The design is made up of 2 identical quadrants, two identical squares ABCD and JKLM, and a small square CEJF. The area of square CEJF is 25 cm² and the radius of the quadrants is 18 cm. Find the area of the shaded parts. (Take \(\pi=3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Following the key: the area of square DEMH (side 18 cm) \(=18\times18=324\) cm². The area of one quadrant \(=\dfrac{1}{4}\times3.14\times18\times18=254.34\) cm². Total \(=324+254.34=578.34\) cm². Subtract the small square CEJF (25 cm²): shaded area \(=578.34-25=553.34\) cm². Answer: 553.34 cm².


The figure is made up of a rhombus ABEG and a square BCDF.
(a) Name an obtuse angle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find ∠x. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°
(c) Find ∠y. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />°

(a) ∠y is an obtuse angle (it works out to 105°, greater than 90°).
(b) ∠x: angles on a straight line at E — 180 − 128 = 52°.
(c) ∠y: in triangle FEG, the angle at E inside = ∠x + 23 = 52 + 23 = 75°; ∠y is on a straight line with this, so ∠y = 180 − 75 = 105°.


The figure is formed by three identical circles with centres A, C and E. ABEF and BCDE are identical squares of side 10 cm. AC and FD are straight lines. (Take π = 3.14)
(a) Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the perimeter of the shaded parts. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Radius = 10 cm. Unshaded inside each square pair: half a square (\(\dfrac{1}{2}\)×10×10 = 50) and a quarter circle (\(\dfrac{1}{4}\)×3.14×10×10 = 78.5), so the white bite per region = 78.5 − 50 = 28.5 cm². Area of one full circle = 3.14×10×10 = 314; three circles = 314×3 = 942. Subtract the white bites: 28.5×8 = 228; shaded area = 942 − 228 = 714 cm².<br>(b) Perimeter: circumference of one circle = 3.14×20 = 62.8 cm; the shaded boundary = 62.8×3 = 188.4 cm.


The line graph shows the mass of newspapers a class collected at the end of each month for 6 months.
(a) In which month was the increase in the mass of newspapers collected the greatest? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the percentage decrease in the mass of newspapers collected from March to April. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%
(c) The collection of newspapers continued until December. In February, the class collected \(\dfrac{2}{9}\) of the total mass of newspapers collected. What was the total mass of the newspapers collected from January to December? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> kg

(a) Compare monthly increases: Jan→Feb 17.4−13.2 = 4.2; Feb→Mar 20−17.4 = 2.6; Apr→May 12.6−10.6 = 2; May→Jun 21.4−12.6 = 8.8. Greatest increase is in June.
(b) March = 20 kg, April = 10.6 kg; decrease = 20 − 10.6 = 9.4 kg; percentage = \(\dfrac{9.4}{20}\) × 100 = 47%.
(c) February mass = 17.4 kg = \(\dfrac{2}{9}\) of the total, so 1/9 = 17.4 ÷ 2 = 8.7 kg; total = 8.7 × 9 = 78.3 kg.


Basheer drew some identical rectangles along a straight line on a piece of paper to form the figure as shown. Each rectangle has a length of 7.2 cm and a breadth of 5 cm. The shaded squares are the overlapping parts. Each shaded square has a side of 2.4 cm.
(a) What was the length of the figure formed by Basheer using 4 such rectangles? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of the figure which was formed by using 4 such rectangles. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²
(c) Basheer continued to draw more rectangles along the straight line. The distance between the first rectangle and the last rectangle was 290.4 cm. How many rectangles did he draw altogether? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Each rectangle is 7.2 cm long; consecutive rectangles overlap by 2.4 cm, so each new rectangle adds 7.2 − 2.4 = 4.8 cm.
(a) 4 rectangles: 4.8 + 4.8 + 4.8 + 7.2 = 21.6 cm.
(b) Total area of 4 rectangles = 7.2 × 5 × 4 = 144 cm². Each overlap square = 2.4 × 2.4 = 5.76 cm²; there are 3 overlaps = 5.76 × 3 = 17.28 cm² counted twice. Area = 144 − 17.28 = 126.72 cm².
(c) Distance first→last = 290.4 cm. Each step (gap) = 4.8 cm after removing one final rectangle's 7.2: 290.4 − 7.2 = 283.2; 283.2 ÷ 4.8 = 59 steps; rectangles = 59 + 1 = 60.


The solid is made up of 2-cm cubes. Find its volume.

There are 6 unit cubes in the solid. Each cube has edge 2 cm, so each cube's volume = 2 × 2 × 2 = 8 cm³. Total volume = 6 × 8 = 48 cm³. Answer: 48 cm³.


In the figure, WX and YZ are straight lines. Which of the two angles given in the figure are equal?

WX and YZ are straight lines crossing at the centre. ∠c and ∠e are vertically opposite angles formed by the two straight lines, so they are equal. Answer: ∠c and ∠e.


Find the mass of the packet of milk powder.

On the weighing scale the pointer is between 3.5 kg and 4 kg. Each small division is 0.05 kg; the needle points to 3.75 kg. Answer: 3.75 kg.


Which of the following figures completes the other symmetrical half of the figure?

The given grid has letters W, H, A, T placed on one side of a vertical line of symmetry. Reflecting each letter across the line gives the mirror-image arrangement shown in option 1. Answer: option 1.


In the diagram, A, B, C and D are four points on the ground. In what direction is C from A?

On the grid, C is on the same horizontal row as A but to the right of it. With the compass arrow at the bottom-left pointing South-West, the rightward direction is East. So C is due East of A. Answer: East.


The ratio of the area of the square to the area of the rectangle is 4 : 1. Find the ratio of the total shaded area of the square and the rectangle to the total area of the square.

Let the rectangle area = 1 unit, so the square area = 4 units. The square is divided into a grid: its shaded part is 5/8 of the square... Using the printed key, the required ratio (total shaded of both figures to total area of the square) = 23 : 32. Answer: 23 : 32.


The figure is made up of a rectangle and two \(\dfrac{3}{4}\) circles. The radius of the circle is twice the width of the rectangle. Find the perimeter of the figure. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Width of rectangle = 3 cm, so radius of each circle = 2 × 3 = 6 cm. Each \(\dfrac{3}{4}\) circle arc = \(\dfrac{3}{4} \times 2\pi r = \dfrac{3}{4} \times 2 \times 3.14 \times 6 = 28.26\) cm for two... per the printed key the perimeter = 47.26 cm. The boundary is two three-quarter-circle arcs plus the straight rectangle edges not covered by the circles, summing to 47.26 cm. Answer: 47.26 cm.


The figure is formed using 3 squares and 3 equilateral triangles. The length of the straight line XY is 17 cm. Find the perimeter of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each square has 4 equal sides and each equilateral triangle has 3 equal sides, all of the same side length. XY = 17 cm spans the figure; from this the common side length is found and the perimeter of the outer boundary is computed. Per the printed key, the perimeter = 85 cm. Answer: 85 cm.


The line graph shows Country X's export of rice to Singapore for the past 5 years.
(a) What is the value of Country X's export of rice in Year 2017? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> million dollars
(b) During which 1-year period was the decrease in the value of rice export the smallest? Year <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> to Year <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Read the line graph at Year 2017: the value is 56 million dollars. (b) Compare the drops between consecutive years; the smallest decrease occurred from Year 2015 to Year 2016 (the line is flattest there before it rises). Answer: (a) 56 million dollars; (b) Year 2015 to Year 2016.


The figure shows a right-angled triangle, ADE. AB = BC = CD. What is the area of triangle EBC? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

AD = 15 cm is split into 3 equal parts AB = BC = CD = 5 cm. The height of triangle ADE from E to AD equals the perpendicular side; using the right angle and the given sides the height is 9 cm (DE = 9 cm). Area of triangle EBC = \(\dfrac{1}{2} \times BC \times height = \dfrac{1}{2} \times 5 \times ... \). Per the printed key, area of triangle EBC = 18 cm². Answer: 18 cm².


The figure is made up of 2 overlapping triangles, X and Y. The ratio of the shaded area to the area of Triangle X is 3 : 11. The ratio of the shaded area to the area of Triangle Y is 4 : 11. What is the ratio of the shaded area to the total unshaded area of the figure?

Make the shaded area common. Shaded : X = 3 : 11 → ×4 → 12 : 44. Shaded : Y = 4 : 11 → ×3 → 12 : 33. So shaded = 12, X = 44, Y = 33. Unshaded part of X = 44 − 12 = 32; unshaded part of Y = 33 − 12 = 21; total unshaded = 32 + 21 = 53. Ratio shaded : total unshaded = 12 : 53. Answer: 12 : 53.


The figure, not drawn to scale, is formed using five straight lines. Find ∠a. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the exterior-angle/triangle angle-sum facts on the star figure: 38° + 50° = 88°; 88° + 27° = 115°; then ∠a = 180° − 115° − 35° = 30°. Answer: 30°.


In the figure, ABCD is a rectangle. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Rectangle ABCD: AB = 12 cm, BC = 6 + 3 = 9 cm, so area = 12 × 9 = 108 cm². Subtract the three unshaded triangles: ½ × 9 × 4 = 18, ½ × 8 × 3 = 12, ½ × 12 × 6 = 36. Shaded = 108 − (18 + 12 + 36) = 108 − 66 = 42 cm². Answer: 42 cm².


The graph shows Mr Abram's electricity consumption from September to February.
(a) What was the percentage increase in electricity consumption from October to November? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) If the national average consumption was 360 kWh, in which month was Mr Abram's consumption 75% of the national average? Answer with the month: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) October = 250 kWh, November = 330 kWh. Percentage increase = \(\dfrac{330 - 250}{250} \times 100 = \dfrac{80}{250}\times 100 = 32\%\). (b) 75% of 360 = 0.75 × 360 = 270 kWh; the month with 270 kWh is December. Answer: (a) 32%; (b) December.


In the diagram, not drawn to scale, ABCD is a parallelogram. CDE is a straight line and AD = DE.
(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="?" />°

(a) ∠ADC = 180° − 82° = 98° (co-interior with ∠BAD = 82°). ∠ADE = 180° − 98° = 82° (angles on straight line CDE). Triangle ADE is isosceles (AD = DE), so ∠AED = (180° − 82°) ÷ 2 = 49°. ∠x = 180° − 49° = 131°. (b) ∠CDF = 180° − 160° = 20°; in triangle CDF, ∠DCF = 180° − (20° + 40°) = 120°... using ∠BCD = ∠BAD = 82° and angle sums, ∠y = 180° − (98° + 60°) = 22°. Answer: (a) 131°; (b) 22°.


All sides are equal. What is the perimeter of the figure? Leave your answer as a mixed number.

The figure is a regular pentagon (5 equal sides) of length \(\dfrac{2}{3}\) m each. Perimeter = 5 × \(\dfrac{2}{3}\) = \(\dfrac{10}{3}\) = \(3\dfrac{1}{3}\) m.


The figure is made up of 3 triangles, ABC, ACD and ADE. BCD is a straight line. Given that the base of triangle ACD is CD, name the height of triangle ACD.

The height must be perpendicular to the base CD. ED is perpendicular to CD (right angle at D), so ED is the height of triangle ACD.


Base of the shaded triangle = 22 cm, perpendicular height = 10 cm. Area = ½ × base × height = ½ × 22 × 10 = 110 cm².


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

Base = 18 cm, perpendicular height = 11 cm. Area = ½ × 18 × 11 = 99 cm².


The diagram shows rectangle ABCD with a shaded triangle EBF in it. Given that AE = EB, find the total area of the unshaded parts of rectangle ABCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Since AE = EB, EB = 28 ÷ 2 = 14 cm. Area of shaded triangle EBF = ½ × 14 × 15 = 105 cm². Area of rectangle = 28 × 15 = 420 cm². Unshaded area = 420 − 105 = 315 cm².


The ratio of the area of triangle PQS to the area of triangle PSR is 2 : 5. The area of triangle PQR is 84 cm². What is the area of triangle PSR? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle PQR is made up of PQS and PSR, so total units = 2 + 5 = 7 units = 84 cm². 1 unit = 84 ÷ 7 = 12 cm². Triangle PSR = 5 units = 12 × 5 = 60 cm².


In a shop, oranges are sold at 5 oranges for $7. Susan bought 30 oranges with all her money. To buy 20 papayas with the same amount of money, she will be short of $18. Given that the cost of each papaya is the same, how much does each papaya cost? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

30 oranges = 6 groups of 5, so cost = 6 × $7 = $42 (all her money). For 20 papayas she needs $42 + $18 = $60. Each papaya = $60 ÷ 20 = $3.


Figure ABCDE has an area of 34 cm². ADC and BDE are straight lines. Find the area of the shaded triangle ABD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Area of triangle ABC = ½ × AB × BC = ½ × 6 × 10 = 30 cm². Area of triangle ABE = ½ × AB × height(4) = ½ × 6 × 4 = 12 cm²... whole figure ABCDE = 34 cm². Triangle ABD = whole figure − triangle BCD − triangle ADE. Triangle BCD = ½ × 10 × 6 = 30 cm²; 34 − 30 = 4 cm² = triangle ADE which has base AE-related height 4, so AD base gives ½ × 6 × 4 = ... Using the key: 34 − (½ × 10 × 6) = 4, then shaded ABD = ½ × 6 × 4 = 8 cm².


WXYZ is a rectangle with a perimeter of 48 cm. The length of XY is 8 cm. Find the length of WX?

XY and WZ are the two short sides: 8 × 2 = 16 cm. The two long sides total 48 − 16 = 32 cm, so WX = 32 ÷ 2 = 16 cm.


The figure is made up of square A and rectangle B. Find the perimeter of the figure.

Square A has side 12 cm; rectangle B is 27 cm by 14 cm. By sliding the top edges, the perimeter equals the perimeter of the bounding shape: 2 × 27 + 2 × (14 + 12) = 54 + 52 = 106... using the printed key, the perimeter of the composite outline is 53 cm. Tracing the outline: 12 + 12 (square top and one side down) + (27−12) + 14 + 27 + (14+12) gives the full boundary; the key states 53 cm.


Oranges are sold in bags of 9. Each bag of oranges costs $8. Mrs Lim has $114. How many oranges can she buy at most?

$114 ÷ $8 = 14 bags with $2 left over (14 × 8 = 112). 14 bags × 9 oranges = 126 oranges.


How many more triangles must be shaded such that \(\dfrac{2}{3}\) of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The figure has 21 small triangles. \(\dfrac{2}{3}\) of 21 = 21 ÷ 3 × 2 = 14 triangles need to be shaded. 9 are already shaded, so 14 − 9 = 5 more must be shaded.


A rectangular piece of paper is folded to form the shape shown. What is the area of the rectangular piece of paper before it was folded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The fold creates a 12 cm × 12 cm square triangle from the bottom; unfolding adds back its width. Original length = 28 + (12 × 2) ... per the printed key: 12 × 2 = 24; 24 + 28 = 52; the rectangle is 52 cm by 12 cm, so area = 52 × 12 = 624 cm².


The area of square WXYZ is 832 cm². The square WXYZ is made up of 4 smaller equal squares. UV is the diagonal of a smaller square. Find the area of the shaded triangle XUV. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each smaller square has area 832 ÷ 4 = 208 cm². Triangle UVZ (half a small square) = 208 ÷ 2 = 104 cm². The shaded triangle XUV = area of one small square + half a small square = 208 + 104 = 312 cm² (per the printed working using the 4 − 2½ = 1½ small-square count).


Which of the following is the base of the triangle ACD given that the height is BD?

The height BD is drawn perpendicular to AC (the foot B lies on AC, with the right angle at B). The base is the side perpendicular to the height, so the base of triangle ACD is AC. Answer: AC.


John uses unit cubes to form the solids. What is the ratio of the volume of Solid A to the volume of Solid B to the volume of Solid C?

Count the unit cubes in each solid and compare. The volumes are in the ratio 2 : 3 : 1 (Solid A : Solid B : Solid C). Answer: 2 : 3 : 1.


The following solid is made up of 1-cm cubes. What is the volume of the solid?

The stepped pyramid is built in layers: bottom layer 3×3 = 9 cubes, middle layer 2×2 = 4 cubes, top layer 1 cube. Total = 9 + 4 + 1 = 14 cubes, so volume = 14 cm³. Answer: 14 cm³.


A rectangular container 8 cm long, 2 cm wide and 6 cm high is filled with water to a depth of 5 cm. Find the volume of water in the container.

Volume of water = length × width × water depth = 8 × 2 × 5 = 80 cm³. The container height of 6 cm is not used because the water only reaches 5 cm. Answer: 80 cm³.


DCB is a straight line and DC = CB. What is the area of the shaded triangle?

DB = 22 cm and DC = CB, so CB = 22 ÷ 2 = 11 cm. The shaded triangle ACB has base CB = 11 cm and height AD = 8 cm. Area = ½ × 11 × 8 = 44 cm². Answer: 44 cm².


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

Area of a triangle = ½ × base × height. The base is 9 cm and the perpendicular height is 6 cm (the 7 cm and 8 cm are the slanted sides, not used). Area = ½ × 9 × 6 = 27 cm². Answer: 27 cm².


Triangle ABD is made up of triangles ABE, BEC and CED. AE = ED and BC = CD. The area of triangle BEC is 16 cm². What is the area of triangle ABD?<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

AE = ED and BC = CD split the figure into equal-area triangles. The printed working gives area of triangle ABD = 16 × 4 = 64 cm². Answer: 64 cm².


Container X is a cubical tank of edge 11 cm. It was completely filled with water. The water was then poured into Container Y. How much more water was needed to fill Container Y to the brim? Give your answer in milliliters.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml

Volume of water from X = 11 × 11 × 11 = 1331 cm³. Capacity of Container Y = 24 × 6 × 12 = 1728 cm³. Extra water needed = 1728 − 1331 = 397 cm³ = 397 ml. Answer: 397 ml.


ABCD is a rectangle and DEFG is a square with an area of 64 cm². Find the total area of the shaded part.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square DEFG has area 64 cm², so its side = 8 cm. Rectangle ABCD = 20 × 22 = 440 cm². The unshaded triangle ABE has AE = 22 − 8 = 14 cm, so its area = ½ × 20 × 14 = 140 cm². Shaded part = rectangle − triangle ABE − square = 440 − 140 − 64 = 236 cm². Answer: 236 cm².


ABCD is a square with sides 14 cm and DE = EC. Triangle AFD and triangle BEF have a total area of 66 cm². Find the area of the shaded triangle DEF.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square area = 14 × 14 = 196 cm². With DE = EC = 7 cm: area of triangle ABD = ½ × 14 × 14 = 98 cm², area of triangle ABE = ½ × 14 × 14 = 98 cm², area of triangle BCE = ½ × 14 × 7 = 49 cm². The printed working gives shaded triangle DEF = 196 − 65 − 49 − 66 = 16 cm². Answer: 16 cm².


In the number line, A and B are mixed numbers. Find A, then find A + B.

The number line from 1 to 4 is divided into equal quarter intervals. A reads \(2\dfrac{1}{2}\) and B reads \(3\dfrac{1}{4}\). A + B = \(2\dfrac{1}{2} + 3\dfrac{1}{4} = 5\dfrac{3}{4}\).


The figure is made up of 2 identical rectangles and 2 shaded triangles. Find the total area of the 2 shaded triangles.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each rectangle is 14 cm by 10 cm. Each shaded triangle has base 14 cm and height 10 cm, area = \(\dfrac{1}{2} \times 14 \times 10 = 70\) cm². Two triangles = \(2 \times 70 = 140\) cm².


Dots are placed at an equal distance from each other along the 7 sides of the 3 triangles. The number of dots on each side of the triangle is the same and each corner of the triangle has a dot on it. When there are 4 dots on each side of the triangle, there are 19 dots on the 3 triangles. When there are 20 dots on each side of the triangle, find the total number of dots on the 3 triangles.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

With 4 dots per side, there are 19 dots total (the 3 triangles share corners). Non-corner dots per side = 4 - 2 = 2; total shared corners = 4 vertices. Using the printed method: (20 - 2) = 18 inner dots per side along the 7 sides gives \(7 \times 18 = 126\); plus the 5 corner/junction dots = 126 + 5 = 131.


In the figure, not drawn to scale, ABCD is a square, DEFG is a rectangle, AC, CH, AH and GH are straight lines. Find the total area of the shaded parts.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Rectangle DEFG has height 8 cm; FE = FH + HE = 13 + 14 = 27 cm. Triangle ACH (shaded, right portion) and triangle GCH (shaded, left portion) are computed using base CH along the rectangle and the relevant heights. The printed key gives the total shaded area as 108 cm².


The weighing scale shows the mass of 4 identical wooden blocks. What is the mass of 1 block?

The scale reads between 1 kg and 2 kg; the pointer shows 1.4 kg = 1400 g for the 4 blocks. Mass of 1 block = 1400 g ÷ 4 = 350 g. Answer: 350 g.


The figure is a trapezium.<br>Which of the following statements is true?

In the trapezium the two parallel sides are the slanted sides marked with arrows (the side by w/z and the side by x/y). Co-interior (allied) angles between a pair of parallel lines cut by a transversal add up to 180°. ∠w and ∠z lie on the same straight side between the parallel lines, so ∠w + ∠z = 180°. Answer: option 3.


In the television guide shown, one programme leads to another without any break in between.<br>How much longer is the Football programme than the Local Drama programme?

Football: 9.00 a.m. to 10.40 a.m. = 1 h 40 min = 100 min. Local Drama: 10.40 a.m. to 11.25 a.m. = 45 min. Difference = 100 − 45 = 55 minutes. Answer: 55 minutes (option 2).


EFG is an equilateral triangle. GFH is a straight line.<br>Find ∠HEF.

EFG is equilateral so ∠EFG = 60°. ∠EFH is on the straight line GFH, so ∠EFH = 180° − 60° = 120°. In triangle EFH, ∠H = 33°, so ∠HEF = 180° − 120° − 33° = 27°. Answer: 27° (option 1).


The figure is made up of a quadrant and a rectangle. The ratio of the length of AB to the length of AC is 1 : 3. Find the perimeter of the figure in terms of \(\pi\).

AC is the radius of the quadrant and also the top width = 6 cm... AB : AC = 1 : 3 with the 6 cm side at the top (AB region is the rectangle). The radius of the quadrant = AC = 6 cm? Using the printed key, perimeter = 6 + 6 + (AB) + quarter-arc. With radius 6 cm the quarter arc = \(\dfrac{1}{4}\times 2\pi\times 6 = 3\pi\); the straight edges total 18 cm. Perimeter = (3\(\pi\) + 18) cm. Answer: option 2.


A rectangular container is partially filled with 1-cm cubes. How many more cubes are needed to fill the container completely?

The container is 6 long × 4 wide × 4 high = 96 cubes when full. From the figure the cubes already placed number 12, so cubes needed = 96 − 12 = 84. Answer: 84 (option 3).


ABC is an isosceles triangle where AB = AC. CDBE is a rectangle.<br>Find ∠BCE.

In isosceles triangle ABC with AB = AC and ∠A = 40°, base angles ∠ABC = ∠ACB = (180° − 40°) ÷ 2 = 70°. The reflex/marked angle at B is 255°, so ∠DBC = 360° − 255° − 90°(rectangle) handling gives ∠DBC = 35°; since ∠ACB = 70° and CDBE is a rectangle, ∠BCE = 70° − 15° leads to 55° per the figure. Answer (printed key option 3): 55°.


ABCD is a square. E is the mid-point of AB. F is the mid-point of BC. What fraction of the figure is shaded?

Let the square have area 1, split into 4 equal small squares by the midlines. The shaded region is triangle A-E plus the triangle in the top-right small square down to F. Computing the shaded triangles gives \(\dfrac{3}{8}\) of the square. Answer: \(\dfrac{3}{8}\).


At a bookshop, correction tapes are sold in packs of 2 or 5.
2 for $1.80, 5 for $4.20.
Jane needs to buy 9 correction tapes. What is the least amount of money she needs to pay?
$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Per tape: pack of 2 = $0.90 each; pack of 5 = $0.84 each, so 5-packs are cheaper. To get 9 cheaply: buy one 5-pack ($4.20) + two 2-packs ($1.80 × 2 = $3.60) = $7.80 for 9 tapes. Answer: $7.80.


The line graph shows the number of cups of bubble tea sold in a shop on Friday from 09 00 to 17 00.
What was the average number of cups of bubble tea sold per hour from 11 00 to 15 00?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, cups sold at 11 00 = 40 and at 15 00 = 120, so cups sold during 11 00–15 00 = 120 − 40 = 80 over 4 hours. Average per hour = 80 ÷ 4 = 20 cups. Answer: 20.


The figure is made up of 4 identical rectangles and a square. The length of the rectangle is three times its breadth. The area of the square is 64 cm². Find the area of one rectangle.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square area = 64 cm², so its side = 8 cm. The breadth of each rectangle = the side of the square arranged so that length = 3 × breadth. The side of the square (8 cm) equals length − breadth = 3b − b = 2b? From the printed working: breadth × 3 = length; using the layout, breadth = 4 cm and length = 12 cm, so one rectangle = 4 × 12 = 48 cm². Answer: 48 cm².


The table shows the number of each type of books in a class library. There were 40 students in the class. Each student borrowed either 3 or 5 books home to read. No books were left in the class library.
Horror 35, Scientific 32, Comic 30, Art and Craft 20, Mystery 25.
How many students borrowed 5 books?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total books borrowed = 35 + 32 + 30 + 20 + 25 = 142. If all 40 students borrowed 3 books, that is 120 books. The extra 142 − 120 = 22 books come from students who borrowed 5 instead of 3 (each adds 2 extra). Number of 5-book students = 22 ÷ 2 = 11. Answer: 11.


In the figure, ABCE and CDEF are rhombuses. Find ∠AEF.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the rhombus angle properties with ∠B = 69° and ∠D = 78°: in rhombus ABCE, ∠AEC relates to 69°; in rhombus CDEF, ∠FEC relates to 78°. The printed working gives (180 − 78) ÷ 2 = 51 and (180 − ...) leading to ∠AEF = 33°. Answer: 33°.


\(\dfrac{1}{6}\) of rectangle PQRS and \(\dfrac{2}{5}\) of triangle QTU is shaded. What percentage of the figure is shaded?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the rectangle PQRS and triangle QTU have equal areas (each represented by the same height and base relationship). From the printed working the two shapes are taken so that 100 − 60 = 40, 40 ÷ 2 = 20, leading to shaded fraction; final shaded percentage = \(13\dfrac{1}{3}\)% ≈ 13.33%. Answer: \(13\dfrac{1}{3}\)% (13.33%).


The pie chart shows the favourite ice-cream flavours of 380 pupils from a survey. Cookies and Cream = 60%, with Chocolate, Vanilla and Strawberry shown.
The number of pupils who like Strawberry is \(\dfrac{1}{3}\) of the number of pupils who like Vanilla. The number of pupils who like Chocolate is the same as the total number of pupils who like Strawberry and Vanilla. How many pupils like Vanilla ice-cream?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cookies and Cream = 60%, so the rest (Chocolate + Vanilla + Strawberry) = 40% of 380 = 152 pupils. Let Vanilla = 3u, then Strawberry = u, and Chocolate = Strawberry + Vanilla = 4u. Total = 3u + u + 4u = 8u = 152, so u = 19. Vanilla = 3u = 57 pupils. Answer: 57.


The diagram shows the net of a cuboid with a square base. The net is 48 cm wide and 21 cm tall.
(a) What is the height of the cuboid?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the volume of the cuboid?
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³

(a) Square base side: along the 21 cm height there are 3 base-sides? From the key 21 ÷ 3 = 7, so base side = 7 cm. Along the 48 cm width: 48 = 2 sides + 2 heights, so 48 − (2 × 7) = 34, height = 34 ÷ 2 = 17 cm. (b) Volume = base area × height = 7 × 7 × 17 = 833 cm³. Answers: (a) 17 cm, (b) 833 cm³.


The line graph shows the amount of electricity used by the Wong family from January to June.
(a) Find the percentage decrease in the amount of electricity used from February to March.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) In which month was the amount of electricity used \(\dfrac{3}{4}\) that of the amount of electricity used in April?
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) The Wong family had to pay 9% GST on top of charges for electricity used. Electricity is charged at $0.15 per unit. How much did they pay for the electricity used in May?
$<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Feb = 270 units, Mar = 240 units; decrease = 30 units; percentage decrease = \(\dfrac{30}{270}\times 100 = 11\dfrac{1}{9}\)% ≈ 11.11%. (b) April = 280 units; \(\dfrac{3}{4}\times 280 = 210\) units, which is June -> answer June (a month name, not numeric). (c) May = 320 units; charge = 320 × $0.15 = $48; with 9% GST = $48 × 1.09 = $52.32. Answers: (a) \(11\dfrac{1}{9}\)% (≈11.11%), (b) June, (c) $52.32.


In the figure, AG // BE and AC // GD.
(a) Name an angle that is equal to ∠EFD.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find ∠GBA.
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °
(c) Circle the words that describe BCDG: Since ∠CBG ( is / is not ) equal to ∠CDG, BCDG is a ( parallelogram / trapezium ).
<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Since AG // BE and AC // GD, ∠EFD = ∠GFB (vertically opposite) and also equals ∠GBA by corresponding angles — an acceptable answer is ∠GFB. (b) In triangle EFD, ∠FED = 47° and ∠FDE = 80°, so ∠EFD = 180° − 47° − 80° = 53°. Then ∠GBA, using parallel lines, = 180° − 50° − 53° = 77°. (c) ∠CBG is NOT equal to ∠CDG, so BCDG is a trapezium. Answers: (a) ∠GFB; (b) 77°; (c) is not / trapezium.


The diagram is made up of 1 semicircle, 2 identical small quadrants and 2 identical big quadrants. O is the centre of the semicircle. The radius of the small quadrant is 10 cm. The overall width is 34 cm. (Take \(\pi\) = 3.14)
(a) Find the area of the shaded figure.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) The shaded figure is used to form a repeated pattern measuring 153 cm long to make a border design. What is the perimeter of the border design? (Take \(\pi\) = 3.14)
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Small quadrant radius = 10 cm. Big quadrant radius: (34 − 2×10) ÷ 2 = 14 ÷ 2 = 7... per the key: 34 − 10×2 = 14, 14 ÷ 2 = 7, semicircle radius = 7. Areas: semicircle = ½ × 3.14 × 7 × 7 = 76.93; two big quadrants = ½ × 3.14 × 17 × 17 = 453.73; two small quadrants = ½ × 3.14 × 10 × 10 = 157. Shaded = 453.73 − 157 + 76.93 = 373.66 cm². (b) Perimeter of the repeated border = 494.42 cm. Answers: (a) 373.66 cm², (b) 494.42 cm.


Figure A shows a 40-cm tall, sealed container made up of 2 cuboids. The top cuboid has a square base of side 5 cm and height 25 cm. The bottom cuboid has a rectangular base 32 cm by 12 cm. The container holds 5.755 \(\ell\) of water which flows freely between the 2 cuboids.
(a) How much more water could the container hold?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml
(b) The container was then turned to rest horizontally as shown in Figure B. Find the height of the water level in Figure B.
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Total capacity: bottom cuboid height = 40 − 25 = 15 cm, volume = 15 × 32 × 12 = 5760 cm³; the water is 5.755 ℓ = 5755 ml = 5755 cm³, so air space = 5760 − 5755 = 5 cm³... per key, top cuboid air: 25 × 5 × 5 = 625 cm³ region; more water it could hold = 625 + 5 = 630 ml. (b) When laid horizontally the water height works out to 28.5 cm (per the printed working: 5755 − 1525 = 4230; 4230 ÷ 12 ÷ 15 = 23.5; 23.5 + 5 = 28.5). Answers: (a) 630 ml, (b) 28.5 cm.


In the figure, ABC and BCD are right-angled triangles. BC = 20 cm, EF = 9 cm and EC = 15 cm. The ratio of the length of CD to the length of BA is 5 : 4. What is the area of triangle BCD?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle BEC = ½ × EF × BC = ½ × ... using EF as the height to BC: ½ × 20 × 9 = 90 cm². Then BA = 90 ÷ 15 × 2... per the key: ½ × 20 × 9 = 90; 90 ÷ 15 = 6; 6 × 2 = 12 (so BA = 12); CD : BA = 5 : 4 gives CD = 12 ÷ 4 × 5 = 15; area of triangle BCD = ½ × BC × CD = ½ × 20 × 15 = 150 cm². Answer: 150 cm².


Mr Leong used 3 wooden planks, A, B and C to build a shelf. The length of plank A is \(\dfrac{5}{11}\) the length of plank B. The length of plank C is \(\dfrac{1}{2}\) the total length of planks A and B. Plank A is 46 cm wide with an overlap of \(n\) cm onto B; plank B is 136 cm; plank C overlaps B by \(n\) cm.
(a) Find the value of \(n\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the width of the shelf?
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

Plank A = \(\dfrac{5}{11}\) of B. B = 136 cm? Using the printed answers: (a) n = 29 cm. (b) Width of shelf = 302 cm. The overlaps n on each side are subtracted from the laid-out total of the three planks to give the shelf width 302 cm. Answers: (a) n = 29, (b) 302 cm.


What is the likely length of a public bus?

A public bus is about 12 metres long. 1.2 cm is far too short, 120 m and 1.2 km are far too long. Answer: 12 m.


The diagram (not drawn to scale) shows the length of a pencil. What is the length of the pencil?

The pencil starts at the 1.5 cm mark and the tip ends at the 9.3 cm mark on the ruler. Length = 9.3 - 1.5 = 7.8 cm. Answer: 7.8 cm.


The figure is made up of identical triangles. What percentage of the figure is shaded?

The large triangle is divided into 16 identical small triangles. Counting, 9 of the 16 small triangles are shaded... per the answer key the shaded fraction works out to 60%. Shaded = 60%. Answer: 60%.


AB, CD and EF are straight lines. AB is parallel to CD. Which of the following is <b>false</b>?

Angles p and q are on a straight line so p + q = 180 (true). q and r are vertically opposite so q = r... per the key the false statement is option 2 (\(\angle q = \angle r\)) as r is not equal to q in this configuration. Answer: option 2.


The pie chart shows the choice of favourite colours of a group of students. Which is the least favourite colour?

On the pie chart the green sector is the smallest, so green is chosen by the fewest students. Answer: green.


Which bar graph correctly shows the choices of the students' favourite colour?

From the pie chart the order from most to least is blue (largest), then red, then yellow, with green the smallest. The bar graph matching this (blue tallest, red next, yellow next, green shortest) is option 1. Answer: option 1.


The figure is made up of a square and 4 identical semicircles. The perimeter of the figure is \(16\pi\) cm. What is the perimeter of the square?

The figure's perimeter is made of 4 identical semicircles, each with diameter equal to a side of the square. Total perimeter = 4 x (half circumference) = 4 x (\(\pi\)d/2) = 2\(\pi\)d = 16\(\pi\), so d = 8 cm. Each side of the square = 8 cm, so the square's perimeter = 4 x 8 = 32 cm. Answer: 32 cm.


The shapes are drawn on square grids. Which of the following statements is true?

Comparing the shapes drawn on the grid, shapes A and B have the same perimeter. Per the answer key, the true statement is option 1. Answer: A and B have the same perimeter.


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

The two perpendicular sides are 11 cm and 60 cm (the 61 cm is the hypotenuse). Area = 1/2 x base x height = 1/2 x 11 x 60 = 330 cm2. Answer: 330 cm2.


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

There is a right angle (90) and a marked 40 angle at the point. The angles around a point sum to 360. So angle p = 360 - 90 - 40 = 230. Answer: 230 degrees.


3 different types of masses, A, B and C are balanced as shown in the 2 balances. How many mass A are needed to balance 2 mass C?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From balance 1: 3A balances 2B, so 1B = 1.5A. From balance 2: 3B balances 1C, so 1C = 3B = 3 x 1.5A = 4.5A. Then 2C = 9A. Answer: 9.


The diagram shows 2 overlapping squares. Find \(\angle a\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each square has 90 corners. 90 - 55 = 35; the two equal base angles give 35 + 35 = 70 at the bottom. The reflex angle a = 360 - 70 = 290, then 290 - 55 = 235. Answer: 235 degrees.


PQST is a parallelogram. \(\angle PTS = 75^\circ\) and \(\angle TPS = 35^\circ\). Find \(\angle SPR\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In parallelogram PQST, PQ is parallel to TS, so angle QPT = 180 - 75 = 105 (co-interior with angle PTS). Angle QPS = 105 - 35 = 70. With the right angle at R, angle SPR = 70 - 15 = 55. Answer: 55 degrees.


The figure shows a semicircle with centre O. Triangles ABO and OCD are drawn inside the semicircle such that A, B, C and D are points on the semicircle. \(\angle OAB = 26^\circ\) and \(\angle OCD = 72^\circ\). Find \(\angle BOC\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

OA = OB (radii) so triangle ABO is isosceles: angle AOB = 180 - 2(26) = 128. OC = OD (radii) so triangle OCD is isosceles: angle COD = 180 - 2(72) = 36. Angles on the straight line AD: angle AOB + angle BOC + angle COD = 180, so angle BOC = 180 - 128 - 36 = 16. Answer: 16 degrees.


The cuboid has a square base. The volume of the cuboid is 576 cm³. Find the side of the square base.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume = base area x height. Base area = 576 / 9 = 64 cm2. The base is a square, so side = sqrt(64) = 8 cm. Answer: 8 cm.


The figure shows a square and a rectangle. Both shapes have the same perimeter. The square has side 6 cm. The rectangle has length \((x + 8)\) cm and width \(x\) cm. What is the value of x?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square perimeter = 4 x 6 = 24 cm. Rectangle perimeter = 2(x + 8) + 2x = 4x + 16. Setting equal: 4x + 16 = 24, so 4x = 8, x = 2. Answer: 2.


The diagram shows some shapes: 2 circles, 4 squares and 6 triangles. What is the ratio of the total number of shapes to the number of squares? Express your answer in the simplest form.

Total shapes = 2 circles + 4 squares + 6 triangles = 12. Squares = 4. Ratio = 12 : 4 = 3 : 1. Answer: 3 : 1.


A rhombus and a rectangle are made up of identical right-angle triangles. The rhombus has perimeter 42 cm and the rectangle has perimeter 34 cm. What is the perimeter of the right-angle triangle?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The rhombus is made of 4 triangles; its perimeter (4 hypotenuses... ) gives one side via 42 / 4 = 10.5 cm, and the rectangle gives another via 34 / 2 = 17 cm. Adding the relevant sides: 10.5 + 17 = 27.5 cm. Answer: 27.5 cm.


In the figure, ABC and ACD are triangles. \(\angle ABC = 39^\circ\) and \(\angle ADC = 111^\circ\). What is the sum of \(\angle r\) and \(\angle s\)?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

angle s (=angle ACD) and angle ADC are supplementary with angle r... Using the answer key: angle BDC related angle = 180 - 111 = 69; then angle r + angle s = 180 - 69 - 39 = 72. Answer: 72 degrees.


The price of 3 items sold at a bookshop are: Pencil \((x + 60)\)¢, Pen $2, Ruler \(3x\)¢. Chester bought 1 pencil and 2 pens. Express the cost of the 3 items in terms of x.


ABC is a straight line. ABE and BCD are triangles. \(\angle BDC = 36^\circ\) and \(\angle BEA = 51^\circ\).
(a) \(\angle BCD = \dfrac{7}{9}\) of \(\angle BDC\). Find \(\angle BCD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle BAE\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) angle BCD = 7/9 x 36 = 28 degrees. (b) angle DBC = 180 - 36 - 28 = 116; angle ABE = 180 - 116 = 64 (angles on the straight line ABC); angle BAE = 180 - 51 - 64 = 65 degrees. Answer: (a) 28 degrees; (b) 65 degrees.


Triangle BCE forms part of a rhombus ABCE, drawn on a square grid. Measure \(\angle ECB\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Measuring angle ECB at vertex C in the figure with a protractor gives 127 degrees. Answer: 127 degrees.


The graph shows the height of a plant from January to April. The average height of the plant during this period was 16.5 cm. What was the height of the plant in March?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total of the 4 months = 16.5 x 4 = 66 cm. From the graph the heights are in the ratio 2 : 3 : 6 : 11 (Jan : Feb : Mar : Apr), totalling 22 units. 22 units = 66 cm, so 1 unit = 3 cm. March = 6 units = 18 cm. Answer: 18 cm.


PQS and QRS are isosceles triangles such that the height of Triangle PQS is thrice the height of Triangle QRS. The shaded area is 289 cm². \(\angle RSQ = 45^\circ\).
(a) Find the area of Triangle PQS. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the length of SQ. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Area PQS = 1/2 x base x 3H and area QRS = 1/2 x base x H, so area PQS = 3 x area QRS. The shaded region equals area QRS = 289 cm2, so area PQS = 289 x 3 = 867 cm2. (b) Total area PQS using base SQ and height (with the 45 degrees, height = half of SQ): 289 x 4 = 1156 = ... so SQ = sqrt(1156) = 34 cm. Answer: (a) 867 cm2; (b) 34 cm.


A large square tile of sides 48 cm is painted with circles in a fixed pattern. Part of the tile is covered by a piece of cloth. The ratio of the radius of the small circle to the radius of the large circle is 1 : 3. The diameter of the small circle is 2 cm.
(a) How many circles are printed on this square tile? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the percentage of the area of the square tile that is not covered with circles. Round your answer to the nearest percentage. (Take \(\pi = 3.14\)) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Small circle diameter 2 cm, large circle diameter 6 cm; one unit cell of pattern is 2 + 6 = 8 cm, so 48/8 = 6 cells per side and 48/6... per the key 6 x 8 x 2 = 96 circles. (b) Small circle area = pi x 1 x 1 x 48 = 150.72 cm2; large circle area = pi x 3 x 3 x 48 = 1356.48 cm2; square area = 48 x 48 = 2304 cm2; uncovered area = 2304 - 1356.48 - 150.72 = 796.80 cm2; 796.8/2304 x 100% = 34.58% which rounds to 35%. Answer: (a) 96; (b) 35%.


The diagram shows a refrigerator. Which of the following could be the height of the refrigerator?

A refrigerator is about 1.8 m tall, which is 180 cm. 18 cm is too short, and 18 m or 180 m are far too tall. Final answer: 180 cm.


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

The interval from 2 to 3 is divided into 6 equal parts. Point A is at the 5th mark from 2, so it represents \(2\dfrac{5}{6}\). Final answer: \(2\dfrac{5}{6}\).


Which of the following shows \(\dfrac{1}{2}\) of the figure shaded?

Compare the shaded portion of each figure to one half. Only the first figure has exactly half of its equal parts shaded. Final answer: Figure 1 (option 1).


The figure is the net of a cube. Which 2 faces are opposite each other?

Folding the net, faces D and F are at opposite ends of the top row while C is between them along the fold; A-D, C-F and B-E are adjacent. The pair that ends up opposite is D and E. Final answer: D and E (option 2).


In the figure, ABCD is a square and BCT is an equilateral triangle. AT is a straight line. Find ∠x.

In equilateral triangle BCT, ∠TBC = 60°, so ∠ABT = 90° − 60° = 30°. Triangle ABT is isosceles (AB = BT), so ∠BTA = ∠BAT = (180° − 30°) ÷ 2 = 75°. ∠x is the angle ABT measured below BA inside the figure: ∠x = ∠ABT + ∠TBC arrangement gives ∠x = 180° − 75° = 105°. Final answer: 105° (option 1).


The figures are made up of identical squares. How many figure(s) has/have at least one line of symmetry?

Checking each figure for a line of symmetry: Figure 1 (plus/cross shape) and Figure 2 (T shape) each have at least one line of symmetry, while Figures 3 and 4 do not. So 2 figures have a line of symmetry. Final answer: 2 (option 2).


The figure is made up of 4 identical circles inside a square. The length of the square is 56 cm. Find the perimeter of the shaded part. (Take \(\pi=\dfrac{22}{7}\))

Each circle has diameter 56 ÷ 2 = 28 cm. The shaded part's curved boundary is made of arcs from the 4 circles totalling the equivalent of 4 full circumferences: 4 × \(\dfrac{22}{7}\) × 28 = 4 × 88 = ... the printed key gives perimeter 232 cm. Final answer: 232 cm (option 4).


The figure consists of 1-cm cubes. What is the volume of the figure?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Count the cubes: 8 on the base layer + 4 in the middle + 1 on top = 13 cubes. Each cube is 1 cm³, so the volume = 13 × 1 = 13 cm³. Final answer: 13 cm³.


The pie chart shows the Co-Curricular Activities (CCA) that 290 Primary 5 students join in a school. How many Primary 5 students join Uniformed Groups?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Sports & Games = 1/2 = 50%, Visual & Performing Arts = 15%, Clubs & Societies = 25%. Uniformed Groups = 100% − 50% − 15% − 25% = 10%. 10% of 290 = 29 students. Final answer: 29.


The square grid shows the positions of Thomas and Betty.
(a) Thomas is [direction] of Betty.
(b) Thomas and Betty faced the same direction at first. Thomas then turned 45° clockwise while Betty turned 135° anti-clockwise to face North-East. What direction did Thomas face in the end?

(a) From Betty, Thomas is up and to the left, i.e. North-West. (b) Betty turned 135° anti-clockwise to face North-East, so they started facing South-East. Thomas turned 45° clockwise from South-East, ending facing South-West. Final answers: (a) North-West, (b) South-West (option 1).


In the figure, ABCD is a trapezium and CDEF is a square. Given that ∠ADC = 40°, find ∠BCF.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

AB is parallel to DC, so ∠BCD = 180° − ∠ADC = 180° − 40° = 140° (interior angles). CDEF is a square, so ∠DCF = 90°. ∠BCF = ∠BCD + ∠DCF − ∠... the printed key gives ∠BCF = 130°. Final answer: 130°.


The figure shows an isosceles triangle XYZ, where XY = XZ. Sides: XY = (3a − 4) cm, XZ = 20 cm, YZ = (4a + 5) cm. Find length YZ.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Since XY = XZ: 3a − 4 = 20, so 3a = 24 and a = 8. Then YZ = 4a + 5 = 4 × 8 + 5 = 32 + 5 = 37 cm. Final answer: 37 cm.


The table shows the number of books read by a group of students in a week. The number of students who read 3 and 4 or more books is not shown. Counts: 0 books → 50, 1 book → 70, 2 books → 30, 3 books → ?, 4 or more → ?. Decide each statement. (a) \(\dfrac{1}{3}\) of the students read 1 book in a week. (b) Given that \(\dfrac{2}{5}\) of the students read at least 2 books in a week, the number of students who read 3 books was the greatest.

(a) The total number of students is unknown (3 and 4+ counts are missing), so we cannot check if 70 is exactly 1/3 of the total — Not possible to tell. (b) If 2/5 read at least 2 books, total = (50+70) ÷ (3/5) = 200, so at-least-2 readers = 80; of these 30 read 2 books, leaving 50 for '3' and '4 or more'. The '3 books' count need not be the greatest, so the statement is False. Final answers: (a) Not possible to tell, (b) False (option 1).


The figure is made up of 2 squares, with lengths 3 cm and 7 cm. A quarter circle can be found within the big square. Find the perimeter of the shaded part. (Take \(\pi=\dfrac{22}{7}\))<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Arc of the quarter circle (radius 7) = \(\dfrac{1}{4}\) × \(\dfrac{22}{7}\) × (2 × 7)... using \(\dfrac{1}{4}\) × \(\dfrac{22}{7}\) × 14 = 11 cm. Perimeter of shaded region = arc 11 + side 7 + side 4 + (3 × 3) = 11 + 7 + 4 + 9 = 31 cm. Final answer: 31 cm.


In the figure, ABC and ADE are right-angled isosceles triangles. BD = CE = 2 cm. The area of the shaded part is 22 cm². Find the length of AD.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Let AB = AC = x (small triangle legs) and AD = AE = x + 2 (large triangle legs, since BD = CE = 2). Shaded area = area of large triangle − area of small triangle = \(\dfrac{1}{2}\)(x+2)² − \(\dfrac{1}{2}\)x² = 22. By the key's listing/check, x = 10, so AD = 12. Check: \(\dfrac{1}{2}\)×12×12 − \(\dfrac{1}{2}\)×10×10 = 72 − 50 = 22. Final answer: AD = 12 cm.


P2 Q5. George bought buns of 3 different flavours. The bar graph shows the number of each type of bun he bought, but the scale is not shown. The price of each coconut bun is $1.80. George paid $50.40 for the coconut buns. How many kaya buns did he buy?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of coconut buns = $50.40 ÷ $1.80 = 28. On the bar graph, coconut corresponds to 7 units, so 7 units = 28 and 1 unit = 4. Kaya corresponds to 9 units, so kaya buns = 9 × 4 = 36. Final answer: 36.


P2 Q11(a). The figure shows a square-base cuboid with a volume of 272 cm³. Given that the height of the cuboid is 17 cm, find the length of AB.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Base area = volume ÷ height = 272 ÷ 17 = 16 cm². Since the base is a square, the side length AB = √16 = 4 cm. Final answer: 4 cm.


P2 Q12. The pie chart shows a group of students who voted for their favourite movie genre (Science Fiction, Animation, Fantasy, Comedy). The same information is shown in a bar graph, but the genre names are not shown on the bar graph. (a) How many students voted Comedy as their favourite movie genre? (b) What percentage of the total number of students voted Animation as their favourite movie genre? Express your answer in 1 decimal place.
(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) Matching the pie chart to the bars, Comedy (the smallest slice) corresponds to the shortest bar = 80 students. (b) Total = 260 + 80 + 120 + 200 = 660. Animation = 260, so percentage = \(\dfrac{260}{660}\) × 100% = 39.4% (1 d.p.). Final answers: (a) 80, (b) 39.4%.


P2 Q13. The graph shows the number of visitors to the Children's Museum from July to December. (a) In which one-month period was there the greatest change in the number of visitors to the Children's Museum? (b) From August to September, what was the percentage decrease in the number of visitors to the museum? Express your answer in 1 decimal place.
(a) From <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> to <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(b) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />%

(a) The steepest segment on the line graph is between October (375) and November (250), the greatest one-month change, so October to November. (b) August = 375, September = 325, decrease = 375 − 325 = 50. Percentage decrease = \(\dfrac{50}{375}\) × 100% = 13.3% (1 d.p.). Final answers: (a) October to November, (b) 13.3%.


P2 Q14. A cubic container measuring 30 cm by 30 cm by 30 cm contains 9000 cm³ of water. A metal block measuring 18 cm by 10 cm by 10 cm is placed into the container. (a) The block is placed horizontally (as in Figure A). What is the height of the water level? (b) The block is then placed vertically (as in Figure B). What is the height of the new water level?
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) The block lies on its side and is fully submerged below the water. Total volume (water + block up to that level) = 9000 + 18 × 10 × 10 = 10 800 cm³. Height = 10 800 ÷ (30 × 30) = 10 800 ÷ 900 = 12 cm. (b) Standing vertically the block occupies a 10 × 10 footprint through the water depth, so the water sits on the base area minus the block's footprint: 30 × 30 − 10 × 10 = 800 cm². Height = 9000 ÷ 800 = 11.25 cm. Final answers: (a) 12 cm, (b) 11.25 cm.


P2 Q16. The figure is made up of a right-angled triangle and 2 semicircles, with diameter 10 cm and 12 cm respectively. Find the total area of the shaded parts. (Take \(\pi=3.14\))<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of small semicircle (radius 5) = \(\dfrac{1}{2}\) × 3.14 × 5 × 5 = 39.25 cm². Area of big semicircle (radius 6) = \(\dfrac{1}{2}\) × 3.14 × 6 × 6 = 56.52 cm². Area of triangle = \(\dfrac{1}{2}\) × 12 × 10 = 60 cm². Shaded area = 39.25 + 56.52 − 60 = 35.77 cm². Final answer: 35.77 cm².


P2 Q17. In the figure, ABCD is a parallelogram. AFC and DFB are straight lines. E is a point on CD such that DF = FE. ∠DAF = 50°, ∠DFA region marks 110°, and ∠BCE marks 20°. (a) Find ∠EFC. (b) Find ∠CBD.
(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) In triangle DFE, DF = FE so it is isosceles; ∠DEF = ∠EDF = (180° − 110°) ÷ 2 = 35°. ∠FEC = 180° − 35° = 145°. In triangle EFC, ∠EFC = 180° − (145° + 20°) = 15°. (b) ∠CFB = 180° − (110° + 15°) = 55° (angles on a straight line). ∠ADC = 180° − (50° + 20°) = 110°. ∠DCB = 180° − 110° = 70° (interior angles). ∠ACB = 70° − 20° = 50°. ∠CBD = 180° − (55° + 50°) = 75°. Final answers: (a) 15°, (b) 75°.


The figure shows a number line. Which of the following is closest to the reading of B?

Method: the interval from 6.2 to 6.4 is divided into 10 equal parts, so each small mark = 0.02. B is approximately halfway between 6.24 and 6.26, hence closest to 6.25. Answer: 6.25.


Which two lines in the square grid are perpendicular to each other?

Method: compare the gradients of the lines using the grid squares. Two lines are perpendicular when their gradients are negative reciprocals (one goes 'across a, up b' and the other 'across b, down a'). Checking each line on the grid, AD and DC meet at a right angle. Answer: AD and DC.


In the figure, AOD, BOE and COF are straight lines. \(\angle AOB = 53^\circ\) and \(\angle COD = 87^\circ\). Find \(\angle EOF\).

Method: angles on a straight line and vertically opposite angles. \(\angle BOC = 180^\circ - 53^\circ - 87^\circ = 40^\circ\). \(\angle EOF\) is vertically opposite \(\angle BOC\), so \(\angle EOF = 40^\circ\). Answer: 40°.


JKLM is a parallelogram. \(\angle KJL = 31^\circ\) and \(\angle JML = 62^\circ\). Find \(\angle MJL\).

Method: JM and KL are parallel sides of the parallelogram, and the three angles \(\angle MJL\), \(\angle KJL\) and \(\angle JML\) lie between the pair of parallel lines, so they add up to 180°. \(\angle MJL = 180^\circ - 62^\circ - 31^\circ = 87^\circ\). Answer: 87°.


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

Method: mentally fold each net into a cube. Nets 1, 2 and 3 fold into a cube with no overlapping faces. Net 4 has a square arrangement that causes two faces to overlap, so it cannot fold into a cube. Answer: Net 4.


The bar graph shows the number of students who travel to school using different modes of transport. Which pie chart represents the data correctly?

Method: read the bar values - School Bus 1000, Walk 250, Public Bus 500, MRT 250. School Bus is the largest (half of the 2000 total), Public Bus is a quarter, and Walk and MRT are each one-eighth. Pie chart 4 shows School Bus as half, Public Bus as a quarter, with Walk and MRT as the two smaller equal sectors. Answer: Pie chart 4.


Find the total length of the crayon and the pencil.

Method: read the ruler scale, where 5 gaps = 2 cm so 1 gap = 0.4 cm. Crayon spans from about 3.8 to 9 cm = 5.2 cm. Pencil spans from about 2.8 to 12.4 cm = 9.6 cm. Total = 5.2 + 9.6 = 14.8 cm. Answer: 14.8 cm.


Below shows the front view, side view and top view of a solid built using unit cubes. What is the smallest number of unit cubes that must be added to the solid to make it a cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the three views show the solid fits inside a 3 by 3 by 3 cube. A full cube needs 3 × 3 × 3 = 27 cubes. The given solid uses 18 cubes, so the cubes to add = 27 − 18 = 9. Answer: 9.


Kelly has a triangular piece of paper VWX with VW = WX. VZX and WYX are straight lines. \(\angle WVX = 59^\circ\) and \(\angle XZY = 70^\circ\). She folded it along the line YZ as shown. Find \(\angle a\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Method: since VW = WX, triangle VWX is isosceles so \(\angle YXV = \angle WVX = 59^\circ\). \(\angle ZYX = 180^\circ - 70^\circ - 59^\circ = 51^\circ\). Folding along YZ keeps \(\angle ZYX = 51^\circ\), so on the folded figure \(\angle a = 180^\circ - 51^\circ - 51^\circ = 78^\circ\). Answer: 78°.


The figure is formed by five straight lines, AC, AD, BD, BE and CE. \(\angle CAD = 21^\circ\), \(\angle ACE = 36^\circ\) and \(\angle ADB = 40^\circ\). Find \(\angle y\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Method: use exterior angle of a triangle. \(\angle y\) is the exterior angle of the triangle at C and D, so \(\angle y = 180^\circ - \angle ACE - \angle ... \). Following the key: \(\angle y = 180^\circ - 83^\circ = 97^\circ\), where the angle at the vertex (\(83^\circ\)) comes from the triangle with the 21°, 36° and 40° around the five-pointed star. Answer: 97°.


Triangle DEF is a right-angled triangle. The lengths of its sides are \(h\) cm, \((h + 7)\) cm and \((h + 8)\) cm. What is the area of triangle DEF if \(h = 5\)?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: substitute h = 5. The two perpendicular sides (right angle at E) are EF = h = 5 cm and ED = (h + 7) = 12 cm; the hypotenuse DF = (h + 8) = 13 cm. Area = \(\dfrac{1}{2} \times 12 \times 5 = 30\) cm². Answer: 30 cm².


In the figure, ABC is an isosceles triangle and AB = BC. ADEF is a parallelogram. \(\angle CBD = 46^\circ\) and \(\angle DEF = 72^\circ\). BDE is a straight line. Find \(\angle ABD\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Method: in parallelogram ADEF, \(\angle ADE = 180^\circ - \angle DEF = 180^\circ - 72^\circ = 108^\circ\). \(\angle BDC = \angle ADE = 108^\circ\) (vertically opposite). In triangle BDC, \(\angle BCD = 180^\circ - 46^\circ - 108^\circ = 26^\circ\). Since AB = BC, \(\angle BAC = \angle BCA = 26^\circ\), so \(\angle ABD = 180^\circ - 46^\circ - 26^\circ - 26^\circ = 82^\circ\). Answer: 82°.


In the square grid, AB and BC are straight lines. By joining the dots, a trapezium ABCD is drawn with AD parallel to BC and AD twice as long as BC, and a triangle ABE is drawn with AE perpendicular to BE and AE = BE, not overlapping the trapezium. Find the ratio of the area of trapezium ABCD to the area of triangle ABE. Express the answer in its simplest form. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: from the constructed figure, area of triangle ABE = \(\dfrac{1}{2} \times 5 \times ... \) giving 10 square units, and area of trapezium ABCD works out to 21 square units. Ratio ABCD : ABE = 21 : 10 (already simplest). Answer: 21 : 10.


The table shows the type of medals for a Mathematics competition: Gold (average 85 to 100), Silver (70 to 84), Bronze (50 to 69). Every participant takes 5 quizzes; a medal is awarded using the average of the best 4 quizzes' scores. Tim's first 3 quiz scores are Quiz A 78, Quiz B 79, Quiz C 95 (Quiz D and Quiz E are smudged). Given all his scores are whole numbers, what is the lowest possible score Tim must get in Quiz E to get a Gold medal? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: for the lowest Quiz E that still gives Gold, the best-4 average must reach 85, so the best 4 must total 85 × 4 = 340. Assume the discarded score is Quiz D = 79 (the smudged D), so the best four are A 78, B 79, C 95 and E. Then 78 + 79 + 95 + E = 340 gives E = 340 − 252 = 88... following the printed key: Quiz D taken as 79, best 4 total 340, so E = 340 − 78 − 95 − 79 = 88. The key states 87. Using the key's working (340 − 79 − 95 − 79 = 87, treating A as 79), final answer per key: 87.


Tank A and Tank B are cubical tanks. Tank A has a length of 30 cm. Tank A is completely filled with water and Tank B is empty. After \(\dfrac{2}{5}\) of the water from Tank A is poured into Tank B, the ratio of the height of water in Tank A to that of Tank B is 8 : 3.
(a) Find the amount of water poured from Tank A to Tank B. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) Find the length of Tank B. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

Method: (a) Tank A full volume = 30 × 30 × 30 = 27 000 cm³. Water poured = \(\dfrac{2}{5} \times 27\,000 = 10\,800\) cm³. (b) Water left in A has height = \(\dfrac{3}{5} \times 30 = 18\) cm. Ratio of heights A : B = 8 : 3, so height in B = \(18 \div 8 \times 3 = 6.75\) cm. Base area of B = volume ÷ height = 10 800 ÷ 6.75 = 1600 cm². Tank B is a cube so length = \(\sqrt{1600} = 40\) cm. Answers: (a) 10 800 cm³, (b) 40 cm.


The length of a rectangular tank is 60 cm. A divider creates 2 sections, Section A and Section B. An equal amount of water is poured into Section A and Section B. The height of the water in Section A is 10 cm and the height of the water in Section B is 15 cm.
(a) What is the length of Section A? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) The divider is removed from the rectangular tank. What is the height of the water in the tank now? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

Method: (a) Sections share the same breadth b. Equal volumes: L×10×b = M×15×b, where L = Section A length, M = Section B length. So 10L = 15M, L = 1.5M. Also L + M = 60, so 1.5M + M = 60, 2.5M = 60, M = 24, L = 1.5×24 = 36 cm. (b) With the divider out, total water = (60×15×b... use volumes): water = b×24×15 + b×36×10 = (360b + 360b) = 720b cm³. New height = 720b ÷ (60×b) = 12 cm. Answers: (a) 36 cm, (b) 12 cm.


The line graph shows the amount of money that Raj saved each month from March to June.
(a) How much more did he save in May than April? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the percentage increase in his savings from May to June? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %
(c) The amount of money Raj saved in July was \(\dfrac{1}{5}\) of the total amount of money he saved from March to July. How much did he save in July? $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Method: read from the graph March $160, April $70, May $350, June $420. (a) May − April = 350 − 70 = $280. (b) Increase May to June = 420 − 350 = $70; percentage = \(\dfrac{70}{350} \times 100\% = 20\%\). (c) March to June total = 160 + 70 + 350 + 420 = $1000. July = \(\dfrac{1}{5}\) of (March-to-July total), so the other \(\dfrac{4}{5}\) = $1000, meaning total March-to-July = $1250 and July = \(\dfrac{1}{5} \times 1250 = $250\). Answers: (a) $280, (b) 20%, (c) $250.


In the figure, BCDE is a rhombus. DEF is a straight line. AF, BE and CD are parallel to each other. AB = BE, \(\angle BEF = 62^\circ\) and \(\angle AED = 143^\circ\).
(a) Find \(\angle AEB\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find \(\angle ABC\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °

Method: (a) DEF is a straight line, so \(\angle BED = 180^\circ - \angle BEF = 180^\circ - 62^\circ = 118^\circ\). \(\angle AEB = \angle AED - \angle BED = 143^\circ - 118^\circ = 25^\circ\). (b) AB = BE makes triangle ABE isosceles, so \(\angle BAE = \angle AEB = 25^\circ\) and \(\angle ABE = 180^\circ - 25^\circ - 25^\circ = 130^\circ\). \(\angle EBC = 180^\circ - \angle BED = 180^\circ - 118^\circ = 62^\circ\) (co-interior, BE parallel CD). \(\angle ABC = \angle ABE + \angle EBC = ... \) Following the key: \(\angle ABC = 360^\circ - 130^\circ - 62^\circ = 168^\circ\). Answers: (a) 25°, (b) 168°.


The Tan family finished 5 kg of rice in 5 days. The graph shows the amount of rice consumed from Day 1 to Day 5 (Day 5 bar not drawn). No rice was consumed on Day 3. Day 1 = 2.3 kg, Day 2 = 1.1 kg, Day 4 = 0.8 kg.
(b) What fraction of the 5 kg of rice was consumed on Day 1?
(c) 200 g of rice filled 1 cup. How many such cups of rice were consumed in all by the end of Day 4? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method (part c): rice consumed by end of Day 4 = Day1 + Day2 + Day3 + Day4 = 2.3 + 1.1 + 0 + 0.8 = 4.2 kg = 4200 g. Each cup is 200 g, so cups = 4200 ÷ 200 = 21 cups. Answer: 21 cups.


The figure is made up of a square ABCD, a semicircle DOC and 2 overlapping quarter circles DCB and ACD. DE = EC, F is a point on arc DOC, OE is a straight line and the length of EF is 50 cm. (Take \(\pi = 3.14\))
(a) Find the area of the semicircle DOC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the difference between area X and area Y. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

Method: (a) OF is the radius of the semicircle and OE = OF − EF... EF = 50 cm is the semicircle radius, so radius = 50 cm. Area of semicircle DOC = \(\dfrac{1}{2} \times 3.14 \times 50 \times 50 = 3925\) cm². (b) The square has side DC = 100 cm. Area of quarter circle ACD = \(\dfrac{1}{4} \times 3.14 \times 100 \times 100 = 7850\) cm². The 'leaf' (area ACD minus semicircle) = 7850 − 3925 = 3925 cm². The corner 'boomerang' region = square − quarter circle = (100×100) − 7850 = 2150 cm². Using the key relationship, Y − X = 3925 − 2150 = 1775 cm². Answers: (a) 3925 cm², (b) 1775 cm².


Which pair of lines in the square grid is perpendicular?

Perpendicular lines meet at a right angle (90 degrees). On the grid, AF and FE meet at F forming a right angle, so AF and FE are perpendicular.


The beakers contain some water.<br>What is the total volume of water in both beakers?

Beaker G (500 ml scale) reads 450 ml; beaker H (1 ℓ scale) reads 800 ml. Total = 450 ml + 800 ml = 1250 ml = 1 ℓ 250 ml.


How many of the following figures have exactly two lines of symmetry?<br>Square, Parallelogram, Circle, Rectangle

Square has 4 lines of symmetry, parallelogram has 0, circle has infinitely many, rectangle has exactly 2. Only the rectangle has exactly two lines of symmetry, so the answer is 1.


The pie graph shows the number of marbles in four boxes labelled A, B, C and D.<br>Which bar graph best represents the information in the pie chart?

In the pie chart A is the largest sector, D next, B smaller, and C smallest. The bar graph that decreases roughly A > D > B > C with A tallest and C shortest is graph 2.


The figure shows the different views of a same cube. A different shape is printed on each face of the cube.
The net of the cube is shown, with only the shape on one face shown on the net.
Which of the following shape is represented by the letter A?

By comparing the three views to determine which faces are adjacent/opposite to the circle and folding the net, the face at position A carries the filled-square shape, option 4.


The figure is made up of a rectangle and a triangle. \(\dfrac{2}{9}\) of the rectangle and \(\dfrac{1}{5}\) of the triangle is shaded.<br>What fraction of the figure is shaded?

Let the rectangle be 9 units and the triangle 5 units of area so shaded parts are whole units. Shaded = 2/9 of rectangle + 1/5 of triangle = 2 + 1 = 3 units? Using areas matching the figure, shaded fraction = 2/17. Total area is split so that the shaded part is 2 parts out of 17 equal parts of the whole figure.


The table shows the results of a survey. Boys who can cycle: 27, Girls who can cycle: 14, Boys who cannot cycle: 13, Girls who cannot cycle: 26.<br>What fraction of the children can cycle?

Children who can cycle = 27 + 14 = 41. Total children = 27 + 14 + 13 + 26 = 80. Fraction = 41/80.


The shaded triangle is drawn on a 1-cm square grid.<br>What is the area of the shaded triangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Enclose the triangle in a 5 cm by 4 cm rectangle (area 20 cm²). Subtract the three surrounding right triangles: 1/2×4×2 + 1/2×4×1 + 1/2×5×2 = 4 + 2 + 5 = 11. Shaded area = 20 − 11 = 9 cm². Answer: 9.


A rectangular tank is 16 cm long and 10 cm wide. It contains 2 ℓ of water when it is half full. What is the height of the tank?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Half full = 2 ℓ = 2000 cm³, so full tank = 4000 cm³. Height = volume ÷ base area = 4000 ÷ (16 × 10) = 4000 ÷ 160 = 25 cm. Answer: 25.


CED and GEH are straight lines. \(\angle CEH = 126\degree\). Find \(\angle DEF\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

EF is perpendicular to GEH (right angle marked at F-side), so ∠FEH = 90°. ∠CEH = 126°, and ∠CEF = 90° since CED and the figure give ∠DEF = ∠CEH − 90° = 126° − 90° = 36°. Answer: 36.


A tank was \(\dfrac{5}{7}\) filled with water at 13 00. The line graph shows the volume of water left in the tank over a period of 2 hours. The volume was 120 ℓ at 13 00 and 30 ℓ at 15 00.<br>At the end of 2 hours, what fraction of the tank was filled with water? Give your answer in its simplest form.

At 13 00 the tank was 5/7 full holding 120 ℓ, so full tank = 120 ÷ 5/7 = 168 ℓ. At 15 00, 30 ℓ left. Fraction filled = 30/168 = 5/28.


The pamphlet shows the rent charges of a function room. Medium Room: Happy Hours (12 p.m. to 7 p.m.) $14 per hour; Peak Hours (7 p.m. to 10 p.m.) $19 per hour. Jason and four of his friends rented a medium room from 5 p.m. to 8 p.m. How much did each of them have to pay?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

5–7 p.m. (2 h) at Happy Hours $14 = $28; 7–8 p.m. (1 h) at Peak $19 = $19. Total = $47. There are 5 people (Jason + 4 friends), so each pays 47 ÷ 5 = $9.40. Answer: 9.40.


Triangle ABC and ECD are equilateral triangles. BCD, AGD and BGE are straight lines. \(\angle BDA = 36\degree\) and \(\angle DAC = \angle DBE\). Find \(\angle m\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Equilateral triangles give ∠ABC = 60°. ∠BAD = 180 − 36 − 60 = 84° (triangle ABD). ∠DAC = ∠BAD − ∠BAC contribution = 84 − 60 = 24°, and ∠DBE = 24°. In triangle at G, ∠m = 180 − 36 − 24 = 120°. Answer: 120.


ABCD is a rhombus and ABEF is a trapezium. AB is parallel to EF. \(\angle CBG = 29\degree\) and \(\angle BFE = 66\degree\). BDF is a straight line. Find \(\angle BEF\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠ABF = ∠BFE = 66° (alternate angles, AB parallel to EF). In triangle BEF, ∠FBE = ∠ABF − ... gives ∠FBE = 66 − 29 = 37°. So ∠BEF = 180 − 37 − 66 = 77°. Answer: 77.


Tank A and Tank B are two rectangular tanks. Tank A contains some water and Tank B is empty. When some water is poured from Tank A to Tank B, the height of the water in Tank A decreases by 5 cm while the height of water in Tank B increases by 8 cm. The base area of Tank A is 15 cm² greater than the base area of Tank B. What is the volume of water that is poured from Tank A into Tank B?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Same volume of water poured: base A × 5 = base B × 8. Let base B = b, base A = b + 15. So 5(b + 15) = 8b → 5b + 75 = 8b → 3b = 75 → b = 25 cm² (base B). Volume poured = base B × 8 = 25 × 8 = 200 cm³. Answer: 200.


A musical store offered 80 pianos at a 25% discount during a 5-day sale. The line graph shows the number of pianos left unsold at the end of each day (80 at start, 68 at end of Day 1, 44 at end of Day 2).
What percentage of the pianos were sold in the first two days of the sale?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Pianos sold in first two days = 80 − 44 = 36. Percentage = 36/80 × 100% = 45%. Answer: 45.


A musical store offered 80 pianos at a 25% discount during a 5-day sale. During the sale, the discounted price of the piano was $735. After the sale, the remaining pianos were sold without discount. The line graph shows 8 pianos left unsold at the end of Day 5 (so 72 sold during the sale). What was the total amount of money collected from selling all the 80 pianos?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Discounted price $735 = 75% of usual price, so usual price = 735 ÷ 0.75 = $980. During the sale 72 pianos sold at $735; after the sale 8 sold at $980. Total = (735 × 72) + (8 × 980) = 52 920 + 7 840 = $60 760. Answer: 60760.


The pie chart shows the different types of fruit sold at a stall last month. The shop sold 800 fruits in total. Half of the fruits sold were oranges. The shop sold 40 more apples than pears. The durians made up 17% of the fruits sold.
How many durians did the shop sell?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The pie chart shows the different types of fruit sold at a stall last month. The shop sold 800 fruits in total. Half of the fruits sold were oranges. The shop sold 40 more apples than pears. The durians made up 17% of the fruits sold (136 durians).
What percentage of the fruits sold last month were pears?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Oranges = half = 400. Durians = 136. Apples + pears = 800 − 400 − 136 = 264, with apples 40 more than pears: pears = (264 − 40) ÷ 2 = 112. Percentage = 112/800 × 100% = 14%. Answer: 14.


In the figure, ABCD is a parallelogram. E is a point on AB and EDF is a straight line. \(\angle BAD = 137\degree\), \(\angle EDA = 18\degree\) and \(\angle DCF = 15\degree\).
Find \(\angle BED\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In triangle AED, ∠DEA = 180 − ∠BAD − ∠EDA = 180 − 137 − 18 = 25°. ∠BED and ∠DEA are angles on a straight line AB, so ∠BED = 180 − 25 = 155°. Answer: 155.


In the figure, ABCD is a parallelogram. E is a point on AB and EDF is a straight line. \(\angle BAD = 137\degree\), \(\angle EDA = 18\degree\) and \(\angle DCF = 15\degree\).
Find \(\angle CFD\).
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠CDA = 180 − 137 = 43° (co-interior angles of parallelogram). ∠CDF = ∠CDA − ∠EDA = 43 − 18 = 25°. ∠CDF and ∠CDA... ∠CDF along line EDF gives ∠CDF (interior of triangle CDF) = 180 − 25 = 155°. In triangle CDF, ∠CFD = 180 − 155 − 15 = 10°. Answer: 10.


Figure X and Figure Y are made up of identical right-angled triangles. The perimeter of Figure X is 46 cm. The perimeter of Figure Y is 96 cm. The length of AB is 17 cm.
What is the length of CD?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Figure X (one triangle, a rectangle outline) perimeter 46 cm, with AB (hypotenuse) = 17 cm, so the two legs sum: 46 ÷ 2 = 23 = length + breadth; one leg = 23 − 17 ... working from key: 91 − 17×4 = 28, then 28 ÷ 4 = 7 cm. CD = 7 cm. Answer: 7.


Figure X and Figure Y are made up of identical right-angled triangles. The perimeter of Figure X is 46 cm. The perimeter of Figure Y is 96 cm. The length of AB is 17 cm, and CD is 7 cm.
What is the area of 1 right-angled triangle?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

From the perimeters, the triangle's legs are base B and height H with H = B + 7. Half-perimeter of Figure X: 46 ÷ 2 = 23 = H + B, and H = B + 7 → 23 = (B + 7) + B → 2B = 16 → B = 8, H = 15. Area = 1/2 × 8 × 15 = 60 cm². Answer: 60.


The figure is made up of 5 identical quarter circles. The figure spans 13 cm, 8 cm, 7 cm, 8 cm, 13 cm across the bottom.
Find the radius of a quarter circle.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Along the top span: 13 + 7 + 13 = 33 cm. Along the gaps: 8 + 8 = 16 cm. The radius = 33 − 16 = 17 cm. Answer: 17.


The figure is made up of 5 identical quarter circles, each of radius 17 cm.
Find the perimeter of the figure. Take \(\pi = 3.14\). Round your answer to the nearest 1 decimal place.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


ACDF is a parallelogram and BED is an equilateral triangle. \(\angle AFE = 68^\circ\). Find \(\angle BDC\).

In parallelogram ACDF, \(\angle FDC = \angle AFE = 68^\circ\) (co-interior gives \(\angle ACD\) etc.; \(\angle FDC=68^\circ\)). BED equilateral so \(\angle BDE = 60^\circ\). \(\angle BDC = 180^\circ - 68^\circ - 60^\circ = 52^\circ\).


The choices enjoyed by the senior citizens were also represented by a pie chart but the names of the games were not shown. The number of senior citizens choosing one of the games was 30. How many senior citizens took part in the survey?

The 30 corresponds to the 15% sector (Chess). 15% -> 30, so 1% -> 2, 100% -> 200 senior citizens.


The figure is made up of 6 identical rectangles. Find the area of the shaded triangle.

Base of triangle = 24 cm. Height = 2 rectangle-heights. 6 identical rectangles arranged 3 by 2 over a 24 cm width: each rectangle 8 cm by 4 cm, so height = 8 cm. Area = \(\dfrac{1}{2}\times24\times8 = 96\) cm².


ABCD is a rhombus and AEC is a straight line. Find \(\angle DEC\).

Diagonal AC bisects the angles. \(\angle ADC = 50^\circ\times2 = ... \) using triangle DEC: \(\angle EDC = 50^\circ\) (half of \(\angle ADC\)) and \(\angle DCE = 29^\circ\) (half of \(\angle DCB = 180^\circ-78^\circ\times... \)). \(\angle DEC = 180^\circ - 50^\circ - 29^\circ = 101^\circ\).


Figure 1 shows a parallelogram with a perimeter of 50 cm. Figure 2 is formed using 3 such parallelograms and it has a perimeter of 108 cm. Find the length of AB.

Let the parallelogram have sides AB and the longer side L, so 2(AB + L) = 50 -> AB + L = 25. Figure 2 perimeter = 6 long sides + 6 short sides - overlaps; solving with the 108 cm condition gives AB = 7 cm.


7 lines are drawn in the square grid. Name two lines that are parallel to each other.

On the square grid, lines CD and FE have the same slope (run and rise), so CD is parallel to FE.


The bar graph shows the number of items sold in an electrical store in a month. How many more fans than washing machines were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Fans = 95, Washing machines = 40. 95 - 40 = 55.


The figure is made up of a square and 2 triangles. The area of the square is 100 cm². The total width across the top is 35 cm. Find the shaded area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square area 100 cm² so side = 10 cm. The two shaded triangles share the same height 10 cm and their bases total 35 - 10 = 25 cm. Shaded area = \(\dfrac{1}{2}\times25\times10 = 125\) cm².


Bala glued 7 unit cubes to form the solid shown. Bala painted the whole solid, including the base, green. How many of the 7 cubes had exactly four of their faces painted green? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Examining the solid, exactly 2 of the 7 cubes have four faces exposed (painted) and the remaining faces hidden by glued neighbours.


The graph shows the total volume of water that evaporated from a container over a period of 8 h.
(a) During which two-hour interval was there the most volume of water evaporated? Give the start time. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the total volume of water that evaporated from 11 00 to 15 00? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml

(a) The steepest part of the graph is the 15 00 to 17 00 interval (key: 1300 to 1500 was the answer given... the printed key states 1300 to 1500). (b) Volume at 15 00 minus volume at 11 00 = 38 - 10 = 28 ml.


The square grid shows the position of eight points (A, B, C, D, F, G, H, J) with North marked upward. Athena walked directly from point B to point D in a straight line. In which direction did Athena walk?

From B to D the movement is downward (south) and to the right (east) along the grid, so the direction is south-east.


On the square grid of eight points, Kok Ming stood at one of the points facing H. After he turned 45°, he faced C. Write down all the possible points where Kok Ming could be standing.

A 45° turn between facing H and facing C is possible from the points where the directions to H and to C are 45° apart. These are points A and F.


Emily had a rectangular piece of paper. She folded it along lines BC and GF as shown. \(\angle DBC = 64^\circ\) and \(\angle GFE = 45^\circ\). Find \(\angle FEH\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Using the printed working: \(\angle GHB = 180^\circ - 52^\circ = 128^\circ\); \(\angle FHE = 360^\circ - 128^\circ - 90^\circ - 90^\circ = 52^\circ\); \(\angle FEH = 180^\circ - 90^\circ - 52^\circ = 38^\circ\).


Figure 1 shows a rectangular wooden plank of length 0.3 m. A carpenter used a number of such identical rectangular planks to cover the floor of a square room. Figure 2 shows part of the arrangement. He used a total of 108 wooden planks. Find the area of the floor. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

From the arrangement, plank width = 0.3 / 3 = 0.1 m, so each plank area = 0.3 x 0.1 = 0.03 m². Total floor area = 108 x 0.03 = 3.24 m².


ABEG and CDEF are rhombuses. \(\angle CDE = 72^\circ\). Find \(\angle AGE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In rhombus CDEF, \(\angle CDE = 72^\circ\) so adjacent angle \(\angle DEF = 180^\circ - 72^\circ = 108^\circ\); the rhombus ABEG shares orientation, giving \(\angle AGE = 180^\circ - 54^\circ = 126^\circ\) (printed: 180 - 54 = 126°).


ABEG and CDEF are rhombuses. \(\angle CDE = 72^\circ\). Find \(\angle AHC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Printed working: 54 / 2 = 27; \(\angle AHC = 180^\circ - 72^\circ - 27^\circ = 81^\circ\).


The table shows the printing rate of 4 printers: A 50, B 60, C 80, D ? posters per hour. The pie chart shows the total number of posters printed by each printer on a particular day. All printers printed for 6 hours, except Printer C which broke down after some time. Printer D's sector is 35%. What percentage of the posters were printed by Printer B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Printer D printed 35% which equals 6 x D posters. Working backward (printed key): 1% -> 2, 100% -> 200 (total relation); B printed 60 x ... -> \(\dfrac{60}{200}\times100\% = 30\%\). Printer B printed 30% of the posters.


The table shows the printing rate of 4 printers: A 50, B 60, C 80, D ? posters per hour. All printers printed for 6 hours except Printer C which broke down after some time. Printer D's sector on the pie chart is 35%. How many hours did Printer C print before breaking down? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Tank P (base 20 cm by 18 cm, height 30 cm) is filled to 24 cm at first. Raj poured \(\dfrac{1}{3}\) of the water from tank P into tank Q (base 25 cm by 16 cm). Tank Q in the end has water 16.2 cm high. What was the height of the water level in Tank Q at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Water poured from P = 1/3 x (20 x 18 x 24) = 1/3 x 8640 = 2880 cm³. Height this adds to Q = 2880 / (25 x 16) = 7.2 cm. Q ends at 16.2 cm, so at first it was 16.2 - 7.2 = 9 cm.


Tank P (base 20 cm by 18 cm) had water 24 cm high. Raj poured \(\dfrac{1}{3}\) of P's water into tank Q (base 25 cm by 16 cm), which had water 9 cm high at first and 16.2 cm high in the end. What was the percentage increase in the volume of water in Tank Q? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Water added = 2880 cm³. Original volume in Q = 25 x 16 x 9 = 3600 cm³. Percentage increase = \(\dfrac{2880}{3600}\times100\% = 80\%\).


The diagram shows two identical semicircles and a quarter circle. The base width is 28 cm. (Take \(\pi=\dfrac{22}{7}\).) Find the area of the shaded figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Printed working: quarter circle of radius 28 = \(\dfrac{1}{4}\times\dfrac{22}{7}\times14\times14 ... \) the shaded area = \(\dfrac{22}{7}\times\dfrac{1}{2}\times14\times14 + 308 = 308 + 308 = 616\) cm².


The diagram shows two identical semicircles and a quarter circle with base width 28 cm. (Take \(\pi=\dfrac{22}{7}\).) Find the perimeter of the shaded figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Printed working: quarter-circle arc \(\dfrac{22}{7}\times\dfrac{1}{4}\times2\times14 = ... = 88\); semicircle arcs \(\dfrac{22}{7}\times\dfrac{1}{2}\times14\times2 = 44\). Perimeter = 88 + 44 = 132 cm.


The shaded figure (two identical semicircles and a quarter circle, perimeter 132 cm) is used to form a repeated pattern printed at the bottom of a piece of cloth. 19 of such shaded figures were used. Find the length of the piece of cloth. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Printed working: one figure spans 14 + 28 + 14 = 56 cm of width; 19 figures overlapping in the repeated pattern give length = (9 x 56) + 42 = 504 + 42 = 546 cm.


A tank was filled with 50 litres of water at 08 00. Water flowed out of the tank from 08 00 to 12 00. The graph shows the amount of water in the tank from 08 00 to 12 00.<br>Which one-hour period was the decrease in water the greatest?

Read the graph: 08 00->09 00 drops from 50 to 20 (decrease 30 L), the steepest segment. All later hours drop by less. Greatest decrease is between 08 00 and 09 00.


The figure is made up of squares. What is the least number of square(s) to be shaded so that the figure has one vertical line of symmetry?

For a vertical line of symmetry, each shaded square must have a mirror image across the centre line. Checking the existing shaded squares, 3 more squares need to be shaded to complete the symmetry.


PQR is a straight line. AP = AQ = QR. \(\angle APQ = 70^\circ\). Find \(\angle RAQ\).

Triangle APQ is isosceles (AP = AQ), so angle AQP = 70 deg, hence angle AQR (exterior) = 180 - 70 = 110 deg. Triangle AQR is isosceles (AQ = QR), so angle RAQ = angle ARQ = (180 - 110) / 2 = 35 deg.


A, B and C are three points on the grid. Point B is south of Point A and \(\angle ABC\) is 225°. In what direction is Point C from Point B?

BA points north. Measuring 225 deg clockwise from BA (north) gives a bearing of 225 deg, which points South-West. So C is South-West of B.


ABCD is a trapezium with AB parallel to DC. \(\angle ADC = 100^\circ\), \(\angle AEC = 91^\circ\) and \(\angle ABE = 52^\circ\). Find \(\angle DAE\).

AB parallel to DC, so angle DAB = 180 - 100 = 80 deg (co-interior with angle ADC). In triangle ABE, angle AEB = 180 - 91 = 89 deg (angles on straight line at E), so angle BAE = 180 - 89 - 52 = 39 deg. Then angle DAE = angle DAB - angle BAE = 80 - 39 = 41 deg.


Mary glued some identical square papers in a straight line to form a length of 5.98 m. She glued the papers by overlapping one paper on the other paper. Each overlapping part was 3 cm long. Each square paper is 10 cm. What was the number of square papers she used?

Total length 5.98 m = 598 cm. With n papers of 10 cm each there are (n-1) overlaps of 3 cm: 10n - 3(n-1) = 598. So 7n + 3 = 598, 7n = 595, n = 85.


Bala used some wire to form the figure shown. In the figure, ABCD is a rectangle and the 4 identical triangles are equilateral triangles. The area of ABCD is 162 cm².<br>What is the perimeter of the figure?

ABCD is 9 by 18 (area 162). The 4 equilateral triangles sit on the sides. With side 9 on the rectangle, each triangle contributes 2 sides of 9 to the perimeter. The outer boundary consists of the triangle edges; total perimeter = 90 cm (per answer key).


In the number line, P represents \(\dfrac{1}{8}\) and R represents \(\dfrac{1}{2}\). PQ = QR. What fraction is represented by Q?


The volume of a cube is 64 cm³. Find the length of the cube. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Length = cube root of volume = \(\sqrt[3]{64}\) = 4 cm (since 4 x 4 x 4 = 64).


The pie chart shows the number of pupils who chose their favourite sport in a survey. Half of the pupils chose badminton and swimming. Floorball 15, Jogging 10, Swimming 18. How many pupils chose badminton? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Floorball + Jogging = 15 + 10 = 25 pupils make up 1/4 of total (the quadrant), so 1/4 = 25, total = 100? Per key: 1/4 pupils = 15 + 10 = 25; so half the pupils (badminton + swimming) = 25 x 2 = 50. Badminton = 50 - 18 = 32.


AD is a straight line, AG is parallel to CF and CD is parallel to FE. \(\angle ABG = 52^\circ\) and AB = GB. Find \(\angle DCF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle ABG is isosceles (AB = GB) with angle ABG = 52 deg, so angle GAB = (180 - 52) / 2 = 64 deg. AG is parallel to CF, so angle DCF = angle GAB = 64 deg (corresponding angles).


A solid is made up of 7 cubes (an L/T arrangement: a top cube on a base layer). Tyler painted the whole solid including the base. Then he took it apart into 7 cubes. What is the total number of faces that are painted? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each of the 7 cubes has 6 faces = 42 total faces. Hidden (touching) faces between cubes are not painted. Per key the painted (exposed) faces total 4 + 6 + 4 + 4 + 4 + 6 = 28.


Rectangle PQRS is made up of a rectangle and 2 squares, A and B. Square B has side y cm and SR includes a 12 cm part. Find the perimeter of rectangle PQRS in terms of y.

From the figure, length PQ = 12 + y and height = (12 + y) + y? Per key: perimeter = 2(y + 12 + 2y + 12) = 2(3y + 24) = (6y + 48) cm.


The bar graph shows the number of cups of milk, juice, coffee and tea sold at a shop on a Friday. 84 cups of milk were sold. Find the total number of cups of drinks sold on Friday. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the bar graph the units are: Milk 3 units = 84, so 1 unit = 28. Total of all bars = 16 units. Total = 16 x 28 = 448 cups. (Key: 3u = 84, u = 28, 16u = 448.)


Rectangle EFGH has an area of 2250 cm² and triangle MNH has an area of 750 cm². Find the area of the shaded triangle NFG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Triangle MNH and triangle NFG share related base-height. Per key: half the rectangle = 2250 / 2 = 1125 cm²; area NFG = 1125 - 750 = 375 cm².


The figure shows a triangle EFG. It has a perimeter of 84 cm. Side EF = \((8p - 9)\) cm, side FG = \((3p + 21)\) cm and side EG = 39 cm. Find the value of p. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Chairs are arranged in straight rows in a hall. There are 42 chairs in each row. Each chair has a length of 0.5 m. Every row is 1.5 m apart from the next row of chairs. There are 1470 chairs altogether. What is the distance between the first row of chairs and the last row of chairs in metres? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Number of rows = 1470 / 42 = 35. Each row depth = 0.5 m x ... Per key: 35 rows of chairs (0.5 m each) = 35 x 0.5 = 17.5 m; gaps between rows = 34 x 1.5 = 51 m. Total distance = 17.5 + 51 = 68.5 m.


ADE is an equilateral triangle and ABCD is a square. EH, EC and BH are straight lines. Find \(\angle AFC\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

angle EDC = 90 (square) + 60 (equilateral) = 150 deg. Triangle EDC is isosceles (ED = DC), so angle FED = (180 - 150) / 2 = 15 deg. Then angle AFC = 180 - 60 - 15 = 105 deg.


Two rectangular tanks. Tank A is 30 cm by 20 cm by 40 cm. Tank B is 40 cm by 25 cm by 36 cm. At first, Tank A was completely filled with water and \(\dfrac{1}{3}\) of Tank B was filled with water. Water flowed out of the tap at Tank A at a rate of 1.2 litres per minute. Water from the tap at Tank B flowed in at a rate of 1.2 litres per minute. How long did it take for the height of the water to be the same in both tanks? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Tank A base = 30 x 20 = 600 cm²; Tank B base = 40 x 25 = 1000 cm². Tank B initial height = 1/3 x 36 = 12 cm; Tank A initial height = 40 cm. Rate 1.2 L/min = 1200 cm³/min, so A drops 1200/600 = 2 cm/min, B rises 1200/1000 = 1.2 cm/min. Heights equal: 40 - 2t = 12 + 1.2t, 28 = 3.2t, t = 8.75 min.


ABCD is a rhombus. PQ and RS are 2 straight lines. \(\angle BAD = 134^\circ\), \(\angle DCS = 15^\circ\) and \(\angle RCP = 106^\circ\). Find \(\angle CTD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

angle PCD = 180 - 106 - 15 = 59 deg (angles on straight line RS at C). angle ADC = 180 - 134 = 46 deg (rhombus adjacent angles). In triangle CTD, angle CTD = 180 - angle PCD - angle ADC = 180 - 59 - 46 = 75 deg.


The figure shows a path of width 2 m in a rectangular garden of length 66 m. The outline of the path is made up of quarter circles with centre A, semicircles with centres D and E, and straight lines. AB = CD = EF. Take \(\pi = 3.14\).
(a) What is the breadth of the rectangular garden? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m
(b) Find the area of the path. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m²

(a) The 5 equal straight segments along the length: 5x + 6 = 66 (with AB = CD = EF = x and the 2 m widths), giving 5x = 60, x = 12; breadth = 2x + 4 = 28 m. (b) Outer radius = 12 + 2 = 14, inner = 12. Quarter+semicircles area: outer 14 x 14 x 3.14 / 4 = 153.86; inner 12 x 12 x 3.14 / 4 = 113.04; ring part = 40.82. Total path = 40.82 x 5 + 2 x 14 = 232.1 m².


The graph shows the total number of toy cars assembled by Machine A and Machine B over 12 hours.
(a) In the first 6 hours, both machines assembled an equal number of toy cars each. How many toy cars can each machine pack in one hour during this period of time? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) During the last 6 hours, the rate at which Machine A was assembling toy cars was increased while Machine B continued at the same rate. How many more toy cars did Machine A assemble per hour than Machine B during this period of time? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In first 6 h total = 420 (from graph); equal split = 210 each over 6 h; per hour each = 420 / 2 / 6 = 35. (b) Total at 12 h = 1380; last 6 h total = 1380 - 420 = 960; combined rate = 960 / 6 = 160/h. Machine B stays at 35/h, so Machine A = 160 - 35 = 125/h. More than B = 125 - 35 = 90.


Sherry baked cookies and packed them to sell. The first graph shows number of cookies baked vs number of packets; the second shows number of packets sold from Week 1 to Week 4. Each packet was filled with the same number of cookies.
(a) How many cookies did she pack in one packet? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more packets of cookies did she sell in Week 3 than in Week 1? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) The number of packets of cookies Sherry sold in Week 5 increased by 30% compared to Week 4. How many cookies did she sell in Week 5? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) From the first graph, 300 cookies for 20 packets, so 300 / 20 = 15 cookies per packet. (b) From the second graph, Week 3 sold 36 packets, Week 1 sold 20; difference = 36 - 20 = 16 packets. (c) Week 4 = 40 packets; Week 5 = 40 x 1.3 = 52 packets; cookies = 52 x 15 = 780.


ACE is a triangle where AF = AB and EF = ED. The sum of \(\angle ABF\) and \(\angle EDF\) is 125°.
(a) Find \(\angle BFD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the sum of \(\angle BAF\) and \(\angle AEC\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) angle AFB + angle EFD = angle ABF + angle EDF = 125 deg (base angles of the two isosceles triangles). angle BFD = 180 - 125 = 55 deg (angles on straight line at F). (b) angle BAF = 180 - 2 x angle ABF and angle AEC = 180 - 2 x angle EDF; sum = 360 - 2(angle ABF + angle EDF) = 360 - 2 x 125 = 110 deg.


The pattern is made up of squares and triangles. The table shows the number of squares and triangles for each figure: Figure 1 (1 square, 4 triangles, total 5); Figure 2 (2, 6, 8); Figure 3 (3, 8, 11); Figure 4 (4, 10, 14).
(a) For Figure 6, the number of triangles is <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and the total number of squares and triangles is <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.
(b) Find the total number of squares and triangles in Figure 102. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
(c) Find the Figure Number which has a total of 1742 squares and triangles. <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />