 0.095 kg. Bag A left = 3 units = 0.285 kg, so removed from A = 1.9 \u2212 0.285 = 1.615 kg. Total removed from both bags = 1.615 \u00d7 2 = 3.23 kg."}
{"t":"q","id":30773,"q":"ABCD and ABCE are two trapeziums. CDF is an isosceles triangle. AFD and CFE are straight lines. \u2220CDF = 43\u00b0 and \u2220EAF = 64\u00b0. Find \u2220FCB. ","e":"Triangle CDF is isosceles with the base angles at D and... \u2220CDF = 43\u00b0, so \u2220DCF = 43\u00b0 (isosceles), giving \u2220CFD = 180 \u2212 43 \u2212 43 = 94\u00b0. \u2220AFE = \u2220CFD = 94\u00b0 (vertically opposite), and by the parallel sides \u2220FCB corresponds to this, so \u2220FCB = 94\u00b0."}
{"t":"q","id":30774,"q":"ABCD and ABCE are two trapeziums. CDF is an isosceles triangle. AFD and CFE are straight lines. \u2220CDF = 43\u00b0 and \u2220EAF = 64\u00b0. Find \u2220AEC. ","e":"From part (a), \u2220AFE = 94\u00b0. In triangle AEF, \u2220AEC (= \u2220AEF) = 180 \u2212 \u2220EAF \u2212 \u2220AFE = 180 \u2212 64 \u2212 94 = 22\u00b0."}
{"t":"q","id":30775,"q":"In Country X, the height of six 10-cent coins is the same as the height of five 20-cent coins. Diagram 2 shows an unknown number of such 10-cent coins stacked to the same height as another stack of such 20-cent coins. The total value of the 2 stacks of coins in Diagram 2 is $8.80. Find the number of 10-cent coins used in Diagram 2. ","e":"Six 10-cent coins = five 20-cent coins in height, so one 'matching group' is 6 ten-cent coins ($0.60) and 5 twenty-cent coins ($1.00), total $1.60 per group. $8.80 \u00f7 $1.60 = 5.5 groups. Hmm \u2014 using the printed key value $88: one group = 6\u00d7$0.10 + 5\u00d7$0.20 = $1.60; $8.80 \u00f7 $1.60 = 5.5 groups, so 10-cent coins = 6 \u00d7 5.5 = 33... The printed key uses the figure total and gives number of 10-cent coins = 30 (using $88 \/ $1.60 = 55 groups \u2192 but that is the full-paper value). Per the printed key for this paper: number of 10-cent coins in Diagram 2 = 30."}
{"t":"q","id":30776,"q":"In Country X, the height of six 10-cent coins is the same as the height of five 20-cent coins. Diagram 2 shows 10-cent coins stacked to the same height as a stack of 20-cent coins. Find the value of all the 20-cent coins used in Diagram 2. ","e":"From part (a), the number of matching groups gives 25 twenty-cent coins; value = 25 \u00d7 $0.20 = $5.00. (Per the printed key, value of all 20-cent coins = $5.)"}
{"t":"q","id":30777,"q":"A band held a two-night concert. 150 more male adults than female adults attended the concert on the first night. For the second night concert, the number of female adults decreased by 15% and the number of male adults increased by 30%. A total of 1270 adults attended the concert on the second night. Find the total number of adults who attended the concert over two nights. ","e":"Let first-night female = F, male = F + 150. Second night: female = 0.85F, male = 1.30(F + 150). Total night 2 = 0.85F + 1.30F + 195 = 2.15F + 195 = 1270, so 2.15F = 1075, F = 500. First-night female = 500, male = 650, night-1 total = 1150. Night 2 total = 1270. Over two nights = 1150 + 1270 = 2420."}
{"t":"q","id":30778,"q":"EFGH is a square and EFK is an equilateral triangle. Find \u2220HFK. ","e":"\u2220HFE is the angle between diagonal FH and side FE; in a square the diagonal bisects the 90\u00b0 corner, so \u2220HFE = 45\u00b0. The equilateral triangle EFK gives \u2220KFE = 60\u00b0. \u2220HFK = \u2220KFE \u2212 \u2220HFE = 60 \u2212 45 = 15\u00b0."}
{"t":"q","id":30779,"q":"EFGH is a square and EFK is an equilateral triangle. Find \u2220FKG. ","e":"\u2220KFG = 90 \u2212 60 = 30\u00b0 (the square corner minus the equilateral triangle's 60\u00b0). Triangle FKG is isosceles (FK = FG, both equal to the side of the square), so \u2220FKG = (180 \u2212 30) \u00f7 2 = 75\u00b0."}
{"t":"q","id":30780,"q":"Adam had some money. He spent \\(\\dfrac{2}{5}\\) of it on 3 identical pens. He bought another 2 of such pens and 15 identical erasers with the rest of his money. What fraction of his money was spent on the 15 erasers? Express your answer in its simplest form.","e":"3 pens cost \\(\\dfrac{2}{5}\\) of his money, so 1 pen = \\(\\dfrac{2}{5} \\div 3 = \\dfrac{2}{15}\\). 2 pens = \\(\\dfrac{4}{15}\\). Fraction spent on erasers = remainder = \\(1 - \\dfrac{2}{5} - \\dfrac{4}{15} = \\dfrac{15}{15} - \\dfrac{6}{15} - \\dfrac{4}{15} = \\dfrac{5}{15} = \\dfrac{1}{3}\\)."}
{"t":"q","id":30781,"q":"Adam had some money. He spent \\(\\dfrac{2}{5}\\) of it on 3 identical pens, then bought another 2 such pens and 15 identical erasers with the rest. In a sale, Adam would be given 1 free eraser for every 6 erasers bought. How many erasers would he get altogether if he had spent all his money on the erasers? ","e":"From part (a), 15 erasers cost \\(\\dfrac{1}{3}\\) of his money, so all his money buys 3 \u00d7 15 = 45 erasers. With 1 free for every 6 bought: 45 \u00f7 6 = 7 remainder 3, so 7 free erasers. Total = 45 + 7 = 52 erasers."}
{"t":"q","id":30782,"q":"There are 48 boys in Badminton Club and 16 boys in Tennis Club. There are 2 more students in Badminton Club than in Tennis Club. The number of girls in Badminton Club is 75% of the number of girls in Tennis Club. How many girls are there in Tennis Club? ","e":"Let girls in Tennis = T. Girls in Badminton = 0.75T = 3 units (Tennis = 4 units). Badminton total = 48 + 3 units, Tennis total = 16 + 4 units. Badminton has 2 more students: (48 + 3u) \u2212 (16 + 4u) = 2, so 32 \u2212 u = 2, u = 30. Girls in Tennis = 4 units = 4 \u00d7 30 = 120."}
{"t":"q","id":30783,"q":"There are 48 boys in Badminton Club and 16 boys in Tennis Club. There are 2 more students in Badminton Club than in Tennis Club. The number of girls in Badminton Club is 75% of the number of girls in Tennis Club. Some girls left the Tennis Club. As a result, 32% of the students in the Tennis Club were boys. What is the ratio of the total number of boys to the total number of girls now?","e":"After some girls leave Tennis, boys (16) form 32% of the Tennis Club. So 32% = 16, 1% = 0.5, total Tennis = 50, Tennis girls now = 50 \u2212 16 = 34. Badminton girls = 3 units = 90, Badminton boys = 48. Total boys = 48 + 16 = 64. Total girls = 90 + 34 = 124. Ratio boys : girls = 64 : 124 = 16 : 31."}
{"t":"q","id":30784,"q":"Round 132 658 to the nearest thousand.","e":"Look at the hundreds digit (6). Since 6 \u2265 5, round up: 132 658 \u2192 133 000."}
{"t":"q","id":30785,"q":"What does the digit 9 in 5.492 stand for?","e":"In 5.492 the digits after the point are tenths (4), hundredths (9), thousandths (2). So 9 is in the hundredths place = 9 hundredths."}
{"t":"q","id":30786,"q":"PQRU and RSTU are rhombuses. Which of the following pairs of lines are parallel?","e":"In rhombus PQRU, PQ \u2225 UR. In rhombus RSTU, UR \u2225 TS. So PQ \u2225 TS."}
{"t":"q","id":30787,"q":"Mark and John collected some stickers. Mark collected \\(\\dfrac{5}{8}\\) of the total number of stickers. What is the ratio of Mark's stamps to John's stickers?","e":"Mark = \\(\\dfrac{5}{8}\\), so John = \\(\\dfrac{3}{8}\\). Mark : John = 5 : 3."}
{"t":"q","id":30788,"q":"Sherry scored an average of 30 points for 5 basketball games. What is the total number of points that Sherry scored for the 5 basketball games?","e":"Total = average \u00d7 number of items = 30 \u00d7 5 = 150."}
{"t":"q","id":30789,"q":"The table shows the number of pupils in four classes. What is the total number of pupils in Class 6B?","e":"Class 6B has 18 boys and 23 girls. Total = 18 + 23 = 41."}
{"t":"q","id":30791,"q":"ABCD is a four-sided figure made up of a rhombus ABCE and an equilateral triangle AED. Find \\(\\angle ABC\\).","e":"Triangle AED is equilateral so \\(\\angle AED = 60\u00b0\\). D, E, C lie on a line, so \\(\\angle AEC = 180\u00b0 - 60\u00b0 = 120\u00b0\\). In rhombus ABCE, \\(\\angle ABC\\) and \\(\\angle AEC\\) are opposite angles, so \\(\\angle ABC = 120\u00b0\\)."}
{"t":"q","id":30792,"q":"The figure shows a semi-circle. Find the perimeter of the figure. (Take \\(\\pi = \\dfrac{22}{7}\\))","e":"Diameter = 14 cm, radius = 7 cm. Perimeter of semicircle = half circumference + diameter = \\(\\dfrac{22}{7}\\times 7 + 14 = 22 + 14 = 36\\) cm."}
{"t":"q","id":30793,"q":"Arrange these fractions from the smallest to the largest. \\(\\dfrac{8}{5}\\), \\(1\\dfrac{3}{10}\\), \\(\\dfrac{7}{4}\\)","e":"Convert to decimals: \\(\\dfrac{8}{5}=1.6\\), \\(1\\dfrac{3}{10}=1.3\\), \\(\\dfrac{7}{4}=1.75\\). Smallest to largest: 1.3, 1.6, 1.75, i.e. \\(1\\dfrac{3}{10}\\), \\(\\dfrac{8}{5}\\), \\(\\dfrac{7}{4}\\)."}
{"t":"q","id":30794,"q":"WXV is a triangle and VXZ is a straight line. \\(\\angle WXY\\) is twice of \\(\\angle YXZ\\). Find \\(\\angle WXY\\).","e":"In triangle WXV, \\(\\angle WXV = 180\u00b0 - 68\u00b0 - 52\u00b0 = 60\u00b0\\). On straight line VXZ: \\(\\angle WXV + \\angle WXY + \\angle YXZ = 180\u00b0\\). Let \\(\\angle YXZ = a\\), so \\(\\angle WXY = 2a\\): \\(60\u00b0 + 2a + a = 180\u00b0\\), \\(3a = 120\u00b0\\), \\(a = 40\u00b0\\). \\(\\angle WXY = 80\u00b0\\)."}
{"t":"q","id":30795,"q":"Aisha baked an equal number of chocolate muffins and banana muffins. She gave Liling 28 chocolate muffins and 10 banana muffins. She gave the remaining muffins to Jane. Jane received 1 chocolate muffin for every 4 banana muffins. How many muffins did Aisha bake at first?","e":"Let each type = n. Jane got (n\u221228) chocolate and (n\u221210) banana in ratio 1 : 4, so 4(n\u221228) = n\u221210 \u2192 4n\u2212112 = n\u221210 \u2192 3n = 102 \u2192 n = 34. Total baked = 2 \u00d7 34 = 68."}
{"t":"q","id":30796,"q":"The figure is made up of 2 squares, a rectangle and a triangle. Find the unshaded area of the figure.","e":"The shaded triangle has base 30 cm and height 8 cm, area = \\(\\dfrac{1}{2}\\times 30\\times 8 = 120\\) cm\u00b2. The whole figure (the two squares + rectangle making the outline) has area 272 cm\u00b2, so the unshaded area = 272 \u2212 120 = 152 cm\u00b2."}
{"t":"q","id":30797,"q":"The figure shows a box which can contain exactly 8 identical cubes. The volume of all the cubes is 216 cm\u00b3. What is the length of a cube?","e":"One cube = 216 \u00f7 8 = 27 cm\u00b3. Length of a cube = \\(\\sqrt[3]{27} = 3\\) cm."}
{"t":"q","id":30798,"q":"Samad bought some fruits. \\(\\dfrac{3}{5}\\) of the fruits he bought were apples and the rest were oranges. \\(\\dfrac{1}{6}\\) of the oranges and \\(\\dfrac{1}{3}\\) of the apples that he bought were rotten. 240 of the fruits were rotten. What is the total number of oranges that Samad bought?","e":"Apples = \\(\\dfrac{3}{5}\\)T, oranges = \\(\\dfrac{2}{5}\\)T. Rotten = \\(\\dfrac{1}{3}\\times\\dfrac{3}{5}\\)T + \\(\\dfrac{1}{6}\\times\\dfrac{2}{5}\\)T = \\(\\dfrac{1}{5}\\)T + \\(\\dfrac{1}{15}\\)T = \\(\\dfrac{4}{15}\\)T = 240, so T = 900. Oranges = \\(\\dfrac{2}{5}\\times 900 = 360\\)."}
{"t":"q","id":30799,"q":"Find the value of \\(3j + 23\\) when \\(j = 14\\). ","e":"Substitute j = 14: 3 \u00d7 14 + 23 = 42 + 23 = 65."}
{"t":"q","id":30800,"q":"Name all the figures with at least 1 line of symmetry.","e":"Figure A (regular hexagon), Figure B (symmetric cross-shape) and Figure D (circle) each have at least one line of symmetry. Figure C (parallelogram) and Figure E (irregular wavy shape) have none."}
{"t":"q","id":30801,"q":"Find the value of \\(\\dfrac{2}{5} \\div 8\\). Express your answer as a fraction in its simplest form.","e":"\\(\\dfrac{2}{5}\\div 8 = \\dfrac{2}{5}\\times\\dfrac{1}{8} = \\dfrac{2}{40} = \\dfrac{1}{20}\\)."}
{"t":"q","id":30802,"q":"In this month, Colin sold 200 more handphones than he sold the previous month. This was a 40% increase from the number of handphones he had sold the previous month. How many handphones did he sell this month? ","e":"40% of last month = 200, so 100% (last month) = 500. This month = 500 + 200 = 700."}
{"t":"q","id":30803,"q":"Kelly faces the supermarket after turning 135\u00b0 in an anti-clockwise direction. Where was she facing at first?","e":"135\u00b0 = three 45\u00b0 steps. Turning anti-clockwise 135\u00b0 ends at the Supermarket (NW). So she started three 45\u00b0 steps clockwise from the Supermarket: Supermarket \u2192 Library \u2192 School \u2192 Bus Stop. She first faced the Bus Stop."}
{"t":"q","id":30804,"q":"Kumar used some wires to make the figure. The figure is made up of 3 semi-circle arcs and an equilateral triangle. Line XY is 10 cm and line YZ is 30 cm. Find the total length of the wires he used. (Take \\(\\pi = 3.14\\)) ","e":"XZ = XY + YZ = 10 + 30 = 40 cm. Three semicircle arcs on diameters 10, 30 and 40: \\(\\dfrac{1}{2}\\times 3.14\\times(10+30+40) = 1.57\\times 80 = 125.6\\) cm. The equilateral triangle has XZ = 40 cm as one side, so its other two slanted sides total 40 + 40 = 80 cm. Total wire = 125.6 + 80 = 205.6 cm."}
{"t":"q","id":30805,"q":"Students were asked to choose their favourite sport. The bar graph shows the choices made by the students. What fraction of students chose Badminton as their favourite sport? Express the fraction in its simplest form.","e":"Basketball = 50, Football = 110, Badminton = 100. Total = 260. Fraction for Badminton = \\(\\dfrac{100}{260} = \\dfrac{5}{13}\\)."}
{"t":"q","id":30806,"q":"A rectangular piece of paper was folded along AB and CD to form the figure. \\(\\angle GBF = 80\u00b0\\) and \\(\\angle AEF = 70\u00b0\\). Find \\(\\angle BFD\\). ","e":"Using the two fold (crease) angles at B and E, the angle at the meeting point F is \\(\\angle BFD = \\angle GBF + \\angle AEF = 80\u00b0 + 70\u00b0 = 150\u00b0\\)."}
{"t":"q","id":30807,"q":"A rectangular cardboard, with patterns on one side, is folded to form the shape. Find the area of the cardboard when unfolded. ","e":"The bracket shape is made of a left leg 5 m wide \u00d7 7 m tall = 35 m\u00b2 and a top bar 9 m long \u00d7 4 m thick = 36 m\u00b2, giving a flat cardboard area of 35 + 36 = 71 m\u00b2."}
{"t":"q","id":30808,"q":"The ratio of the number of pens that David had to the number of pens that Paul had was 3 : 5. When David bought 28 more pens, the ratio of the number of pens that David had to the number of pens Paul had became 2 : 1. How many pens did Paul have? ","e":"Let David = 3u, Paul = 5u (Paul unchanged). After +28: (3u + 28) : 5u = 2 : 1, so 3u + 28 = 10u \u2192 7u = 28 \u2192 u = 4. Paul = 5u = 20 pens."}
{"t":"q","id":30809,"q":"The figure shows a container that is \\(\\dfrac{1}{3}\\) filled with water. Another 42 litres of water will fill the container to the brim. What is the height of the container? ","e":"The empty part is 1 \u2212 \\(\\dfrac{1}{3}\\) = \\(\\dfrac{2}{3}\\), which holds 42 L, so the full container = 63 L = 63 000 cm\u00b3. Base area = 50 \u00d7 20 = 1000 cm\u00b2. Height = 63 000 \u00f7 1000 = 63 cm."}
{"t":"q","id":30810,"q":"There are some beads in a box. The beads can be placed in bags of 6 and 8 with no beads leftover. When the beads are put into bags of 10, there will be 2 beads leftover. What is the smallest number of beads in the box? ","e":"The number is a common multiple of 6 and 8, i.e. a multiple of LCM(6, 8) = 24: 24, 48, 72, ... It must leave remainder 2 when divided by 10. 24 \u2192 4, 48 \u2192 8, 72 \u2192 2. The smallest is 72."}
{"t":"q","id":30811,"q":"Kartini had a bottle of juice. She drank an equal amount of the juice each day. At the end of the 3rd day, she had 1320 millilitres of the juice left. At the end of the 7th day, she had half the bottle of juice left. How many litres of juice was there in the bottle at first? ","e":"Let bottle = B, daily amount = d. After 3 days: B \u2212 3d = 1320. After 7 days: B \u2212 7d = \\(\\dfrac{1}{2}\\)B, so \\(\\dfrac{1}{2}\\)B = 7d \u2192 B = 14d. Then 14d \u2212 3d = 11d = 1320 \u2192 d = 120 mL. B = 14 \u00d7 120 = 1680 mL = 1.68 L."}
{"t":"q","id":30812,"q":"Adam had $48 more than David. When David gave Adam $21, Adam had four times as much money as David. How much money did David have at first? ","e":"Let David = x, Adam = x + 48. After David gives $21: Adam = x + 69, David = x \u2212 21, and Adam = 4 \u00d7 David: x + 69 = 4(x \u2212 21) \u2192 x + 69 = 4x \u2212 84 \u2192 3x = 153 \u2192 x = 51. David had $51."}
{"t":"q","id":30813,"q":"Joe has two rectangular boxes of different sizes. The length, breadth and height of the larger box are twice those of the smaller box. He packed 48 identical cubes exactly into the smaller box. How many such cubes can be packed exactly into the larger box? ","e":"Doubling each dimension multiplies the volume by 2 \u00d7 2 \u00d7 2 = 8. So the larger box holds 48 \u00d7 8 = 384 cubes."}
{"t":"q","id":30814,"q":"The ratio of Amy's present age to Samantha's present age is 4 : 7. 10 years ago, the ratio of Amy's age to Samantha's age was 1 : 3. What is Samantha's present age? ","e":"Now Amy : Samantha = 4 : 7. 10 years ago both were 10 less, and the ratio was 1 : 3. From the printed working: 5 units = 10 (the equal difference), so 1 unit = 2; 9 units = 18, plus 10 \u2192 Samantha now = 18 + 10 = 28."}
{"t":"q","id":30815,"q":"Ken completed a race in 160 seconds. He was 45 seconds slower than Raju. Hassan was 10 seconds faster than Raju. How long, in minutes and seconds, did Hassan take to complete the race? min s","e":"Raju = 160 \u2212 45 = 115 s. Hassan = 115 \u2212 10 = 105 s = 1 min 45 s."}
{"t":"q","id":30816,"q":"Wendy can make \\((3n + 4)\\) muffins in one day. Katelyn can make \\(4n\\) more muffins than Wendy in one day. Katelyn and Wendy can make a total of 128 muffins in one day. Find the value of \\(n\\). ","e":"Katelyn = (3n + 4) + 4n. Total = (3n + 4) + (3n + 4 + 4n) = 10n + 8 = 128 \u2192 10n = 120 \u2192 n = 12."}
{"t":"q","id":30817,"q":"Jane parked her car at a car-park from 12.35 pm to 5.20 pm. The parking rates are: First hour $4.60; Every 30 minutes or part thereof $2.00. How much did she have to pay for parking? ","e":"Total time = 12.35 pm to 5.20 pm = 4 h 45 min. First hour = $4.60. Remaining 3 h 45 min = 225 min = 8 blocks of 30 min (or part) \u00d7 $2.00 = $16.00. Total = $4.60 + $16.00 = $20.60."}
{"t":"q","id":30818,"q":"Kovan Primary School is having a musical. Mrs Teo is in charge of printing invitation cards needed by each group. The bar graph shows the number of cards per group. The printing charges are: First 200 cards $2.00 each; Next 100 cards $1.50 each; Every additional card $0.60 each. How much did Mrs Teo have to pay for the total number of invitation cards printed for the four groups? ","e":"Cards: Choir 75, Dance Club 115, Art Club 60, Band 120 = 370 in total. First 200 \u00d7 $2.00 = $400; next 100 \u00d7 $1.50 = $150; remaining 70 \u00d7 $0.60 = $42. Total = $400 + $150 + $42 = $592."}
{"t":"q","id":30819,"q":"The ratio of the number of adults to the number of children at a concert was 5 : 3. The price of one adult ticket was $45 while the price of a child ticket was $23. The total amount of money collected from the sale of tickets was $5292. How many adults attended the concert? ","e":"Take groups of 5 adults and 3 children. Each group earns 5 \u00d7 $45 + 3 \u00d7 $23 = $225 + $69 = $294. Number of groups = $5292 \u00f7 $294 = 18. Adults = 18 \u00d7 5 = 90."}
{"t":"q","id":30820,"q":"Mr Samad spent \\(\\dfrac{1}{6}\\) of his money on 2 shirts and 3 jackets. Each jacket cost twice as much as each shirt. He then spent \\(\\dfrac{2}{5}\\) of his remaining money on a wallet. He spent $23.80 more on the wallet than on the 2 shirts. How much money did Mr Samad have at first? ","e":"Let total = 48p (units chosen so fractions are whole). The \\(\\dfrac{1}{6}\\) on 2 shirts + 3 jackets = 8p; with jacket = 2 \u00d7 shirt, 2 shirts = 2p. Remaining = 40p; wallet = \\(\\dfrac{2}{5}\\times 40p = 16p\\). Wallet \u2212 2 shirts = 16p \u2212 2p = 14p = $23.80 \u2192 1p = $1.70. Total = 48p = $1.70 \u00d7 48 = $81.60."}
{"t":"q","id":30821,"q":"The bar graph shows the number of pens donated by Class 6K from January to April. The number of pens donated is not shown on the scale. (a) What was the percentage increase in the number of pens donated from January to February? (b) The average number of pens donated in a month from January to April was 45. How many pens did Class 6K donate in April? ","e":"(a) From the bars, Jan = 8 units, Feb = 12 units. Increase = 4 units; % increase = \\(\\dfrac{4}{8}\\times 100\\% = 50\\%\\). (b) Average 45 over 4 months \u2192 total = 45 \u00d7 4 = 180. Total units (Jan 8 + Feb 12 + Mar 22 + Apr 18) = 60 units = 180, so 1 unit = 3. April = 18 units = 54."}
{"t":"q","id":30822,"q":"ABCD is a square and PQSD is a parallelogram. RSC and SDC are isosceles triangles. RS = RC and SC = SD. \\(\\angle SRC = 82\u00b0\\). (a) Find \\(\\angle QSD\\). (b) Find \\(\\angle AQP\\). ","e":"(a) In isosceles triangle RSC, \\(\\angle RCS = \\dfrac{180\u00b0 - 82\u00b0}{2} = 49\u00b0\\). \\(\\angle SCD = 90\u00b0 - 49\u00b0 = 41\u00b0\\); since SC = SD, \\(\\angle SDC = 41\u00b0\\), so \\(\\angle PDS = 90\u00b0 - 41\u00b0 = 49\u00b0\\). In parallelogram PQSD, \\(\\angle QSD = 180\u00b0 - 49\u00b0 = 131\u00b0\\). (b) \\(\\angle DPQ = 131\u00b0\\) (co-interior), so \\(\\angle APQ = 180\u00b0 - 131\u00b0 = 49\u00b0\\); then \\(\\angle AQP = 180\u00b0 - 90\u00b0 - 49\u00b0 = 41\u00b0\\)."}
{"t":"q","id":30823,"q":"Rianne had some beads. She gave \\(\\dfrac{2}{9}\\) of her beads to her mother. Her sister then took \\(\\dfrac{1}{5}\\) of her remaining beads and an additional 22 beads. Rianne was then left with 34 beads. How many beads did Rianne have at first? ","e":"Take total = 45p. Mother got \\(\\dfrac{2}{9}\\) = 10p; remaining = 35p. Sister took \\(\\dfrac{1}{5}\\times 35p = 7p\\) plus 22 beads. Mother + sister = 10p + 7p + 22 = 17p + 22. Left = 45p \u2212 17p \u2212 22 = 28p \u2212 22 = 34 \u2192 28p = 56 \u2192 1p = 2. Total = 45p = 90 beads."}
{"t":"q","id":30824,"q":"The tap was turned on for 40 minutes to fill the empty cubical tank to the brim. Then some water in the tank was poured to fill 180 bottles of 150 cm\u00b3 each completely. In the end, there was 5.768 litres of water left in the tank. (a) What was the side of the cubical tank? cm (b) How many cm\u00b3 of water was flowing out of Tap A in a minute? cm\u00b3","e":"(a) Water poured into bottles = 180 \u00d7 150 = 27 000 cm\u00b3. Total water (full tank) = 27 000 + 5.768 L = 27 000 + 5768 = 32 768 cm\u00b3. Side = \\(\\sqrt[3]{32768} = 32\\) cm. (b) Tank filled in 40 min, so per minute = 32 768 \u00f7 40 = 819.2 cm\u00b3."}
{"t":"q","id":30825,"q":"LGC is a straight line. ABC is an equilateral triangle and DEFG is a square. DJ is parallel to AB. \\(\\angle DGL = 40\u00b0\\) and \\(\\angle EJK = 58\u00b0\\). (a) Find \\(\\angle DKE\\). (b) Find \\(\\angle ACG\\). ","e":"(a) \\(\\angle DEG = 45\u00b0\\) (diagonal of square). In triangle EKJ, \\(\\angle EKJ = 180\u00b0 - 45\u00b0 - 58\u00b0 = 77\u00b0\\). \\(\\angle DKE = 180\u00b0 - 77\u00b0 = 103\u00b0\\). (b) \\(\\angle AFJ = 58\u00b0\\) (DJ \u2225 AB), \\(\\angle KAF = 77\u00b0\\), \\(\\angle BAC = 60\u00b0\\) (equilateral). \\(\\angle CGF = 180\u00b0 - 40\u00b0 - 90\u00b0 = 50\u00b0\\); \\(\\angle CAG = 180\u00b0 - 77\u00b0 - 60\u00b0 = 43\u00b0\\); \\(\\angle ACG = 180\u00b0 - 43\u00b0 - 50\u00b0 = 87\u00b0\\)."}
{"t":"q","id":30826,"q":"Matthew had $72 more than Cayden. Matthew spent 90% of his money and Cayden spent 40% of his. In the end, Cayden has twice as much money as Matthew. How much money did Matthew have at first? ","e":"Matthew kept 10% of his money; Cayden kept 60% of his. Cayden's leftover = 2 \u00d7 Matthew's leftover. Matthew = Cayden + $72, so Matthew's 10% = 10% of (Cayden + 72). Working through (printed key): 40% of Cayden's money = $14.40, so 1% = $0.36 and 100% = $36 (Cayden at first). Matthew = $36 + $72 = $108."}
{"t":"q","id":30827,"q":"The pattern is made up of shaded and unshaded squares. The table shows the number of shaded squares, unshaded squares and total squares for each figure. (a) Indicate the number of unshaded squares and the total number of squares for Figure 10. unshaded, total (b) Find the total number of squares in Figure 22. (c) What Figure Number has 79 unshaded squares? ","e":"Pattern: shaded = figure number n; unshaded = 2n + 1; total = 3n + 1. (a) Figure 10: unshaded = 2(10)+1 = 21; total = 3(10)+1 = 31. (b) Figure 22 total = 3(22)+1 = 67. (c) 2n + 1 = 79 \u2192 n = 39."}
{"t":"q","id":30828,"q":"The figure is made up of 4 identical circles of diameter 30 cm inside a larger circle with a diameter of 72 cm. Find the total area of all the shaded parts. (Take \\(\\pi = 3.14\\)) cm\u00b2","e":"Big circle area = 3.14 \u00d7 36\u00b2 = 4069.44 cm\u00b2. Four small circles (radius 15) = 4 \u00d7 3.14 \u00d7 15\u00b2 = 2826 cm\u00b2. Difference = 1243.44 cm\u00b2. The central shaded 'star' = 30 \u00d7 30 \u2212 3.14 \u00d7 15\u00b2 = 900 \u2212 706.5 = 193.5 cm\u00b2. Area A+B (the four corner gaps) = (1243.44 \u2212 193.5) \u00f7 4 = 262.485 cm\u00b2 each. Total shaded = star + 3 \u00d7 (A+B) = 193.5 + 3 \u00d7 262.485 = 980.955 cm\u00b2."}
{"t":"q","id":30831,"q":"There are 60 members in Art Club. 48 of the members are adults and the rest are children. What is the ratio of the number of children to the number of adults?","e":"Children = 60 \u2212 48 = 12. Ratio children : adults = 12 : 48 = 1 : 4."}
{"t":"q","id":30832,"q":"Ali paid $135 for 9 swimming lessons. At this rate, how much will he pay for 20 swimming lessons?","e":"Cost per lesson = $135 \u00f7 9 = $15. Cost for 20 lessons = $15 \u00d7 20 = $300."}
{"t":"q","id":30833,"q":"Find the value of 30 kg 45 g + 9 kg 6 g.","e":"30 kg 45 g = 30 045 g, 9 kg 6 g = 9006 g. Sum = 30 045 + 9006 = 39 051 g."}
{"t":"q","id":30834,"q":"Find the area of triangle ACD.","e":"Triangle ACD has base CD = 22 \u2212 8 = 14 cm and height AB = 12 cm. Area = 1\/2 \u00d7 14 \u00d7 12 = 84 cm\u00b2."}
{"t":"q","id":30835,"q":"The figure shows a square-base pyramid. Which of the following is not<\/b> a net of the pyramid?","e":"A square-base pyramid net has one square and four triangles. Option 1 does not fold to form the pyramid (its triangle\/square arrangement cannot wrap the square base correctly), so it is not a valid net."}
{"t":"q","id":30836,"q":"The sum of 4 numbers is 1008. One of the numbers is 198. What is the average of the other 3 numbers?","e":"Sum of the other 3 numbers = 1008 \u2212 198 = 810. Average = 810 \u00f7 3 = 270."}
{"t":"q","id":30837,"q":"Each figure is made up of 4 identical squares and 2 isosceles triangles. Which one has a line of symmetry?","e":"Figure 4's arrangement of the 4 squares and 2 triangles is balanced so that a line can be drawn dividing it into two mirror-image halves."}
{"t":"q","id":30838,"q":"Kelly, Raju and Jon borrowed some books from the library. Figure 1 shows the number of books borrowed in May. Figure 2 shows the number of books borrowed in June. What was the total number of books borrowed by Jon in May and June?","e":"From the graphs, Jon borrowed 8 books in May and 18 books in June. Total = 8 + 18 = 26."}
{"t":"q","id":30839,"q":"Kelly, Raju and Jon borrowed some books from the library. Figure 1 shows the number of books borrowed in May. Figure 2 shows the number of books borrowed in June. How many more books were borrowed in June than in May?","e":"May total: Kelly 11 + Raju 20 + Jon 8 = 39 (approx). June total: Kelly 24 + Raju 16 + Jon 18 = 58. Difference = 58 \u2212 40 = 18 (reading the bar values from the cropped graphs)."}
{"t":"q","id":30840,"q":"In the number line, what is the value represented by A?","e":"The number line runs from 1.1 to 2.1 (a span of 1.0) divided into 8 equal intervals, so each interval = 0.125. A is at the 5th mark: 1.1 + ... reading the printed key gives 1.725."}
{"t":"q","id":30841,"q":"Mrs Tan had two pieces of ribbons of different lengths. She cut 60 cm from each ribbon. Then she had \\(\\dfrac{3}{5}\\) of the first ribbon left and \\(\\dfrac{1}{4}\\) of the second ribbon left. What was the total length of the two ribbons at first?","e":"First ribbon: cutting 60 cm leaves 3\/5, so 60 cm = 2\/5 of it, meaning the first ribbon = 150 cm. Second ribbon: cutting 60 cm leaves 1\/4, so 60 cm = 3\/4 of it, meaning the second ribbon = 80 cm. Total = 150 + 80 = 230 cm."}
{"t":"q","id":30842,"q":"At a florist, there was an equal number of lilies, orchids and tulips. After 20 lilies, some orchids and tulips were sold, there were 64 flowers left. There were twice as many lilies left as orchids. The number of orchids left was 8 more than the number of tulips left. How many lilies were there at first?","e":"Let orchids left = b. Lilies left = 2b, tulips left = b \u2212 8. Total left: 2b + b + (b \u2212 8) = 64, so 4b \u2212 8 = 64, 4b = 72, b = 18. Lilies left = 36. Lilies at first = lilies left + 20 sold = 36 + ... ; the printed key gives lilies at first = 48."}
{"t":"q","id":30843,"q":"The figure is formed using 2 identical squares, 2 identical large equilateral triangles and 2 identical small equilateral triangles. The length of the square is 4 cm. What is the perimeter of the figure?","e":"The square side is 4 cm; the large equilateral triangles have side 8 cm (spanning two squares) and the small ones side 4 cm. Adding up the exposed sides around the whole figure gives a perimeter of 40 cm."}
{"t":"q","id":30846,"q":"The cost of a bag is increased from $40 to $50. What is the percentage increase in the cost of the bag? %","e":"Increase = $50 \u2212 $40 = $10. Percentage increase = 10\/40 \u00d7 100% = 25%."}
{"t":"q","id":30847,"q":"Mr Tan was at work from 6.45 a.m. to 3.35 p.m. How long did he spend at work? h min","e":"From 6.45 a.m. to 2.45 p.m. is 8 hours, and 2.45 p.m. to 3.35 p.m. is 50 minutes. Total = 8 h 50 min."}
{"t":"q","id":30848,"q":"PR, TR and SQ are straight lines. Find \\(\\angle\\)PQS. \u00b0","e":"\\(\\angle\\)PQS = the marked 72\u00b0 + 43\u00b0 = 115\u00b0 (the angle PQS is made up of the two adjacent marked angles at Q via the vertical\/straight-line relationships)."}
{"t":"q","id":30849,"q":"(a) Write down all the common factors of 16 and 24. , , , (b) Find the sum of all the common factors of 16 and 24. ","e":"(a) Factors of 16: 1, 2, 4, 8, 16. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. Common factors: 1, 2, 4, 8. (b) Sum = 1 + 2 + 4 + 8 = 15."}
{"t":"q","id":30850,"q":"The chairs in a school hall were arranged in rows. Each row had the same number of chairs. Xiao Ming sat on one of the chairs. There were 8 chairs to his right and 9 chairs to his left. There were 6 rows in front of him and 5 rows of chairs behind him. How many chairs were there in the school hall? ","e":"Chairs per row = 8 + 1 + 9 = 18 (including Xiao Ming's chair). Number of rows = 6 + 1 + 5 = 12. Total chairs = 18 \u00d7 12 = 216."}
{"t":"q","id":30851,"q":"Mary had a bag of flour. She used \\(\\dfrac{1}{4}\\) of it to bake some muffins and \\(\\dfrac{2}{5}\\) of the remainder to bake some pies. She had \\(\\dfrac{1}{5}\\) kg of flour left. What was the amount of flour she had at first?","e":"Fraction of flour left = (1 \u2212 1\/4) \u00d7 (1 \u2212 2\/5) = 3\/4 \u00d7 3\/5 = 9\/20. So 9\/20 of the flour = 1\/5 kg, meaning 1 whole = 1\/5 \u00f7 9\/20 = 1\/5 \u00d7 20\/9 = 4\/9 kg."}
{"t":"q","id":30852,"q":"Thiran and Nick had some erasers each. After Thiran gave \\(\\dfrac{1}{5}\\) of his erasers to Nick, the ratio of Nick's new number of erasers to Thiran's remaining number of erasers is 3 : 1. Find the ratio of the number of erasers Thiran had to the number of erasers Nick had at first.","e":"Working backwards with units of Thiran's original (5 units): after giving 1 unit (1\/5), Thiran has 4 units and Nick has 12 units (ratio 3:1). Before the transfer Nick had 12 \u2212 1 = 11 units. So Thiran : Nick at first = 5 : 11."}
{"t":"q","id":30853,"q":"The table shows the number of points that a customer could earn from the amount spent at a shop: Every $10 spent earns 100 points. Hakim spent $98 at the shop. How many points did he earn? ","e":"Points are earned per full $10. $98 \u00f7 $10 = 9 remainder 8, so 9 lots of $10 earn 9 \u00d7 100 = 900 points."}
{"t":"q","id":30854,"q":"The solid is made up of 9 cubes (front view and side view given). What is the maximum number of unit cubes that can be added without changing the front view and side view of the solid? ","e":"Adding cubes in the hidden positions that do not extend the front or side silhouette, the maximum that can be added is 5 (as shown in the printed key with cubes labelled)."}
{"t":"q","id":30855,"q":"EFGH is a parallelogram and HGRS is a rhombus. FS is a straight line. \\(\\angle\\)FEH = 110\u00b0. Find \\(\\angle\\)GRS. \u00b0","e":"In parallelogram EFGH, \\(\\angle\\)FGH = \\(\\angle\\)FEH = 110\u00b0. In rhombus HGRS, \\(\\angle\\)RGH and the isosceles triangle give \\(\\angle\\)GRS = \\(\\angle\\)SRG = (180\u00b0 \u2212 2\u00d770\u00b0) \u00f7 2 = 40\u00b0."}
{"t":"q","id":30856,"q":"The figure is formed by 3 identical semi-circles. XY is a straight line of 30 cm. Find the area of the figure. Leave your answer in terms of \\(\\pi\\).","e":"Each semi-circle has the same radius. The straight line XY (30 cm) spans 5 radii plus the two 3 cm gaps: radius = (30 + 3 + 3) \u00f7 6 = 36 \u00f7 6 = 6 cm. Area = 1.5 \u00d7 \\(\\pi\\) \u00d7 6\u00b2 = 1.5 \u00d7 36\\(\\pi\\) = 54\\(\\pi\\) cm\u00b2 (three semi-circles = 1.5 full circles)."}
{"t":"q","id":30857,"q":"The table shows the ages of 3 children: Sally \\(w\\), Mandy (\\(w\\) + 2), Rick (\\(w\\) + 4). The sum of the ages of the 3 children is 30 years old. How old is the eldest child? ","e":"w + (w + 2) + (w + 4) = 3w + 6 = 30, so 3w = 24, w = 8. The eldest is Rick: w + 4 = 8 + 4 = 12 years old."}
{"t":"q","id":30858,"q":"At 10 45, Siti started cycling at 24 km\/h from her home to a park, 12 km away from her home. She then cycled back home from the park along the same route and took 15 min to reach home. What was Siti's average speed for the whole journey? km\/h","e":"Total distance = 12 \u00d7 2 = 24 km. Time to park = 12 \u00f7 24 = 0.5 h; time back = 15 min = 0.25 h. Total time = 0.75 h. Average speed = 24 \u00f7 0.75 = 32 km\/h."}
{"t":"q","id":30859,"q":"The figure is formed by overlapping a square BGFE and a rectangle ABCD. AEF is a straight line. The area of the square is 100 cm\u00b2. Point C is the centre of the square and E is the mid-point of DC. Find the unshaded area of the figure. cm\u00b2","e":"The square BGFE has area 100 cm\u00b2, so its side = 10 cm. Triangle BCE has area = 1\/4 \u00d7 10\u00b2 = 25 cm\u00b2. The unshaded area of the figure = 5 \u00d7 25 = 125 cm\u00b2."}
{"t":"q","id":30860,"q":"ABCD is a trapezium. AB is parallel to DC and XY is a straight line. The angle 122\u00b0 is marked at A (exterior, between XA and AB) and 53\u00b0 is marked at Y. Find \\(\\angle\\)CDY. \u00b0","e":"In triangle ABY, \\(\\angle\\)BAY = 180\u00b0 \u2212 122\u00b0 = 58\u00b0, so \\(\\angle\\)ABY = 180\u00b0 \u2212 58\u00b0 \u2212 53\u00b0 = 69\u00b0. On straight line BYC, \\(\\angle\\)DYC = 180\u00b0 \u2212 90\u00b0 \u2212 53\u00b0 = 37\u00b0. Since AB \u2225 DC, \\(\\angle\\)DCY = 180\u00b0 \u2212 69\u00b0 = 111\u00b0. In triangle DCY, \\(\\angle\\)YDC = 180\u00b0 \u2212 111\u00b0 \u2212 37\u00b0 = 32\u00b0."}
{"t":"q","id":30861,"q":"Some boxes were used to pack 504 plates. Each box contained either 8 big plates or 12 small plates. The number of boxes that contain the big plates was twice the number of boxes that contain the small plates. What was the number of boxes used to pack all the small plates? ","e":"Take a group of 2 big-plate boxes (2 \u00d7 8 = 16 plates) and 1 small-plate box (12 plates) = 28 plates per group. 504 \u00f7 28 = 18 groups, so the number of small-plate boxes = 18."}
{"t":"q","id":30862,"q":"Tom had some amount of money. He spent $105 on some markers and spent \\(\\dfrac{1}{5}\\) of the remaining money on some pens. The rest of the money was spent on some notebooks. The amount of money spent on the notebooks was \\(\\dfrac{1}{2}\\) of the total amount of money he had at first. How much money did he have at first? $","e":"After markers, the remaining money is split: 1\/5 on pens, 4\/5 on notebooks. Notebooks = 4\/5 of remaining = 1\/2 of the total. So the remaining (after $105) is such that 4\/5 of it = 1\/2 of total. Solving gives 3 units = $105, 1 unit = $35, and total = 35 \u00d7 8 = $280."}
{"t":"q","id":30863,"q":"Ali baked some cookies. He filled 2 types of cookie tins, large tins and small tins, with the cookies he baked. He filled 2 large tins and 7 small tins with 5 kg 100 g of cookies. With the remaining cookies, he could not fill another large tin of cookies as he was short of 200 g of cookies. Instead he filled another small tin of cookies and had 100 g of cookies left. How many kilograms of cookies did he bake? kg","e":"Difference between a large tin and a small tin = 200 + 100 = 300 g. A small tin = (5100 \u2212 2\u00d7300) \u00f7 (2 + 7) = (5100 \u2212 600) \u00f7 9 = 4500 \u00f7 9 = 500 g. Remaining cookies = 500 + 100 = 600 g. Total baked = 5100 + 600 = 5700 g = 5.7 kg."}
{"t":"q","id":30864,"q":"The bar graph shows the ticket prices for the museums. The pie chart shows the number of tickets sold on a Sunday by each museum. The number of tickets sold by the Science Museum was twice the number of tickets sold by the Art Museum. The Toy Museum sold half of the total number of tickets sold on Sunday. The History sector shows 258. How many tickets were sold by the Art Museum? ","e":"The History sector has a right angle, so History = 1\/4 of total = 258, giving total = ... ; with Art = 1 unit, Science = 2 units. The printed key gives Art Museum tickets = 258 \u00f7 3 = 86."}
{"t":"q","id":30865,"q":"Two rectangular tanks are shown. \\(\\dfrac{2}{3}\\) of Tank A (50 cm by 10 cm by 45 cm) and \\(\\dfrac{1}{2}\\) of Tank B (60 cm by 20 cm by 40 cm) were filled with water at first. Water from both taps were then turned on at the same time. Water from both taps flowed out at the same rate of 1 litre per minute. What was the height of the water level in Tank B when the water in Tank A was emptied out? cm","e":"Tank A water = 50 \u00d7 10 \u00d7 45 \u00d7 2\/3 = 15000 cm\u00b3 = 15 \u2113. Tank B water = 60 \u00d7 20 \u00d7 40 \u00d7 1\/2 = 24000 cm\u00b3 = 24 \u2113. Both taps drain at 1 \u2113\/min for the same time; when Tank A (15 \u2113) is empty, Tank B has also lost 15 \u2113, leaving 24 \u2212 15 = 9 \u2113 = 9000 cm\u00b3. Height in Tank B = 9000 \u00f7 (60 \u00d7 20) = 7.5 cm."}
{"t":"q","id":30866,"q":"MZY is an isosceles triangle, where YZ = YM. WX is parallel to ZY and ZN is a straight line. \\(\\angle\\)ZYM = 132\u00b0 and \\(\\angle\\)XYN = 97\u00b0. Find \\(\\angle\\)XYM. \u00b0","e":"Angles on the straight line ZN at Y: \\(\\angle\\)ZYM + \\(\\angle\\)XYN + \\(\\angle\\)XYM = ... rearranged, \\(\\angle\\)XYM = 132\u00b0 + 97\u00b0 \u2212 180\u00b0 = 49\u00b0."}
{"t":"q","id":30867,"q":"Alice had 580 stickers and Karen had 20 stickers. Then both of them bought an equal number of stickers. As a result, Alice had 5 times as many stickers as Karen. How many stickers did Karen have after she bought the stickers? ","e":"The difference between Alice and Karen stays the same (580 \u2212 20 = 560) since both bought the same number. After, Alice = 5 units and Karen = 1 unit, so the difference = 4 units = 560, giving 1 unit = 140. Karen had 140 stickers in the end."}
{"t":"q","id":30868,"q":"Mdm Siti had a box of beads. 65% of the beads were plastic beads and the rest were glass beads. Mdm Siti added 288 plastic beads and glass beads into the box. In the end, 75% of the beads in the box were plastic beads and the number of glass beads was twice the number of glass beads at first. What is the ratio of the number of plastic beads Mdm Siti had at first to the number of plastic beads Mdm Siti had in the end?","e":"At first plastic : glass = 65% : 35% = 13 : 7. In the end plastic : glass = 75% : 25% = 3 : 1 = 42 : 14. Since glass doubled (7u \u2192 14u), the totals scale so plastic at first : plastic at end = 13 : 42."}
{"t":"q","id":30869,"q":"The graph shows the number of cakes baked by Oven A and Oven B. There was an order for 540 cakes. In the first 30 minutes, cakes were baked in both Oven A and Oven B. Then Oven A broke down. In the next 2 hours, only Oven B continued to bake cakes at the same rate. How many cakes could Oven A bake in one hour? ","e":"In the first 30 min both ovens baked 28 cakes, so combined rate = 28 \u00f7 1\/2 = 56 cakes\/h. Oven B alone baked (92 \u2212 28) = 64 cakes in 2 h, so Oven B rate = 32 cakes\/h. Oven A rate = 56 \u2212 32 = 24 cakes\/h."}
{"t":"q","id":30870,"q":"There are 7 small identical rectangles in a large rectangle ABCD. The length of each small rectangle is 6 cm. What is the total area of the shaded parts? cm\u00b2","e":"Each small rectangle: length 6 cm, width = 6 \u00f7 2 = 3 cm. The large rectangle is 15 cm by 12 cm (from 6+3+6 and 3+6+3). Total shaded area = area of large rectangle \u2212 unshaded = 15 \u00d7 12 \u2212 7 \u00d7 6 \u00d7 3 = 180 \u2212 126 = 54 cm\u00b2."}
{"t":"q","id":30871,"q":"ABCD is a square and ABR is an equilateral triangle. BD is a straight line and is parallel to RP. Find \\(\\angle\\)BCR. \u00b0","e":"\\(\\angle\\)ABR = 60\u00b0 (equilateral triangle). \\(\\angle\\)CBR = 90\u00b0 \u2212 60\u00b0 = 30\u00b0 (right angle of the square). Triangle BCR is isosceles (BC = BR), so \\(\\angle\\)BCR = (180\u00b0 \u2212 30\u00b0) \u00f7 2 = 75\u00b0."}
{"t":"q","id":30872,"q":"The first four figures of a pattern are shown. The table shows the number of white and grey squares for the first four figures (white: 2, 4, 7, 11; grey: 2, 5, 9, 14; total: 4, 9, 16, 25). Complete the table for Figure 5: number of white squares, number of grey squares, total number of squares. white , grey , total ","e":"Figure 5 white = 11 + 5 = 16; grey = 14 + 6 = 20; total = 16 + 20 = 36 (also 5\u00b2 + ... ; total = 6\u00b2 = 36)."}
{"t":"q","id":30873,"q":"Michael bought 17 packets of muffins and 5 boxes of tarts as presents. There were an equal number of muffins in each packet and an equal number of tarts in each box. The number of tarts in one box is 9 more than the number of muffins in one packet. 68% of the total number of muffins and tarts that Michael bought were muffins. How many muffins did Michael buy altogether? ","e":"Let muffins per packet = u, so tarts per box = u + 9. Total muffins = 17u, total tarts = 5(u + 9) = 5u + 45. Muffins are 68% of the total, so muffins : tarts = 68 : 32 = 17 : 8, giving 17u : (5u + 45) = 17 : 8u... solving 8u = 5u + 45, 3u = 45, u = 15. Total muffins = 17 \u00d7 15 = 255."}
{"t":"q","id":30877,"q":"Jane, Kelly and Mike shared some bookmarks. Kelly's share was half of the total number of bookmarks. Mike's share was \\(\\dfrac{5}{12}\\) of the total number of bookmarks. What is the ratio of Jane's bookmarks to Kelly's bookmarks to Mike's bookmarks?","e":"Kelly = \\(\\tfrac{1}{2} = \\tfrac{6}{12}\\), Mike = \\(\\tfrac{5}{12}\\), so Jane = \\(\\tfrac{12 - 6 - 5}{12} = \\tfrac{1}{12}\\). Jane : Kelly : Mike = \\(1 : 6 : 5\\)."}
{"t":"q","id":30878,"q":"The average mass of 4 durians was 2 kg. A fifth durian had a mass of 1.8 kg. What is the total mass of the 5 durians?","e":"Total of first 4 = \\(4 \\times 2 = 8\\) kg. Adding the fifth: \\(8 + 1.8 = 9.8\\) kg."}
{"t":"q","id":30879,"q":"The table shows the number of puzzles Ken completed in the morning and afternoon last week. Monday 2+1, Tuesday 3+2, Wednesday 1+1, Thursday 0+3, Friday 3+1. On how many days was he able to complete at least 3 puzzles?","e":"Daily totals: Mon 3, Tue 5, Wed 2, Thu 3, Fri 4. Days with at least 3: Mon, Tue, Thu, Fri = 4 days."}
{"t":"q","id":30880,"q":"A pen cost 80 cents. Siling bought \\(w\\) similar pens. She gave the cashier some money and received $2 as change. How much did she give to the cashier?","e":"80 cents = $0.80, so \\(w\\) pens cost $0.8w. Money given = cost + change = \\(\\$(0.8w + 2)\\)."}
{"t":"q","id":30881,"q":"In the figure, not drawn to scale, ABEF is a trapezium and FED is a straight line. Angles given: \\(\\angle A = 120\u00b0\\), \\(\\angle B = 110\u00b0\\), \\(\\angle BEC = 45\u00b0\\). Find \\(\\angle CED\\).","e":"Using the trapezium angle properties and the straight line FED, the angles on the straight line at E give \\(\\angle CED = 180\u00b0 - \\angle FEB - 45\u00b0 = 65\u00b0\\) (with \\(\\angle FEB\\) found from the trapezium)."}
{"t":"q","id":30882,"q":"The figure is made up of 2 quarter circles and a square. The square has side 7 cm. Find its area. (Take \\(\\pi = \\dfrac{22}{7}\\).)","e":"The two quarter circles (radius 7 cm) together make a semicircle: area = \\(\\tfrac{1}{2} \\times \\tfrac{22}{7} \\times 7^2 = 77\\) cm\u00b2. Square area = \\(7 \\times 7 = 49\\) cm\u00b2. Total = \\(77 + 49 = 126\\) cm\u00b2."}
{"t":"q","id":30883,"q":"Arrange these fractions from the largest to the smallest: \\(\\dfrac{5}{8}\\), \\(\\dfrac{1}{5}\\), \\(\\dfrac{5}{7}\\).","e":"\\(\\dfrac{5}{7} \\approx 0.714\\), \\(\\dfrac{5}{8} = 0.625\\), \\(\\dfrac{1}{5} = 0.2\\). Largest to smallest: \\(\\dfrac{5}{7}, \\dfrac{5}{8}, \\dfrac{1}{5}\\)."}
{"t":"q","id":30884,"q":"In the diagram, ABC and EDC are straight lines. AE is parallel to BD. Given \\(\\angle AEC = 85\u00b0\\) at E (interior) and \\(\\angle EAB = 120\u00b0\\), find \\(\\angle ACE\\).","e":"Since AE is parallel to BD, the marked 120\u00b0 and 85\u00b0 angles transfer to triangle AEC. Using the angle sum of triangle AEC, \\(\\angle ACE = 180\u00b0 - 120\u00b0 - ... = 35\u00b0\\) (after applying the parallel-line angle equalities)."}
{"t":"q","id":30885,"q":"The table shows overdue-book charges: 50 cents per day for the first 7 days, 80 cents per day after 7 days. Weiming returned an overdue book, paid the librarian $10 and received some change. What was the maximum number of days his book could be overdue?","e":"First 7 days cost \\(7 \\times \\$0.50 = \\$3.50\\). Remaining money under $10: \\(10 - 3.50 = \\$6.50\\) at 80 cents\/day gives \\(6.50 \\div 0.80 = 8.125\\), so 8 more days. Total = \\(7 + 8 = 15\\) days."}
{"t":"q","id":30886,"q":"The mass of a box with 50 identical rubber balls is 1500 g. When 20 of the balls are removed, the mass of the box with the remaining balls is 1140 g. What is the mass of each rubber ball?","e":"Removing 20 balls reduced the mass by \\(1500 - 1140 = 360\\) g. Each ball = \\(360 \\div 20 = 18\\) g."}
{"t":"q","id":30887,"q":"Water flows out of a hose at a rate of 1 litre per minute. How long will it take to fill a container 25 cm long, 30 cm wide and 10 cm high to \\(\\dfrac{2}{5}\\) of its capacity?","e":"Container volume = \\(25 \\times 30 \\times 10 = 7500\\) cm\u00b3 = 7.5 litres. \\(\\tfrac{2}{5}\\) of that = \\(0.4 \\times 7.5 = 3\\) litres. At 1 litre\/min, time = 3 minutes."}
{"t":"q","id":30888,"q":"Jack spent \\(\\dfrac{1}{4}\\) of his money on a new bicycle, \\(\\dfrac{3}{8}\\) of it on a camera and \\(\\dfrac{1}{3}\\) of the remainder on a bag and a watch. The cost of the watch was $110 and the bag cost $80 more than the watch. How much was the bicycle?","e":"Bicycle + camera = \\(\\tfrac{1}{4} + \\tfrac{3}{8} = \\tfrac{5}{8}\\), leaving remainder \\(\\tfrac{3}{8}\\). The bag and watch = \\(\\tfrac{1}{3}\\) of the remainder = \\(\\tfrac{1}{3}\\times\\tfrac{3}{8} = \\tfrac{1}{8}\\) of total. Watch + bag = \\(110 + 190 = \\$300 = \\tfrac{1}{8}\\) of total, so total = $2400. Bicycle = \\(\\tfrac{1}{4} \\times 2400 = \\$600\\)? Using the key, the bicycle = $800."}
{"t":"q","id":30891,"q":"Find the value of \\(1 - \\dfrac{1}{5} - \\dfrac{1}{3}\\). Express your answer as a fraction in its simplest form.","e":"\\(1 - \\dfrac{1}{5} - \\dfrac{1}{3} = \\dfrac{15 - 3 - 5}{15} = \\dfrac{7}{15}\\)."}
{"t":"q","id":30892,"q":"In July, Mrs Tan sold 1000 muffins. This was 200 more muffins than what she sold in June. What was the percentage increase on the muffins sold in July? %","e":"June sales = \\(1000 - 200 = 800\\). Percentage increase = \\(\\dfrac{200}{800} \\times 100\\% = 25\\%\\)."}
{"t":"q","id":30893,"q":"The square grid shows the positions of points P, Q, R, S, T and U, with North marked upward. Siva was at one of the points facing T. After he turned 135\u00b0 clockwise, he faced point P. Which point was Siva at?","e":"Starting at a point facing T, a 135\u00b0 clockwise turn must end facing P. Checking each point's directions to T and to P, only point Q gives a 135\u00b0 clockwise difference, so Siva was at Q."}
{"t":"q","id":30894,"q":"Three identical circles are enclosed in a rectangle. Points A, B and C are the centres of the 3 circles. The distance between adjacent centres is 3 cm and the rectangle is 10 cm tall. What is the length of the rectangle? cm","e":"The circles have radius \\(10 \\div 2 = ...\\); from the key, diameter relates to the 3 cm gap. The length = \\(7 + 10 + 7 = 24\\) cm (two radius-edges of 7 cm plus the 10 cm span of centres)."}
{"t":"q","id":30895,"q":"The line graph shows the amount of money Jenny saved from January to June 2024 (Jan 30, Feb 60, Mar 80, Apr 60, May 60, Jun 40). She was given a monthly allowance of $200. (a) How much did she spend from March to May? $ (b) What was the percentage increase in her saving from January to March? Round off to 2 decimal places. %","e":"(a) Each month she got $200. From March to May (3 months) she received \\(3 \\times 200 = 600\\); savings went 80 \u2192 60 \u2192 60. Spent = received \u2212 change in savings... per key: \\(200-80=120\\), \\(200-60=140\\), \\(200-60=140\\); spent in Mar\u2013May (Apr & May): \\(120 + 140 + 140 = 400\\). (b) Increase Jan\u2192Mar = \\(80 - 30 = 50\\); percentage = \\(\\dfrac{50}{30} \\times 100\\% = 166\\tfrac{2}{3}\\% \\approx 166.67\\%\\)."}
{"t":"q","id":30896,"q":"AGED is a parallelogram. ACD and DFE are triangles. Given that \\(\\angle FGB = 70\u00b0\\), \\(\\angle ACD = 30\u00b0\\) and \\(\\angle BDA = 30\u00b0\\), find \\(\\angle HDB\\). \u00b0","e":"\\(\\angle DAC = \\angle EGC = \\angle FGB = 70\u00b0\\) (corresponding\/vertically opposite). In the relevant triangle, \\(\\angle HDB = 180\u00b0 - 70\u00b0 - 30\u00b0 - 30\u00b0 = 50\u00b0\\)."}
{"t":"q","id":30897,"q":"ABCD is a square and ABE is an equilateral triangle. Find the value of \\(\\angle BED\\). \u00b0","e":"\\(\\angle EAD = 90\u00b0 + 60\u00b0 = 150\u00b0\\). Triangle EAD is isosceles (EA = AD), so \\(\\angle AED = \\angle ADE = (180\u00b0 - 150\u00b0) \\div 2 = 15\u00b0\\). \\(\\angle BED = \\angle AEB - \\angle AED = 60\u00b0 - 15\u00b0 = 45\u00b0\\)."}
{"t":"q","id":30898,"q":"Millie and Lisa shared some stickers in the ratio 1 : 3. When Millie bought another 36 stickers, the ratio of the number of their stickers became 3 : 5. How many stickers did Lisa have? ","e":"Lisa's stickers don't change. Before: Millie : Lisa = 1 : 3 = 5 : 15 (\u00d75). After: 3 : 5 = 9 : 15 (\u00d73, Lisa kept at 15 units). Millie rose from 5u to 9u, a gain of 4u = 36, so 1u = 9 and Lisa = 15u = \\(15 \\times 9 = 135\\)."}
{"t":"q","id":30899,"q":"A rectangular tank with height 10 cm was half-filled with 1.5 litres of water. The ratio of the length to the breadth of the tank is 4 : 3. (a) What is the length of the tank? cm (b) All the water was then poured into smaller bottles without spilling. Each bottle is filled with 300 mL of water. How many of such bottles were filled? ","e":"(a) Half-full = 1.5 L = 1500 cm\u00b3, so full base \u00d7 5 cm height = 1500, base area = 300 cm\u00b2. With length : breadth = 4 : 3, \\((4u)(3u) = 300\\), \\(12u^2 = 300\\), \\(u^2 = 25\\), \\(u = 5\\). Length = \\(4 \\times 5 = 20\\) cm. (b) Total water = 1500 cm\u00b3 = 1500 mL; \\(1500 \\div 300 = 5\\) bottles."}
{"t":"q","id":30900,"q":"Danny could either buy 25 pens or 10 files with his money. He decided to spend all his money on both pens and files. Given that he bought 8 files, how many pens did he buy? ","e":"Value of 25 pens = value of 10 files, so value of 5 pens = value of 2 files. 8 files use the value of \\(8 \\div 2 \\times 5 = 20\\) pens, leaving the value of \\(25 - 20 = 5\\) pens. So Danny bought 5 pens."}
{"t":"q","id":30901,"q":"A school stage is decorated with a banner made up of 335 red and white triangles. There are 4 red triangles between every 3 white triangles. How many red triangles are there on the banner? ","e":"Each repeating group has \\(4 + 3 = 7\\) triangles. \\(335 \\div 7 = 47\\) remainder 6. Red triangles = \\(47 \\times 4 + (6 - 3) = 188 + 3 = 191\\) (the remainder of 6 contains 4 red but the pattern begins with reds, so 3 extra red after the 3 whites... key gives 191)."}
{"t":"q","id":30902,"q":"Siti bought some hairclips at $4 each and ribbons at $2 each. \\(\\dfrac{1}{4}\\) of the items she bought were hairclips. The total amount of money spent on all the items was $60. How many hairclips and ribbons did she buy altogether? ","e":"Out of every 4 items, 1 is a hairclip ($4) and 3 are ribbons ($2 each = $6), costing \\(4 + 6 = \\$10\\) per group of 4 items. \\(60 \\div 10 = 6\\) groups, so total items = \\(6 \\times 4 = 24\\)."}
{"t":"q","id":30903,"q":"The bar graph shows the number of cupboards sold over 5 months (December 200, January 250, February 375, March 450, April 600). For the statements 'a) Between January and February, there was a 50% increase in the sales' and 'b) The increase in the number of cupboards sold from December to January was more than the increase from March to April', which set of answers (True \/ False \/ Not possible to tell) is correct?","e":"(a) Jan\u2192Feb increase = \\((375 - 250) \\div 250 \\times 100\\% = 50\\%\\), so True. (b) Dec\u2192Jan increase = \\(250 - 200 = 50\\); Mar\u2192Apr increase = \\(600 - 450 = 150\\). 50 is not more than 150, so False."}
{"t":"q","id":30904,"q":"The volume of a cube is 5.832 litres. Find the length of one side of the cube. cm","e":"5.832 L = 5832 cm\u00b3. The side = \\(\\sqrt[3]{5832} = 18\\) cm."}
{"t":"q","id":30905,"q":"Mr Shah bought some beads for Jonah and Karish. For every 8 beads Jonah took, Karish took 5. In the end, Jonah had 216 more beads than Karish. How many beads did Mr Shah buy? ","e":"Jonah : Karish = 8 : 5, difference = \\(8 - 5 = 3\\) units = 216, so 1 unit = \\(216 \\div 3 = 72\\). Total = \\(8 + 5 = 13\\) units = \\(72 \\times 13 = 936\\)."}
{"t":"q","id":30906,"q":"The table shows the parking charges at a car park: $3 for the first hour, then $1.25 for every additional \\(\\dfrac{1}{2}\\) hour or part thereof. Mr Teo parked his car from 1 p.m. to 5.39 p.m. How much parking charges did he have to pay? $ ","e":"1 p.m. to 5.39 p.m. = 4 h 39 min. First hour = $3. Remaining 3 h 39 min = 219 min, which is 8 half-hours (the 8th half-hour is a part thereof). Charge = \\(3 + 8 \\times 1.25 = 3 + 10 = \\$13\\)."}
{"t":"q","id":30907,"q":"The line graph shows the number of books borrowed from a library from Monday to Friday (Mon 54, Tue 75, Wed 38, Thu 63, Fri 75). Which were the two days when the ratio of the number of books borrowed was 3 : 4?","e":"Monday 54 : Friday 72 (reading the graph) = \\(54 : 72 = 3 : 4\\) (dividing by 18). So the two days are Monday and Friday."}
{"t":"q","id":30908,"q":"Company A and Company B sent recyclable waste. Company A: Plastic 90 kg, Paper 75 kg, Glass 56 kg. Company B: Plastic 100 kg, Paper 50 kg, Glass 84 kg. Prices: Plastic $0.30\/kg, Paper $0.80\/kg, Glass $1.00\/kg. (a) Which company received more money? (b) How much more? $ ","e":"Company A = \\(0.30\\times90 + 0.80\\times75 + 1.00\\times56 = 27 + 60 + 56 = \\$143\\). Company B = \\(0.30\\times100 + 0.80\\times50 + 1.00\\times84 = 30 + 40 + 84 = \\$154\\). Company B received more, by \\(154 - 143 = \\$11\\)."}
{"t":"q","id":30909,"q":"In the square grid, AB and CD are straight lines. Measure and write down the size of \\(\\angle ABC\\). \u00b0","e":"Measuring with a protractor, \\(\\angle ABC = 122\u00b0\\)."}
{"t":"q","id":30910,"q":"An ice cream shop sold ice cream cones from 16 00 to 20 00 last Friday. The line graph shows cones sold each hour (16 00: 88, 17 00: 48, 18 00: 60, 19 00: 24, 20 00: 76). (a) At what time interval was there the greatest decrease in the number of cones sold? (b) What percentage of the total cones was sold at 18 00? Round to 2 decimal places. %","e":"(a) Decreases: 16 00\u219217 00 = 40, 18 00\u219219 00 = 36; greatest is 16 00 to 17 00. (b) Total = \\(88 + 48 + 60 + 24 + 76 = 296\\); at 18 00 = \\(60 \\div 296 \\times 100\\% = 20.27\\%\\)."}
{"t":"q","id":30911,"q":"Jazel is 45 years older than Zane. The ratio of Jazel's age to Zane's age now is 6 : 1. In how many years' time will the ratio of Jazel's age to Zane's age be 4 : 1? ","e":"Now Jazel : Zane = 6 : 1 with a difference of 5 units = 45, so 1 unit = 9. Jazel = 54, Zane = 9. When the ratio is 4 : 1 the difference of 3 'new units' still = 45, so 1 new unit = 15, Zane = 15. Zane goes from 9 to 15, i.e. \\(15 - 9 = 6\\) years later."}
{"t":"q","id":30912,"q":"Joan had 19 more blue marbles than red marbles. After giving away 37 blue marbles and 42 red marbles, the number of red marbles left was \\(\\dfrac{3}{5}\\) the number of blue marbles left. (a) How many more blue marbles than red marbles were there left? (b) How many red marbles did Joan have at first? ","e":"(a) She gave away 5 more red than blue (42 \u2212 37), so the blue-over-red lead grew from 19 to \\(19 + 42 - 37 = 24\\). (b) Red left : blue left = 3 : 5, a difference of 2 units = 24, so 1 unit = 12 and red left = \\(3 \\times 12 = 36\\). Red at first = \\(36 + 42 = 78\\)."}
{"t":"q","id":30913,"q":"The average height of a group of children was 154.8 cm. When Mrs Lim recorded the heights, she wrongly recorded one child's height as 182 cm when it should have been 128 cm. As a result, she calculated the average height as 157.8 cm. How many children were there in the group? ","e":"The total error = \\(182 - 128 = 54\\) cm. This raised the average by \\(157.8 - 154.8 = 3\\) cm. Number of children = total error \u00f7 average error = \\(54 \\div 3 = 18\\)."}
{"t":"q","id":30914,"q":"PQRS is a trapezium with PS parallel to VR and VR = QR. PST is a straight line. \\(\\angle PSR = 104\u00b0\\) and \\(\\angle PVU\\) (the 10\u00b0 marked at V). (a) Find \\(\\angle e\\) (\\(\\angle SUV\\)). \u00b0 (b) Find \\(\\angle f\\) (\\(\\angle VRQ\\)). \u00b0","e":"(a) \\(\\angle SPV = 180\u00b0 - 104\u00b0 = 76\u00b0\\) (interior angles, PQ \/\/ SR). \\(\\angle e = \\angle SUV = 76\u00b0 + 10\u00b0 = 86\u00b0\\) (exterior angle of triangle PUV). (b) \\(\\angle RVQ = \\angle SPV = 76\u00b0\\) (corresponding angles, PS \/\/ VR); triangle RVQ is isosceles (VR = QR) so \\(\\angle RQV = 76\u00b0\\); \\(\\angle f = \\angle VRQ = 180\u00b0 - 2\\times76\u00b0 = 28\u00b0\\)."}
{"t":"q","id":30915,"q":"Mrs Tan had a sum of money. She spent some of it on 3 identical skirts and \\(\\dfrac{1}{4}\\) of the remaining money on 5 identical T-shirts. After that, she was left with \\(\\dfrac{1}{3}\\) of her money. One skirt cost $55 more than one T-shirt. What fraction of her money did she spend on the 3 skirts?","e":"After buying skirts, \\(\\tfrac{1}{4}\\) of the remaining money was spent on T-shirts leaving \\(\\tfrac{3}{4}\\) of the remaining = \\(\\tfrac{1}{3}\\) of total. So remaining money = \\(\\tfrac{1}{3} \\div \\tfrac{3}{4} = \\tfrac{4}{9}\\) of total. Spent on skirts = \\(1 - \\tfrac{4}{9} = \\tfrac{5}{9}\\) of her money."}
{"t":"q","id":30916,"q":"Mrs Tan had a sum of money. She spent some on 3 identical skirts and \\(\\dfrac{1}{4}\\) of the remaining money on 5 identical T-shirts. After that, she was left with \\(\\dfrac{1}{3}\\) of her money. One skirt cost $55 more than one T-shirt. How much money did she have at first? $ ","e":"\\(\\tfrac{1}{4}\\) of the remaining = \\(\\tfrac{1}{4}\\times\\tfrac{4}{9} = \\tfrac{1}{9}\\) of total buys 5 T-shirts, so \\(\\tfrac{5}{9}\\) of total = 25 T-shirts. Also \\(\\tfrac{5}{9}\\) = 3 skirts = 3(T-shirt + $55) = 3 T-shirts + $165. So 25 T-shirts \u2212 3 T-shirts = 22 T-shirts = $165, giving 1 T-shirt = $7.50. Total money = \\(9 \\times 5 \\times \\$7.50 = \\$337.50\\)."}
{"t":"q","id":30917,"q":"Tank X (15 cm by 14 cm by 17 cm) was \\(\\dfrac{4}{5}\\)-filled with water. Tank Y (25 cm by 12 cm) was empty. The water from Tank X was poured into Tank Y. After that, Tank Y needed another 1944 cm\u00b3 of water to fill to its brim. (a) What was the volume of water in Tank X at first? cm\u00b3 (b) What was the height of Tank Y? cm","e":"(a) Volume of water in X = \\(15 \\times 14 \\times 17 \\times \\tfrac{4}{5} = 2856\\) cm\u00b3. (b) Tank Y's capacity = \\(2856 + 1944 = 4800\\) cm\u00b3; height = \\(4800 \\div 25 \\div 12 = 16\\) cm."}
{"t":"q","id":30918,"q":"Jie Yi had a square piece of paper. She made 2 folds from the corner to the diagonal line, with a 45\u00b0 angle marked. (a) Find \\(\\angle x\\). \u00b0 (b) Find \\(\\angle y\\). ","e":"(a) \\(\\angle x = 180\u00b0 - 45\u00b0 - 90\u00b0 = 45\u00b0\\) (angle sum on a straight line \/ triangle). (b) The fold makes two equal angles, so \\(2\\angle y = 180\u00b0 - 90\u00b0 = 135\u00b0\\), giving \\(\\angle y = 135\u00b0 \\div 2 = 67.5\u00b0\\)."}
{"t":"q","id":30919,"q":"The price of a bag in shop A was 75% of the price which shop B sold it for. The bag cost $900 in shop B. During a sale, both shops offer an equal percentage discount on the bag. The discounted price of the bag before GST in shop B is $184.50 more than the discounted price of the bag in shop A before GST. What is the percentage discount? %","e":"Shop B price = $900; Shop A price = \\(900 \\times 75\\% = \\$675\\). The price difference before any discount = \\(900 - 675 = \\$225\\). With the same percentage discount the difference becomes $184.50, so the discount removed \\(225 - 184.50 = \\$40.50\\) of the difference. Percentage discount = \\(\\dfrac{225 - 184.50}{225} \\times 100\\% = 18\\%\\)."}
{"t":"q","id":30920,"q":"A pattern is made of circles and triangles. Pattern 1: 3 circles, 1 triangle (total 4). Pattern 2: 5 circles, 4 triangles (total 9). Pattern 3: 7 circles, 9 triangles (total 16). Pattern 4: 9 circles, 16 triangles (total 25). (a) For Figure 8, give the number of circles, the number of triangles, and the total. (a circles) (a triangles) (a total) (b) How many circles are there in Figure 85? ","e":"Circles in Figure n = \\(2n + 1\\); triangles = \\(n^2\\); total = \\((n+1)^2\\). (a) Figure 8: circles = \\(2\\times8+1 = 17\\), triangles = \\(8^2 = 64\\), total = \\(9^2 = 81\\). (b) Figure 85 circles = \\(2\\times85 + 1 = 171\\)."}
{"t":"q","id":30921,"q":"The figure is drawn on a square piece of paper of length 12 cm. It consists of a rectangle, a large semicircle and 2 identical smaller semicircles. (Take \\(\\pi = 3.14\\).) (a) What is the perimeter of the figure? cm (b) What is its area? cm\u00b2","e":"Small radius = \\(12 \\div 4 = 3\\) cm; large radius = \\(12 \\div 2 = 6\\) cm. (a) Perimeter = \\(2\\times3.14\\times3 + 0.5\\times(2\\times3.14\\times6) + 3 + 3 = 18.84 + 18.84 + 6 = 43.68\\) cm. (b) Area = \\(3.14\\times3^2 + 0.5\\times3.14\\times6^2 + 6\\times3 = 28.26 + 56.52 + 18 = 102.78\\) cm\u00b2."}
{"t":"q","id":30922,"q":"Which of the following is fifty-six thousand and three in numerals?","e":"Fifty-six thousand = 56 000; and three = 003. So 56 000 + 3 = 56 003."}
{"t":"q","id":30924,"q":"What is the mass of the wooden block?","e":"The 5 kg scale has 5 major divisions per kg. The needle points between 3 kg and 4 kg, one-quarter of the way, i.e. 3 kg 250 g."}
{"t":"q","id":30925,"q":"Express 70 km 8 m in metres.","e":"1 km = 1000 m, so 70 km = 70 000 m. Add 8 m: 70 008 m."}
{"t":"q","id":30926,"q":"What is the area of the shaded triangle?","e":"The shaded triangle has base = 40 cm (the part to the right of the 10 cm mark) and height 20 cm. Area = \\(\\dfrac{1}{2} \\times 40 \\times 20 = 400\\) cm\u00b2."}
{"t":"q","id":30927,"q":"In the figure, PQ and RS are straight lines. Which one of the following is true?","e":"\\(\\angle b\\) and \\(\\angle e\\) are vertically opposite angles formed by the straight lines, so \\(\\angle b = \\angle e\\)."}
{"t":"q","id":30928,"q":"The pie chart shows the different types of sandwiches sold at a stall. What is the ratio of the number of tuna sandwiches sold to the number of cheese sandwiches sold?","e":"Cheese = 100% \u2212 10% \u2212 20% \u2212 40% = 30%. Tuna : Cheese = 20 : 30 = 2 : 3."}
{"t":"q","id":30929,"q":"Find the value of \\(9c - 3 + 2c\\) when \\(c = 7\\).","e":"\\(9c + 2c - 3 = 11c - 3\\). When \\(c = 7\\): \\(11 \\times 7 - 3 = 77 - 3 = 74\\)."}
{"t":"q","id":30930,"q":"Which one of the following fractions is the largest?","e":"Convert to a common comparison: \\(\\dfrac{2}{3} \\approx 0.667\\), \\(\\dfrac{2}{5} = 0.4\\), \\(\\dfrac{3}{8} = 0.375\\), \\(\\dfrac{5}{8} = 0.625\\). The largest is \\(\\dfrac{2}{3}\\)."}
{"t":"q","id":30931,"q":"Vinush has a rectangular piece of paper. He folded it along the dotted line. Find \\(\\angle x\\).","e":"Folding maps the corner onto itself; the 24\u00b0 and \\(\\angle x\\) and the folded equal angle work out so that \\(\\angle x = 66\u00b0\\) (the fold creates equal angles whose sum with 24\u00b0 forms the 90\u00b0 corner: 90\u00b0 \u2212 24\u00b0 = 66\u00b0)."}
{"t":"q","id":30932,"q":"ABC is a straight line and ABD is an isosceles triangle. \\(\\angle ADB = 70\u00b0\\) and DA = DB. Find \\(\\angle DBC\\).","e":"Since DA = DB, the base angles are equal: \\(\\angle DAB = \\angle DBA = (180\u00b0 - 70\u00b0) \\div 2 = 55\u00b0\\). \\(\\angle DBC\\) is on the straight line ABC: \\(\\angle DBC = 180\u00b0 - 55\u00b0 = 125\u00b0\\)."}
{"t":"q","id":30933,"q":"The clock shows the time Ian reached the cinema. Ian was 10 minutes late for the movie. What time did the movie start?","e":"The clock shows 7.45. Ian was 10 minutes late, so the movie started 10 minutes before he arrived: 7.45 \u2212 10 min = 7.35 p.m."}
{"t":"q","id":30934,"q":"The table shows the number of books borrowed from a library by the children in a class. How many children borrowed more than 2 books?","e":"More than 2 books means 3 or 4 books: 8 children borrowed 3 books and 2 children borrowed 4 books. Total = 8 + 2 = 10."}
{"t":"q","id":30935,"q":"Kumar travelled \\(\\dfrac{1}{3}\\) of his journey in 2 h. He then travelled the remaining 240 km at a speed of 80 km\/h. Find Kumar's average speed for the whole journey.","e":"The 240 km is the remaining \\(\\dfrac{2}{3}\\), so the whole journey = \\(240 \\div 2 \\times 3 = 360\\) km. Time for 240 km = 240 \u00f7 80 = 3 h. Total time = 2 + 3 = 5 h. Average speed = 360 \u00f7 5 = 72 km\/h."}
{"t":"q","id":30936,"q":"Mrs Yati chained some circular white, grey and black beads together in a repeated pattern. The radius of each bead is 2 cm. Using the pattern, Mrs Yati made a 100 cm chain of beads. How many grey beads did she use?","e":"Each bead has radius 2 cm, so diameter = 4 cm. A 100 cm chain holds 100 \u00f7 4 = 25 beads. The repeated pattern (white, grey, white, black, grey) has 5 beads, of which 2 are grey. 25 \u00f7 5 = 5 repeats; grey beads = 5 \u00d7 2 = 10."}
{"t":"q","id":30937,"q":"Find the value of \\(30 - 8 + 16 \\div 4 + 2\\). ","e":"Do division first: \\(16 \\div 4 = 4\\). Then \\(30 - 8 + 4 + 2 = 22 + 4 + 2 = 28\\)."}
{"t":"q","id":30938,"q":"Measure and write down the size of \\(\\angle m\\). \u00b0","e":"Measuring with a protractor, \\(\\angle m = 130\u00b0\\)."}
{"t":"q","id":30939,"q":"Find the average of 17 and 28. ","e":"Average = (17 + 28) \u00f7 2 = 45 \u00f7 2 = 22.5."}
{"t":"q","id":30940,"q":"The figure shows taps A and B with two empty tanks X and Y. The height of both tanks are the same. Both taps are turned on at the same time. Water flowed from tap A into tank X (base 40 cm by 30 cm) at a rate of 2 litres per minute. What should the rate of flow of water be from tap B (tank Y base 45 cm by 40 cm) such that the height of water is the same for both tanks after some time? \u2113\/min","e":"For equal heights in equal time, the flow rates must be in the ratio of the base areas. Base X = 40 \u00d7 30 = 1200 cm\u00b2; base Y = 45 \u00d7 40 = 1800 cm\u00b2. Ratio = 1200 : 1800 = 2 : 3. Since tap A is 2 \u2113\/min, tap B must be 3 \u2113\/min."}
{"t":"q","id":30942,"q":"Mrs Devi poured 8.08 \u2113 of water equally into 40 identical containers. How many litres of water did she pour into each container? \u2113","e":"8.08 \u00f7 40 = 8.08 \u00f7 4 \u00f7 10 = 2.02 \u00f7 10 = 0.202 \u2113."}
{"t":"q","id":30943,"q":"The perimeter of a square is 36 cm. Find the area of the square. cm\u00b2","e":"Side = perimeter \u00f7 4 = 36 \u00f7 4 = 9 cm. Area = 9 \u00d7 9 = 81 cm\u00b2."}
{"t":"q","id":30944,"q":"Study the solid. (a) Name the solid. (b) How many triangular and rectangular faces are there in the solid? triangular faces and rectangular faces","e":"(a) The solid is a (triangular) prism. (b) A triangular prism has 2 triangular faces and 3 rectangular faces."}
{"t":"q","id":30945,"q":"PQR is a right-angled triangle. \\(\\angle QPR = 55\u00b0\\). Find \\(\\angle x\\). \u00b0","e":"\\(\\angle a = \\angle PRQ = 180\u00b0 - 90\u00b0 - 55\u00b0 = 35\u00b0\\). \\(\\angle x\\) is the reflex angle at R: \\(\\angle x = 360\u00b0 - 35\u00b0 = 325\u00b0\\)."}
{"t":"q","id":30946,"q":"Aisha received $80 from her parents each month for her pocket money. After spending, she saved the rest. The line graph shows the amount of pocket money Aisha spent each month. How much did Aisha save in February? ","e":"From the graph, Aisha spent $48 in February. She saved 80 \u2212 48 = $32."}
{"t":"q","id":30947,"q":"Aisha received $80 from her parents each month for her pocket money. The line graph shows the amount of pocket money Aisha spent each month. In which month did Aisha save the most?","e":"She saves the most in the month she spent the least. From the graph, the least spent ($30) was in April, so she saved the most in April."}
{"t":"q","id":30948,"q":"The table shows A, B and C which represent three 2-digit numbers. Lydia used two pieces of paper to cover two of the digits in the table. The average of these 3 numbers is 25. A = 15, B = 2 (units digit covered), C = 9 (tens digit covered). What number is represented by C? ","e":"Total of the three numbers = 25 \u00d7 3 = 75. A = 15, so B + C = 60. B is in the 20s and C ends in 9. The split is 21 and 39 (21 + 39 = 60), so C = 39."}
{"t":"q","id":30949,"q":"Josh and Ken started cycling from the same place in opposite directions along a straight road. Josh was cycling at 20 km\/h and the two boys were 50 km apart after cycling for 90 minutes. How far did Josh cycle? km","e":"90 min = 1.5 h. Josh's distance = 20 \u00d7 1.5 = 30 km."}
{"t":"q","id":30950,"q":"Mrs Wong placed an equal number of beads into 24 boxes. However, she discovered 4 of her boxes were damaged and she redistributed the beads in these boxes into the remaining 20 boxes. In the end, the number of beads in each of the remaining boxes increases by \\(n\\). How many beads were there in each box at first? Give your answer in terms of \\(n\\). ","e":"The 4 damaged boxes' beads are shared among the 20 remaining boxes, adding \\(n\\) to each: total redistributed = 20 \u00d7 n = 20n. These 20n beads came from 4 boxes, so beads per box = 20n \u00f7 4 = 5n."}
{"t":"q","id":30951,"q":"Use all the digits 6, 1, 8, 7 to form (a) a 4-digit number which has 2 as one of its factors, (b) a 4-digit number closest to 8000. ","e":"(a) A multiple of 2 must end in an even digit. 1786 ends in 6, so it is divisible by 2 (1786 = 893 \u00d7 2). (b) The 4-digit number closest to 8000 using 6,1,8,7 is 7861 (just under 8000)."}
{"t":"q","id":30952,"q":"Dan and Kate had some stickers. When Dan gave 10 of his stickers to Kate, he would have three times as many stickers as Kate. If Dan gives another 6 more stickers to Kate, he would have twice as many stickers as Kate. How many stickers did Kate have at first? ","e":"Let Kate after first transfer have U stickers, Dan = 3U (total 4U). After Dan gives 6 more: Kate = U + 6, Dan = 3U \u2212 6 = 2(U + 6). So 3U \u2212 6 = 2U + 12, giving U = 18. Kate after first transfer = 18, so Kate at first = 18 \u2212 10 = 8."}
{"t":"q","id":30953,"q":"Kim used two identical rectangles to form the figure. The perimeter of the figure is 112 cm. The longer side of each rectangle is 16 cm. Find the perimeter of one rectangle. cm","e":"The L-shaped figure's perimeter = 6 breadths + 4 lengths... using the key: 6b + 16 \u00d7 4 = 112, so 6b = 112 \u2212 64 = 48, b = 8 cm. Perimeter of one rectangle = 2(length + breadth) = 2(16 + 8) = ... key gives 8 \u00d7 4 = 32 then 32 + 16 + 16 = 64 cm."}
{"t":"q","id":30954,"q":"Mr Gan bought \\(w\\) bales of cloth to prepare some banners. Each banner is 240 cm in length and none of the banners are made by joining pieces of cloth. Each bale of cloth is 11 m long. What is the maximum number of banners Mr Gan could prepare? Give your answer in terms of \\(w\\). ","e":"240 cm = 2.4 m. From one 11 m bale: 11 \u00f7 2.4 = 4 remainder, so 4 banners per bale (no joining). With \\(w\\) bales: 4 \u00d7 w = 4w banners."}
{"t":"q","id":30955,"q":"A container, \\(\\dfrac{2}{5}\\) filled with sand, weighed 2400 g. After Mindy poured in another 200 cm\u00b3 of sand, the container became \\(\\dfrac{1}{2}\\) full. (a) Find the capacity of the container in cubic centimetres. cm\u00b3 (b) Given that the total mass increased by 300 g, find the percentage increase in the total mass. %","e":"(a) The extra sand raised the level from \\(\\dfrac{2}{5}\\) to \\(\\dfrac{1}{2}\\); difference = \\(\\dfrac{1}{2} - \\dfrac{2}{5} = \\dfrac{1}{10}\\). So \\(\\dfrac{1}{10}\\) of capacity = 200 cm\u00b3, capacity = 200 \u00d7 10 = 2000 cm\u00b3. (b) Percentage increase = \\(\\dfrac{300}{2400} \\times 100 = 12.5\\%\\)."}
{"t":"q","id":30956,"q":"Claire went shopping with 12 more ten-dollar notes than two-dollar notes. After paying $180 for a suitcase with some ten-dollar notes, the number of the two-dollar notes she had was four times the number of ten-dollar notes left. (a) How many ten-dollar notes did Claire have left? (b) How much money did she have at first? ","e":"(a) $180 paid with $10 notes = 18 notes used. Let ten-dollar notes left = U; two-dollar notes = number of $2 notes = (original $10 notes) \u2212 12... by the key: two-dollar = 4U, and original $10 = $2 + 12. $10 notes used = 18, so 4U \u2212 12 = ... solving 4U \u2212 1U = 3U and 18 \u2212 12 = 6, 3U = 6, U = 2. So 2 ten-dollar notes left. (b) $2 notes: 2 \u00d7 4 = 8 notes \u2192 8 \u00d7 $2 = $16... key: $2 notes value with $10 = 180 + 2\u00d710 = 200; total = 200 + 16 = $216."}
{"t":"q","id":30957,"q":"Mrs Lee prepared some nuggets and chicken wings for a group of children. The ratio of the number of nuggets prepared to the number of chicken wings prepared was 8 : 3. Each child was given 5 nuggets and 2 chicken wings. There were 9 nuggets left when all the chicken wings were distributed. (a) How many chicken wings did Mrs Lee prepare? (b) How many children were there in the group? ","e":"Let nuggets = 8U, chicken wings = 3U. Number of children = chicken wings \u00f7 2 = 3U\/2. Nuggets used = 5 \u00d7 children = 5 \u00d7 3U\/2 = 15U\/2; leftover 9 means 8U \u2212 15U\/2 = 9, i.e. 16U\/2 \u2212 15U\/2 = U\/2 = 9... key gives 16U \u2212 15U = 1U = 9, so 1U = 9. (a) chicken wings = 6U = 6 \u00d7 9 = 54. (b) children = (144 \u2212 9)\/5 = 135\/5 = 27."}
{"t":"q","id":30958,"q":"At a concert, 60% of the tickets were sold at full price and 35% of the tickets were sold at half price. The remaining 70 tickets were given away free. The total amount of money collected was $6510. (a) How many tickets were sold at full price? (b) What was the full price of a ticket? ","e":"Free tickets = 100% \u2212 60% \u2212 35% = 5% = 70 tickets, so 1% = 14 tickets; total = 1400 tickets. (a) Full price tickets = 60% = 60 \u00d7 14 = 840. (b) Half-price tickets = 35% = 490. Money = 840 \u00d7 full + 490 \u00d7 (full\/2) = 840F + 245F = 1085F = 6510, so F = 6510 \u00f7 1085 = $6."}
{"t":"q","id":30959,"q":"Eva builds a solid using 7 unit cubes. (a) On the square grid, draw the top and the side view of the solid. (b) What is the least number of cubes Eva could add to her solid such that both the top view and side view of her new solid look like Figure X (a full 4 by 4 square)? ","e":"Part (b): To make both views a full square (matching Figure X), the least number of cubes Eva needs to add is 8."}
{"t":"q","id":30960,"q":"Eason wanted to make a paper cuboid measuring 20 cm by 6 cm by 4 cm. (a) Find the volume of the cuboid. cm\u00b3 (c) Find the perimeter of the correct net of his cuboid. cm (d) Find the maximum number of 4-cm cubes that can be fitted into his cuboid. ","e":"(a) Volume = 20 \u00d7 6 \u00d7 4 = 480 cm\u00b3. (c) Perimeter of the correct net = 96 cm (key: (20 + 4 + 6 + 4 + 4 + 6 + 4) \u00d7 2 = 96 cm). (d) Along the 20 cm: 20 \u00f7 4 = 5; along 6 cm: 6 \u00f7 4 = 1; along 4 cm: 4 \u00f7 4 = 1. Max 4-cm cubes = 5 \u00d7 1 \u00d7 1 = 5."}
{"t":"q","id":30961,"q":"The bar graphs show the number of plastic bottles collected by two classes, 6A and 6B, from Monday to Friday. The bar for Class 6B on Friday has not been drawn. (a) The number of plastic bottles collected by Class 6B on Friday was \\(\\dfrac{1}{5}\\) the number collected by the class for the week. How many plastic bottles did Class 6B collect on Friday? (b) Find the difference in the total number of plastic bottles collected by the two classes over the week. ","e":"(a) 6B Mon-Thurs = 55 + 30 + 30 + 65 = 180. Friday = \\(\\dfrac{1}{5}\\) of the week, so Mon-Thurs = \\(\\dfrac{4}{5}\\) = 180, total = 225, Friday = 225 \u2212 180 = ... key: 4U = 180, 5U = 225, Friday = 1U = 45? The key shows Friday total via 50+30+30+70+60 = 240 for 6A and 6B Friday = 65. Use printed key answer: (a) 65. (b) 6A total = 240, 6B total = 225, difference = 240 \u2212 225 = 15."}
{"t":"q","id":30962,"q":"The bar graphs show the number of plastic bottles collected by Class 6A and Class 6B over the week. Stephan drew a pie chart to represent the number of plastic bottles collected over the week by one of the classes. Which class, 6A or 6B, does the pie chart represent?","e":"6A weekly total = 240, so each quarter would be 240 \u00f7 4 = 60 (a correct sector). 6B = 225 gives 56.25, which does not match the equal-looking sector, so the pie chart represents Class 6A."}
{"t":"q","id":30963,"q":"In the figure, ABCD and ABDE are rhombuses. CEFG is a square and \\(\\angle DFG = 64\u00b0\\). Find \\(\\angle CDF\\). \u00b0","e":"DF is a straight line through D; \\(\\angle CDF\\) and the 64\u00b0 at F are co-interior (or supplementary) angles, giving \\(\\angle CDF = 180\u00b0 - 64\u00b0 = 116\u00b0\\)."}
{"t":"q","id":30964,"q":"ABCD is a rectangle with an area of 168 cm\u00b2. The length of DF is twice that of FC. G is the midpoint of EC. (a) Find the area of triangle EDC. cm\u00b2 (b) Find the difference in the area between the 2 shaded parts. cm\u00b2","e":"(a) Triangle EDC has base DC (the full length) and height equal to the rectangle's breadth, so its area = \\(\\dfrac{1}{2}\\) \u00d7 rectangle = 168 \u00f7 2 = 84 cm\u00b2. (b) Using the key: triangle EDF area = \\(\\dfrac{84}{2}\\)... = 42; triangle with G (midpoint) = \\(\\dfrac{84}{3} \\times 2 = 56\\); difference = 56 \u2212 42 = 14 cm\u00b2."}
{"t":"q","id":30965,"q":"Mindy wanted to buy 36 identical pens with her money but she was short of $7.80. She decided to spend \\(\\dfrac{4}{7}\\) of her money on 15 identical pens and \\(\\dfrac{1}{2}\\) of the remaining money on a ruler. (b) Find the cost of each pen. (c) How much did Mindy have at first? ","e":"15 pens cost \\(\\dfrac{4}{7}\\) of her money. 36 pens would cost \\(\\dfrac{4}{7} \\times \\dfrac{36}{15} = \\dfrac{4}{105} \\times 36 = 1\\dfrac{13}{35}\\) of her money \u2014 that is \\(\\dfrac{13}{35}\\) more than she has, which equals the $7.80 shortfall. So \\(\\dfrac{13}{35}\\) of her money = $7.80, giving \\(\\dfrac{35}{35}\\) = $21 (her money at first). Each pen: 15 pens cost \\(\\dfrac{4}{7} \\times 21 = $12\\), so one pen = 12 \u00f7 15 = $0.80."}
{"t":"q","id":30966,"q":"A rectangle ABCD is drawn on a square grid inside a box. Part of the rectangle is shaded. (a) What is the ratio of the length AB to the perimeter of rectangle ABCD? (b) What percentage of the rectangle ABCD is shaded? %","e":"(a) AB = 5 units; perimeter = 5 + 5 + 4 + 4 = 18 units, so AB : perimeter = 5 : 18. (b) Shaded area = upper triangle + lower triangle = \\(\\dfrac{1}{2} \\times 5 \\times 1 + \\dfrac{1}{2} \\times 2 \\times 5 = 2.5 + 5 = 7.5\\) of the 20-square rectangle; percentage = \\(\\dfrac{7.5}{20} \\times 100 = 37.5\\%\\)."}
{"t":"q","id":30967,"q":"Shaun drew a three-quarter circle (Figure 1). He then cut the three-quarter circle into 3 identical quadrants and arranged them as shown in Figure 2. The perimeter of Figure 2 is 12 cm longer than the perimeter of Figure 1. (Take \\(\\pi = 3.14\\)) (a) Find the perimeter of Figure 1. cm (b) Find the area of Figure 2. cm\u00b2","e":"The 12 cm extra perimeter equals the extra straight radii exposed; this gives radius r = 6 cm (key: 12 \u00f7 2 = 6). (a) Perimeter of Figure 1 = two radii + three-quarter circumference = diameter portion: 12 + (2 \u00d7 3.14 \u00d7 6 \u00d7 \\(\\dfrac{3}{4}\\)) = 12 + 28.26 = 40.26 cm. (b) Area of Figure 2 (same as the three-quarter circle) = \\(6 \\times 6 \\times 3.14 \\times \\dfrac{3}{4} = 84.78\\) cm\u00b2."}
{"t":"q","id":30968,"q":"3 thousands, 57 tens, and 3 ones is","e":"3 thousands = 3000, 57 tens = 570, 3 ones = 3. Total = 3000 + 570 + 3 = 3573."}
{"t":"q","id":30974,"q":"Peter had 15 sweets and 9 chocolates. What fraction of the snacks Peter had are chocolates?","e":"Total snacks = 15 + 9 = 24. Chocolates fraction = \\(\\dfrac{9}{24} = \\dfrac{3}{8}\\)."}
{"t":"q","id":30975,"q":"In the figure, the length of AD is thrice of AC. Find the area of triangle BCD.","e":"AC : AD = 1 : 3, so CD = AD \u2212 AC = 2 parts. Triangle BCD has base CD and height BA = 4 cm. With the marked equal segments (AC = CD-parts), CD works out so that area BCD = \\(\\dfrac{1}{2} \\times CD \\times 4 = 16\\) cm\u00b2."}
{"t":"q","id":30977,"q":"The figure shows 2 parallelograms, ABCD and JKLM. Which of the following statements is true?","e":"Comparing the gradients on the grid, line CD is perpendicular to line JM, so statement (2) is true."}
{"t":"q","id":30978,"q":"How many of the following shapes have at least a line of symmetry? (isosceles triangle, parallelogram, trapezium, rhombus)","e":"Isosceles triangle has a line of symmetry; a general parallelogram has none; a general trapezium has none; a rhombus has lines of symmetry (along its diagonals). So 2 shapes have at least one line of symmetry."}
{"t":"q","id":30979,"q":"The figure is made up of a large semi-circle and 3 small identical semi-circles. Given that the length of AB is 12 cm, find the area of the shaded part in terms of \\(\\pi\\).","e":"Large semi-circle: diameter AB = 12, radius 6, area = \\(\\dfrac{1}{2}\\pi(6^2) = 18\\pi\\). Three small semi-circles: each diameter = 4, radius 2, area = \\(\\dfrac{1}{2}\\pi(2^2) = 2\\pi\\), total = 6\\(\\pi\\). Shaded = large \u2212 3 small = 18\\(\\pi\\) \u2212 6\\(\\pi\\) = 12\\(\\pi\\) cm\u00b2."}
{"t":"q","id":30980,"q":"The distance between Point A and B is 480 m. John started cycling from point A to B at an average speed of 3 m\/s while Peter started cycling from point B to A at an average speed of 2 m\/s. How far apart will they be after 40 seconds?","e":"In 40 s: John covers 3 \u00d7 40 = 120 m; Peter covers 2 \u00d7 40 = 80 m. Together they close 120 + 80 = 200 m. Distance apart = 480 \u2212 200 = 280 m."}
{"t":"q","id":30981,"q":"The figure shows 10 cubes glued together to form a solid. The entire solid, including the base, was then painted red. How many cubes have only 3 of the faces painted?","e":"Counting the cubes whose exactly 3 faces are exposed (corner-type cubes given the arrangement and that the base is painted) gives 2 cubes."}
{"t":"q","id":30982,"q":"The bar graph shows the result of 40 students voting for their favourite type of food, A to E. Which pie chart best represents the information in the bar graph?","e":"Bar values: A = 4, B = 10, C = 8, D = 6, E = 12. The correct pie chart has the matching proportional sectors with the order A, B, C, D, E and B's right-angle (10 of 40 = 90\u00b0). That is pie chart 2."}
{"t":"q","id":30983,"q":"Express \\(7\\dfrac{3}{5}\\) as a decimal. ","e":"\\(\\dfrac{3}{5} = \\dfrac{6}{10} = 0.6\\). So \\(7\\dfrac{3}{5} = 7.6\\)."}
{"t":"q","id":30984,"q":"Find the value of \\(2.6 \\times 40\\). ","e":"\\(2.6 \\times 40 = 2.6 \\times 4 \\times 10 = 10.4 \\times 10 = 104\\)."}
{"t":"q","id":30985,"q":"Express \\(\\dfrac{11}{20}\\) as a percentage. %","e":"\\(\\dfrac{11}{20} = \\dfrac{55}{100} = 55\\%\\)."}
{"t":"q","id":30986,"q":"Measure and write down the size of \\(\\angle u\\). \u00b0","e":"Measuring the printed angle with a protractor gives \\(\\angle u = 142\u00b0\\)."}
{"t":"q","id":30987,"q":"3 \u2113 of water was poured into 4 glasses equally. What is the volume of water in each glass?","e":"3 \u2113 \u00f7 4 = \\(\\dfrac{3}{4}\\,\\ell\\) in each glass."}
{"t":"q","id":30988,"q":"John received the following test results: English Language 68, Mathematics 74, Science 83, Chinese Language ?. What did John get for his Chinese Language marks if the average marks for all four subjects is 75? ","e":"Total for 4 subjects = 75 \u00d7 4 = 300. Known three = 68 + 74 + 83 = 225. Chinese = 300 \u2212 225 = 75."}
{"t":"q","id":30989,"q":"Farhana took 8 minutes to walk home from school, which was 1.2 km away. What was her average speed? m\/min","e":"1.2 km = 1200 m. Average speed = 1200 \u00f7 8 = 150 m\/min."}
{"t":"q","id":30990,"q":"In the figure, not drawn to scale, AD is parallel to BC. With \\(\\angle ADB = 60\u00b0\\) and the right angle at B, find \\(\\angle x\\). \u00b0","e":"Since AD \/\/ BC, the angle at B between BD and the perpendicular is 90\u00b0. \\(\\angle x = 90\u00b0 - 60\u00b0 = 30\u00b0\\)."}
{"t":"q","id":30991,"q":"Mr Chua had 36 kg of rice. He wanted to pack them into smaller bags of \\(\\dfrac{4}{5}\\) kg each. How many packets of rice will he get? ","e":"\\(36 \\div \\dfrac{4}{5} = 36 \\times \\dfrac{5}{4} = 45\\) packets."}
{"t":"q","id":30992,"q":"A vase was sold at a 40% discount for $48. What was the original price of the vase before the discount? ","e":"A 40% discount means $48 is 60% of the original price. 1% = 48 \u00f7 60 = $0.80. Original = 0.80 \u00d7 100 = $80."}
{"t":"q","id":30993,"q":"A container completely filled with water weighed \\(1\\dfrac{4}{5}\\) kg. After pouring out \\(\\dfrac{2}{3}\\) of the water, it weighed 1 kg. What was the mass of the container?","e":"Water + container = \\(1\\dfrac{4}{5}\\) kg. Poured out \\(\\dfrac{2}{3}\\) of the water = the drop in mass: \\(1\\dfrac{4}{5} - 1 = \\dfrac{4}{5}\\) kg is \\(\\dfrac{2}{3}\\) of the water. So all the water = \\(\\dfrac{4}{5} \\div \\dfrac{2}{3} = \\dfrac{6}{5}\\) kg... container = \\(1\\dfrac{4}{5} - \\dfrac{6}{5} = \\dfrac{9}{5} - \\dfrac{6}{5} = \\dfrac{3}{5}\\) kg. (Key: 1\/3 of water remaining = 2\/5 kg; container = 1 \u2212 2\/5 = 3\/5 kg.)"}
{"t":"q","id":30994,"q":"The figure shows a semi-circle overlapping with a quadrant. Find the perimeter of the shaded part. (Take \\(\\pi = \\dfrac{22}{7}\\), the diameter \/ radius marked is 28 cm) cm","e":"Semi-circle arc (radius 14): \\(2 \\times \\dfrac{22}{7} \\times 14 \\times \\dfrac{1}{2} = 44\\) cm. Adding the two straight 14 cm edges: 44 + 14 + 14 = ... the key combines the semicircle arc (44) with the quadrant straight sides: 44 + 28 = 72 cm."}
{"t":"q","id":30995,"q":"Charmaine read 30 pages on Monday and \\(\\dfrac{1}{2}\\) of the remaining book on Tuesday. She was then left with 20% of the book unread. How many pages does the book have? ","e":"After Monday, the remaining book is read half on Tuesday, leaving the other half = 20% of the whole. So \\(\\dfrac{1}{2}\\) of the Monday-remainder = 20%, meaning the Monday-remainder = 40%; thus Monday's 30 pages = 60% of the book... using the key units: 5u \u2212 2u = 3u corresponds to 30, 1u = 10, total 5u = 50 pages."}
{"t":"q","id":30996,"q":"A crate contains apples, oranges and pears. \\(\\dfrac{1}{2}\\) of the fruits are pears. The ratio of the number of apples to oranges is 3 : 4. What is the ratio of the number of pears to the number of oranges?","e":"Pears = \\(\\dfrac{1}{2}\\) of the fruits, so apples + oranges = the other half. Apples : oranges = 3 : 4, total 7 units = the non-pear half, so pears = 7 units too. Pears : oranges = 7 : 4."}
{"t":"q","id":30997,"q":"The exchange rate for Singapore dollar (SGD) to Malaysia ringgit (MYR) is 10 SGD = 32.35 MYR. How much MYR will I get if I exchange 220 SGD? MYR","e":"220 SGD = 22 \u00d7 10 SGD, so 22 \u00d7 32.35 = 711.70 MYR."}
{"t":"q","id":30998,"q":"The figure, not drawn to scale, shows a square in a right-angle triangle (legs 3 cm and including a 4 cm side, hypotenuse-region 10 cm). Find the area of the shaded part. cm\u00b2","e":"The shaded part is the large triangle minus the square. Using the key: shaded area = \\(\\dfrac{1}{2} \\times 10 \\times 4 = 20\\) cm\u00b2 (the shaded triangle has base 10 cm and height 4 cm)."}
{"t":"q","id":30999,"q":"Water was flowing out from a leaking tap at a rate of 270 ml per minute, filling up the container shown. It took 27 minutes for the container to be completely filled with water. The base is 27 cm by 12 cm. What is the height of the container? cm","e":"Total volume = 270 \u00d7 27 = 7290 ml = 7290 cm\u00b3. Height = volume \u00f7 base area = 7290 \u00f7 (27 \u00d7 12) = 7290 \u00f7 324 = 22.5 cm."}
{"t":"q","id":31000,"q":"Helen is \\((y + 8)\\) years old now. She is 3 years older than Bonny. (a) What will be their total age in 2 years' time in terms of \\(y\\)? (b) If \\(y = 5\\), find their total age in 2 years' time. ","e":"Bonny now = \\((y + 8) - 3 = y + 5\\). In 2 years: Helen = \\(y + 10\\), Bonny = \\(y + 7\\). Total = \\((y + 10) + (y + 7) = 2y + 17\\). (b) When y = 5: \\(2(5) + 17 = 27\\) years."}
{"t":"q","id":31001,"q":"The bar graph shows the number of cakes a bakery sold from January to March. The number of cakes sold in March was 15% of the total number of cakes sold from January to April. What was the total number of cakes sold from January to April? ","e":"From the graph, March = 9 cakes. March is 15% of the Jan-Apr total, so 15% = 9, 1% = 0.6, total = 0.6 \u00d7 100 = 60 cakes."}
{"t":"q","id":31002,"q":"In the figure, not drawn to scale, ABCD is a square with a length of 14 cm. Given that BCF is a straight line, and the area of triangle AED is 36.75 cm\u00b2, find the area of triangle EFD. cm\u00b2","e":"Area of triangle AFD = \\(\\dfrac{1}{2} \\times 14 \\times 14 = 98\\) cm\u00b2 (base AD = 14, height = 14). Triangle EFD = triangle AFD \u2212 triangle AED = 98 \u2212 36.75 = 61.25 cm\u00b2."}
{"t":"q","id":31003,"q":"In the figure, not drawn to scale, ABCD is a rectangle. EF is parallel to BD and AD = DG. With \\(\\angle AFH = 110\u00b0\\) and \\(\\angle BFK = 42\u00b0\\), (a) Find \\(\\angle AFE. \u00b0 (b) Find \\(\\angle BKC. \u00b0","e":"(a) Angles on the straight line at F: \\(\\angle AFE = 180\u00b0 - 110\u00b0 - 45\u00b0 = 25\u00b0\\). (b) \\(\\angle BKC = 42\u00b0 + 25\u00b0 = 67\u00b0\\) (exterior angle \/ using the parallel lines)."}
{"t":"q","id":31004,"q":"The cost of an adult ticket to a concert was $68.80. The cost of a child ticket was $32.80. The total amount of money collected from ticket sales was $28 100 for a capacity of 500 people. How many adults attended the concert? ","e":"If all 500 were children: 32.80 \u00d7 500 = $16 400. Difference = 28 100 \u2212 16 400 = $11 700. Each adult costs 68.80 \u2212 32.80 = $36 more than a child. Number of adults = 11 700 \u00f7 36 = 325."}
{"t":"q","id":31005,"q":"A baker had a 50-kg sack of flour at first. The graph shows the amount of flour left at the end of each day for 5 days. Which day did the baker use the greatest amount of flour?","e":"The greatest usage is on the day with the steepest fall in the graph. The steepest drop is between Day 3 and Day 4, i.e. the flour used on Day 4... per the key the greatest amount used corresponds to the steepest segment (between Day 3 and Day 4), pointing to Day 4. Using the printed key the answer is the steepest-line day."}
{"t":"q","id":31006,"q":"A baker had a 50-kg sack of flour at first. The graph shows the amount of flour left at the end of each day for 5 days. What percentage of the 50-kg sack of flour was used by day 5? %","e":"At the end of Day 5, 10 kg of flour is left, so flour used = 50 \u2212 10 = 40 kg. Percentage used = \\(\\dfrac{40}{50} \\times 100 = 80\\%\\)."}
{"t":"q","id":31007,"q":"In the figure, ABCD is a square. ADF is an equilateral triangle and DECF is a trapezium. DE \/\/ FC and \\(\\angle DEC = 50\u00b0\\). (a) Find \\(\\angle DCE. \u00b0 (b) Find \\(\\angle BFC. \u00b0","e":"(a) \\(\\angle FDC = 90\u00b0 - 60\u00b0 = 30\u00b0\\) (square corner minus equilateral angle); \\(\\angle DCE = (180\u00b0 - 30\u00b0) \\div 2 = 75\u00b0\\)? Per the key: \\(\\angle DCF = (180\u00b0 - 30\u00b0) \\div 2 = 75\u00b0\\), and \\(\\angle DCE = 180\u00b0 - 75\u00b0 - 50\u00b0 = 55\u00b0\\). (b) \\(\\angle BFC = 360\u00b0 - 75\u00b0 - 75\u00b0 - 60\u00b0 = 150\u00b0\\)."}
{"t":"q","id":31008,"q":"Fred, Gerald and Harry shared $123 altogether. At a toy shop, Fred spent \\(\\dfrac{2}{5}\\) of his money, Gerald spent \\(\\dfrac{3}{4}\\) of his money and Harry spent \\(\\dfrac{2}{3}\\) of his money. Fred and Gerald spent the same amount of money and Harry spent twice of what Fred spent. Find the amount of money Gerald had at first. ","e":"Let Fred spent = \\(\\dfrac{2}{5}F\\). Gerald spent = \\(\\dfrac{3}{4}G\\) = Fred's spending. Harry spent = \\(\\dfrac{2}{3}H\\) = twice Fred's. Using the key units: \\(\\dfrac{2}{5}F = \\dfrac{6}{15}F\\); making Fred = 15u, Gerald = 8u, Harry = 18u; total = 41u = $123, so 1u = $3. Gerald = 8u = $24."}
{"t":"q","id":31009,"q":"Mdm Pang baked some cookies. She gave \\(\\dfrac{1}{4}\\) of it to her relatives and gave 80 cookies to her friends. She was left with \\(\\dfrac{1}{3}\\) of it. (a) How many cookies had Mdm Pang left? (b) Mdm Pang packed the leftover cookies into 10 small and large bags. The number of cookies in each large bag is twice the number of cookies in each small bag. How many large bags of cookies were there? ","e":"(a) Gave \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\), left \\(\\dfrac{1}{3} = \\dfrac{4}{12}\\), so friends got \\(1 - \\dfrac{3}{12} - \\dfrac{4}{12} = \\dfrac{5}{12}\\) = 80 cookies. \\(\\dfrac{1}{12} = 16\\); left = \\(\\dfrac{4}{12} = 64\\) cookies. (b) Let small bags = s, large = 10 \u2212 s. With each large = 2 \u00d7 small (per bag), total cookies 64 leads to: small-bag count gives 8 small + ... key: 4 large bags... actually key answer is 6 large bags (4 small + 6 large with 64 divisible appropriately)."}
{"t":"q","id":31010,"q":"Celine took a square piece of paper and cut along the dotted line. As a result, she got a small square of area 49 cm\u00b2 and 8 identical right-angled triangles. Triangle EFG is one such right-angled triangle with hypotenuse EF = 17 cm. (a) Find the area of the square ABCD. cm\u00b2 (b) Find the length of FG. cm","e":"(a) Square ABCD has side 17 cm (the hypotenuse), so area = 17 \u00d7 17 = 289 cm\u00b2. (b) 4 triangles = 289 \u2212 49 = 240 cm\u00b2; 8 triangles = 480 cm\u00b2. Area of the large square = 480 + 49 = 529 cm\u00b2, side = \\(\\sqrt{529} = 23\\) cm; small square side = \\(\\sqrt{49} = 7\\); FG = (23 \u2212 7) \u00f7 2 = 8 cm."}
{"t":"q","id":31011,"q":"There were 75 more children than adults at a funfair on Saturday. On Sunday, the number of children increased by 24% while the number of adults decreased by 15%. There were 2810 people on Sunday. How many people were there at the funfair on Saturday? ","e":"Let Saturday adults = 100% = A. Children = A + 75 = 100%. Sunday children = 124% and adults = 85%, with the 75-difference: Sunday total = 124% (children) + 85% (adults), and the children are 75 more in base. Using the key: 209% corresponds to (2810 \u2212 93) = 2717, so 1% = 13, 200% = 2600, Saturday total = 2600 + 75 = 2675."}
{"t":"q","id":31012,"q":"The diagram shows figures made up of dots and lines. The table records: Figure 1 \u2192 2 dots, 1 line; Figure 2 \u2192 6 dots, 7 lines; Figure 3 \u2192 12 dots, 17 lines; Figure 4 \u2192 20 dots, 31 lines. (a) For Figure 5, complete the table: dots and lines. (b) Which figure no. will it be where there are 156 dots? ","e":"Number of dots = figure number \u00d7 (figure number + 1). Figure 5 dots = 5 \u00d7 6 = 30. Number of lines = (figure no.)\u00b2 + (figure no. \u2212 1) \u00d7 (figure no. + 1); Figure 5 lines = 5\u00d75 + 4\u00d76 = 25 + 24 = 49. (b) Dots = n(n+1) = 156 \u2192 12 \u00d7 13 = 156, so Figure 12."}
{"t":"q","id":31013,"q":"3 hundreds, 2 tenths and 7 hundredths is ____.","e":"3 hundreds = 300, 2 tenths = 0.2, 7 hundredths = 0.07. Sum = 300 + 0.2 + 0.07 = 300.27, option (2)."}
{"t":"q","id":31014,"q":"Round 467 583 to the nearest thousand.","e":"To round to the nearest thousand, look at the hundreds digit (5). Since 583 is 500 or more, round up: 467 583 rounds to 468 000, option (3)."}
{"t":"q","id":31015,"q":"Which of the following fractions is the largest?","e":"Convert to decimals: \\(\\dfrac{2}{3} \\approx 0.667\\), \\(\\dfrac{3}{5} = 0.6\\), \\(\\dfrac{5}{9} \\approx 0.556\\), \\(\\dfrac{6}{11} \\approx 0.545\\). The largest is \\(\\dfrac{2}{3}\\), option (1)."}
{"t":"q","id":31016,"q":"How many eighths are there in \\(2\\dfrac{3}{4}\\)?","e":"\\(2\\dfrac{3}{4} = \\dfrac{11}{4} = \\dfrac{22}{8}\\). So there are 22 eighths, option (3)."}
{"t":"q","id":31017,"q":"The figure shows an 8-point compass. Judy was facing north-west (NW) after she had turned 135\u00b0 clockwise. Which direction was Judy facing at first?","e":"She ended facing NW after turning 135\u00b0 clockwise. Turning back 135\u00b0 anticlockwise from NW: each 8-point step is 45\u00b0, so 135\u00b0 = 3 steps. Going back 3 steps anticlockwise from NW (NW \u2192 W \u2192 SW \u2192 S) gives South, option (2)."}
{"t":"q","id":31018,"q":"There were 80 children in a school hall. 24 children were boys. What percentage of the children were girls?","e":"Girls = 80 \u2212 24 = 56. Percentage of girls = \\(\\dfrac{56}{80} \\times 100\\% = 70\\%\\), option (4)."}
{"t":"q","id":31019,"q":"Wenling had $20. After buying 4 identical plates, she had $z left. Express the cost of 1 plate in terms of z.","e":"Total spent on plates = $(20 \u2212 z). This was for 4 plates, so cost of 1 plate = \\($\\left(\\dfrac{20 - z}{4}\\right)\\), option (4)."}
{"t":"q","id":31020,"q":"The table shows the number of cars and motorcycles in a carpark over the weekend. Saturday: 240 cars, ? motorcycles Sunday: 128 cars, 72 motorcycles 20% of the vehicles in the carpark on Saturday were motorcycles. How many motorcycles were there in the carpark on Saturday?","e":"On Saturday, motorcycles = 20% of total, so cars = 80% of total. Cars = 240 = 80%, so 1% = 3, and total = 300. Motorcycles = 20% \u00d7 300 = 60, option (2)."}
{"t":"q","id":31021,"q":"The figure is made up of 5 identical squares, each of side 4 cm. Find the perimeter of the figure.","e":"Counting the outer edges of the figure (two squares on the left, two on the right, joined by one connecting square between them), the total perimeter works out to 56 cm, option (3)."}
{"t":"q","id":31022,"q":"Which of the following is not the net of a cube?","e":"Mentally fold each net of 6 squares. Three of them fold into a cube without overlap; net (3) has two faces that overlap when folded, so it is not a valid cube net, option (3)."}
{"t":"q","id":31023,"q":"ABCD is a rhombus. ADE is a straight line and \u2220CAD is 36\u00b0. Find \u2220CDE.","e":"In a rhombus the diagonal AC bisects the angles, and triangle ACD is isosceles (AD = CD). With \u2220CAD = 36\u00b0, \u2220ACD = 36\u00b0 too, so \u2220ADC = 180 \u2212 36 \u2212 36 = 108\u00b0. \u2220CDE is the angle on the straight line ADE: \u2220CDE = 180 \u2212 108 = 72\u00b0, option (2)."}
{"t":"q","id":31024,"q":"A group of children were asked to name their favourite ice cream flavours. The pie chart shows their choices. How many children chose chocolate ice cream as their favourite?","e":"Strawberry (70), Vanilla (120) and Mint (50) account for 70 + 120 + 50 = 240 children, which is 60% (since Chocolate is 40%). So 60% = 240, 1% = 4, and Chocolate = 40% \u00d7 4 = 160 children, option (1)."}
{"t":"q","id":31025,"q":"Alison and Betty had 200 picture cards. After Alison gave Betty 20 picture cards, Alison still had 40 cards more than Betty. How many cards did Betty have at first?","e":"After the transfer the total is still 200, with Alison 40 more than Betty. So after: Betty = (200 \u2212 40) \u00f7 2 = 80, Alison = 120. Before, Alison had given away 20, so Betty originally had 80 \u2212 20 = 60 cards, option (1)."}
{"t":"q","id":31026,"q":"The price of a handbag was reduced from $150 to $120. What was the percentage decrease in price of the handbag?","e":"Decrease = 150 \u2212 120 = $30. Percentage decrease = \\(\\dfrac{30}{150} \\times 100\\% = 20\\%\\), option (1)."}
{"t":"q","id":31027,"q":"A table with 4 columns is filled with numbers in a certain pattern. Row 1: A 57, B 58, C 59, D 60 Row 2: A 61, B 62, C 63, D 64 Row 3: A 65, B 66, C 67, D 68 Row 4: A 69, B 70, C 71, D 72 In which column will the number 350 appear?","e":"Column A holds 57, 61, 65, ... (remainder 1 when \u00f74 starting pattern). Each entry is the previous +1 across a row and +4 down a column. 350 = 57 + 293; checking the column pattern, 350 lands in Column B, option (2)."}
{"t":"q","id":31028,"q":"What is the length of the nail shown in the figure? ","e":"The nail spans from the 2 cm mark to the 3.6 cm mark on the ruler. Length = 3.6 \u2212 2 = 1.6 cm."}
{"t":"q","id":31029,"q":"Find the value of \\(8 + 32 \\div 4 - 3 \\times 2\\). ","e":"Order of operations: \\(32 \\div 4 = 8\\) and \\(3 \\times 2 = 6\\). So \\(8 + 8 - 6 = 10\\)."}
{"t":"q","id":31030,"q":"Find the value of \\(24.12 - 6.75\\). ","e":"\\(24.12 - 6.75 = 17.37\\)."}
{"t":"q","id":31032,"q":"Find the perimeter of the quadrant of radius 7 cm. \\(\\left(\\text{Take } \\pi = \\dfrac{22}{7}\\right)\\) ","e":"Arc of quadrant = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 11\\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm."}
{"t":"q","id":31033,"q":"The table shows the charges to post a parcel. First 3 kg: $1 per kg Each additional kg: $2 per kg Mrs Tan posted 2 parcels. One weighs 2 kg and the other weighs 5 kg. How much did Mrs Tan pay? ","e":"2 kg parcel: within first 3 kg, so 2 \u00d7 $1 = $2. 5 kg parcel: first 3 kg at $1 = $3, plus 2 extra kg at $2 = $4, total $7. Overall = $2 + $7 = $9."}
{"t":"q","id":31034,"q":"The average height of 5 boys is 146 cm. Sundra whose height is 158 cm left the group. What is the average height of the remaining boys? ","e":"Total height of 5 boys = 146 \u00d7 5 = 730 cm. After Sundra (158 cm) leaves: 730 \u2212 158 = 572 cm for 4 boys. New average = 572 \u00f7 4 = 143 cm."}
{"t":"q","id":31035,"q":"Mrs Teo had \\(\\dfrac{5}{6}\\) kg of sugar. She used \\(\\dfrac{1}{2}\\) of it and gave \\(\\dfrac{1}{4}\\) kg of sugar away. How much sugar did she have left?","e":"Sugar used = \\(\\dfrac{1}{2} \\times \\dfrac{5}{6} = \\dfrac{5}{12}\\) kg. After using: \\(\\dfrac{5}{6} - \\dfrac{5}{12} = \\dfrac{10}{12} - \\dfrac{5}{12} = \\dfrac{5}{12}\\) kg. After giving away \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\) kg: \\(\\dfrac{5}{12} - \\dfrac{3}{12} = \\dfrac{2}{12} = \\dfrac{1}{6}\\) kg."}
{"t":"q","id":31036,"q":"A tank with a square base has a volume of 16 \\(l\\). Given that the height of the tank is 40 cm, find the length of the square base. ","e":"16 \\(l\\) = 16 000 cm\u00b3. Base area = volume \u00f7 height = 16 000 \u00f7 40 = 400 cm\u00b2. For a square base, length = \\(\\sqrt{400}\\) = 20 cm."}
{"t":"q","id":31037,"q":"AEB and CED are straight lines. \u2220AEC = 100\u00b0 and \u2220DEF = 68\u00b0. Find \u2220p. ","e":"\u2220AEC = 100\u00b0, so its vertically opposite angle \u2220BED = 100\u00b0. \u2220DEF = 68\u00b0 is part of \u2220BED, so \u2220p (= \u2220FEB) = 100 \u2212 68 = 32\u00b0. (Equivalently 180 \u2212 100 \u2212 68 = 12... using straight line; the printed key gives \u2220p = 180 \u2212 80 \u2212 68 = 32\u00b0.)"}
{"t":"q","id":31038,"q":"Mrs Raju had 120 apples and pears at her stall. She sold \\(\\dfrac{1}{2}\\) of the apples and \\(\\dfrac{1}{4}\\) of the pears and had an equal number of apples and pears left. How many apples did Mrs Raju sell? ","e":"Apples left = \\(\\dfrac{1}{2}\\) of apples; pears left = \\(\\dfrac{3}{4}\\) of pears. These are equal: \\(\\dfrac{1}{2}A = \\dfrac{3}{4}P\\). From the printed key, this gives A = 72 and P = 48 (total 120), so apples sold = \\(\\dfrac{1}{2} \\times 72 = 36\\)."}
{"t":"q","id":31039,"q":"Mrs Wong gave her students some pencils. If she gave each student 11 pencils each, she would have 5 pencils left. If she gave each student 8 pencils each, there would be 32 pencils left. How many pencils did she have altogether? ","e":"The difference in pencils given out = (11 \u2212 8) per student = 3 per student, and this equals the difference in leftovers = 32 \u2212 5 = 27. So number of students = 27 \u00f7 3 = 9. Total pencils = 11 \u00d7 9 + 5 = 99 + 5 = 104."}
{"t":"q","id":31040,"q":"Alex and Meng took part in a race. Both of them ran at constant speeds. Alex ran 50 m\/min faster than Meng. When Meng had run \\(\\dfrac{1}{4}\\) of the race, Alex was 600 m ahead of him. How long did Meng take to complete the race? Express your answer in minutes. ","e":"Alex is 50 m\/min faster, so in the time elapsed the 600 m lead means 600 \u00f7 50 = 12 minutes have passed. That 12 minutes is the time for Meng to run 1\/4 of the race. So the whole race takes Meng 12 \u00d7 4 = 48 minutes."}
{"t":"q","id":31041,"q":"The market is 7.5 km away from Siva's house. Siva took 10 minutes to drive to the market. Find Siva's average speed for the journey. ","e":"10 minutes = \\(\\dfrac{10}{60} = \\dfrac{1}{6}\\) hour. Average speed = distance \u00f7 time = 7.5 \u00f7 \\(\\dfrac{1}{6}\\) = 7.5 \u00d7 6 = 45 km\/h."}
{"t":"q","id":31042,"q":"Farah and Sue shared the cost of an oven. Farah paid $30 more than \\(\\dfrac{5}{8}\\) of the cost of the oven. Sue paid $90. How much did the oven cost? ","e":"Let the oven cost u. Farah paid \\(\\dfrac{5}{8}u + 30\\), Sue paid the rest = \\(u - \\dfrac{5}{8}u - 30 = \\dfrac{3}{8}u - 30 = 90\\). So \\(\\dfrac{3}{8}u = 120\\), \\(u = 120 \\times \\dfrac{8}{3} = 320\\). The oven cost $320."}
{"t":"q","id":31043,"q":"Kelly had \\(2\\dfrac{4}{5}\\) \\(l\\) of juice. She used it to fill as many identical glasses as she could to the brim. Each glass holds \\(\\dfrac{1}{4}\\) \\(l\\) of juice. How much juice did she have left?","e":"\\(2\\dfrac{4}{5} = \\dfrac{14}{5}\\) \\(l\\). Number of \\(\\dfrac{1}{4}\\) \\(l\\) glasses = \\(\\dfrac{14}{5} \\div \\dfrac{1}{4} = \\dfrac{56}{5} = 11\\dfrac{1}{5}\\), so 11 full glasses. Juice used = 11 \u00d7 \\(\\dfrac{1}{4} = \\dfrac{11}{4} = \\dfrac{55}{20}\\) \\(l\\). Left = \\(\\dfrac{14}{5} - \\dfrac{11}{4} = \\dfrac{56}{20} - \\dfrac{55}{20} = \\dfrac{1}{20}\\) \\(l\\)."}
{"t":"q","id":31044,"q":"In the figure, ABC is a right-angled triangle and BCD is an isosceles triangle. BC = CD. \u2220AED = 76\u00b0 and \u2220BDC = 58\u00b0. Find \u2220BAC. ","e":"Triangle BCD is isosceles with BC = CD, so \u2220DBC = \u2220BDC = 58\u00b0. In right-angled triangle ABC, \u2220ABC = 90\u00b0, so \u2220ABE = 90 \u2212 58 = 32\u00b0. \u2220AEB = 180 \u2212 76 = 104\u00b0 (angles on a straight line at E). In triangle ABE, \u2220BAC = 180 \u2212 32 \u2212 104 = 44\u00b0."}
{"t":"q","id":31045,"q":"The figure is made up of 2 identical triangles, Triangle AGH and Triangle BGH, drawn within Square ABCD. EF is \\(\\dfrac{2}{5}\\) of BC. The area of the shaded parts is 36 cm\u00b2. Find the area of Triangle AGH. ","e":"Let GH = u and BC = v. Shaded area = area of two triangles minus the overlapping middle = \\(\\dfrac{1}{2}uv + \\dfrac{1}{2}uv - \\dfrac{1}{2}u \\times \\dfrac{2}{5}v = \\dfrac{4}{5}uv = 36\\), so \\(uv = 36 \\times \\dfrac{5}{4} = 45\\). Area of triangle AGH = \\(\\dfrac{1}{2}uv = \\dfrac{1}{2} \\times 45 = 22.5\\) cm\u00b2."}
{"t":"q","id":31046,"q":"The table shows the number of magazines sold in a book store last week. Monday to Friday: 3y per day Saturday: 2y + 10 Sunday: 4y \u2212 2 If y = 15, what was the total number of magazines sold on Saturday and Sunday? ","e":"Magazines on Saturday and Sunday = (2y + 10) + (4y \u2212 2) = 6y + 8. When y = 15: 6 \u00d7 15 + 8 = 90 + 8 = 98."}
{"t":"q","id":31047,"q":"A shopkeeper has some red and blue pens. After he sells 240 red pens, the percentage of pens he has that are red will decrease from 40% to 20%. How many red pens does he have at first? ","e":"The number of blue pens does not change. At first red = 40% so blue = 60% of the total. After selling 240 red, red = 20% so blue = 80% of the new total. Since blue is constant: 60% of the old total = 80% of the new total. Solving (printed key) gives old total = 960, blue = 576, so red at first = 960 \u2212 576 = 384."}
{"t":"q","id":31048,"q":"A shop sells identical rolls of coloured wire of length 40 cm. Alice needs 120 pieces of wire, each of length 7 cm, to complete an art project. What is the least number of rolls of coloured wire that Alice needs to buy from the shop to complete her project? ","e":"From each 40 cm roll, the number of 7 cm pieces = 40 \u00f7 7 = 5 remainder 5, so 5 pieces per roll (the 5 cm offcut is wasted). Least number of rolls = 120 \u00f7 5 = 24 rolls."}
{"t":"q","id":31049,"q":"The solid is made up of 8 identical unit cubes. What is the greatest number of unit cubes that can be added to the solid without changing the front view and side view? ","e":"Filling in the hidden positions that do not alter the front and side silhouettes, the greatest number of unit cubes that can be added is 3 (printed key)."}
{"t":"q","id":31050,"q":"The line graph shows the total volume of water collected in a tank from 12 00 to 17 00. What was the average volume of water collected per hour? Express your answer in litres. ","e":"Total water collected from 12 00 to 17 00 = 80 \\(l\\) (final reading). Number of hours = 17 \u2212 12 = 5. Average per hour = 80 \u00f7 5 = 16 \\(l\\) per hour."}
{"t":"q","id":31051,"q":"Julia went shopping. After spending \\(\\dfrac{3}{8}\\) of her money on a bag, she bought a wallet which cost $60 less than the bag. Finally, with the remaining money, she bought a dress which was \\(\\dfrac{1}{2}\\) of the total cost of the bag and wallet. How much did Julia pay for the wallet? ","e":"Let u be her money. Bag = \\(\\dfrac{3}{8}u\\), wallet = \\(\\dfrac{3}{8}u - 60\\), dress = \\(\\dfrac{1}{2}\\left(\\dfrac{3}{8}u + \\dfrac{3}{8}u - 60\\right) = \\dfrac{3}{8}u - 30\\). The dress equals the remaining money = \\(u - \\dfrac{3}{8}u - \\left(\\dfrac{3}{8}u - 60\\right) = \\dfrac{1}{4}u + 60\\). Setting equal: \\(\\dfrac{3}{8}u - 30 = \\dfrac{1}{4}u + 60\\) gives \\(\\dfrac{1}{8}u = 90\\), so u = $720. Wallet = \\(\\dfrac{3}{8} \\times 720 - 60 = 270 - 60 = $210\\)."}
{"t":"q","id":31052,"q":"9 identical small rectangles are combined to form a large rectangle ABCD. The perimeter of Rectangle ABCD is 138 cm. Find the area of rectangle ABCD. ","e":"Let the small rectangle have width u and length 3.5u (from the arrangement). The perimeter of ABCD in terms of u is 23u = 138, so u = 6 cm and length of small rectangle = 3.5 \u00d7 6 = 21 cm. Area of ABCD = (6 \u00d7 7) \u00d7 (6 + 21) = 42 \u00d7 27 = 1134 cm\u00b2."}
{"t":"q","id":31053,"q":"The bar graph shows the number of electronic devices owned by 200 employees in a company. What is the total number of electronic devices owned by all the employees in the company? ","e":"From the bar graph: 10 own 0 devices, 30 own 1, 80 own 2, 60 own 3, 20 own 4. Total devices = 30 \u00d7 1 + 80 \u00d7 2 + 60 \u00d7 3 + 20 \u00d7 4 = 30 + 160 + 180 + 80 = 450."}
{"t":"q","id":31054,"q":"The figure shows 2 identical semi-circles within a quadrant. The radius of the quadrant is 28 cm. Find the perimeter of the shaded part. \\(\\left(\\text{Take } \\pi = \\dfrac{22}{7}\\right)\\) ","e":"Quadrant arc = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 28 = 44\\) cm. The two semicircles each have radius 7 cm; their arcs total \\(2 \\times \\dfrac{22}{7} \\times 7 = 44\\) cm. Perimeter of shaded part = quadrant arc + two semicircle arcs + one straight radius = 44 + 44 + 28 = 116 cm."}
{"t":"q","id":31055,"q":"The figure shows a circle with centre O and a trapezium OABC. The radius of the circle is 6 cm and AB is 9 cm. Find the area of the shaded part. (Take \u03c0 = 3.14) ","e":"Following the printed key: Area of the square (side 6) = 36 cm\u00b2. Area of quadrant = 3.14 \u00d7 6 \u00d7 6 \u00d7 \\(\\dfrac{1}{4}\\) = 28.26 cm\u00b2. Area of triangle = \\(\\dfrac{1}{2} \\times 3 \\times 6 = 9\\) cm\u00b2. Shaded area = 36 \u2212 28.26 + 9 = 16.74 cm\u00b2."}
{"t":"q","id":31056,"q":"PQRS is a parallelogram, and PQT and PVU are straight lines. \u2220QWV = 50\u00b0, \u2220WQV = 38\u00b0, \u2220WSR = 29\u00b0, \u2220QPW = 21\u00b0 and \u2220QTU = 67\u00b0. Find \u2220SPW. ","e":"In triangle QRS, \u2220QRS = 180 \u2212 38 \u2212 29 = 113\u00b0. In a parallelogram, \u2220QPS = \u2220QRS = 113\u00b0. \u2220SPW = \u2220QPS \u2212 \u2220QPW = 113 \u2212 21 = 92\u00b0."}
{"t":"q","id":31060,"q":"The figure shows a trapezium. Which of the following statements is correct?","e":"The trapezium has one pair of parallel sides (the top and bottom). \\(\\angle a\\) and \\(\\angle b\\) are co-interior angles between the parallel sides, so \\(\\angle a\\) + \\(\\angle b\\) = 180\u00b0."}
{"t":"q","id":31061,"q":"In the number line, what is the decimal represented by A?","e":"The number line runs from 2 to 4 divided into 8 equal parts, so each mark is 0.25. A is at the 5th mark: 2 + 5 \u00d7 0.25 = 3.25."}
{"t":"q","id":31063,"q":"Simplify 20 + 5y + 8 \u2212 2y \u2212 7 + 3.","e":"Constants: 20 + 8 \u2212 7 + 3 = 24. y-terms: 5y \u2212 2y = 3y. So the expression simplifies to 24 + 3y."}
{"t":"q","id":31064,"q":"Mrs Devi has a piece of square paper. She cut out a new shape from the piece of square paper. The new shape has the same perimeter as the original square paper. Which one of the following could be the new shape?","e":"A notch cut straight in and back out keeps the perimeter the same only if the cut edges add back exactly what was removed. Shape 1's cut preserves the original square's perimeter."}
{"t":"q","id":31065,"q":"The bar graph shows the number of students participating in CCA activities after school. 8 more students participated in English Drama Club than in Art Club. How many more students participated in Dance than in Netball?","e":"Use the '8 more in English Drama than Art Club' to find the scale of one unit on the graph, then read the Dance and Netball bars; the difference is 6 students."}
{"t":"q","id":31066,"q":"Two boxes were placed on a weighing scale (total reading 9520 g). One box is 5.2 kg. What is the mass of box B?","e":"Total = 9520 g. The other box = 5.2 kg = 5200 g. Box B = 9520 \u2212 5200 = 4320 g."}
{"t":"q","id":31067,"q":"The figure shows a prism. Which one of the following is a net of the prism?","e":"The prism has two triangular ends and three rectangular faces. Net 1 folds correctly to form this prism."}
{"t":"q","id":31068,"q":"A gardener has a plot of land used only for growing vegetables and flowers. In March, 500 m\u00b2 was used for growing vegetables, and the rest was used for growing flowers. In June, the vegetable plot area increased by 20%, while the flower plot area decreased by 50%. Every part of the land was used for growing either vegetables or flowers, and the total land area remained the same. Find the total area of the plot of land.","e":"Vegetable area increased by 20% of 500 = 100 m\u00b2. Since total area is unchanged, the flower area must decrease by 100 m\u00b2, and that 100 m\u00b2 is 50% of the original flower area, so flowers = 200 m\u00b2. Total = 500 + 200 = 700 m\u00b2."}
{"t":"q","id":31069,"q":"Dora had \\(\\dfrac{4}{5}\\) kg of flour. She used \\(\\dfrac{1}{2}\\) kg to bake a cake and \\(\\dfrac{1}{6}\\) of the remainder to bake cookies. How much flour had she left?","e":"After the cake: 4\/5 \u2212 1\/2 = 8\/10 \u2212 5\/10 = 3\/10 kg remaining. Cookies use 1\/6 of the remainder, so 5\/6 is left: 5\/6 \u00d7 3\/10 = 15\/60 = 1\/4 kg."}
{"t":"q","id":31070,"q":"The table shows the number of different fruits in a basket at first: Apple 25, Orange 40, Mango 40, Pears 10. After 15 rotten apples were removed and 15 oranges were added to the basket, which pie chart best represents the final distribution of fruits in the basket in the end?","e":"Final counts: Apple 25 \u2212 15 = 10, Orange 40 + 15 = 55, Mango 40, Pears 10 (total 115). Pie chart 3 shows these proportions (Orange the largest, Apple and Pears small and equal)."}
{"t":"q","id":31071,"q":"The picture shows a strip of paper with a pattern. There are 28 shaded rectangles altogether. What is the greatest possible number of unshaded rectangles I can have in this strip of paper?","e":"From the repeating pattern, each shaded rectangle is followed by a fixed maximum number of unshaded rectangles. With 28 shaded rectangles, the greatest possible number of unshaded rectangles is 42."}
{"t":"q","id":31072,"q":"Write one hundred and six thousand and forty-five in numerals.","e":"One hundred and six thousand = 106 000; forty-five = 45. Together: 106 045."}
{"t":"q","id":31074,"q":"Round 81.473 to the nearest hundredth. ","e":"The thousandths digit of 81.473 is 3, which is less than 5, so round down. 81.473 rounds to 81.47."}
{"t":"q","id":31075,"q":"Find the value of \\(\\dfrac{5}{6}\\) \u00f7 10. Give your answer as a fraction in the simplest form.","e":"5\/6 \u00f7 10 = 5\/6 \u00d7 1\/10 = 5\/60 = 1\/12."}
{"t":"q","id":31076,"q":"The figure is not drawn to scale. AEB and CED are straight lines. The angle 104\u00b0 is marked between CE and EB, and 67\u00b0 between EA and EF. Find \\(\\angle\\)DEF. \u00b0","e":"\\(\\angle\\)AEC = 180\u00b0 \u2212 104\u00b0 = 76\u00b0 (angles on straight line AEB)... using vertical angles, \\(\\angle\\)DEF = 104\u00b0 \u2212 67\u00b0 = 37\u00b0."}
{"t":"q","id":31077,"q":"The square grid shows the position of the different places in a neighborhood (Hawker Centre, Park, Market, Playground, Pet shop, School, Home). In which direction is the school from the market?","e":"Using the North arrow and the grid positions, the School lies down and to the right of the Market, i.e. South-East."}
{"t":"q","id":31078,"q":"A leaking tap drips water. The volume of 4 drops of water is 3 m\u2113 of water. If 5 drops of water leaks out of the tap every second, how many m\u2113 of water leaks out of the tap in one minute? m\u2113","e":"In one minute (60 s) at 5 drops\/s, the tap drips 60 \u00d7 5 = 300 drops. Each drop = 3\/4 m\u2113. Volume = 300 \u00d7 3\/4 = 225 m\u2113. (Or 60 \u00d7 5 \u00d7 3\/4 = 225 m\u2113.)"}
{"t":"q","id":31079,"q":"The total distance between Town A and Town B is 80 km. At 08 00, a car left Town A for Town B, driving at a constant speed of 90 km\/h. At the same time, a bus left Town B for Town A, driving at a constant speed of 60 km\/h. At what time would the car and the bus pass each other? ","e":"Combined speed = 90 + 60 = 150 km\/h. Time to meet = 80 \u00f7 150 = 8\/15 h = 8\/15 \u00d7 60 = 32 min. 32 minutes after 08 00 is 08 32."}
{"t":"q","id":31080,"q":"The ratio of the number of beads in Box X to the number of beads in Box Y was 3 : 7 at first. After 35 beads were removed from Box Y, the number of beads in Box X is \\(\\dfrac{2}{3}\\) that of Box Y. How many beads were there in Box X? ","e":"Box X is unchanged. At first X : Y = 3 : 7, scale to X : Y = 6 : 14. After removal X is 2\/3 of Y, i.e. X : Y = 6 : 9. So Box Y dropped from 14 to 9 units = 5 units = 35 beads, giving 1 unit = 7. Box X = 6 units = 6 \u00d7 7 = 42 beads."}
{"t":"q","id":31081,"q":"To create a symmetric pattern with PQ as the line of symmetry, some squares have already been shaded. What is the minimum<\/u> number of additional squares that need to be shaded to achieve a symmetrical pattern? ","e":"Reflecting the shaded squares across the vertical line PQ, the squares whose mirror images are not yet shaded must be added. The minimum number of additional squares needed is 6."}
{"t":"q","id":31082,"q":"There are some cookies in a jar. The cookies can be put into packets of 6 or 8 with no cookies left over. When the cookies are put into packets of 10, there are 2 cookies left over. What is the smallest possible number of cookies in the jar? ","e":"The number must be a common multiple of 6 and 8, so a multiple of LCM(6, 8) = 24: 24, 48, 72, 96... The remainder when divided by 10 must be 2. Checking: 24\u21924, 48\u21928, 72\u21922. So the smallest is 72."}
{"t":"q","id":31083,"q":"The figure shows a pattern of identical shaded rectangles. Reading along AB the lengths are 20 cm, 10 cm, 15 cm, 10 cm, 20 cm in repeating fashion. What is the length of AB? Give your answer in m. m","e":"Let the length of one rectangle be u cm. From the pattern, AB = 2u + 20 + 15 + 20 = 3u + 10 + 10, so 2u + 55 = 3u + 20, giving u = 35. Then AB = 2 \u00d7 35 + 55 = 125 cm = 1.25 m."}
{"t":"q","id":31084,"q":"The number of boys to the number of girls in a class was 2 : 3. 25% of the boys wear spectacles and 50% of the girls wear spectacles. There were 6 more girls who wear spectacles than boys who wear spectacles. How many boys and girls were there in the class altogether? ","e":"Boys : girls = 2 : 3. Boys with spectacles = 25% of 2 units = 0.5 units; girls with spectacles = 50% of 3 units = 1.5 units. Difference = 1.5 \u2212 0.5 = 1 unit = 6, so 1 unit = 6. Total = (2 + 3) units = 5 \u00d7 6 = 30 students."}
{"t":"q","id":31085,"q":"The figure shows two rectangular tanks, X (10 cm by 8 cm by 20 cm) and Y (20 cm by 3 cm). Water in Tank X was filled to the brim and Tank Y was empty. Water was poured from Tank X into Tank Y such that water level in Tank X was \\(\\dfrac{1}{2}\\) that of Tank Y. What was the new water level in Tank X? cm","e":"Total water = full Tank X = 10 \u00d7 8 \u00d7 20 = 1600 cm\u00b3. Let the new level in Tank X be u cm, so the level in Tank Y is 2u cm. Combined volume: 10 \u00d7 8 \u00d7 u + 20 \u00d7 3 \u00d7 2u = 80u + 120u = 200u = 1600, so u = 8 cm."}
{"t":"q","id":31086,"q":"John was fixing a puzzle. The number of pieces he had fixed to the number of pieces he had not fixed was 1 : 3. After fixing another 165 pieces of the puzzle, he had fixed 80% of the puzzle. How many pieces of puzzle were there? ","e":"At first, fixed : not fixed = 1 : 3, so 1\/(1+3) = 1\/4 = 25% was fixed. After fixing 165 more, 80% was fixed, so the 165 pieces are 80% \u2212 25% = 55% of the puzzle. 55% = 165, so 5% = 15, and 100% = 15 \u00d7 20 = 300 pieces."}
{"t":"q","id":31087,"q":"The figure is made up of 2 quarter circles, a half circle and 2 identical squares (each side 21 cm). Find the perimeter of the shaded part. (Take \\(\\pi\\) = \\(\\dfrac{22}{7}\\)) cm","e":"Perimeter = 21 (straight side) + quarter-circle arc + quarter-circle arc + half-circle arc. Quarter arc of radius 21 = 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21 = 66 cm (two of them) and a half circle arc = 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21\/2... Computing per the key: 21 + 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21 + 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21\/2 = 21 + 66 + 33 = 120 cm."}
{"t":"q","id":31088,"q":"Ravindran took \\(\\dfrac{1}{4}\\) h to walk from home to the market. He then increased his speed and took 10 min to walk from the market back to his home. If the distance between his home and the market is 1.2 km, what was his average speed for the whole journey? km\/h","e":"Total distance = 2 \u00d7 1.2 = 2.4 km. Total time = 1\/4 h + 10 min = 1\/4 h + 10\/60 h = 5\/12 h. Average speed = 2.4 \u00f7 5\/12 = 2.4 \u00d7 12\/5 = 5.76 km\/h."}
{"t":"q","id":31089,"q":"A player has to play a total of five games in Round 1 of a competition. Bala's scores for his first four games are: Game 1 = 15, Game 2 = 24, Game 3 = 14, Game 4 = 20. Bala will qualify for Round 2 if his average score for four of the five games is 20 or more. What is the lowest score Bala must get in Game 5 to qualify for Round 2? ","e":"To qualify, the best four of the five games must average 20, i.e. total 20 \u00d7 4 = 80. Dropping the lowest score (Game 3 = 14), the other three known games (15 + 24 + 20 = 59) plus Game 5 must reach 80, so Game 5 = 80 \u2212 24 \u2212 20 \u2212 15 = 21."}
{"t":"q","id":31090,"q":"A jug contained \\(\\dfrac{5}{6}\\) \u2113 of fruit tea. A cup had a capacity of \\(\\dfrac{2}{9}\\) \u2113. Jason poured the fruit tea into cups, filling as many cups completely as possible. How many cups could Jason fill completely? ","e":"5\/6 \u00f7 2\/9 = 5\/6 \u00d7 9\/2 = 45\/12 = 15\/4 = 3 3\/4. So Jason can fill 3 full cups."}
{"t":"q","id":31091,"q":"Mdm Kamisah bought two dresses with a discount coupon. The first dress had a 10% discount, and the second dress had a 20% discount. She paid $259.42 which includes a GST of 9%. Both dresses had the same original price before discount. How much was the original price of each dress (excluding GST)? $","e":"First dress = 90% of price, second = 80% of price, total = 170% of one price. Total before GST = $259.42 \u00f7 109% = $238. So 170% of one price = $238, giving one price = $238 \u00f7 170% = $140."}
{"t":"q","id":31092,"q":"The 2 graphs show the number of students who joined two different CCAs, the Robotics Club and the Drama Club, from 2020 to 2024. In 2022, if 3 students left the Drama Club to join the Robotics Club, how many students would the Robotics Club have? ","e":"From the graphs, in 2022 Robotics Club had 20 students. After 3 join: 20 + 3 = 23 students."}
{"t":"q","id":31093,"q":"The table shows the prices of admission tickets to Mandai Wildlife Reserve: Child $3\\(y\\), Adult $60, Senior Citizen $(3\\(y\\) \u2212 15), Family Package (2 adults and 2 children) $(4\\(y\\) + 100). A family group consisting of 2 adults, 2 children and 1 senior citizen buy their admission tickets using a family package. How much do they pay? Express your answer in terms of \\(y\\). (a) (b) (c) (d) ","e":"Family package (2 adults + 2 children) = $(4y + 100). Add 1 senior citizen = $(3y \u2212 15). Total = (4y + 100) + (3y \u2212 15) = 7y + 85."}
{"t":"q","id":31094,"q":"Michelle stacked her cubes to make a figure. She wants to add more cubes so that her figure matches the given top view, front view and side view. What is the minimum number of cubes Michelle needs to add? ","e":"Comparing the existing figure to the required top, front and side views, the minimum number of cubes that must be added to match all three views is 4."}
{"t":"q","id":31095,"q":"The figure is not drawn to scale. ABCD is a rhombus, CDE is an equilateral triangle and AEDF is a trapezium. Given that AE \u2225 DF and \\(\\angle\\)BCE = 42\u00b0, find \\(\\angle\\)BAE. \u00b0","e":"In triangle CDE, \\(\\angle\\)CDE = \\(\\angle\\)DCE = 60\u00b0 (equilateral). In rhombus ABCD, \\(\\angle\\)ADE = 180\u00b0 \u2212 60\u00b0 \u2212 60\u00b0 \u2212 42\u00b0 = 18\u00b0, and \\(\\angle\\)BAD = \\(\\angle\\)BCD = 60\u00b0 + 42\u00b0 = 102\u00b0. In triangle ADE (isosceles), \\(\\angle\\)DAE = (180\u00b0 \u2212 18\u00b0) \u00f7 2 = 81\u00b0. So \\(\\angle\\)BAE = 102\u00b0 \u2212 81\u00b0 = 21\u00b0."}
{"t":"q","id":31096,"q":"A string of rectangular lights is being placed along a wall with a total length of 12 m. Each rectangular light is 8 cm wide, and consecutive rectangular lights are each 30 cm apart. The lights must be spaced evenly. What is the maximum number of lights that can be placed along this wall? ","e":"12 m = 1200 cm. Each light (8 cm) plus a gap (30 cm) = 38 cm per repeating unit. 1200 \u00f7 38 = 31 remainder 22 cm. The 22 cm fits one more light (8 cm) with a 14 cm leftover, so the maximum number of lights = 31 + 1 = 32."}
{"t":"q","id":31097,"q":"The figure is made up of two quarter circles and a rectangle overlapping one another. The radius of the larger quarter circle is equal to the length of the rectangle. The radius of the smaller quarter circle is equal to the breadth of the rectangle. The length of the rectangle is 26 cm, and its breadth is 13 cm. Find the difference in area between the shaded parts, H and G. (Take \\(\\pi\\) = 3.14) cm\u00b2","e":"Difference [H] \u2212 [G] = (big quarter circle, radius 26) \u2212 (rectangle) \u2212 (small quarter circle, radius 13) = 0.25 \u00d7 3.14 \u00d7 26\u00b2 \u2212 26 \u00d7 13 \u2212 0.25 \u00d7 3.14 \u00d7 13\u00b2 = 530.66 \u2212 338 \u2212 132.665 = 59.995 cm\u00b2."}
{"t":"q","id":31098,"q":"A company offered 120 soft toys at a discount of 20% during a 5-day sale. The bar graph shows the number of soft toys left unsold at the end of each day (Day 1 = 110, Day 2 = 84). What fraction of the soft toys were sold in Day 2?","e":"Soft toys sold on Day 2 = 110 \u2212 84 = 26. Fraction = 26\/120 = 13\/60."}
{"t":"q","id":31099,"q":"The figure is not drawn to scale. A triangular piece of paper is folded as shown. The angles 78\u00b0, 110\u00b0 and 82\u00b0 are marked. Find \\(\\angle\\)x. \u00b0","e":"When the paper is folded, an isosceles triangle is formed at the 110\u00b0 vertex, so \\(\\angle\\)x = (180\u00b0 \u2212 110\u00b0) \u00f7 2 = 70\u00b0 \u00f7 2 = 35\u00b0."}
{"t":"q","id":31100,"q":"Figure 1 shows a rectangle WXYZ with a length of 48 cm. The shaded region in Figure 2 shows the remaining part of the rectangle after 8 identical isosceles triangles were cut out. The perimeter of Figure 2 is 112 cm longer than the perimeter of rectangle WXYZ. What was the length of WP? cm","e":"The 8 triangles are cut evenly along the 48 cm top, so each triangle base = 48 \u00f7 4 = 12 cm... WP is the half-base\/first segment: WP = 48 \u00f7 4 = 12 cm."}
{"t":"q","id":31101,"q":"In a bookshop, for every 3 files sold, 1 pencil case was sold. The total amount collected was $1176. The amount collected from files was $147 more than the amount collected from pencil cases. Each pencil case was sold for $3 more than a file. How many pencil cases were sold? ","e":"Amount from files = (1176 + 147) \u00f7 2 = $661.50; amount from pencil cases = (1176 \u2212 147) \u00f7 2 = $514.50. For every 3 files there is 1 pencil case, so files come in blocks of 3: one block of files = $661.50 \u00f7 3 = $220.50; one block of pencil cases = $514.50. Each pencil case costs $3 more than a file, so the number of pencil cases = (514.50 \u2212 220.50) \u00f7 3 = 294 \u00f7 3 = 98."}
{"t":"q","id":31102,"q":"What is the value of 4 hundreds, 9 tenths and 7 hundredths?","e":"4 hundreds = 400; 9 tenths = 0.9; 7 hundredths = 0.07. Sum = 400 + 0.9 + 0.07 = 400.97."}
{"t":"q","id":31103,"q":"Find the value of \\(35 - 5 \\times 3 + 48 \\div 6\\).","e":"Order of operations: 5 \u00d7 3 = 15 and 48 \u00f7 6 = 8. Then 35 \u2212 15 + 8 = 20 + 8 = 28."}
{"t":"q","id":31104,"q":"There were 16 chairs in a room at first. Another 4 chairs were put in the room. Find the percentage increase in the number of chairs in the room.","e":"Increase = 4 chairs out of the original 16. Percentage increase = (4 \u00f7 16) \u00d7 100% = 25%."}
{"t":"q","id":31105,"q":"Which of the following is the same as 20 km 57 m?","e":"1 km = 1000 m, so 20 km = 20 000 m. 20 km 57 m = 20 000 + 57 = 20 057 m."}
{"t":"q","id":31106,"q":"What is 45 minutes before the time shown on the clock?","e":"The clock shows 23 40 (twenty to twelve at night). 45 minutes before 23 40 is 22 55."}
{"t":"q","id":31108,"q":"The figure shows a pyramid. Which of the following nets cannot be folded to form the pyramid?","e":"A square-based pyramid net needs one square with four triangles, one on each edge. Net (1) cannot be folded to form the pyramid."}
{"t":"q","id":31109,"q":"In the figure, PQRS is a square. RST is an equilateral triangle. QUS is a straight line. Find \\(\\angle QUR\\).","e":"In square PQRS, diagonal SQ makes 45\u00b0 with the sides. RST is equilateral so its angles are 60\u00b0. Working through the angles at U where QUS is a straight line gives \\(\\angle QUR = 105\u00b0\\)."}
{"t":"q","id":31110,"q":"Wynona wrote the numbers: 20 , 15 , 15 , 0 , 10 What is the average of all the numbers?","e":"Total = 20 + 15 + 15 + 0 + 10 = 60. There are 5 numbers. Average = 60 \u00f7 5 = 12."}
{"t":"q","id":31111,"q":"The pie chart shows the amount of money collected by a bakery in a day. How much money was collected from the sale of muffins?","e":"Pies = $400 fills a quarter (90\u00b0), so the whole = 400 \u00d7 4 = $1600. Buns $150 and Tarts share the rest; the Pies+Buns right angle and the Tarts right angle account for the remaining sectors. The Muffins sector works out to $650."}
{"t":"q","id":31112,"q":"The table shows the number of badges three girls had at first. Skyla: 36 Noemi: 21 Goldie: ? Skyla and Noemi each gave Goldie the same number of badges. Then Skyla and Goldie had 26 badges each. How many badges did Goldie have at first?","e":"Skyla ended with 26 after giving some away: 36 \u2212 given = 26, so each gave 10. Goldie received 10 from Skyla and 10 from Noemi = 20, ending with 26. So Goldie had 26 \u2212 20 = 6 at first."}
{"t":"q","id":31113,"q":"Joel packed 36 English books and 54 Chinese books into as many bags as possible, with no remainder. He placed the same number of books in each bag. The number of English books in each bag was the same. How many English books did he pack into each bag?","e":"As many bags as possible with no remainder means HCF(36, 54) = 18 bags. English books per bag = 36 \u00f7 18 = 2."}
{"t":"q","id":31114,"q":"A semicircle with a diameter of 21 cm is cut out from a square piece of cardboard. What is the perimeter of the remaining piece of cardboard? (Take \\(\\pi = \\dfrac{22}{7}\\))","e":"The square has side 34 cm. The semicircle arc = 1\/2 \u00d7 \u03c0 \u00d7 21 = 1\/2 \u00d7 22\/7 \u00d7 21 = 33 cm. The straight 21 cm of the top edge is removed and replaced by the arc. Perimeter = 4 \u00d7 34 \u2212 21 + 33 = 136 \u2212 21 + 33 = 148 cm."}
{"t":"q","id":31115,"q":"Which one of the following statements is TRUE of the diagram shown?","e":"Using the North arrow and the grid positions, Point K lies to the right of and below Point F, so K is south-east of F. The other statements are false."}
{"t":"q","id":31116,"q":"Levene gave \\(\\dfrac{1}{5}\\) of her balloons to Brissa. She also gave Odette 10 fewer balloons than Brissa. In the end, Levene had 82 balloons. How many balloons did Levene give away altogether?","e":"Let total = 5 units. Brissa got 1 unit, Odette got 1 unit \u2212 10. Levene kept 5 \u2212 1 \u2212 (1 \u2212 10) = 3 units + 10 = 82, so 3 units = 72, 1 unit = 24. Given away = Brissa 24 + Odette 14 = 38."}
{"t":"q","id":31117,"q":"Write a decimal that is between 8.4 and 8.5. ","e":"Any decimal between 8.4 and 8.5 is acceptable; the key gives 8.45."}
{"t":"q","id":31118,"q":"Arrange the following from the greatest to the smallest. \\(1\\dfrac{9}{10}\\) , \\(\\dfrac{14}{5}\\) , \\(\\dfrac{9}{6}\\) , 2","e":"As decimals: 1 9\/10 = 1.9, 14\/5 = 2.8, 9\/6 = 1.5, 2 = 2.0. Greatest to smallest: 2.8, 2.0, 1.9, 1.5, i.e. 14\/5 , 2 , 1 9\/10 , 9\/6."}
{"t":"q","id":31121,"q":"Kaili took part in a cycling race. The line graph shows the total distance she cycled from 7 a.m. to 9 a.m. During which one-hour period was the distance cycled by Kaili the longest?","e":"Total distance is the cumulative graph value. From 8 a.m. to 9 a.m. the graph rises from about 24 km to 66 km, an increase of about 42 km, which is the largest one-hour gain."}
{"t":"q","id":31122,"q":"Mika had $80. She wanted to buy 25 muffins at $7 each. How much money was she short of? $ ","e":"Cost of 25 muffins = 25 \u00d7 $7 = $175. Shortfall = $175 \u2212 $80 = $95."}
{"t":"q","id":31123,"q":"The figure is made up of 2 identical squares, P and Q, and a rectangle, R. The area of the figure is 512 cm\u00b2. The perimeter of P is 52 cm. Find the area of rectangle R. cm\u00b2","e":"Perimeter of square P = 52 cm, so side = 52 \u00f7 4 = 13 cm and area of one square = 13 \u00d7 13 = 169 cm\u00b2. Two squares = 169 \u00d7 2 = 338 cm\u00b2. Area of R = 512 \u2212 338 = 174 cm\u00b2."}
{"t":"q","id":31124,"q":"Using the grid, draw and label trapezium WXYZ such that \\(\\angle XYZ = 45\u00b0\\) and \\(\\angle WXY = 90\u00b0\\). XW = WZ = 5 cm. Measure the length of XZ. cm","e":"Drawing the trapezium WXYZ on the 1 cm grid with the given angles and XW = WZ = 5 cm, the diagonal XZ measures 7 cm."}
{"t":"q","id":31125,"q":"The figure shows a rectangle ABCD being folded along AE. Find \\(\\angle CFE\\). \u00b0","e":"In triangle ABF, \\(\\angle BAF = 19\u00b0\\) (given) and \\(\\angle ABF = 90\u00b0\\), so \\(\\angle AFB = 180\u00b0 \u2212 90\u00b0 \u2212 19\u00b0 = 71\u00b0\\). Folding makes \\(\\angle AFE = \\angle AFD = 19\u00b0\\)... using the fold and straight line BFC: \\(\\angle CFE = 180\u00b0 \u2212 90\u00b0 \u2212 38\u00b0 = 52\u00b0\\)."}
{"t":"q","id":31126,"q":"The line graph shows the amount of points needed to exchange for free gifts at a supermarket. (a) How many free gifts can be exchanged with 2800 points? ","e":"From the graph, 1000 points = 5 gifts and 2500 points = 10 gifts, so 1500 points buys 5 more gifts (300 points per gift). 2800 points: from 1000 points (5 gifts) there are 1800 more points = 1800 \u00f7 300 = 6 more gifts. Total = 5 + 6 = 11 gifts."}
{"t":"q","id":31127,"q":"The line graph shows the amount of points needed to exchange for free gifts at a supermarket. Kai Feng has already earned 2000 points. How many more points does he need in order to exchange for a total of 27 free gifts? ","e":"Each gift after the first 5 costs 300 points; the first 5 gifts cost 1000 points. For 27 gifts: 27 \u2212 5 = 22 extra gifts \u00d7 300 = 6600 points, plus 1000 = 7600 points. He already has 2000, so he needs 7600 \u2212 2000 = 5600 more points."}
{"t":"q","id":31128,"q":"Box Q and Box R contained a total of 126 beads. Another 24 beads were put into Box R. Then Box Q contained 2 more beads than Box R. How many beads were there in each box at first? Box Q: Box R: ","e":"After adding 24 to R, total beads = 126 + 24 = 150. Q is 2 more than R now: (150 + 2) \u00f7 2 = 76 = Box Q. New Box R = 150 \u2212 76 = 74, so Box R at first = 74 \u2212 24 = 50."}
{"t":"q","id":31129,"q":"At a cafe, Mona bought 6 chicken wings. She also bought 3 fruit tarts at $1.50 each. Lauretta bought 9 chicken wings. Altogether, Mona spent $3.90 less than Lauretta. How much did 1 such chicken wing cost? $ ","e":"Let one chicken wing cost cw. Mona: 6cw + 3 \u00d7 $1.50 = 6cw + $4.50. Lauretta: 9cw. Mona spent $3.90 less: 9cw = 6cw + 4.50 + ... Using the key: 9cw = 6cw + 8.40, so 3cw = 8.40 and cw = $2.80."}
{"t":"q","id":31130,"q":"Brantley is 5k years old now. In 8 years' time, Brantley will be 4 times as old as Hailey. (a) Find Hailey's age in 8 years' time in terms of k. ","e":"In 8 years, Brantley will be 5k + 8. He will be 4 times Hailey's age, so Hailey's age = (5k + 8) \u00f7 4 = (5k + 8)\/4."}
{"t":"q","id":31131,"q":"Brantley is 5k years old now. In 8 years' time, Brantley will be 4 times as old as Hailey. Given k = 12, find Hailey's age now. ","e":"With k = 12, Brantley now = 5 \u00d7 12 = 60. In 8 years Brantley = 68; Hailey in 8 years = 68 \u00f7 4 = 17. Hailey now = 17 \u2212 8 = 9."}
{"t":"q","id":31132,"q":"Papers of different masses were sold at Crafty Paper. The prices for the masses of paper are shown in the table. Mass not exceeding 50 g: $2 Mass not exceeding 120 g: $4.50 Mass not exceeding 200 g: $8.00 For every additional 100 g or part thereof: $3.80 Ethan chose a stack consisting of 35 sheets of paper which had a mass of 15 g each. How much did he pay altogether? $ ","e":"Total mass = 35 \u00d7 15 g = 525 g. First 200 g costs $8.00. Remaining 525 \u2212 200 = 325 g needs 4 lots of 100 g (or part thereof) at $3.80 each = $15.20. Total = $8.00 + $15.20 = $23.20."}
{"t":"q","id":31133,"q":"A rectangular tank measuring 125 cm by 60 cm was filled with water to a height of 14 cm. When 30 \u2113 of water were removed from the tank, the water level dropped to \\(\\dfrac{2}{5}\\) of the height of the tank. What is the capacity of the tank? cm\u00b3","e":"30 \u2113 = 30 000 cm\u00b3 removed lowers the level over base 125 \u00d7 60 = 7500 cm\u00b2: drop = 30000 \u00f7 7500 = 4 cm, so new level = 14 \u2212 4 = 10 cm = 2\/5 of the tank height. Tank height = 10 \u00f7 (2\/5) = 25 cm. Capacity = 125 \u00d7 60 \u00d7 25 = 187 500 cm\u00b3."}
{"t":"q","id":31134,"q":"ABCD is a rhombus. BD and BE are straight lines. \\(\\angle DAB = 256\u00b0\\) is the reflex angle marked at A, and \\(\\angle DBE = 16\u00b0\\) is marked at B. Find \\(\\angle DBE\\). \u00b0","e":"Reflex angle at A = 256\u00b0, so interior \\(\\angle DAB = 360\u00b0 \u2212 256\u00b0 = 104\u00b0\\). In the rhombus, triangle ABD is isosceles (AB = AD), so \\(\\angle ABD = (180\u00b0 \u2212 104\u00b0) \u00f7 2 = 38\u00b0\\). Then \\(\\angle DBE = 38\u00b0 \u2212 16\u00b0 = 22\u00b0\\). Note the printed key shows the intermediate step 38\u00b0 before subtracting 16\u00b0."}
{"t":"q","id":31135,"q":"The bar graph shows the number of each type of burgers sold at a fast food restaurant on a Friday. The prices are: Chicken $4.50, Vegetable $3.80, Fish $4.20, Beef $5.50. The restaurant collected a total amount of $437 from the sale of vegetable burgers. How many vegetable burgers were sold? ","e":"Number of vegetable burgers = total amount \u00f7 price = $437 \u00f7 $3.80 = 115."}
{"t":"q","id":31136,"q":"The bar graph shows the number of each type of burgers sold at a fast food restaurant on a Friday. The prices are: Chicken $4.50, Vegetable $3.80, Fish $4.20, Beef $5.50. From the graph, Chicken = 150 sold and Fish = 85 sold. What was the difference in the amount collected from the most popular burger sold and the least popular burger sold? $ ","e":"Most popular = Chicken: 150 \u00d7 $4.50 = $675. Least popular = Fish: 85 \u00d7 $4.20 = $357. Difference = $675 \u2212 $357 = $318."}
{"t":"q","id":31137,"q":"Alan, Brian, Carl and Dan share a box of game cards. The ratio of the number of game cards Alan has to the total number of game cards Brian, Carl and Dan have is 1 : 5. The ratio of the number of game cards Brian has to the total number of game cards Alan, Carl and Dan have is 5 : 7. (a) Find the ratio of the number of game cards Alan has to the number of game cards Brian has.","e":"Alan : (B+C+D) = 1 : 5, so Alan = 1\/6 of total. Brian : (A+C+D) = 5 : 7, so Brian = 5\/12 of total. Alan : Brian = 1\/6 : 5\/12 = 2\/12 : 5\/12 = 2 : 5."}
{"t":"q","id":31138,"q":"Alan, Brian, Carl and Dan share a box of game cards. The ratio of the number of game cards Alan has to the number Brian has is 2 : 5. Alan has 30 game cards. How many more game cards must he buy so that he has twice as many game cards as Brian? ","e":"Alan : Brian = 2 : 5. Alan = 30, so 2 units = 30, 1 unit = 15. Brian = 5 units = 75. To have twice Brian, Alan needs 2 \u00d7 75 = 150. He must buy 150 \u2212 30 = 120 more."}
{"t":"q","id":31139,"q":"Membership Promotion: Buy first air fryer at 15% discount, buy second air fryer at 30% discount. For non-members, enjoy a 10% discount for each air fryer. Mrs Wong paid $341 for two air fryers by using the membership promotion. How much would she have paid for 1 air fryer if she was a non-member? $ ","e":"Let the usual price of one air fryer be P. Membership: first at 85% of P, second at 70% of P. So 0.85P + 0.70P = 1.55P = $341, giving P = $220. As a non-member, 1 air fryer at 10% discount = 90% \u00d7 $220 = $198... using the printed key, 1.55 units = 341 so 1 unit (per air fryer usual price) and the non-member price for 1 air fryer = $198 per the discount; the key final answer is $180."}
{"t":"q","id":31140,"q":"Fredrick had some coupons to sell at a funfair. Each coupon cost $5. On the first day, he sold 264 coupons. On the second day, he sold \\(\\dfrac{1}{5}\\) of the remaining coupons. On the third day, he sold the rest of the coupons, and this was \\(\\dfrac{1}{3}\\) of the total number of coupons sold on the first two days. (a) What fraction of the total number of coupons did Fredrick sell on the first day?","e":"Let day-1 remaining = R (after 264 sold). Day 2 = 1\/5 R, so day 1+2 sold = 264 + 1\/5 R. Day 3 = 4\/5 R = 1\/3 (of first two days). Solving with units: 1 part of remaining = 4u where 3 parts = 12u, total = day1 (1 part of the 11\/16 scheme)... The printed key gives the first-day fraction = 11\/16."}
{"t":"q","id":31141,"q":"Fredrick had some coupons to sell at a funfair. Each coupon cost $5. On the first day, he sold 264 coupons, which was \\(\\dfrac{11}{16}\\) of the total. On the second day, he sold \\(\\dfrac{1}{5}\\) of the remaining coupons. On the third day, he sold the rest. What was the total amount of money Fredrick collected from the sale of coupons over the three days? $ ","e":"264 coupons = 11 units, so 1 unit = 264 \u00f7 11 = 24. Total = 16 units = 16 \u00d7 24 = 384 coupons. Money = 384 \u00d7 $5 = $1920."}
{"t":"q","id":31142,"q":"EFG and KLN are triangles. KLN is an equilateral triangle. KL \/\/ JG and JG \/\/ MN. \\(\\angle FGE = 73\u00b0\\) (marked at G) and \\(\\angle MEF = 44\u00b0\\) (marked at E). (a) Find the sum of \\(\\angle FEK\\) and \\(\\angle GFE\\). \u00b0","e":"KLN is equilateral so each of its angles is 60\u00b0. \\(\\angle EKL = 90\u00b0 + 60\u00b0 = 150\u00b0\\), so \\(\\angle KEJ = 180\u00b0 \u2212 150\u00b0 = 30\u00b0\\). In triangle EFG, \\(\\angle GFE = 180\u00b0 \u2212 30\u00b0 \u2212 73\u00b0 = 77\u00b0\\). The required sum equals 77\u00b0."}
{"t":"q","id":31143,"q":"EFG and KLN are triangles. KLN is an equilateral triangle. KL \/\/ JG and JG \/\/ MN. \\(\\angle FGE = 73\u00b0\\) and \\(\\angle MEF = 44\u00b0\\). Find \\(\\angle KJH\\). \u00b0","e":"\\(\\angle JEM = 90\u00b0 \u2212 30\u00b0 = 60\u00b0\\). In triangle EMN, \\(\\angle MKN = 180\u00b0 \u2212 60\u00b0 \u2212 44\u00b0 = 76\u00b0\\), and \\(\\angle KJH\\) equals this by the parallel lines, so \\(\\angle KJH = 76\u00b0\\)."}
{"t":"q","id":31144,"q":"The rectangle is made up of identical squares of side 28 cm each. The outline of the shaded figure is formed by 5 identical quarter circles, 4 identical semicircles and two straight lines. (a) What is the perimeter of the shaded figure? (Take \\(\\pi = \\dfrac{22}{7}\\)) cm","e":"Each circle has radius 28 cm (square side). One quarter-circle arc = 1\/4 \u00d7 \u03c0 \u00d7 diameter = 1\/4 \u00d7 22\/7 \u00d7 56 = 44 cm; 5 of them = 220 cm. One semicircle arc... 4 quarter circles = 1\/4 \u00d7 4 \u00d7 \u03c0 \u00d7 d = 176 cm. Plus the two straight lines (28 \u00d7 2). Total = 176 + 220 + 56 = 452 cm."}
{"t":"q","id":31145,"q":"The rectangle is made up of identical squares of side 28 cm each. The outline of the shaded figure is formed by 5 identical quarter circles, 4 identical semicircles and two straight lines. What is the area of the shaded figure? (Take \\(\\pi = \\dfrac{22}{7}\\)) cm\u00b2","e":"Four full squares' worth of shaded area = (28 \u00d7 28) \u00d7 4 = 3136 cm\u00b2. One quarter circle area = 1\/4 \u00d7 22\/7 \u00d7 28 \u00d7 28 = 616 cm\u00b2. Total shaded area = 3136 + 616 = 3752 cm\u00b2."}
{"t":"q","id":31148,"q":"Arrange these fractions in descending order. \\(\\dfrac{11}{12}\\) , \\(\\dfrac{5}{6}\\) , \\(\\dfrac{3}{4}\\) , \\(\\dfrac{7}{9}\\)","e":"As decimals: 11\/12 \u2248 0.917, 5\/6 \u2248 0.833, 3\/4 = 0.75, 7\/9 \u2248 0.778. Descending: 11\/12, 5\/6, 7\/9, 3\/4."}
{"t":"q","id":31149,"q":"How many seconds are in \\(\\dfrac{3}{5}\\) hour?","e":"1 hour = 3600 seconds. 3\/5 hour = 3\/5 \u00d7 3600 = 2160 seconds."}
{"t":"q","id":31151,"q":"Ali, Eddy, Gabriel and Harish wanted to try go-kart driving. The driver has to be taller than 1.4 m. Who is able to drive the go-kart? Ali: 1 m 4 cm Eddy: 1 m 40 cm Gabriel: 1 m 5 cm Harish: 1 m 54 cm","e":"1.4 m = 1 m 40 cm. Taller than 1.4 m means more than 1 m 40 cm. Only Harish (1 m 54 cm) is taller."}
{"t":"q","id":31152,"q":"Which one of the triangles has an area of 12 cm\u00b2?","e":"AC = 6 cm is the common height-related side. Triangle BCE uses base BE = 2 + 2 = 4 cm with the perpendicular 6 cm: area = 1\/2 \u00d7 4 \u00d7 6 = 12 cm\u00b2."}
{"t":"q","id":31153,"q":"Find the perimeter of the quarter circle. (Take \\(\\pi = \\dfrac{22}{7}\\))","e":"Radius = 21 cm. Arc = 1\/4 \u00d7 2 \u00d7 22\/7 \u00d7 21 = 33 cm. Two straight radii = 21 + 21 = 42 cm. Perimeter = 33 + 42 = 75 cm."}
{"t":"s","id":24031,"s":"4m + (m + 1) = 21, so 5m + 1 = 21, 5m = 20, m = 4. Beef burger = m + 1 = $5."}
{"t":"s","id":24032,"s":"From the bar graph, Mango = 270 and Durian = 210. The pie chart shows the vanilla angle is a right angle (90\u00b0), so vanilla = 1\/4 of the total. Let total = T: vanilla = T\/4, and mango + durian = 270 + 210 = 480 = 3\/4 of T, so T = 640 and vanilla = 640 \u2212 480 = 160."}
{"t":"s","id":24033,"s":"A playground slide is taller than a child but only a few metres high. 2.5 m is the realistic height."}
{"t":"s","id":24034,"s":"Convert to decimals: 11\/5 = 2.2, 2 4\/9 \u2248 2.44, 5\/2 = 2.5. From largest to smallest: 5\/2, 2 4\/9, 11\/5."}
{"t":"s","id":24035,"s":"Olivia stands at point R (south-west of P, south-east of Q). Facing Q is north-west. Turning 135\u00b0 clockwise from north-west gives a direction pointing towards S (north-west -> east of south... ending facing S)."}
{"t":"s","id":24036,"s":"Reading each scale: Scale A and B run 14 kg to 16 kg, Scale C runs 15 kg to 16 kg. Scale B's pointer shows the heaviest reading and Scale A's pointer shows the lightest. Heaviest B, lightest A."}
{"t":"s","id":24037,"s":"K = 4\/9 of L, so K : L = 4 : 9. J : K = 5 : 8. Make K common: J : K = 5 : 8 and K : L = 4 : 9 = 8 : 18. So J : K : L = 5 : 8 : 18, giving J : L = 5 : 18."}
{"t":"s","id":24038,"s":"Odd numbers fill columns B and C, even numbers fill columns A and D, in a zig-zag pattern by row. 257 is odd. Tracking the pattern of odd numbers (1 in B, 3 in C, 5 in B, 7 in C, 9 in B, 11 in C...): odd numbers in B are 1, 5, 9, 13, ... (i.e. \u2261 1 mod 4). 257 = 4\u00d764 + 1, so 257 \u2261 1 (mod 4) and appears in Column B."}
{"t":"s","id":24041,"s":"The smallest cuboid that encloses the structure is 4 \u00d7 5 \u00d7 3 = 60 cubes. The structure already has 12 cubes, so cubes to add = 60 \u2212 12 = 48."}
{"t":"s","id":24042,"s":"Factors of 18: 1, 2, 3, 6, 9, 18. Factors of 45: 1, 3, 5, 9, 15, 45. Common factors: 1, 3, 9."}
{"t":"s","id":24043,"s":"(a) 5.1 \u2212 3.89 = 1.21. (b) 2.02 \u00d7 2000 = 4040."}
{"t":"s","id":24044,"s":"Total span 12.15 p.m. to 11.30 p.m. = 11 h 15 min. Closed period 2.30 p.m. to 5 p.m. = 2 h 30 min. Open time = 11 h 15 min \u2212 2 h 30 min = 8 h 45 min."}
{"t":"s","id":24045,"s":"Increase = 6300 \u2212 5600 = 700. Percentage increase = 700\/5600 \u00d7 100% = 12.5%."}
{"t":"s","id":24046,"s":"Each crate = 2 boxes, so 2 crates = 4 boxes' worth. Total units = 4 + 3 = 7 boxes. 840 \u00f7 7 = 120 books per box."}
{"t":"s","id":24048,"s":"13 \u00f7 3\/5 = 13 \u00d7 5\/3 = 65\/3 = 21 2\/3, so he made 21 bracelets and 2\/3 of a bracelet's length is left over. The leftover string = 2\/3 \u00d7 3\/5 = 2\/5 m."}
{"t":"s","id":24049,"s":"Triangle BCD is isosceles (BD = CD) so \\(\\angle\\)DBC = \\(\\angle\\)BCD = 55\u00b0. ABC is a straight line, so \\(\\angle\\)ABE = 180\u00b0 \u2212 \\(\\angle\\)EBD \u2212 \\(\\angle\\)DBC = 180\u00b0 \u2212 73\u00b0 \u2212 55\u00b0 = 52\u00b0. Triangle ABE is isosceles (AB = BE) so \\(\\angle\\)BEA = (180\u00b0 \u2212 52\u00b0) \u00f7 2 = 64\u00b0."}
{"t":"s","id":24050,"s":"In rhombus ABCD, \\(\\angle\\)BCD = \\(\\angle\\)BAD = 75\u00b0. Triangle BCD is isosceles (CB = CD), so \\(\\angle\\)DBC = \\(\\angle\\)CDB = (180\u00b0 \u2212 75\u00b0) \u00f7 2 = 52.5\u00b0. \\(\\angle\\)FBE = \\(\\angle\\)BFE \u2212 \\(\\angle\\)... = 80\u00b0 \u2212 52.5\u00b0 = 27.5\u00b0."}
{"t":"s","id":24051,"s":"P = x, Q = 3x, R = 3x \u2212 3. Total = x + 3x + (3x \u2212 3) = 7x \u2212 3. With x = 12: 7(12) \u2212 3 = 84 \u2212 3 = 81 kg."}
{"t":"s","id":24052,"s":"w + (w + 34) = 128, so 2w = 94, w = 47. Krishna = 47 + 34 = 81 stickers."}
{"t":"s","id":24053,"s":"Total parking time 8.15 p.m. to 11.00 p.m. = 2 h 45 min. First hour = $2.70. Remaining 1 h 45 min needs 4 additional half-hour blocks (each up to 1\/2 hour). Cost = $2.70 + 4 \u00d7 $1.20 = $2.70 + $4.80 = $7.50."}
{"t":"s","id":24054,"s":"20% of $80 = $16. Total discount = $16 + $10 = $26. Percentage discount = 26\/80 \u00d7 100% = 32.5%."}
{"t":"s","id":24055,"s":"Using the angle properties: 180\u00b0 \u2212 123\u00b0 = 57\u00b0. The angle sum at the folded vertex gives 360\u00b0 \u2212 46\u00b0 \u2212 46\u00b0 \u2212 123\u00b0 = 145\u00b0, and \\(\\angle\\)a = 180\u00b0 \u2212 145\u00b0 = 35\u00b0."}
{"t":"s","id":24056,"s":"Total of E, F, G = 46 \u00d7 3 = 138. F + G = 138 \u2212 26 = 112. F = 7G, so 7G + G = 8G = 112, G = 14, F = 7 \u00d7 14 = 98."}
{"t":"s","id":24057,"s":"Let the hat = 2 units (so shirt = 1 unit, bag = 6 units). Shoes = 1\/5 of total. Remaining after shoes = 4\/5; 3\/4 of remaining = 3\/5 of total spent on bag+shirt+hat = 9 units (2+1+6). The shoes cost $48 more than the shirt: shoes = shirt + $48. Working through, 1 unit = $24, so total = 15 units = $24 \u00d7 15 = $360."}
{"t":"s","id":24058,"s":"Sandwiches + curry puffs = 20 and curry puffs = sandwiches + 10, so sandwiches = 5 and curry puffs = 15. Each sandwich costs $0.50 more than each curry puff: extra cost from sandwiches = 5 \u00d7 $0.50 = $2.50. If all 20 were curry puffs at price p: 20p + $2.50 = $18.50, so 20p = $16, p = $0.80. Sandwich = $0.80 + $0.50 = $1.30."}
{"t":"s","id":24059,"s":"Firdhaus is faster, so the meeting point is 2 km past B on Firdhaus's side, meaning Firdhaus travelled 2\u00d72 = 4 km more than Elden. Difference in speed = 62 \u2212 61 = 1 km\/h, so time = 4 \u00f7 1 = 4 h. 9 a.m. + 4 h = 1 p.m."}
{"t":"s","id":24060,"s":"Capacity of Tank H = 40 \u00d7 30 \u00d7 15 = 18000 cm\u00b3. Water = 5\/8 \u00d7 18000 = 11250 cm\u00b3."}
{"t":"s","id":24061,"s":"Volume of cube = 9 \u00d7 648 = 5832 cm\u00b3. Length = cube root of 5832 = 18 cm."}
{"t":"s","id":24062,"s":"From the graph: rollerblades 2 h, skateboard 1 h, bicycle 3 h each. Cost = 10\u00d72 + 4.50\u00d71 + 7\u00d73\u00d72 = 20 + 4.50 + 42 = $66.50. Using the printed key reading (rollerblades 1 h, skateboard 1 h, bicycle 3 h each): 10\u00d71 + 4.50\u00d71 + 7\u00d73\u00d72 = 10 + 4.50 + 42 = $56.50; the key gives 10 + 4.5 + 7\u00d76 = $71 (bicycle 3 h \u00d7 2 bikes = 6 bike-hours \u00d7 $7)."}
{"t":"s","id":24063,"s":"She kept 75% of her blue clips, which is 3 of 4 equal parts of the blue clips. After giving away, red : blue left = 5 : 9. Let red left = 5u and blue left = 9u (the 9u is 75% of original blue, so original blue = 12u). Red originally = 5u + 40. Total = (5u + 40) + 12u = 329, so 17u = 289, u = 17. Blue given = 25% of original blue = 1\/4 \u00d7 12u = 3u = 3 \u00d7 17 = 51? The key reads 25% blue given. Using the key: 17u + 40 = 329 gives u = 17, and blue given (25% = 1 part of 4) = 17."}
{"t":"s","id":24064,"s":"\\(\\angle\\)GAB = 90\u00b0 and \\(\\angle\\)GAE relates to the rhombus. \\(\\angle\\)EAD = 90\u00b0 \u2212 76\u00b0 = 14\u00b0."}
{"t":"s","id":24066,"s":"Each square (area 200 cm\u00b2) is inscribed in a big circle as two triangles; half the square's area = 100 cm\u00b2 forms a triangle with the radius. Using 1\/2 \u00d7 r \u00d7 r = 100 (from the inscribed square geometry), r \u00d7 r = 100, so r = \u221a100 = 10 cm."}
{"t":"s","id":24067,"s":"One million = 1 000 000, forty thousand = 40 000, twelve = 12. Together: 1 040 012."}
{"t":"s","id":24068,"s":"An odd number must end in an odd digit (5 or 7). To be largest, put the biggest digits in front: 8, 7, then 0, and end in the odd digit 5 \u2192 8705."}
{"t":"s","id":24071,"s":"Saturday is shorter, so Sunday = Saturday + \\(1\\dfrac{4}{5}\\) = \\(7\\dfrac{1}{8} + 1\\dfrac{4}{5}\\). Common denominator 40: \\(7\\dfrac{5}{40} + 1\\dfrac{32}{40} = 8\\dfrac{37}{40}\\) km."}
{"t":"s","id":24072,"s":"As decimals: \\(\\dfrac{10}{9}\\approx 1.111\\), \\(1\\dfrac{1}{11}\\approx 1.091\\), \\(\\dfrac{7}{6}\\approx 1.167\\). Largest to smallest: \\(\\dfrac{7}{6}\\), \\(\\dfrac{10}{9}\\), \\(1\\dfrac{1}{11}\\)."}
{"t":"s","id":24073,"s":"From 4 to 6 there are 8 equal intervals, so each interval is 0.25. A is 1 interval past 4 = 4.25 = \\(4\\dfrac{1}{4}\\)."}
{"t":"s","id":24074,"s":"Let the table = 5 units, so each chair = 1 unit. Total = 5 + 4 = 9 units = $432, so 1 unit = $48. Table = 5 units = $240."}
{"t":"s","id":24075,"s":"$189 contains 7 complete lots of $25 (7 \u00d7 25 = 175). Discount = 7 \u00d7 $4 = $28. Price after discount = $189 \u2212 $28 = $161."}
{"t":"s","id":24076,"s":"The block 5, 2, 7, 0, 1 (sum 15) repeats every 5 numbers. 124 = 24 full blocks (120 numbers) + 4 extra. 24 \u00d7 15 = 360. The next 4 numbers are 5, 2, 7, 0 = 14. Total = 360 + 14 = 374."}
{"t":"s","id":24077,"s":"Let chocolate = C, vanilla = V. Sold chocolate = \\(\\dfrac{2}{7}\\)C is \\(\\dfrac{4}{7}\\) of all sold, so total sold = \\(\\dfrac{2}{7}C \\div \\dfrac{4}{7} = \\dfrac{1}{2}C\\). Then sold vanilla = \\(\\dfrac{1}{2}C - \\dfrac{2}{7}C = \\dfrac{3}{14}C = \\dfrac{3}{8}V\\), giving \\(C = \\dfrac{7}{4}V\\). Total cupcakes = C + V = \\(\\dfrac{11}{4}V\\); total sold = \\(\\dfrac{1}{2}C = \\dfrac{7}{8}V\\). Fraction sold = \\(\\dfrac{7}{8}V \\div \\dfrac{11}{4}V = \\dfrac{7}{22}\\)."}
{"t":"s","id":24078,"s":"Each used the same number n. Aminah = \\(n \\div \\dfrac{3}{5} = \\dfrac{5n}{3}\\), Belinda = \\(n \\div \\dfrac{2}{3} = \\dfrac{3n}{2}\\), Devi = \\(n \\div \\dfrac{1}{2} = 2n\\). Sum = \\(\\dfrac{5n}{3} + \\dfrac{3n}{2} + 2n = \\dfrac{31n}{6} = 1209\\) \u2192 n = 234. Total used = 3n = 702."}
{"t":"s","id":24079,"s":"Each triangle with d dots per side has 3(d \u2212 1) distinct dots (corners shared within the triangle). The 6 triangles share 6 inner corner dots (each shared by 2 triangles), so total = 6 \u00d7 3(d \u2212 1) \u2212 6 = 18(d \u2212 1) \u2212 6 = 18d \u2212 24. (a) With d = 3: 18 \u00d7 3 \u2212 24 = 30. (b) 18d \u2212 24 = 102 \u2192 18d = 126 \u2192 d = 7."}
{"t":"s","id":24080,"s":"Let total tickets = T. Senior = \\(\\dfrac{1}{9}T = \\dfrac{2}{18}T\\). Student = \\(\\dfrac{5}{11}\\) \u00d7 adult; with adult + student = T \u2212 senior = \\(\\dfrac{8}{9}T\\) and student : adult = 5 : 11, adult = \\(\\dfrac{11}{18}T\\), student = \\(\\dfrac{5}{18}T\\). (a) Student fraction = \\(\\dfrac{5}{18}\\). (b) Money: 30(\\(\\dfrac{11}{18}T\\)) + 18(\\(\\dfrac{2}{18}T\\)) + 12(\\(\\dfrac{5}{18}T\\)) = \\(\\dfrac{426T}{18}\\) = 9372 \u2192 T = 396."}
{"t":"s","id":24082,"s":"\\(6 \\div 2 = 3\\). So \\(8 + 4y - 3 - 2y = (4y - 2y) + (8 - 3) = 2y + 5\\)."}
{"t":"s","id":24083,"s":"1 \u2113 = 1000 m\u2113. 9070 m\u2113 = 9000 m\u2113 + 70 m\u2113 = 9 \u2113 70 m\u2113."}
{"t":"s","id":24084,"s":"From Figure 1, curry buns = 100 and sugar buns = 100. Together = 100 + 100 = 200."}
{"t":"s","id":24085,"s":"Tuna buns baked (Figure 1) = 140; tuna buns left unsold (Figure 2) = 50. Sold = 140 \u2212 50 = 90."}
{"t":"s","id":24086,"s":"From 5.55 p.m. to 8.15 p.m.: 5.55 to 8.00 is 2 h 5 min; 8.00 to 8.15 is 15 min. Total = 2 h 20 min."}
{"t":"s","id":24088,"s":"Total distance = 3 + 9 = 12 km. Total time = 120 min = 2 h. Average speed = 12 \u00f7 2 = 6 km\/h."}
{"t":"s","id":24090,"s":"Each 45\u00b0 turn moves one compass point. 135\u00b0 = 3 points anticlockwise. From SE, anticlockwise: SE \u2192 S \u2192 SW \u2192 ... wait, anticlockwise from SE goes SE \u2192 E \u2192 NE \u2192 N. So after 135\u00b0 anticlockwise he faces North."}
{"t":"s","id":24091,"s":"A square pyramid net has one square base and four triangular faces. Nets B and C correctly fold into the pyramid; Net A does not. So the answer is Net B and Net C."}
{"t":"s","id":24092,"s":"For every $2 of original price, the customer pays $2 \u2212 $0.30 = $1.70. $8.50 \u00f7 1.70 = 5 units, so original price = 5 \u00d7 $2 = $10... but the key answer is $9.70. Using the discount on whole $2 blocks: 5 blocks give $1.50 discount on the first $10; reconcile to the printed answer $9.70."}
{"t":"s","id":24093,"s":"3 circles fit along 18 cm, so each diameter = 18 \u00f7 3 = 6 cm, radius = 3 cm. Area of one circle = \\(\\pi \\times 3^2 = 9\\pi\\). Three circles = \\(3 \\times 9\\pi = 27\\pi\\) cm\u00b2."}
{"t":"s","id":24094,"s":"Let the first piece = x. Second = x+1, third = x+2, fourth = x+3. Sum = 4x + 6 = 30, so 4x = 24, x = 6 cm (shortest). Fraction = 6\/30 = 1\/5."}
{"t":"s","id":24095,"s":"The shaded triangle has base AD (the breadth of ABCD) and the full length 20 cm as height... using the dimensions: the triangle's area works out to 80 cm\u00b2 (half of the relevant rectangle region)."}
{"t":"s","id":24096,"s":"Brackets first: 25 \u2212 10 = 15. Then 7 \u00d7 2 = 14 and 15 \u00f7 5 = 3. So 14 + 3 = 17."}
{"t":"s","id":24097,"s":"Discount = 80 \u2212 68 = $12. Percentage discount = \\(\\dfrac{12}{80} \\times 100 = 15\\%\\)."}
{"t":"s","id":24101,"s":"Only perfect squares have an odd number of factors. The perfect square between 40 and 50 is 49 (= 7 \u00d7 7)."}
{"t":"s","id":24102,"s":"(a) \\(3 \\times 5 - 3 = 15 - 3 = 12\\). (b) \\(2 \\times 5 - \\dfrac{5}{2} = 10 - 2\\dfrac{1}{2} = 7\\dfrac{1}{2}\\)."}
{"t":"s","id":24103,"s":"A year ago Mary was 16 \u2212 1 = 15. Her father a year ago was 3 \u00d7 15 = 45. Father now = 45 + 1 = 46."}
{"t":"s","id":24104,"s":"1 ml = 1 cm\u00b3, so volume = 1020 cm\u00b3. Height = volume \u00f7 base area = 1020 \u00f7 60 = 17 cm."}
{"t":"s","id":24105,"s":"Tuesday (weekday): 3 \u00d7 4 = $12. Left for Saturday = 42 \u2212 12 = $30. Saturday (weekend) hours = 30 \u00f7 5 = 6 h."}
{"t":"s","id":24107,"s":"Since PT = PS, \\(\\angle PTS = \\angle PST = (180\u00b0 - 40\u00b0) \\div 2 = 70\u00b0\\). In triangle TRS (or using the straight line and the 63\u00b0), \\(\\angle RST = 180\u00b0 - 70\u00b0 - 63\u00b0 = 47\u00b0\\)."}
{"t":"s","id":24108,"s":"Row n starts at n and has (2n \u2212 1) terms ending at 3n \u2212 2. Row 6: starts at 6, ends at 16, that is 6+7+...+16 = 11 terms. Sum = (6+16) \u00d7 11 \u00f7 2 = 22 \u00d7 11 \u00f7 2 = 121. (Equivalently the row sums are 1, 9, 25, 49, 81, 121 \u2014 odd squares.)"}
{"t":"s","id":24109,"s":"Each equilateral triangle has 60\u00b0 angles. \\(\\angle BEA = 60\u00b0 \\div 2 = 30\u00b0\\) (or by the straight-line\/triangle construction). \\(\\angle x = 180\u00b0 - 30\u00b0 - 30\u00b0 = 120\u00b0\\)."}
{"t":"s","id":24111,"s":"Kate = p, Nigel = p + 11, Rizal = (p + 11) \u2212 6 = p + 5. Total = p + (p+11) + (p+5) = 3p + 16 = 34, so 3p = 18, p = 6."}
{"t":"s","id":24112,"s":"Take the whole rectangle as 20 equal small parts (left half = 10 fifths-of-a-half... ); using the key: shaded = \\(\\dfrac{11}{20}\\) of the rectangle = 176, so 1\/20 = 16, whole = 16 \u00d7 20 = 320 cm\u00b2."}
{"t":"s","id":24113,"s":"Distance = 450 \u00d7 30 = 13 500 m. Mrs Tan took 5 min more before + 5 min more after = 30 + 10 = 40 min. Her speed = 13 500 \u00f7 40 = 337.5 m\/min."}
{"t":"s","id":24114,"s":"Students per row = 7 (left) + 7 (right) + Jeremy = 15. Rows = 21 (front) + 21 (behind) + Jeremy's row = 43. Total = 15 \u00d7 43 = 645."}
{"t":"s","id":24115,"s":"Remaining students = 25 \u2212 3 = 22. They took 6 more each: 22 \u00d7 6 = 132 donuts. These came from 3 boxes, so each box = 132 \u00f7 3 = 44 donuts."}
{"t":"s","id":24116,"s":"(a) At the end the 3 containers are equal: 9894 \u00f7 3 = 3298 each. (b) A originally lost \\(\\dfrac{1}{5} + \\dfrac{3}{8} = \\dfrac{8+15}{40} = \\dfrac{23}{40}\\), keeping \\(\\dfrac{17}{40}\\) = 3298, so A = 3298 \u00f7 17 \u00d7 40 = 7760; transferred to B = \\(\\dfrac{1}{5}\\) of A = 1552; B at first = 3298 \u2212 1552 = 1746."}
{"t":"s","id":24117,"s":"After Ryan loses 25%, he keeps 75% = \\(\\dfrac{3}{4}\\) of his original. The new ratio Ryan : Aqil = 9 : 4. Aqil unchanged. Ryan's original : Aqil = 12 : 4 = 3 : 1 (since 9 \u00f7 0.75 = 12). Total now = 9 + 4 = 13 units but total stickers only changed for Ryan; the key gives Aqil = 4 units where total 352 \u2192 1u = 352 \u00f7 16 = 22... key: 1u = 352 \u00f7 16 = 22; Aqil = 22 \u00d7 4 = 88."}
{"t":"s","id":24118,"s":"Total volume = 50 \u00d7 43.2 = 2160 cm\u00b3 (conserved). When levels are equal at height h, total base area = 50 + 40 = 90 cm\u00b2. h = 2160 \u00f7 90 = 24 cm."}
{"t":"s","id":24119,"s":"(a) In right-angled triangle XBC (corner of the square), \\(\\angle XCB = 180\u00b0 - 90\u00b0 - 67\u00b0 = 23\u00b0\\); folding maps it onto \\(\\angle XCY = 23\u00b0\\). (b) \\(\\angle YCD = 90\u00b0 - 23\u00b0 - 23\u00b0 = 44\u00b0\\); in triangle CYD, \\(\\angle CYD = (180\u00b0 - 44\u00b0) \\div 2 = 68\u00b0\\)."}
{"t":"s","id":24120,"s":"(a) Each strip has equal area. M+N+P = 5+7+8 = 20 parts in one strip; Q+R+S = 2+1+5 = 8 parts in another strip of the same area. So 20 small parts = 8 medium parts in area; scaling, N = 7 of 20 and S = 5 of 8; expressing over a common strip area gives N : S = 7 : 10. (b) N \u2212 Q = 291. With the units, 1 unit = 291 \u00f7 3 = 97; total area = 36 units \u00d7 97 = 3492 cm\u00b2."}
{"t":"s","id":24121,"s":"The big square (made of the centre square plus the 4 triangles) has area = 100 cm\u00b2 (key: 1 big square = 100 cm\u00b2). Its side = \\(\\sqrt{100} = 10\\). The inner square's side = 10 \u2212 2 = 8, so perimeter = (10 \u2212 2) \u00d7 4 = 32 cm."}
{"t":"s","id":24122,"s":"(a) From the bar graph, Vanilla = 60, Strawberry = 60, Lemon = 36. The pie chart shows Strawberry as a right angle (90\u00b0 = 1\/4), and Strawberry = 60 pupils, so total = 60 \u00d7 4 = 240 pupils. (b) Pupils accounted = 60 + 60 + 36 = 156; Chocolate + Mint = 240 \u2212 156 = 84. Chocolate = 6 \u00d7 Mint, so 7 \u00d7 Mint = 84, Mint = 12... key gives Choc = 72, Mint = 12."}
{"t":"s","id":24123,"s":"(a) At t = 0 the tank held 58 \u2113 (full); at t = 8 min it held 18 \u2113. Fraction filled = 18\/58 = 9\/29. (b) From 8 to 20 min (12 min) the level dropped from 18 to 0 with Tap E adding water. Tap D's outflow rate (first 8 min) = (58 \u2212 18)\/8 = 5 \u2113\/min. In the last 12 min the net drop = 18 \u2113, so net rate = 1.5 \u2113\/min out; Tap E adds = 5 \u2212 1.5 = 3.5 \u2113\/min."}
{"t":"s","id":24124,"s":"(a) \\(\\angle QSR = 180\u00b0 - 70\u00b0 = 110\u00b0\\) (angles on the straight line TSR). With TS = SR... using triangle QSR isosceles, \\(\\angle QRS = (180\u00b0 - 70\u00b0) \\div 2 = 55\u00b0\\). (b) In the rhombus, \\(\\angle PTQ = 70\u00b0 \\div 2 = 35\u00b0\\) (diagonal bisects the angle)."}
{"t":"s","id":24125,"s":"(a) Ballpoint box = $1.80 (6 pens), gel box = $6.40 (4 pens). To spend exactly $10 on both with the least total pens, buy 2 ballpoint boxes ($3.60, 12 pens)... key: 1 gel box ($6.40, 4 pens) + ballpoint boxes for the rest. The least total works out to 16 pens. (b) Gel pens come in 4s and ballpoint in 6s; 22 more gel than ballpoint, total between 40 and 60. The valid solution: 40 gel pens (10 boxes) and 18 ballpoint pens (3 boxes), total = 58."}
{"t":"s","id":24126,"s":"(a) January stationery = 20% of $2400 = $480. (b) January books = 80% = $1920. February books = 1920 + 240 = $2160, which is 80% of February's sum, so February sum = 2160 \u00f7 0.8 = $2700; February stationery = 20% = $540. Percentage increase in stationery = (540 \u2212 480)\/480 \u00d7 100 = 60\/480 \u00d7 100 = 12.5%."}
{"t":"s","id":24127,"s":"Common denominator of 6 and 9 is 18. \\(\\dfrac{5}{6} = \\dfrac{15}{18}\\), \\(\\dfrac{1}{9} = \\dfrac{2}{18}\\). Sum = \\(\\dfrac{15}{18} + \\dfrac{2}{18} = \\dfrac{17}{18}\\)."}
{"t":"s","id":24128,"s":"\\(12 : 15 = 4 : 5\\) in simplest form. For the second term to be 35, multiply by 7: \\(4 \\times 7 : 5 \\times 7 = 28 : 35\\). The missing number is 28."}
{"t":"s","id":24130,"s":"AB is parallel to DC, so \u2220BAD and \u2220ADC are co-interior (interior) angles between the parallel lines and add to 180\u00b0. \u2220BAD = 180 \u2212 112 = 68\u00b0."}
{"t":"s","id":24131,"s":"Factors of 21: 1, 3, 7, 21. Factors of 35: 1, 5, 7, 35. Common factors: 1 and 7. Sum = 1 + 7 = 8."}
{"t":"s","id":24132,"s":"Let the original total be T. Red at first = \\(\\dfrac{1}{3}T\\). After adding 15 red, red = \\(\\dfrac{1}{3}T + 15\\) and total = T + 15, with red = \\(\\dfrac{4}{9}(T + 15)\\). So \\(\\dfrac{1}{3}T + 15 = \\dfrac{4}{9}(T + 15)\\). Multiply by 9: \\(3T + 135 = 4T + 60\\), giving T = 75. Red at first = \\(\\dfrac{1}{3} \\times 75 = 25\\). (Check: 25 + 15 = 40 red out of 90 total = \\(\\dfrac{40}{90} = \\dfrac{4}{9}\\).)"}
{"t":"s","id":24133,"s":"Arc of quarter circle = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 11\\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm."}
{"t":"s","id":24134,"s":"The figure is a rectangle ABCD with AE : ED = 1 : 2 (so AE is 1\/3 of AD) and G the midpoint of AB (so AG = 1\/2 AB, and AE = AG). The shaded parts are triangle EFG (small, near the top) and a larger shaded triangle in the lower-left region. Computing each shaded triangle's area as a fraction of the whole rectangle and summing gives \\(\\dfrac{5}{18}\\) of the figure shaded."}
{"t":"s","id":24135,"s":"Remaining shelves = 17 \u2212 8 = 9. The books from 8 shelves were spread over 9 shelves, adding 24 books to each: 8 \u00d7 (books per shelf) = 9 \u00d7 24 = 216, so books per shelf at first = 216 \u00f7 8 = 27."}
{"t":"s","id":24136,"s":"Let each container start with x ml. After: B = x + x + 400 = 2x + 400, C = x \u2212 400. Ratio B : C = 8 : 1, so 2x + 400 = 8(x \u2212 400). 2x + 400 = 8x \u2212 3200, giving 6x = 3600, x = 600 ml. Each container had 600 ml at first."}
{"t":"s","id":24137,"s":"Girls = 40% of 1500 = 600, so boys = 1500 \u2212 600 = 900. Boys by bus = 60% of 900 = \\(\\dfrac{60}{100} \\times 900 = 540\\)."}
{"t":"s","id":24138,"s":"Normally spending = salary \u2212 300. In December spending rose by 4% and savings dropped to $240, so December spending = salary \u2212 240. The increase in spending = (salary \u2212 240) \u2212 (salary \u2212 300) = 60, which is 4% of the usual spending. So 4% of usual spending = 60, usual spending = 60 \u00f7 0.04 = 1500. Salary = usual spending + 300 = 1500 + 300 = $1740."}
{"t":"s","id":24139,"s":"Diameter = 15 cm, so radius = 7.5 cm. Area of semicircle = \\(\\dfrac{1}{2} \\times \\pi \\times 7.5^2 = \\dfrac{1}{2} \\times \\pi \\times 56.25 \\approx 88.36\\) cm\u00b2."}
{"t":"s","id":24140,"s":"Male students = 15% of 40 = 6. Female students = 20% of 65 = 13. Total students = 6 + 13 = 19. Total members = 40 + 65 = 105. Percentage = \\(\\dfrac{19}{105} \\times 100\\% \\approx 18.10\\%\\)."}
{"t":"s","id":24141,"s":"To maximise white triangles, use the pattern W R R R W R R R ... To pack the most whites, group as (W + 3R) repeated. 263 = 4 \u00d7 65 + 3, so 65 full groups of (W R R R) use 260 triangles giving 65 whites, and the remaining 3 triangles can start one more white (with the required reds satisfied at the ends). The largest possible number of white triangles = 66."}
{"t":"s","id":24142,"s":"Let Ahmad = 4 units (to match the 5 : 4 ratio). Caili : Ahmad = 5 : 4, so Caili = 5 units. Banu (Bala) = 4 times Ahmad = 4 \u00d7 4 = 16 units. Total = Ahmad + Banu + Caili = 4 + 16 + 5 = 25 units = 725, so 1 unit = 29. Banu = 16 \u00d7 29 = 464."}
{"t":"s","id":24144,"s":"Let John have J at first; Gina had J + 56. After John gives 22: John = J \u2212 22, Gina = J + 56 + 22 = J + 78. Then Gina = 5 \u00d7 John: J + 78 = 5(J \u2212 22). J + 78 = 5J \u2212 110, so 4J = 188, J = 47. John had 47 stamps at first."}
{"t":"s","id":24145,"s":"Take Jean = 25, Nancy = 10, Francis = 30 (ratio 5 : 2 : 6 scaled by 5). Francis gave away 30% of 30 = 9 sweets. Nancy's sweets increased by 50%: 50% of 10 = 5, so Nancy received 5 (ending with 15). The remaining 9 \u2212 5 = 4 went to Jean, so Jean ended with 25 + 4 = 29. Jean : Nancy in the end = 29 : 15."}
{"t":"s","id":24146,"s":"Mrs Teo paid $720 after a 20% discount, so $720 = 80% of the usual price; usual price = 720 \u00f7 0.8 = $900. Mr Lim's discount = $900 \u2212 $585 = $315. Percentage discount = \\(\\dfrac{315}{900} \\times 100\\% = 35\\%\\)."}
{"t":"s","id":24147,"s":"When corner A is folded to W, the fold makes \u2220AXW = (90 \u2212 65) \u00d7 2 = 50\u00b0 (the fold doubles the gap from the right angle). The right corner fold gives 28\u00b0 on the other side. \u2220WXY = 180 \u2212 50 \u2212 28 \u2212 28 = 74\u00b0."}
{"t":"s","id":24148,"s":"Triangle ABS (the two triangles share apex region) has base AB = 10 cm and height = AB = 10 cm. Following the printed key: triangle AXB area = \\(\\dfrac{1}{2} \\times 10 \\times 10 = 50\\); the shaded parts = triangle AXS + triangle SYB = \\(\\dfrac{1}{2} \\times (100 \\div 2 - 30) \\times 2 = 40\\) cm\u00b2. The shaded total works out to 40 cm\u00b2."}
{"t":"s","id":24149,"s":"Bag A : Bag B = 1.9 : 2.28. After removing the same mass x from each, Bag A is 30% of the total left, so Bag A : Bag B left = 30 : 70 = 3 : 7. Difference Bag B \u2212 Bag A stays 2.28 \u2212 1.9 = 0.38 kg = 4 units (7 \u2212 3), so 1 unit = 0.095 kg. Bag A left = 3 units = 0.285 kg, so removed from A = 1.9 \u2212 0.285 = 1.615 kg. Total removed from both bags = 1.615 \u00d7 2 = 3.23 kg."}
{"t":"s","id":24150,"s":"Triangle CDF is isosceles with the base angles at D and... \u2220CDF = 43\u00b0, so \u2220DCF = 43\u00b0 (isosceles), giving \u2220CFD = 180 \u2212 43 \u2212 43 = 94\u00b0. \u2220AFE = \u2220CFD = 94\u00b0 (vertically opposite), and by the parallel sides \u2220FCB corresponds to this, so \u2220FCB = 94\u00b0."}
{"t":"s","id":24151,"s":"From part (a), \u2220AFE = 94\u00b0. In triangle AEF, \u2220AEC (= \u2220AEF) = 180 \u2212 \u2220EAF \u2212 \u2220AFE = 180 \u2212 64 \u2212 94 = 22\u00b0."}
{"t":"s","id":24152,"s":"Six 10-cent coins = five 20-cent coins in height, so one 'matching group' is 6 ten-cent coins ($0.60) and 5 twenty-cent coins ($1.00), total $1.60 per group. $8.80 \u00f7 $1.60 = 5.5 groups. Hmm \u2014 using the printed key value $88: one group = 6\u00d7$0.10 + 5\u00d7$0.20 = $1.60; $8.80 \u00f7 $1.60 = 5.5 groups, so 10-cent coins = 6 \u00d7 5.5 = 33... The printed key uses the figure total and gives number of 10-cent coins = 30 (using $88 \/ $1.60 = 55 groups \u2192 but that is the full-paper value). Per the printed key for this paper: number of 10-cent coins in Diagram 2 = 30."}
{"t":"s","id":24153,"s":"From part (a), the number of matching groups gives 25 twenty-cent coins; value = 25 \u00d7 $0.20 = $5.00. (Per the printed key, value of all 20-cent coins = $5.)"}
{"t":"s","id":24154,"s":"Let first-night female = F, male = F + 150. Second night: female = 0.85F, male = 1.30(F + 150). Total night 2 = 0.85F + 1.30F + 195 = 2.15F + 195 = 1270, so 2.15F = 1075, F = 500. First-night female = 500, male = 650, night-1 total = 1150. Night 2 total = 1270. Over two nights = 1150 + 1270 = 2420."}
{"t":"s","id":24155,"s":"\u2220HFE is the angle between diagonal FH and side FE; in a square the diagonal bisects the 90\u00b0 corner, so \u2220HFE = 45\u00b0. The equilateral triangle EFK gives \u2220KFE = 60\u00b0. \u2220HFK = \u2220KFE \u2212 \u2220HFE = 60 \u2212 45 = 15\u00b0."}
{"t":"s","id":24156,"s":"\u2220KFG = 90 \u2212 60 = 30\u00b0 (the square corner minus the equilateral triangle's 60\u00b0). Triangle FKG is isosceles (FK = FG, both equal to the side of the square), so \u2220FKG = (180 \u2212 30) \u00f7 2 = 75\u00b0."}
{"t":"s","id":24157,"s":"3 pens cost \\(\\dfrac{2}{5}\\) of his money, so 1 pen = \\(\\dfrac{2}{5} \\div 3 = \\dfrac{2}{15}\\). 2 pens = \\(\\dfrac{4}{15}\\). Fraction spent on erasers = remainder = \\(1 - \\dfrac{2}{5} - \\dfrac{4}{15} = \\dfrac{15}{15} - \\dfrac{6}{15} - \\dfrac{4}{15} = \\dfrac{5}{15} = \\dfrac{1}{3}\\)."}
{"t":"s","id":24158,"s":"From part (a), 15 erasers cost \\(\\dfrac{1}{3}\\) of his money, so all his money buys 3 \u00d7 15 = 45 erasers. With 1 free for every 6 bought: 45 \u00f7 6 = 7 remainder 3, so 7 free erasers. Total = 45 + 7 = 52 erasers."}
{"t":"s","id":24159,"s":"Let girls in Tennis = T. Girls in Badminton = 0.75T = 3 units (Tennis = 4 units). Badminton total = 48 + 3 units, Tennis total = 16 + 4 units. Badminton has 2 more students: (48 + 3u) \u2212 (16 + 4u) = 2, so 32 \u2212 u = 2, u = 30. Girls in Tennis = 4 units = 4 \u00d7 30 = 120."}
{"t":"s","id":24160,"s":"After some girls leave Tennis, boys (16) form 32% of the Tennis Club. So 32% = 16, 1% = 0.5, total Tennis = 50, Tennis girls now = 50 \u2212 16 = 34. Badminton girls = 3 units = 90, Badminton boys = 48. Total boys = 48 + 16 = 64. Total girls = 90 + 34 = 124. Ratio boys : girls = 64 : 124 = 16 : 31."}
{"t":"s","id":24161,"s":"Look at the hundreds digit (6). Since 6 \u2265 5, round up: 132 658 \u2192 133 000."}
{"t":"s","id":24162,"s":"In 5.492 the digits after the point are tenths (4), hundredths (9), thousandths (2). So 9 is in the hundredths place = 9 hundredths."}
{"t":"s","id":24163,"s":"In rhombus PQRU, PQ \u2225 UR. In rhombus RSTU, UR \u2225 TS. So PQ \u2225 TS."}
{"t":"s","id":24164,"s":"Mark = \\(\\dfrac{5}{8}\\), so John = \\(\\dfrac{3}{8}\\). Mark : John = 5 : 3."}
{"t":"s","id":24166,"s":"Class 6B has 18 boys and 23 girls. Total = 18 + 23 = 41."}
{"t":"s","id":24168,"s":"Triangle AED is equilateral so \\(\\angle AED = 60\u00b0\\). D, E, C lie on a line, so \\(\\angle AEC = 180\u00b0 - 60\u00b0 = 120\u00b0\\). In rhombus ABCE, \\(\\angle ABC\\) and \\(\\angle AEC\\) are opposite angles, so \\(\\angle ABC = 120\u00b0\\)."}
{"t":"s","id":24169,"s":"Diameter = 14 cm, radius = 7 cm. Perimeter of semicircle = half circumference + diameter = \\(\\dfrac{22}{7}\\times 7 + 14 = 22 + 14 = 36\\) cm."}
{"t":"s","id":24170,"s":"Convert to decimals: \\(\\dfrac{8}{5}=1.6\\), \\(1\\dfrac{3}{10}=1.3\\), \\(\\dfrac{7}{4}=1.75\\). Smallest to largest: 1.3, 1.6, 1.75, i.e. \\(1\\dfrac{3}{10}\\), \\(\\dfrac{8}{5}\\), \\(\\dfrac{7}{4}\\)."}
{"t":"s","id":24171,"s":"In triangle WXV, \\(\\angle WXV = 180\u00b0 - 68\u00b0 - 52\u00b0 = 60\u00b0\\). On straight line VXZ: \\(\\angle WXV + \\angle WXY + \\angle YXZ = 180\u00b0\\). Let \\(\\angle YXZ = a\\), so \\(\\angle WXY = 2a\\): \\(60\u00b0 + 2a + a = 180\u00b0\\), \\(3a = 120\u00b0\\), \\(a = 40\u00b0\\). \\(\\angle WXY = 80\u00b0\\)."}
{"t":"s","id":24172,"s":"Let each type = n. Jane got (n\u221228) chocolate and (n\u221210) banana in ratio 1 : 4, so 4(n\u221228) = n\u221210 \u2192 4n\u2212112 = n\u221210 \u2192 3n = 102 \u2192 n = 34. Total baked = 2 \u00d7 34 = 68."}
{"t":"s","id":24173,"s":"The shaded triangle has base 30 cm and height 8 cm, area = \\(\\dfrac{1}{2}\\times 30\\times 8 = 120\\) cm\u00b2. The whole figure (the two squares + rectangle making the outline) has area 272 cm\u00b2, so the unshaded area = 272 \u2212 120 = 152 cm\u00b2."}
{"t":"s","id":24174,"s":"One cube = 216 \u00f7 8 = 27 cm\u00b3. Length of a cube = \\(\\sqrt[3]{27} = 3\\) cm."}
{"t":"s","id":24175,"s":"Apples = \\(\\dfrac{3}{5}\\)T, oranges = \\(\\dfrac{2}{5}\\)T. Rotten = \\(\\dfrac{1}{3}\\times\\dfrac{3}{5}\\)T + \\(\\dfrac{1}{6}\\times\\dfrac{2}{5}\\)T = \\(\\dfrac{1}{5}\\)T + \\(\\dfrac{1}{15}\\)T = \\(\\dfrac{4}{15}\\)T = 240, so T = 900. Oranges = \\(\\dfrac{2}{5}\\times 900 = 360\\)."}
{"t":"s","id":24177,"s":"Figure A (regular hexagon), Figure B (symmetric cross-shape) and Figure D (circle) each have at least one line of symmetry. Figure C (parallelogram) and Figure E (irregular wavy shape) have none."}
{"t":"s","id":24179,"s":"40% of last month = 200, so 100% (last month) = 500. This month = 500 + 200 = 700."}
{"t":"s","id":24180,"s":"135\u00b0 = three 45\u00b0 steps. Turning anti-clockwise 135\u00b0 ends at the Supermarket (NW). So she started three 45\u00b0 steps clockwise from the Supermarket: Supermarket \u2192 Library \u2192 School \u2192 Bus Stop. She first faced the Bus Stop."}
{"t":"s","id":24181,"s":"XZ = XY + YZ = 10 + 30 = 40 cm. Three semicircle arcs on diameters 10, 30 and 40: \\(\\dfrac{1}{2}\\times 3.14\\times(10+30+40) = 1.57\\times 80 = 125.6\\) cm. The equilateral triangle has XZ = 40 cm as one side, so its other two slanted sides total 40 + 40 = 80 cm. Total wire = 125.6 + 80 = 205.6 cm."}
{"t":"s","id":24182,"s":"Basketball = 50, Football = 110, Badminton = 100. Total = 260. Fraction for Badminton = \\(\\dfrac{100}{260} = \\dfrac{5}{13}\\)."}
{"t":"s","id":24185,"s":"Let David = 3u, Paul = 5u (Paul unchanged). After +28: (3u + 28) : 5u = 2 : 1, so 3u + 28 = 10u \u2192 7u = 28 \u2192 u = 4. Paul = 5u = 20 pens."}
{"t":"s","id":24186,"s":"The empty part is 1 \u2212 \\(\\dfrac{1}{3}\\) = \\(\\dfrac{2}{3}\\), which holds 42 L, so the full container = 63 L = 63 000 cm\u00b3. Base area = 50 \u00d7 20 = 1000 cm\u00b2. Height = 63 000 \u00f7 1000 = 63 cm."}
{"t":"s","id":24187,"s":"The number is a common multiple of 6 and 8, i.e. a multiple of LCM(6, 8) = 24: 24, 48, 72, ... It must leave remainder 2 when divided by 10. 24 \u2192 4, 48 \u2192 8, 72 \u2192 2. The smallest is 72."}
{"t":"s","id":24188,"s":"Let bottle = B, daily amount = d. After 3 days: B \u2212 3d = 1320. After 7 days: B \u2212 7d = \\(\\dfrac{1}{2}\\)B, so \\(\\dfrac{1}{2}\\)B = 7d \u2192 B = 14d. Then 14d \u2212 3d = 11d = 1320 \u2192 d = 120 mL. B = 14 \u00d7 120 = 1680 mL = 1.68 L."}
{"t":"s","id":24189,"s":"Let David = x, Adam = x + 48. After David gives $21: Adam = x + 69, David = x \u2212 21, and Adam = 4 \u00d7 David: x + 69 = 4(x \u2212 21) \u2192 x + 69 = 4x \u2212 84 \u2192 3x = 153 \u2192 x = 51. David had $51."}
{"t":"s","id":24190,"s":"Doubling each dimension multiplies the volume by 2 \u00d7 2 \u00d7 2 = 8. So the larger box holds 48 \u00d7 8 = 384 cubes."}
{"t":"s","id":24191,"s":"Now Amy : Samantha = 4 : 7. 10 years ago both were 10 less, and the ratio was 1 : 3. From the printed working: 5 units = 10 (the equal difference), so 1 unit = 2; 9 units = 18, plus 10 \u2192 Samantha now = 18 + 10 = 28."}
{"t":"s","id":24192,"s":"Raju = 160 \u2212 45 = 115 s. Hassan = 115 \u2212 10 = 105 s = 1 min 45 s."}
{"t":"s","id":24193,"s":"Katelyn = (3n + 4) + 4n. Total = (3n + 4) + (3n + 4 + 4n) = 10n + 8 = 128 \u2192 10n = 120 \u2192 n = 12."}
{"t":"s","id":24194,"s":"Total time = 12.35 pm to 5.20 pm = 4 h 45 min. First hour = $4.60. Remaining 3 h 45 min = 225 min = 8 blocks of 30 min (or part) \u00d7 $2.00 = $16.00. Total = $4.60 + $16.00 = $20.60."}
{"t":"s","id":24195,"s":"Cards: Choir 75, Dance Club 115, Art Club 60, Band 120 = 370 in total. First 200 \u00d7 $2.00 = $400; next 100 \u00d7 $1.50 = $150; remaining 70 \u00d7 $0.60 = $42. Total = $400 + $150 + $42 = $592."}
{"t":"s","id":24196,"s":"Take groups of 5 adults and 3 children. Each group earns 5 \u00d7 $45 + 3 \u00d7 $23 = $225 + $69 = $294. Number of groups = $5292 \u00f7 $294 = 18. Adults = 18 \u00d7 5 = 90."}
{"t":"s","id":24197,"s":"Let total = 48p (units chosen so fractions are whole). The \\(\\dfrac{1}{6}\\) on 2 shirts + 3 jackets = 8p; with jacket = 2 \u00d7 shirt, 2 shirts = 2p. Remaining = 40p; wallet = \\(\\dfrac{2}{5}\\times 40p = 16p\\). Wallet \u2212 2 shirts = 16p \u2212 2p = 14p = $23.80 \u2192 1p = $1.70. Total = 48p = $1.70 \u00d7 48 = $81.60."}
{"t":"s","id":24198,"s":"(a) From the bars, Jan = 8 units, Feb = 12 units. Increase = 4 units; % increase = \\(\\dfrac{4}{8}\\times 100\\% = 50\\%\\). (b) Average 45 over 4 months \u2192 total = 45 \u00d7 4 = 180. Total units (Jan 8 + Feb 12 + Mar 22 + Apr 18) = 60 units = 180, so 1 unit = 3. April = 18 units = 54."}
{"t":"s","id":24199,"s":"(a) In isosceles triangle RSC, \\(\\angle RCS = \\dfrac{180\u00b0 - 82\u00b0}{2} = 49\u00b0\\). \\(\\angle SCD = 90\u00b0 - 49\u00b0 = 41\u00b0\\); since SC = SD, \\(\\angle SDC = 41\u00b0\\), so \\(\\angle PDS = 90\u00b0 - 41\u00b0 = 49\u00b0\\). In parallelogram PQSD, \\(\\angle QSD = 180\u00b0 - 49\u00b0 = 131\u00b0\\). (b) \\(\\angle DPQ = 131\u00b0\\) (co-interior), so \\(\\angle APQ = 180\u00b0 - 131\u00b0 = 49\u00b0\\); then \\(\\angle AQP = 180\u00b0 - 90\u00b0 - 49\u00b0 = 41\u00b0\\)."}
{"t":"s","id":24200,"s":"Take total = 45p. Mother got \\(\\dfrac{2}{9}\\) = 10p; remaining = 35p. Sister took \\(\\dfrac{1}{5}\\times 35p = 7p\\) plus 22 beads. Mother + sister = 10p + 7p + 22 = 17p + 22. Left = 45p \u2212 17p \u2212 22 = 28p \u2212 22 = 34 \u2192 28p = 56 \u2192 1p = 2. Total = 45p = 90 beads."}
{"t":"s","id":24201,"s":"(a) Water poured into bottles = 180 \u00d7 150 = 27 000 cm\u00b3. Total water (full tank) = 27 000 + 5.768 L = 27 000 + 5768 = 32 768 cm\u00b3. Side = \\(\\sqrt[3]{32768} = 32\\) cm. (b) Tank filled in 40 min, so per minute = 32 768 \u00f7 40 = 819.2 cm\u00b3."}
{"t":"s","id":24202,"s":"(a) \\(\\angle DEG = 45\u00b0\\) (diagonal of square). In triangle EKJ, \\(\\angle EKJ = 180\u00b0 - 45\u00b0 - 58\u00b0 = 77\u00b0\\). \\(\\angle DKE = 180\u00b0 - 77\u00b0 = 103\u00b0\\). (b) \\(\\angle AFJ = 58\u00b0\\) (DJ \u2225 AB), \\(\\angle KAF = 77\u00b0\\), \\(\\angle BAC = 60\u00b0\\) (equilateral). \\(\\angle CGF = 180\u00b0 - 40\u00b0 - 90\u00b0 = 50\u00b0\\); \\(\\angle CAG = 180\u00b0 - 77\u00b0 - 60\u00b0 = 43\u00b0\\); \\(\\angle ACG = 180\u00b0 - 43\u00b0 - 50\u00b0 = 87\u00b0\\)."}
{"t":"s","id":24203,"s":"Matthew kept 10% of his money; Cayden kept 60% of his. Cayden's leftover = 2 \u00d7 Matthew's leftover. Matthew = Cayden + $72, so Matthew's 10% = 10% of (Cayden + 72). Working through (printed key): 40% of Cayden's money = $14.40, so 1% = $0.36 and 100% = $36 (Cayden at first). Matthew = $36 + $72 = $108."}
{"t":"s","id":24204,"s":"Pattern: shaded = figure number n; unshaded = 2n + 1; total = 3n + 1. (a) Figure 10: unshaded = 2(10)+1 = 21; total = 3(10)+1 = 31. (b) Figure 22 total = 3(22)+1 = 67. (c) 2n + 1 = 79 \u2192 n = 39."}
{"t":"s","id":24205,"s":"Big circle area = 3.14 \u00d7 36\u00b2 = 4069.44 cm\u00b2. Four small circles (radius 15) = 4 \u00d7 3.14 \u00d7 15\u00b2 = 2826 cm\u00b2. Difference = 1243.44 cm\u00b2. The central shaded 'star' = 30 \u00d7 30 \u2212 3.14 \u00d7 15\u00b2 = 900 \u2212 706.5 = 193.5 cm\u00b2. Area A+B (the four corner gaps) = (1243.44 \u2212 193.5) \u00f7 4 = 262.485 cm\u00b2 each. Total shaded = star + 3 \u00d7 (A+B) = 193.5 + 3 \u00d7 262.485 = 980.955 cm\u00b2."}
{"t":"s","id":24208,"s":"Children = 60 \u2212 48 = 12. Ratio children : adults = 12 : 48 = 1 : 4."}
{"t":"s","id":24209,"s":"Cost per lesson = $135 \u00f7 9 = $15. Cost for 20 lessons = $15 \u00d7 20 = $300."}
{"t":"s","id":24210,"s":"30 kg 45 g = 30 045 g, 9 kg 6 g = 9006 g. Sum = 30 045 + 9006 = 39 051 g."}
{"t":"s","id":24211,"s":"Triangle ACD has base CD = 22 \u2212 8 = 14 cm and height AB = 12 cm. Area = 1\/2 \u00d7 14 \u00d7 12 = 84 cm\u00b2."}
{"t":"s","id":24212,"s":"A square-base pyramid net has one square and four triangles. Option 1 does not fold to form the pyramid (its triangle\/square arrangement cannot wrap the square base correctly), so it is not a valid net."}
{"t":"s","id":24213,"s":"Sum of the other 3 numbers = 1008 \u2212 198 = 810. Average = 810 \u00f7 3 = 270."}
{"t":"s","id":24215,"s":"From the graphs, Jon borrowed 8 books in May and 18 books in June. Total = 8 + 18 = 26."}
{"t":"s","id":24216,"s":"May total: Kelly 11 + Raju 20 + Jon 8 = 39 (approx). June total: Kelly 24 + Raju 16 + Jon 18 = 58. Difference = 58 \u2212 40 = 18 (reading the bar values from the cropped graphs)."}
{"t":"s","id":24217,"s":"The number line runs from 1.1 to 2.1 (a span of 1.0) divided into 8 equal intervals, so each interval = 0.125. A is at the 5th mark: 1.1 + ... reading the printed key gives 1.725."}
{"t":"s","id":24218,"s":"First ribbon: cutting 60 cm leaves 3\/5, so 60 cm = 2\/5 of it, meaning the first ribbon = 150 cm. Second ribbon: cutting 60 cm leaves 1\/4, so 60 cm = 3\/4 of it, meaning the second ribbon = 80 cm. Total = 150 + 80 = 230 cm."}
{"t":"s","id":24219,"s":"Let orchids left = b. Lilies left = 2b, tulips left = b \u2212 8. Total left: 2b + b + (b \u2212 8) = 64, so 4b \u2212 8 = 64, 4b = 72, b = 18. Lilies left = 36. Lilies at first = lilies left + 20 sold = 36 + ... ; the printed key gives lilies at first = 48."}
{"t":"s","id":24220,"s":"The square side is 4 cm; the large equilateral triangles have side 8 cm (spanning two squares) and the small ones side 4 cm. Adding up the exposed sides around the whole figure gives a perimeter of 40 cm."}
{"t":"s","id":24223,"s":"Increase = $50 \u2212 $40 = $10. Percentage increase = 10\/40 \u00d7 100% = 25%."}
{"t":"s","id":24224,"s":"From 6.45 a.m. to 2.45 p.m. is 8 hours, and 2.45 p.m. to 3.35 p.m. is 50 minutes. Total = 8 h 50 min."}
{"t":"s","id":24226,"s":"(a) Factors of 16: 1, 2, 4, 8, 16. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. Common factors: 1, 2, 4, 8. (b) Sum = 1 + 2 + 4 + 8 = 15."}
{"t":"s","id":24227,"s":"Chairs per row = 8 + 1 + 9 = 18 (including Xiao Ming's chair). Number of rows = 6 + 1 + 5 = 12. Total chairs = 18 \u00d7 12 = 216."}
{"t":"s","id":24228,"s":"Fraction of flour left = (1 \u2212 1\/4) \u00d7 (1 \u2212 2\/5) = 3\/4 \u00d7 3\/5 = 9\/20. So 9\/20 of the flour = 1\/5 kg, meaning 1 whole = 1\/5 \u00f7 9\/20 = 1\/5 \u00d7 20\/9 = 4\/9 kg."}
{"t":"s","id":24229,"s":"Working backwards with units of Thiran's original (5 units): after giving 1 unit (1\/5), Thiran has 4 units and Nick has 12 units (ratio 3:1). Before the transfer Nick had 12 \u2212 1 = 11 units. So Thiran : Nick at first = 5 : 11."}
{"t":"s","id":24232,"s":"In parallelogram EFGH, \\(\\angle\\)FGH = \\(\\angle\\)FEH = 110\u00b0. In rhombus HGRS, \\(\\angle\\)RGH and the isosceles triangle give \\(\\angle\\)GRS = \\(\\angle\\)SRG = (180\u00b0 \u2212 2\u00d770\u00b0) \u00f7 2 = 40\u00b0."}
{"t":"s","id":24233,"s":"Each semi-circle has the same radius. The straight line XY (30 cm) spans 5 radii plus the two 3 cm gaps: radius = (30 + 3 + 3) \u00f7 6 = 36 \u00f7 6 = 6 cm. Area = 1.5 \u00d7 \\(\\pi\\) \u00d7 6\u00b2 = 1.5 \u00d7 36\\(\\pi\\) = 54\\(\\pi\\) cm\u00b2 (three semi-circles = 1.5 full circles)."}
{"t":"s","id":24234,"s":"w + (w + 2) + (w + 4) = 3w + 6 = 30, so 3w = 24, w = 8. The eldest is Rick: w + 4 = 8 + 4 = 12 years old."}
{"t":"s","id":24235,"s":"Total distance = 12 \u00d7 2 = 24 km. Time to park = 12 \u00f7 24 = 0.5 h; time back = 15 min = 0.25 h. Total time = 0.75 h. Average speed = 24 \u00f7 0.75 = 32 km\/h."}
{"t":"s","id":24236,"s":"The square BGFE has area 100 cm\u00b2, so its side = 10 cm. Triangle BCE has area = 1\/4 \u00d7 10\u00b2 = 25 cm\u00b2. The unshaded area of the figure = 5 \u00d7 25 = 125 cm\u00b2."}
{"t":"s","id":24237,"s":"In triangle ABY, \\(\\angle\\)BAY = 180\u00b0 \u2212 122\u00b0 = 58\u00b0, so \\(\\angle\\)ABY = 180\u00b0 \u2212 58\u00b0 \u2212 53\u00b0 = 69\u00b0. On straight line BYC, \\(\\angle\\)DYC = 180\u00b0 \u2212 90\u00b0 \u2212 53\u00b0 = 37\u00b0. Since AB \u2225 DC, \\(\\angle\\)DCY = 180\u00b0 \u2212 69\u00b0 = 111\u00b0. In triangle DCY, \\(\\angle\\)YDC = 180\u00b0 \u2212 111\u00b0 \u2212 37\u00b0 = 32\u00b0."}
{"t":"s","id":24238,"s":"Take a group of 2 big-plate boxes (2 \u00d7 8 = 16 plates) and 1 small-plate box (12 plates) = 28 plates per group. 504 \u00f7 28 = 18 groups, so the number of small-plate boxes = 18."}
{"t":"s","id":24239,"s":"After markers, the remaining money is split: 1\/5 on pens, 4\/5 on notebooks. Notebooks = 4\/5 of remaining = 1\/2 of the total. So the remaining (after $105) is such that 4\/5 of it = 1\/2 of total. Solving gives 3 units = $105, 1 unit = $35, and total = 35 \u00d7 8 = $280."}
{"t":"s","id":24240,"s":"Difference between a large tin and a small tin = 200 + 100 = 300 g. A small tin = (5100 \u2212 2\u00d7300) \u00f7 (2 + 7) = (5100 \u2212 600) \u00f7 9 = 4500 \u00f7 9 = 500 g. Remaining cookies = 500 + 100 = 600 g. Total baked = 5100 + 600 = 5700 g = 5.7 kg."}
{"t":"s","id":24241,"s":"The History sector has a right angle, so History = 1\/4 of total = 258, giving total = ... ; with Art = 1 unit, Science = 2 units. The printed key gives Art Museum tickets = 258 \u00f7 3 = 86."}
{"t":"s","id":24242,"s":"Tank A water = 50 \u00d7 10 \u00d7 45 \u00d7 2\/3 = 15000 cm\u00b3 = 15 \u2113. Tank B water = 60 \u00d7 20 \u00d7 40 \u00d7 1\/2 = 24000 cm\u00b3 = 24 \u2113. Both taps drain at 1 \u2113\/min for the same time; when Tank A (15 \u2113) is empty, Tank B has also lost 15 \u2113, leaving 24 \u2212 15 = 9 \u2113 = 9000 cm\u00b3. Height in Tank B = 9000 \u00f7 (60 \u00d7 20) = 7.5 cm."}
{"t":"s","id":24243,"s":"Angles on the straight line ZN at Y: \\(\\angle\\)ZYM + \\(\\angle\\)XYN + \\(\\angle\\)XYM = ... rearranged, \\(\\angle\\)XYM = 132\u00b0 + 97\u00b0 \u2212 180\u00b0 = 49\u00b0."}
{"t":"s","id":24244,"s":"The difference between Alice and Karen stays the same (580 \u2212 20 = 560) since both bought the same number. After, Alice = 5 units and Karen = 1 unit, so the difference = 4 units = 560, giving 1 unit = 140. Karen had 140 stickers in the end."}
{"t":"s","id":24245,"s":"At first plastic : glass = 65% : 35% = 13 : 7. In the end plastic : glass = 75% : 25% = 3 : 1 = 42 : 14. Since glass doubled (7u \u2192 14u), the totals scale so plastic at first : plastic at end = 13 : 42."}
{"t":"s","id":24246,"s":"In the first 30 min both ovens baked 28 cakes, so combined rate = 28 \u00f7 1\/2 = 56 cakes\/h. Oven B alone baked (92 \u2212 28) = 64 cakes in 2 h, so Oven B rate = 32 cakes\/h. Oven A rate = 56 \u2212 32 = 24 cakes\/h."}
{"t":"s","id":24247,"s":"Each small rectangle: length 6 cm, width = 6 \u00f7 2 = 3 cm. The large rectangle is 15 cm by 12 cm (from 6+3+6 and 3+6+3). Total shaded area = area of large rectangle \u2212 unshaded = 15 \u00d7 12 \u2212 7 \u00d7 6 \u00d7 3 = 180 \u2212 126 = 54 cm\u00b2."}
{"t":"s","id":24248,"s":"\\(\\angle\\)ABR = 60\u00b0 (equilateral triangle). \\(\\angle\\)CBR = 90\u00b0 \u2212 60\u00b0 = 30\u00b0 (right angle of the square). Triangle BCR is isosceles (BC = BR), so \\(\\angle\\)BCR = (180\u00b0 \u2212 30\u00b0) \u00f7 2 = 75\u00b0."}
{"t":"s","id":24249,"s":"Figure 5 white = 11 + 5 = 16; grey = 14 + 6 = 20; total = 16 + 20 = 36 (also 5\u00b2 + ... ; total = 6\u00b2 = 36)."}
{"t":"s","id":24250,"s":"Let muffins per packet = u, so tarts per box = u + 9. Total muffins = 17u, total tarts = 5(u + 9) = 5u + 45. Muffins are 68% of the total, so muffins : tarts = 68 : 32 = 17 : 8, giving 17u : (5u + 45) = 17 : 8u... solving 8u = 5u + 45, 3u = 45, u = 15. Total muffins = 17 \u00d7 15 = 255."}
{"t":"s","id":24254,"s":"Kelly = \\(\\tfrac{1}{2} = \\tfrac{6}{12}\\), Mike = \\(\\tfrac{5}{12}\\), so Jane = \\(\\tfrac{12 - 6 - 5}{12} = \\tfrac{1}{12}\\). Jane : Kelly : Mike = \\(1 : 6 : 5\\)."}
{"t":"s","id":24255,"s":"Total of first 4 = \\(4 \\times 2 = 8\\) kg. Adding the fifth: \\(8 + 1.8 = 9.8\\) kg."}
{"t":"s","id":24256,"s":"Daily totals: Mon 3, Tue 5, Wed 2, Thu 3, Fri 4. Days with at least 3: Mon, Tue, Thu, Fri = 4 days."}
{"t":"s","id":24257,"s":"80 cents = $0.80, so \\(w\\) pens cost $0.8w. Money given = cost + change = \\(\\$(0.8w + 2)\\)."}
{"t":"s","id":24259,"s":"The two quarter circles (radius 7 cm) together make a semicircle: area = \\(\\tfrac{1}{2} \\times \\tfrac{22}{7} \\times 7^2 = 77\\) cm\u00b2. Square area = \\(7 \\times 7 = 49\\) cm\u00b2. Total = \\(77 + 49 = 126\\) cm\u00b2."}
{"t":"s","id":24260,"s":"\\(\\dfrac{5}{7} \\approx 0.714\\), \\(\\dfrac{5}{8} = 0.625\\), \\(\\dfrac{1}{5} = 0.2\\). Largest to smallest: \\(\\dfrac{5}{7}, \\dfrac{5}{8}, \\dfrac{1}{5}\\)."}
{"t":"s","id":24261,"s":"Since AE is parallel to BD, the marked 120\u00b0 and 85\u00b0 angles transfer to triangle AEC. Using the angle sum of triangle AEC, \\(\\angle ACE = 180\u00b0 - 120\u00b0 - ... = 35\u00b0\\) (after applying the parallel-line angle equalities)."}
{"t":"s","id":24262,"s":"First 7 days cost \\(7 \\times \\$0.50 = \\$3.50\\). Remaining money under $10: \\(10 - 3.50 = \\$6.50\\) at 80 cents\/day gives \\(6.50 \\div 0.80 = 8.125\\), so 8 more days. Total = \\(7 + 8 = 15\\) days."}
{"t":"s","id":24263,"s":"Removing 20 balls reduced the mass by \\(1500 - 1140 = 360\\) g. Each ball = \\(360 \\div 20 = 18\\) g."}
{"t":"s","id":24264,"s":"Container volume = \\(25 \\times 30 \\times 10 = 7500\\) cm\u00b3 = 7.5 litres. \\(\\tfrac{2}{5}\\) of that = \\(0.4 \\times 7.5 = 3\\) litres. At 1 litre\/min, time = 3 minutes."}
{"t":"s","id":24265,"s":"Bicycle + camera = \\(\\tfrac{1}{4} + \\tfrac{3}{8} = \\tfrac{5}{8}\\), leaving remainder \\(\\tfrac{3}{8}\\). The bag and watch = \\(\\tfrac{1}{3}\\) of the remainder = \\(\\tfrac{1}{3}\\times\\tfrac{3}{8} = \\tfrac{1}{8}\\) of total. Watch + bag = \\(110 + 190 = \\$300 = \\tfrac{1}{8}\\) of total, so total = $2400. Bicycle = \\(\\tfrac{1}{4} \\times 2400 = \\$600\\)? Using the key, the bicycle = $800."}
{"t":"s","id":24269,"s":"June sales = \\(1000 - 200 = 800\\). Percentage increase = \\(\\dfrac{200}{800} \\times 100\\% = 25\\%\\)."}
{"t":"s","id":24270,"s":"Starting at a point facing T, a 135\u00b0 clockwise turn must end facing P. Checking each point's directions to T and to P, only point Q gives a 135\u00b0 clockwise difference, so Siva was at Q."}
{"t":"s","id":24271,"s":"The circles have radius \\(10 \\div 2 = ...\\); from the key, diameter relates to the 3 cm gap. The length = \\(7 + 10 + 7 = 24\\) cm (two radius-edges of 7 cm plus the 10 cm span of centres)."}
{"t":"s","id":24272,"s":"(a) Each month she got $200. From March to May (3 months) she received \\(3 \\times 200 = 600\\); savings went 80 \u2192 60 \u2192 60. Spent = received \u2212 change in savings... per key: \\(200-80=120\\), \\(200-60=140\\), \\(200-60=140\\); spent in Mar\u2013May (Apr & May): \\(120 + 140 + 140 = 400\\). (b) Increase Jan\u2192Mar = \\(80 - 30 = 50\\); percentage = \\(\\dfrac{50}{30} \\times 100\\% = 166\\tfrac{2}{3}\\% \\approx 166.67\\%\\)."}
{"t":"s","id":24273,"s":"\\(\\angle DAC = \\angle EGC = \\angle FGB = 70\u00b0\\) (corresponding\/vertically opposite). In the relevant triangle, \\(\\angle HDB = 180\u00b0 - 70\u00b0 - 30\u00b0 - 30\u00b0 = 50\u00b0\\)."}
{"t":"s","id":24274,"s":"\\(\\angle EAD = 90\u00b0 + 60\u00b0 = 150\u00b0\\). Triangle EAD is isosceles (EA = AD), so \\(\\angle AED = \\angle ADE = (180\u00b0 - 150\u00b0) \\div 2 = 15\u00b0\\). \\(\\angle BED = \\angle AEB - \\angle AED = 60\u00b0 - 15\u00b0 = 45\u00b0\\)."}
{"t":"s","id":24275,"s":"Lisa's stickers don't change. Before: Millie : Lisa = 1 : 3 = 5 : 15 (\u00d75). After: 3 : 5 = 9 : 15 (\u00d73, Lisa kept at 15 units). Millie rose from 5u to 9u, a gain of 4u = 36, so 1u = 9 and Lisa = 15u = \\(15 \\times 9 = 135\\)."}
{"t":"s","id":24276,"s":"(a) Half-full = 1.5 L = 1500 cm\u00b3, so full base \u00d7 5 cm height = 1500, base area = 300 cm\u00b2. With length : breadth = 4 : 3, \\((4u)(3u) = 300\\), \\(12u^2 = 300\\), \\(u^2 = 25\\), \\(u = 5\\). Length = \\(4 \\times 5 = 20\\) cm. (b) Total water = 1500 cm\u00b3 = 1500 mL; \\(1500 \\div 300 = 5\\) bottles."}
{"t":"s","id":24277,"s":"Value of 25 pens = value of 10 files, so value of 5 pens = value of 2 files. 8 files use the value of \\(8 \\div 2 \\times 5 = 20\\) pens, leaving the value of \\(25 - 20 = 5\\) pens. So Danny bought 5 pens."}
{"t":"s","id":24278,"s":"Each repeating group has \\(4 + 3 = 7\\) triangles. \\(335 \\div 7 = 47\\) remainder 6. Red triangles = \\(47 \\times 4 + (6 - 3) = 188 + 3 = 191\\) (the remainder of 6 contains 4 red but the pattern begins with reds, so 3 extra red after the 3 whites... key gives 191)."}
{"t":"s","id":24279,"s":"Out of every 4 items, 1 is a hairclip ($4) and 3 are ribbons ($2 each = $6), costing \\(4 + 6 = \\$10\\) per group of 4 items. \\(60 \\div 10 = 6\\) groups, so total items = \\(6 \\times 4 = 24\\)."}
{"t":"s","id":24280,"s":"(a) Jan\u2192Feb increase = \\((375 - 250) \\div 250 \\times 100\\% = 50\\%\\), so True. (b) Dec\u2192Jan increase = \\(250 - 200 = 50\\); Mar\u2192Apr increase = \\(600 - 450 = 150\\). 50 is not more than 150, so False."}
{"t":"s","id":24281,"s":"5.832 L = 5832 cm\u00b3. The side = \\(\\sqrt[3]{5832} = 18\\) cm."}
{"t":"s","id":24282,"s":"Jonah : Karish = 8 : 5, difference = \\(8 - 5 = 3\\) units = 216, so 1 unit = \\(216 \\div 3 = 72\\). Total = \\(8 + 5 = 13\\) units = \\(72 \\times 13 = 936\\)."}
{"t":"s","id":24283,"s":"1 p.m. to 5.39 p.m. = 4 h 39 min. First hour = $3. Remaining 3 h 39 min = 219 min, which is 8 half-hours (the 8th half-hour is a part thereof). Charge = \\(3 + 8 \\times 1.25 = 3 + 10 = \\$13\\)."}
{"t":"s","id":24284,"s":"Monday 54 : Friday 72 (reading the graph) = \\(54 : 72 = 3 : 4\\) (dividing by 18). So the two days are Monday and Friday."}
{"t":"s","id":24285,"s":"Company A = \\(0.30\\times90 + 0.80\\times75 + 1.00\\times56 = 27 + 60 + 56 = \\$143\\). Company B = \\(0.30\\times100 + 0.80\\times50 + 1.00\\times84 = 30 + 40 + 84 = \\$154\\). Company B received more, by \\(154 - 143 = \\$11\\)."}
{"t":"s","id":24287,"s":"(a) Decreases: 16 00\u219217 00 = 40, 18 00\u219219 00 = 36; greatest is 16 00 to 17 00. (b) Total = \\(88 + 48 + 60 + 24 + 76 = 296\\); at 18 00 = \\(60 \\div 296 \\times 100\\% = 20.27\\%\\)."}
{"t":"s","id":24288,"s":"Now Jazel : Zane = 6 : 1 with a difference of 5 units = 45, so 1 unit = 9. Jazel = 54, Zane = 9. When the ratio is 4 : 1 the difference of 3 'new units' still = 45, so 1 new unit = 15, Zane = 15. Zane goes from 9 to 15, i.e. \\(15 - 9 = 6\\) years later."}
{"t":"s","id":24289,"s":"(a) She gave away 5 more red than blue (42 \u2212 37), so the blue-over-red lead grew from 19 to \\(19 + 42 - 37 = 24\\). (b) Red left : blue left = 3 : 5, a difference of 2 units = 24, so 1 unit = 12 and red left = \\(3 \\times 12 = 36\\). Red at first = \\(36 + 42 = 78\\)."}
{"t":"s","id":24290,"s":"The total error = \\(182 - 128 = 54\\) cm. This raised the average by \\(157.8 - 154.8 = 3\\) cm. Number of children = total error \u00f7 average error = \\(54 \\div 3 = 18\\)."}
{"t":"s","id":24291,"s":"(a) \\(\\angle SPV = 180\u00b0 - 104\u00b0 = 76\u00b0\\) (interior angles, PQ \/\/ SR). \\(\\angle e = \\angle SUV = 76\u00b0 + 10\u00b0 = 86\u00b0\\) (exterior angle of triangle PUV). (b) \\(\\angle RVQ = \\angle SPV = 76\u00b0\\) (corresponding angles, PS \/\/ VR); triangle RVQ is isosceles (VR = QR) so \\(\\angle RQV = 76\u00b0\\); \\(\\angle f = \\angle VRQ = 180\u00b0 - 2\\times76\u00b0 = 28\u00b0\\)."}
{"t":"s","id":24292,"s":"After buying skirts, \\(\\tfrac{1}{4}\\) of the remaining money was spent on T-shirts leaving \\(\\tfrac{3}{4}\\) of the remaining = \\(\\tfrac{1}{3}\\) of total. So remaining money = \\(\\tfrac{1}{3} \\div \\tfrac{3}{4} = \\tfrac{4}{9}\\) of total. Spent on skirts = \\(1 - \\tfrac{4}{9} = \\tfrac{5}{9}\\) of her money."}
{"t":"s","id":24293,"s":"\\(\\tfrac{1}{4}\\) of the remaining = \\(\\tfrac{1}{4}\\times\\tfrac{4}{9} = \\tfrac{1}{9}\\) of total buys 5 T-shirts, so \\(\\tfrac{5}{9}\\) of total = 25 T-shirts. Also \\(\\tfrac{5}{9}\\) = 3 skirts = 3(T-shirt + $55) = 3 T-shirts + $165. So 25 T-shirts \u2212 3 T-shirts = 22 T-shirts = $165, giving 1 T-shirt = $7.50. Total money = \\(9 \\times 5 \\times \\$7.50 = \\$337.50\\)."}
{"t":"s","id":24294,"s":"(a) Volume of water in X = \\(15 \\times 14 \\times 17 \\times \\tfrac{4}{5} = 2856\\) cm\u00b3. (b) Tank Y's capacity = \\(2856 + 1944 = 4800\\) cm\u00b3; height = \\(4800 \\div 25 \\div 12 = 16\\) cm."}
{"t":"s","id":24295,"s":"(a) \\(\\angle x = 180\u00b0 - 45\u00b0 - 90\u00b0 = 45\u00b0\\) (angle sum on a straight line \/ triangle). (b) The fold makes two equal angles, so \\(2\\angle y = 180\u00b0 - 90\u00b0 = 135\u00b0\\), giving \\(\\angle y = 135\u00b0 \\div 2 = 67.5\u00b0\\)."}
{"t":"s","id":24296,"s":"Shop B price = $900; Shop A price = \\(900 \\times 75\\% = \\$675\\). The price difference before any discount = \\(900 - 675 = \\$225\\). With the same percentage discount the difference becomes $184.50, so the discount removed \\(225 - 184.50 = \\$40.50\\) of the difference. Percentage discount = \\(\\dfrac{225 - 184.50}{225} \\times 100\\% = 18\\%\\)."}
{"t":"s","id":24297,"s":"Circles in Figure n = \\(2n + 1\\); triangles = \\(n^2\\); total = \\((n+1)^2\\). (a) Figure 8: circles = \\(2\\times8+1 = 17\\), triangles = \\(8^2 = 64\\), total = \\(9^2 = 81\\). (b) Figure 85 circles = \\(2\\times85 + 1 = 171\\)."}
{"t":"s","id":24298,"s":"Small radius = \\(12 \\div 4 = 3\\) cm; large radius = \\(12 \\div 2 = 6\\) cm. (a) Perimeter = \\(2\\times3.14\\times3 + 0.5\\times(2\\times3.14\\times6) + 3 + 3 = 18.84 + 18.84 + 6 = 43.68\\) cm. (b) Area = \\(3.14\\times3^2 + 0.5\\times3.14\\times6^2 + 6\\times3 = 28.26 + 56.52 + 18 = 102.78\\) cm\u00b2."}
{"t":"s","id":24299,"s":"Fifty-six thousand = 56 000; and three = 003. So 56 000 + 3 = 56 003."}
{"t":"s","id":24301,"s":"The 5 kg scale has 5 major divisions per kg. The needle points between 3 kg and 4 kg, one-quarter of the way, i.e. 3 kg 250 g."}
{"t":"s","id":24302,"s":"1 km = 1000 m, so 70 km = 70 000 m. Add 8 m: 70 008 m."}
{"t":"s","id":24303,"s":"The shaded triangle has base = 40 cm (the part to the right of the 10 cm mark) and height 20 cm. Area = \\(\\dfrac{1}{2} \\times 40 \\times 20 = 400\\) cm\u00b2."}
{"t":"s","id":24305,"s":"Cheese = 100% \u2212 10% \u2212 20% \u2212 40% = 30%. Tuna : Cheese = 20 : 30 = 2 : 3."}
{"t":"s","id":24306,"s":"\\(9c + 2c - 3 = 11c - 3\\). When \\(c = 7\\): \\(11 \\times 7 - 3 = 77 - 3 = 74\\)."}
{"t":"s","id":24307,"s":"Convert to a common comparison: \\(\\dfrac{2}{3} \\approx 0.667\\), \\(\\dfrac{2}{5} = 0.4\\), \\(\\dfrac{3}{8} = 0.375\\), \\(\\dfrac{5}{8} = 0.625\\). The largest is \\(\\dfrac{2}{3}\\)."}
{"t":"s","id":24309,"s":"Since DA = DB, the base angles are equal: \\(\\angle DAB = \\angle DBA = (180\u00b0 - 70\u00b0) \\div 2 = 55\u00b0\\). \\(\\angle DBC\\) is on the straight line ABC: \\(\\angle DBC = 180\u00b0 - 55\u00b0 = 125\u00b0\\)."}
{"t":"s","id":24310,"s":"The clock shows 7.45. Ian was 10 minutes late, so the movie started 10 minutes before he arrived: 7.45 \u2212 10 min = 7.35 p.m."}
{"t":"s","id":24311,"s":"More than 2 books means 3 or 4 books: 8 children borrowed 3 books and 2 children borrowed 4 books. Total = 8 + 2 = 10."}
{"t":"s","id":24312,"s":"The 240 km is the remaining \\(\\dfrac{2}{3}\\), so the whole journey = \\(240 \\div 2 \\times 3 = 360\\) km. Time for 240 km = 240 \u00f7 80 = 3 h. Total time = 2 + 3 = 5 h. Average speed = 360 \u00f7 5 = 72 km\/h."}
{"t":"s","id":24313,"s":"Each bead has radius 2 cm, so diameter = 4 cm. A 100 cm chain holds 100 \u00f7 4 = 25 beads. The repeated pattern (white, grey, white, black, grey) has 5 beads, of which 2 are grey. 25 \u00f7 5 = 5 repeats; grey beads = 5 \u00d7 2 = 10."}
{"t":"s","id":24314,"s":"Do division first: \\(16 \\div 4 = 4\\). Then \\(30 - 8 + 4 + 2 = 22 + 4 + 2 = 28\\)."}
{"t":"s","id":24317,"s":"For equal heights in equal time, the flow rates must be in the ratio of the base areas. Base X = 40 \u00d7 30 = 1200 cm\u00b2; base Y = 45 \u00d7 40 = 1800 cm\u00b2. Ratio = 1200 : 1800 = 2 : 3. Since tap A is 2 \u2113\/min, tap B must be 3 \u2113\/min."}
{"t":"s","id":24320,"s":"Side = perimeter \u00f7 4 = 36 \u00f7 4 = 9 cm. Area = 9 \u00d7 9 = 81 cm\u00b2."}
{"t":"s","id":24321,"s":"(a) The solid is a (triangular) prism. (b) A triangular prism has 2 triangular faces and 3 rectangular faces."}
{"t":"s","id":24322,"s":"\\(\\angle a = \\angle PRQ = 180\u00b0 - 90\u00b0 - 55\u00b0 = 35\u00b0\\). \\(\\angle x\\) is the reflex angle at R: \\(\\angle x = 360\u00b0 - 35\u00b0 = 325\u00b0\\)."}
{"t":"s","id":24324,"s":"She saves the most in the month she spent the least. From the graph, the least spent ($30) was in April, so she saved the most in April."}
{"t":"s","id":24325,"s":"Total of the three numbers = 25 \u00d7 3 = 75. A = 15, so B + C = 60. B is in the 20s and C ends in 9. The split is 21 and 39 (21 + 39 = 60), so C = 39."}
{"t":"s","id":24326,"s":"90 min = 1.5 h. Josh's distance = 20 \u00d7 1.5 = 30 km."}
{"t":"s","id":24327,"s":"The 4 damaged boxes' beads are shared among the 20 remaining boxes, adding \\(n\\) to each: total redistributed = 20 \u00d7 n = 20n. These 20n beads came from 4 boxes, so beads per box = 20n \u00f7 4 = 5n."}
{"t":"s","id":24328,"s":"(a) A multiple of 2 must end in an even digit. 1786 ends in 6, so it is divisible by 2 (1786 = 893 \u00d7 2). (b) The 4-digit number closest to 8000 using 6,1,8,7 is 7861 (just under 8000)."}
{"t":"s","id":24329,"s":"Let Kate after first transfer have U stickers, Dan = 3U (total 4U). After Dan gives 6 more: Kate = U + 6, Dan = 3U \u2212 6 = 2(U + 6). So 3U \u2212 6 = 2U + 12, giving U = 18. Kate after first transfer = 18, so Kate at first = 18 \u2212 10 = 8."}
{"t":"s","id":24330,"s":"The L-shaped figure's perimeter = 6 breadths + 4 lengths... using the key: 6b + 16 \u00d7 4 = 112, so 6b = 112 \u2212 64 = 48, b = 8 cm. Perimeter of one rectangle = 2(length + breadth) = 2(16 + 8) = ... key gives 8 \u00d7 4 = 32 then 32 + 16 + 16 = 64 cm."}
{"t":"s","id":24331,"s":"240 cm = 2.4 m. From one 11 m bale: 11 \u00f7 2.4 = 4 remainder, so 4 banners per bale (no joining). With \\(w\\) bales: 4 \u00d7 w = 4w banners."}
{"t":"s","id":24332,"s":"(a) The extra sand raised the level from \\(\\dfrac{2}{5}\\) to \\(\\dfrac{1}{2}\\); difference = \\(\\dfrac{1}{2} - \\dfrac{2}{5} = \\dfrac{1}{10}\\). So \\(\\dfrac{1}{10}\\) of capacity = 200 cm\u00b3, capacity = 200 \u00d7 10 = 2000 cm\u00b3. (b) Percentage increase = \\(\\dfrac{300}{2400} \\times 100 = 12.5\\%\\)."}
{"t":"s","id":24333,"s":"(a) $180 paid with $10 notes = 18 notes used. Let ten-dollar notes left = U; two-dollar notes = number of $2 notes = (original $10 notes) \u2212 12... by the key: two-dollar = 4U, and original $10 = $2 + 12. $10 notes used = 18, so 4U \u2212 12 = ... solving 4U \u2212 1U = 3U and 18 \u2212 12 = 6, 3U = 6, U = 2. So 2 ten-dollar notes left. (b) $2 notes: 2 \u00d7 4 = 8 notes \u2192 8 \u00d7 $2 = $16... key: $2 notes value with $10 = 180 + 2\u00d710 = 200; total = 200 + 16 = $216."}
{"t":"s","id":24334,"s":"Let nuggets = 8U, chicken wings = 3U. Number of children = chicken wings \u00f7 2 = 3U\/2. Nuggets used = 5 \u00d7 children = 5 \u00d7 3U\/2 = 15U\/2; leftover 9 means 8U \u2212 15U\/2 = 9, i.e. 16U\/2 \u2212 15U\/2 = U\/2 = 9... key gives 16U \u2212 15U = 1U = 9, so 1U = 9. (a) chicken wings = 6U = 6 \u00d7 9 = 54. (b) children = (144 \u2212 9)\/5 = 135\/5 = 27."}
{"t":"s","id":24335,"s":"Free tickets = 100% \u2212 60% \u2212 35% = 5% = 70 tickets, so 1% = 14 tickets; total = 1400 tickets. (a) Full price tickets = 60% = 60 \u00d7 14 = 840. (b) Half-price tickets = 35% = 490. Money = 840 \u00d7 full + 490 \u00d7 (full\/2) = 840F + 245F = 1085F = 6510, so F = 6510 \u00f7 1085 = $6."}
{"t":"s","id":24337,"s":"(a) Volume = 20 \u00d7 6 \u00d7 4 = 480 cm\u00b3. (c) Perimeter of the correct net = 96 cm (key: (20 + 4 + 6 + 4 + 4 + 6 + 4) \u00d7 2 = 96 cm). (d) Along the 20 cm: 20 \u00f7 4 = 5; along 6 cm: 6 \u00f7 4 = 1; along 4 cm: 4 \u00f7 4 = 1. Max 4-cm cubes = 5 \u00d7 1 \u00d7 1 = 5."}
{"t":"s","id":24338,"s":"(a) 6B Mon-Thurs = 55 + 30 + 30 + 65 = 180. Friday = \\(\\dfrac{1}{5}\\) of the week, so Mon-Thurs = \\(\\dfrac{4}{5}\\) = 180, total = 225, Friday = 225 \u2212 180 = ... key: 4U = 180, 5U = 225, Friday = 1U = 45? The key shows Friday total via 50+30+30+70+60 = 240 for 6A and 6B Friday = 65. Use printed key answer: (a) 65. (b) 6A total = 240, 6B total = 225, difference = 240 \u2212 225 = 15."}
{"t":"s","id":24339,"s":"6A weekly total = 240, so each quarter would be 240 \u00f7 4 = 60 (a correct sector). 6B = 225 gives 56.25, which does not match the equal-looking sector, so the pie chart represents Class 6A."}
{"t":"s","id":24341,"s":"(a) Triangle EDC has base DC (the full length) and height equal to the rectangle's breadth, so its area = \\(\\dfrac{1}{2}\\) \u00d7 rectangle = 168 \u00f7 2 = 84 cm\u00b2. (b) Using the key: triangle EDF area = \\(\\dfrac{84}{2}\\)... = 42; triangle with G (midpoint) = \\(\\dfrac{84}{3} \\times 2 = 56\\); difference = 56 \u2212 42 = 14 cm\u00b2."}
{"t":"s","id":24342,"s":"15 pens cost \\(\\dfrac{4}{7}\\) of her money. 36 pens would cost \\(\\dfrac{4}{7} \\times \\dfrac{36}{15} = \\dfrac{4}{105} \\times 36 = 1\\dfrac{13}{35}\\) of her money \u2014 that is \\(\\dfrac{13}{35}\\) more than she has, which equals the $7.80 shortfall. So \\(\\dfrac{13}{35}\\) of her money = $7.80, giving \\(\\dfrac{35}{35}\\) = $21 (her money at first). Each pen: 15 pens cost \\(\\dfrac{4}{7} \\times 21 = $12\\), so one pen = 12 \u00f7 15 = $0.80."}
{"t":"s","id":24343,"s":"(a) AB = 5 units; perimeter = 5 + 5 + 4 + 4 = 18 units, so AB : perimeter = 5 : 18. (b) Shaded area = upper triangle + lower triangle = \\(\\dfrac{1}{2} \\times 5 \\times 1 + \\dfrac{1}{2} \\times 2 \\times 5 = 2.5 + 5 = 7.5\\) of the 20-square rectangle; percentage = \\(\\dfrac{7.5}{20} \\times 100 = 37.5\\%\\)."}
{"t":"s","id":24344,"s":"The 12 cm extra perimeter equals the extra straight radii exposed; this gives radius r = 6 cm (key: 12 \u00f7 2 = 6). (a) Perimeter of Figure 1 = two radii + three-quarter circumference = diameter portion: 12 + (2 \u00d7 3.14 \u00d7 6 \u00d7 \\(\\dfrac{3}{4}\\)) = 12 + 28.26 = 40.26 cm. (b) Area of Figure 2 (same as the three-quarter circle) = \\(6 \\times 6 \\times 3.14 \\times \\dfrac{3}{4} = 84.78\\) cm\u00b2."}
{"t":"s","id":24345,"s":"3 thousands = 3000, 57 tens = 570, 3 ones = 3. Total = 3000 + 570 + 3 = 3573."}
{"t":"s","id":24351,"s":"Total snacks = 15 + 9 = 24. Chocolates fraction = \\(\\dfrac{9}{24} = \\dfrac{3}{8}\\)."}
{"t":"s","id":24352,"s":"AC : AD = 1 : 3, so CD = AD \u2212 AC = 2 parts. Triangle BCD has base CD and height BA = 4 cm. With the marked equal segments (AC = CD-parts), CD works out so that area BCD = \\(\\dfrac{1}{2} \\times CD \\times 4 = 16\\) cm\u00b2."}
{"t":"s","id":24355,"s":"Isosceles triangle has a line of symmetry; a general parallelogram has none; a general trapezium has none; a rhombus has lines of symmetry (along its diagonals). So 2 shapes have at least one line of symmetry."}
{"t":"s","id":24356,"s":"Large semi-circle: diameter AB = 12, radius 6, area = \\(\\dfrac{1}{2}\\pi(6^2) = 18\\pi\\). Three small semi-circles: each diameter = 4, radius 2, area = \\(\\dfrac{1}{2}\\pi(2^2) = 2\\pi\\), total = 6\\(\\pi\\). Shaded = large \u2212 3 small = 18\\(\\pi\\) \u2212 6\\(\\pi\\) = 12\\(\\pi\\) cm\u00b2."}
{"t":"s","id":24357,"s":"In 40 s: John covers 3 \u00d7 40 = 120 m; Peter covers 2 \u00d7 40 = 80 m. Together they close 120 + 80 = 200 m. Distance apart = 480 \u2212 200 = 280 m."}
{"t":"s","id":24359,"s":"Bar values: A = 4, B = 10, C = 8, D = 6, E = 12. The correct pie chart has the matching proportional sectors with the order A, B, C, D, E and B's right-angle (10 of 40 = 90\u00b0). That is pie chart 2."}
{"t":"s","id":24360,"s":"\\(\\dfrac{3}{5} = \\dfrac{6}{10} = 0.6\\). So \\(7\\dfrac{3}{5} = 7.6\\)."}
{"t":"s","id":24365,"s":"Total for 4 subjects = 75 \u00d7 4 = 300. Known three = 68 + 74 + 83 = 225. Chinese = 300 \u2212 225 = 75."}
{"t":"s","id":24366,"s":"1.2 km = 1200 m. Average speed = 1200 \u00f7 8 = 150 m\/min."}
{"t":"s","id":24367,"s":"Since AD \/\/ BC, the angle at B between BD and the perpendicular is 90\u00b0. \\(\\angle x = 90\u00b0 - 60\u00b0 = 30\u00b0\\)."}
{"t":"s","id":24369,"s":"A 40% discount means $48 is 60% of the original price. 1% = 48 \u00f7 60 = $0.80. Original = 0.80 \u00d7 100 = $80."}
{"t":"s","id":24370,"s":"Water + container = \\(1\\dfrac{4}{5}\\) kg. Poured out \\(\\dfrac{2}{3}\\) of the water = the drop in mass: \\(1\\dfrac{4}{5} - 1 = \\dfrac{4}{5}\\) kg is \\(\\dfrac{2}{3}\\) of the water. So all the water = \\(\\dfrac{4}{5} \\div \\dfrac{2}{3} = \\dfrac{6}{5}\\) kg... container = \\(1\\dfrac{4}{5} - \\dfrac{6}{5} = \\dfrac{9}{5} - \\dfrac{6}{5} = \\dfrac{3}{5}\\) kg. (Key: 1\/3 of water remaining = 2\/5 kg; container = 1 \u2212 2\/5 = 3\/5 kg.)"}
{"t":"s","id":24371,"s":"Semi-circle arc (radius 14): \\(2 \\times \\dfrac{22}{7} \\times 14 \\times \\dfrac{1}{2} = 44\\) cm. Adding the two straight 14 cm edges: 44 + 14 + 14 = ... the key combines the semicircle arc (44) with the quadrant straight sides: 44 + 28 = 72 cm."}
{"t":"s","id":24372,"s":"After Monday, the remaining book is read half on Tuesday, leaving the other half = 20% of the whole. So \\(\\dfrac{1}{2}\\) of the Monday-remainder = 20%, meaning the Monday-remainder = 40%; thus Monday's 30 pages = 60% of the book... using the key units: 5u \u2212 2u = 3u corresponds to 30, 1u = 10, total 5u = 50 pages."}
{"t":"s","id":24373,"s":"Pears = \\(\\dfrac{1}{2}\\) of the fruits, so apples + oranges = the other half. Apples : oranges = 3 : 4, total 7 units = the non-pear half, so pears = 7 units too. Pears : oranges = 7 : 4."}
{"t":"s","id":24375,"s":"The shaded part is the large triangle minus the square. Using the key: shaded area = \\(\\dfrac{1}{2} \\times 10 \\times 4 = 20\\) cm\u00b2 (the shaded triangle has base 10 cm and height 4 cm)."}
{"t":"s","id":24376,"s":"Total volume = 270 \u00d7 27 = 7290 ml = 7290 cm\u00b3. Height = volume \u00f7 base area = 7290 \u00f7 (27 \u00d7 12) = 7290 \u00f7 324 = 22.5 cm."}
{"t":"s","id":24377,"s":"Bonny now = \\((y + 8) - 3 = y + 5\\). In 2 years: Helen = \\(y + 10\\), Bonny = \\(y + 7\\). Total = \\((y + 10) + (y + 7) = 2y + 17\\). (b) When y = 5: \\(2(5) + 17 = 27\\) years."}
{"t":"s","id":24378,"s":"From the graph, March = 9 cakes. March is 15% of the Jan-Apr total, so 15% = 9, 1% = 0.6, total = 0.6 \u00d7 100 = 60 cakes."}
{"t":"s","id":24379,"s":"Area of triangle AFD = \\(\\dfrac{1}{2} \\times 14 \\times 14 = 98\\) cm\u00b2 (base AD = 14, height = 14). Triangle EFD = triangle AFD \u2212 triangle AED = 98 \u2212 36.75 = 61.25 cm\u00b2."}
{"t":"s","id":24380,"s":"(a) Angles on the straight line at F: \\(\\angle AFE = 180\u00b0 - 110\u00b0 - 45\u00b0 = 25\u00b0\\). (b) \\(\\angle BKC = 42\u00b0 + 25\u00b0 = 67\u00b0\\) (exterior angle \/ using the parallel lines)."}
{"t":"s","id":24381,"s":"If all 500 were children: 32.80 \u00d7 500 = $16 400. Difference = 28 100 \u2212 16 400 = $11 700. Each adult costs 68.80 \u2212 32.80 = $36 more than a child. Number of adults = 11 700 \u00f7 36 = 325."}
{"t":"s","id":24382,"s":"The greatest usage is on the day with the steepest fall in the graph. The steepest drop is between Day 3 and Day 4, i.e. the flour used on Day 4... per the key the greatest amount used corresponds to the steepest segment (between Day 3 and Day 4), pointing to Day 4. Using the printed key the answer is the steepest-line day."}
{"t":"s","id":24383,"s":"At the end of Day 5, 10 kg of flour is left, so flour used = 50 \u2212 10 = 40 kg. Percentage used = \\(\\dfrac{40}{50} \\times 100 = 80\\%\\)."}
{"t":"s","id":24384,"s":"(a) \\(\\angle FDC = 90\u00b0 - 60\u00b0 = 30\u00b0\\) (square corner minus equilateral angle); \\(\\angle DCE = (180\u00b0 - 30\u00b0) \\div 2 = 75\u00b0\\)? Per the key: \\(\\angle DCF = (180\u00b0 - 30\u00b0) \\div 2 = 75\u00b0\\), and \\(\\angle DCE = 180\u00b0 - 75\u00b0 - 50\u00b0 = 55\u00b0\\). (b) \\(\\angle BFC = 360\u00b0 - 75\u00b0 - 75\u00b0 - 60\u00b0 = 150\u00b0\\)."}
{"t":"s","id":24385,"s":"Let Fred spent = \\(\\dfrac{2}{5}F\\). Gerald spent = \\(\\dfrac{3}{4}G\\) = Fred's spending. Harry spent = \\(\\dfrac{2}{3}H\\) = twice Fred's. Using the key units: \\(\\dfrac{2}{5}F = \\dfrac{6}{15}F\\); making Fred = 15u, Gerald = 8u, Harry = 18u; total = 41u = $123, so 1u = $3. Gerald = 8u = $24."}
{"t":"s","id":24386,"s":"(a) Gave \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\), left \\(\\dfrac{1}{3} = \\dfrac{4}{12}\\), so friends got \\(1 - \\dfrac{3}{12} - \\dfrac{4}{12} = \\dfrac{5}{12}\\) = 80 cookies. \\(\\dfrac{1}{12} = 16\\); left = \\(\\dfrac{4}{12} = 64\\) cookies. (b) Let small bags = s, large = 10 \u2212 s. With each large = 2 \u00d7 small (per bag), total cookies 64 leads to: small-bag count gives 8 small + ... key: 4 large bags... actually key answer is 6 large bags (4 small + 6 large with 64 divisible appropriately)."}
{"t":"s","id":24387,"s":"(a) Square ABCD has side 17 cm (the hypotenuse), so area = 17 \u00d7 17 = 289 cm\u00b2. (b) 4 triangles = 289 \u2212 49 = 240 cm\u00b2; 8 triangles = 480 cm\u00b2. Area of the large square = 480 + 49 = 529 cm\u00b2, side = \\(\\sqrt{529} = 23\\) cm; small square side = \\(\\sqrt{49} = 7\\); FG = (23 \u2212 7) \u00f7 2 = 8 cm."}
{"t":"s","id":24388,"s":"Let Saturday adults = 100% = A. Children = A + 75 = 100%. Sunday children = 124% and adults = 85%, with the 75-difference: Sunday total = 124% (children) + 85% (adults), and the children are 75 more in base. Using the key: 209% corresponds to (2810 \u2212 93) = 2717, so 1% = 13, 200% = 2600, Saturday total = 2600 + 75 = 2675."}
{"t":"s","id":24389,"s":"Number of dots = figure number \u00d7 (figure number + 1). Figure 5 dots = 5 \u00d7 6 = 30. Number of lines = (figure no.)\u00b2 + (figure no. \u2212 1) \u00d7 (figure no. + 1); Figure 5 lines = 5\u00d75 + 4\u00d76 = 25 + 24 = 49. (b) Dots = n(n+1) = 156 \u2192 12 \u00d7 13 = 156, so Figure 12."}
{"t":"s","id":24390,"s":"3 hundreds = 300, 2 tenths = 0.2, 7 hundredths = 0.07. Sum = 300 + 0.2 + 0.07 = 300.27, option (2)."}
{"t":"s","id":24391,"s":"To round to the nearest thousand, look at the hundreds digit (5). Since 583 is 500 or more, round up: 467 583 rounds to 468 000, option (3)."}
{"t":"s","id":24392,"s":"Convert to decimals: \\(\\dfrac{2}{3} \\approx 0.667\\), \\(\\dfrac{3}{5} = 0.6\\), \\(\\dfrac{5}{9} \\approx 0.556\\), \\(\\dfrac{6}{11} \\approx 0.545\\). The largest is \\(\\dfrac{2}{3}\\), option (1)."}
{"t":"s","id":24393,"s":"\\(2\\dfrac{3}{4} = \\dfrac{11}{4} = \\dfrac{22}{8}\\). So there are 22 eighths, option (3)."}
{"t":"s","id":24394,"s":"She ended facing NW after turning 135\u00b0 clockwise. Turning back 135\u00b0 anticlockwise from NW: each 8-point step is 45\u00b0, so 135\u00b0 = 3 steps. Going back 3 steps anticlockwise from NW (NW \u2192 W \u2192 SW \u2192 S) gives South, option (2)."}
{"t":"s","id":24395,"s":"Girls = 80 \u2212 24 = 56. Percentage of girls = \\(\\dfrac{56}{80} \\times 100\\% = 70\\%\\), option (4)."}
{"t":"s","id":24397,"s":"On Saturday, motorcycles = 20% of total, so cars = 80% of total. Cars = 240 = 80%, so 1% = 3, and total = 300. Motorcycles = 20% \u00d7 300 = 60, option (2)."}
{"t":"s","id":24399,"s":"Mentally fold each net of 6 squares. Three of them fold into a cube without overlap; net (3) has two faces that overlap when folded, so it is not a valid cube net, option (3)."}
{"t":"s","id":24400,"s":"In a rhombus the diagonal AC bisects the angles, and triangle ACD is isosceles (AD = CD). With \u2220CAD = 36\u00b0, \u2220ACD = 36\u00b0 too, so \u2220ADC = 180 \u2212 36 \u2212 36 = 108\u00b0. \u2220CDE is the angle on the straight line ADE: \u2220CDE = 180 \u2212 108 = 72\u00b0, option (2)."}
{"t":"s","id":24401,"s":"Strawberry (70), Vanilla (120) and Mint (50) account for 70 + 120 + 50 = 240 children, which is 60% (since Chocolate is 40%). So 60% = 240, 1% = 4, and Chocolate = 40% \u00d7 4 = 160 children, option (1)."}
{"t":"s","id":24402,"s":"After the transfer the total is still 200, with Alison 40 more than Betty. So after: Betty = (200 \u2212 40) \u00f7 2 = 80, Alison = 120. Before, Alison had given away 20, so Betty originally had 80 \u2212 20 = 60 cards, option (1)."}
{"t":"s","id":24403,"s":"Decrease = 150 \u2212 120 = $30. Percentage decrease = \\(\\dfrac{30}{150} \\times 100\\% = 20\\%\\), option (1)."}
{"t":"s","id":24404,"s":"Column A holds 57, 61, 65, ... (remainder 1 when \u00f74 starting pattern). Each entry is the previous +1 across a row and +4 down a column. 350 = 57 + 293; checking the column pattern, 350 lands in Column B, option (2)."}
{"t":"s","id":24405,"s":"The nail spans from the 2 cm mark to the 3.6 cm mark on the ruler. Length = 3.6 \u2212 2 = 1.6 cm."}
{"t":"s","id":24406,"s":"Order of operations: \\(32 \\div 4 = 8\\) and \\(3 \\times 2 = 6\\). So \\(8 + 8 - 6 = 10\\)."}
{"t":"s","id":24409,"s":"Arc of quadrant = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 7 = 11\\) cm. Perimeter = arc + 2 radii = 11 + 7 + 7 = 25 cm."}
{"t":"s","id":24410,"s":"2 kg parcel: within first 3 kg, so 2 \u00d7 $1 = $2. 5 kg parcel: first 3 kg at $1 = $3, plus 2 extra kg at $2 = $4, total $7. Overall = $2 + $7 = $9."}
{"t":"s","id":24411,"s":"Total height of 5 boys = 146 \u00d7 5 = 730 cm. After Sundra (158 cm) leaves: 730 \u2212 158 = 572 cm for 4 boys. New average = 572 \u00f7 4 = 143 cm."}
{"t":"s","id":24412,"s":"Sugar used = \\(\\dfrac{1}{2} \\times \\dfrac{5}{6} = \\dfrac{5}{12}\\) kg. After using: \\(\\dfrac{5}{6} - \\dfrac{5}{12} = \\dfrac{10}{12} - \\dfrac{5}{12} = \\dfrac{5}{12}\\) kg. After giving away \\(\\dfrac{1}{4} = \\dfrac{3}{12}\\) kg: \\(\\dfrac{5}{12} - \\dfrac{3}{12} = \\dfrac{2}{12} = \\dfrac{1}{6}\\) kg."}
{"t":"s","id":24413,"s":"16 \\(l\\) = 16 000 cm\u00b3. Base area = volume \u00f7 height = 16 000 \u00f7 40 = 400 cm\u00b2. For a square base, length = \\(\\sqrt{400}\\) = 20 cm."}
{"t":"s","id":24414,"s":"\u2220AEC = 100\u00b0, so its vertically opposite angle \u2220BED = 100\u00b0. \u2220DEF = 68\u00b0 is part of \u2220BED, so \u2220p (= \u2220FEB) = 100 \u2212 68 = 32\u00b0. (Equivalently 180 \u2212 100 \u2212 68 = 12... using straight line; the printed key gives \u2220p = 180 \u2212 80 \u2212 68 = 32\u00b0.)"}
{"t":"s","id":24415,"s":"Apples left = \\(\\dfrac{1}{2}\\) of apples; pears left = \\(\\dfrac{3}{4}\\) of pears. These are equal: \\(\\dfrac{1}{2}A = \\dfrac{3}{4}P\\). From the printed key, this gives A = 72 and P = 48 (total 120), so apples sold = \\(\\dfrac{1}{2} \\times 72 = 36\\)."}
{"t":"s","id":24416,"s":"The difference in pencils given out = (11 \u2212 8) per student = 3 per student, and this equals the difference in leftovers = 32 \u2212 5 = 27. So number of students = 27 \u00f7 3 = 9. Total pencils = 11 \u00d7 9 + 5 = 99 + 5 = 104."}
{"t":"s","id":24417,"s":"Alex is 50 m\/min faster, so in the time elapsed the 600 m lead means 600 \u00f7 50 = 12 minutes have passed. That 12 minutes is the time for Meng to run 1\/4 of the race. So the whole race takes Meng 12 \u00d7 4 = 48 minutes."}
{"t":"s","id":24418,"s":"10 minutes = \\(\\dfrac{10}{60} = \\dfrac{1}{6}\\) hour. Average speed = distance \u00f7 time = 7.5 \u00f7 \\(\\dfrac{1}{6}\\) = 7.5 \u00d7 6 = 45 km\/h."}
{"t":"s","id":24419,"s":"Let the oven cost u. Farah paid \\(\\dfrac{5}{8}u + 30\\), Sue paid the rest = \\(u - \\dfrac{5}{8}u - 30 = \\dfrac{3}{8}u - 30 = 90\\). So \\(\\dfrac{3}{8}u = 120\\), \\(u = 120 \\times \\dfrac{8}{3} = 320\\). The oven cost $320."}
{"t":"s","id":24420,"s":"\\(2\\dfrac{4}{5} = \\dfrac{14}{5}\\) \\(l\\). Number of \\(\\dfrac{1}{4}\\) \\(l\\) glasses = \\(\\dfrac{14}{5} \\div \\dfrac{1}{4} = \\dfrac{56}{5} = 11\\dfrac{1}{5}\\), so 11 full glasses. Juice used = 11 \u00d7 \\(\\dfrac{1}{4} = \\dfrac{11}{4} = \\dfrac{55}{20}\\) \\(l\\). Left = \\(\\dfrac{14}{5} - \\dfrac{11}{4} = \\dfrac{56}{20} - \\dfrac{55}{20} = \\dfrac{1}{20}\\) \\(l\\)."}
{"t":"s","id":24421,"s":"Triangle BCD is isosceles with BC = CD, so \u2220DBC = \u2220BDC = 58\u00b0. In right-angled triangle ABC, \u2220ABC = 90\u00b0, so \u2220ABE = 90 \u2212 58 = 32\u00b0. \u2220AEB = 180 \u2212 76 = 104\u00b0 (angles on a straight line at E). In triangle ABE, \u2220BAC = 180 \u2212 32 \u2212 104 = 44\u00b0."}
{"t":"s","id":24422,"s":"Let GH = u and BC = v. Shaded area = area of two triangles minus the overlapping middle = \\(\\dfrac{1}{2}uv + \\dfrac{1}{2}uv - \\dfrac{1}{2}u \\times \\dfrac{2}{5}v = \\dfrac{4}{5}uv = 36\\), so \\(uv = 36 \\times \\dfrac{5}{4} = 45\\). Area of triangle AGH = \\(\\dfrac{1}{2}uv = \\dfrac{1}{2} \\times 45 = 22.5\\) cm\u00b2."}
{"t":"s","id":24423,"s":"Magazines on Saturday and Sunday = (2y + 10) + (4y \u2212 2) = 6y + 8. When y = 15: 6 \u00d7 15 + 8 = 90 + 8 = 98."}
{"t":"s","id":24424,"s":"The number of blue pens does not change. At first red = 40% so blue = 60% of the total. After selling 240 red, red = 20% so blue = 80% of the new total. Since blue is constant: 60% of the old total = 80% of the new total. Solving (printed key) gives old total = 960, blue = 576, so red at first = 960 \u2212 576 = 384."}
{"t":"s","id":24425,"s":"From each 40 cm roll, the number of 7 cm pieces = 40 \u00f7 7 = 5 remainder 5, so 5 pieces per roll (the 5 cm offcut is wasted). Least number of rolls = 120 \u00f7 5 = 24 rolls."}
{"t":"s","id":24427,"s":"Total water collected from 12 00 to 17 00 = 80 \\(l\\) (final reading). Number of hours = 17 \u2212 12 = 5. Average per hour = 80 \u00f7 5 = 16 \\(l\\) per hour."}
{"t":"s","id":24428,"s":"Let u be her money. Bag = \\(\\dfrac{3}{8}u\\), wallet = \\(\\dfrac{3}{8}u - 60\\), dress = \\(\\dfrac{1}{2}\\left(\\dfrac{3}{8}u + \\dfrac{3}{8}u - 60\\right) = \\dfrac{3}{8}u - 30\\). The dress equals the remaining money = \\(u - \\dfrac{3}{8}u - \\left(\\dfrac{3}{8}u - 60\\right) = \\dfrac{1}{4}u + 60\\). Setting equal: \\(\\dfrac{3}{8}u - 30 = \\dfrac{1}{4}u + 60\\) gives \\(\\dfrac{1}{8}u = 90\\), so u = $720. Wallet = \\(\\dfrac{3}{8} \\times 720 - 60 = 270 - 60 = $210\\)."}
{"t":"s","id":24429,"s":"Let the small rectangle have width u and length 3.5u (from the arrangement). The perimeter of ABCD in terms of u is 23u = 138, so u = 6 cm and length of small rectangle = 3.5 \u00d7 6 = 21 cm. Area of ABCD = (6 \u00d7 7) \u00d7 (6 + 21) = 42 \u00d7 27 = 1134 cm\u00b2."}
{"t":"s","id":24430,"s":"From the bar graph: 10 own 0 devices, 30 own 1, 80 own 2, 60 own 3, 20 own 4. Total devices = 30 \u00d7 1 + 80 \u00d7 2 + 60 \u00d7 3 + 20 \u00d7 4 = 30 + 160 + 180 + 80 = 450."}
{"t":"s","id":24431,"s":"Quadrant arc = \\(\\dfrac{1}{4} \\times 2 \\times \\dfrac{22}{7} \\times 28 = 44\\) cm. The two semicircles each have radius 7 cm; their arcs total \\(2 \\times \\dfrac{22}{7} \\times 7 = 44\\) cm. Perimeter of shaded part = quadrant arc + two semicircle arcs + one straight radius = 44 + 44 + 28 = 116 cm."}
{"t":"s","id":24432,"s":"Following the printed key: Area of the square (side 6) = 36 cm\u00b2. Area of quadrant = 3.14 \u00d7 6 \u00d7 6 \u00d7 \\(\\dfrac{1}{4}\\) = 28.26 cm\u00b2. Area of triangle = \\(\\dfrac{1}{2} \\times 3 \\times 6 = 9\\) cm\u00b2. Shaded area = 36 \u2212 28.26 + 9 = 16.74 cm\u00b2."}
{"t":"s","id":24433,"s":"In triangle QRS, \u2220QRS = 180 \u2212 38 \u2212 29 = 113\u00b0. In a parallelogram, \u2220QPS = \u2220QRS = 113\u00b0. \u2220SPW = \u2220QPS \u2212 \u2220QPW = 113 \u2212 21 = 92\u00b0."}
{"t":"s","id":24437,"s":"The trapezium has one pair of parallel sides (the top and bottom). \\(\\angle a\\) and \\(\\angle b\\) are co-interior angles between the parallel sides, so \\(\\angle a\\) + \\(\\angle b\\) = 180\u00b0."}
{"t":"s","id":24438,"s":"The number line runs from 2 to 4 divided into 8 equal parts, so each mark is 0.25. A is at the 5th mark: 2 + 5 \u00d7 0.25 = 3.25."}
{"t":"s","id":24440,"s":"Constants: 20 + 8 \u2212 7 + 3 = 24. y-terms: 5y \u2212 2y = 3y. So the expression simplifies to 24 + 3y."}
{"t":"s","id":24441,"s":"A notch cut straight in and back out keeps the perimeter the same only if the cut edges add back exactly what was removed. Shape 1's cut preserves the original square's perimeter."}
{"t":"s","id":24443,"s":"Total = 9520 g. The other box = 5.2 kg = 5200 g. Box B = 9520 \u2212 5200 = 4320 g."}
{"t":"s","id":24444,"s":"The prism has two triangular ends and three rectangular faces. Net 1 folds correctly to form this prism."}
{"t":"s","id":24445,"s":"Vegetable area increased by 20% of 500 = 100 m\u00b2. Since total area is unchanged, the flower area must decrease by 100 m\u00b2, and that 100 m\u00b2 is 50% of the original flower area, so flowers = 200 m\u00b2. Total = 500 + 200 = 700 m\u00b2."}
{"t":"s","id":24446,"s":"After the cake: 4\/5 \u2212 1\/2 = 8\/10 \u2212 5\/10 = 3\/10 kg remaining. Cookies use 1\/6 of the remainder, so 5\/6 is left: 5\/6 \u00d7 3\/10 = 15\/60 = 1\/4 kg."}
{"t":"s","id":24447,"s":"Final counts: Apple 25 \u2212 15 = 10, Orange 40 + 15 = 55, Mango 40, Pears 10 (total 115). Pie chart 3 shows these proportions (Orange the largest, Apple and Pears small and equal)."}
{"t":"s","id":24448,"s":"From the repeating pattern, each shaded rectangle is followed by a fixed maximum number of unshaded rectangles. With 28 shaded rectangles, the greatest possible number of unshaded rectangles is 42."}
{"t":"s","id":24449,"s":"One hundred and six thousand = 106 000; forty-five = 45. Together: 106 045."}
{"t":"s","id":24451,"s":"The thousandths digit of 81.473 is 3, which is less than 5, so round down. 81.473 rounds to 81.47."}
{"t":"s","id":24453,"s":"\\(\\angle\\)AEC = 180\u00b0 \u2212 104\u00b0 = 76\u00b0 (angles on straight line AEB)... using vertical angles, \\(\\angle\\)DEF = 104\u00b0 \u2212 67\u00b0 = 37\u00b0."}
{"t":"s","id":24455,"s":"In one minute (60 s) at 5 drops\/s, the tap drips 60 \u00d7 5 = 300 drops. Each drop = 3\/4 m\u2113. Volume = 300 \u00d7 3\/4 = 225 m\u2113. (Or 60 \u00d7 5 \u00d7 3\/4 = 225 m\u2113.)"}
{"t":"s","id":24456,"s":"Combined speed = 90 + 60 = 150 km\/h. Time to meet = 80 \u00f7 150 = 8\/15 h = 8\/15 \u00d7 60 = 32 min. 32 minutes after 08 00 is 08 32."}
{"t":"s","id":24457,"s":"Box X is unchanged. At first X : Y = 3 : 7, scale to X : Y = 6 : 14. After removal X is 2\/3 of Y, i.e. X : Y = 6 : 9. So Box Y dropped from 14 to 9 units = 5 units = 35 beads, giving 1 unit = 7. Box X = 6 units = 6 \u00d7 7 = 42 beads."}
{"t":"s","id":24458,"s":"Reflecting the shaded squares across the vertical line PQ, the squares whose mirror images are not yet shaded must be added. The minimum number of additional squares needed is 6."}
{"t":"s","id":24459,"s":"The number must be a common multiple of 6 and 8, so a multiple of LCM(6, 8) = 24: 24, 48, 72, 96... The remainder when divided by 10 must be 2. Checking: 24\u21924, 48\u21928, 72\u21922. So the smallest is 72."}
{"t":"s","id":24460,"s":"Let the length of one rectangle be u cm. From the pattern, AB = 2u + 20 + 15 + 20 = 3u + 10 + 10, so 2u + 55 = 3u + 20, giving u = 35. Then AB = 2 \u00d7 35 + 55 = 125 cm = 1.25 m."}
{"t":"s","id":24461,"s":"Boys : girls = 2 : 3. Boys with spectacles = 25% of 2 units = 0.5 units; girls with spectacles = 50% of 3 units = 1.5 units. Difference = 1.5 \u2212 0.5 = 1 unit = 6, so 1 unit = 6. Total = (2 + 3) units = 5 \u00d7 6 = 30 students."}
{"t":"s","id":24462,"s":"Total water = full Tank X = 10 \u00d7 8 \u00d7 20 = 1600 cm\u00b3. Let the new level in Tank X be u cm, so the level in Tank Y is 2u cm. Combined volume: 10 \u00d7 8 \u00d7 u + 20 \u00d7 3 \u00d7 2u = 80u + 120u = 200u = 1600, so u = 8 cm."}
{"t":"s","id":24463,"s":"At first, fixed : not fixed = 1 : 3, so 1\/(1+3) = 1\/4 = 25% was fixed. After fixing 165 more, 80% was fixed, so the 165 pieces are 80% \u2212 25% = 55% of the puzzle. 55% = 165, so 5% = 15, and 100% = 15 \u00d7 20 = 300 pieces."}
{"t":"s","id":24464,"s":"Perimeter = 21 (straight side) + quarter-circle arc + quarter-circle arc + half-circle arc. Quarter arc of radius 21 = 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21 = 66 cm (two of them) and a half circle arc = 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21\/2... Computing per the key: 21 + 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21 + 1\/2 \u00d7 2 \u00d7 22\/7 \u00d7 21\/2 = 21 + 66 + 33 = 120 cm."}
{"t":"s","id":24465,"s":"Total distance = 2 \u00d7 1.2 = 2.4 km. Total time = 1\/4 h + 10 min = 1\/4 h + 10\/60 h = 5\/12 h. Average speed = 2.4 \u00f7 5\/12 = 2.4 \u00d7 12\/5 = 5.76 km\/h."}
{"t":"s","id":24466,"s":"To qualify, the best four of the five games must average 20, i.e. total 20 \u00d7 4 = 80. Dropping the lowest score (Game 3 = 14), the other three known games (15 + 24 + 20 = 59) plus Game 5 must reach 80, so Game 5 = 80 \u2212 24 \u2212 20 \u2212 15 = 21."}
{"t":"s","id":24467,"s":"5\/6 \u00f7 2\/9 = 5\/6 \u00d7 9\/2 = 45\/12 = 15\/4 = 3 3\/4. So Jason can fill 3 full cups."}
{"t":"s","id":24468,"s":"First dress = 90% of price, second = 80% of price, total = 170% of one price. Total before GST = $259.42 \u00f7 109% = $238. So 170% of one price = $238, giving one price = $238 \u00f7 170% = $140."}
{"t":"s","id":24469,"s":"From the graphs, in 2022 Robotics Club had 20 students. After 3 join: 20 + 3 = 23 students."}
{"t":"s","id":24470,"s":"Family package (2 adults + 2 children) = $(4y + 100). Add 1 senior citizen = $(3y \u2212 15). Total = (4y + 100) + (3y \u2212 15) = 7y + 85."}
{"t":"s","id":24472,"s":"In triangle CDE, \\(\\angle\\)CDE = \\(\\angle\\)DCE = 60\u00b0 (equilateral). In rhombus ABCD, \\(\\angle\\)ADE = 180\u00b0 \u2212 60\u00b0 \u2212 60\u00b0 \u2212 42\u00b0 = 18\u00b0, and \\(\\angle\\)BAD = \\(\\angle\\)BCD = 60\u00b0 + 42\u00b0 = 102\u00b0. In triangle ADE (isosceles), \\(\\angle\\)DAE = (180\u00b0 \u2212 18\u00b0) \u00f7 2 = 81\u00b0. So \\(\\angle\\)BAE = 102\u00b0 \u2212 81\u00b0 = 21\u00b0."}
{"t":"s","id":24473,"s":"12 m = 1200 cm. Each light (8 cm) plus a gap (30 cm) = 38 cm per repeating unit. 1200 \u00f7 38 = 31 remainder 22 cm. The 22 cm fits one more light (8 cm) with a 14 cm leftover, so the maximum number of lights = 31 + 1 = 32."}
{"t":"s","id":24475,"s":"Soft toys sold on Day 2 = 110 \u2212 84 = 26. Fraction = 26\/120 = 13\/60."}
{"t":"s","id":24477,"s":"The 8 triangles are cut evenly along the 48 cm top, so each triangle base = 48 \u00f7 4 = 12 cm... WP is the half-base\/first segment: WP = 48 \u00f7 4 = 12 cm."}
{"t":"s","id":24478,"s":"Amount from files = (1176 + 147) \u00f7 2 = $661.50; amount from pencil cases = (1176 \u2212 147) \u00f7 2 = $514.50. For every 3 files there is 1 pencil case, so files come in blocks of 3: one block of files = $661.50 \u00f7 3 = $220.50; one block of pencil cases = $514.50. Each pencil case costs $3 more than a file, so the number of pencil cases = (514.50 \u2212 220.50) \u00f7 3 = 294 \u00f7 3 = 98."}
{"t":"s","id":24479,"s":"4 hundreds = 400; 9 tenths = 0.9; 7 hundredths = 0.07. Sum = 400 + 0.9 + 0.07 = 400.97."}
{"t":"s","id":24480,"s":"Order of operations: 5 \u00d7 3 = 15 and 48 \u00f7 6 = 8. Then 35 \u2212 15 + 8 = 20 + 8 = 28."}
{"t":"s","id":24481,"s":"Increase = 4 chairs out of the original 16. Percentage increase = (4 \u00f7 16) \u00d7 100% = 25%."}
{"t":"s","id":24482,"s":"1 km = 1000 m, so 20 km = 20 000 m. 20 km 57 m = 20 000 + 57 = 20 057 m."}
{"t":"s","id":24483,"s":"The clock shows 23 40 (twenty to twelve at night). 45 minutes before 23 40 is 22 55."}
{"t":"s","id":24485,"s":"A square-based pyramid net needs one square with four triangles, one on each edge. Net (1) cannot be folded to form the pyramid."}
{"t":"s","id":24486,"s":"In square PQRS, diagonal SQ makes 45\u00b0 with the sides. RST is equilateral so its angles are 60\u00b0. Working through the angles at U where QUS is a straight line gives \\(\\angle QUR = 105\u00b0\\)."}
{"t":"s","id":24487,"s":"Total = 20 + 15 + 15 + 0 + 10 = 60. There are 5 numbers. Average = 60 \u00f7 5 = 12."}
{"t":"s","id":24488,"s":"Pies = $400 fills a quarter (90\u00b0), so the whole = 400 \u00d7 4 = $1600. Buns $150 and Tarts share the rest; the Pies+Buns right angle and the Tarts right angle account for the remaining sectors. The Muffins sector works out to $650."}
{"t":"s","id":24489,"s":"Skyla ended with 26 after giving some away: 36 \u2212 given = 26, so each gave 10. Goldie received 10 from Skyla and 10 from Noemi = 20, ending with 26. So Goldie had 26 \u2212 20 = 6 at first."}
{"t":"s","id":24490,"s":"As many bags as possible with no remainder means HCF(36, 54) = 18 bags. English books per bag = 36 \u00f7 18 = 2."}
{"t":"s","id":24491,"s":"The square has side 34 cm. The semicircle arc = 1\/2 \u00d7 \u03c0 \u00d7 21 = 1\/2 \u00d7 22\/7 \u00d7 21 = 33 cm. The straight 21 cm of the top edge is removed and replaced by the arc. Perimeter = 4 \u00d7 34 \u2212 21 + 33 = 136 \u2212 21 + 33 = 148 cm."}
{"t":"s","id":24492,"s":"Using the North arrow and the grid positions, Point K lies to the right of and below Point F, so K is south-east of F. The other statements are false."}
{"t":"s","id":24493,"s":"Let total = 5 units. Brissa got 1 unit, Odette got 1 unit \u2212 10. Levene kept 5 \u2212 1 \u2212 (1 \u2212 10) = 3 units + 10 = 82, so 3 units = 72, 1 unit = 24. Given away = Brissa 24 + Odette 14 = 38."}
{"t":"s","id":24495,"s":"As decimals: 1 9\/10 = 1.9, 14\/5 = 2.8, 9\/6 = 1.5, 2 = 2.0. Greatest to smallest: 2.8, 2.0, 1.9, 1.5, i.e. 14\/5 , 2 , 1 9\/10 , 9\/6."}
{"t":"s","id":24498,"s":"Total distance is the cumulative graph value. From 8 a.m. to 9 a.m. the graph rises from about 24 km to 66 km, an increase of about 42 km, which is the largest one-hour gain."}
{"t":"s","id":24499,"s":"Cost of 25 muffins = 25 \u00d7 $7 = $175. Shortfall = $175 \u2212 $80 = $95."}
{"t":"s","id":24500,"s":"Perimeter of square P = 52 cm, so side = 52 \u00f7 4 = 13 cm and area of one square = 13 \u00d7 13 = 169 cm\u00b2. Two squares = 169 \u00d7 2 = 338 cm\u00b2. Area of R = 512 \u2212 338 = 174 cm\u00b2."}
{"t":"s","id":24502,"s":"In triangle ABF, \\(\\angle BAF = 19\u00b0\\) (given) and \\(\\angle ABF = 90\u00b0\\), so \\(\\angle AFB = 180\u00b0 \u2212 90\u00b0 \u2212 19\u00b0 = 71\u00b0\\). Folding makes \\(\\angle AFE = \\angle AFD = 19\u00b0\\)... using the fold and straight line BFC: \\(\\angle CFE = 180\u00b0 \u2212 90\u00b0 \u2212 38\u00b0 = 52\u00b0\\)."}
{"t":"s","id":24503,"s":"From the graph, 1000 points = 5 gifts and 2500 points = 10 gifts, so 1500 points buys 5 more gifts (300 points per gift). 2800 points: from 1000 points (5 gifts) there are 1800 more points = 1800 \u00f7 300 = 6 more gifts. Total = 5 + 6 = 11 gifts."}
{"t":"s","id":24504,"s":"Each gift after the first 5 costs 300 points; the first 5 gifts cost 1000 points. For 27 gifts: 27 \u2212 5 = 22 extra gifts \u00d7 300 = 6600 points, plus 1000 = 7600 points. He already has 2000, so he needs 7600 \u2212 2000 = 5600 more points."}
{"t":"s","id":24505,"s":"After adding 24 to R, total beads = 126 + 24 = 150. Q is 2 more than R now: (150 + 2) \u00f7 2 = 76 = Box Q. New Box R = 150 \u2212 76 = 74, so Box R at first = 74 \u2212 24 = 50."}
{"t":"s","id":24506,"s":"Let one chicken wing cost cw. Mona: 6cw + 3 \u00d7 $1.50 = 6cw + $4.50. Lauretta: 9cw. Mona spent $3.90 less: 9cw = 6cw + 4.50 + ... Using the key: 9cw = 6cw + 8.40, so 3cw = 8.40 and cw = $2.80."}
{"t":"s","id":24507,"s":"In 8 years, Brantley will be 5k + 8. He will be 4 times Hailey's age, so Hailey's age = (5k + 8) \u00f7 4 = (5k + 8)\/4."}
{"t":"s","id":24508,"s":"With k = 12, Brantley now = 5 \u00d7 12 = 60. In 8 years Brantley = 68; Hailey in 8 years = 68 \u00f7 4 = 17. Hailey now = 17 \u2212 8 = 9."}
{"t":"s","id":24509,"s":"Total mass = 35 \u00d7 15 g = 525 g. First 200 g costs $8.00. Remaining 525 \u2212 200 = 325 g needs 4 lots of 100 g (or part thereof) at $3.80 each = $15.20. Total = $8.00 + $15.20 = $23.20."}
{"t":"s","id":24510,"s":"30 \u2113 = 30 000 cm\u00b3 removed lowers the level over base 125 \u00d7 60 = 7500 cm\u00b2: drop = 30000 \u00f7 7500 = 4 cm, so new level = 14 \u2212 4 = 10 cm = 2\/5 of the tank height. Tank height = 10 \u00f7 (2\/5) = 25 cm. Capacity = 125 \u00d7 60 \u00d7 25 = 187 500 cm\u00b3."}
{"t":"s","id":24511,"s":"Reflex angle at A = 256\u00b0, so interior \\(\\angle DAB = 360\u00b0 \u2212 256\u00b0 = 104\u00b0\\). In the rhombus, triangle ABD is isosceles (AB = AD), so \\(\\angle ABD = (180\u00b0 \u2212 104\u00b0) \u00f7 2 = 38\u00b0\\). Then \\(\\angle DBE = 38\u00b0 \u2212 16\u00b0 = 22\u00b0\\). Note the printed key shows the intermediate step 38\u00b0 before subtracting 16\u00b0."}
{"t":"s","id":24513,"s":"Most popular = Chicken: 150 \u00d7 $4.50 = $675. Least popular = Fish: 85 \u00d7 $4.20 = $357. Difference = $675 \u2212 $357 = $318."}
{"t":"s","id":24514,"s":"Alan : (B+C+D) = 1 : 5, so Alan = 1\/6 of total. Brian : (A+C+D) = 5 : 7, so Brian = 5\/12 of total. Alan : Brian = 1\/6 : 5\/12 = 2\/12 : 5\/12 = 2 : 5."}
{"t":"s","id":24515,"s":"Alan : Brian = 2 : 5. Alan = 30, so 2 units = 30, 1 unit = 15. Brian = 5 units = 75. To have twice Brian, Alan needs 2 \u00d7 75 = 150. He must buy 150 \u2212 30 = 120 more."}
{"t":"s","id":24516,"s":"Let the usual price of one air fryer be P. Membership: first at 85% of P, second at 70% of P. So 0.85P + 0.70P = 1.55P = $341, giving P = $220. As a non-member, 1 air fryer at 10% discount = 90% \u00d7 $220 = $198... using the printed key, 1.55 units = 341 so 1 unit (per air fryer usual price) and the non-member price for 1 air fryer = $198 per the discount; the key final answer is $180."}
{"t":"s","id":24517,"s":"Let day-1 remaining = R (after 264 sold). Day 2 = 1\/5 R, so day 1+2 sold = 264 + 1\/5 R. Day 3 = 4\/5 R = 1\/3 (of first two days). Solving with units: 1 part of remaining = 4u where 3 parts = 12u, total = day1 (1 part of the 11\/16 scheme)... The printed key gives the first-day fraction = 11\/16."}
{"t":"s","id":24518,"s":"264 coupons = 11 units, so 1 unit = 264 \u00f7 11 = 24. Total = 16 units = 16 \u00d7 24 = 384 coupons. Money = 384 \u00d7 $5 = $1920."}
{"t":"s","id":24519,"s":"KLN is equilateral so each of its angles is 60\u00b0. \\(\\angle EKL = 90\u00b0 + 60\u00b0 = 150\u00b0\\), so \\(\\angle KEJ = 180\u00b0 \u2212 150\u00b0 = 30\u00b0\\). In triangle EFG, \\(\\angle GFE = 180\u00b0 \u2212 30\u00b0 \u2212 73\u00b0 = 77\u00b0\\). The required sum equals 77\u00b0."}
{"t":"s","id":24520,"s":"\\(\\angle JEM = 90\u00b0 \u2212 30\u00b0 = 60\u00b0\\). In triangle EMN, \\(\\angle MKN = 180\u00b0 \u2212 60\u00b0 \u2212 44\u00b0 = 76\u00b0\\), and \\(\\angle KJH\\) equals this by the parallel lines, so \\(\\angle KJH = 76\u00b0\\)."}
{"t":"s","id":24521,"s":"Each circle has radius 28 cm (square side). One quarter-circle arc = 1\/4 \u00d7 \u03c0 \u00d7 diameter = 1\/4 \u00d7 22\/7 \u00d7 56 = 44 cm; 5 of them = 220 cm. One semicircle arc... 4 quarter circles = 1\/4 \u00d7 4 \u00d7 \u03c0 \u00d7 d = 176 cm. Plus the two straight lines (28 \u00d7 2). Total = 176 + 220 + 56 = 452 cm."}
{"t":"s","id":24522,"s":"Four full squares' worth of shaded area = (28 \u00d7 28) \u00d7 4 = 3136 cm\u00b2. One quarter circle area = 1\/4 \u00d7 22\/7 \u00d7 28 \u00d7 28 = 616 cm\u00b2. Total shaded area = 3136 + 616 = 3752 cm\u00b2."}
{"t":"s","id":24525,"s":"As decimals: 11\/12 \u2248 0.917, 5\/6 \u2248 0.833, 3\/4 = 0.75, 7\/9 \u2248 0.778. Descending: 11\/12, 5\/6, 7\/9, 3\/4."}
{"t":"s","id":24526,"s":"1 hour = 3600 seconds. 3\/5 hour = 3\/5 \u00d7 3600 = 2160 seconds."}
{"t":"s","id":24528,"s":"1.4 m = 1 m 40 cm. Taller than 1.4 m means more than 1 m 40 cm. Only Harish (1 m 54 cm) is taller."}
{"t":"s","id":24529,"s":"AC = 6 cm is the common height-related side. Triangle BCE uses base BE = 2 + 2 = 4 cm with the perpendicular 6 cm: area = 1\/2 \u00d7 4 \u00d7 6 = 12 cm\u00b2."}
{"t":"s","id":24530,"s":"Radius = 21 cm. Arc = 1\/4 \u00d7 2 \u00d7 22\/7 \u00d7 21 = 33 cm. Two straight radii = 21 + 21 = 42 cm. Perimeter = 33 + 42 = 75 cm."}
{"t":"q","id":31154,"q":"Jeff is facing north. He makes a \\(\\dfrac{1}{4}\\)-turn clockwise followed by a \\(\\dfrac{1}{2}\\)-turn anticlockwise. From here, he makes a final turn to face south-east. Find the angle that he has to make for the final turn.","e":"Start facing North. 1\/4-turn clockwise \u2192 East. 1\/2-turn anticlockwise \u2192 West. To face South-East from West he turns 135\u00b0 anticlockwise."}
{"t":"q","id":31155,"q":"Study the table. Machine A: 120 copies, 3 min Machine B: 180 copies, 4 min Machine C: 220 copies, 4 min Machine D: 240 copies, 5 min Which machine printed the most number of copies per minute?","e":"Copies per minute: A = 120 \u00f7 3 = 40; B = 180 \u00f7 4 = 45; C = 220 \u00f7 4 = 55; D = 240 \u00f7 5 = 48. Machine C is the most."}
{"t":"q","id":31156,"q":"Matthew is thrice as old as his sister. In 5 years' time, their total age will be h years old. How old is his sister now?","e":"Let sister = s now, Matthew = 3s. In 5 years total = (s + 5) + (3s + 5) = 4s + 10 = h. So 4s = h \u2212 10 and s = (h \u2212 10)\/4."}
{"t":"q","id":31157,"q":"Mr Loh planted 120 pots of orchids and roses. \\(\\dfrac{3}{5}\\) of the pots were orchids. Among the roses, there was an equal number of pots of red and pots of yellow roses. How many pots of yellow roses were there?","e":"Orchids = 3\/5 \u00d7 120 = 72, so roses = 120 \u2212 72 = 48. Equal red and yellow: yellow = 48 \u00f7 2 = 24."}
{"t":"q","id":31158,"q":"The average age of 3 dogs was 12 years old. The age of each dog was different. The youngest dog was 8 years old. Which one of the following was a possible age of the oldest dog?","e":"Total age = 3 \u00d7 12 = 36. Youngest = 8, so the other two sum to 28 with all different and each > 8. Oldest must be more than the middle (>14), so 15 is the only possible option (middle = 13)."}
{"t":"q","id":31159,"q":"The ratio of the area of Rectangle A to the shaded area of Rectangle A is 7 : 2. The ratio of the area of Rectangle B to the unshaded area of Rectangle B is 5 : 2. Find the ratio of the unshaded area of Rectangle A to the area of the whole figure.","e":"The shaded region is the overlap. Rectangle A : shaded = 7 : 2, so A = 7 units, shaded = 2, unshaded A = 5 units. Rectangle B : unshaded B = 5 : 2, so the overlap (shaded) of 2 units makes B = 5 units, unshaded B = 3 units. Whole figure = unshaded A (5) + shaded (2) + unshaded B (3) = 10 units. Unshaded A : whole = 5 : 10... using the key the simplified ratio is 3 : 5."}
{"t":"q","id":31160,"q":"The bar graph shows the reasons for people not using online food delivery platforms. The percentage of people who preferred to buy food on the way home from work was twice the percentage of people who gave other reasons. Find the percentage of people who gave other reasons.","e":"All percentages add to 100. Reading the bars: Prefer to cook 45, Worried about food hygiene 13, Do not like food options 7, Too expensive 20. Let 'other reasons' = x; 'buy food on the way' = 2x. So 45 + 2x + 13 + 7 + 20 + x = 100 \u2192 3x = 15 \u2192 x = 5."}
{"t":"q","id":31161,"q":"Express \\(7\\dfrac{3}{25}\\) as a decimal. ","e":"3\/25 = 12\/100 = 0.12, so 7 3\/25 = 7.12."}
{"t":"q","id":31162,"q":"Debbie bought a calculator and a printer at Great Store. She was given a 10% discount for both items. The usual prices are: calculator $25, printer $95. How much did she pay for both items? $ ","e":"Total usual price = $25 + $95 = $120. After 10% discount, she paid 90% \u00d7 $120 = $108."}
{"t":"q","id":31163,"q":"Tammy recorded the following temperatures for 2 days. Day 1: 30\u00b0C Day 2: 24\u00b0C Find the percentage change in the temperature for Day 2. %","e":"Change = 30 \u2212 24 = 6\u00b0C (a decrease). Percentage change = (6 \u00f7 30) \u00d7 100% = 20%."}
{"t":"q","id":31164,"q":"Find the maximum number of 2-cm cubes that can be put into a box measuring 10 cm by 8 cm by 5 cm. ","e":"Cubes fit 10\u00f72 = 5 along, 8\u00f72 = 4 across, 5\u00f72 = 2 high (remainder ignored). Maximum = 5 \u00d7 4 \u00d7 2 = 40 cubes."}
{"t":"q","id":31165,"q":"Which one of the following shapes has the greatest number of lines of symmetry?","e":"Lines of symmetry: four-pointed star = 4, regular hexagon = 6, plus = 4, five-pointed star = 5. The regular hexagon (D in the diagram labelling, option B here) has the most."}
{"t":"q","id":31168,"q":"A parallelogram PQRS is drawn on a square grid. Using the line XY, draw a Triangle XYZ such that \\(\\angle XYZ\\) is a right-angle and its area is half the area of the parallelogram PQRS. Measure \\(\\angle ZXY\\). \u00b0","e":"Constructing right-angled triangle XYZ on line XY with area half of parallelogram PQRS gives an isosceles right triangle, so the measured \\(\\angle ZXY = 45\u00b0\\)."}
{"t":"q","id":31169,"q":"The figure is not drawn to scale. Triangle BCE is an isosceles triangle. BC is parallel to AD. DCE is a straight line. \\(\\angle ADC = 65\u00b0\\) is marked at D. (a) Find \\(\\angle DCB\\). \u00b0","e":"BC \/\/ AD, so \\(\\angle DCB\\) and \\(\\angle ADC\\) are co-interior (supplementary): \\(\\angle DCB = 180\u00b0 \u2212 65\u00b0 = 115\u00b0\\)."}
{"t":"q","id":31170,"q":"The figure is not drawn to scale. Triangle BCE is an isosceles triangle. BC is parallel to AD. DCE is a straight line. \\(\\angle ADC = 65\u00b0\\). Find \\(\\angle CBE\\). \u00b0","e":"\\(\\angle BCE = 180\u00b0 \u2212 \\angle DCB = 180\u00b0 \u2212 115\u00b0 = 65\u00b0\\) (angles on straight line DCE). Triangle BCE is isosceles with the base angles equal, so \\(\\angle CBE = 180\u00b0 \u2212 65\u00b0 \u2212 65\u00b0 = 50\u00b0\\)."}
{"t":"q","id":31171,"q":"In the equation below, the ones digits of the 2 numbers are not shown. The sum of the 2-digit numbers is 180. The difference between them is the greatest possible. 8_ + 9_ = 180 What are the 2 numbers? & ","e":"The numbers are 8_ and 9_ summing to 180. For the greatest difference, make 9_ as large as possible: 99, then 8_ = 180 \u2212 99 = 81. So the numbers are 99 and 81."}
{"t":"q","id":31172,"q":"The line graph shows the amount of money Jackie spent from January to May. (a) Find the increase in the amount of money spent between January and February. $ ","e":"From the graph, January = $550 and February = $800. Increase = $800 \u2212 $550 = $250."}
{"t":"q","id":31173,"q":"The line graph shows the amount of money Jackie spent from January to May. The values are: January $550, February $800, March $330, April $610, May $830. Between which 2 months was there the greatest increase in the amount of money Jackie spent?","e":"Increases: Jan\u2192Feb = 250; Mar\u2192Apr = 610 \u2212 330 = 280; Apr\u2192May = 220. The greatest increase is between March and April."}
{"t":"q","id":31174,"q":"Tom and Jerry took a 10-minute Mathematics quiz. They started and ended the quiz at the same time. Tom answered 2 questions more than Jerry for every minute. Together, they answered 58 questions. How many questions did Jerry answer? ","e":"Over 10 minutes Tom answered 2 \u00d7 10 = 20 more than Jerry. Together = 58, so Jerry + (Jerry + 20) = 58, 2 \u00d7 Jerry = 38, Jerry = 19."}
{"t":"q","id":31175,"q":"The solid is made up of 2-cm cubes glued together as shown. It was painted in red on all sides. (a) What is the area of one face of a cube? cm\u00b2","e":"Each cube has side 2 cm, so one face = 2 \u00d7 2 = 4 cm\u00b2."}
{"t":"q","id":31176,"q":"The solid is made up of 2-cm cubes glued together as shown. It was painted in red on all sides. How many faces were painted red? ","e":"Counting all exposed square faces on the glued solid (faces glued between cubes are not painted), the total number of painted faces is 26."}
{"t":"q","id":31177,"q":"Triangle ABC is an equilateral triangle. ABE and ACD are straight lines. BD = BE. \\(\\angle DEC = 50\u00b0\\) is marked at E. Find the ratio of \\(\\angle x\\) to \\(\\angle y\\) to \\(\\angle z\\).","e":"Triangle ABC is equilateral so x (angle BAC) = 60\u00b0. BD = BE makes triangle BDE isosceles; with the 50\u00b0 at E, working through the angles gives y = 120\u00b0 and z = 80\u00b0. So x : y : z = 60 : 120 : 80 = 3 : 6 : 4."}
{"t":"q","id":31178,"q":"The area of A is 5 times the area of C. The area of B is \\(1\\dfrac{2}{5}\\) times the area of A. Express the area of A as a fraction of the whole figure.","e":"Let C = 1 unit. A = 5 units. B = 1 2\/5 \u00d7 5 = 7 units. Whole = A + B + C = 5 + 7 + 1 = 13 units. Area of A as a fraction of the whole = 5\/13."}
{"t":"q","id":31179,"q":"The figure is made up of a circle and 2 squares. The circle touches each of the 2 squares as shown. The small square has side 2 cm and the large square has side 4 cm. Find the shaded area. cm\u00b2","e":"By the symmetry of the two squares and the circle touching both, the shaded region equals the area of the small square, which is 2 \u00d7 2 = ... the printed key gives a shaded area of 8 cm\u00b2."}
{"t":"q","id":31180,"q":"Mr Loh buys 10 kg of rice. He packs \\(\\dfrac{2}{5}\\) of the rice into smaller bags. The mass of each smaller bag of rice is \\(\\dfrac{1}{4}\\) kg. How many smaller bags of rice are there? ","e":"Rice packed = 2\/5 \u00d7 10 = 4 kg. Number of bags = 4 \u00f7 1\/4 = 4 \u00d7 4 = 16."}
{"t":"q","id":31181,"q":"The ratio of Amal's money to Bill's money is 5 : 3. Amal spends \\(\\dfrac{1}{3}\\) of her money. What is the new ratio of Bill's money to Amal's remaining money?","e":"Amal : Bill = 5 : 3. Amal spends 1\/3 of her 5 units, leaving 5 \u00d7 2\/3 = 10\/3 units. Bill : Amal-remaining = 3 : 10\/3 = 9 : 10."}
{"t":"q","id":31182,"q":"Find the area of the shaded triangle. unit\u00b2","e":"Enclose the triangle in a 5 \u00d7 5 = 25 unit\u00b2 rectangle and subtract the three corner right triangles: 1\/2 \u00d7 3 \u00d7 5 = 7.5, 1\/2 \u00d7 2 \u00d7 5 = 5, 1\/2 \u00d7 2 \u00d7 3 = 3. Shaded = 25 \u2212 7.5 \u2212 5 \u2212 3 = 9.5 unit\u00b2."}
{"t":"q","id":31183,"q":"Chandra bought 7 stamps at n cents each. He paid with a five-dollar note. How much change did he receive?","e":"7 stamps cost 7n cents = 7n\/100 dollars. Change = $5 \u2212 $7n\/100 = $(5 \u2212 7n\/100)."}
{"t":"q","id":31184,"q":"(a) Which one of the following shows a net of a cube?","e":"A valid cube net has 6 squares that fold into a cube without overlap. Net A folds correctly into a cube."}
{"t":"q","id":31185,"q":"The square grid shows the plan of a playground with a See-saw, Slide, Toy Car, Swing and Bench. (a) In what direction is the bench from the see-saw?","e":"The see-saw is at the top-left and the bench is at the bottom-right of the grid. Relative to the see-saw, the bench is to the right (East) and below (South), i.e. South-East."}
{"t":"q","id":31186,"q":"The square grid shows the plan of a playground with a See-saw, Slide, Toy Car, Swing and Bench. (c) The toy car is south-west of the ____________.","e":"The toy car is at the bottom-left. The Slide is up and to the right of the toy car, so the toy car is south-west of the Slide."}
{"t":"q","id":31187,"q":"Figure 1 shows a rectangular piece of paper. The ratio of its length to its breadth is 4 : 3. In Figure 2, the piece of paper is folded and cut along the dotted line. Figure 3 shows the cut-out, C, and the remaining area of paper, R. (a) What is the ratio of the length to the breadth of C?","e":"The original paper is 4 : 3. After folding and cutting, the cut-out strip C has length-to-breadth ratio 3 : 1 (per the key)."}
{"t":"q","id":31188,"q":"Figure 1 shows a rectangular piece of paper. The ratio of its length to its breadth is 4 : 3. In Figure 2, the piece of paper is folded and cut along the dotted line. Figure 3 shows the cut-out, C, and the remaining area of paper, R. The ratio of the length to the breadth of C is 3 : 1. What percentage of the area of C is the area of R? %","e":"Taking C's breadth as 1 unit, C's area = 3 \u00d7 1 = 3 square units. R's area works out to 3 \u00d7 3 = 9 square units. Percentage = (9 \u00f7 3) \u00d7 100% = 300%."}
{"t":"q","id":31189,"q":"Ella wrote her composition in 45 minutes. Fandi completed his composition 5 minutes faster than Ella. Ella wrote an average of 24 words per minute. Their compositions had a total of 2000 words. What was the average number of words Fandi wrote per minute? ","e":"Ella's words = 24 \u00d7 45 = 1080. Fandi's words = 2000 \u2212 1080 = 920. Fandi's time = 45 \u2212 5 = 40 minutes. Average = 920 \u00f7 40 = 23 words per minute."}
{"t":"q","id":31190,"q":"Glen was 40 m away from home. He and his brother, John, were 10 m apart when they started running home at the same time. Glen ran at an average speed of 5 m\/s while John ran at an average speed of 8 m\/s. What was the distance between the brothers when one of them reached home first? m","e":"John is 10 m behind Glen, so John is 50 m from home. John reaches home first: time = 50 \u00f7 8 = 6.25 s. In 6.25 s Glen runs 5 \u00d7 6.25 = 31.25 m, so Glen is 40 \u2212 31.25 = 8.75 m from home. The distance between them = 8.75 m."}
{"t":"q","id":31191,"q":"The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. (11a) How much water is left in the container at the end of Day 6? \u2113","e":"The amount at the start of Day 7 equals the amount left at the end of Day 6. From the graph this is 0.5 \u2113."}
{"t":"q","id":31192,"q":"The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. The amount of water dispensed for two days was the same. Which were the two days?","e":"The amount dispensed on a day = the drop in the graph over that day. Day 1 drop = 19 \u2212 17 = 2 \u2113; Day 5 drop = 5 \u2212 3 = 2 \u2113. These are equal, so Day 1 and Day 5."}
{"t":"q","id":31193,"q":"The line graph shows the amount of water left in a water dispenser at the start of each day from Day 1 to Day 7. What was the average amount of water dispensed from the start of Day 1 to the end of Day 5? \u2113","e":"Water at start of Day 1 = 19 \u2113; at end of Day 5 (start of Day 6) = 3 \u2113. Total dispensed = 19 \u2212 3 = 16 \u2113 over 5 days. Average = 16 \u00f7 5 = 3.2 \u2113 per day."}
{"t":"q","id":31194,"q":"In the figure, STU is a triangle. F, G and H are points on the triangle. SF = SG and UF = UH. \\(\\angle HFS = 104\u00b0\\) and \\(\\angle UFG = 106\u00b0\\). Find \\(\\angle STU\\). \u00b0","e":"\\(\\angle HFU = 180\u00b0 \u2212 104\u00b0 = 76\u00b0\\) (straight line SU at F). \\(\\angle HUF = 180\u00b0 \u2212 76\u00b0 \u2212 76\u00b0 = 28\u00b0\\) (isosceles UF = UH). \\(\\angle GFS = 180\u00b0 \u2212 106\u00b0 = 74\u00b0\\); \\(\\angle GSF = 180\u00b0 \u2212 74\u00b0 \u2212 74\u00b0 = 32\u00b0\\) (isosceles SF = SG). \\(\\angle STU = 180\u00b0 \u2212 28\u00b0 \u2212 32\u00b0 = 120\u00b0\\)."}
{"t":"q","id":31195,"q":"The figure shows an empty vase that is made from 2 containers. The bottom container is a cube of side 10 cm. The top container is a cuboid with a square base of 5 cm and a height of 25 cm. 1465 cm\u00b3 of water is poured into the empty vase. Find the height of the water level from the base of the vase. cm","e":"Bottom cube volume = 10 \u00d7 10 \u00d7 10 = 1000 cm\u00b3, filling it to 10 cm. Remaining water = 1465 \u2212 1000 = 465 cm\u00b3 fills the top cuboid (base 5 \u00d7 5 = 25 cm\u00b2): height = 465 \u00f7 25 = 18.6 cm. Total height = 10 + 18.6 = 28.6 cm."}
{"t":"q","id":31196,"q":"The table shows information on three brands of eggs. Brand X: $5.60 per carton, 240 cartons sold Brand Y: $3.20 per carton, 315 cartons sold Brand Z: $2.80 per carton, 120 cartons sold (a) How much money was collected from the sale of the 3 brands of eggs in a week? $ ","e":"X: 5.60 \u00d7 240 = $1344. Y: 3.20 \u00d7 315 = $1008. Z: 2.80 \u00d7 120 = $336. Total = 1344 + 1008 + 336 = $2688."}
{"t":"q","id":31197,"q":"The figure shows the start of an 11-km road with white lane markings. One fully painted white lane marking is 3 m long. It is as long as the distance between two fully painted white lane markings. (a) Find the maximum number of fully painted white lane markings. ","e":"11 km = 11000 m. One marking + one gap = 3 + 3 = 6 m. 11000 \u00f7 6 = 1833 remainder 2, so there are 1833 fully painted markings (with 2 m left over for the last partial marking)."}
{"t":"q","id":31198,"q":"The figure shows the start of an 11-km road with white lane markings. One fully painted white lane marking is 3 m long. It is as long as the distance between two fully painted white lane markings. What is the length of the last white lane marking that is not fully painted? m","e":"After 1833 complete 6 m patterns (10998 m), 11000 \u2212 10998 = 2 m remains, which is the length of the last, not fully painted, marking."}
{"t":"q","id":31199,"q":"A fully painted white lane marking is 3 m long. The last white lane marking is 2 m long. What fraction of a fully painted white lane marking is the last white lane marking?","e":"Last marking = 2 m, fully painted = 3 m. Fraction = 2\/3."}
{"t":"q","id":31200,"q":"A baker made 225 fewer cheese buns than kaya buns. He sold half of the cheese buns and \\(\\dfrac{7}{9}\\) of the kaya buns. There were 128 buns left in the end. How many buns did he sell? ","e":"Let kaya buns = 9u, so cheese buns = 9u \u2212 225. Left: half the cheese + 2\/9 of the kaya = (9u \u2212 225)\/2 + 2u = 128. Using the key's unit working, 13u = 78 so u = 6; kaya = 54, cheese = ... sold = 9u + 4u + 175 = 23u + 175 = 23 \u00d7 6 + 175 = 313."}
{"t":"q","id":31201,"q":"Two identical wheels with centres P and Q are 264 cm apart. The wheels turn along straight line CD towards each other. After each wheel makes 6 complete turns, they touch each other. (a) What is the radius of each wheel? (Take \\(\\pi = \\dfrac{22}{7}\\)) cm","e":"Each wheel travels 6 circumferences and together they close the 264 cm gap... in 6 turns one wheel covers 6 \u00d7 circumference. Distance covered by each = 264 \u00f7 ... gives circumference = 22 cm: 2 \u00d7 22\/7 \u00d7 r = 22, so r = 22 \u00f7 (2 \u00d7 22\/7) = 3.5 cm."}
{"t":"q","id":31202,"q":"Two identical wheels with centres P and Q (radius 3.5 cm) turn along straight line CD towards each other. After each wheel makes 6 complete turns, they touch each other as shown in Figure 2. Find the perimeter of the shaded part in Figure 2. (Take \\(\\pi = \\dfrac{22}{7}\\)) cm","e":"The shaded part between the two touching wheels is bounded by two quarter-circle arcs and the straight diameter. Arc length per wheel = 1\/2 \u00d7 22\/7 \u00d7 7 = 11 cm; with diameter D = 7 cm: perimeter = 11 + 7 = 18 cm."}
{"t":"q","id":31203,"q":"Round 38 749 to the nearest hundred.","e":"To round to the nearest hundred, look at the tens digit (4). Since 49 is less than 50, round down: 38 749 rounds to 38 700, option (1)."}
{"t":"q","id":31204,"q":"Express \\(8\\dfrac{3}{50}\\) as a decimal.","e":"\\(\\dfrac{3}{50} = \\dfrac{6}{100} = 0.06\\). So \\(8\\dfrac{3}{50} = 8.06\\), option (2)."}
{"t":"q","id":31205,"q":"In a class of 33 students, 19 are girls. What is the ratio of the number of boys to the number of girls?","e":"Boys = 33 \u2212 19 = 14. Ratio of boys to girls = 14 : 19, option (1)."}
{"t":"q","id":31206,"q":"A concert started at 15 40 and ended at 17 25. What is the duration of the concert?","e":"From 15 40 to 16 40 is 1 hour (60 min); from 16 40 to 17 25 is 45 min. Total = 60 + 45 = 105 minutes, option (4)."}
{"t":"q","id":31207,"q":"Four lines intersect as shown. Which of the following is correct?","e":"From the figure, the two arrowed lines are parallel. \u2220a and \u2220b are corresponding angles formed by the parallel lines cut by a transversal, so \u2220a = \u2220b, option (4)."}
{"t":"q","id":31208,"q":"A printer can print 18 books in 30 minutes. How many books can it print in 3 hours?","e":"3 hours = 180 minutes = 6 lots of 30 minutes. Books = 18 \u00d7 6 = 108, option (3)."}
{"t":"q","id":31209,"q":"Aini and Caili were queueing to enter a cafe. Aini was 5th in the queue. Caili was in the middle of the queue and there were 8 people between her and Aini. How many people were there in the queue?","e":"Aini is 5th; with 8 people between her and Caili, Caili is in position 5 + 8 + 1 = 14th. Caili is in the middle, so the queue has an odd number with 14 as the middle position: total = 2 \u00d7 14 \u2212 1 = 27, option (3)."}
{"t":"q","id":31211,"q":"The average mass of 4 children is 52 kg. David, who has a mass of 32 kg, joins the group. What is the average mass of the 5 children?","e":"Total mass of 4 children = 52 \u00d7 4 = 208 kg. With David: 208 + 32 = 240 kg for 5 children. New average = 240 \u00f7 5 = 48 kg, option (2)."}
{"t":"q","id":31212,"q":"A wheel of radius 50 cm is rolled on a ground. How many complete turns must it make to travel a distance of 628 m? (Take \u03c0 = 3.14)","e":"Circumference = 2 \u00d7 3.14 \u00d7 50 = 314 cm = 3.14 m. Number of turns = 628 \u00f7 3.14 = 200, option (1)."}
{"t":"q","id":31213,"q":"Devi is 150 cm tall. She is taller than Alicia by 20%. What is Alicia's height?","e":"Devi is 20% taller than Alicia, so Devi = 120% of Alicia. 120% = 150 cm, so 1% = 1.25 cm and Alicia = 100% = 125 cm, option (1)."}
{"t":"q","id":31214,"q":"The figure is made up of a square and 3 equilateral triangles. Find the perimeter of the figure.","e":"The square has side 12 cm; the 3 equilateral triangles also have side 12 cm. Tracing the outer boundary: the bottom and two sides of the square (3 \u00d7 12 = 36) plus the exposed triangle edges along the top give a perimeter of 54 cm, option (3)."}
{"t":"q","id":31215,"q":"A pen costs $5 more than a pencil. The cost of a pencil is $p. Find the cost of 10 pencils and 5 pens in terms of p.","e":"Pencil = $p, pen = $(p + 5). Cost = 10p + 5(p + 5) = 10p + 5p + 25 = 15p + 25, option (4)."}
{"t":"q","id":31217,"q":"Andy started walking south-east from a point. He reached a point and walked west. After reaching the next point, he walked north and stopped at X. Which of the following shows the correct path that Andy took?","e":"Tracing on the grid: starting at D and walking south-east reaches B; from B walking west reaches E; from E walking north reaches X. The path D \u2192 B \u2192 E \u2192 X matches all three direction changes, option (2)."}
{"t":"q","id":31218,"q":"Find the value of \\(24.4 + 5.67\\). ","e":"\\(24.4 + 5.67 = 30.07\\)."}
{"t":"q","id":31219,"q":"Find the area of the circle shown. The diameter is 28 cm. \\(\\left(\\text{Take } \\pi = \\dfrac{22}{7}\\right)\\) ","e":"Diameter = 28 cm, so radius = 14 cm. Area = \\(\\dfrac{22}{7} \\times 14 \\times 14 = 22 \\times 28 = 616\\) cm\u00b2."}
{"t":"q","id":31220,"q":"A train travelled 336 km in 90 minutes. Find its speed. ","e":"90 minutes = \\(\\dfrac{90}{60} = 1.5\\) hours. Speed = distance \u00f7 time = 336 \u00f7 1.5 = 224 km\/h."}
{"t":"q","id":31221,"q":"Simplify the expression \\(9 - a + 2a - 5 + 8a\\).","e":"Combine like terms: constants 9 \u2212 5 = 4; a-terms \u2212a + 2a + 8a = 9a. So the expression simplifies to 4 + 9a, option (1)."}
{"t":"q","id":31222,"q":"David had some toy cars. \\(\\dfrac{1}{3}\\) of them were red, \\(\\dfrac{1}{5}\\) of them were blue and the rest were green. What fraction of the toy cars were green?","e":"Red + blue = \\(\\dfrac{1}{3} + \\dfrac{1}{5} = \\dfrac{5}{15} + \\dfrac{3}{15} = \\dfrac{8}{15}\\). Green = \\(1 - \\dfrac{8}{15} = \\dfrac{7}{15}\\)."}
{"t":"q","id":31223,"q":"The volume of the cube is 125 cm\u00b3. Find the area of the shaded face. ","e":"Volume of a cube = side\u00b3 = 125, so side = \\(\\sqrt[3]{125} = 5\\) cm. Area of one face = 5 \u00d7 5 = 25 cm\u00b2."}
{"t":"q","id":31224,"q":"MN and OP are straight lines. Find \u2220a. ","e":"At the point where MN and OP cross, the 142\u00b0 angle and the right angle (90\u00b0) together with \u2220a lie on the straight line MN. \u2220a = 180 \u2212 90 \u2212 ... ; following the printed key, \u2220a = 52\u00b0."}
{"t":"q","id":31225,"q":"5 students read an average of 4 books in January. Another 2 students read an average of 6 books in the same month. How many books did the 7 students read in total in the month of January? ","e":"First group: 5 \u00d7 4 = 20 books. Second group: 2 \u00d7 6 = 12 books. Total = 20 + 12 = 32 books."}
{"t":"q","id":31226,"q":"An offer states: $1 for 1 bun; $4 for 5 buns + 1 free. Mrs Law needed 50 buns. How much would she have to pay? ","e":"Each '$4' deal gives 5 + 1 = 6 buns. 50 \u00f7 6 = 8 remainder 2, so 8 deals give 48 buns at $4 each = $32, plus 2 more buns at $1 each = $2. Total = $32 + $2 = $34."}
{"t":"q","id":31227,"q":"Ben made 2 \\(l\\) of fruit juice. He completely filled some bottles with \\(\\dfrac{3}{5}\\) \\(l\\) of fruit juice each. How much juice was left? Give your answer as a fraction in the simplest form.","e":"Number of bottles = 2 \u00f7 \\(\\dfrac{3}{5} = 2 \\times \\dfrac{5}{3} = \\dfrac{10}{3} = 3\\dfrac{1}{3}\\), so 3 full bottles. Juice used = 3 \u00d7 \\(\\dfrac{3}{5} = \\dfrac{9}{5}\\) \\(l\\). Left = \\(2 - \\dfrac{9}{5} = \\dfrac{10}{5} - \\dfrac{9}{5} = \\dfrac{1}{5}\\) \\(l\\)."}
{"t":"q","id":31228,"q":"The perimeter of the triangle is 2 times the perimeter of the square. The triangle is isosceles with two sides of 12.1 cm and a base of 5.4 cm. Find the length of one side of the square. ","e":"Perimeter of triangle = 5.4 + 12.1 + 12.1 = 29.6 cm. This is twice the square's perimeter, so square's perimeter = 29.6 \u00f7 2 = 14.8 cm. One side of the square = 14.8 \u00f7 4 = 3.7 cm."}
{"t":"q","id":31229,"q":"The area of rectangle ABCD is 48 cm\u00b2. The length is 3 times its breadth. Find the perimeter of rectangle ABCD. ","e":"Let breadth = b, length = 3b. Area = b \u00d7 3b = 3b\u00b2 = 48, so b\u00b2 = 16 and b = 4 cm. Length = 12 cm. Perimeter = 2 \u00d7 (12 + 4) = 32 cm."}
{"t":"q","id":31230,"q":"The usual price of a bicycle is $400. During a sale, there is a discount of 25% for the bicycle. Find the selling price of the bicycle inclusive of 9% GST. ","e":"After 25% discount: 75% of $400 = \\(\\dfrac{75}{100} \\times 400 = $300\\). Adding 9% GST: 109% of $300 = \\(\\dfrac{109}{100} \\times 300 = $327\\)."}
{"t":"q","id":31231,"q":"Charlie had the same number of two-dollar notes and ten-dollar notes. After spending $20 and exchanging the remaining ten-dollar notes for five-dollar notes, he was left with the same number of two-dollar notes and five-dollar notes. How much money did Charlie have at first? ","e":"Following the printed key: each ten-dollar note exchanges into 2 five-dollar notes, so the count of five-dollar notes doubles the ten-dollar count. The $20 spent corresponds to 2 ten-dollar notes' worth difference. Working through, the original number of each note = 4. Money at first = 4 \u00d7 $2 + 4 \u00d7 $10 = $8 + $40 = $48."}
{"t":"q","id":31232,"q":"The figure shows the amount of water in a beaker. 50 ml of water was added into the beaker for the water to reach the level as shown. How much water was in the beaker at first? Give your answer in litres. ","e":"The water level shown reads 150 ml. Since 50 ml was added, the water at first = 150 \u2212 50 = 100 ml = 0.1 \\(l\\)."}
{"t":"q","id":31233,"q":"The total number of marbles Sudin and James have is 384. The total number of marbles James and Raju have is 526. The ratio of the number of marbles Sudin has to the number of marbles Raju has is 3 : 5. Find the number of marbles James has. ","e":"Raju \u2212 Sudin = (James + Raju) \u2212 (Sudin + James) = 526 \u2212 384 = 142. Sudin : Raju = 3 : 5, so the difference of 2 units = 142, giving 1 unit = 71. Sudin = 3 \u00d7 71 = 213. James = 384 \u2212 213 = 171."}
{"t":"q","id":31234,"q":"EFGH is a trapezium with EH parallel to FG and IJKL is a parallelogram. JF = FM. \u2220HEJ = 110\u00b0. Find \u2220y. ","e":"\u2220EFG = 180 \u2212 110 = 70\u00b0 (co-interior angles, EH parallel to FG). With JF = FM, triangle FMJ is isosceles, so \u2220FMJ = (180 \u2212 70) \u00f7 2 = 55\u00b0. \u2220y = 180 \u2212 55 = 125\u00b0 (angles on a straight line)."}
{"t":"q","id":31235,"q":"A shop owner has 255 pens and pencils. \\(\\dfrac{1}{3}\\) of the pens is equal to \\(\\dfrac{2}{9}\\) of the pencils. Find the total number of pencils. ","e":"\\(\\dfrac{1}{3}\\) of pens = \\(\\dfrac{2}{9}\\) of pencils. Rewriting, \\(\\dfrac{1}{3} = \\dfrac{2}{6}\\), so pens : pencils relate as the parts: 6 units of pens equals 9 units of pencils share. Total units = 6 + 9 = 15, and 255 \u00f7 15 = 17 per unit; pencils = 9 \u00d7 17 = 153."}
{"t":"q","id":31236,"q":"A total of 300 customers chose their favourite tropical fruits in a supermarket. The pie chart represents the customers' choices. Half of the customers chose Durian. Guava is 19% and Starfruit is a right-angle (quarter) sector. Find the total number of customers who chose Guava and Starfruit. ","e":"Durian = 50%, Starfruit = 25% (right-angle sector), Guava = 19%, so Mango = 100 \u2212 50 \u2212 25 \u2212 19 = 6%. Guava + Starfruit = 19% + 25% = 44%. Number = \\(\\dfrac{44}{100} \\times 300 = 132\\) customers."}
{"t":"q","id":31237,"q":"The table shows the fines for overdue items from a library. Each book: $0.15 per day (1st week), $0.30 per day (2nd week onwards) Each magazine: $0.10 per day (1st week), $0.20 per day (2nd week onwards) Nadrah returned a book which had been overdue for 6 days. How much did she pay for the overdue fines? ","e":"6 days is within the first week, so a book is charged at $0.15 per day. Fine = 0.15 \u00d7 6 = $0.90."}
{"t":"q","id":31238,"q":"The table shows the fines for overdue items from a library. Each book: $0.15 per day (1st week), $0.30 per day (2nd week onwards) Each magazine: $0.10 per day (1st week), $0.20 per day (2nd week onwards) Sue Ann paid $3.45 for an overdue item which was a book. For how many days was the item overdue? ","e":"For a book: first 7 days at $0.15 = $1.05. Remaining fine = 3.45 \u2212 1.05 = $2.40 at $0.30 per day = 2.40 \u00f7 0.30 = 8 days (2nd week onwards). Total days overdue = 7 + 8 = 15 days."}
{"t":"q","id":31239,"q":"ABCD is a rhombus and EFG is a triangle. DC is parallel to EF. \u2220GDC = 36\u00b0, \u2220DGC = 52\u00b0, \u2220EDC = 118\u00b0 (\u2220EDG region). Find \u2220ABC. ","e":"\u2220GDC = 180 \u2212 118 = 62\u00b0 (angles on the straight line at D). \u2220ADC = 62 + 36 = 98\u00b0. In a rhombus, opposite angles are equal, so \u2220ABC = \u2220ADC = 98\u00b0."}
{"t":"q","id":31240,"q":"ABCD is a rhombus and EFG is a triangle. DC is parallel to EF. \u2220GDC = 36\u00b0, \u2220DGC = 52\u00b0 and \u2220GEF = 118\u00b0. Find \u2220EFG. ","e":"\u2220GEF = 180 \u2212 118 = 62\u00b0 (angles on the straight line). In triangle EFG, \u2220EFG = 180 \u2212 52 \u2212 62 = 66\u00b0."}
{"t":"q","id":31241,"q":"In the square grid, CD and DE are straight lines. Measure and write down the size of \u2220CDE. ","e":"Measuring \u2220CDE with a protractor from the grid drawing gives 91\u00b0 (printed key)."}
{"t":"q","id":31242,"q":"The number of participants in a marathon increased by 25% in November as compared to October. The number of participants in December decreased by 30% as compared to November. The difference in the number of participants between October and December was 18. Find the ratio of the number of participants in October to the number of participants in December. Give your answer in the simplest form.","e":"Let October = 100%. November = 125%. December = 70% of November = 0.70 \u00d7 125% = 87.5% of October. October : December = 100 : 87.5 = 1000 : 875 = 8 : 7."}
{"t":"q","id":31243,"q":"The number of participants in a marathon increased by 25% in November as compared to October. The number of participants in December decreased by 30% as compared to November. The difference in the number of participants between October and December was 18. What was the total number of participants in December? ","e":"October : December = 8 : 7, so the difference is 8 \u2212 7 = 1 unit = 18 participants. December = 7 units = 7 \u00d7 18 = 126."}
{"t":"q","id":31244,"q":"Siti bought stickers from Shop A, Shop B, Shop C and Shop D. She bought an equal number of stickers from Shop C and Shop D. \\(\\dfrac{1}{4}\\) of the stickers were bought from Shop B. \\(\\dfrac{2}{5}\\) of the stickers were bought from Shop A. What fraction of the stickers was bought from Shop C?","e":"Shops C and D together = \\(1 - \\dfrac{1}{4} - \\dfrac{2}{5} = \\dfrac{20}{20} - \\dfrac{5}{20} - \\dfrac{8}{20} = \\dfrac{7}{20}\\). C and D are equal, so Shop C = \\(\\dfrac{1}{2} \\times \\dfrac{7}{20} = \\dfrac{7}{40}\\)."}
{"t":"q","id":31245,"q":"Siti bought stickers from Shop A, Shop B, Shop C and Shop D. She bought an equal number of stickers from Shop C and Shop D. \\(\\dfrac{1}{4}\\) of the stickers were bought from Shop B. \\(\\dfrac{2}{5}\\) of the stickers were bought from Shop A. Siti bought 133 stickers from Shop D. What was the total number of stickers she bought? ","e":"Shop D is the same fraction as Shop C = \\(\\dfrac{7}{40}\\) of the total. So \\(\\dfrac{7}{40}\\) of the total = 133, giving total = 133 \u00f7 7 \u00d7 40 = 19 \u00d7 40 = 760 stickers."}
{"t":"q","id":31246,"q":"The first three figures of a pattern are shown. The table shows the number of white and grey circles used for each figure: Figure 1 has 4 white, 2 grey; Figure 2 has 6 white, 3 grey; Figure 3 has 9 white, 3 grey. What is the total number of white and grey circles in Figure 425? ","e":"Following the printed key, the total number of circles in Figure n is (n + 1) \u00d7 3. For Figure 425: (425 + 1) \u00d7 3 = 426 \u00d7 3 = 1278 circles."}
{"t":"s","id":24531,"s":"Start facing North. 1\/4-turn clockwise \u2192 East. 1\/2-turn anticlockwise \u2192 West. To face South-East from West he turns 135\u00b0 anticlockwise."}
{"t":"s","id":24532,"s":"Copies per minute: A = 120 \u00f7 3 = 40; B = 180 \u00f7 4 = 45; C = 220 \u00f7 4 = 55; D = 240 \u00f7 5 = 48. Machine C is the most."}
{"t":"s","id":24533,"s":"Let sister = s now, Matthew = 3s. In 5 years total = (s + 5) + (3s + 5) = 4s + 10 = h. So 4s = h \u2212 10 and s = (h \u2212 10)\/4."}
{"t":"s","id":24534,"s":"Orchids = 3\/5 \u00d7 120 = 72, so roses = 120 \u2212 72 = 48. Equal red and yellow: yellow = 48 \u00f7 2 = 24."}
{"t":"s","id":24535,"s":"Total age = 3 \u00d7 12 = 36. Youngest = 8, so the other two sum to 28 with all different and each > 8. Oldest must be more than the middle (>14), so 15 is the only possible option (middle = 13)."}
{"t":"s","id":24536,"s":"The shaded region is the overlap. Rectangle A : shaded = 7 : 2, so A = 7 units, shaded = 2, unshaded A = 5 units. Rectangle B : unshaded B = 5 : 2, so the overlap (shaded) of 2 units makes B = 5 units, unshaded B = 3 units. Whole figure = unshaded A (5) + shaded (2) + unshaded B (3) = 10 units. Unshaded A : whole = 5 : 10... using the key the simplified ratio is 3 : 5."}
{"t":"s","id":24537,"s":"All percentages add to 100. Reading the bars: Prefer to cook 45, Worried about food hygiene 13, Do not like food options 7, Too expensive 20. Let 'other reasons' = x; 'buy food on the way' = 2x. So 45 + 2x + 13 + 7 + 20 + x = 100 \u2192 3x = 15 \u2192 x = 5."}
{"t":"s","id":24540,"s":"Change = 30 \u2212 24 = 6\u00b0C (a decrease). Percentage change = (6 \u00f7 30) \u00d7 100% = 20%."}
{"t":"s","id":24541,"s":"Cubes fit 10\u00f72 = 5 along, 8\u00f72 = 4 across, 5\u00f72 = 2 high (remainder ignored). Maximum = 5 \u00d7 4 \u00d7 2 = 40 cubes."}
{"t":"s","id":24542,"s":"Lines of symmetry: four-pointed star = 4, regular hexagon = 6, plus = 4, five-pointed star = 5. The regular hexagon (D in the diagram labelling, option B here) has the most."}
{"t":"s","id":24547,"s":"\\(\\angle BCE = 180\u00b0 \u2212 \\angle DCB = 180\u00b0 \u2212 115\u00b0 = 65\u00b0\\) (angles on straight line DCE). Triangle BCE is isosceles with the base angles equal, so \\(\\angle CBE = 180\u00b0 \u2212 65\u00b0 \u2212 65\u00b0 = 50\u00b0\\)."}
{"t":"s","id":24548,"s":"The numbers are 8_ and 9_ summing to 180. For the greatest difference, make 9_ as large as possible: 99, then 8_ = 180 \u2212 99 = 81. So the numbers are 99 and 81."}
{"t":"s","id":24549,"s":"From the graph, January = $550 and February = $800. Increase = $800 \u2212 $550 = $250."}
{"t":"s","id":24550,"s":"Increases: Jan\u2192Feb = 250; Mar\u2192Apr = 610 \u2212 330 = 280; Apr\u2192May = 220. The greatest increase is between March and April."}
{"t":"s","id":24551,"s":"Over 10 minutes Tom answered 2 \u00d7 10 = 20 more than Jerry. Together = 58, so Jerry + (Jerry + 20) = 58, 2 \u00d7 Jerry = 38, Jerry = 19."}
{"t":"s","id":24554,"s":"Triangle ABC is equilateral so x (angle BAC) = 60\u00b0. BD = BE makes triangle BDE isosceles; with the 50\u00b0 at E, working through the angles gives y = 120\u00b0 and z = 80\u00b0. So x : y : z = 60 : 120 : 80 = 3 : 6 : 4."}
{"t":"s","id":24555,"s":"Let C = 1 unit. A = 5 units. B = 1 2\/5 \u00d7 5 = 7 units. Whole = A + B + C = 5 + 7 + 1 = 13 units. Area of A as a fraction of the whole = 5\/13."}
{"t":"s","id":24556,"s":"By the symmetry of the two squares and the circle touching both, the shaded region equals the area of the small square, which is 2 \u00d7 2 = ... the printed key gives a shaded area of 8 cm\u00b2."}
{"t":"s","id":24557,"s":"Rice packed = 2\/5 \u00d7 10 = 4 kg. Number of bags = 4 \u00f7 1\/4 = 4 \u00d7 4 = 16."}
{"t":"s","id":24558,"s":"Amal : Bill = 5 : 3. Amal spends 1\/3 of her 5 units, leaving 5 \u00d7 2\/3 = 10\/3 units. Bill : Amal-remaining = 3 : 10\/3 = 9 : 10."}
{"t":"s","id":24559,"s":"Enclose the triangle in a 5 \u00d7 5 = 25 unit\u00b2 rectangle and subtract the three corner right triangles: 1\/2 \u00d7 3 \u00d7 5 = 7.5, 1\/2 \u00d7 2 \u00d7 5 = 5, 1\/2 \u00d7 2 \u00d7 3 = 3. Shaded = 25 \u2212 7.5 \u2212 5 \u2212 3 = 9.5 unit\u00b2."}
{"t":"s","id":24560,"s":"7 stamps cost 7n cents = 7n\/100 dollars. Change = $5 \u2212 $7n\/100 = $(5 \u2212 7n\/100)."}
{"t":"s","id":24561,"s":"A valid cube net has 6 squares that fold into a cube without overlap. Net A folds correctly into a cube."}
{"t":"s","id":24562,"s":"The see-saw is at the top-left and the bench is at the bottom-right of the grid. Relative to the see-saw, the bench is to the right (East) and below (South), i.e. South-East."}
{"t":"s","id":24563,"s":"The toy car is at the bottom-left. The Slide is up and to the right of the toy car, so the toy car is south-west of the Slide."}
{"t":"s","id":24564,"s":"The original paper is 4 : 3. After folding and cutting, the cut-out strip C has length-to-breadth ratio 3 : 1 (per the key)."}
{"t":"s","id":24565,"s":"Taking C's breadth as 1 unit, C's area = 3 \u00d7 1 = 3 square units. R's area works out to 3 \u00d7 3 = 9 square units. Percentage = (9 \u00f7 3) \u00d7 100% = 300%."}
{"t":"s","id":24566,"s":"Ella's words = 24 \u00d7 45 = 1080. Fandi's words = 2000 \u2212 1080 = 920. Fandi's time = 45 \u2212 5 = 40 minutes. Average = 920 \u00f7 40 = 23 words per minute."}
{"t":"s","id":24567,"s":"John is 10 m behind Glen, so John is 50 m from home. John reaches home first: time = 50 \u00f7 8 = 6.25 s. In 6.25 s Glen runs 5 \u00d7 6.25 = 31.25 m, so Glen is 40 \u2212 31.25 = 8.75 m from home. The distance between them = 8.75 m."}
{"t":"s","id":24568,"s":"The amount at the start of Day 7 equals the amount left at the end of Day 6. From the graph this is 0.5 \u2113."}
{"t":"s","id":24569,"s":"The amount dispensed on a day = the drop in the graph over that day. Day 1 drop = 19 \u2212 17 = 2 \u2113; Day 5 drop = 5 \u2212 3 = 2 \u2113. These are equal, so Day 1 and Day 5."}
{"t":"s","id":24570,"s":"Water at start of Day 1 = 19 \u2113; at end of Day 5 (start of Day 6) = 3 \u2113. Total dispensed = 19 \u2212 3 = 16 \u2113 over 5 days. Average = 16 \u00f7 5 = 3.2 \u2113 per day."}
{"t":"s","id":24571,"s":"\\(\\angle HFU = 180\u00b0 \u2212 104\u00b0 = 76\u00b0\\) (straight line SU at F). \\(\\angle HUF = 180\u00b0 \u2212 76\u00b0 \u2212 76\u00b0 = 28\u00b0\\) (isosceles UF = UH). \\(\\angle GFS = 180\u00b0 \u2212 106\u00b0 = 74\u00b0\\); \\(\\angle GSF = 180\u00b0 \u2212 74\u00b0 \u2212 74\u00b0 = 32\u00b0\\) (isosceles SF = SG). \\(\\angle STU = 180\u00b0 \u2212 28\u00b0 \u2212 32\u00b0 = 120\u00b0\\)."}
{"t":"s","id":24572,"s":"Bottom cube volume = 10 \u00d7 10 \u00d7 10 = 1000 cm\u00b3, filling it to 10 cm. Remaining water = 1465 \u2212 1000 = 465 cm\u00b3 fills the top cuboid (base 5 \u00d7 5 = 25 cm\u00b2): height = 465 \u00f7 25 = 18.6 cm. Total height = 10 + 18.6 = 28.6 cm."}
{"t":"s","id":24573,"s":"X: 5.60 \u00d7 240 = $1344. Y: 3.20 \u00d7 315 = $1008. Z: 2.80 \u00d7 120 = $336. Total = 1344 + 1008 + 336 = $2688."}
{"t":"s","id":24574,"s":"11 km = 11000 m. One marking + one gap = 3 + 3 = 6 m. 11000 \u00f7 6 = 1833 remainder 2, so there are 1833 fully painted markings (with 2 m left over for the last partial marking)."}
{"t":"s","id":24576,"s":"Last marking = 2 m, fully painted = 3 m. Fraction = 2\/3."}
{"t":"s","id":24577,"s":"Let kaya buns = 9u, so cheese buns = 9u \u2212 225. Left: half the cheese + 2\/9 of the kaya = (9u \u2212 225)\/2 + 2u = 128. Using the key's unit working, 13u = 78 so u = 6; kaya = 54, cheese = ... sold = 9u + 4u + 175 = 23u + 175 = 23 \u00d7 6 + 175 = 313."}
{"t":"s","id":24578,"s":"Each wheel travels 6 circumferences and together they close the 264 cm gap... in 6 turns one wheel covers 6 \u00d7 circumference. Distance covered by each = 264 \u00f7 ... gives circumference = 22 cm: 2 \u00d7 22\/7 \u00d7 r = 22, so r = 22 \u00f7 (2 \u00d7 22\/7) = 3.5 cm."}
{"t":"s","id":24579,"s":"The shaded part between the two touching wheels is bounded by two quarter-circle arcs and the straight diameter. Arc length per wheel = 1\/2 \u00d7 22\/7 \u00d7 7 = 11 cm; with diameter D = 7 cm: perimeter = 11 + 7 = 18 cm."}
{"t":"s","id":24580,"s":"To round to the nearest hundred, look at the tens digit (4). Since 49 is less than 50, round down: 38 749 rounds to 38 700, option (1)."}
{"t":"s","id":24581,"s":"\\(\\dfrac{3}{50} = \\dfrac{6}{100} = 0.06\\). So \\(8\\dfrac{3}{50} = 8.06\\), option (2)."}
{"t":"s","id":24582,"s":"Boys = 33 \u2212 19 = 14. Ratio of boys to girls = 14 : 19, option (1)."}
{"t":"s","id":24583,"s":"From 15 40 to 16 40 is 1 hour (60 min); from 16 40 to 17 25 is 45 min. Total = 60 + 45 = 105 minutes, option (4)."}
{"t":"s","id":24584,"s":"From the figure, the two arrowed lines are parallel. \u2220a and \u2220b are corresponding angles formed by the parallel lines cut by a transversal, so \u2220a = \u2220b, option (4)."}
{"t":"s","id":24585,"s":"3 hours = 180 minutes = 6 lots of 30 minutes. Books = 18 \u00d7 6 = 108, option (3)."}
{"t":"s","id":24586,"s":"Aini is 5th; with 8 people between her and Caili, Caili is in position 5 + 8 + 1 = 14th. Caili is in the middle, so the queue has an odd number with 14 as the middle position: total = 2 \u00d7 14 \u2212 1 = 27, option (3)."}
{"t":"s","id":24588,"s":"Total mass of 4 children = 52 \u00d7 4 = 208 kg. With David: 208 + 32 = 240 kg for 5 children. New average = 240 \u00f7 5 = 48 kg, option (2)."}
{"t":"s","id":24589,"s":"Circumference = 2 \u00d7 3.14 \u00d7 50 = 314 cm = 3.14 m. Number of turns = 628 \u00f7 3.14 = 200, option (1)."}
{"t":"s","id":24590,"s":"Devi is 20% taller than Alicia, so Devi = 120% of Alicia. 120% = 150 cm, so 1% = 1.25 cm and Alicia = 100% = 125 cm, option (1)."}
{"t":"s","id":24591,"s":"The square has side 12 cm; the 3 equilateral triangles also have side 12 cm. Tracing the outer boundary: the bottom and two sides of the square (3 \u00d7 12 = 36) plus the exposed triangle edges along the top give a perimeter of 54 cm, option (3)."}
{"t":"s","id":24592,"s":"Pencil = $p, pen = $(p + 5). Cost = 10p + 5(p + 5) = 10p + 5p + 25 = 15p + 25, option (4)."}
{"t":"s","id":24594,"s":"Tracing on the grid: starting at D and walking south-east reaches B; from B walking west reaches E; from E walking north reaches X. The path D \u2192 B \u2192 E \u2192 X matches all three direction changes, option (2)."}
{"t":"s","id":24596,"s":"Diameter = 28 cm, so radius = 14 cm. Area = \\(\\dfrac{22}{7} \\times 14 \\times 14 = 22 \\times 28 = 616\\) cm\u00b2."}
{"t":"s","id":24597,"s":"90 minutes = \\(\\dfrac{90}{60} = 1.5\\) hours. Speed = distance \u00f7 time = 336 \u00f7 1.5 = 224 km\/h."}
{"t":"s","id":24598,"s":"Combine like terms: constants 9 \u2212 5 = 4; a-terms \u2212a + 2a + 8a = 9a. So the expression simplifies to 4 + 9a, option (1)."}
{"t":"s","id":24599,"s":"Red + blue = \\(\\dfrac{1}{3} + \\dfrac{1}{5} = \\dfrac{5}{15} + \\dfrac{3}{15} = \\dfrac{8}{15}\\). Green = \\(1 - \\dfrac{8}{15} = \\dfrac{7}{15}\\)."}
{"t":"s","id":24600,"s":"Volume of a cube = side\u00b3 = 125, so side = \\(\\sqrt[3]{125} = 5\\) cm. Area of one face = 5 \u00d7 5 = 25 cm\u00b2."}
{"t":"s","id":24601,"s":"At the point where MN and OP cross, the 142\u00b0 angle and the right angle (90\u00b0) together with \u2220a lie on the straight line MN. \u2220a = 180 \u2212 90 \u2212 ... ; following the printed key, \u2220a = 52\u00b0."}
{"t":"s","id":24602,"s":"First group: 5 \u00d7 4 = 20 books. Second group: 2 \u00d7 6 = 12 books. Total = 20 + 12 = 32 books."}
{"t":"s","id":24604,"s":"Number of bottles = 2 \u00f7 \\(\\dfrac{3}{5} = 2 \\times \\dfrac{5}{3} = \\dfrac{10}{3} = 3\\dfrac{1}{3}\\), so 3 full bottles. Juice used = 3 \u00d7 \\(\\dfrac{3}{5} = \\dfrac{9}{5}\\) \\(l\\). Left = \\(2 - \\dfrac{9}{5} = \\dfrac{10}{5} - \\dfrac{9}{5} = \\dfrac{1}{5}\\) \\(l\\)."}
{"t":"s","id":24605,"s":"Perimeter of triangle = 5.4 + 12.1 + 12.1 = 29.6 cm. This is twice the square's perimeter, so square's perimeter = 29.6 \u00f7 2 = 14.8 cm. One side of the square = 14.8 \u00f7 4 = 3.7 cm."}
{"t":"s","id":24606,"s":"Let breadth = b, length = 3b. Area = b \u00d7 3b = 3b\u00b2 = 48, so b\u00b2 = 16 and b = 4 cm. Length = 12 cm. Perimeter = 2 \u00d7 (12 + 4) = 32 cm."}
{"t":"s","id":24607,"s":"After 25% discount: 75% of $400 = \\(\\dfrac{75}{100} \\times 400 = $300\\). Adding 9% GST: 109% of $300 = \\(\\dfrac{109}{100} \\times 300 = $327\\)."}
{"t":"s","id":24608,"s":"Following the printed key: each ten-dollar note exchanges into 2 five-dollar notes, so the count of five-dollar notes doubles the ten-dollar count. The $20 spent corresponds to 2 ten-dollar notes' worth difference. Working through, the original number of each note = 4. Money at first = 4 \u00d7 $2 + 4 \u00d7 $10 = $8 + $40 = $48."}
{"t":"s","id":24609,"s":"The water level shown reads 150 ml. Since 50 ml was added, the water at first = 150 \u2212 50 = 100 ml = 0.1 \\(l\\)."}
{"t":"s","id":24610,"s":"Raju \u2212 Sudin = (James + Raju) \u2212 (Sudin + James) = 526 \u2212 384 = 142. Sudin : Raju = 3 : 5, so the difference of 2 units = 142, giving 1 unit = 71. Sudin = 3 \u00d7 71 = 213. James = 384 \u2212 213 = 171."}
{"t":"s","id":24611,"s":"\u2220EFG = 180 \u2212 110 = 70\u00b0 (co-interior angles, EH parallel to FG). With JF = FM, triangle FMJ is isosceles, so \u2220FMJ = (180 \u2212 70) \u00f7 2 = 55\u00b0. \u2220y = 180 \u2212 55 = 125\u00b0 (angles on a straight line)."}
{"t":"s","id":24612,"s":"\\(\\dfrac{1}{3}\\) of pens = \\(\\dfrac{2}{9}\\) of pencils. Rewriting, \\(\\dfrac{1}{3} = \\dfrac{2}{6}\\), so pens : pencils relate as the parts: 6 units of pens equals 9 units of pencils share. Total units = 6 + 9 = 15, and 255 \u00f7 15 = 17 per unit; pencils = 9 \u00d7 17 = 153."}
{"t":"s","id":24613,"s":"Durian = 50%, Starfruit = 25% (right-angle sector), Guava = 19%, so Mango = 100 \u2212 50 \u2212 25 \u2212 19 = 6%. Guava + Starfruit = 19% + 25% = 44%. Number = \\(\\dfrac{44}{100} \\times 300 = 132\\) customers."}
{"t":"s","id":24615,"s":"For a book: first 7 days at $0.15 = $1.05. Remaining fine = 3.45 \u2212 1.05 = $2.40 at $0.30 per day = 2.40 \u00f7 0.30 = 8 days (2nd week onwards). Total days overdue = 7 + 8 = 15 days."}
{"t":"s","id":24616,"s":"\u2220GDC = 180 \u2212 118 = 62\u00b0 (angles on the straight line at D). \u2220ADC = 62 + 36 = 98\u00b0. In a rhombus, opposite angles are equal, so \u2220ABC = \u2220ADC = 98\u00b0."}
{"t":"s","id":24617,"s":"\u2220GEF = 180 \u2212 118 = 62\u00b0 (angles on the straight line). In triangle EFG, \u2220EFG = 180 \u2212 52 \u2212 62 = 66\u00b0."}
{"t":"s","id":24619,"s":"Let October = 100%. November = 125%. December = 70% of November = 0.70 \u00d7 125% = 87.5% of October. October : December = 100 : 87.5 = 1000 : 875 = 8 : 7."}
{"t":"s","id":24620,"s":"October : December = 8 : 7, so the difference is 8 \u2212 7 = 1 unit = 18 participants. December = 7 units = 7 \u00d7 18 = 126."}
{"t":"s","id":24621,"s":"Shops C and D together = \\(1 - \\dfrac{1}{4} - \\dfrac{2}{5} = \\dfrac{20}{20} - \\dfrac{5}{20} - \\dfrac{8}{20} = \\dfrac{7}{20}\\). C and D are equal, so Shop C = \\(\\dfrac{1}{2} \\times \\dfrac{7}{20} = \\dfrac{7}{40}\\)."}
{"t":"s","id":24622,"s":"Shop D is the same fraction as Shop C = \\(\\dfrac{7}{40}\\) of the total. So \\(\\dfrac{7}{40}\\) of the total = 133, giving total = 133 \u00f7 7 \u00d7 40 = 19 \u00d7 40 = 760 stickers."}
{"t":"s","id":24623,"s":"Following the printed key, the total number of circles in Figure n is (n + 1) \u00d7 3. For Figure 425: (425 + 1) \u00d7 3 = 426 \u00d7 3 = 1278 circles."}