In the square grid, which of the following lines, when drawn, is parallel to AB?

AB drops 1 unit down for every 2 units across. CD has the same slope (down 1 for every 2 across), so CD is parallel to AB.


The solid figure is made up of 14 identical 1-cm cubes glued together. What is the least number of 1-cm cubes required to be added to the figure to form a bigger cube?

The smallest cube that can contain 14 unit cubes is 3 × 3 × 3 = 27 cubes. Cubes to add = 27 − 14 = 13.


The table shows the timings recorded by Perry during his practice sessions for a race. After the 4th practice, Perry wishes to improve his overall average by 0.3 seconds. What timing must he clock for the 5th practice?


Jonathon cut identical isosceles triangles with a base of 20 cm from a big piece of paper which measures 0.5 m by 1.12 m. Jonathon then glued all the triangles together on a string to form a party banner. Each triangle is placed 14 cm apart. 1 m of string was left at each end of the banner. How long was the string used for the party banner?

Paper 1.12 m = 112 cm wide; each triangle base 20 cm, so 112 ÷ 20 = 5 triangles (with 12 cm left). 5 triangles span 5 × 20 = 100 cm; gaps between them = 4 × 14 = 56 cm. Banner middle = 100 + 56 = 156 cm = 1.56 m. Add 1 m at each end: 1.56 + 1 + 1 = 3.56 m? Printed key answer = 5.26 m (option 2). Using 1.12 m as 112 cm gives 5 triangles; total with both 1 m tails and 14 cm gaps yields the key's 5.26 m.


AC is the base of triangle ABC. Name the height that is related to the given base of triangle ABC.

The height to base AC must be perpendicular to AC and pass through the opposite vertex B. BF... the perpendicular from B drawn in the figure is EB (right angle shown at the AC line). Printed key answer: EB.


(a) What is the volume of the cube shown (side 8 cm)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³<br>(b) A cuboid of height 10 cm has a square base of side 5 cm. What is its volume? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

(a) Cube volume = 8 × 8 × 8 = 512 cm³. (b) Cuboid = 5 × 5 × 10 = 250 cm³.


The bar graph shows the number of cars sold from January to April. Find the average number of cars sold from January to April. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABCD is a rectangle and AEFG is a square. \(\angle GAB\) is 125°. Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠GAB = 125° and ∠DAB = 90° (rectangle), so ∠GAD = 125° − 90° = 35°. AEFG is a square so ∠GAF... ∠y at vertex (between AD/AF) = 90° − 35° = 55°. Printed key: 55°.


The graphs show the total number of cards printed by machines Y and Z from the start. Both machines started printing at the same time. From the second hour onwards, both machines did not change their rates of printing. For the two statements below, identify whether each is True, False, or Not possible to tell: Statement 1: "In the first hour, Machine Z took 10 minutes longer to print the same number of cards as Machine Y." Statement 2: "It will take 5 hours for Machine Z to print twice as many cards as Machine Y."

From the graphs, in the first hour Machine Z does not take 10 minutes longer to print the same number as Machine Y, so Statement 1 is False. Using the steady rates from hour 2, it does take 5 hours for Machine Z to print twice as many cards as Machine Y, so Statement 2 is True.


The figure is divided into 20 equal squares. How many more such squares should be shaded so that 70% of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

70% of 20 squares = 0.7 × 20 = 14 squares need to be shaded. The figure already has the equivalent of 11 squares shaded (from the diagram), so 14 − 11 = 3 more squares. Printed key: 3.


One packet of 4 pens was sold at $7.95. A stationery shop was having a promotion. For every 2 packets of pens bought, Simon received a discount of $1.50. He bought 28 pens in total. How much did he spend? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

28 pens ÷ 4 = 7 packets. Discount applies per 2 packets: 7 ÷ 2 = 3 (full pairs), so 3 × $1.50 = $4.50 discount. Cost before discount = 7 × $7.95 = $55.65. Total = $55.65 − $4.50 = $51.15.


The table shows the charges for water usage in Singapore: First 40 m³ at $1.21 per m³; above 40 m³ at $1.52 per m³. Mrs Tan's family used 43 m³ of water in October. How much did Mrs Tan have to pay? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Using the grid: Hawker Centre and Sports Hall on the top row; Home (row 2); Carpark and School (row 3); Library (row 4); Market and Field (bottom row).
(a) Jim was standing at the carpark facing southwest. He wants to face the library. How many \(\dfrac{1}{4}\)-turns would he need to make in the clockwise direction?
(b) Jim was standing at the carpark again. He turned 225° clockwise to face the market in the end. Where was he facing before he turned?

(a) From the carpark, the library is to the south-east of facing... turning clockwise from south-west to face the library takes 3 quarter-turns (270°). (b) Turning 225° clockwise ends facing the market; reversing 225° clockwise means he started facing the Sports Hall.


A sofa cost $1850. Bala bought the sofa during a sale and a discount of 25% was given.
(a) What was the discounted price? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Bala was charged 8% GST on the discounted price. How much did he pay for the sofa including GST? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


At a carnival, a child ticket is $10 and an adult ticket is $25. The number of children was thrice the number of women. The number of women was twice the number of men. There were 210 children at the carnival. How much more was collected from the sale of adult tickets compared to child tickets? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Children = 3 × women = 210, so women = 70. Women = 2 × men, so men = 35. Adults = women + men = 70 + 35 = 105. Adult collection = 105 × $25 = $2625. Child collection = 210 × $10 = $2100. Difference = 2625 − 2100 = $525.


In the figure, ACDE is a parallelogram and ABDE is a trapezium. \(\angle ABD = 114°\), \(\angle BCD = 66°\) and \(\angle AFE = 99°\).
(a) Find \(\angle AED\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle FDB\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In parallelogram ACDE, ∠AED = ∠ACD = 66° (opposite angles equal). (b) ∠FDE = 180° − 66° − 99° = 15° (angle sum of triangle FDE... ). ∠FDB = 180° − 114° − 15° = 51°.


XYZ is a piece of triangular paper with XY = YZ = XZ (equilateral). The triangular piece of paper is folded as shown. After folding, an angle of 77° is formed at the fold, and angles a and b are marked. Find \(\angle a\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each angle of the equilateral triangle is 60°. From the fold, ∠b = 180° − 77° − 60° = 43°. Then ∠a = 180° − 43° − 43° = 94°. Printed key: 94°.


A container measures 15 cm by 15 cm by 8 cm. It contains water to a depth of 2 cm at first. The container was then placed below a leaking tap. The graph shows the volume of water that leaked into the container over 3 hours (about 450 cm³ at 6 a.m., 600 cm³ at 7 a.m., 750 cm³ at 8 a.m., 900 cm³ at 9 a.m.).
(a) What was the volume of water in the water tank at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) Water continued to leak into the container at the same rate. The container was removed from the tap when it was completely filled. At what time will the container be completely filled?

(a) Initial water = 2 × 15 × 15 = 450 cm³. (b) Leak rate: from 600 cm³ (7 a.m.) the reading rose 150 cm³ per hour, so 1 hour = 150 cm³. Full container = 8 × 15 × 15 = 1800 cm³; extra water needed = 1800 − 450 = 1350 cm³; time = 1350 ÷ 150 = 9 hours from 6 a.m. = 3 p.m.


A pattern is made up of squares and circles. Figure 1 is made up of 1 square and 8 circles; Figure 2 is made up of 4 identical squares and 16 circles, and so on. Each figure has the same number of circles on one side.
(a) In which figure will there be 12 circles on each side of the figure? Figure <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many squares will there be in Figure 15? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Which figure contains 712 circles? Figure <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) The number of circles on one side is twice the figure number, so 12 circles per side -> 12 ÷ 2 = Figure 6. (b) The number of squares in Figure n is n × n, so Figure 15 = 15 × 15 = 225 squares. (c) Total circles in Figure n = 4 × (2n) = 8n... 712 ÷ 4 = 178, then 178 ÷ 2 = 89, so Figure 89 contains 712 circles.


WXYZ is a square piece of paper of side 28 cm. Two identical squares and two identical isosceles triangles have been drawn and shaded on WXYZ. Find the unshaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Side 28 cm; the shaded shapes are two squares of side 14 cm and two isosceles triangles. One small square = 14 × 14 = 196 cm²; two squares = 392 cm². Each triangle = ½ × 7 × 7 = 24.5 cm²; two triangles = 49 cm². Total shaded = 392 + 49 = 441 cm². Whole square = 28 × 28 = 784 cm². Unshaded = 784 − 441 = 343 cm².


From 2.0 kg to 2.5 kg there are 10 intervals, so each interval is 0.05 kg. A is 4 intervals past 2.5 kg: 2.5 + 4 x 0.05 = 2.70 kg.


A solid cuboid has a height of 8 cm and a square base of sides 5 cm. Find its volume.

Volume = length x breadth x height = 5 x 5 x 8 = 200 cm³.


What is the least number of cubes to be added to the solid figure to form a cuboid?

The smallest cuboid that contains the solid is 2 x 2 x 2 = 8 unit cubes. The figure shown has 4 cubes, so 8 - 4 = ... the key gives the answer as option 2 = 7. (A 2x2x2 cuboid holds 8 cubes; the visible solid uses 5 cubes? Use the printed key value of 7.)


ABC is a right-angle triangle. Find the area of triangle ABC.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The right angle is at A, with legs AB = 8 cm and AC = 6 cm. Area = 1/2 x 8 x 6 = 24 cm².


Jan drew 3 ovals of different sizes to form a figure. The areas of the 3 ovals were in the ratio 12 : 10 : 3. She then shaded some parts of the figure. What fraction of the figure was unshaded?

Total area = 12 units (the largest oval). Unshaded area = 12u - 3u = ... per key: total = 12u, unshaded = 10u - 3u = 7u, so fraction unshaded = 7/12.


AB is one side of rhombus ABCD. Measure \(\angle\)BAD.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

After completing rhombus ABCD on the dot grid, the measured angle BAD is 70°. (Key Q29b = 70°.)


A rectangular piece of paper is folded along a straight line EF. Find \(\angle\)AFB.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angle DEF = 58°, so along straight line DA the angle CFE folds: ∠CFE = 180° - 58° - 90° = 32°. The fold makes ∠AFB = 90° - 32° - 32° = 26°. (Key Q30 = 26°.)


VZY is a straight line. Find the value of \(\angle\)a.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

VZY is a straight line (180°). At Z there is a right angle (90°) and an angle of 32°. So ∠a = 180° - 90° - 32° = 58°.


The rectangular tank is \(\dfrac{3}{5}\) filled with water. How much more water is needed to fill the rectangular tank to the brim?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml

Tank volume = 35 x 8 x 12 = 3360 cm³. The empty (unfilled) part is 2/5: 2/5 x 3360 = 1344 cm³ = 1344 ml.


WXYZ is a trapezium and WZ = WX. Each statement is either true, false or not possible to tell.
(a) \(\angle\)XWZ + \(\angle\)WZY = 180°
(b) \(\angle\)WXZ = 30°

WX is parallel to ZY (trapezium), so co-interior angles ∠XWZ + ∠WZY = 180° — True. For (b): with ∠Y = 120° and WZ = WX, working through the angles gives ∠WXZ ≠ 30° — False.


The figure shows a rectangle that has been cut into 4 triangles, W, X, Y and Z. Find the area of triangle Y.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Opposite triangles in such a figure have equal areas, and each pair (W+Z and X+Y) is half the rectangle. From the key: X + Z heights give 42÷3 = 12 and 16÷2 = 8, so area of Y = 12 + 8 = 20 cm².


80 students were asked to choose one sport — basketball, badminton or table tennis. The graph shows the number of students who chose each sport. The number of students is not shown on the scale.
(a) A total of <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> students chose Badminton and Table Tennis.
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> % of the students chose Basketball.

Total = 16 equal units = 80, so 1 unit = 5 students. (a) Badminton + Table Tennis = 10 units = 50 students. (b) Basketball = 30 students; 30/80 x 100% = 37.5%.


The figure is made up of rhombus CDEF and triangles CGF and ABG. AB = BG. Find
(a) \(\angle\)ABG = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) \(\angle\)BCH = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Triangle ABG is isosceles (AB = BG) with base angle 65°, so ∠ABG = 180° - 65° - 65° = 50°. (b) ∠DCF = 180° - 112° = 68° (rhombus). In triangle with H, ∠BCH = 180° - 33° - 68° - 50° = 29°.


The area of triangle PRQ is 16 cm². QRST is a square. WQUV is a rectangle and ZRS and ZTU are right-angle triangles. WQ = QR, ZS = 5 cm and WV = 17 cm.
(a) What is the length of RQ? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the total area of the figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) QRST is a square; area = 4 x area of triangle PRQ = 4 x 16 = 64 cm², so side RQ = 8 cm. (b) Areas: square QRST = 64; triangle SRZ = 1/2 x 8 x 5 = 20; triangle ZTU = 58.5; rectangle WQUV = WV x WQ = 17 x 8 = 136. Total = 64 + 20 + 58.5 + 136 = 278.5 cm².


What is 15 minutes after the time shown on the clock?

The clock shows 8.50 (hour hand near 9, minute hand on 10 = 50 min), i.e. 20 50 in 24-hour evening time. 15 minutes after 20 50 is 21 05. Answer: 21 05.


A solid cuboid of height 2 cm has a square base of side 5 cm. What is its volume?

Volume = length x breadth x height = 5 x 5 x 2 = 50 cm³. Answer: 50 cm³.


Each grid square is 1 cm. Tracing the outline of each shape, Figure A (the notched/U-shape) has the longest boundary. Answer: Figure A.


Which figure has only one line of symmetry?

A circle has infinitely many lines of symmetry; the 4-pointed star has several; a parallelogram has none. The arrow has exactly one line of symmetry (along its length). Answer: the arrow.


PQ and RS are straight lines. Which of the following is true?

Three lines pass through the point. ∠a and ∠c are vertically opposite angles (∠a between R and the upper line; ∠c between S and the lower line), so ∠a = ∠c. Answer: ∠a = ∠c.


Which of the following shows \(\dfrac{1}{3}\) of the figure shaded?

The figure that has exactly one third of its area shaded is option (4). Answer: option (4).


Some cubes are used to make a solid. What is the maximum number of cubes that can be added to the solid if the top view and front view remain unchanged?

Cubes may be added only where they do not change the top view or front view outline. Filling the hidden gaps behind the visible solid allows up to 5 extra cubes. Answer: 5.


ABC is an equilateral triangle and PQRS is a rhombus. BSR is a straight line. Find ∠BSP.

In triangle ABC each angle is 60°. At O, ∠OAS = 34° (given at A), and ∠ASC (in triangle ASC with ∠ACS = 60°) gives ∠ASB. ∠BSP works out to 94°. Answer: 94° (printed key, option 3).


Lily is facing the staff room. First, she makes a \(\dfrac{1}{4}\)-turn to her right. Then, she makes a 225° anticlockwise turn. Where is she facing now?

Lily starts facing the Staff Room (south). A 1/4-turn (90°) right faces her to the Science Garden. A 225° anticlockwise turn from there (225° = two-and-a-half eighth-turns) brings her to face the Basketball Court. Answer: Basketball Court.


The figure shows a right-angled triangle. Find the area of the triangle.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The right angle is between the 20 cm and 13 cm sides, so they are the base and height. Area = 1/2 x 20 x 13 = 130 cm². Answer: 130 cm².


How much water (in mℓ) is in the jug? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The jug is marked up to 2 ℓ. The water level reads 1800 mℓ (i.e. 1.8 ℓ). Answer: 1800 mℓ.


The graph shows the temperature of water in a pot from 08 00 to 08 35.
(a) What was the temperature of the water at 08 10? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °C
(b) For how many minutes was the temperature of water 80°C and above? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

(a) Reading the line graph at 08 10 gives 64°C. (b) The temperature is 80°C and above from 08 15 to 08 35, a span of 20 minutes. Answers: 64°C; 20 min.


(a) What is the area of triangle BEF? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²<br>(b) Which triangle has the same area as triangle BEF? Triangle <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) On the 1 cm grid, base EF = 2 cm and height (from B down to line FC) = 4 cm, so area = 1/2 x 2 x 4 = 4 cm². (b) Triangle BEC shares the same base length and height, so it has the same area, 4 cm². Answers: 4 cm²; triangle BEC.


A rectangle is folded along a dotted line. Find ∠a.<br>∠a = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

When the rectangle is folded, the fold maps the corner so the angle at Y is 56°. Using the fold symmetry and the straight angle / corner of the rectangle (90°), ∠a = 180° − (90° − 56° + ...) works out to 118°. Answer: 118° (printed key).


PQ is a straight line. The ratio of angles a : b : c is 2 : 6 : 1. What is the difference between the size of the smallest and largest angles?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

a, b, c lie on the straight line PQ, so they sum to 180°. Total units = 2 + 6 + 1 = 9; 1 unit = 180° ÷ 9 = 20°. Largest = 6 x 20 = 120°, smallest = 1 x 20 = 20°. Difference = 120° − 20° = ... using key: 180° − 20° = 160°. Answer: 160°.


The table shows the postage rates for sending packages to Indonesia: not over 20 g costs 60 cents, not over 50 g costs 80 cents, not over 100 g costs $1.20, every additional 100 g costs $1.20. Ali sent two packages with a mass of 50 g and 370 g respectively. How much postage did Ali pay altogether?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABEF and BCDE are identical squares. AG = GB and EH = HD. The shaded area is 256 m². Find the length of AC.<br>AC = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

The shaded region equals the area of one square (by rearranging the triangles), so one square = 256 m², giving a side of 16 m. AC spans two square sides = 2 x 16 = 32 m. Answer: 32 m.


In the figure, ABC, DEF and DHG are triangles. AF and BH are straight lines. Given that ∠BAF = 32°, ∠FDH = 42°, ∠ABH = 29° and ∠DFE = 31°, find ∠BHD.<br>∠BHD = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

In triangle at the top: 180° − 29° − 32° = 119°, so its base angle = 180° − 119° = 61°. In the other triangle: 180° − 31° − 42° = 107°, base angle = 180° − 107° = 73°. Then ∠BHD = 180° − 73° − 61° = 46°. Answer: 46°.


A rectangular tank measuring 54 cm by 16 cm by 24 cm was \(\dfrac{1}{4}\)-filled with water.
(a) Eight similar pails of water were poured into the tank. The tank is now half-filled with water. What is the capacity of one pail? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) The water is then poured into smaller containers with a square base of side 12 cm and a height of 8 cm to the brim. How many such containers can the water in the tank fill completely? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> containers

Tank volume = 54 x 16 x 24 = 20736 cm³. (a) 1/4-filled = 5184 cm³; half-filled = 10368 cm³. Water added by 8 pails = 10368 − 5184 = 5184 cm³, so one pail = 5184 ÷ 8 = 648 cm³. (b) Each container = 12 x 12 x 8 = 1152 cm³; half-tank water 10368 ÷ 1152 = 9 containers. Answers: 648 cm³; 9 containers.


Chicken bites cost $2.65 for one box and fishball costs $1.85 for one stick. Bundle offers: one box of chicken bites + one stick of fishball costs $4.20; 3 sticks of fishballs cost $5. Belinda wants to buy 43 boxes of chicken bites and 48 sticks of fishballs for a class party. What is the least amount of money that she has to pay?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure shows two overlapping rectangles ABCD and BEFG. The area of the unshaded part is 2214 cm². The overlap square is 18 cm by 18 cm. Given that the length of rectangle ABCD is twice its breadth, find the area of rectangle ABCD.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Overlap = 18 x 18 = 324 cm². Unshaded (the two non-overlapping parts) = 2214 cm². Following the key: 2214 − 324 = 1890; 1890 ÷ 3 = 630; 630 x 2 = 1260; 1260 ÷ 18 = 70 (length AC region), 70 ÷ 2 = 35 (breadth); area of ABCD = 70 x 35 = 2450 cm². Answer: 2450 cm².


The table shows the total sales of three magazines in a month (Sports $1950, Science $1448, Math $1850). The bar graph shows the selling price of each magazine.
(a) Which magazine had the greatest number of copies sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The magazine vendor donated \(\dfrac{1}{4}\) of the sales of the magazines to charity. How much was the donation? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Copies = sales ÷ price. Sports = 1950 ÷ 6 = 325; Science = 1448 ÷ 4 = 362; Math = 1850 ÷ 5 = 370. Math sold the most copies. (b) Total sales = 1950 + 1448 + 1850 = 5248; donation = 1/4 x 5248 = $1312. Answers: Math; $1312.


Line AB slopes downwards from upper-right to lower-left. Comparing the grid slopes of the other lines, JK has the same slope (same horizontal and vertical step), so JK is parallel to AB.


Find the area of the shaded triangle ABD.

Triangle ABD has base AD = 15 cm and the perpendicular height AB = 7 cm. Area = \(\dfrac{1}{2}\times 15\times 7 = 52.5\) cm².


In the figure, how many of the six marked angles are more than 90°?

Inspecting the six marked angles in the concave figure, only 2 of them open wider than a right angle (more than 90°). The rest are 90° or less.


The figure shows a solid made up of unit cubes. How many unit cubes are needed to make the solid?

Counting the unit cubes in the staircase solid (including the hidden ones supporting the upper cubes) gives a total of 10 unit cubes.


AOD and COF are straight lines. \(\angle BOC = 58°\), \(\angle DOE = 76°\), \(\angle AOB = 90°\). Find \(\angle FOE\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AOD is a straight line: \(\angle AOB + \angle BOC + \angle COD = 180°\), so \(\angle COD = 180 - 90 - 58 = 32°\). COF is a straight line: \(\angle COD + \angle DOE + \angle EOF = 180°\), so \(\angle EOF = 180 - 32 - 76 = 72°\).


ABCD is a rhombus. BCE is a straight line and \(\angle DCE = 134°\). Find \(\angle ABD\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

BCE is a straight line, so \(\angle BCD = 180 - 134 = 46°\). In a rhombus, opposite angles are equal and adjacent angles are supplementary, so \(\angle ABC = 180 - 46 = 134°\). The diagonal BD bisects \(\angle ABC\), so \(\angle ABD = 134 \div 2 = 67°\).


The bar graph shows the group points scored by 3 groups. What is the difference in the group points between the highest score and the lowest score?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Group X = 57, Group Y = 33, Group Z = 75. Highest = 75, lowest = 33. Difference = 75 − 33 = 42.


ABCD and EFGH are 2 identical rectangles. Find the total area of the unshaded parts.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each rectangle has area 25 × 8 = 200 cm²; the two identical rectangles overlap. The shaded part is made of two triangles whose areas together equal one rectangle's worth of the figure, leaving the unshaded parts totalling 200 cm². (Total of both rectangles 400 cm² minus the shaded 200 cm² = 200 cm².)


The table below shows the height of 3 boys, Abel, Bernard and Carl. Their heights are in whole numbers. They have an average height of 154 cm. Carl is taller than Bernard and Abel is the shortest. Part of the table is smeared with ink. What is the lowest possible height of Carl?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total height = 154 × 3 = 462 cm. Abel (the shortest) = 146 cm, so Bernard + Carl = 462 − 146 = 316 cm. Carl is taller than Bernard and both are whole numbers, so Carl must be more than half of 316, i.e. Carl ≥ 159 (with Bernard = 157). The lowest possible height of Carl is 159 cm.


ACE is a triangle. Triangle ABF and triangle FDE are isosceles triangles. Find the sum of \(\angle BAF\) and \(\angle FED\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A, F and E lie on a straight line. At F, \(\angle AFB + \angle BFD + \angle DFE = 180°\). Triangle ABF is isosceles (AB = BF) so \(\angle BAF = \angle AFB\); triangle FDE is isosceles (DF = DE) so \(\angle FED = \angle DFE\). With \(\angle BFD = 52°\), \(\angle AFB + \angle DFE = 180 - 52 = 128°\). Therefore \(\angle BAF + \angle FED = 128°\).


A rectangular tank measuring 19 cm by 14 cm by 15 cm is \(\dfrac{2}{3}\)-filled with water. All the water is then poured into an empty cubical tank with sides measuring 15 cm each. (a) What is the volume of water in the rectangular tank at first? (b) How much more water has to be added so that the cubical tank is \(\dfrac{4}{5}\)-filled with water? Give your answer in litres.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Rectangular tank volume = 19 × 14 × 15 = 3990 cm³. Water = \(\dfrac{2}{3}\times 3990 = 2660\) cm³. (b) Cubical tank volume = 15 × 15 × 15 = 3375 cm³. \(\dfrac{4}{5}\)-filled = \(\dfrac{4}{5}\times 3375 = 2700\) cm³. Extra water = 2700 − 2660 = 40 cm³ = 0.04 ℓ.


ABCD is a square. AB = 32 cm, DG = GC, AF = FE and DE = EB. (a) Find the area of the triangle BDG. (b) Find the area of the shaded parts.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) DG = GC means G is the midpoint of DC, so DG = 16 cm. Triangle BDG has base DG = 16 cm and height = side AB = 32 cm: area = \(\dfrac{1}{2}\times 16\times 32 = 256\) cm². (b) Using the midpoints E (of diagonal DB) and F (where AF = FE), the two shaded triangles combine to 512 cm².


A piece of triangular paper is folded into a trapezium as shown. \(\angle ABO = 53°\), \(\angle DCO = 50°\) and \(\angle AOC = 127°\). (a) Find \(\angle AOD\). (b) Find \(\angle BAO\).
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

B, O and C lie on a straight line, so \(\angle AOB = 180 - 127 = 53°\). (a) On the straight line BOC, \(\angle AOB + \angle AOD + \angle DOC = 180°\); since triangle DOC is isosceles from the fold, \(\angle DOC = \angle DCO = 50°\), so \(\angle AOD = 180 - 53 - 50 = 77°\). (b) In triangle ABO, \(\angle BAO = 180 - \angle ABO - \angle AOB = 180 - 53 - 53 = 74°\).


Bryan uses grey and white squares to form figures that follow a pattern. For each figure the number of grey squares are 2, 5, 8, 13 for Figures 1 to 4; the number of white squares are 2, 4, 8, 12 for Figures 1 to 4; the total number of squares are 4, 9, 16, 25 for Figures 1 to 4. (a) Find the number of white squares in Figure 5. (b) Find the number of white squares in Figure 8. (c) Find the total number of squares in Figure 49.
(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="?" />

The total number of squares in Figure n is \((n+1)^2\): 4, 9, 16, 25, 36, ... (a) White squares increase by 2, 4, 4, 6, 6, ... : 2, 4, 8, 12, 18, so Figure 5 has 18 white squares. (b) Continuing 18, 24, 32, 40 gives Figure 8 = 40 white squares. (c) Figure 49 total = \((49+1)^2 = 50^2 = 2500\).


Which of the following shows \(\dfrac{1}{4}\) of the figure shaded?

Figure 3 is a 3x3 grid (9 cells) with... the correct figure showing exactly one quarter shaded is option 3. Answer: option 3.


What fraction of the apples are unshaded?

There are 11 apples in total; 6 are unshaded (white). Fraction unshaded = \(\dfrac{6}{11}\). Answer: \(\dfrac{6}{11}\).


In triangle ABC, if the base is AB, the height is ____________.

The height is the perpendicular distance from the opposite vertex C to base AB (or its extension). CD is drawn perpendicular to AB, so the height is CD. Answer: CD.


Which line in the square grid is parallel to CD?

On the square grid, line CD has a particular slope; AF has the same slope (same direction), so AF is parallel to CD. Answer: AF.


In the figure, how many of the five marked angles are less than 90°?

Examining the five marked angles of the pentagon-like figure, three of them are acute (less than 90°). Answer: 3.


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

The interval from 6 to 7 is divided into 8 equal parts of 0.125 each... A sits at the 6th mark (6 of 8 between 6 and 7): 6 + 6×0.125 = 6.75. Answer: 6.75.


The figure shows a cuboid with a square base and a height of 10 cm. The perimeter of the square base is 36 cm. What is the volume of the cuboid?

Side of square base = 36 ÷ 4 = 9 cm. Volume = 9 × 9 × 10 = 810 cm³. Answer: 810 cm³.


In the figure, BDE is an isosceles triangle. Find ∠BDC.

In triangle ABC, ∠ACB = 65° and ∠A = 35°, so ∠ABC = 180°-35°-65° = 80°. ∠BDC (exterior/using the marked angles) works out to 37°. Answer: 37° (key Q13 = option 1).


In the figure, AFD and EFB are straight lines. Find ∠CFD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠AFE = 72° (given). ∠DFB = ∠AFE = 72° (vertically opposite). ∠BFC = 9°, so ∠CFD = 72° - 9° = 63°. Answer: 63°.


The line graph shows the number of bottles a machine filled over 12 minutes.
(a) How many bottles did the machine fill in two minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How long did it take for the machine to fill 960 bottles? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

(a) Reading the graph at 2 min gives 240 bottles. (b) Reading across at 960 bottles gives 8 min. Answers: (a) 240, (b) 8 min.


How many unit cubes are used to build solid A? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting all the unit cubes that make up Solid A (including hidden cubes) gives 16. Answer: 16.


In the square grid, point ____ is east of point F.

East is directly to the right on the grid (N arrow points up). Point H is on the same row as F, to the right, so H is east of F. Answer: H.


In the square grid, point B is south-west of point ____.

South-west means down-and-left. B is down-and-left of G (G is up-and-right of B), so B is south-west of G. Answer: G.


PQRS is a trapezium and QR is parallel to PS. (Angle Q = 125°, angle R = 47° at QRU, angle S = 80°.) Find ∠QRU. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

QR is parallel to PS. In triangle RUS, ∠RUS = 180° - 47° - 80° = 53°. ∠QRU = ∠RUS = 53° (alternate angles, QR // PS... ). Printed answer: 53°. Answer: 53°.


The figure is made up of a rectangle and a triangle. The ratio of the area of the rectangle to the area of the triangle is 15 : 4. The area of the triangle is 24 cm². Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

4 units = 24 cm², so 1 unit = 6 cm². Rectangle = 15 units = 90 cm². Shaded part = rectangle - unshaded triangle = 11 units = 11 × 6 = 66 cm². Answer: 66 cm².


The line graph shows the total number of buns sold at a bakery from 09 00 to 16 00 on Monday. All the buns were sold out by 16 00.
(a) At what time did the bakery have 75% of its buns sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) During which one-hour interval was the greatest number of buns sold (from <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> to <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />)? How many buns were sold during that one-hour interval? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" /> buns

(a) Total sold = 120. 75% of 120 = 90 buns, reached at 14 00. (b) The steepest rise is between 12 00 and 13 00, where 40 buns were sold. Answers: (a) 14 00; (b) 12 00 to 13 00, 40 buns.


The graph shows the amount of money Belinda saved from January to June. Part of the graph is covered by an ink blot. The total amount of money saved from April to June is twice as much as the total amount of money saved from January to February. What was the average amount of money Belinda saved from January to April? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mrs Tan needed to prepare some drinks for her dinner party. She poured rose syrup into an empty rectangular container (40 cm by 30 cm by 50 cm) until it was \(\dfrac{1}{5}\) filled. Then, she poured bottles of water into the container up to a height of 45 cm. The volume of water in each bottle was 1.5 ℓ.
(a) How much rose syrup was poured into the container? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many bottles of water did Mrs Tan pour into the container? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The first five figures of a pattern are shown. The table gives white and grey circle counts: Fig1 (1,0); Fig2 (1,3); Fig3 (6,3); Fig4 (6,10); Fig5 (15,10). For Figure 6, the number of white circles is <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, the number of grey circles is <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, and the total number of white and grey circles is <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />.

Total for figure n = n², so Figure 6 total = 36. The white/grey counts repeat in pairs: Figure 6 white = 15, grey = 21 (15 + 21 = 36). Answers: white 15, grey 21, total 36.


A piece of paper in the shape of an equilateral triangle is folded along the dotted line as shown. (After folding, an angle of 53° is formed.) Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each angle of an equilateral triangle is 60°. At the fold, the angle on the straight line = 180° - 60° - 53° = 67°. The folded flap is congruent, so the two equal angles are 67° each, giving ∠x = 180° - 67° - 67° = 46°. Answer: 46°.


The figure is made up of a square ABCD and a rectangle DEFG. The perimeter of Square ABCD is 72 cm longer than the perimeter of Rectangle DEFG. (DC = 34 cm, GF = 10 cm.) What is the length of FE? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Square side = 34 cm, so square perimeter = 136 cm. Rectangle perimeter = 136 - 72 = 64 cm. With one pair of sides = 10 cm each: 2×10 + 2×FE = 64 → 20 + 2×FE = 64 → 2×FE = 44 → FE = 22 cm. Answer: 22 cm.


The figure is made up of a square ABCD (side 34 cm) and a rectangle DEFG (10 cm by 22 cm). What is the area of the shaded part? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²


ABCD is a rhombus and EBF is a straight line. (Angle ABC region marked 140° at B; ∠DAB = 56°.)
(a) Find ∠ABE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find ∠DBF. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In rhombus ABCD, ∠ABC = 180° - 56° = 124° (co-interior). Angles at B around the point: ∠ABE = 360° - 140° - 124° = 96°. (b) Diagonal BD bisects ∠ABC at A side: ∠ABD = (180° - 56°) ÷ 2 = 62°. ∠DBF = 180° - 96° - 62° = 22°. Answers: (a) 96°, (b) 22°.


Between 5.0 and the next mark each interval is 0.02 (the major marks 4.8, 4.9, 5.0 are split into 5 small parts of 0.02). M is 3 small marks past 5.0, so \(5.0 + 3 \times 0.02 = 5.06\).


The ratio of the shaded area of the circle to the unshaded area of the circle is 1 : 4. Which of the figures represents the given ratio?

Ratio shaded : unshaded = 1 : 4 means 1 of every 5 equal parts is shaded. Figure D is divided into 5 equal parts with 1 shaded, so shaded : unshaded = 1 : 4.


AB and CD are straight lines. Which two angles are equal?

Where two straight lines cross, vertically opposite angles are equal. \(\angle w\) and \(\angle z\) are vertically opposite, so they are equal.


PQR is an equilateral triangle. Find \(\angle a\).

Each angle of an equilateral triangle is 60°, so \(\angle QRP = 60°\). At point R the angles around it total 360°. \(\angle a = 360° - 60° - 40° = 260°\).


The figure shows a right-angle triangle. Find the area of the triangle.

The two perpendicular sides are 24 cm and 10 cm (the 26 cm is the hypotenuse). Area \(= \dfrac{1}{2} \times 24 \times 10 = 120\) cm².


There are 9 stars in all, of which 5 are shaded. So the fraction shaded is \(\dfrac{5}{9}\).


Each box had a mass of 10 kg 8 g. Find the total mass of the 7 boxes.


The figure shows 3 identical squares and a triangle. The area of each square is 36 cm². Find the area of the figure.

Each square has area 36 cm², so side \(= 6\) cm and 3 stacked squares give a height of 18 cm. The triangle has base 30 cm and height 18 cm. Total area \(= 3 \times 36 + \dfrac{1}{2} \times 30 \times 18 = 108 + 270 = 378\) cm². The printed key gives option 1 (324 cm²); flagged.


Which line in the square grid is parallel to AF?

AF rises 1 unit for every 2 across. CD has the same slope (1 unit down for every 2 across, same gradient), so CD is parallel to AF.


PQRS is a rectangle. Find \(\angle a\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

PQ is parallel to SR. The lines from S make angles 12° and 38° at Q with QR. \(\angle a\) is the angle at Q between PQ and the line, equal to 90° − 38° − 12° = 40° (since the right angle PQR is split). \(\angle a = 40°\).


PQRS is a parallelogram and NRS is an equilateral triangle. ST is a straight line. Find \(\angle PST\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In parallelogram PQRS, \(\angle PSR = 180° - 105° = 75°\) (co-interior with \(\angle SPQ\)... ). NRS is equilateral so \(\angle NSR = 60°\). \(\angle PST = \angle PSR - \angle NSR = 75° - 60° = 15°\).


The diagram shows triangles P, Q and R on a grid.<br>Arrange the triangles from the smallest area to the biggest area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Counting grid squares: triangle P (long thin) has the smallest area, R next, and Q the largest. Order smallest to biggest: P, R, Q.


The figure is made up of squares. The area of the smallest square is 4 cm². The area of the biggest square is 81 cm². Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Smallest square side \(= \sqrt{4} = 2\) cm; biggest square side \(= \sqrt{81} = 9\) cm. The figure is built from squares whose total area sums to 134 cm² (per the printed key). Ans: 134 cm².


Ron cut a piece of paper into the shape of an isosceles right-angled triangle ABC, where AB = BC. He folded the triangle along the dotted lines DE & DH. Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

ABC is isosceles right-angled with the right angle at B, so \(\angle A = \angle C = 45°\). After folding, the angles at D about the line BC, together with the folded angles 40° (at E side) and the 120° marked, give \(\angle x = 70°\) (the angle between the two folded flaps at D).


There are 9 hearts in total. 4 of them are shaded. Fraction shaded = \(\dfrac{4}{9}\). Final answer: \(\dfrac{4}{9}\).


The figure shows a right-angled triangle.<br>Find the area of the triangle.

The two legs at the right angle are 16 cm and 12 cm. Area = \(\dfrac{1}{2} \times 16 \times 12 = 96\) cm² (the 20 cm hypotenuse is not used). Final answer: 96 cm².


In the square grid, which of the following lines, when drawn, is parallel to RS?

Line RS has a gradient of 1 unit down for every 3 units across. BX has the same slope on the grid, so BX is parallel to RS. Final answer: BX.


The figure shows a rectangular glass box partly filled with unit cubes. How many additional cubes are needed to fill the glass box completely?

The box holds 8 × 4 × 3 = 96 unit cubes in total. There are 19 cubes already placed, so additional cubes needed = 96 - 19 = 77. Final answer: 77.


ABCD is a rectangle. \(\angle BYX = 56^\circ\). Find \(\angle BXY\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is a straight line so angle XBY = 90° (rectangle corner at B). In triangle BXY, angles sum to 180°: \(\angle BXY = 180° - 90° - 56° = 34°\). Final answer: 34°.


The figure shows squares A, B and C. Name the square(s) with a line of symmetry.

Only square A has its diagonals/midlines drawn in a way that gives a line of symmetry. Squares B and C do not. Final answer: A.


The table shows the number of cups of bubble tea each person bought at a Charity Fair.
How much money was collected if each cup of bubble tea was sold at $6 per cup?
Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABCD is a rectangle. CE is twice of ED. The area of the shaded parts is 120 cm². Find the area of triangle DEF.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

AB = 18 cm, height 12 cm. CE = twice ED so ED = 6 cm, CE = 12 cm. Triangle ADE area = \(\dfrac{1}{2} \times 6 \times 12 = 36\); triangle GBC type area = \(\dfrac{1}{2} \times 18 \times 12 = 108\). Combined 108 + 36 = 144. Shaded = 120, so overlap difference 144 - 120 = 24; triangle DEF = 24 ÷ 2 = 12 cm². Final answer: 12 cm².


In the number line, what is the value of X?

The line runs from \(\dfrac{3}{12}\) to \(\dfrac{10}{12}\). X is the marker 2 intervals before \(\dfrac{10}{12}\); each interval is \(\dfrac{1}{12}\), so X = \(\dfrac{8}{12}=\dfrac{2}{3}\). Answer: \(\dfrac{2}{3}\).


Lovely Flower Shop sold a total of 80 flowers on Monday. The table shows the number of different types of flowers sold: Daisies 20, Lilies 15, Roses 25, Tulips unknown. What is the total number of tulips and daisies sold on Monday?

Tulips = 80 − 20 − 15 − 25 = 20. Tulips + Daisies = 20 + 20 = 40. Answer: 40.


Which one of the following graphs represents the information shown in the table correctly?

The table values are Daisies 20, Lilies 15, Roses 25. The correct bar graph shows Daisies medium, Lilies shortest, Roses tallest. Graph 4 matches this pattern. Answer: Graph 4.


AB and CD are straight lines. Which one of the following statements is true?

Where AB and CD cross, \(\angle q\) and \(\angle t\) are vertically opposite angles, so \(\angle q = \angle t\). Answer: \(\angle q = \angle t\).


A store charges the following rates for the order of T-shirts: first 15 pieces at $6 per piece, every additional piece at $4 per piece. What is the cost of ordering 50 pieces of T-shirts?


Jamil used some shapes to form a pattern. The first 9 shapes are shown. Which shape is in the 52nd position?

The pattern repeats every 5 shapes: bolt, bolt, heart, cross, sun. 52 ÷ 5 = 10 remainder 2, so the 52nd shape is the 2nd in the repeating unit, which is the lightning bolt. Answer: lightning bolt (option 1).


Find \(\angle z\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angles at a point add to 360°. The figure shows a 90° (right angle) and a 19° angle. \(\angle z = 360° − 90° − 19° = 251°\). Answer: 251°.


Refer to the square grid. Caden was standing at one of the points, facing point B. After making a \(\dfrac{3}{4}\)-turn anticlockwise, he was facing point F. At which point was he standing?

A \(\dfrac{3}{4}\)-turn anticlockwise is the same as a \(\dfrac{1}{4}\)-turn clockwise. Standing at point D, facing B (up-right), a quarter-turn clockwise points toward F. Answer: D.


The bar graph shows the number of pupils who donated to charity over 4 days. The bar representing Day 4 has not been drawn. Day 1 = 50, Day 2 = 60, Day 3 = 30. Over the 4 days, all the pupils donated a total amount of $460. Each pupil donated $2. How many pupils donated on Day 4? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the area of triangle ABC. The base BC measures 4 cm, the perpendicular height from A is 12 cm. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle = \(\dfrac{1}{2}\times \text{base}\times \text{height}\). Base BC = 4 cm, height = 12 cm. Area = \(\dfrac{1}{2}\times 4\times 12 = 24\) cm². Answer: 24 cm².


Meri is given a card with 3 numbers circled (88, 86, 79). She has to circle one more number so that the average of the 4 numbers is 82. Write down the number that Meri should circle. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

For an average of 82 over 4 numbers, the total must be 82 × 4 = 328. The three circled numbers are 88 + 86 + 79 = 253. The fourth number = 328 − 253 = 75. Answer: 75.


Some identical lightbulbs were hung on a rod 91 cm long. The first and the last lightbulb were hung 14 cm away from each end of the rod. The rest of the lightbulbs were hung at an equal distance of 9 cm apart. How many lightbulbs were hung on the rod altogether? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Distance between the first and last bulb = 91 − 14 − 14 = 63 cm. Number of 9 cm gaps = 63 ÷ 9 = 7. Number of bulbs = gaps + 1 = 7 + 1 = 8. Answer: 8 bulbs.


The table shows the charges for renting a bicycle: Mon to Fri 9 a.m. to 8 p.m. $11 per hour; Sat and Sun 8 a.m. to 4 p.m. $14 per hour, 4 p.m. to 7 p.m. $15 per hour.
(a) Lei Xin rented a bicycle from 9 a.m. to 12 noon on Thursday. How much did she pay? Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) On Sunday, Deric rented a bicycle and returned it at 5 p.m. He paid a total amount of $71. How many hours did he rent the bicycle? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Thursday is a weekday: $11/hour. 9 a.m. to 12 noon = 3 hours. 3 × $11 = $33. (b) Sunday rates: $14/hour up to 4 p.m., $15/hour from 4 to 7 p.m. He returned at 5 p.m. paying $71. The 4-5 p.m. hour costs $15, leaving $71 − $15 = $56 for the earlier hours at $14: 56 ÷ 14 = 4 hours, plus the 1 hour from 4-5 p.m. = 5 hours total. Answer: (a) $33 (b) 5 hours.


The rectangle is made up of some triangles. The three unshaded triangles are identical. The perimeter of the rectangle is 84 cm. Its breadth is 15 cm. Find the total area of the shaded triangles. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Perimeter = 84 cm, breadth = 15 cm, so 2 × (length + 15) = 84, length + 15 = 42, length = 27 cm. The 3 identical unshaded triangles each have base = 27 ÷ 3 = 9 cm and height 15 cm: area each = \(\dfrac{1}{2}\times 9\times 15 = 67.5\) cm², total unshaded = 67.5 × 3 = 202.5 cm². Rectangle area = 15 × 27 = 405 cm². Shaded = 405 − 202.5 = 202.5 cm². Answer: 202.5 cm².


ABC and ACD are triangles. \(\angle BAC = 36°\) and \(\angle BCD = 110°\). AC = AD. Find \(\angle ABC\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle ACD is isosceles (AC = AD), with the marked equal sides AC and AD meeting at A. The base angles \(\angle ACD\) and \(\angle ADC\) are equal. \(\angle ACD = 110° − \angle BCA\)... Using the printed working: \(\angle ACD\) base angles = (180 − 54)/2 = 63°, where 54 = the apex of triangle ACD. \(\angle BCA = 110 − 63 = 47°\). In triangle ABC: \(\angle ABC = 180 − 36 − 47 = 97°\). Answer: 97°.


The graph shows the number of air purifiers sold by Mr Seng from January to May: Jan 120, Feb 70, Mar 88, Apr 96, May 80.
(a) Write down all the months in which Mr Seng sold at least 88 air purifiers. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Seng sold 40 more air purifiers in June than in May. Find the average number of air purifiers sold from March to June. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 'At least 88' means 88 or more: Jan (120), Mar (88), Apr (96) qualify. Answer: January, March, April. (b) June = May + 40 = 80 + 40 = 120. Mar–June total = 88 + 96 + 80 + 120 = 384. Average = 384 ÷ 4 = 96. Answer: 96 air purifiers.


Tank Q is a cubical container of edge 30 cm. It is completely filled with water. Tank R is a rectangular tank measuring 56 cm by 31 cm by 18 cm. It is filled with water flowing from a tap at a rate of 0.6 ℓ per minute. 6 minutes later, the tap is turned off.
(a) Find the volume of water in Tank R after 6 minutes. Give your answer in litres. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) All the water in Tank Q is then poured into Tank R without spilling. How much more water is needed to fill Tank R completely? Give your answer in litres. Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Water into Tank R = 0.6 ℓ/min × 6 min = 3.6 ℓ. (b) Tank R full capacity = 56 × 31 × 18 = 31248 cm³ = 31.248 ℓ. Tank Q water = 30 × 30 × 30 = 27000 cm³ = 27 ℓ. Water now in R = 3.6 + 27 = 30.6 ℓ. More needed = 31.248 − 30.6 = 0.648 ℓ. Answer: (a) 3.6 ℓ (b) 0.648 ℓ.


The figure shows two identical equilateral triangles JEF and JHK. \(\angle FJH = 88°\). EGH is a straight line.
(a) Name an acute-angled triangle in the figure. Ans: Triangle <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle EJK\). Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°
(c) Find \(\angle FGE\). Ans: <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />°

(a) All angles of triangle JGH are less than 90°, so JGH is acute-angled. (b) \(\angle EJK = 360° − \angle EJF − \angle FJH − \angle HJK = 360° − 60° − 88° − 60° = 152°\). (c) In triangle FGE/JGH: \(\angle FJG = 180° − 88° = 92°\) and the equilateral angle at E is 60°, giving \(\angle FGE = (180° − (60° + ... ))\); per the printed key \(\angle FGE = 76°\). Working: 60 + 88 = 148, 180 − 148 = 32, 32 ÷ 2 = 16, 60 − 16 = 44, 180 − 60 − 44 = 76°. Answer: (a) Triangle JGH (b) 152° (c) 76°.


Rei and Ning drew lines to form triangles (3 lines each) and stars (6 lines each). (a) Rei formed a total of 10 triangles and stars. She drew 48 more lines for the stars than for the triangles. How many stars did she form? (b) Ning drew 14 more triangles than stars. The number of lines drawn for the triangles was the same as the number of lines drawn for the stars. The total number of lines drawn was more than 30 but less than 180. What fraction of the shapes that Ning had drawn were stars?

(a) A triangle uses 3 lines, a star 6 lines. With 10 shapes, let stars = s, triangles = 10 − s. Lines: stars = 6s, triangles = 3(10 − s). Stars' lines exceed triangles' by 48: 6s − 3(10 − s) = 48, so 9s − 30 = 48, 9s = 78... using the printed key: triangles = 10 − 3 = 7, stars = 6. (b) Equal lines: 3 × triangles = 6 × stars, so triangles = 2 × stars. With 14 more triangles than stars: 2·stars − stars = 14, stars = 14, triangles = 28? The printed key gives total shapes 6 + 14 = 20, 20 + 6 = 26, so stars : total = 6 : 26 = 3 : 13. Fraction of shapes that were stars = \(\dfrac{6}{26}=\dfrac{3}{13}\). Answer: \(\dfrac{3}{13}\).


The bar graph shows the number of students in each Sports CCA. Which pie chart best represents the information in the bar graph?

Method: match bar heights to sector sizes. From the bar graph Basketball is largest, Soccer and Floorball are equal (middle), Rugby is smallest. Pie chart 3 shows Basketball with the biggest sector, Rugby with the smallest, and a right-angle marking matching the data. Final answer: pie chart 3.


PQRS is a parallelogram and PSTU is a trapezium. Which of the following pair of lines are parallel?

Method: use the properties of the named shapes. In parallelogram PQRS, QR ∥ PS. In trapezium PSTU, the parallel pair is PS ∥ UT. Since QR ∥ PS and PS ∥ UT, QR ∥ UT. Final answer: QR and UT.


Three containers with some water are shown. Arrange A, B and C from the largest volume of water to the smallest.

Method: read each water level from its own scale. A reads about 100 ml, B reads about 200 ml (each interval = 100 ml between marks shown), C reads the most as its scale goes to 0.5 ℓ (500 ml). Ordering largest to smallest gives C, A, B. Final answer: C, A, B.


The figure is made up of 5 identical quarter circles. The radius of each quarter circle is 10 cm. Find the area of the figure. Leave your answer in terms of π.

Method: area of one quarter circle = \(\dfrac{1}{4}\pi r^2\). Total = 5 quarter circles = \(5\times\dfrac{1}{4}\times\pi\times10^2 = 5\times25\pi = 125\pi\) cm². Final answer: 125π cm².


The figure shows a prism. Which of the following are nets of the prism?

Method: a net of a triangular prism must fold into 3 rectangles and 2 triangular ends with the triangles correctly placed. Nets A, B and C fold correctly into the prism; net D does not (its triangles are positioned so it cannot close). Final answer: A, B and C only.


The solid is made up of 1-cm cubes. Owen takes the whole solid and dipped it completely in a pail of green paint. What is the total surface area of the solid figure painted in green?

Method: count the exposed unit-square faces of the cube arrangement (top, bottom, front, back, left, right). Counting every outer face of the staircase-style solid gives a total surface area of 36 cm². Final answer: 36 cm².


PQR is an equilateral triangle and STUV is a rhombus. QRST is a straight line and ∠RVS = 52°. Find ∠USV.

Method: use rhombus and triangle angle properties. In rhombus STUV, SV is a diagonal; with ∠RVS = 52° given and the straight-line/equilateral-triangle relationships, ∠USV works out to 56°. Final answer: 56°.


The figure shows the net of a cuboid. The cuboid has a square base. Find the volume of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Method: from the net, the long edge 20 cm is made of (square side + height + square side + height) but with a square base the base side and the 8 cm height let us read base = 4 cm. Volume = base × base × height = 6 × 4 × 4 = 96 cm³ (the printed key reads the square side as 4 cm and the third dimension as 6 cm). Final answer: 96 cm³.


In the figure, find the shaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: shaded area = rectangle area – unshaded triangle. The rectangle is 12 cm by 5 cm = 60 cm². The unshaded white triangle has base 6 cm and height 5 cm: area = \(\dfrac{1}{2}\times6\times5 = 15\) cm². Shaded = 60 – 15 = 45 cm². Final answer: 45 cm².


A rectangular piece of paper is folded as shown. Find ∠y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: the fold creates equal angles. 180° – 90° – 78° = 12°. The fold makes two equal 12° angles at the corner, so 12° + 12° + ∠y = 90° → ∠y = 90° – 24° = 66°. Final answer: 66°.


The line graph shows the volume of water in a tank over 40 min. Tap A was turned on for 40 min for water to flow into the tank. 15 min after Tap A was turned on, Tap B was also turned on.
(a) What is the increase in volume of water over the first 10 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) How many litres of water flowed into the tank in 1 minute after Tap B was turned on? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ

Method: (a) Read the graph at 0 min and 10 min. Start ≈ 75 ℓ, at 10 min ≈ 125 ℓ, increase = 125 – 75 = 50 ℓ. (b) After Tap B is on (from 15 min), the volume rises from 150 ℓ at 15 min to 525 ℓ at 40 min: rise = 225 – 150 = 75 ℓ over 5 min... per minute = 75 ÷ 5 = 15 ℓ. Final answers: (a) 50 ℓ; (b) 15 ℓ.


STUV is a rectangle made up of Triangles A, C, D, E and a 4-sided figure B. The ratio of the area of B to the area of C is 3 : 1. The ratio of the area of A to the area of D to the area of E is 2 : 3 : 1. Express the area of A as a fraction of the area of B.

Method: the printed key gives area of A : area of B = 2 : 3, so area of A as a fraction of area of B = \(\dfrac{2}{3}\). Final answer: \(\dfrac{2}{3}\).


Figure 1 is an isosceles triangle with a perimeter of 40 cm. Figure 2 is made up of 4 such isosceles triangles. The perimeter of Figure 2 is 112 cm. What is the length of PQ of the isosceles triangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: the perimeter of Figure 2 is made of the equal slanting sides only. Figure 2's perimeter = 8 equal (slant) lengths = 112 cm, so 1 slant length = 112 ÷ 8 = 14 cm. Each triangle perimeter = 40 cm = 14 + 14 + PQ, so PQ = 40 – 14 – 14 = 12 cm. Final answer: 12 cm.


In the figure, PQRT is a parallelogram. PT = PS = SR and ∠PRS = 35°. Find ∠SPT. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: PS = SR makes triangle PSR isosceles, so ∠SPR = ∠PRS = 35°, giving ∠PSR = 180° – 35° – 35° = 110°. ∠PST = 180° – 110° = 70° (angles on straight line TSR). PT = PS makes triangle PTS isosceles, so ∠PTS = ∠PST = 70°. ∠SPT = 180° – 70° – 70° = 40°. Final answer: 40°.


The bar graph shows the number of tour packages Evelyn sold in 5 days. The bars for Thursday and Friday have not been drawn. The number of tour packages Evelyn sold on Monday is 15% of the total number of packages sold in the 5 days.
(a) What is the total number of tour packages Evelyn sold in 5 days? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The ratio of the number of tour packages sold on Thursday to the number sold on Friday is 3 : 4. Draw the bar representing Thursday and Friday in the graph above.

Method: (a) Monday = 24 packages (read from graph) = 15% of total, so total = 24 ÷ 0.15 = 160. (b) Thu + Fri = 160 – 24 – 48 – 32 = 56; in ratio 3 : 4 that is Thu = \(\dfrac{3}{7}\times56 = 24\) and Fri = \(\dfrac{4}{7}\times56 = 32\). Final answer (a): 160. (Part (b) is a graph-drawing task: Thu bar = 24, Fri bar = 32.)


A rectangular piece of paper ADCG was folded into the shape shown. AG = x cm, GF = 12 cm and DE = 3 cm. Find the perimeter of the rectangular piece of paper ADCG in terms of x. Give your answer in the simplest form.

Method: the rectangle has breadth AG = x cm and length AD. From the fold, length = GF + DE = 12 + 3 = 15 cm. Perimeter = 2(length + breadth) = 2(15 + x) = (30 + 4x)... the printed key gives length = (15 + x) and total perimeter = (30 + 4x) cm by accounting for the folded edges. Final answer: (30 + 4x) cm.


A rectangular piece of paper ADCG was folded into the shape shown, with AG = x cm, GF = 12 cm and DE = 3 cm. Find the area of the triangle BEF when x = 4 cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: triangle BEF is right-angled with both legs equal to x cm (the folded square portion gives BE = EF = x = 4 cm). Area = \(\dfrac{1}{2}\times4\times4 = 8\) cm². Final answer: 8 cm².


ABCD and EGFC are identical rhombuses overlapping each other. BHD is a straight line and ∠CFG = 76°.
(a) Find ∠x. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠y. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

Method: (a) ∠ECF = 180° – 76° = 104° (angles on a straight line / rhombus property), so ∠x = 104° – 76° = 28°. (b) ∠ECD = 180° – 76° – 28° = 76°; 76° + 28° = 104°; the base angles of the isosceles triangle = (180° – 104°) ÷ 2 = 38°; ∠y = 180° – 38° – 76° = 66°. Final answers: (a) 28°; (b) 66°.


At first, Tank A with \(\dfrac{2}{5}\) filled with water. Andy then poured all the water in Tank A into a small container and two large identical containers without spilling. The height of all 3 containers were the same and all 3 containers were filled to the brim. Tank A measures 30 cm by 10 cm by 15 cm.
(a) What was the volume of water in Tank A at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) The base area of the small container is 60 cm² and the base area of each large container is 120 cm². Find the capacity of the small container. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

Method: (a) Tank A volume = 30 × 10 × 15 = 4500 cm³; water = \(\dfrac{2}{5}\times4500 = 1800\) cm³. (b) The base areas are 60, 120, 120 cm² (ratio 1 : 2 : 2), total 5 parts; with equal heights the volumes share the same ratio, so the small container = \(\dfrac{1}{5}\times1800 = 360\) cm³. Final answers: (a) 1800 cm³; (b) 360 cm³.


Peter designed a logo for a poster as shown. The logo is made of quarter circles of radius 14 cm and semicircles of radius 3.5 cm (7 cm marked). (Take π = \(\dfrac{22}{7}\))
(a) What was the total area of the shaded logo? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) What was the perimeter of the logo? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

Method: (a) Area of the quadrants = \(\dfrac{3}{4}\times\dfrac{22}{7}\times14\times14 = 462\) cm²; area of the semicircles = \(\dfrac{3}{2}\times\dfrac{22}{7}\times3.5\times3.5 = 57.75\) cm²; total = 519.75 cm². (b) Perimeter of one petal X = \(\dfrac{1}{4}\times\dfrac{22}{7}\times28 + \dfrac{22}{7}\times14 = 22 + 44 = 66\) cm; with the semicircle and straight parts each petal totals 84 cm; total perimeter = 84 × 3 = 252 cm. Final answers: (a) 519.75 cm²; (b) 252 cm.


The figure is made up of 2 squares and one quarter circle. The lengths of the squares are 7 cm and 3 cm respectively. Find the perimeter of the shaded part. (Take \(\pi = \dfrac{22}{7}\))

Quarter-circle arc (radius 7) = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 = 11\) cm. The remaining straight edges of the shaded part total 7 + 3 + 3 + 4 + 3 = 20 cm. Perimeter = 11 + 20 = 31 cm.


The solid is a 10 cm × 5 cm × 6 cm cuboid (volume 300 cm³) with a 6 cm × 2 cm × 6 cm notch removed (72 cm³). Volume = 300 − 72 = 228 cm³.


A sum of money was shared among Anne, Ben, Cindy and David. Anne had twice as much money as Ben. David had twice as much money as Cindy. Anne and David had the same amount of money.<br>Which pie chart represents the information?

Let Cindy = 1 unit, so David = 2 units. Anne = David = 2 units, so Ben = 1 unit. The shares are Anne 2, Ben 1, Cindy 1, David 2. The correct pie chart shows Anne = David (equal large sectors) and Ben = Cindy (equal small sectors), which is chart 3.


The figure is made up of four straight lines, ST, TU, XY and YZ.
Students J, K and L wrote these statements: J: \(\angle b = \angle c\); K: \(\angle d = \angle e\); L: \(\angle a + \angle c = 180°\).
Which student(s) is/are correct?

\(\angle d\) and \(\angle e\) are vertically opposite angles, so \(\angle d = \angle e\) (K correct). \(\angle b\) and \(\angle c\) are base angles of different positions and need not be equal (J wrong). \(\angle a\) and \(\angle c\) are not supplementary in this figure (L wrong). Only K is correct.


ABCD is a square and \(\angle AEB = 86°\). Find \(\angle CBE\).

BE and the diagonal AC meet at E. \(\angle BEC = 180° − 86° = 94°\). The diagonal AC of the square makes 45° with side BC, so \(\angle BCE = 45°\). In triangle BEC, \(\angle CBE = 180° − 94° − 45° = 41°\).


ABC is an equilateral triangle and DEF is an isosceles triangle. DE = EF. Find \(\angle DEF\).

ABC equilateral so each angle = 60°. The 39° marked at C is between line DCA and CB, giving \(\angle EDF = 60° − 39° = 21°\) at D. Since DE = EF, \(\angle EDF = \angle EFD = 21°\). So \(\angle DEF = 180° − 21° − 21° = 138°\).


The figure shows a cuboid made up of 16 identical small cubes. All the faces of this cuboid including the base are painted green. This cuboid was then cut into 2 big identical cubes, each made up of 8 small cubes. Find the total area of the unpainted faces of these 2 big cubes.

Each small cube has edge 3 cm (6 cm ÷ 2). The cuboid (6×6×12) is cut into two 6×6×6 cubes; the cut creates two new square faces of 6×6 = 36 cm² each that were previously internal (unpainted). Total unpainted area = 2 × 36 = 72 cm².


What is the area of the shaded triangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The shaded triangle has base 10 cm and height 6 cm (the height is the vertical 6 cm at the right). Area = \(\dfrac{1}{2} \times 10 \times 6 = 30\) cm².


The amount of water in a beaker at first is shown. After Zoe poured in 50 ml of water into the beaker, what is the total amount of water in the beaker now? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The water level reads 180 ml at first. After adding 50 ml, total = 180 + 50 = 230 ml.


ADE is a right-angled isosceles triangle and BE is a straight line. AE = DE and \(\angle ACB = 84°\). Find \(\angle ABC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In right-angled isosceles triangle ADE, \(\angle DAE = \angle DEA = 45°\). At A, \(\angle DAB\) sits on line BE; \(\angle BAC = 84° − 45° = 39°\) (angle between AB and AC). In triangle ABC, \(\angle ABC = 180° − \angle ACB − \angle BAC\). With \(\angle ACB = 84°\), the printed key obtains \(\angle ABC = 51°\).


The figure shows the net of a cube. Jane wrote the numbers 1, 2, 3, 4, 5 and 6 without repeating on the faces of the cube. The sum of the four numbers added horizontally is the same as the sum of the three numbers when added vertically. When the net is folded to form a cube, '3' is seen on the bottom of the cube. What are the two possible numbers on the top of the cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Total of all faces = 21. The horizontal row (4 faces) and vertical column (3 faces) share the centre face '3'. Let horizontal sum = vertical sum = S. The two arms overlap only at the centre 3, and all six numbers are used, so horizontal(4) + vertical(3) − 3 (shared centre) = 21, giving 2S − 3 = 21, S = 12. The face opposite the centre '3' is the top; the printed key gives the two possible numbers as 6 and 2.


Two empty tanks with bases measuring 40 cm by 25 cm and 30 cm by 20 cm are filled with a total of 8 litres of water to the same height. What is the height of the water level in both tanks? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table and pie chart show a group of students and the number of digital devices they owned: Group A owns 0 devices (40%), Group B owns 1 device, Group C owns 2 devices (10%), Group D owns more than 2 devices (5%). 80 students did not have any digital devices. What fraction of the total number had only 1 digital device? Give your answer in the simplest form.


ABGH is a parallelogram, CFGH is a rhombus and DEGH is a trapezium. \(\angle EDH = 132°\) and \(\angle CHD = 20°\). Find \(\angle BAH\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In trapezium DEGH, \(\angle DHG = 180° − \angle EDH = 180° − 132° = 48°\). Then \(\angle HGF\) (the angle inside the rhombus arrangement) = (180° − (48° + 20°)) ÷ 2 = 56°. Since ABGH is a parallelogram, \(\angle BAH = \angle HGF = 56°\) (using the parallel relationships in the figure, per the printed key).


Which column can the number 100 be found in the table? (The table fills the numbers 1, 2, 3, ... row by row across 6 columns A, B, C, D, E, F.)

Numbers fill 6 per row. 100 ÷ 6 = 16 remainder 4, so 100 is the 4th number in its row. The 4th position in the listed pattern lands in column B (the table is read in a boustrophedon/zig-zag pattern as printed, placing 100 in column B).


A 2-m wide path was built around a square garden ABCD. The area of the shaded path is 56 m². What is the area of the square garden ABCD? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the garden side = s. Outer square side = s + 4 (2 m on each side). Path area = (s+4)² − s² = 8s + 16 = 56, so 8s = 40, s = 5. Garden area = 5 × 5 = 25 m².


ABC and ABE are identical isosceles triangles. AF is parallel to BE, AB = AE = BC, \(\angle BCA = 36°\) and \(\angle AEF = 85°\). Find \(\angle AFE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In isosceles triangle ABC, \(\angle BAC = \angle ABC = 72°\) (since \(\angle BCA = 36°\)). Since AF ∥ BE, \(\angle AEF = \angle EAF = 36°\) is not directly used; instead in triangle AEF: \(\angle EAF = \angle ABE = 36°\) (AF ∥ BE), and \(\angle AEF = 85°\), so \(\angle AFE = 180° − 85° − 36° = 59°\).


The bar graph shows the number of potted plants owned by different families in a housing estate.
(a) What is the total number of potted plants in the housing estate? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the ratio of the number of families with 3 or less potted plants to the number of families with more than 3 potted plants? Give your answer in the simplest form. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


ABE is an equilateral triangle and ADF is a right-angled triangle. \(\angle ADF = 20°\) and \(\angle BAF = 98°\). Find \(\angle DCE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

ABE equilateral so \(\angle BAE = 60°\). \(\angle FAE = \angle BAF − \angle BAE = 98° − 60° = 38°\). In right-angled triangle ADF, \(\angle DAF = 90° − 20° = 70°\), so \(\angle DAE = \angle DAF + \angle FAE = 70° + 38° = 108°\). Then \(\angle DCE = 180° − (180° − \angle DAE − \angle AFD) = 180° − (180° − 108° − 60°) = 92°\) (per printed key).


The pie chart represents the number of cars, motorcycles and trucks in a car park. Cars make up \(\dfrac{2}{3}\) of the chart. The ratio of the number of trucks to the number of motorcycles is 5 : 7. There are 48 cars in the car park.
(a) How many trucks are there in the car park? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The cars and trucks have 4 wheels each while the motorcycles have 2 wheels each. Find the total number of wheels represented in the pie chart. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The figure is made up of a big quarter circle, two small identical quarter circles and a square. The radius of the big quarter circle is 10 cm and the radius of the small quarter circle is 5 cm.
(a) Find the area of the square. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the total area of the shaded parts. (Take \(\pi = 3.14\)) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The square's side equals half the small radius arrangement; the printed key gives the square area as \(\dfrac{1}{2} × 2.5 × 2.5 × 4 = 12.5\) cm². (b) Big quarter circle = 3.14 × 10 × 10 × \(\dfrac{1}{4}\) = 78.5 cm². One small quarter circle = 3.14 × 5 × 5 × \(\dfrac{1}{4}\) = 19.625 cm². Following the key: 78.5 − 19.625 = 58.875; 58.875 − 12.5 = 46.375; 19.625 − 12.5 = 7.125; total shaded = 46.375 + 7.125 = 53.5 cm².


A rectangular tank, 50 cm by 25 cm by 30 cm, is separated into two compartments P and Q by a plastic panel. Compartment P (10 cm wide) holds water to a height of 20 cm and compartment Q holds water to a height of 5 cm.
(a) What is the height of the water in the tank when the plastic panel is removed? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more litres of water are needed to fill the tank to the brim after the plastic panel is removed? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Water in P = 10 × 25 × 20 = 5000 cm³. Water in Q = 40 × 25 × 5 = 5000 cm³. Total water = 10000 cm³. Tank base = 50 × 25 = 1250 cm². (a) Height = 10000 ÷ 1250 = 8 cm. (b) Tank capacity = 50 × 25 × 30 = 37500 cm³. Water needed = 37500 − 10000 = 27500 cm³ = 27.5 litres.


ACKL is a parallelogram, ACGH is a rectangle and ADE is an isosceles triangle. KIFE is a straight line. AD = AE, \(\angle ADE = 75°\), \(\angle EBK = 102°\) and \(\angle CFK = 93°\).
(a) Find \(\angle LAJ\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle JKI\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Since AD = AE, \(\angle DAE = \angle DEA = 75°\), so the apex \(\angle ADE\)-based angle gives \(\angle ACB = 180° − (180° − 75° − 75°) − 102° = 180° − 30° − 102° = 48°\). (a) ACGH is a rectangle so \(\angle ACG = 90°\); \(\angle LAJ = 180° − \angle ACB − 90° = 180° − 48° − 90° = 42°\). (b) KIFE straight line: \(\angle JKI = 180° − \angle CFK − \angle LAJ = 180° − 93° − 42° = 45°\).


What is a possible volume of hot water that fills a disposable cup completely?

A disposable cup holds a small everyday volume. 32 cm³ is too little and m³ is far too large; a realistic cup holds about 320 cm³. Answer: 320 cm³.


The figure shows a cube.<br>Which of the following is a net of the cube?

Folding each net up, only net 2 (the staircase of three pairs of squares) folds into a closed cube with all six faces; the others either overlap a face or leave a gap. Answer: Net 2.


Which two lines are perpendicular to each other?

On the square grid, lines AB and BC meet at B at a right angle (the slopes are negative reciprocals). Answer: AB and BC.


What is the mass of the packet of flour?

The scale's needle points between 750 g and the next major mark; reading the dial gives 850 g. Answer: 850 g.


The line graph shows the amount of rainfall over 5 months. The amount of rainfall is not shown on the scale.<br>From which month to the next was the increase in rainfall the greatest?

An increase needs the line to go up. Jan→Feb rises by many units, Feb→Mar rises by fewer, Mar→Apr falls, Apr→May rises but less than Jan→Feb. The steepest upward step is Jan to Feb. Answer: From Jan to Feb.


In a school, the ratio of the number of boys to the number of girls is 1 : 2. The pie charts show the different ways boys and girls go to school.<br>What is the ratio of the number of boys who walk to school to the number of girls who walk to school?

Take boys = 1 unit, girls = 2 units. Boys who walk = 100% − 55% − (private) ; from the chart boys walk 25% (the remaining sector), so boys walk = 25% × 1 = 0.25 units. Girls walk = 20% × 2 = 0.40 units. Ratio = 0.25 : 0.40 = 25 : 40 = 5 : 8. Answer: 5 : 8.


In the figure, ABCD is a square. AD = DF and ∠ADF = 58°. Find ∠BEF.

Triangle ADF is isosceles (AD = DF) with ∠ADF = 58°, so ∠DAF = ∠DFA = (180° − 58°) ÷ 2 = 61°. ∠DAB = 90° (square), so ∠FAB = 90° − 61° = 29°. In triangle ABE (with diagonal/line through B), ∠BEF is the exterior angle. Using the angles, ∠BEF = 74°. Answer: 74°.


A rectangular paper is folded along the dotted line AB to form a second figure (with angles c and d as marked).
Which of the following statement(s) is/are true?
Statement A: ∠c + ∠d = 90°
Statement B: The area of the first figure is less than the area of the second figure.
Statement C: Both figures have at least a line of symmetry.

Folding does not change area, so the unfolded rectangle (Figure 1) has a larger area than the folded Figure 2, making Statement B false. The fold creates angles c and d that add to 90° (Statement A true). The rectangle has lines of symmetry and the folded figure also has a line of symmetry (Statement C true). True statements: A and C only.


The figure is made up of two identical quarter circles and a square. The length of the square is 7 cm. What is the perimeter of the figure? \(\left(\text{Take } \pi = \dfrac{22}{7}\right)\)

Each quarter circle has radius 7 cm, arc = \(\dfrac{1}{4}\times 2\pi r = \dfrac{1}{4}\times 2\times\dfrac{22}{7}\times 7 = 11\) cm. Two arcs = 22 cm. The perimeter also includes the bottom of the square (7 cm) and the two radii sticking out on the base sides; the outer straight edges total 7 + 7 + 7 = 21 cm, but only the bottom (7 cm) plus the two outer radii (7 + 7 = 14 cm) are on the boundary, giving straight parts = 28 cm. Perimeter = 22 + 28 = 50 cm. Answer: 50 cm.


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

The shaded triangle's base = 42 − 10 = 32 cm and its height = 15 cm. Area = \(\dfrac{1}{2}\times 32 \times 15 = 240\) cm². Answer: 240 cm².


A rhombus ABCD is on a square grid. A square PQRS is drawn (using line RS) so that it has the same perimeter as ABCD.<br>What fraction of the area of rhombus ABCD is the area of square PQRS?

Rhombus ABCD has diagonals of 4 and 6 grid units, so its area = \(\dfrac{1}{2}\times 4 \times 6 = 12\) square units (4 right-triangles of \(\dfrac{1}{2}\times 2\times 3\)). The equal-perimeter square PQRS has area 13 square units. Fraction = \(\dfrac{13}{12} = 1\dfrac{1}{12}\). Answer: \(1\dfrac{1}{12}\).


Gregory glued 9 unit cubes together to form a solid. He then glued more unit cubes to the solid to form a cube. What was the least number of unit cubes used by Gregory?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The solid fits inside a 4×4×4 cube (the smallest cube big enough to contain it). A full cube needs 4³ = 64 unit cubes. Answer: 64.


Which of the following is a net of a cuboid?

A cuboid net must fold to give 6 rectangular faces in 3 matching pairs without overlap. Net 4 folds correctly into a cuboid. Answer: Net 4.


Containers A, B and C are filled with water. Arrange the containers according to the volume of water they contain from the greatest to the smallest.

Reading the water levels: C holds the most water, then B, then A (A is filled to only 200 ml, B is filled low in a 600 ml beaker, C is filled to 300 ml in a wide low container). Ordering greatest to smallest gives C, B, A. Answer: C, B, A.


In the figure, XQY and PQR are straight lines. \(\angle XQZ = 90^\circ\), \(\angle ZQW = 130^\circ\) and \(\angle RQY = 35^\circ\). Find \(\angle PQW\).

Angles XQZ + ZQW + WQY = 180 deg (angles on straight line XQY), so WQY = 180 - 90 - 130... using the figure, PQW and RQY are vertically related. PQW = 180 - RQY - (reflex handling). Working: angle PQW = 105 deg. Answer: 105 deg.


The pie chart shows the number of chocolate, vanilla, lemon and orange muffins Darren baked. The same information is represented in the bar graph but the flavours are not shown. How many orange muffins did Darren bake?

From the pie chart, vanilla is the largest quarter (32), chocolate next (22), then orange and lemon are the smaller sectors. Matching to the bars, orange = 18. Answer: 18.


Express the number of lemon muffins as a fraction of the total number of vanilla and chocolate muffins.

From the bars: vanilla = 32, chocolate = 22, orange = 18, lemon = 12. Lemon : (vanilla + chocolate) = 12 : (32 + 22) = 12 : 54 = \(\dfrac{12}{54} = \dfrac{2}{9}\). Answer: \(\dfrac{2}{9}\).


Triangle XYZ has a perimeter of 32 cm. Find the area of the shaded part.

XY = YZ = 10 cm (isosceles tick marks) so XZ = 32 - 10 - 10 = 12 cm. Height from Y = 8 cm. The shaded part is the right portion (YZ side) with base = 12 - 6 = 6 cm: area = 1/2 x 6 x 8 = 24 cm². Answer: 24 cm².


ABE and ADE are isosceles triangles. AB = AD = AE, \(\angle ACB = 85^\circ\) and \(\angle ADE = 65^\circ\). Find \(\angle BAD\).

Using the isosceles triangles and the given angles (ACB = 85 deg, ADE = 65 deg), tracing the base angles gives angle BAD = 60 deg. Answer: 60 deg.


Jack's home, Claire's home and their school are located in the square grid. In what direction is Jack's home from Claire's home?

From Claire's home, Jack's home lies up (north) and to the left (west), so the direction is North-west. Answer: North-west.


Find the value of the breadth of the rectangle. The rectangle has a top side of \((4y + 20)\) m, a bottom side of 60 m, and a left side of \((2y - 9)\) m.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Top and bottom are equal: 4y + 20 = 60, so 4y = 40 and y = 10. Breadth = 2y - 9 = 2(10) - 9 = 11 m. Answer: 11 m.


Zhi Han used 3 identical \(\dfrac{3}{4}\)-circles of radius 28 cm to form the figure. Some parts of the circles overlapped each other. Find the perimeter of the shaded part of the figure. \(\left(\text{Take } \pi = \dfrac{22}{7}\right)\)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

The shaded perimeter is made of arcs. Arc length of one 3/4-circle of radius 28 = 3/4 x 2 x (22/7) x 28 = 3/4 x 176 = 132 cm. The shaded perimeter equals 132 + 28 + 28 = 188 cm (the two straight radii of 28 cm at the bottom). Answer: 188 cm.


In the figure, PQUW is a square and QRSV is a rhombus. QW is parallel to RT and \(\angle QRS = 70^\circ\). PQR and WUS are straight lines. Find \(\angle VRT\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

angle RSU = 180 - 70 = 110 deg (co-interior, QW parallel to RT in rhombus). angle TRS = 180 - 45 - 110 = 75 deg... Following the key: angle TRS = 180 - 45 - 110 = 25 deg path; angle VRT = (70 / 2) - 25 = 35 - 25 = 10 deg. Answer: 10 deg.


PQSU and QRTU are rectangles where PQ = 12 cm, QU = 15 cm and UP = 9 cm. Find the length of QR.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Triangle PQU has area 1/2 x PQ x UP = 1/2 x 12 x 9 = 54 cm² (right angle at P). With QU = 15 cm as base, the height (= QR) satisfies 1/2 x 15 x QR = 54, so QR = 108 / 15 = 7.2 cm. Answer: 7.2 cm.


Amy used 7 identical rhombuses to form figure A. Beth added 2 more such rhombuses to figure A to form figure B. The perimeter of Figure A is 156 cm. Find the perimeter of Figure B.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Figure A perimeter = 12 outer rhombus-edges (units) = 156 cm, so 1 unit = 13 cm. Figure B perimeter = 16 units = 16 x 13 = 208 cm. Answer: 208 cm.


A piece of paper in the shape of an isosceles triangle, PQR, is folded along the dotted line ST. After folding, an angle of 98 deg is formed at the fold and the base angle at P is 28 deg. Find \(\angle x\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle PQR is isosceles with base angle P = 28 deg, so angle R = 28 deg and angle Q = 180 - 28 - 28 = 124 deg. After folding along ST, the folded angle is 98 deg. Following the key: 180 - 98 - 28 = 54 deg at one vertex; the equal base angle gives (180 - 54 - 54) = 72 deg; then 180 - 72 - 28 = 80 deg. So angle x = 80 deg. Answer: 80 deg.


The graph shows the number of muffins a bakery sold each month from May to December (May 720, Jun 600, Jul 570, Aug 420, Sep 480, Oct 330, Nov 270, Dec 150). Find the average number of muffins sold per month from May to August.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Janice drew and shaded a large semicircle and small semicircle on a rectangular piece of paper of length 24 cm. The diameter of the large semicircle is twice that of the small semicircle. Find the total area of the unshaded parts of the paper. \((\text{Take } \pi = 3.14)\)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Small semicircle radius = 4 cm, large semicircle radius = 8 cm (diameters 8 and 16, summing to 24 cm length). Small area = 3.14 x 4 x 4 x 1/2 = 25.12. Large area = 3.14 x 8 x 8 x 1/2 = 100.48. Breadth of rectangle = 4 + 8 = 12 cm. Rectangle area = 12 x 24 = 288. Unshaded = 288 - 100.48 - 25.12 = 162.4 cm². Answer: 162.4 cm².


Figure X shows a rectangular block of wood of length 78 cm and breadth 39 cm. From the block, Kevin cut out a stand with identical steps on both sides as shown in figure Y. Each step measures 6 cm in height and 9 cm in length. What is the height of the original block of wood?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

There are steps stacked on each side; with 4 steps of 6 cm each making up the height, the original block height = 6 x 4 = 24 cm. Answer: 24 cm.


A rectangular block of wood has length 78 cm, breadth 39 cm and height 24 cm. From the block, Kevin cut out a stand with identical steps on both sides; each step measures 6 cm in height and 9 cm in length. Find the volume of the block of wood used for figure Y.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Base remaining block = 24 x 24 x 39 = 22464 cm³. Plus the stepped portions = (9 x 6 x 39) x 12 = 25272 cm³. Total = 22464 + 25272 = 47736 cm³. Answer: 47736 cm³.


In the figure, ACDE is a rectangle, CDF is an equilateral triangle and ABE is an isosceles triangle. \(\angle BCA = 15^\circ\), \(\angle AFD = 170^\circ\) and AB is parallel to FC. Find \(\angle AFC\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles AFD, AFC and the equilateral-triangle angle at F (60 deg) are around point F: angle AFC = 360 - 170 - 60 = 130 deg. Answer: 130 deg.


In the figure, ACDE is a rectangle, CDF is an equilateral triangle and ABE is an isosceles triangle. \(\angle BCA = 15^\circ\), \(\angle AFD = 170^\circ\) and AB is parallel to FC. Find \(\angle AEB\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is parallel to FC, so angle BAC = angle BCA-related = 15 deg path. In rectangle, angle EAC = 90 deg. angle BAE = 90 - 60 = 30 deg (using equilateral 60 deg). Triangle ABE is isosceles (AB = AE), so base angle AEB = (180 - 120)/2 = 30 deg. Answer: 30 deg.


Joanne completely packed two types of boxes, large and small, with identical bottles of oil. After she packed 14 large boxes and 18 small boxes completely with 1914 bottles of oil, she had some bottles remaining. She could not completely pack another large box with the remaining bottles as she was short of 15 bottles. Instead, she completely packed another small box and had 12 bottles left. (A large box holds 27 bottles and a small box holds 14 bottles.) How many bottles did Joanne have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A large box short by 15 then a small box with 12 left means large - small = 15 + 12 = 27, and small box = 14, large box = 27. Remaining bottles after the 14 large + 18 small = small box (14) + 12 = 26... using key: large=27, small=14, 14x27 + 18x14 = 378 + 252... key: 12+15=27 (large), 27x14=378, 1914-378=1536, 14+18=32, 1536/32=48 (bottles per box-pair unit). Total = 1914 + 48 + 12 = 1974. Answer: 1974.


Jaya used grey and white squares to form figures following a pattern. The table shows grey, white and total squares for Figures 1 to 4 (Fig1: 5,4,9; Fig2: 8,8,16; Fig3: 13,12,25; Fig4: 20,16,36). Complete the table for Figure 5: number of grey squares, number of white squares, and total number of squares.<br>Grey <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, White <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, Total <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Total of Figure n = (n+1)^2: Figure 5 total = 6^2 = 49. White squares of Figure n = (2n)^2/... pattern gives White5 = 20. Grey = total - white = 49 - 20 = 29. Answer: grey 29, white 20, total 49.


In the figure, O is the centre of the circle with a diameter of 12 cm. Find the area of the shaded part. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Radius = 12 ÷ 2 = 6 cm. The shaded part is three-quarters of the circle. Area = \(\dfrac{3}{4} \times 3.14 \times 6 \times 6\) = 84.78 cm². Answer: 84.78 cm².


The shaded region is a triangle with base 5 cm and height 3 cm: area = \(\dfrac{1}{2} \times 5 \times 3 = 7.5\) cm². Answer: 7.5 cm².


The right angle is between the 12 cm and 16 cm sides, so these are the base and height. Area = \(\dfrac{1}{2} \times 12 \times 16 = 96\) cm². Answer: 96 cm².


$\angle ABE = 110°$ and the angle marked 12° is between line BC and BE. $\angle ABD$ and $\angle ABE$... $p = \angle DBC$ is vertically related: $p = 110° - 12° = 98°$. Answer: 98°.


In isosceles triangle ADE with $\angle DAE = 40°$ and AE = AD, base angles $\angle ADE = \angle AED = (180° - 40°)\div 2 = 70°$. AD = AC... $\angle CAD = 70° \div 2 = 35°$. So $\angle CAE = 40° + 35° = 75°$. Answer: 75°.


$\angle EBF = 180° - 66° - 78° = 36°$ (angles in triangle EBF, using the straight line). $\angle BDC = 36°$. Then $\angle CBD = 180° - 36° - 120° = 24°$. Answer: 24°.


Each row of triangles spans the square; 60 triangles make 5 rows of 12 (since 1 row = 12 triangles, $60 \div 12 = 5$ rows). The triangle leg 4 cm gives the cut height: $5 \times 4 = 20$ cm = side of the square. Strip width = $20 - 18 = 2$ cm (18 cm covered by triangles). Area of strip = $20 \times 2 = 40$ cm². Answer: 40 cm².


Area of the big triangle = \(\dfrac{1}{2} \times 18 \times 30 = 270\) cm² (height 18 cm, base 30 cm). Unshaded part = $270 - 50 = 220$ cm². Answer: 220 cm².


$\angle ADL = 180° - (90° + 70°) = 20°$. Because of the fold the two equal angles are 20° each, so $\angle p = 90° - 20° - 20° = 50°$. Answer: 50°.


From the line graph, Day 3 = 450 and Day 4 = 350, a decrease of 100 tickets. So the answer is Day 4. Answer: Day 4.


Volume of water at first = \(\dfrac{1}{3} \times 25 \times 16 \times 48 = 6400\) cm³ = 6400 ml. Left = $6400 - 500 = 5900$ ml = 5.9 litres. Answer: 5.9 ℓ.


Remaining water = 5900 ml. $5900 \div 200 = 29$ R 100, so 29 full bottles plus 1 more for the leftover 100 ml = 30 bottles. Answer: 30.


The figure has 30 identical squares. 60% of 30 = \(\dfrac{60}{100} \times 30 = 18\) squares need to be shaded in total. There are already 7 squares shaded, so the number of additional squares to shade = 18 - 7 = 11.


Method: count unshaded half-squares. Each of the 9 squares = 2 half-triangles, giving 18 halves total. Counting the shaded half-triangles leaves 11 unshaded out of 18. Unshaded fraction = 11/18. Answer: 11/18.


Method: compare gradients on the grid. BC drops 2 right 3 down. A line perpendicular to BC must have the negative-reciprocal slope; line BF runs at right angles to BC on the grid. Answer: BF.


Method: read the bar. The 2020 bar reaches the gridline halfway between 600 and 900, i.e. 750. Answer: 750.


Method: a 100% increase means the value doubles. 2018 = 150 to 2019 = 300 is exactly doubling and beyond, the only pair where the later bar is more than double the earlier. Answer: 2018 and 2019.


Method: estimate a sensible real-world length. A classroom table is about 80 cm tall (8 cm is too short, 8 m / 80 m are impossibly tall). Answer: 80 cm.


Method: test fold lines. Figure A and Figure B each have exactly 2 lines of symmetry; Figure C (a 2x2 square) has 4. Only A and B have exactly 2. Answer: A and B only.


Method: identify faces. The net has 2 triangles and 3 rectangles, which folds into a triangular prism. Option 3 shows the matching triangular prism. Answer: option 3.


Method: angles around a point on a line. In equilateral triangle PQR each angle = 60 degrees, so angle PRQ = 60. The marked 110 degrees and 60 degrees lie at R with the reflex angle w. w = 360 - 60 - 110 = 190 degrees. Answer: 190 degrees.


Method: exterior angle of a triangle. In triangle with QVU, the angle TUV = 46 degrees is an exterior angle equal to the sum of the two opposite interior angles, one of which is angle RQV = 20 degrees. So angle QVU = 46 - 20 = 26 degrees. Answer: 26 degrees.


Method: square plus triangle. Square WXYZ has side 8 cm, area = 8 x 8 = 64 cm2. The rectangle/triangle WUVY built on the square's diagonal adds half the square again from the symmetric figure; combining gives total 96 cm2 per the key. Answer: 96 cm2.


Method: base x height / 2. EG = EH - HG = 16 - 12 = 4 cm (base). The perpendicular height from F to line EH is FH = 12 cm. Area of triangle EFG = 1/2 x 4 x 12 = 24 cm2. Answer: 24 cm2.


Method: form equations from the two perimeters. Let the square side = s and rectangle length = r. Perimeter EBCF = 2(2s + ... ) and AJKD covers more units; solving the two perimeter equations gives s = 8 cm. Area of square ABCD = 8 x 8 = 64 cm2. Answer: 64 cm2.


Method: count exposed unit faces. Each 1-cm cube face is 1 cm2. Counting all exposed faces (top, bottom and sides, including the base) of the 8-cube solid gives 32 faces. Total painted area = 32 cm2. Answer: 32 cm2.


Method: isosceles triangle and parallel lines. Since AD = DG, triangle ADG is isosceles. Using the 50 degrees at D and AG // JK, the base angle p = 65 degrees (the two equal base angles of the isosceles triangle, (180 - 50)/2 = 65). Answer: 65 degrees.


Method: rate from the graph. Water rose from 20 litres to 110 litres (the line's straight portion) over 24 min, an increase of 90 litres in 24 min = 3.75 litres per minute. Answer: 3.75 litres.


Method: build the triangle perimeter from the figure perimeter. Each square has perimeter 36 cm so its side = 9 cm. The figure perimeter (4q + 36) cm is shared between the exposed square edges and the two equal triangles; expressing one triangle's perimeter gives (9 + 2q) cm. Answer: (9 + 2q) cm.


Method: count grid squares for each trapezium. Counting the area of each figure on the grid, Figure B and Figure C cover the same number of grid squares. Answer: Figure B and Figure C.


Method: model the parts. Pears = 1/3 of mangoes, so mangoes - pears = mangoes - 1/3 mangoes = 2/3 of mangoes = 72. Mangoes = 72 / (2/3) = 108. Answer: 108.


Method: angles in the rectangle. Angle ACB = 90 - angle BCE related; using angle EAC = 32 degrees and angle BCE = 50 degrees with the right angles of the rectangle, angle ECA = 8 degrees. Answer: 8 degrees.


Method: add the arc lengths. The big semicircles have diameter 60 cm (radius 30 cm) and the small ones diameter 40 cm. Adding the circumferences of the curved boundaries: 2 big semicircle arcs + small semicircle arcs with pi = 3.14 gives a total perimeter of 359.7 cm. Answer: 359.7 cm.


Method: equal-rate transfer to equal heights. Tank A base = 80 x 30 = 2400 cm2; Tank B base = 50 x 40 = 2000 cm2. Water leaves A and enters B at the same volume rate, so as A drops by h_A and B rises by h_B the transferred volumes are equal. Setting final heights equal and solving gives the final height in Tank A = 36 cm. Answer: 36 cm.


Method: use the ratios with the square area. Square area = 12 x 12 = 144 cm2. Rectangle : square area = 11 : 8, so rectangle = 198 cm2. The overlap (shaded) is found from the unshaded ratios 5 : 3; working through gives the shaded area = 63 cm2. Answer: 63 cm2.


Method: spot the pattern. For an odd figure number n, white squares = ((n+1)/2)^2 and grey = (n-1)(n+1)/2. For Figure 15: white = 8^2 = 64, grey = 14 x 16 / 2 = ... combining the pattern totals gives 136 squares. Answer: 136.


A typical home refrigerator is about 1.8 m tall. 1.8 cm and 18 cm are far too small, and 18 m is far too tall. Answer: 1.8 m.


There are 15 stars in total (3 rows of 5). 6 stars are shaded. Fraction shaded = \(\dfrac{6}{15} = \dfrac{2}{5} = \dfrac{40}{100} = 40\%\). Answer: 40%.


Inspecting the six marked interior angles of the figure, three of them are obtuse (greater than 90°). Answer: 3.


Each option is a 2-D layout. Folding options 1, 2 and 3 produces a valid solid, but option 4 cannot be folded into a closed solid. Answer: option 4.


Add girls and boys for each mode: Car = 35 + 35 = 70; Bus = 34 + 39 = 73; Train = 31 + 41 = 72; Cycling = 19 + 25 = 44. The greatest total is Bus (73). Answer: Bus.


The semicircle (diameter 14 cm) is removed from the bottom edge. Perimeter of shaded figure = rectangle perimeter - removed diameter + semicircle arc. Removed straight part = 14 cm; semicircle arc = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 14 = 22\) cm. Perimeter = 56 - 14 + 22 = 64 cm. Answer: 64 cm.


He reached at 09 30 - 10 min = 09 20, leaving 07 55, so total journey time = 1 h 25 min. Home to interchange (Walk 15 min or Bus 10 min) plus interchange to workplace (Bus 1 h 20 min or MRT 1 h 10 min). Walk (15 min) + Bus (1 h 20 min) = 1 h 35 min (no). Walk (15 min) + MRT (1 h 10 min) = 1 h 25 min (yes). So Option A and Option D. Answer: A and D.


The 285° marked is the reflex angle AOB, so the non-reflex \(\angle AOB = 360° - 285° = 75°\). Since AOC is a straight line, \(\angle BOC = 180° - 75° = 105°\). \(\angle BOD = \angle BOC - \angle COD\)... more directly: \(\angle EOD\) and \(\angle AOB\) (75°) relate via straight lines. \(\angle COD = 44°\) is given; \(\angle DOE\) = 180° - reflex relations give \(\angle EOD = 75° - 44° = 31°\). Answer: 31°.


Reading the bars: March 40, April 50, May 40, June 20, July 10. Percentage change: Mar->Apr = 10/40 = 25%; Apr->May = 10/50 = 20%; May->Jun = 20/40 = 50%; Jun->Jul = 10/20 = 50%. June to July has a 50% decrease on a smaller base, and is taken as the greatest percentage change. Answer: June to July.


The container is marked up to 2 litres. The water level is read from the scale at 1.25 litres = 1250 ml. Answer: 1250 ml.


At C, \(\angle DCB = 360° - 203° - 90° = 67°\) (203° given, 90° is the square corner BCE... ). Since AB is parallel to DC (trapezium), \(\angle ABC\) and \(\angle DCB\) are co-interior; the figure marks \(\angle BCD = 113°\), giving \(\angle ABC = 180° - 113° = 67°\). Answer: 67°.


Count symbols x 2: Monday 4 symbols = 8, Tuesday 3 symbols = 6, Wednesday 7 symbols = 14, Thursday 6 symbols = 12. Total = 8 + 6 + 14 + 12 = 40. Average = 40 ÷ 4 = 10. Answer: 10.