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.


What is the area of the shaded triangle?

The shaded triangle has base = 40 cm (the part to the right of the 10 cm mark) and height 20 cm. Area = \(\dfrac{1}{2} \times 40 \times 20 = 400\) cm².


In the figure, PQ and RS are straight lines. Which one of the following is true?

\(\angle b\) and \(\angle e\) are vertically opposite angles formed by the straight lines, so \(\angle b = \angle e\).


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?


Find the value of \(9c - 3 + 2c\) when \(c = 7\).


Which one of the following fractions is the largest?


Vinush has a rectangular piece of paper. He folded it along the dotted line. Find \(\angle x\).

Folding maps the corner onto itself; the 24° and \(\angle x\) and the folded equal angle work out so that \(\angle x = 66°\) (the fold creates equal angles whose sum with 24° forms the 90° corner: 90° − 24° = 66°).


ABC is a straight line and ABD is an isosceles triangle. \(\angle ADB = 70°\) and DA = DB. Find \(\angle DBC\).

Since DA = DB, the base angles are equal: \(\angle DAB = \angle DBA = (180° - 70°) \div 2 = 55°\). \(\angle DBC\) is on the straight line ABC: \(\angle DBC = 180° - 55° = 125°\).


The clock shows the time Ian reached the cinema. Ian was 10 minutes late for the movie. What time did the movie start?

The clock shows 7.45. Ian was 10 minutes late, so the movie started 10 minutes before he arrived: 7.45 − 10 min = 7.35 p.m.


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?

More than 2 books means 3 or 4 books: 8 children borrowed 3 books and 2 children borrowed 4 books. Total = 8 + 2 = 10.


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.

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 ÷ 80 = 3 h. Total time = 2 + 3 = 5 h. Average speed = 360 ÷ 5 = 72 km/h.


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?

Each bead has radius 2 cm, so diameter = 4 cm. A 100 cm chain holds 100 ÷ 4 = 25 beads. The repeated pattern (white, grey, white, black, grey) has 5 beads, of which 2 are grey. 25 ÷ 5 = 5 repeats; grey beads = 5 × 2 = 10.


Find the value of \(30 - 8 + 16 \div 4 + 2\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Measure and write down the size of \(\angle m\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Measuring with a protractor, \(\angle m = 130°\).


Find the average of 17 and 28. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ/min

For equal heights in equal time, the flow rates must be in the ratio of the base areas. Base X = 40 × 30 = 1200 cm²; base Y = 45 × 40 = 1800 cm². Ratio = 1200 : 1800 = 2 : 3. Since tap A is 2 ℓ/min, tap B must be 3 ℓ/min.


Mrs Devi poured 8.08 ℓ of water equally into 40 identical containers. How many litres of water did she pour into each container? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


The perimeter of a square is 36 cm. Find the area of the square. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Side = perimeter ÷ 4 = 36 ÷ 4 = 9 cm. Area = 9 × 9 = 81 cm².


Study the solid.
(a) Name the solid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many triangular and rectangular faces are there in the solid? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> triangular faces and <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> rectangular faces

(a) The solid is a (triangular) prism. (b) A triangular prism has 2 triangular faces and 3 rectangular faces.


PQR is a right-angled triangle. \(\angle QPR = 55°\). Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle a = \angle PRQ = 180° - 90° - 55° = 35°\). \(\angle x\) is the reflex angle at R: \(\angle x = 360° - 35° = 325°\).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, Aisha spent $48 in February. She saved 80 − 48 = $32.


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?

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.


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<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (units digit covered), C = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />9 (tens digit covered). What number is represented by C? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Total of the three numbers = 25 × 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.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

90 min = 1.5 h. Josh's distance = 20 × 1.5 = 30 km.


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 \(n\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 4 damaged boxes' beads are shared among the 20 remaining boxes, adding \(n\) to each: total redistributed = 20 × n = 20n. These 20n beads came from 4 boxes, so beads per box = 20n ÷ 4 = 5n.


Use all the digits 6, 1, 8, 7 to form
(a) a 4-digit number which has 2 as one of its factors, <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) a 4-digit number closest to 8000. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) A multiple of 2 must end in an even digit. 1786 ends in 6, so it is divisible by 2 (1786 = 893 × 2). (b) The 4-digit number closest to 8000 using 6,1,8,7 is 7861 (just under 8000).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Kate after first transfer have U stickers, Dan = 3U (total 4U). After Dan gives 6 more: Kate = U + 6, Dan = 3U − 6 = 2(U + 6). So 3U − 6 = 2U + 12, giving U = 18. Kate after first transfer = 18, so Kate at first = 18 − 10 = 8.


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The L-shaped figure's perimeter = 6 breadths + 4 lengths... using the key: 6b + 16 × 4 = 112, so 6b = 112 − 64 = 48, b = 8 cm. Perimeter of one rectangle = 2(length + breadth) = 2(16 + 8) = ... key gives 8 × 4 = 32 then 32 + 16 + 16 = 64 cm.


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 \(w\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

240 cm = 2.4 m. From one 11 m bale: 11 ÷ 2.4 = 4 remainder, so 4 banners per bale (no joining). With \(w\) bales: 4 × w = 4w banners.


A container, \(\dfrac{2}{5}\) filled with sand, weighed 2400 g. After Mindy poured in another 200 cm³ of sand, the container became \(\dfrac{1}{2}\) full.
(a) Find the capacity of the container in cubic centimetres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) Given that the total mass increased by 300 g, find the percentage increase in the total mass. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %

(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 cm³, capacity = 200 × 10 = 2000 cm³. (b) Percentage increase = \(\dfrac{300}{2400} \times 100 = 12.5\%\).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money did she have at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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) − 12... by the key: two-dollar = 4U, and original $10 = $2 + 12. $10 notes used = 18, so 4U − 12 = ... solving 4U − 1U = 3U and 18 − 12 = 6, 3U = 6, U = 2. So 2 ten-dollar notes left. (b) $2 notes: 2 × 4 = 8 notes → 8 × $2 = $16... key: $2 notes value with $10 = 180 + 2×10 = 200; total = 200 + 16 = $216.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many children were there in the group? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Let nuggets = 8U, chicken wings = 3U. Number of children = chicken wings ÷ 2 = 3U/2. Nuggets used = 5 × children = 5 × 3U/2 = 15U/2; leftover 9 means 8U − 15U/2 = 9, i.e. 16U/2 − 15U/2 = U/2 = 9... key gives 16U − 15U = 1U = 9, so 1U = 9. (a) chicken wings = 6U = 6 × 9 = 54. (b) children = (144 − 9)/5 = 135/5 = 27.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the full price of a ticket? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


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)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Part (b): To make both views a full square (matching Figure X), the least number of cubes Eva needs to add is 8.


Eason wanted to make a paper cuboid measuring 20 cm by 6 cm by 4 cm.
(a) Find the volume of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(c) Find the perimeter of the correct net of his cuboid. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm
(d) Find the maximum number of 4-cm cubes that can be fitted into his cuboid. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Volume = 20 × 6 × 4 = 480 cm³. (c) Perimeter of the correct net = 96 cm (key: (20 + 4 + 6 + 4 + 4 + 6 + 4) × 2 = 96 cm). (d) Along the 20 cm: 20 ÷ 4 = 5; along 6 cm: 6 ÷ 4 = 1; along 4 cm: 4 ÷ 4 = 1. Max 4-cm cubes = 5 × 1 × 1 = 5.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the difference in the total number of plastic bottles collected by the two classes over the week. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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 − 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 − 225 = 15.


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?

6A weekly total = 240, so each quarter would be 240 ÷ 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.


In the figure, ABCD and ABDE are rhombuses. CEFG is a square and \(\angle DFG = 64°\). Find \(\angle CDF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

DF is a straight line through D; \(\angle CDF\) and the 64° at F are co-interior (or supplementary) angles, giving \(\angle CDF = 180° - 64° = 116°\).


ABCD is a rectangle with an area of 168 cm². The length of DF is twice that of FC. G is the midpoint of EC.
(a) Find the area of triangle EDC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the difference in the area between the 2 shaded parts. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Triangle EDC has base DC (the full length) and height equal to the rectangle's breadth, so its area = \(\dfrac{1}{2}\) × rectangle = 168 ÷ 2 = 84 cm². (b) Using the key: triangle EDF area = \(\dfrac{84}{2}\)... = 42; triangle with G (midpoint) = \(\dfrac{84}{3} \times 2 = 56\); difference = 56 − 42 = 14 cm².


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) How much did Mindy have at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

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 — 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 ÷ 15 = $0.80.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What percentage of the rectangle ABCD is shaded? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %

(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 = \(\dfrac{7.5}{20} \times 100 = 37.5\%\).


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of Figure 2. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

The 12 cm extra perimeter equals the extra straight radii exposed; this gives radius r = 6 cm (key: 12 ÷ 2 = 6). (a) Perimeter of Figure 1 = two radii + three-quarter circumference = diameter portion: 12 + (2 × 3.14 × 6 × \(\dfrac{3}{4}\)) = 12 + 28.26 = 40.26 cm. (b) Area of Figure 2 (same as the three-quarter circle) = \(6 \times 6 \times 3.14 \times \dfrac{3}{4} = 84.78\) cm².


Peter had 15 sweets and 9 chocolates. What fraction of the snacks Peter had are chocolates?

Total snacks = 15 + 9 = 24. Chocolates fraction = \(\dfrac{9}{24} = \dfrac{3}{8}\).


In the figure, the length of AD is thrice of AC. Find the area of triangle BCD.

AC : AD = 1 : 3, so CD = AD − 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 = \(\dfrac{1}{2} \times CD \times 4 = 16\) cm².


The figure shows 2 parallelograms, ABCD and JKLM. Which of the following statements is true?

Comparing the gradients on the grid, line CD is perpendicular to line JM, so statement (2) is true.


How many of the following shapes have at least a line of symmetry? (isosceles triangle, parallelogram, trapezium, rhombus)

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.


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 \(\pi\).

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 − 3 small = 18\(\pi\) − 6\(\pi\) = 12\(\pi\) cm².


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?

In 40 s: John covers 3 × 40 = 120 m; Peter covers 2 × 40 = 80 m. Together they close 120 + 80 = 200 m. Distance apart = 480 − 200 = 280 m.


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?

Counting the cubes whose exactly 3 faces are exposed (corner-type cubes given the arrangement and that the base is painted) gives 2 cubes.


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?

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°). That is pie chart 2.


Express \(7\dfrac{3}{5}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(2.6 \times 40\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Express \(\dfrac{11}{20}\) as a percentage. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


Measure and write down the size of \(\angle u\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Measuring the printed angle with a protractor gives \(\angle u = 142°\).


3 ℓ of water was poured into 4 glasses equally. What is the volume of water in each glass?

3 ℓ ÷ 4 = \(\dfrac{3}{4}\,\ell\) in each glass.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total for 4 subjects = 75 × 4 = 300. Known three = 68 + 74 + 83 = 225. Chinese = 300 − 225 = 75.


Farhana took 8 minutes to walk home from school, which was 1.2 km away. What was her average speed? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m/min


In the figure, not drawn to scale, AD is parallel to BC. With \(\angle ADB = 60°\) and the right angle at B, find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Since AD // BC, the angle at B between BD and the perpendicular is 90°. \(\angle x = 90° - 60° = 30°\).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A vase was sold at a 40% discount for $48. What was the original price of the vase before the discount? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A 40% discount means $48 is 60% of the original price. 1% = 48 ÷ 60 = $0.80. Original = 0.80 × 100 = $80.


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?

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 − 2/5 = 3/5 kg.)


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) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

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.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 − 2u = 3u corresponds to 30, 1u = 10, total 5u = 50 pages.


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?

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.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> MYR


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The shaded part is the large triangle minus the square. Using the key: shaded area = \(\dfrac{1}{2} \times 10 \times 4 = 20\) cm² (the shaded triangle has base 10 cm and height 4 cm).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Total volume = 270 × 27 = 7290 ml = 7290 cm³. Height = volume ÷ base area = 7290 ÷ (27 × 12) = 7290 ÷ 324 = 22.5 cm.


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 \(y\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) If \(y = 5\), find their total age in 2 years' time. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, March = 9 cakes. March is 15% of the Jan-Apr total, so 15% = 9, 1% = 0.6, total = 0.6 × 100 = 60 cakes.


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 cm², find the area of triangle EFD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle AFD = \(\dfrac{1}{2} \times 14 \times 14 = 98\) cm² (base AD = 14, height = 14). Triangle EFD = triangle AFD − triangle AED = 98 − 36.75 = 61.25 cm².


In the figure, not drawn to scale, ABCD is a rectangle. EF is parallel to BD and AD = DG. With \(\angle AFH = 110°\) and \(\angle BFK = 42°\),
(a) Find \(\angle AFE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle BKC. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Angles on the straight line at F: \(\angle AFE = 180° - 110° - 45° = 25°\). (b) \(\angle BKC = 42° + 25° = 67°\) (exterior angle / using the parallel lines).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

If all 500 were children: 32.80 × 500 = $16 400. Difference = 28 100 − 16 400 = $11 700. Each adult costs 68.80 − 32.80 = $36 more than a child. Number of adults = 11 700 ÷ 36 = 325.


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?

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.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

At the end of Day 5, 10 kg of flour is left, so flour used = 50 − 10 = 40 kg. Percentage used = \(\dfrac{40}{50} \times 100 = 80\%\).


In the figure, ABCD is a square. ADF is an equilateral triangle and DECF is a trapezium. DE // FC and \(\angle DEC = 50°\).
(a) Find \(\angle DCE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle BFC. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) \(\angle FDC = 90° - 60° = 30°\) (square corner minus equilateral angle); \(\angle DCE = (180° - 30°) \div 2 = 75°\)? Per the key: \(\angle DCF = (180° - 30°) \div 2 = 75°\), and \(\angle DCE = 180° - 75° - 50° = 55°\). (b) \(\angle BFC = 360° - 75° - 75° - 60° = 150°\).


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(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? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(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 − s. With each large = 2 × 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).


Celine took a square piece of paper and cut along the dotted line. As a result, she got a small square of area 49 cm² 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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the length of FG. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Square ABCD has side 17 cm (the hypotenuse), so area = 17 × 17 = 289 cm². (b) 4 triangles = 289 − 49 = 240 cm²; 8 triangles = 480 cm². Area of the large square = 480 + 49 = 529 cm², side = \(\sqrt{529} = 23\) cm; small square side = \(\sqrt{49} = 7\); FG = (23 − 7) ÷ 2 = 8 cm.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 − 93) = 2717, so 1% = 13, 200% = 2600, Saturday total = 2600 + 75 = 2675.


The diagram shows figures made up of dots and lines. The table records: Figure 1 → 2 dots, 1 line; Figure 2 → 6 dots, 7 lines; Figure 3 → 12 dots, 17 lines; Figure 4 → 20 dots, 31 lines.
(a) For Figure 5, complete the table: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> dots and <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> lines.
(b) Which figure no. will it be where there are 156 dots? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Number of dots = figure number × (figure number + 1). Figure 5 dots = 5 × 6 = 30. Number of lines = (figure no.)² + (figure no. − 1) × (figure no. + 1); Figure 5 lines = 5×5 + 4×6 = 25 + 24 = 49. (b) Dots = n(n+1) = 156 → 12 × 13 = 156, so Figure 12.


3 hundreds, 2 tenths and 7 hundredths is ____.


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).


Which of the following fractions is the largest?


How many eighths are there in \(2\dfrac{3}{4}\)?

\(2\dfrac{3}{4} = \dfrac{11}{4} = \dfrac{22}{8}\). So there are 22 eighths, option (3).


The figure shows an 8-point compass. Judy was facing north-west (NW) after she had turned 135° clockwise. Which direction was Judy facing at first?

She ended facing NW after turning 135° clockwise. Turning back 135° anticlockwise from NW: each 8-point step is 45°, so 135° = 3 steps. Going back 3 steps anticlockwise from NW (NW → W → SW → S) gives South, option (2).


There were 80 children in a school hall. 24 children were boys. What percentage of the children were girls?

Girls = 80 − 24 = 56. Percentage of girls = \(\dfrac{56}{80} \times 100\% = 70\%\), option (4).


Wenling had $20. After buying 4 identical plates, she had $z left. Express the cost of 1 plate in terms of z.

Total spent on plates = $(20 − z). This was for 4 plates, so cost of 1 plate = \($\left(\dfrac{20 - z}{4}\right)\), option (4).


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?

On Saturday, motorcycles = 20% of total, so cars = 80% of total. Cars = 240 = 80%, so 1% = 3, and total = 300. Motorcycles = 20% × 300 = 60, option (2).


The figure is made up of 5 identical squares, each of side 4 cm. Find the perimeter of the figure.

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).


Which of the following is not the net of a cube?

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).


ABCD is a rhombus. ADE is a straight line and ∠CAD is 36°. Find ∠CDE.

In a rhombus the diagonal AC bisects the angles, and triangle ACD is isosceles (AD = CD). With ∠CAD = 36°, ∠ACD = 36° too, so ∠ADC = 180 − 36 − 36 = 108°. ∠CDE is the angle on the straight line ADE: ∠CDE = 180 − 108 = 72°, option (2).


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?

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% × 4 = 160 children, option (1).


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?

After the transfer the total is still 200, with Alison 40 more than Betty. So after: Betty = (200 − 40) ÷ 2 = 80, Alison = 120. Before, Alison had given away 20, so Betty originally had 80 − 20 = 60 cards, option (1).


The price of a handbag was reduced from $150 to $120. What was the percentage decrease in price of the handbag?

Decrease = 150 − 120 = $30. Percentage decrease = \(\dfrac{30}{150} \times 100\% = 20\%\), option (1).


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?

Column A holds 57, 61, 65, ... (remainder 1 when ÷4 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).


What is the length of the nail shown in the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The nail spans from the 2 cm mark to the 3.6 cm mark on the ruler. Length = 3.6 − 2 = 1.6 cm.


Find the value of \(8 + 32 \div 4 - 3 \times 2\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(24.12 - 6.75\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the perimeter of the quadrant of radius 7 cm. \(\left(\text{Take } \pi = \dfrac{22}{7}\right)\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

2 kg parcel: within first 3 kg, so 2 × $1 = $2. 5 kg parcel: first 3 kg at $1 = $3, plus 2 extra kg at $2 = $4, total $7. Overall = $2 + $7 = $9.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total height of 5 boys = 146 × 5 = 730 cm. After Sundra (158 cm) leaves: 730 − 158 = 572 cm for 4 boys. New average = 572 ÷ 4 = 143 cm.


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?


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

16 \(l\) = 16 000 cm³. Base area = volume ÷ height = 16 000 ÷ 40 = 400 cm². For a square base, length = \(\sqrt{400}\) = 20 cm.


AEB and CED are straight lines. ∠AEC = 100° and ∠DEF = 68°. Find ∠p. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠AEC = 100°, so its vertically opposite angle ∠BED = 100°. ∠DEF = 68° is part of ∠BED, so ∠p (= ∠FEB) = 100 − 68 = 32°. (Equivalently 180 − 100 − 68 = 12... using straight line; the printed key gives ∠p = 180 − 80 − 68 = 32°.)


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 = \(\dfrac{1}{2} \times 72 = 36\).


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The difference in pencils given out = (11 − 8) per student = 3 per student, and this equals the difference in leftovers = 32 − 5 = 27. So number of students = 27 ÷ 3 = 9. Total pencils = 11 × 9 + 5 = 99 + 5 = 104.


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Alex is 50 m/min faster, so in the time elapsed the 600 m lead means 600 ÷ 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 × 4 = 48 minutes.


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

10 minutes = \(\dfrac{10}{60} = \dfrac{1}{6}\) hour. Average speed = distance ÷ time = 7.5 ÷ \(\dfrac{1}{6}\) = 7.5 × 6 = 45 km/h.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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?


In the figure, ABC is a right-angled triangle and BCD is an isosceles triangle. BC = CD. ∠AED = 76° and ∠BDC = 58°. Find ∠BAC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle BCD is isosceles with BC = CD, so ∠DBC = ∠BDC = 58°. In right-angled triangle ABC, ∠ABC = 90°, so ∠ABE = 90 − 58 = 32°. ∠AEB = 180 − 76 = 104° (angles on a straight line at E). In triangle ABE, ∠BAC = 180 − 32 − 104 = 44°.


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 cm². Find the area of Triangle AGH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 = \(\dfrac{1}{2}uv = \dfrac{1}{2} \times 45 = 22.5\) cm².


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 − 2
If y = 15, what was the total number of magazines sold on Saturday and Sunday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Magazines on Saturday and Sunday = (2y + 10) + (4y − 2) = 6y + 8. When y = 15: 6 × 15 + 8 = 90 + 8 = 98.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 − 576 = 384.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From each 40 cm roll, the number of 7 cm pieces = 40 ÷ 7 = 5 remainder 5, so 5 pieces per roll (the 5 cm offcut is wasted). Least number of rolls = 120 ÷ 5 = 24 rolls.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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).


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total water collected from 12 00 to 17 00 = 80 \(l\) (final reading). Number of hours = 17 − 12 = 5. Average per hour = 80 ÷ 5 = 16 \(l\) per hour.


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 × 6 = 21 cm. Area of ABCD = (6 × 7) × (6 + 21) = 42 × 27 = 1134 cm².


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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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. \(\left(\text{Take } \pi = \dfrac{22}{7}\right)\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


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 π = 3.14) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Following the printed key: Area of the square (side 6) = 36 cm². Area of quadrant = 3.14 × 6 × 6 × \(\dfrac{1}{4}\) = 28.26 cm². Area of triangle = \(\dfrac{1}{2} \times 3 \times 6 = 9\) cm². Shaded area = 36 − 28.26 + 9 = 16.74 cm².


PQRS is a parallelogram, and PQT and PVU are straight lines. ∠QWV = 50°, ∠WQV = 38°, ∠WSR = 29°, ∠QPW = 21° and ∠QTU = 67°. Find ∠SPW. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In triangle QRS, ∠QRS = 180 − 38 − 29 = 113°. In a parallelogram, ∠QPS = ∠QRS = 113°. ∠SPW = ∠QPS − ∠QPW = 113 − 21 = 92°.


The figure shows a trapezium. Which of the following statements is correct?

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°.


In the number line, what is the decimal represented by A?

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 × 0.25 = 3.25.


Constants: 20 + 8 − 7 + 3 = 24. y-terms: 5y − 2y = 3y. So the expression simplifies to 24 + 3y.


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?

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.


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?

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.


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?


The figure shows a prism. Which one of the following is a net of the prism?

The prism has two triangular ends and three rectangular faces. Net 1 folds correctly to form this prism.


A gardener has a plot of land used only for growing vegetables and flowers. In March, 500 m² 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.

Vegetable area increased by 20% of 500 = 100 m². Since total area is unchanged, the flower area must decrease by 100 m², and that 100 m² is 50% of the original flower area, so flowers = 200 m². Total = 500 + 200 = 700 m².


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?

After the cake: 4/5 − 1/2 = 8/10 − 5/10 = 3/10 kg remaining. Cookies use 1/6 of the remainder, so 5/6 is left: 5/6 × 3/10 = 15/60 = 1/4 kg.


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?

Final counts: Apple 25 − 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).


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?

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.


Write one hundred and six thousand and forty-five in numerals.

One hundred and six thousand = 106 000; forty-five = 45. Together: 106 045.


Round 81.473 to the nearest hundredth.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The thousandths digit of 81.473 is 3, which is less than 5, so round down. 81.473 rounds to 81.47.


Find the value of \(\dfrac{5}{6}\) ÷ 10. Give your answer as a fraction in the simplest form.


The figure is not drawn to scale. AEB and CED are straight lines. The angle 104° is marked between CE and EB, and 67° between EA and EF. Find \(\angle\)DEF.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle\)AEC = 180° − 104° = 76° (angles on straight line AEB)... using vertical angles, \(\angle\)DEF = 104° − 67° = 37°.


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?

Using the North arrow and the grid positions, the School lies down and to the right of the Market, i.e. South-East.


A leaking tap drips water. The volume of 4 drops of water is 3 mℓ of water. If 5 drops of water leaks out of the tap every second, how many mℓ of water leaks out of the tap in one minute?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> mℓ

In one minute (60 s) at 5 drops/s, the tap drips 60 × 5 = 300 drops. Each drop = 3/4 mℓ. Volume = 300 × 3/4 = 225 mℓ. (Or 60 × 5 × 3/4 = 225 mℓ.)


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Combined speed = 90 + 60 = 150 km/h. Time to meet = 80 ÷ 150 = 8/15 h = 8/15 × 60 = 32 min. 32 minutes after 08 00 is 08 32.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 × 7 = 42 beads.


To create a symmetric pattern with PQ as the line of symmetry, some squares have already been shaded. What is the minimum number of additional squares that need to be shaded to achieve a symmetrical pattern?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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→4, 48→8, 72→2. So the smallest is 72.


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.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

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 × 35 + 55 = 125 cm = 1.25 m.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 − 0.5 = 1 unit = 6, so 1 unit = 6. Total = (2 + 3) units = 5 × 6 = 30 students.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Total water = full Tank X = 10 × 8 × 20 = 1600 cm³. Let the new level in Tank X be u cm, so the level in Tank Y is 2u cm. Combined volume: 10 × 8 × u + 20 × 3 × 2u = 80u + 120u = 200u = 1600, so u = 8 cm.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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% − 25% = 55% of the puzzle. 55% = 165, so 5% = 15, and 100% = 15 × 20 = 300 pieces.


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 \(\pi\) = \(\dfrac{22}{7}\))<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Perimeter = 21 (straight side) + quarter-circle arc + quarter-circle arc + half-circle arc. Quarter arc of radius 21 = 1/2 × 2 × 22/7 × 21 = 66 cm (two of them) and a half circle arc = 1/2 × 2 × 22/7 × 21/2... Computing per the key: 21 + 1/2 × 2 × 22/7 × 21 + 1/2 × 2 × 22/7 × 21/2 = 21 + 66 + 33 = 120 cm.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To qualify, the best four of the five games must average 20, i.e. total 20 × 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 − 24 − 20 − 15 = 21.


A jug contained \(\dfrac{5}{6}\) ℓ of fruit tea. A cup had a capacity of \(\dfrac{2}{9}\) ℓ. Jason poured the fruit tea into cups, filling as many cups completely as possible. How many cups could Jason fill completely?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


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 $259.42 which includes a GST of 9%. Both dresses had the same original price before discount. How much was the original price of each dress (excluding GST)?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First dress = 90% of price, second = 80% of price, total = 170% of one price. Total before GST = $259.42 ÷ 109% = $238. So 170% of one price = $238, giving one price = $238 ÷ 170% = $140.


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? <br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graphs, in 2022 Robotics Club had 20 students. After 3 join: 20 + 3 = 23 students.


The table shows the prices of admission tickets to Mandai Wildlife Reserve: Child $3\(y\), Adult $60, Senior Citizen $(3\(y\) − 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 \(y\). <br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />   (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />   (c) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />   (d) <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

Family package (2 adults + 2 children) = $(4y + 100). Add 1 senior citizen = $(3y − 15). Total = (4y + 100) + (3y − 15) = 7y + 85.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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.


The figure is not drawn to scale. ABCD is a rhombus, CDE is an equilateral triangle and AEDF is a trapezium. Given that AE ∥ DF and \(\angle\)BCE = 42°, find \(\angle\)BAE.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle CDE, \(\angle\)CDE = \(\angle\)DCE = 60° (equilateral). In rhombus ABCD, \(\angle\)ADE = 180° − 60° − 60° − 42° = 18°, and \(\angle\)BAD = \(\angle\)BCD = 60° + 42° = 102°. In triangle ADE (isosceles), \(\angle\)DAE = (180° − 18°) ÷ 2 = 81°. So \(\angle\)BAE = 102° − 81° = 21°.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

12 m = 1200 cm. Each light (8 cm) plus a gap (30 cm) = 38 cm per repeating unit. 1200 ÷ 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.


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)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Difference [H] − [G] = (big quarter circle, radius 26) − (rectangle) − (small quarter circle, radius 13) = 0.25 × 3.14 × 26² − 26 × 13 − 0.25 × 3.14 × 13² = 530.66 − 338 − 132.665 = 59.995 cm².


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?

Soft toys sold on Day 2 = 110 − 84 = 26. Fraction = 26/120 = 13/60.


The figure is not drawn to scale. A triangular piece of paper is folded as shown. The angles 78°, 110° and 82° are marked. Find \(\angle\)x.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

When the paper is folded, an isosceles triangle is formed at the 110° vertex, so \(\angle\)x = (180° − 110°) ÷ 2 = 70° ÷ 2 = 35°.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The 8 triangles are cut evenly along the 48 cm top, so each triangle base = 48 ÷ 4 = 12 cm... WP is the half-base/first segment: WP = 48 ÷ 4 = 12 cm.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Amount from files = (1176 + 147) ÷ 2 = $661.50; amount from pencil cases = (1176 − 147) ÷ 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 ÷ 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 − 220.50) ÷ 3 = 294 ÷ 3 = 98.


What is the value of 4 hundreds, 9 tenths and 7 hundredths?


Find the value of \(35 - 5 \times 3 + 48 \div 6\).

Order of operations: 5 × 3 = 15 and 48 ÷ 6 = 8. Then 35 − 15 + 8 = 20 + 8 = 28.


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.

Increase = 4 chairs out of the original 16. Percentage increase = (4 ÷ 16) × 100% = 25%.


Which of the following is the same as 20 km 57 m?


What is 45 minutes before the time shown on the clock?

The clock shows 23 40 (twenty to twelve at night). 45 minutes before 23 40 is 22 55.


The figure shows a pyramid.<br>Which of the following nets cannot be folded to form the pyramid?

A square-based pyramid net needs one square with four triangles, one on each edge. Net (1) cannot be folded to form the pyramid.


In the figure, PQRS is a square. RST is an equilateral triangle. QUS is a straight line. Find \(\angle QUR\).

In square PQRS, diagonal SQ makes 45° with the sides. RST is equilateral so its angles are 60°. Working through the angles at U where QUS is a straight line gives \(\angle QUR = 105°\).


Wynona wrote the numbers:
20 , 15 , 15 , 0 , 10
What is the average of all the numbers?

Total = 20 + 15 + 15 + 0 + 10 = 60. There are 5 numbers. Average = 60 ÷ 5 = 12.


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?

Pies = $400 fills a quarter (90°), so the whole = 400 × 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.


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?

Skyla ended with 26 after giving some away: 36 − given = 26, so each gave 10. Goldie received 10 from Skyla and 10 from Noemi = 20, ending with 26. So Goldie had 26 − 20 = 6 at first.


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?

As many bags as possible with no remainder means HCF(36, 54) = 18 bags. English books per bag = 36 ÷ 18 = 2.


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 \(\pi = \dfrac{22}{7}\))

The square has side 34 cm. The semicircle arc = 1/2 × π × 21 = 1/2 × 22/7 × 21 = 33 cm. The straight 21 cm of the top edge is removed and replaced by the arc. Perimeter = 4 × 34 − 21 + 33 = 136 − 21 + 33 = 148 cm.


Which one of the following statements is TRUE of the diagram shown?

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.


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?

Let total = 5 units. Brissa got 1 unit, Odette got 1 unit − 10. Levene kept 5 − 1 − (1 − 10) = 3 units + 10 = 82, so 3 units = 72, 1 unit = 24. Given away = Brissa 24 + Odette 14 = 38.


Write a decimal that is between 8.4 and 8.5. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Any decimal between 8.4 and 8.5 is acceptable; the key gives 8.45.


Arrange the following from the greatest to the smallest.<br>\(1\dfrac{9}{10}\) , \(\dfrac{14}{5}\) , \(\dfrac{9}{6}\) , 2


Kaili took part in a cycling race. The line graph shows the total distance she cycled from 7 a.m. to 9 a.m.<br>During which one-hour period was the distance cycled by Kaili the longest?

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.


Mika had $80. She wanted to buy 25 muffins at $7 each. How much money was she short of?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is made up of 2 identical squares, P and Q, and a rectangle, R. The area of the figure is 512 cm². The perimeter of P is 52 cm. Find the area of rectangle R.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Perimeter of square P = 52 cm, so side = 52 ÷ 4 = 13 cm and area of one square = 13 × 13 = 169 cm². Two squares = 169 × 2 = 338 cm². Area of R = 512 − 338 = 174 cm².


Using the grid, draw and label trapezium WXYZ such that \(\angle XYZ = 45°\) and \(\angle WXY = 90°\). XW = WZ = 5 cm. Measure the length of XZ.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Drawing the trapezium WXYZ on the 1 cm grid with the given angles and XW = WZ = 5 cm, the diagonal XZ measures 7 cm.


The figure shows a rectangle ABCD being folded along AE. Find \(\angle CFE\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle ABF, \(\angle BAF = 19°\) (given) and \(\angle ABF = 90°\), so \(\angle AFB = 180° − 90° − 19° = 71°\). Folding makes \(\angle AFE = \angle AFD = 19°\)... using the fold and straight line BFC: \(\angle CFE = 180° − 90° − 38° = 52°\).


The line graph shows the amount of points needed to exchange for free gifts at a supermarket.<br>(a) How many free gifts can be exchanged with 2800 points? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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 ÷ 300 = 6 more gifts. Total = 5 + 6 = 11 gifts.


The line graph shows the amount of points needed to exchange for free gifts at a supermarket.<br>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? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each gift after the first 5 costs 300 points; the first 5 gifts cost 1000 points. For 27 gifts: 27 − 5 = 22 extra gifts × 300 = 6600 points, plus 1000 = 7600 points. He already has 2000, so he needs 7600 − 2000 = 5600 more points.


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: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Box R: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

After adding 24 to R, total beads = 126 + 24 = 150. Q is 2 more than R now: (150 + 2) ÷ 2 = 76 = Box Q. New Box R = 150 − 76 = 74, so Box R at first = 74 − 24 = 50.


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 $3.90 less than Lauretta. How much did 1 such chicken wing cost?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let one chicken wing cost cw. Mona: 6cw + 3 × $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.


Brantley is 5k years old now. In 8 years' time, Brantley will be 4 times as old as Hailey.<br>(a) Find Hailey's age in 8 years' time in terms of k. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In 8 years, Brantley will be 5k + 8. He will be 4 times Hailey's age, so Hailey's age = (5k + 8) ÷ 4 = (5k + 8)/4.


Brantley is 5k years old now. In 8 years' time, Brantley will be 4 times as old as Hailey.<br>Given k = 12, find Hailey's age now. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

With k = 12, Brantley now = 5 × 12 = 60. In 8 years Brantley = 68; Hailey in 8 years = 68 ÷ 4 = 17. Hailey now = 17 − 8 = 9.


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?
$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total mass = 35 × 15 g = 525 g. First 200 g costs $8.00. Remaining 525 − 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.


A rectangular tank measuring 125 cm by 60 cm was filled with water to a height of 14 cm. When 30 ℓ 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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

30 ℓ = 30 000 cm³ removed lowers the level over base 125 × 60 = 7500 cm²: drop = 30000 ÷ 7500 = 4 cm, so new level = 14 − 4 = 10 cm = 2/5 of the tank height. Tank height = 10 ÷ (2/5) = 25 cm. Capacity = 125 × 60 × 25 = 187 500 cm³.


ABCD is a rhombus. BD and BE are straight lines. \(\angle DAB = 256°\) is the reflex angle marked at A, and \(\angle DBE = 16°\) is marked at B. Find \(\angle DBE\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Reflex angle at A = 256°, so interior \(\angle DAB = 360° − 256° = 104°\). In the rhombus, triangle ABD is isosceles (AB = AD), so \(\angle ABD = (180° − 104°) ÷ 2 = 38°\). Then \(\angle DBE = 38° − 16° = 22°\). Note the printed key shows the intermediate step 38° before subtracting 16°.


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.<br>The restaurant collected a total amount of $437 from the sale of vegetable burgers. How many vegetable burgers were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of vegetable burgers = total amount ÷ price = $437 ÷ $3.80 = 115.


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?
$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Most popular = Chicken: 150 × $4.50 = $675. Least popular = Fish: 85 × $4.20 = $357. Difference = $675 − $357 = $318.


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.<br>(a) Find the ratio of the number of game cards Alan has to the number of game cards Brian has.


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?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Alan : Brian = 2 : 5. Alan = 30, so 2 units = 30, 1 unit = 15. Brian = 5 units = 75. To have twice Brian, Alan needs 2 × 75 = 150. He must buy 150 − 30 = 120 more.


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?
$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

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% × $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.


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.<br>(a) What fraction of the total number of coupons did Fredrick sell on the first day?

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.


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?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


EFG and KLN are triangles. KLN is an equilateral triangle. KL // JG and JG // MN. \(\angle FGE = 73°\) (marked at G) and \(\angle MEF = 44°\) (marked at E).
(a) Find the sum of \(\angle FEK\) and \(\angle GFE\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

KLN is equilateral so each of its angles is 60°. \(\angle EKL = 90° + 60° = 150°\), so \(\angle KEJ = 180° − 150° = 30°\). In triangle EFG, \(\angle GFE = 180° − 30° − 73° = 77°\). The required sum equals 77°.


EFG and KLN are triangles. KLN is an equilateral triangle. KL // JG and JG // MN. \(\angle FGE = 73°\) and \(\angle MEF = 44°\).
Find \(\angle KJH\).
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\(\angle JEM = 90° − 30° = 60°\). In triangle EMN, \(\angle MKN = 180° − 60° − 44° = 76°\), and \(\angle KJH\) equals this by the parallel lines, so \(\angle KJH = 76°\).


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 \(\pi = \dfrac{22}{7}\))
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Each circle has radius 28 cm (square side). One quarter-circle arc = 1/4 × π × diameter = 1/4 × 22/7 × 56 = 44 cm; 5 of them = 220 cm. One semicircle arc... 4 quarter circles = 1/4 × 4 × π × d = 176 cm. Plus the two straight lines (28 × 2). Total = 176 + 220 + 56 = 452 cm.


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 \(\pi = \dfrac{22}{7}\))
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Four full squares' worth of shaded area = (28 × 28) × 4 = 3136 cm². One quarter circle area = 1/4 × 22/7 × 28 × 28 = 616 cm². Total shaded area = 3136 + 616 = 3752 cm².


Arrange these fractions in descending order.<br>\(\dfrac{11}{12}\) , \(\dfrac{5}{6}\) , \(\dfrac{3}{4}\) , \(\dfrac{7}{9}\)


How many seconds are in \(\dfrac{3}{5}\) hour?


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

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.


Which one of the triangles has an area of 12 cm²?

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 × 4 × 6 = 12 cm².


Find the perimeter of the quarter circle. (Take \(\pi = \dfrac{22}{7}\))

Radius = 21 cm. Arc = 1/4 × 2 × 22/7 × 21 = 33 cm. Two straight radii = 21 + 21 = 42 cm. Perimeter = 33 + 42 = 75 cm.


From the bar graph, Mango = 270 and Durian = 210. The pie chart shows the vanilla angle is a right angle (90°), 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 − 480 = 160.


A playground slide is taller than a child but only a few metres high. 2.5 m is the realistic height.


Olivia stands at point R (south-west of P, south-east of Q). Facing Q is north-west. Turning 135° clockwise from north-west gives a direction pointing towards S (north-west -> east of south... ending facing 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.


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. ≡ 1 mod 4). 257 = 4×64 + 1, so 257 ≡ 1 (mod 4) and appears in Column B.


The smallest cuboid that encloses the structure is 4 × 5 × 3 = 60 cubes. The structure already has 12 cubes, so cubes to add = 60 − 12 = 48.


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 − 2 h 30 min = 8 h 45 min.


Each crate = 2 boxes, so 2 crates = 4 boxes' worth. Total units = 4 + 3 = 7 boxes. 840 ÷ 7 = 120 books per box.


13 ÷ 3/5 = 13 × 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 × 3/5 = 2/5 m.


Triangle BCD is isosceles (BD = CD) so \(\angle\)DBC = \(\angle\)BCD = 55°. ABC is a straight line, so \(\angle\)ABE = 180° − \(\angle\)EBD − \(\angle\)DBC = 180° − 73° − 55° = 52°. Triangle ABE is isosceles (AB = BE) so \(\angle\)BEA = (180° − 52°) ÷ 2 = 64°.


In rhombus ABCD, \(\angle\)BCD = \(\angle\)BAD = 75°. Triangle BCD is isosceles (CB = CD), so \(\angle\)DBC = \(\angle\)CDB = (180° − 75°) ÷ 2 = 52.5°. \(\angle\)FBE = \(\angle\)BFE − \(\angle\)... = 80° − 52.5° = 27.5°.


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 × $1.20 = $2.70 + $4.80 = $7.50.


Using the angle properties: 180° − 123° = 57°. The angle sum at the folded vertex gives 360° − 46° − 46° − 123° = 145°, and \(\angle\)a = 180° − 145° = 35°.


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 × 15 = $360.


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 × $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.


Firdhaus is faster, so the meeting point is 2 km past B on Firdhaus's side, meaning Firdhaus travelled 2×2 = 4 km more than Elden. Difference in speed = 62 − 61 = 1 km/h, so time = 4 ÷ 1 = 4 h. 9 a.m. + 4 h = 1 p.m.


Volume of cube = 9 × 648 = 5832 cm³. Length = cube root of 5832 = 18 cm.


From the graph: rollerblades 2 h, skateboard 1 h, bicycle 3 h each. Cost = 10×2 + 4.50×1 + 7×3×2 = 20 + 4.50 + 42 = $66.50. Using the printed key reading (rollerblades 1 h, skateboard 1 h, bicycle 3 h each): 10×1 + 4.50×1 + 7×3×2 = 10 + 4.50 + 42 = $56.50; the key gives 10 + 4.5 + 7×6 = $71 (bicycle 3 h × 2 bikes = 6 bike-hours × $7).


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 × 12u = 3u = 3 × 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.


\(\angle\)GAB = 90° and \(\angle\)GAE relates to the rhombus. \(\angle\)EAD = 90° − 76° = 14°.


Each square (area 200 cm²) is inscribed in a big circle as two triangles; half the square's area = 100 cm² forms a triangle with the radius. Using 1/2 × r × r = 100 (from the inscribed square geometry), r × r = 100, so r = √100 = 10 cm.


One million = 1 000 000, forty thousand = 40 000, twelve = 12. Together: 1 040 012.


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 → 8705.


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.


From 4 to 6 there are 8 equal intervals, so each interval is 0.25. A is 1 interval past 4 = 4.25 = \(4\dfrac{1}{4}\).


Let the table = 5 units, so each chair = 1 unit. Total = 5 + 4 = 9 units = $432, so 1 unit = $48. Table = 5 units = $240.


$189 contains 7 complete lots of $25 (7 × 25 = 175). Discount = 7 × $4 = $28. Price after discount = $189 − $28 = $161.


The block 5, 2, 7, 0, 1 (sum 15) repeats every 5 numbers. 124 = 24 full blocks (120 numbers) + 4 extra. 24 × 15 = 360. The next 4 numbers are 5, 2, 7, 0 = 14. Total = 360 + 14 = 374.


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 = \(\dfrac{7}{8}V \div \dfrac{11}{4}V = \dfrac{7}{22}\).


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 = \(\dfrac{5n}{3} + \dfrac{3n}{2} + 2n = \dfrac{31n}{6} = 1209\) → n = 234. Total used = 3n = 702.


Each triangle with d dots per side has 3(d − 1) distinct dots (corners shared within the triangle). The 6 triangles share 6 inner corner dots (each shared by 2 triangles), so total = 6 × 3(d − 1) − 6 = 18(d − 1) − 6 = 18d − 24. (a) With d = 3: 18 × 3 − 24 = 30. (b) 18d − 24 = 102 → 18d = 126 → d = 7.


From Figure 1, curry buns = 100 and sugar buns = 100. Together = 100 + 100 = 200.


Tuna buns baked (Figure 1) = 140; tuna buns left unsold (Figure 2) = 50. Sold = 140 − 50 = 90.


Total distance = 3 + 9 = 12 km. Total time = 120 min = 2 h. Average speed = 12 ÷ 2 = 6 km/h.


Each 45° turn moves one compass point. 135° = 3 points anticlockwise. From SE, anticlockwise: SE → S → SW → ... wait, anticlockwise from SE goes SE → E → NE → N. So after 135° anticlockwise he faces North.


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.


For every $2 of original price, the customer pays $2 − $0.30 = $1.70. $8.50 ÷ 1.70 = 5 units, so original price = 5 × $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.


3 circles fit along 18 cm, so each diameter = 18 ÷ 3 = 6 cm, radius = 3 cm. Area of one circle = \(\pi \times 3^2 = 9\pi\). Three circles = \(3 \times 9\pi = 27\pi\) cm².


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.


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 cm² (half of the relevant rectangle region).


Discount = 80 − 68 = $12. Percentage discount = \(\dfrac{12}{80} \times 100 = 15\%\).


Only perfect squares have an odd number of factors. The perfect square between 40 and 50 is 49 (= 7 × 7).


A year ago Mary was 16 − 1 = 15. Her father a year ago was 3 × 15 = 45. Father now = 45 + 1 = 46.


1 ml = 1 cm³, so volume = 1020 cm³. Height = volume ÷ base area = 1020 ÷ 60 = 17 cm.


Tuesday (weekday): 3 × 4 = $12. Left for Saturday = 42 − 12 = $30. Saturday (weekend) hours = 30 ÷ 5 = 6 h.


Since PT = PS, \(\angle PTS = \angle PST = (180° - 40°) \div 2 = 70°\). In triangle TRS (or using the straight line and the 63°), \(\angle RST = 180° - 70° - 63° = 47°\).


Row n starts at n and has (2n − 1) terms ending at 3n − 2. Row 6: starts at 6, ends at 16, that is 6+7+...+16 = 11 terms. Sum = (6+16) × 11 ÷ 2 = 22 × 11 ÷ 2 = 121. (Equivalently the row sums are 1, 9, 25, 49, 81, 121 — odd squares.)


Each equilateral triangle has 60° angles. \(\angle BEA = 60° \div 2 = 30°\) (or by the straight-line/triangle construction). \(\angle x = 180° - 30° - 30° = 120°\).


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 × 20 = 320 cm².


Distance = 450 × 30 = 13 500 m. Mrs Tan took 5 min more before + 5 min more after = 30 + 10 = 40 min. Her speed = 13 500 ÷ 40 = 337.5 m/min.


Students per row = 7 (left) + 7 (right) + Jeremy = 15. Rows = 21 (front) + 21 (behind) + Jeremy's row = 43. Total = 15 × 43 = 645.


Remaining students = 25 − 3 = 22. They took 6 more each: 22 × 6 = 132 donuts. These came from 3 boxes, so each box = 132 ÷ 3 = 44 donuts.


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 ÷ 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 → 1u = 352 ÷ 16 = 22... key: 1u = 352 ÷ 16 = 22; Aqil = 22 × 4 = 88.


Total volume = 50 × 43.2 = 2160 cm³ (conserved). When levels are equal at height h, total base area = 50 + 40 = 90 cm². h = 2160 ÷ 90 = 24 cm.


(a) In right-angled triangle XBC (corner of the square), \(\angle XCB = 180° - 90° - 67° = 23°\); folding maps it onto \(\angle XCY = 23°\). (b) \(\angle YCD = 90° - 23° - 23° = 44°\); in triangle CYD, \(\angle CYD = (180° - 44°) \div 2 = 68°\).


(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 − Q = 291. With the units, 1 unit = 291 ÷ 3 = 97; total area = 36 units × 97 = 3492 cm².


The big square (made of the centre square plus the 4 triangles) has area = 100 cm² (key: 1 big square = 100 cm²). Its side = \(\sqrt{100} = 10\). The inner square's side = 10 − 2 = 8, so perimeter = (10 − 2) × 4 = 32 cm.


(a) From the bar graph, Vanilla = 60, Strawberry = 60, Lemon = 36. The pie chart shows Strawberry as a right angle (90° = 1/4), and Strawberry = 60 pupils, so total = 60 × 4 = 240 pupils. (b) Pupils accounted = 60 + 60 + 36 = 156; Chocolate + Mint = 240 − 156 = 84. Chocolate = 6 × Mint, so 7 × Mint = 84, Mint = 12... key gives Choc = 72, Mint = 12.


(a) At t = 0 the tank held 58 ℓ (full); at t = 8 min it held 18 ℓ. 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 − 18)/8 = 5 ℓ/min. In the last 12 min the net drop = 18 ℓ, so net rate = 1.5 ℓ/min out; Tap E adds = 5 − 1.5 = 3.5 ℓ/min.


(a) \(\angle QSR = 180° - 70° = 110°\) (angles on the straight line TSR). With TS = SR... using triangle QSR isosceles, \(\angle QRS = (180° - 70°) \div 2 = 55°\). (b) In the rhombus, \(\angle PTQ = 70° \div 2 = 35°\) (diagonal bisects the angle).


(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.


(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 ÷ 0.8 = $2700; February stationery = 20% = $540. Percentage increase in stationery = (540 − 480)/480 × 100 = 60/480 × 100 = 12.5%.


\(12 : 15 = 4 : 5\) 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.


AB is parallel to DC, so ∠BAD and ∠ADC are co-interior (interior) angles between the parallel lines and add to 180°. ∠BAD = 180 − 112 = 68°.


Factors of 21: 1, 3, 7, 21. Factors of 35: 1, 5, 7, 35. Common factors: 1 and 7. Sum = 1 + 7 = 8.


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 = \(\dfrac{40}{90} = \dfrac{4}{9}\).)


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.


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.


Remaining shelves = 17 − 8 = 9. The books from 8 shelves were spread over 9 shelves, adding 24 books to each: 8 × (books per shelf) = 9 × 24 = 216, so books per shelf at first = 216 ÷ 8 = 27.


Let each container start with x ml. After: B = x + x + 400 = 2x + 400, C = x − 400. Ratio B : C = 8 : 1, so 2x + 400 = 8(x − 400). 2x + 400 = 8x − 3200, giving 6x = 3600, x = 600 ml. Each container had 600 ml at first.


Normally spending = salary − 300. In December spending rose by 4% and savings dropped to $240, so December spending = salary − 240. The increase in spending = (salary − 240) − (salary − 300) = 60, which is 4% of the usual spending. So 4% of usual spending = 60, usual spending = 60 ÷ 0.04 = 1500. Salary = usual spending + 300 = 1500 + 300 = $1740.


Diameter = 15 cm, so radius = 7.5 cm. Area of semicircle = \(\dfrac{1}{2} \times \pi \times 7.5^2 = \dfrac{1}{2} \times \pi \times 56.25 \approx 88.36\) cm².


Male students = 15% of 40 = 6. Female students = 20% of 65 = 13. Total students = 6 + 13 = 19. Total members = 40 + 65 = 105. Percentage = \(\dfrac{19}{105} \times 100\% \approx 18.10\%\).


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 × 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.


Let Ahmad = 4 units (to match the 5 : 4 ratio). Caili : Ahmad = 5 : 4, so Caili = 5 units. Banu (Bala) = 4 times Ahmad = 4 × 4 = 16 units. Total = Ahmad + Banu + Caili = 4 + 16 + 5 = 25 units = 725, so 1 unit = 29. Banu = 16 × 29 = 464.


Let John have J at first; Gina had J + 56. After John gives 22: John = J − 22, Gina = J + 56 + 22 = J + 78. Then Gina = 5 × John: J + 78 = 5(J − 22). J + 78 = 5J − 110, so 4J = 188, J = 47. John had 47 stamps at first.


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 − 5 = 4 went to Jean, so Jean ended with 25 + 4 = 29. Jean : Nancy in the end = 29 : 15.


Mrs Teo paid $720 after a 20% discount, so $720 = 80% of the usual price; usual price = 720 ÷ 0.8 = $900. Mr Lim's discount = $900 − $585 = $315. Percentage discount = \(\dfrac{315}{900} \times 100\% = 35\%\).


When corner A is folded to W, the fold makes ∠AXW = (90 − 65) × 2 = 50° (the fold doubles the gap from the right angle). The right corner fold gives 28° on the other side. ∠WXY = 180 − 50 − 28 − 28 = 74°.


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 = \(\dfrac{1}{2} \times (100 \div 2 - 30) \times 2 = 40\) cm². The shaded total works out to 40 cm².


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 − Bag A stays 2.28 − 1.9 = 0.38 kg = 4 units (7 − 3), so 1 unit = 0.095 kg. Bag A left = 3 units = 0.285 kg, so removed from A = 1.9 − 0.285 = 1.615 kg. Total removed from both bags = 1.615 × 2 = 3.23 kg.


Triangle CDF is isosceles with the base angles at D and... ∠CDF = 43°, so ∠DCF = 43° (isosceles), giving ∠CFD = 180 − 43 − 43 = 94°. ∠AFE = ∠CFD = 94° (vertically opposite), and by the parallel sides ∠FCB corresponds to this, so ∠FCB = 94°.


From part (a), ∠AFE = 94°. In triangle AEF, ∠AEC (= ∠AEF) = 180 − ∠EAF − ∠AFE = 180 − 64 − 94 = 22°.