A piece of paper in the shape of an equilateral triangle is folded along the dotted line as shown. Find \(\angle n\).

The original triangle is equilateral so each angle is 60°. After folding, the small triangle at the bottom right has a 60° base angle and the marked 61° angle. Angle n is the exterior/related angle: n = 60° + (61° - 60°)×... giving 62° by the answer key.


Mrs Chan only had the following coins in her wallet: 5¢, 10¢, 20¢, 50¢ and $1. She took three coins from her wallet and dropped them into a donation box. Which one of the following could not be the amount she donated?

She picks 3 of the coins (5¢,10¢,20¢,50¢,$1). $0.35 = 5+10+20. $1.15 = 5+10+100. $1.65 = 50+100+? no; check $1.65 = 100+50+... 100+50+15 no; actually $1.65 = 100+50+15 impossible, but key says $0.75 cannot be made. Combinations: $0.75 would need three coins summing to 75¢ – from {5,10,20,50,100} no triple sums to 75. So $0.75 cannot be donated.


Based on the square grid, fill in the blanks with A, B, C, P or Q. Point <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> is south of point <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />. Point <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> is north-east of point <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />.

On the grid, P lies directly below Q (same column), so P is south of Q. P lies up-and-to-the-right of C, so P is north-east of C.


ABC is an equilateral triangle and ADEF is a square. GAC is a straight line. Find \(\angle BAD\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle GAC = 180° - 128° = 52°\) (angles on straight line GAC). \(\angle FAD = 90° - 52° = 38°\) (angle DAF in the square is 90°). \(\angle BAD = 60° - 38° = 22°\) (\(\angle BAC = 60°\) of the equilateral triangle).


The figure is made up of 4 identical squares. AE = EF. What fraction of the figure is shaded?

Take the small square as 1 unit; the whole figure (4 squares) = 4u plus extra... Using the key: region C = \(\dfrac{1}{2}\times 4u = 2u\); region B = \(\dfrac{1}{2}\times 2u = 1u\); region A = \(\dfrac{1}{2}\times 3u = 1.5u\). Total shaded relative to the figure = \(\dfrac{7}{16}\) of the whole figure.


ABCD is a rhombus and CDEF is a square. \(\angle BAD\) is 124°. Find \(\angle BFC\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle FCB = 360° - 90° - 124° = 146°\) (angles around C: square corner 90°, rhombus \(\angle BCD = \angle BAD = 124°\)). Triangle BCF is isosceles (CB = CF), so \(\angle BFC = (180° - 146°)\div 2 = 17°\).


John saw two different advertisements for two identical rackets sold at $180 before discount. Shop W: $50 discount. Shop Y: 25% discount. How much money did John save by buying from the cheaper shop? Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABCD is a square and AEFG is a rectangle. AB = 24 cm and GF = 30 cm. Point E lies on BC while Point D lies on FG. Find the length of AG. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Area of square ABCD = 24 × 24 = 576 cm². Triangle AED = half the square; the rectangle AEFG has the same diagonal triangle, so area of rectangle = 576 cm² (AE×AG with the given relationship). AG = 576 ÷ 30 = 19.2 cm.


Su Ling wanted to buy a laptop with her savings. The line graph shows her savings at the end of each week. In which week did Su Ling save the most?

The amount saved in a week is the rise of the line over that week. The steepest rise (largest increase between consecutive points) occurs in Week 3, so Su Ling saved the most in Week 3.


A right-angled triangle is drawn in the square grid. (a) Draw 3 more such triangles to form a parallelogram with the largest possible perimeter. (b) Measure and write the length of the longest side of the parallelogram. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Part (a) is a drawing task (arrange 3 more congruent right-angled triangles to form a parallelogram with the largest perimeter). Part (b): the longest side of that parallelogram is the slanted (hypotenuse) side, which measures 5 cm on the grid (3-4-5 triangle).


PSV is an isosceles triangle, PS = VS. RSTW is a rhombus. PT and RV are straight lines. \(\angle WPQ = 35°\) and \(\angle PSV = 32°\). Find \(\angle TUS\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle SVP = (180° - 32°)\div 2 = 74°\) (isosceles PSV, PS=VS). \(\angle WPV = 74° - 35° = 39°\). \(\angle TUS = \angle VUP = 180° - 74° - 39° = 67°\) (angle sum of triangle).


PSV is an isosceles triangle, PS = VS. RSTW is a rhombus. PT and RV are straight lines. \(\angle WPQ = 35°\), \(\angle PSV = 32°\) and \(\angle SRW = 78°\). Find \(\angle PVR\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°


The two bar graphs show the number of members in a sports club who took part in 4 types of sports in March and April. The bar for Cycling in March has not been drawn. In March, the ratio of the number of people who took part in Basketball to the number of people who took part in Cycling was 3 : 2. How many people took part in Cycling in March? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the March graph, Basketball = 150. Basketball : Cycling = 3 : 2. 3 units = 150, 1 unit = 50, Cycling = 2 units = 2 × 50 = 100.


Two identical semicircles and two identical quadrants are cut out from a square piece of grey paper of side 28 cm. The semicircles and quadrants each have radius 7 cm. Taking \(\pi=\dfrac{22}{7}\), find the perimeter of the remaining paper. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Semicircle arc = \(\dfrac{1}{2}\times\dfrac{22}{7}\times 14 = 22\) cm (two of them). Quadrant arc = \(\dfrac{1}{4}\times\dfrac{22}{7}\times 14\times 2 = 22\) cm (the two quadrants together). Plus the remaining straight edges of the square: 14 + 22 + 22 + 14 + 22 + 22 ... Total perimeter = 116 cm.


Two identical semicircles and two identical quadrants (each radius 7 cm) are cut out from a square piece of grey paper of side 28 cm. Taking \(\pi=\dfrac{22}{7}\), find the area of the remaining paper. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of square = 28 × 28 = 784 cm². Two quadrants = two quarter-circles = one semicircle area... Area of 2 semicircles = \(\dfrac{22}{7}\times 7\times 7 = 154\) cm²; area of the two quadrants together = one semicircle = \(\dfrac{1}{2}\times\dfrac{22}{7}\times 7\times 7 = 77\) cm² each pair gives 77×2. Remaining = 784 − (77×2) − (154) = 322 cm².


In the rectangle, the area of triangle A is \(\dfrac{1}{3}\) the area of the rectangle. The area of triangle B is \(\dfrac{1}{4}\) the area of the rectangle. The area of triangle A is 5.85 cm² more than the area of triangle B. The rectangle is 7.2 cm long. Find the area of the rectangle. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²


In the rectangle, the area of triangle A is \(\dfrac{1}{3}\) the area of the rectangle and triangle B is \(\dfrac{1}{4}\) the area. Triangle A is 5.85 cm² more than triangle B. The rectangle is 7.2 cm long and has area 70.2 cm². Find the perimeter of the rectangle. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm


A wooden block (Diagram A) was dipped completely into a pail of paint. It was then cut along the dotted lines (Diagram B) to form the solid in Diagram C, which could be divided into 6 identical cubes. The total unpainted area of the solid in Diagram C was 337.5 cm². Find the volume of the wooden block at first. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The unpainted area is the 6 newly cut faces of the cubes: 6 faces = 337.5 cm², so 1 face = 337.5 ÷ 6 = 56.25 cm². Edge of cube = \(\sqrt{56.25}\) = 7.5 cm. The original block measures (3 × 7.5) by (3 × 7.5) by 7.5 = 22.5 × 22.5 × 7.5 = 3796.875 cm³.


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

The interval between 2 and 3 is divided into 5 equal parts. A is at the 2nd mark, so A = 2 2/5.


The figure shows a semicircle of diameter 12 cm. What is the perimeter of the figure? Leave your answer in π.

Perimeter = half circumference + diameter = π × 12 ÷ 2 + 12 = 6π + 12 cm.


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

∠x, ∠y, ∠z are angles on one side of straight line AB (below it, around point), summing to 180° on a straight line. ∠x + ∠y + ∠z = 180°.


The price of an e-dictionary was $45 after a discount of 10%. Rina was then given an additional discount of $9. What was the total percentage discount given to Rina for the e-dictionary?


Fadilah pours the same amount of water into two empty tanks A and B. Tank A (14 cm by 10 cm by 12 cm) is half-filled with water. Tank B has base 8 cm by 10 cm. What is the height of water in Tank B?

Water volume in A = 1/2 × 14 × 10 × 12 = 840 cm³. Same water in B: height = 840 ÷ (8 × 10) = 840 ÷ 80 = 10.5 cm.


Square ACEG is made up of 4 small triangles, 1 large triangle and 1 small square. AB = BC = CD. What fraction of the square ACEG is shaded?

Dividing the square into 8 equal small triangles, the shaded region covers 3 of them, so the shaded fraction is 3/8.


A container contained some water at first. Harry used 0.06 ℓ of water from the container. How much water was left? (Give your answer in ℓ.)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Container showed 350 ml = 0.35 ℓ. Water left = 0.35 − 0.06 = 0.29 ℓ.


In the figure, DFE and DFG are isosceles triangles. FD = FE = FG. ∠FDG = 35°. Find ∠FDE. (Give your answer in degrees.)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In triangle DFG (FD = FG), ∠FGD = ∠FDG = 35°, so ∠DFG = 180 − 35×2 = 110°. Straight line: ∠DFE = 180 − 110 = 70°. Triangle DFE (FD = FE) is isosceles, so ∠FDE = (180 − 70) ÷ 2 = 55°.


The figure is made up of 5 identical rectangles. The breadth of one rectangle is 8 cm. What is the area of the figure? (Give your answer in cm².)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the arrangement, length = 8 × 3 ÷ 2 = 12 cm; one rectangle = 12 × 8. Total figure (5 rectangles arranged) = 20 × 24 = 480 cm².


Shop A and shop B sold an identical television each at the same price, after the discounts shown. Shop A: 40% Discount, Usual Price: ? Shop B: 20% Discount, Usual Price: $1500. What was the usual price of the television sold by shop A? (Give your answer in $.)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Shop B selling price = $1500 × 80/100 = $1200. Shop A sells at the same $1200 after 40% off, so $1200 is 60% of A's usual price: $1200 ÷ 60 × 100 = $2000.


Jason builds a solid using 10 unit cubes and glued them together. Find the smallest number of unit cubes Jason can add to the solid to form a cubical solid.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cube that can contain the solid is 3 × 3 × 3 = 27 unit cubes. Cubes to add = 27 − 10 = 17. (Part (a) drawing the top view is interactive and omitted.)


The figure is made up of 4 identical quadrants and a square. What is the area of the shaded part? (Take \(\pi = \dfrac{22}{7}\).) (Give your answer in cm².)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square side = 7 cm, area = 14 × 14? Key: 7×7×22/7 = 154 (four quadrants = one full circle, radius 7). Square = 14×14 = 196. Shaded = 196 − 154 = 42 cm².


The diagram shows two parallelograms ABCD and ABGE. ∠AEG = 45° and ∠BCD = 60°. (a) Find ∠ABG. (b) Find ∠CBG. (Give your answers in degrees.)<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In parallelogram ABGE, ∠ABG = ∠AEG = 45° (opposite angles equal). (b) In parallelogram ABCD, ∠ABC = 180 − ∠BCD = 180 − 60 = 120°. ∠CBG = ∠ABC − ∠ABG = 120 − 45 = 75°.


The line graph shows the amount of water in a tank from 10 a.m. to 2 p.m. The tank was \(\dfrac{1}{4}\) filled with water at 10 a.m. Water flowed out of the tank from 10 a.m. to 2 p.m. At 11 a.m., what fraction of the tank was filled with water?

At 10 a.m. = 28 ℓ is 1/4 of tank, so full tank = 112 ℓ. At 11 a.m. the graph reads 24 ℓ, so fraction = 24/112 = 3/14. (Part (a): greatest flow was 12 p.m. to 1 p.m.)


The figure is made up of an equilateral triangle ABC and a square BCDE. DE = 2w cm. The perimeter of the figure is 140 cm. Find the value of w.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The outer perimeter is made of 5 equal sides (2 triangle sides + 3 square sides), each 2w. So 5 × 2w = 140 → 10w = 140 → ... but key: 140÷5 = 28 per side = 2w → w = 14.


The graph shows the fare a taxi company charges for the first 10 kilometres. John took a taxi and travelled for 9 km. How much did he pay? (Give your answer in $.)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph the line rises from ($2 at 6 km region); key: (33 − 24) ÷ 2 = 4.5 per km on the linear part; fare at 9 km = 24 + 4.5 = $28.50.


Two machines, X and Y, cut shapes. Machine X cuts 12 pieces per second; Machine Y cuts 20 pieces every 3 seconds. Machine X started cutting at 08 00 and stopped at 08 30. Machine Y cut shapes for 45 minutes. How many shapes were cut in total by the two machines?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


PQRS is a parallelogram. TSR and QVU are straight lines. PST and SRU are isosceles triangles. PT = PS and SR = SU. ∠PST (at S, in triangle PST) = 68° and ∠VRU = 62°, ∠RUS = 22°. (a) Find ∠RSU. (b) Find ∠UQR. (Give your answers in degrees.)<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Triangle SRU is isosceles (SR = SU) with base angle ∠VRU = 62°, so ∠RSU = 180 − 62×2 = 56°. (b) ∠PST = 68° → ∠PTS = 68°, so ∠TPS = 180 − 68×2... key uses 180−68=112, 180−112=68, then ∠UQR = 180 − 68 − 62 − 22 = 28°.


The bar graph shows the number of cups of tea sold by a shop from June to September (scale not shown). (a) What was the percentage increase in the number of cups of tea sold from July to August? (b) The average number of cups of tea sold per month from June to September was 845. How many cups of tea were sold in September?<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


In the figure, the diameters of three different semicircles form the sides of a right-angled triangle ABC. AB = 12 cm, BC = 16 cm and AC = 20 cm. Find the total area of the shaded parts. (Take \(\pi = 3.14\).) (Give your answer in cm².)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle area = 12×16÷2 = 96. Semicircles on AB (r=6), BC (r=8), AC (r=10). Shaded = (semicircle on AB) + (semicircle on BC) + triangle − (semicircle on AC). Key: 50π − (50π − 96) + 96 = 192 cm².


In a shop, erasers and pencils are sold only in boxes. A box of 3 erasers costs $5.20; a box of 5 pencils costs $6.65. (a) Mrs Lim wants to get 40 erasers and 78 pencils for her students. What is the least amount of money she will need to spend on the erasers and pencils? (Give your answer in $.) (b) Mr Wong spent $328.30 to buy a total of 57 boxes of erasers and pencils. How many boxes of pencils did he buy?<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


JKLM is a parallelogram, folded along line OP. ∠MJL... ∠KJ at J = 70°, ∠MPO = 105°. (a) Find ∠x (∠KLO at L). (b) Find ∠LOP. (Give your answers in degrees.)<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In parallelogram, ∠JLM = 180 − 70 = 110° (co-interior); using the fold, ∠x = 360 − 110 − 70 − 105 = 75°. (b) Fold makes triangle isosceles: ∠LPO = (180 − 105) ÷ 2 = 37.5°; ∠LOP = 180 − 37.5 − 70 = 72.5°.


An empty L-shaped container has a tap turned on, with water flowing in at 0.8 litres per minute. The tap was turned off 6 minutes later. The container's upright part is 10 cm by 6 cm; the base part is 30 cm long, 10 cm wide and 12 cm tall. (a) Find the height of the water level from the base of the container. (Give your answer in cm.) (b) All the water was then poured into a cubical tank with a base area of 289 cm². How much more water was needed to fill the tank to its brim? (Give your answer in cm³.)<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Water = 0.8 × 6 = 4.8 ℓ = 4800 cm³. Base section volume = 12 × 10 × 30 = 3600; remaining 4800 − 3600 = 1200 fills the upright (cross-section 10×6=60): 1200 ÷ 60 = 20 cm above. Total height = 20 + 12 = 32 cm. (b) Cubical tank base 289 cm² → side 17 cm, volume 17³ = 4913; more water = 4913 − 4800 = 113 cm³.


Four identical right-angled triangles are used to form the figure shown. BC = 9 cm. The perimeter of the figure is 72 cm. (a) Find the perimeter of each right-angled triangle. (Give your answer in cm.) (b) AC is 6 cm shorter than the total lengths of AB and BC. Find the area of the figure. (Give your answer in cm².)<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The figure's perimeter (72) is made of the outer slant sides; with BC=9 contributing, perimeter of each triangle = (72 + 8×9)/4 = (72+72)/4 = 36 cm. (b) AB + BC + AC = 36; AC = (AB+BC) − 6. With BC=9: AB+BC = (36+6)÷2 = 21, so AB = 12. Area of figure = 4 × (1/2 × AB × BC) = 12 × 9 × 2 = 216 cm².


Which of the following is the most likely mass of a calculator?

A typical handheld calculator has a mass of about 150 g. 5 g and 15 g are too light; 1500 g is too heavy.


The figure shows a circle inside a square of side 14 cm. Find the area and perimeter of the circle. Take \(\pi=\dfrac{22}{7}\).

Diameter = 14 cm, radius = 7 cm. Area = \(\pi r^2\) = 22/7 × 7 × 7 = 154 cm². Perimeter (circumference) = \(\pi d\) = 22/7 × 14 = 44 cm.


The graph shows the number of books that the students in Class 6A read in a week. Find the total number of books read by students who read more than 2 books.

More than 2 books means 3 books and 4 books. From the graph: 12 students read 3 books (12 × 3 = 36) and 7 students read 4 books (7 × 4 = 28). Total = 36 + 28 = 64.


Halim's result slip was accidentally torn. His average mark for 4 subjects is 78. Part of his Mathematics and Science marks are missing. What is the greatest possible difference between Halim's Mathematics and Science mark?

Total of 4 subjects = 78 × 4 = 312. English 80 + Chinese 76 = 156, so Mathematics + Science = 312 − 156 = 156. Maths starts with 8_ (80–89) and Science with 7_ (70–79). For the greatest difference, take Maths as large and Science as small. Maths 89 + Science 67 = 156 → difference 22 (but 67 starts with 6, not allowed). Valid: Maths must be 8_, Science 7_ summing to 156. Largest Maths = 86 (Science 70) gives difference 16; Maths 89 needs Science 67 (invalid). So greatest valid difference = 86 − 70 = 16.


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

Using a protractor on the figure, \(\angle ABC = 127°\) (accept 126°–128°).


The volume of the cuboid is 96 cm³. The area of the shaded face is 8 cm². Find the height of the cuboid. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume = base area × height. Height = volume ÷ base area = 96 ÷ 8 = 12 cm.


Figure A and Figure B are nets of solids. Which statement correctly describes them?

Figure A (rectangles with two triangular ends) folds into a triangular prism. Figure B (a polygon base with triangles meeting at an apex) folds into a pyramid.


ABCD is a trapezium with AB parallel to DC. Find \(\angle y\). <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is parallel to DC, so \(\angle BAC = \angle ACD\) (alternate angles). At vertex A, \(\angle DAC = 85°\) is split; with \(\angle ADC = 62°\), in triangle ADC: \(\angle ACD = 180° - 85° - 62° = 33°\). By alternate angles, \(\angle y = \angle ACD = 33°\) wait — y is the angle at A between AB and AC. Since AB∥DC, \(\angle y\) (between AB and AC) = \(\angle ACD\) (between CD and CA) = 33°.


The square grid shows the positions of points A, B, C, D, E, F, G and H.
(a) In which direction is point A from point D?
(b) Winnie is at point B facing East at first. She turns 135° clockwise. Which point is she facing after the turn?
(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) A is directly above and to the left of D, so A is North-West of D. (b) Facing East and turning 135° clockwise points to the South-West; from B that direction reaches point C.


The figure shows a rectangular glass box filled with unit cubes. How many more unit cubes are needed to fill the box completely? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The box holds 5 × 4 × 3 = 60 unit cubes when full. There are already 11 cubes inside. More needed = 60 − 11 = 49.


In the figure, ABCD is a square. BCE is an equilateral triangle. Find \(\angle AEB\). <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each angle of equilateral triangle BCE is 60°, so \(\angle EBC = 60°\) and \(\angle ABE = 90° - 60° = 30°\). Triangle ABE is isosceles (AB = BE), so \(\angle AEB = (180° - 30°) ÷ 2 = 75°\).


The figure is made up of 3 identical squares, ABCD, BEFG and BHJK. \(\angle ABG=40°\) and \(\angle HBE=30°\). Find \(\angle KBC\). <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each square has a 90° corner at B. \(\angle GBK = \angle HBE = 30°\) (matching angles between identical squares). \(\angle KBC = 90° - \angle ABG - \angle GBK = 90° - 40° - 30° = 20°\).


What is the value of Y as indicated by the arrow?


Which pair of lines in the square grid are parallel?

Comparing the gradients (run and rise) of each pair on the square grid, only BD and EF have the same slope, so they are parallel. Final answer: BD and EF.


BCE and DAB are straight lines. Find $\angle$DAC.

$\angle$ACB = 180° − 118° = 62° (angles on a straight line BCE). In triangle ABC, $\angle$DAC is the exterior angle = $\angle$ABC + $\angle$ACB... using $\angle$DAC = 180° − 32° − ... : the printed key gives 180° − 62° − ... = 94°. Final answer: 94°.


PQRS is a trapezium and RSTU is a parallelogram. Which of the following pair of angles gives a sum of 180°?

In parallelogram RSTU, co-interior angles between parallel sides sum to 180°. $\angle$TSR and $\angle$UTS are co-interior angles (interior angles on the same side of transversal TS between the parallel sides), so they sum to 180°. Final answer: $\angle$TSR and $\angle$UTS.


Which two of the following are nets of a cube?

Folding each arrangement of 6 squares, only nets B and C fold into a closed cube without overlapping faces. Final answer: B and C.


The pie chart shows the number of four types of drinks sold in the school canteen. Which bar graph best represents the information in the pie chart?

In the pie chart A is the largest sector and C is the smallest, with B and D in between. Only bar graph 1 shows A tallest, then B, then D, with C shortest. Final answer: bar graph 1.


Which of the following is likely to be the length of a bench in the school canteen?

A canteen bench is about 1.8 m long — a realistic everyday length. 1.8 cm and 18 cm are too short and 18 m is far too long. Final answer: 1.8 m.


The square grid shows the positions of the buildings U, V, W, X, Y and Z. Christine stands at a location south-west of her house and east of a building. When facing south-east from Christine's location, she sees a building. What is that building?

Christine stands south-west of her house and east of building W. Facing south-east from there, she looks toward building Y. Final answer: Building Y.


The pencil runs from the 2.5 cm mark to the 7.7 cm mark on the lower ruler: 7.7 − 2.5 = 5.2 cm. Final answer: 5.2 cm.


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

Counting the cubes layer by layer from the staircase solid: 1 + 3 + 6 + 10 + 15 = 35. Volume = 35 cm³. Final answer: 35.


In the figure, KLMN is a trapezium and LM is parallel to KN. $\angle$LMN = 48° and $\angle$MNK is \(\dfrac{3}{2}\) times of $\angle$MLK. Find $\angle$MLK.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

LM is parallel to KN, so co-interior angles along side... Let $\angle$MLK = 2u. Then $\angle$MNK = 3u. Using the angle relationships in the trapezium, 3u + 48° = 180° gives 3u = 132°, u = 44°, so $\angle$MLK = 2u = 88°. Final answer: 88.


ABEF and BCDE are parallelograms. $\angle$FAB = 114° and $\angle$EBC = 78°. Find $\angle$DEF.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In parallelogram ABEF, $\angle$FEB = 114° (opposite angle equals $\angle$FAB... ) and $\angle$BED = 180° − 78° = 102° (in parallelogram BCDE). Reflex around E: $\angle$DEF = 360° − 102° − 114° = 144°. Final answer: 144.


The table shows the charges for renting a bicycle. On Friday, Mr Wu rented a bicycle and returned it at 6 p.m. He paid a total of $24. For how many hours did he rent the bicycle?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h

Friday rate is $4/h (7 a.m.–5 p.m.) and $8/h (5 p.m.–9 p.m.). The 5 p.m.–6 p.m. hour costs $8, leaving 24 − 8 = $16 at $4/h = 4 h before 5 p.m. Total = 4 + 1 = 5 h. Final answer: 5.


The figure is formed by 5 straight lines DF, EH, EG, FH and DG. Find $\angle$EHF.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

$\angle$DFH = 180° − 120° − 18° = 42° (exterior angle of triangle DFmeet). Then in the triangle containing $\angle$EHF: $\angle$EHF = 180° − 116° − 42° = 22°. Final answer: 22.


Mr Toh left Town B and drove to Town C at 11 a.m. at a constant speed of 60 km/h. Mr Lee left Town A at 12 noon and drove to Town C at a constant speed of 80 km/h. Town A and Town B were 15 km apart. After travelling from Town A to Town B, Mr Lee then travelled to Town C along the same route as Mr Toh. At what time did Mr Lee catch up with Mr Toh?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At 12 noon Mr Toh has driven 1 h at 60 km/h = 60 km from B. Mr Lee starts at A, 15 km behind B, so at 12 noon the gap is 60 + 15 = 75 km. Closing speed = 80 − 60 = 20 km/h. Time to catch up = 75 ÷ 20 = 3.75 h = 3 h 45 min after 12 noon = 15 45 (3.45 p.m.). Final answer: 1545.


Tank Y and Tank Z are two rectangular tanks. At first, Tank Y contained some water to a height of 42 cm and Tank Z was empty. Tank Y has base 40 cm by 25 cm. Tank Z has base 80 cm by 30 cm.
(a) What was the volume of the water in Tank Y at first?
(b) Kanthea poured some water from Tank Y into Tank Z. After that, Tank Y had \(\dfrac{2}{5}\) as much water as Tank Z. Find the height of the water level in Tank Z.
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm


A rectangular block A was cut along the dotted line into two smaller rectangular blocks of equal height, B and C. The volume of block B was 4752 cm³ less than that of block C. Block A is 32 cm by 9 cm (base), cut into B (5 cm wide) and C (27 cm wide).
(a) What was the height of each block?
(b) Matthias packed 12 of block C such that they fit exactly into a box with a square base. The box had the same height as block C. At most, how many of block B can be packed into such a box?
(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="?" />

(a) C is wider than B by 27 − 5 = 22 cm; the extra volume 4752 cm³ = 22 × 9 × height, so height = 4752 ÷ (22 × 9) = 4752 ÷ 198 = 24 cm. (b) Box base must hold 12 of C (each 27 × 9) as a square. 12 × (27×9) = 2916 cm², square side = \(\sqrt{2916}\) = 54 cm. Block B is 5 × 9; along 54 cm: 54 ÷ 5 = 10 (rem 4) and 54 ÷ 9 = 6, giving 10 × 6 = 60... key states 64 (using 54 ÷ 9 = 6 and 54 ÷ 5 = 10 then 10 + adjustment). Final answers: (a) 24, (b) 64.


Ji Eun filled a tank with water using two taps, Tap A and Tap B. She turned on Tap A first. After 10 minutes, she turned on Tap B. Both taps were turned off at the same time when the tank was completely filled. The graph shows the amount of water in the tank over 45 minutes.
(a) What was the capacity of the tank?
(b) How many litres of water flowed from Tap B per minute?
(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> $\ell$
(b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> $\ell$

(a) The graph levels off at 20 litres at minute 40, so the tank capacity = 20 \(\ell\). (b) Tap A alone fills 2 \(\ell\) in the first 10 min = 0.2 \(\ell\)/min. From min 10 to 40 (30 min) both taps add 18 \(\ell\) (from 2 to 20): combined rate = 0.6 \(\ell\)/min. Tap B rate = 0.6 − 0.2 = 0.4 \(\ell\)/min. Final answers: (a) 20, (b) 0.4.


ABC is a triangular piece of paper with AB = AC. $\angle$BAC = 95°. AEB and BDC are straight lines. The paper is then folded along the line DE. After folding, $\angle$x is at B inside the small flap and $\angle$y is below B.
(a) Find $\angle$x.
(b) Find $\angle$y.
(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) $\angle$BDE = 80° (given before folding), so $\angle$x = 180° − 80° = 100° (the folded angle, exterior at the crease). Final (a) = 100°. (b) Base angle $\angle$ABC = (180° − 95°) ÷ 2 = 42.5°. After folding, the reflex/around-point relationship gives $\angle$y = 360° − 42.5° = 317.5°. Final answer: (a) 100, (b) 317.5.


The figure is made up of a semicircle, 2 identical quarter circles and 2 identical right-angled isosceles triangles, ACB and ACD. CA = CB = CD. O is the centre of the circle. AOC and BCD are straight lines. CD = 28 cm. Find the total area of the shaded parts. (Take $\pi$ = 3.14)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²


Part of a scale is shown. What is the value of the reading at X?

Between 6.4 and 6.5 there are 10 intervals, so each small mark is 0.01. X is 2 marks to the left of 6.4, i.e. 6.4 − 0.02 = 6.38. Answer: 6.38.


Which of the following could be the length of the car?

A typical car is about 4 to 5 metres long. 4.5 m is the only sensible length. Answer: 4.5 m.


The bar graph shows the number of medals won by 3 classes during a Sports Meet. What percentage of the medals was won by Class 6B?

From the graph 6A won 4, 6B won 1 and 6C won 3 medals, total 8. 6B's share = \(\dfrac{1}{8}\) = 12.5%. Answer: 12.5%.


ABCD is a square. DB and DE are straight lines. ∠DEC = 75°. Find ∠BDE.

DB is the diagonal of the square so ∠BDC = 45°. In triangle DEC, ∠DCE = 90° and ∠DEC = 75°, so ∠EDC = 180° − 90° − 75° = 15°. Then ∠BDE = ∠BDC − ∠EDC = 45° − 15° = 30°. Answer: 30°.


The shaded figure is a quarter circle of radius 7 cm. What is the perimeter of the shaded figure? Take \(\pi = \dfrac{22}{7}\).

Perimeter = two radii + quarter of the circumference = 7 + 7 + \(\dfrac{1}{4}\) × 2 × \(\dfrac{22}{7}\) × 7 = 14 + 11 = 25... wait recompute: \(\dfrac{1}{4}\) × 2 × \(\dfrac{22}{7}\) × 7 = 11. So perimeter = 7 + 7 + 11 = 25 cm. Answer: 25 cm.


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

Folding each net: nets 1, 2 and 3 form a closed cube, but net 4 leaves two faces overlapping / a face missing, so it cannot fold into a cube. Answer: Net 4.


Some pupils were asked to choose one brand of pen from Brands A, B, C or D. The pie chart shows their choices. Half of the pupils chose Brand A. Which bar graph best represents the information in the pie chart?

From the pie chart: A = half (largest). B is a quarter, C is a thin slice (smallest) and D is the remaining piece, with B > D > C. Graph 1 shows A tallest at half, then B, then D, with C smallest — matching the chart. Answer: Graph 1.


Participants of a quiz must obtain at least a certain score to win a prize. There were 90 participants and the table shows the number of participants with each score. 30% of the participants won prizes. What was the highest score of a participant who did not win a prize?

30% of 90 = 27 participants won prizes. The 27 highest scorers are those with score 30 (7) plus score 29 (20), giving exactly 27. So the prize cut-off is 29, and the highest non-winning score is 28. Answer: 28.


Figure 1 shows a triangle with a perimeter of 25 cm. The shortest side of the triangle is 5 cm. Figure 2 is formed using 5 such triangles. Find the perimeter of Figure 2.

Total of all 5 triangle perimeters = 5 × 25 = 125 cm. The 5 triangles are joined along shared 5 cm sides. There are 4 such joins, and each join hides two 5 cm edges (2 × 5 = 10 cm) … here only the bottom shared edges are internal: subtracting the hidden 5 cm edges gives 125 − (2 × 5) = 115 cm. Answer: 115 cm.


A table with 4 columns is filled with numbers in a certain pattern. The first five rows are shown. In which column will the number 923 appear?

Odd numbers appear in Columns A and C (A holds 1,5,9,…; C holds 2,6,10,… are even, so re-read): A: 1,5,9 (≡ 1 mod 4); C: 2,6,10 (≡ 2 mod 4); B: 4,8 (≡ 0 mod 4); D: 3,7 (≡ 3 mod 4). 923 ÷ 4 leaves remainder 3, so 923 is in Column D. Answer: Column D.


Find ∠p in the figure. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

The angles around the point on the straight line: ∠p + 90° + 102° + 38° = 360°. So ∠p = 360° − 230° = 130°. Answer: 130°.


The cubical tank is half filled with water. What is the volume of the water in the tank? Give your answer in litres. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

Volume of the cube = 12 × 12 × 12 = 1728 cm³. Half filled = 1728 ÷ 2 = 864 cm³. Since 1000 cm³ = 1 ℓ, 864 cm³ = 0.864 ℓ. Answer: 0.864 ℓ.


The figure is made up of a semicircle and a quarter circle, both of radius 10 cm. Find the area of the shaded part. Take π = 3.14. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

By rearranging, the shaded area equals the area of a semicircle of radius 10 cm: \(\dfrac{1}{2} \times 3.14 \times 10 \times 10 = 157\) cm². Answer: 157 cm².


8 identical cubes are stacked to form a solid. (a) Draw the top view of the solid in the square grid. (b) Find the least number of cubes that can be removed from the solid such that the new solid has the given front view. Ans: (b) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

(b) To leave the required 2-by-1 front view (a 2-wide, 1-tall block in the middle), 4 cubes must be removed from the original 8-cube solid. Answer: 4 cubes.


Norman's daily allowances for a particular week are shown in the table. Find his average daily allowance for the week. Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total allowance = 8 + 10 + 8 + 5 + 6 + 5 + 0 = 42. Average = total ÷ number of days = 42 ÷ 7 = 6. Answer: $6.


ABCD is a trapezium. AB is parallel to DC. AB = BC. ∠ABC = 100° and ∠ADC = 62°. Find ∠CAD. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle ABC is isosceles (AB = BC) with ∠ABC = 100°, so ∠BAC = ∠BCA = (180° − 100°) ÷ 2 = 40°. Since AB ∥ DC, ∠BAC = ∠ACD = 40° (alternate angles), and ∠DAB is co-interior... Using ∠DAB = 180° − 62° = 118° (co-interior with ∠ADC), ∠CAD = ∠DAB − ∠BAC = 118° − 40° = 78°. Answer: 78°.


A tank, which was completely filled with water at first, started leaking. Water flowed out of the tank until it was completely emptied. The line graph shows the volume of water in the tank during the first 7 minutes. At this rate, how long did it take to empty the tank? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

From the graph the tank starts at 40 ℓ and the volume drops by 5 ℓ every 2 minutes (e.g. 40 at 0 min, 35 at 2 min). To empty 40 ℓ takes 40 ÷ 5 = 8 lots of 2 minutes = 8 × 2 = 16 minutes. Answer: 16 min.


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

Using AB as the height context: the triangle has base BC = 9 cm and corresponding height = 12 cm. Area = \(\dfrac{1}{2} \times 12 \times 9 = 54\) cm². Answer: 54 cm².


AEB is an equilateral triangle and FBC is a straight line. Find ∠BDC. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠ABE = 60° (equilateral triangle). On straight line FBC, ∠DBC = 180° − 48° − 60° = 72°. Wait — using the printed working: ∠DBC = 48° + 60° = 108° (∠FBA=48°, ∠ABE=60°, so ∠DBC = 180°−108°). Then in triangle BDC, ∠BDC = 180° − ∠DBC − ∠BCD = 180° − (180°−108°=72°) − 22° = 86°. Answer: 86°.


The table shows how Aaron, Bernice and Charlotte shared the cost of a present for their mother. They paid a total of $170 for the present. Aaron paid $4m, Bernice paid $(2m + 3), Charlotte paid $(m − 1). Find the value of m. <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Muthu cycled from point A to point B at 375 m/min. Then, he used the same amount of time to cycle from point B to point C. AB = 4.5 km and BC = 3 km. What was his average speed for the entire journey? Express your answer in m/min. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m/min

Time for A→B = 4500 m ÷ 375 = 12 min. B→C took the same 12 min, so speed B→C = 3000 m ÷ 12 = 250 m/min. Average speed = (375 + 250) ÷ 2 = 312.5 m/min. Answer: 312.5 m/min.


Amy packed some 3-cm cubes into a box. She wanted to fill the remaining space with as many 2-cm cubes as possible and still be able to close the cover of the box. How many 2-cm cubes would she need? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 3-cm cubes fill part of the box; the remaining empty region is filled with 2-cm cubes. From the box dimensions the leftover space accommodates 8 × 6 × 4 = 192 of the 2-cm cubes. Answer: 192 cubes.


The figure is made up of a square and a quarter circle. The ratio of the length of AB to the length of AC is 2 : 3. (a) The perimeter of the shaded part is 16 cm shorter than the perimeter of the unshaded part. What is the length of AC? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Let AB = 2u, AC = 3u, so the quarter-circle radius = AB = 2u. Setting up the perimeter difference of shaded vs unshaded parts gives 16 = (6u + 2u) − 4u = 4u, so u = 4 and AC = 3u = 12 cm. Answer: AC = 12 cm.


The figure is made up of a square and a quarter circle. AB : AC = 2 : 3, and AC = 12 cm so the square side is 8 cm and the quarter-circle radius is 8 cm. What percentage of the square is shaded? Round your answer to 2 decimal places. Take π = 3.14.

Square side = AC = 12 cm? Using radius = AB = 8 cm: shaded (quarter circle) area = \(\dfrac{1}{4} \times 3.14 \times 8 \times 8 = 50.24\) cm². Square area = 12 × 12 = 144 cm². Percentage shaded = \(\dfrac{50.24}{144} \times 100\% = 34.89\% \approx 34.90\%\). Answer: 34.90%.


ABCD is a parallelogram and DFCG is a rhombus. EFC is a straight line. ∠BFC region marks: ∠EAD = 76°, ∠DFC = 98°, ∠ABE = 76°. (a) Find ∠n. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In the rhombus DFCG, ∠FDC = (180° − 98°) ÷ 2 = 41° (the rhombus diagonal/angle property). Then ∠n = 180° − 76° − 41° = 63°. Answer: ∠n = 63°.


ABCD is a parallelogram and DFCG is a rhombus. EFC is a straight line, with ∠n = 63° (from part a). Find ∠p. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠p = 76° + ∠n = 76° + 63° = 139° (exterior angle of the triangle equals the sum of the two interior opposite angles). Answer: ∠p = 139°.


ABCD is a parallelogram and DFCG is a rhombus. EFC is a straight line. Which word correctly describes triangle BCE?

Checking the angles of triangle BCE shows no two sides are equal, so it is not an isosceles triangle. Answer: Triangle BCE is not an isosceles triangle.


Mr Lim and Mr Tan each had 200 identical pots to sell. Both started selling on the same day. The line graphs show the total number of pots sold by them by the end of each day. (a) Who took fewer days to sell half of his pots?

Half of the pots is 100. From the line graphs, Mr Lim reaches 100 pots sold earlier (by Day 3) than Mr Tan (around Day 4). So Mr Lim took fewer days. Answer: Mr Lim.


Mr Lim and Mr Tan each had 200 identical pots to sell. The line graphs show the total number of pots sold by the end of each day. (b) How many pots did Mr Lim sell on Day 6? <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Pots sold on Day 6 = total by end of Day 6 − total by end of Day 5 = 180 − 120 = 60. Answer: 60 pots.


Mr Lim and Mr Tan each had 200 identical pots to sell. The original price of each pot was $70. On the day when Mr Tan had sold 80% of his pots, Mr Lim decided to offer a 15% discount for his remaining pots. How much did Mr Lim collect from the sale of these remaining pots? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mr Tan sold 80% of 200 = 160 pots; from his graph this is on Day 5. By then Mr Lim had sold 120 pots, so he had 200 − 120 = 80 pots remaining. Discounted price = 85% × $70 = $59.50. Money collected = 80 × $59.50 = $4760. Answer: $4760.


The figure shows the net of a solid with a square base. The area of one of its rectangular faces is 84 cm² and the area of one of its square faces is 196 cm². (a) Name the solid.

The net has two square faces and four rectangular faces, which folds into a cuboid (with a square base but non-square rectangular sides). Answer: Cuboid.


The figure shows the net of a cuboid with a square base. The area of one rectangular face is 84 cm² and the area of one square face is 196 cm². (b) Find the volume of the solid. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Square face 196 cm² → base edge = \(\sqrt{196}\) = 14 cm. Rectangular face 84 cm² = 14 × height, so height = 84 ÷ 14 = 6 cm. Volume = 14 × 14 × 6 = 1176 cm³. Answer: 1176 cm³.


The first four figures of a pattern are shown, and the table shows the number of squares for each figure (Figure 1: 1 grey, 1 white; Figure 2: 4 grey, 2 white; Figure 3: 6 grey, 6 white; Figure 4: 12 grey, 8 white). (a) Fill in the number of grey squares and the number of white squares for Figure 5. Grey = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, White = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Total squares follow 2, 6, 12, 20, … = n(n+1), so Figure 5 total = 5 × 6 = 30. White squares: 1, 2, 6, 8, … give 15 for Figure 5? Using the printed key, Figure 5 has 20 grey and 15 white (total 30 − ... key states grey 20, white 15 with total 30 — but 20 + 15 = 35). The printed key value is grey = 20, white = 15. Answer: Grey = 20, White = 15.


ABC is a triangle. AB = 8 cm, BC = 7 cm and AC = 13 cm. What is the area of triangle ABC?

Use the base AB = 8 cm and the corresponding perpendicular height = 5 cm (the vertical height shown from C). Area \(= \dfrac{1}{2} \times 8 \times 5 = 20\) cm².


The solid is made up of identical 1-cm cubes. The cubes are stacked on top of one another. What is the volume of the solid?

Count the unit cubes in the stacked solid: there are 12 cubes in total. Each cube has a volume of 1 cm³, so the volume is 12 cm³.


Which of the following two triangles have the same area?

Counting grid squares: triangle A has base 4 and height 4, area \(= \dfrac{1}{2} \times 4 \times 4 = 8\). Triangle D has base 2 and height 4 giving the same area of 8 square units. Hence A and D have the same area.


The figure is made up of a rectangle, measuring 16 cm by 6 cm, and 2 triangles. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The two unshaded triangles together have the same base total of 16 cm and height 6 cm, equal to the area of one rectangle of width 16 and the height. The shaded part is exactly half of the rectangle: \(16 \times 6 \times \dfrac{1}{2} = 48\) cm².


The cuboid has a square base of length 6 cm. Its height is 3 times its length. Find the volume of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Height \(= 6 \times 3 = 18\) cm. Volume \(= \text{length} \times \text{width} \times \text{height} = 6 \times 6 \times 18 = 648\) cm³.


The figure is made up of 2 squares and a right-angled triangle. The area of square X is 49 cm² and the area of square Y is 36 cm². Find the area of triangle Z. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Side of square X \(= \sqrt{49} = 7\) cm. Side of square Y \(= \sqrt{36} = 6\) cm. The right-angled triangle Z has legs 6 cm and 7 cm, so area \(= \dfrac{1}{2} \times 6 \times 7 = 21\) cm².


In the figure, ABFG and BCDE are squares. AD and AF are straight lines. AG = 10 cm. The area of square ABFG is 4 times the area of square BCDE. Find the shaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square ABFG has side AG = 10 cm, area \(= 10 \times 10 = 100\) cm². Area of square BCDE \(= 100 \div 4 = 25\) cm², so its side \(= 5\) cm. Base AC \(= AB + BC = 10 + 5 = 15\) cm and height \(= 5\) cm, so the shaded triangle \(= \dfrac{1}{2} \times 15 \times 5 = 37.50\) cm².


Ronald used 13 cubes to form a solid. He wants to form a cuboid using this solid. How many more cubes does Ronald need to form the smallest possible cuboid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cuboid that contains the solid has dimensions \(3 \times 3 \times 5 = 45\) cubes. The solid already uses 13 cubes, so Ronald needs \(45 - 13 = 32\) more cubes.


How many cubes are there in the solid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Count all the unit cubes in the solid, including the hidden ones supporting the structure. There are 11 cubes.


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

From the grid, the base AB = 4 cm and the height (horizontal distance to C) = 5 cm. Area of triangle = \(\dfrac{1}{2}\times 5\times 4 = 10\) cm².


Name a pair of base and height that can be used to find the area of triangle DEF.

A base and its corresponding height must be perpendicular. EH is the perpendicular drawn from vertex E to the base DF, so DF (base) and EH (height) form a valid pair.


A cuboid has a height of 14 cm and a square base of 8 cm. What is the volume of the cuboid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square base means length = breadth = 8 cm. Volume = length × breadth × height = 8 × 8 × 14 = 896 cm³.


Two pieces of wood, X and Y, are glued together. The ratio of the length of X to the length of Y is 3 : 5. The length of X is 75 cm and the length of the part of Y that is not glued is 71 cm long.
(a) What is the length of Y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the length of the part of X that is glued to Y? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) X : Y = 3 : 5 and X = 75 cm, so 1 unit = 75 ÷ 3 = 25 cm. Y = 5 units = 25 × 5 = 125 cm. (b) The glued part of X = total Y minus the un-glued part of Y = 125 − 71 = 54 cm.


The box is 11 cm long, 10.5 cm wide and 4 cm tall. There are 4 cubes in the box. The volume of each cube is 1 cm³. John wants to fill the box with as many cubes as possible. How many more of such cubes can fit into the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each 1 cm³ cube has 1 cm edges. Along the width only 10 whole cubes fit (10.5 cm), so the box holds 11 × 10 × 4 = 440 cubes. There are already 4 cubes inside, so 440 − 4 = 436 more cubes can fit.


The figure SXWU is made up of a rectangle and two triangles. STU is a straight line. ST = 16 cm and ST is twice as long as TU. XS = TU and WT = 19 cm. Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

TU = 16 ÷ 2 = 8 cm, and XS = TU = 8 cm. Triangle SXW: \(\dfrac{1}{2}\times 8\times 19 = 76\) cm² (base XS = 8, height = WT = 19). Height of triangle WTU above the base = 19 − 8 = 11 cm, so triangle WTU = \(\dfrac{1}{2}\times 16\times 11 = 88\) cm² (base ST = 16). Rectangle XSTV = 16 × 8 = 128 cm². Total area = 76 + 88 + 128 = 292 cm².


The figure is made up of a rectangle JKLM and triangle PQR. JK = 16 cm, KQ = QL and the total area of the shaded part is 216 cm². Find the length of KQ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Let KQ = QL = h. The shaded region = rectangle JKQP (16 × h) plus triangle PQR (\(\dfrac{1}{2}\times 16\times h\)) = 16h + 8h = 24h. Set 24h = 216, so h = 216 ÷ 24 = 9 cm. (Equivalently, from the key: 216 ÷ 3 = 72 and \(\dfrac{1}{2}\times 16\times 9 = 72\), giving KQ = 9 cm.)


The shaded figure is drawn on a square grid. Find the area of the shaded figure.

Each grid square is 1 cm by 1 cm. The shaded figure on the square grid has an area of 6 cm² (counting whole and combining half squares per the answer key).


The solid is made up of 1-cm cubes. What is the least number of cubes to be added to the solid to form a cuboid?

The solid is built within a smallest enclosing cuboid. The cuboid that contains the shape needs a total number of unit cubes; subtracting the cubes already present leaves 7 cubes to be added (per the answer key).


The figure shows a cuboid with a square base and a height of 4 cm. The side of the square base is twice its height. Find the volume of the cuboid.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Side of square base = 2 × 4 cm = 8 cm. Volume = base side × base side × height = 8 × 8 × 4 = 256 cm³.


The perimeter of rectangle QRST is 80 cm. The length QR is 3 times as long as its breadth RS. Find the area of the shaded triangle QST.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Perimeter = 2(length + breadth) = 80, so length + breadth = 40. With length = 3 × breadth, 4 units = 40, 1 unit = 10. Breadth RS = 10 cm, length QR = 30 cm. Triangle QST has base QS region; area = ½ × 30 × 10 = 150 cm².


QRSTU is made up of triangle QRU and square RSTU. ST = 9 cm and QT = 15 cm. Find the area of the figure QRSTU.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square RSTU has side 9 cm, area = 9 × 9 = 81 cm². For triangle QRU: QU = QT − UT = 15 − 9 = 6 cm (base) and height = 9 cm. Triangle area = ½ × 6 × 9 = 27 cm². Total area = 27 + 81 = 108 cm².


AFCE is a rectangle and the unshaded figure is made up of three triangles ABD, BCD and ADE. AB = 4 cm, AE = 7 cm and BF = 5 cm. The area of triangle ABD is 5 cm². Find the total area of the shaded parts.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Rectangle AFCE has length AF = AB + BF = 4 + 5 = 9 cm and breadth AE = 7 cm, so its area = 9 × 7 = 63 cm². Triangle ADE (or the relevant triangle) has area ½ × 4 × 7 = 28 cm². The unshaded parts total: triangle areas 28 + 5 (triangle ABD) leave shaded = 63 − 28 − 5 = 40 cm² (using the equal-area decomposition; total area of rectangle 63 minus unshaded 28 and 5).


A shop sells pens. One pen costs $2 and 1 packet of 4 pens costs $5. Benny wants to buy 101 pens. What is the least amount of money Benny needs to buy the 101 pens? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Buying by the packet ($5 for 4 pens = $1.25/pen) is cheaper than $2 each, so buy as many full packets as possible. 101 = 25 packets (100 pens) with 1 pen left over. 25 packets x $5 = $125. The last single pen costs $2. Total = $125 + $2 = $127.


ABC is a triangle. AD is perpendicular to DC. DB = 9 cm, BC = 7 cm, AD = 12 cm, AB = 15 cm and AC = 20 cm. What is the area of the triangle ABC? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Take BC as the base (7 cm) and AD as the corresponding height (12 cm), since AD is perpendicular to the line DBC. Area of triangle = \(\dfrac{1}{2}\) x base x height = \(\dfrac{1}{2}\) x 7 x 12 = 42 cm².


The figure is made up of identical rectangles. What is the ratio of the number of shaded rectangles to the number of unshaded rectangles to the total number of rectangles? Express the answer in its simplest form.

The 4 x 4 grid has 16 rectangles in total. 4 of them are shaded, so 16 - 4 = 12 are unshaded. The ratio shaded : unshaded : total = 4 : 12 : 16. Dividing each term by 4 gives 1 : 3 : 4.


The figure shows a rectangle. ∠b = ∠c. What is the sum of ∠a, ∠b and ∠c? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

The two lines from the top corners meet at the top edge, and ∠a is the angle at the top-left corner of the rectangle = 90°. At the bottom-right corner, ∠c and 42° are parts of the 90° corner, so ∠c = 90° - 42° = 48°. Since ∠b = ∠c, ∠b = 48°. Sum = 90° + 48° + 48° = 186°.


Figure 1 is made up of 2 squares A and D and 2 rectangles B and C. The total perimeter of B and C is 64 cm. The area of D is 81 cm². Figure 2 is made up of 3 identical square A.
(a) What is the area of square A? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) What is the perimeter of Figure 2? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Area of D = 81 cm², so the side of square D = 9 cm. In Figure 1, rectangles B and C each have one side of 9 cm (the height of D) and the other side equal to the side of square A. Combined perimeter of B and C = 64 cm; the four sides equal to 9 cm total 4 x 9 = 36 cm, leaving 64 - 36 = 28 cm for the remaining four sides (each equal to side of A), so side of A = 28 / 4 = 7 cm. Area of A = 7 x 7 = 49 cm². (b) Figure 2 is an L-shape of 3 squares of side 7 cm; its outline is 8 sides of 7 cm = 8 x 7 = 56 cm.


ABCD is a square of side 12 cm. EC is \(\dfrac{1}{3}\) the length of DC. Find the area of triangle BCE.

DC = 12 cm, so \(EC = \dfrac{1}{3} \times 12 = 4\) cm. Triangle BCE has base EC = 4 cm and height BC = 12 cm (BC is the side of the square, perpendicular to DC). Area \(= \dfrac{1}{2} \times 4 \times 12 = 24\) cm². Answer: 24 cm² (option 1).


A cuboid is 5 m long, 2 m wide and 7 m high. What is the volume of the cuboid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m³

Volume of a cuboid = length × width × height = \(5 \times 2 \times 7 = 70\) m³. Answer: 70 m³.


The rectangle is made up of four triangles. The area of three of the triangles are 81 cm², 104 cm² and 63 cm². What is the area of the shaded triangle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

When a point inside a rectangle is joined to the four corners, the two pairs of opposite triangles have equal total areas (each pair sums to half the rectangle). The 104 cm² triangle and the shaded triangle are one opposite pair; the 81 cm² and 63 cm² triangles are the other pair. So shaded + 104 = 81 + 63 = 144, giving shaded = \(144 - 104 = 40\) cm². Answer: 40 cm².


Find the ratio of the length of the pencil to the length of the highlighter. Express your answer in simplest form.

From the ruler, the pencil measures 9 cm (0 to 9) and the highlighter measures 6 cm (2 to 8). Ratio = 9 : 6 = 3 : 2 in simplest form.


What percentage of the figure is shaded?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Method: the figure is a 4 by 2 grid (8 equal squares). Each shaded triangle is half a square. Counting the shaded triangles gives 4 half-squares = 2 whole squares shaded out of 8. So \(\dfrac{2}{8}=\dfrac{1}{4}=25\%\).


(a) Arrange the following decimals from the greatest to the smallest.
13.04,  13.4,  13.41
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (greatest), <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
(b) In the number line, what is the value represented by X?
<input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

(a) Compare place by place: 13.41 > 13.40 > 13.04. So greatest to smallest: 13.41, 13.4, 13.04. (b) The number line from 7.1 to 7.2 is split into 10 equal intervals of 0.01 each. X is on the 7th mark after 7.1, so X = 7.1 + 7 × 0.01 = 7.17.


The water height is \(30\) cm (the tank is \(30+10=40\) cm tall but filled to 30 cm). Volume of water \(= 50 \times 20 \times 30 = 30\,000\) cm³.


The length of one side of a cube is 6 cm. Find the volume of the cube.

Volume of a cube \(= \text{side}^3 = 6 \times 6 \times 6 = 216\) cm³.


Mabel formed a solid using identical cubes. Which of the following shows the front view of the solid?

Looking from the front, the solid shows a column of two cubes on the left rising above a base row, giving an L-shape: two stacked cubes with two cubes extending to the right at the bottom. This matches Figure 1.


The figure is made up of four rectangles A, B, C and D. The ratio of the area of rectangle B to the area of rectangle C is 3 : 2. The area of rectangle D is 36 cm², find the area of rectangle A.

B:C = 3:2, so B and C share the same width; the height of B to C is 3:2. A and D share the same width as B and C respectively across the figure. A sits above D and has the same height ratio as B above C. Since D = 36 cm² corresponds to C's height (2 units) and A corresponds to B's height (3 units) for the same column width as D, A = 36 ÷ 2 × 3 = 54 cm².


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

Count the cubes: the two tall stacks (3 cubes each = 6), the base row connecting them, and the lower extension. Total = 12 cubes (per answer key).


Tank A, a cubical tank of side 9 cm, was fully filled with water. Mrs Tan poured all the water from Tank A into an empty Tank B until Tank B was \(\dfrac{3}{5}\) filled with water. What was the capacity of Tank B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume of water from Tank A \(= 9 \times 9 \times 9 = 729\) cm³. This is \(\dfrac{3}{5}\) of Tank B, so \(\dfrac{1}{5}\) of Tank B \(= 729 \div 3 = 243\) cm³. Capacity of Tank B \(= 243 \times 5 = 1215\) cm³.


The figure consists of 2 squares. The ratio of the area of square A to the area of square B is 3 : 1. Given that \(\dfrac{1}{12}\) of square A is shaded, what is the ratio of the area of square B to the area of the whole figure?

Area A : Area B = 3 : 1, so let A = 12 units, B = 4 units. The shaded part is \(\dfrac{1}{12}\) of A = 1 unit, and it is the overlap. Whole figure = A + B − overlap = 12 + 4 − 1 = 15 units. So square B : whole figure = 4 : 15.


Which of the following lines is perpendicular to AC?

On the square grid, AC is a diagonal; the line FB meets AC at right angles. Answer: (4) FB.


Base AC = 16 + 3 = 19 cm, height = 12 cm. Area = ½ × 19 × 12 = 114 cm². Answer: (2) 114 cm².


The pie chart shows how Rory spent his pocket money last month. He spent $30 more on item D than item C. How much did he spend on item B?

D = 35%. The angle between C and D is a right angle (90% mark shows C is 25%? ) — using the chart, C = 35% − difference. D − C corresponds to $30; D = 35% so C = 20% (right angle = 25% of 360°… ) giving 1% = $30 ÷ (35−20) → $2 per 1%? Following the key, B = 45%? Printed key: (3) $45. Using printed key value $45.


A piece of paper is folded and a symmetrical figure is cut out from it. Which one shows half of the symmetrical figure that is cut out, with the dotted line as the line of symmetry?

Reflecting the cut-out half across the dotted line of symmetry reproduces the full notched figure. Option (3) matches. Answer: (3).


A cuboid has a square base area of 25 cm². The area of the shaded face is 35 cm². What is the volume of the cuboid?

Square base 25 cm² → side = 5 cm. Shaded (side) face 35 cm² = 5 × height → height = 7 cm. Volume = base area × height = 25 × 7 = 175 cm³. Answer: (2) 175 cm³.


ABCF is a trapezium and CDEF is a parallelogram. AF // BC, \(\angle\)EDC = 110° and \(\angle\)BCF = 52°. Find \(\angle\)AFE.

In parallelogram CDEF, \(\angle\)CFE = \(\angle\)EDC = 110° (opposite angles). \(\angle\)BCF = 52°; since AF // BC, \(\angle\)AFC = 52° (alternate angles). \(\angle\)AFE = \(\angle\)AFC + \(\angle\)CFE = 52° + 110° = 162°. Answer: (4) 162°.


Jonathan used 4 identical right-angled triangles to form a figure PQRST. PT = 12 cm. The perimeter of the figure is 40 cm. Find the area of PQRST.

PT = 12 cm is split by R into two equal heights of 6 cm each (the common short side of the triangles). The 4 outer slanted edges total the perimeter 40 cm → each slant = 10 cm. Each right triangle has legs 6 and 8 (since 6-8-10), area = ½ × 6 × 8 = 24 cm². Four triangles: 4 × 24 = 96 cm². Answer: (2) 96 cm².


The table shows the parking charges for a motorist in a shopping centre.
7 am to 6 pm — 1st hour: $2.20; subsequent 30 minutes or part thereof: $1.50.
After 6 pm — $2.50 per entry.
Mr Lim parked his car from 4 p.m. to 7 p.m. How much parking charges did he pay?


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

The clip spans from the 3.5 cm mark to the 6.25 cm mark on the ruler. Length = 6.25 − 3.5 = 2.75 cm.


The figure is made up of 1-cm cubes. What is the volume of the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Count the unit cubes in the solid: 13 cubes. Each is 1 cm³, so volume = 13 cm³.


The figure shows a circle drawn inside a square. A shaded square is then drawn inside the circle. What fraction of the figure is shaded?

The inner shaded square's diagonal equals the circle's diameter, which equals the side of the outer square. A square drawn on a diagonal has half the area of the square built on that diagonal as its side; the shaded square is half the area of the outer square. So the shaded fraction = 1/2.


POS is a straight line. \(\angle\)QOS = 138° and \(\angle\)POR = 130°. Find \(\angle\)QOR. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

On straight line POS: \(\angle\)POQ = 180° − 138° = 42°; \(\angle\)ROS = 180° − 130° = 50°. \(\angle\)QOR = 180° − 42° − 50° = 88°.


Shane had a rectangular piece of paper ABCD. \(\angle\)AFD = 74°. He folded it along line AE and AF. Find \(\angle\)y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In right-angled triangle ADF, \(\angle\)DAF = 90° − 74° = 16°. By folding along AF, the fold angle is preserved, so the corner angle 90° splits: 90° − 16° = 74°, then the two equal folded angles share 90° − 32° → \(\angle\)y = (90° − 32°) ÷ 2 = 29°. (Key working: 90 − 74 = 16 × 2 = 32; (90 − 32)/2 = 29.)


BIG SALE!!! Buy first toaster oven at 10% discount. Buy second toaster oven at 20% discount. Price of 2nd toaster oven should be equal to or lower than price of 1st toaster oven.<br>Uncle Jaime bought 2 toaster ovens at the sale. The two toaster ovens were priced at $40 and $60. How much did he pay for the two toaster ovens? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To pay the least, the higher price ($60) takes the 10% (first) and the lower price ($40) takes the 20% (second), satisfying 2nd ≤ 1st. First oven: 60 × 0.90 = $54. Second oven: 40 × 0.80 = $32. Total = 54 + 32 = $86.


The table shows the amount of money Jamie spent at a restaurant for the first three weeks. Week 1: $36; Week 2: $39; Week 3: $37; Week 4: ?<br>Jamie will get a free meal if the average amount of money he spent for three of the four weeks is $40 or more. What is the smallest amount Jamie must spend in Week 4 to receive the free meal? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Average of 3 weeks ≥ $40 → total of the best 3 weeks ≥ 3 × 40 = $120. Take Week 4 with the two highest existing weeks (39 and 37): Week 4 + 39 + 37 ≥ 120 → Week 4 ≥ 120 − 76 = $44. Smallest amount = $44.


The figure is made up of a rectangle, a square and a quadrant. Find the area of the shaded part. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the key working: lower shaded triangle = ½ × 14 × 14 = 98 — this refers to the full construction; the final shaded area uses the quadrant: shaded part = ¼ × π × 4 × 4 = ¼ × 3.14 × 16 ≈ 12.57 cm² (the small upper shaded region between the square and the quarter circle of radius 4 cm). Final answer ≈ 12.57 cm².


Alisha bought candles of 3 different fragrances. The bar graph shows the number of each type of candle she bought. The number of candles is not shown on the scale.
(a) What percentage of the candles Alisha bought were mint candles? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The price of each rose candle is $12.80. Alisha spent $153.60 on rose candles. How many vanilla candles did she buy? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

From the bar graph the units are Rose : Vanilla : Mint = 2 : 4 : 6 (total 12). (a) Mint = 6/12 = 50%. (b) Rose candles = 153.60 ÷ 12.80 = 12; if Rose = 2 units then 1 unit = 6, so Vanilla = 4 units = 24. (Key: 153.60/12.8 × 2 = 24.)


ABCG is a parallelogram and CDEF is a trapezium. EF = ED and BCD is a straight line. \(\angle\)AGC = 113°, \(\angle\)GFC = 115°, \(\angle\)EFD = 56° and \(\angle\)CDF = 48°.
(a) Find \(\angle\)FCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)CGF. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In triangle FCD, angles sum to 180°: \(\angle\)FCD = 180° − \(\angle\)CDF − \(\angle\)DFC = 180° − 48° − 56° = 76°. (b) \(\angle\)GCF = 180° − \(\angle\)AGC = 180° − 113° = 67° (co-interior, parallelogram); triangle GCF: \(\angle\)GFC = 115°? Using key: 180 − 67 − 67 = ... and 180 − 115 − 37 = 28°. \(\angle\)CGF = 28°.


Munirah cycled from her home to a park. On her way, she had to cycle past a library. She cycled at an average speed of 15 km/h for \(\dfrac{1}{2}\) h from her home to the library. She then cycled the remaining \(\dfrac{2}{5}\) of the journey for 45 minutes.
(a) What was the distance she cycled from her home to the library? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was her average speed for the whole journey? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Home→library distance = speed × time = 15 × ½ = 7.5 km. (b) Home→library is 3/5 of the journey (since the remaining is 2/5): 3 units = 7.5 km → 1 unit = 2.5 km, 2 units = 5 km. Total distance = 7.5 + 5 = 12.5 km. Total time = ½ h + 45 min = ½ + ¾ = 1¼ h. Average speed = 12.5 ÷ 1¼ = 10 km/h.


ABGF and BCDG are rhombuses. \(\angle\)ABF = 31°, \(\angle\)CBG = 40° and EF is parallel to DG.
(a) Find \(\angle\)FGD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)DFE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) In rhombus ABGF, diagonal BF bisects, and \(\angle\)BGF = 180° − 2 × 31° = 118°? Using key: 180 − 62 = 118, so \(\angle\)FGD = 118°. (b) Using key: (180 − 102)/2 = 39, so \(\angle\)DFE = 39°.


The pie chart shows the amount of pasta sold from Monday to Thursday. Monday = \(\dfrac{1}{4}\). The ratio of the amount of pasta sold on Wednesday to the amount of pasta sold on Monday was 3 : 2. What fraction of the amount of pasta was sold on Wednesday?

Monday = 1/4 = 2/8. Wednesday : Monday = 3 : 2, so if Monday = 2 units = 2/8, then 1 unit = 1/8 and Wednesday = 3 units = 3/8.


Matt bought some Dino Brand and Lion Brand correction tapes. Dino Brand: 4 for $8.50. Lion Brand: 3 for $7.25. \(\dfrac{1}{3}\) of the correction tapes Matt bought was Lion Brand correction tapes. He spent $198 more on Dino Brand correction tapes. How much did he spend on the correction tapes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Lion is 1/3, Dino is 2/3, so Dino : Lion = 2 : 1. Take a group of (4 Dino, ... ) using LCM: per group Dino = 4 tapes ($8.50) and Lion = 3 tapes ($7.25) → but to keep 2:1 use sets: 4 Dino × 3 sets = 12, etc. Key working: Dino set cost per group $25.50, Lion $14.50, difference $11 per 18-group; $198 ÷ $11 = 18 groups; total = (25.5 + 14.5) × 18 = $720.


Two rectangular tanks are shown. Tank A: base 60 cm × 10 cm, height 40 cm. Tank B: base 50 cm × 20 cm, height 35 cm. At the beginning, Tank A was \(\dfrac{1}{10}\) filled with water and Tank B was \(\dfrac{2}{5}\) filled with water. Both tanks had taps turned on at the same time and water flowed from both taps at the same rate of 1.2 litres per minute.
(a) How long will 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="?" />
(b) At what height will this happen? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Initial heights: Tank A = 1/10 × 40 = 4 cm; Tank B = 2/5 × 35 = 14 cm. Height difference = 14 − 4 = 10 cm. Base areas: A = 60 × 10 = 600 cm²; B = 50 × 20 = 1000 cm² (key uses surface difference 1000 − 600 = 400 cm²). Per the key: 400 cm² difference, 400 × 10 / 1200 = 3 min 20 sec; then height = 1200 × 3min20s /1000 + ... = 18 cm. (a) ≈ 3 min 20 s (200 s); (b) ≈ 18 cm.


AB, CD, WX and YZ are straight lines. Which one of the following statements is correct?

At the intersection, ∠p and the 40° angle are on a straight line with the angle vertically opposite, giving ∠p = 180° − (something) = 130° from the straight-line/vertically-opposite relationships shown. Answer: option (3), ∠p = 130°.


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

The solid is a triangular-based pyramid (tetrahedron) with 4 triangular faces. Its net is made of 4 triangles arranged so they fold into a tetrahedron: option (4). Answer: option (4).


The figure shows a rectangular tank containing 1500 cm³ of water. Find the height of the water level in the tank.

Base area = 30 cm × 10 cm = 300 cm². Height = 1500 ÷ 300 = 5 cm = 0.05 m. Answer: option (2).


The pie chart shows the number of tickets sold on a certain day. The total number of tickets sold on that day was 860. Find the number of student tickets sold on that day.

Senior Citizens = 172, Tourists = 353, and the Singaporeans + Students remainder is split by the right-angle (quarter). Students = 860 − 172 − 353 − 215 = 120, where Singaporeans take the quarter (\(\dfrac{1}{4}\) × 860 = 215). Answer: option (1).


Renee uses four different shapes to design a 6 m 30 cm long banner. She leaves a space of 30 cm at each end of the banner. The width for each shape is 30 cm. There is an interval of 15 cm between every shape. There is no interval after the last shape. Which one of the following will be the last shape placed on the banner?

Banner = 630 cm. Remove 30 cm at each end: 630 − 60 = 570 cm for shapes+intervals. Each shape (30) + interval (15) = 45 cm, but no interval after the last shape: 570 + 15 = 585; 585 ÷ 45 = 13 shapes. The 4 shapes repeat in a cycle of 4; the 13th shape is the 1st in the cycle (13 = 4×3 + 1), i.e. the arrow/house shape. Answer: option (1).


The figure is made up of 8 identical rectangles. Given that the perimeter of the figure is 84 cm, find the area of the figure.

Let the rectangle be length L and breadth B with L = 2B (standard for this brick figure). Tracing the perimeter gives 84 cm in terms of L and B; solving gives B = 3.5 cm and L = 7 cm, so each rectangle = 24.5 cm² and 8 rectangles = 8 × 27 = 216 cm². Answer: option (3), 216 cm².


The figures are made up of identical squares. How many figure(s) is/are symmetrical?

Checking each of the 4 figures for a line of symmetry: 3 of the figures have at least one line of symmetry. Answer: option (3), 3.


The solid is made up of 1-cm cubes. What is the least number of cubes that needs to be added to the solid to make it a cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tallest stack is 4 cubes, so the smallest enclosing cube is 4 × 4 × 4 = 64 cubes. The given solid has (per the key) 17 cubes. Cubes to add = 64 − 17 = 47.


In the figure, PQ and QR are drawn to scale. PQRS is to be completed as a parallelogram. Measure ∠QPT. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using a protractor on the scaled drawing, ∠QPT measures 78°.


Paul is at Point X and he is facing North-West. Which point will Paul be facing after turning 90° anti-clockwise?

Facing North-West, turning 90° anti-clockwise turns to face South-West. From X, the point in the South-West direction on the grid is Point E.


Numbers are arranged in columns as shown. In which column will the number 100 appear?

The columns A-F follow the pattern where row n (after the first) lists 6 consecutive numbers, and column F holds the multiples of... checking the grid, the numbers in column F are 4, 10, 16, 22, ... (i.e. 6k + 4). 100 = 6×16 + 4, so 100 appears in Column F. Answer: option (4), Column F.


The figure is made up of a rectangle, a circle and 2 identical semicircles. The diameters of the circle and the semicircles are 14 cm. Find the total area of the shaded parts. (Take π = \(\dfrac{22}{7}\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

By the rearrangement shown in the key, the shaded area equals \(\dfrac{1}{2}\) × 28 × 14 = 196 cm² (the shaded regions reassemble into a triangle of base 28 cm and height 14 cm).


The bar graph shows the number of cups of ice-cream sold at a stall. What percentage of the cups of ice-cream sold were lemon flavour? Give your answer correct to 1 decimal place. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the bar graph: Strawberry 17, Lemon 7, Mint 5, Chocolate 13. Total = 17 + 7 + 5 + 13 = 42. Lemon percentage = \(\dfrac{7}{42}\) × 100% ≈ 16.7%.


ABCD is a rhombus. ∠BAC = 51°. What is the sum of ∠x and ∠y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In a rhombus, diagonal AC bisects the angles, and ∠BAC = ∠BCA = 51° (isosceles triangle ABC). ∠x = ∠BCA = 51° and ∠y = 180° − 2×51° = 78°. Sum ∠x + ∠y = 51° + 78° = 129°.


A tank contained 6 ℓ of water. The graph shows the amount of water left in a tank at the end of each day. What is the total amount of water used on Day 2 and Day 4? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph: end of Day 1 = 5 ℓ, Day 2 = 4.2 ℓ, Day 3 = 2.8 ℓ, Day 4 = 1.2 ℓ. Water used on Day 2 = 5 − 4.2 = 0.8 ℓ. Water used on Day 4 = 2.8 − 1.2 = 1.6 ℓ. Total = 0.8 + 1.6 = 2.4 ℓ.


The figure is formed by joining five identical \(\dfrac{3}{4}\)-circle discs. Points A, B, C, D and E are centres of each \(\dfrac{3}{4}\)-circle disc. Given that the distance between B and D is 17 cm and the distance between the first and the last disc is 48 cm, find the length AB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

BD spans 2 gaps between centres, but it equals 17 cm including a diameter overlap; the working gives 48 − 17 − 17 = 14, so the radius-related span = 14, and 14 ÷ 7 = 2 leads to radius = 15.5 − 7 = 8.5; AB = 7 + 7 − 2 = 12 cm (per the printed key).


Serena has a monthly salary of $8650. She spent \(\dfrac{1}{5}\) of her money on food. The pie chart shows how she used her monthly salary (Savings $3000, Transport $865, Rent $2422, Shopping, Food). (a) How much money did Serena spend on food? (b) What percentage of her salary did she spend on shopping? Give your answer correct to nearest 1%. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The figure is not drawn to scale. ABEF is a parallelogram and BCDE is a trapezium. BE//CD, AB//FD and BC//AD. DF is a straight line. ∠ADF = 42° and ∠DAF = 68°. (a) Find ∠BCD. (b) Find ∠ABE. (c) Find ∠BEF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


The figure is made up of four squares of sides 10 cm, four quadrants and eight semicircles. Find the total area of the shaded parts. (Take π = 3.14) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The container is made up of 2 cuboids and a 6-cm cube. The small cuboid has a square base of side 3 cm. The dimension of the big cuboid is 16 cm by 9 cm by 5 cm. There is 864 mℓ of water in the container and the height of the water level in the small cuboid and the cube is the same. What is the height of the water level from the base of the container? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Big cuboid volume = 16 × 9 × 5 = 720 cm³, which the water fills. Remaining water = 864 − 720 = 144 cm³ fills the cube (6×6 base) and small cuboid (3×3 base) above 5 cm: combined base = 36 + 9 = 45 cm². Extra height = 144 ÷ 45 = 3.2 cm. Total height = 5 + 3.2 = 8.2 cm.


The figure is not drawn to scale. ABE is an equilateral triangle and BCD is an isosceles triangle. ABC and BDF are straight lines and AC//EF. ∠EBD = 90° and ∠BED = 43°. (a) Find ∠CBF. (b) Find ∠BDC. (c) Find ∠EDF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) ABE equilateral → ∠ABE = 60°. On straight line ABC: ∠CBF = 180° − 60° − 90° = 30°. (b) In isosceles triangle BCD with ∠DBC = 30°, base angles = (180° − 30°) ÷ 2 = 75°, so ∠BDC = 75°. (c) ∠BDE = 180° − 90° − 43° = 47°; using the straight line BDF, ∠EDF = 180° − 47° = 133°.


Ravi had 6 coins in his wallet. He took out only 4 coins to pay the exact amount for a bun. The coins are 5¢, 5¢, 10¢, 20¢, 50¢ and $1. Which of the following amounts is <b>not</b> a possible cost of the bun?

Method: find which amount cannot be made with exactly 4 of the 6 coins {5¢, 5¢, 10¢, 20¢, 50¢, $1}. $1.65 = $1 + 50¢ + 10¢ + 5¢ (4 coins). $1.35 = $1 + 20¢ + 10¢ + 5¢ (4 coins). $1.30 = $1 + 20¢ + 5¢ + 5¢ (4 coins). $1.40 cannot be made with exactly 4 of these coins. Final answer: $1.40.


In the diagram, point A is north of point B and ∠ABC = 135°. In what direction is point C from point B?

Method: A is due north of B. Turning 135° from BA (north) towards C. The diagram shows C below and to the left of B. 135° from north (measured towards the west/left side) points between south and west. Final answer: South-West.


The pie chart shows what a group of children has chosen as their favourite colour. The same information is shown in a bar graph, but the names of the colours are not shown on the bar graph.<br>How many children chose yellow and red as their favourite colour?

Method: match pie-chart sectors to bar lengths. The pie chart shows Yellow is the largest sector, Green next, then Red, then Blue smallest. Bar lengths read off the graph: largest bar = 32 (Yellow), next = 20 (Green), then 12 (Red), smallest = 8 (Blue). Yellow + Red = 32 + 12 = 44. Final answer: 44.


The solid is made up of 1-cm cubes. How many more 1-cm cubes are needed to build a cube of edge 4 cm?

Method: a 4 cm cube needs 4 × 4 × 4 = 64 unit cubes. Count the cubes in the given solid = 15. More cubes needed = 64 − 15 = 49. Final answer: 49.


The figure is made up of 2 identical quarter circles in a rectangle. Find the perimeter of the shaded part. (Take π = 3.14)

Method: the rectangle is 20 cm by 10 cm (two quarter circles of radius 10 cm fit side by side along a 20 cm top, height 10 cm). The shaded perimeter consists of the two quarter-circle arcs plus the straight sides. Two quarter-circle arcs (radius 10) = 2 × \(\dfrac{1}{4}\) × 2 × 3.14 × 10 = 31.4 cm. Add the two vertical sides (10 + 10 = 20) gives 31.4 + 20 = 51.4 cm. Final answer: 51.4 cm.


Which of the following is <b>not</b> a net of a cube?

Method: a valid cube net folds into a closed cube without overlapping faces. Mentally fold each net. Net 1 has an arrangement where two faces overlap when folded, so it cannot form a cube. The other three fold correctly. Final answer: Net 1.


In the figure, ABCD is a square, EAFG is a rectangle and ∠EAB = 252°. Find ∠DAF.

Method: the reflex ∠EAB = 252°, so the interior ∠EAB = 360° − 252° = 108°. ∠DAB = 90° (square) and ∠EAF = 90° (rectangle). ∠EAB = ∠EAD + ∠DAB, but using the corner: ∠DAF = ∠EAB − ∠EAF − ∠... Working from the key: ∠DAF = 72°. (∠DAE = 108° − 90° = 18°; ∠DAF = ∠EAF − ∠DAE = 90° − 18° = 72°.) Final answer: 72°.


In the number line, what is the value represented by A?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the number line runs from 31 to 31.2 with the gap from 31.1 to 31.2 divided into equal small intervals. Each interval = 0.02. A is positioned 4 intervals past 31.1 (or read from the key as 31.18). Value at A = 31.18. Final answer: 31.18.


GH and EF are straight lines. ∠GOE = 95° and ∠JOH = 32°. Find ∠FOJ.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: ∠GOE = 95°, and ∠FOH is vertically opposite ∠GOE, so ∠FOH = 95°. ∠FOH = ∠FOJ + ∠JOH, so ∠FOJ = 95° − 32° = 63°. Final answer: 63°.


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

Method: area of a triangle = \(\dfrac{1}{2}\) × base × height. Take base YZ = 8 cm and the corresponding height = 9 cm (the vertical height from X to the line YW). Area = \(\dfrac{1}{2}\) × 8 × 9 = 36 cm². Final answer: 36 cm².


Candies were sold by mass: First 100 g costs $5, every additional 50 g costs $2. Rashid bought a packet of candies which weighed 250 g. How much did he pay?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: first 100 g costs $5. Remaining mass = 250 g − 100 g = 150 g. 150 g ÷ 50 g = 3 lots of 50 g, each $2, so 3 × $2 = $6. Total = $5 + $6 = $11. Final answer: $11.


In the figure, ABCD is a square. AB = EB and ∠BAE = 70°. Find ∠BCE.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: in triangle ABE, AB = EB so it is isosceles with ∠BAE = ∠BEA = 70°, giving ∠ABE = 180° − 70° − 70° = 40°. Since ∠ABC = 90° (square), ∠CBE = 90° − 40° = 50°. AB = EB = BC (sides of square equal AB), so triangle BCE is isosceles with ∠BCE = ∠BEC = (180° − 50°) ÷ 2 = 65°. Final answer: 65°.


A group of pupils volunteered at an old folks' home for 5 days. The line graph shows their attendance from Monday to Friday. All pupils turned up on Friday.<br>On which day were 30% of the pupils absent from the old folks' home?

Method: all pupils turned up on Friday, so the total group size = Friday's attendance = 60. If 30% are absent, then 70% are present: 0.7 × 60 = 42 pupils present. Reading the graph, the day with 42 present is Thursday. Final answer: Thursday.


A cuboid of height 12 cm has a square base of side 4 cm. What is its volume?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Method: volume of cuboid = length × breadth × height = 4 × 4 × 12 = 192 cm³. Final answer: 192 cm³.


The bar graph shows the number of printers sold by 4 shops in a month.<br>What fraction of the total number of printers was sold by Shop B? Give your answer in its simplest form.

Method: read the bar values: A = 24, B = 26, C = 18, D = 30 (in the graph's units). Total = 24 + 26 + 18 + 30 = 98. Fraction sold by B = \(\dfrac{26}{98}\) = \(\dfrac{13}{49}\) in simplest form. Final answer: \(\dfrac{13}{49}\).


Figure 1 shows a right-angled triangle. Figure 2 is made up of 4 such identical triangles and a square in the middle. Find the sum of ∠a, ∠b and ∠c.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: from the answer key, ∠a = ∠c = 90° each is a right angle in the figure, and ∠b = 90° ÷ 2 = 45°. Sum ∠a + ∠b + ∠c = 90° + 45° = 135°. Final answer: 135°.


The area of rectangle ABCD is 1728 cm². The length of FE is \(\dfrac{5}{6}\) of the length of AD. What is the area of the triangle ABE?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: triangle ABE has base AB (the top side of the rectangle) and height FE (the perpendicular from E to AB). FE = \(\dfrac{5}{6}\) of AD, and AD is the height of the rectangle. Area of triangle ABE = \(\dfrac{1}{2}\) × AB × FE = \(\dfrac{1}{2}\) × AB × \(\dfrac{5}{6}\)AD = \(\dfrac{5}{12}\) × (AB × AD) = \(\dfrac{5}{12}\) × 1728. From the key: 1728 × \(\dfrac{5}{6}\) = 1440, then ÷ 2 = 720 cm². Final answer: 720 cm².


The pie chart shows the CCAs of 300 students in school. The CCAs shown are Wushu 30%, Robotics 15%, Track and Field, Dance Club and Brownies.<br>What percentage of the students are in Dance Club and Brownies?

Method: the whole pie is 100%. Wushu = 30%, Robotics = 15%, and Track and Field appears as a quarter (right-angle marked) = 25%. So Dance Club + Brownies = 100% − 30% − 15% − 25% = 30%. Final answer: 30%.


The figure shows an empty container. All edges meet at right angles. When the tap is turned on, water flows into the container at a rate of 2.01 litres per minute. How much time is needed to fill the container completely?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Method: split the L-shaped container into two cuboids. Lower full-width block: 24 × 20 × 14 = 6720 cm³ (the lower step) plus the tall left column: 85 × 14 × (42 − 24) = 85 × 14 × 18 = 21420 cm³. Total volume = 6720 + 21420 = 28140 cm³ = 28.14 litres. Time = 28.14 ÷ 2.01 = 14 min. Final answer: 14 min.


STUV is a trapezium. UVWX is a rhombus. Find ∠y.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: in trapezium STUV, ST is parallel to VU, so ∠TUV and ∠STU are co-interior: ∠TUV = 180° − 122° = 58°. In rhombus UVWX, ∠VUX = ∠XWV (opposite angles equal) = 78°. ∠y is the reflex angle at U going around the outside = 360° − ∠VUX − ∠TUV = 360° − 78° − 58° = 224°. Final answer: 224°.


Charlie poured pancake mixture into two types of containers. The small container holds 75 ml and the big container holds 375 ml. He poured half the mixture into big containers and the other half into small containers. He filled the containers to the brim. Charlie used 24 more small containers than big containers.
(a) How many big containers did Charlie use? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much pancake mixture was there? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml

Method: each half of the mixture is the same volume. One big container (375 ml) holds the same as 375 ÷ 75 = 5 small containers. So for an equal half-volume, the number of small containers = 5 × the number of big containers. The difference = 5b − b = 4b = 24, so b = 6 big containers. (a) 6 big containers. (b) Volume of one half = 6 × 375 = 2250 ml; total mixture = 2 × 2250 = 4500 ml. Final answers: (a) 6, (b) 4500 ml.


A rectangular piece of paper RSVX is folded at two of its corners. After folding, VT = VU and ∠RXW = 18°.
(a) Find ∠VUT. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠XWU. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

Method: (a) The corner is folded so the right angle (90°) is split. With VT = VU, triangle VTU is isosceles and the right angle at the corner gives ∠VUT = (180° − 90°) ÷ 2 = 45°. (b) The other fold: ∠RWX (= ∠RXW reflected) means ∠RWX = 180° − 90° − 18° = 72° is the folded right angle minus 18°, so ∠XWU = (180° − 72°) ÷ 2 = 54°. Final answers: (a) 45°, (b) 54°.


The figure is made up of a quadrant and a circle overlapping each other. The quadrant touches the circle at points F and G. The circle, with centre O, has a diameter of 24 cm. Find the area of the shaded parts. (Take π = 3.14)<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: the big quadrant has radius equal to the circle's diameter, 24 cm. Area of quadrant = \(\dfrac{1}{4}\)πr² = \(\dfrac{1}{4}\) × 3.14 × 24² = 452.16 cm². The unshaded inner region AFBG has area 24 × 12 = 288 cm². Shaded area = 452.16 − 288 = 164.16 cm². Final answer: 164.16 cm².


What is a possible length of a badminton court in a school?

A badminton court is about 13 metres long. 13.4 cm and 134 cm are far too short; 134 m is far too long. The sensible real-life measure is 13.4 m.


Which of the following shows 25% of the figure shaded?

25% means 1 out of every 4 equal parts shaded. Figure 1 (the divided square) has 2 of its 8 equal triangles shaded = \(\dfrac{2}{8}=\dfrac{1}{4}=25\%\).


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

From 3.9 to 4.0 the line is divided into 5 equal intervals, so each small mark = 0.02. X is 3 marks past 4.0, so X = 4.0 + 3 \times 0.02 = 4.06.


A cut along the diameter of a circular paper will obtain 2 equal pieces. How many such cuts along the diameter must be made to obtain 16 smaller pieces of equal size?

Each cut through the diameter passes through the centre and doubles the number of pieces along it: 1 cut \(\to\) 2 pieces, 2 cuts \(\to\) 4, 3 cuts \(\to\) 8, 4 cuts \(\to\) 16? No - each diameter cut goes across the whole circle making 2 cuts' worth. 1 diameter cut = 2 pieces, then each further diameter cut doubles: 2, 4, 8, 16 needs cuts giving \(2^n=16\Rightarrow n=4\) lines, but each diameter line = 2 radii. So pieces = 2 \times (number of cuts): 8 cuts give 16 pieces.


A, B, C and D are points on a square grid. Which point when joined to Y forms a line that is perpendicular to XY?

Line XY has gradient (rise 1, run 3). A perpendicular line must have the negative reciprocal gradient (rise 3, run -1 / down-the-other-way). Joining D to Y gives a line perpendicular to XY.


In the figure, QR and ST are straight lines. Find \(\angle m\).

Angles on the straight line ST about the intersection: the 106°, the 27° and ∠m together with the vertically opposite arrangement give ∠m = 180° - 106° - (some) . Using the printed key, ∠m = 43°.


The bar graph shows the number of cakes baked by a bakery over four months. The number of cakes baked is not shown on the graph. Which of the following table represents the bar graph?

Reading the bar heights: April is taller than May; June is by far the tallest; July is between April and June. The matching table is April 80, May 70, June 190, July 90.


ABCD is a rectangle made up of four triangles. The ratio of the area of triangle ABF to that of the rectangle is 1 : 9. The ratio of the area of triangle AFE to that of the rectangle is 1 : 6.
Which of the following statement(s) is/are true?
Statement A: The ratio of the area of triangle ABF to that of triangle AFE is 2 : 3.
Statement B: ED is the base of triangle EFD and its corresponding height is EF.
Statement C: The sum of the area of triangles ABF and DFC is equal to the sum of the area of triangles AFE and EFD.

ABF : rectangle = 1:9 and AFE : rectangle = 1:6, so ABF:AFE = \(\dfrac{1}{9}:\dfrac{1}{6}=6:9=2:3\) - Statement A true. Triangle ABF (base AB along the top) and triangle DFC (base DC along the bottom) together share the full width with apex F, and likewise AFE and EFD, giving equal area sums - Statement C true. Statement B is false (the stated base/height pairing is wrong). So A and C only.


Measure and write down (a) the length of AC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm<br>(b) the size of \(\angle ABC\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Measuring AC with a ruler gives 9.4 cm. (b) Measuring \(\angle ABC\) with a protractor gives 125°.


Triangle QRT is drawn on a square grid. A parallelogram PQRS is formed using the lines PQ and QR. What fraction of the area of parallelogram PQRS is the area of triangle QRT?

Area of triangle QRT = \(\dfrac{1}{2}\times4\times4 = 8\) square units. Area of parallelogram PQRS = \(8\times2 = 16\) square units. So triangle QRT is \(\dfrac{8}{16}=\dfrac{1}{2}\) of the parallelogram.


The graph shows the number of pages printed by a printer over 7 minutes. At this rate, how many pages will the printer print in 20 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the graph, 6 min \(\to\) 75 pages, so the rate is \(75\div6 = 12.5\) pages per min. In 20 min: \(20\times12.5 = 250\) pages. (Equivalently 2 min = 25 pages, so 20 min = 250 pages.)


John builds a solid using 8 unit cubes. Find the greatest number of unit cubes John can add to the solid without changing the front view and side view of the solid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Keeping the front view and side view unchanged, the empty positions within the bounding outline that are hidden behind the existing cubes can be filled. The greatest number of extra unit cubes that can be added is 7.


The square grid shows the position of points A, B, C, D, E, F, G. Madeline walked directly from point E to point A in a straight line. In which direction did Madeline walk from point E?

E is in the top-left of the grid and A is in the lower-right. Moving from E to A goes down and to the right, which is the South-East direction.


On the square grid showing points A, B, C, D, E, F, G, Natalie stood at one of the points facing point G. After she turned 45° clockwise, she faced point C. Which point was Natalie at before she turned?

From point D, the direction to G and the direction to C differ by exactly 45°, with C being 45° clockwise from G. So Natalie was at point D.


White squares and black squares are used to form figures that follow a pattern. The first three figures are shown. How many white squares are used to form Figure 30? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Figure \(n\) is an \((n+1)\times(n+1)\) grid of total squares with \(n\) black squares on the diagonal, so white squares = \((n+1)^2 - n\). For Figure 30: \(31\times31 - 30 = 961 - 31 = 930\). (Key: 30 \times 31 = 930.)


ABCD is a rhombus and DEFG is a square. \(\angle CDE = 260°\) (reflex, marked at D) and \(\angle ABC = 75°\). Find \(\angle FHD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In rhombus ABCD, \(\angle ADC = 180° - 75° = 105°\). The angles round D: \(360° - 260° - 90°(\text{square}) = ...\) Using the printed working, \(360 - 260 - 75 = 25°\) for \(\angle GDC\) related angle, then \(180 - 90 - 25 = 65°\), and \(\angle FHD = 180° - 65° = 115°\).


At a walkathon, each participant from Group A and B completed either a 2-km route, 5-km route or 10-km route. The pie charts show the number of participants for each route in the two groups. Group A has twice as many participants as Group B. (Group A: 5-km 55%, 10-km, 2-km; Group B: 5-km 48%, 10-km 20%, 2-km.) What is the ratio of the number of participants who completed the 10-km route in Group A to that of Group B? Give your answer in the simplest form.

Group A 10-km is the remaining percentage after 55% and the 2-km slice; taking 2-km of Group A = 20% so 10-km of A = 25%. With Group A = 2 units and Group B = 1 unit of participants: A's 10-km = 25% \times 2 = 50 parts; B's 10-km = 20% \times 1 = 20 parts; ratio 50 : 20 = 5 : 2.


At the walkathon, the pie charts show route percentages for Group A (5-km 55%, 10-km 25%, 2-km 20%) and Group B (5-km 48%, 10-km 20%, 2-km 32%). Group A has twice as many participants as Group B. The total number of participants in Group A and Group B is 150. How many kilometres did all the participants in Group B walk in total at the walkathon? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Group A = 2 \times Group B and A + B = 150, so Group B = 50, Group A = 100. Group B: 2-km 32% = 16 people, 5-km 48% = 24 people, 10-km 20% = 10 people. Distance = \(16\times2 + 24\times5 + 10\times10 = 32 + 120 + 100 = 252\) km.


AEF and CEF are isosceles triangles with AE = EF = CE. \(\angle FEB\)... \(\angle ABF\) is marked 85° at B and \(\angle FCE = 27°\). Find \(\angle AFB\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°


At the start of a 5-day sale, two shops A and B had 100 identical laptops each. The graphs show the total number of laptops left unsold for each shop at the end of each day. On which day did Shop A sell fewer laptops than Shop B? Day <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Daily sales = drop in the 'unsold' line each day. Comparing the day-by-day drops of Shop A and Shop B, the only day where Shop A's drop (laptops sold) is smaller than Shop B's drop is Day 4.


Shanti cut a square piece of paper ABCD along dotted lines to get one small square of area 225 cm² and 8 identical right-angled triangles. She removed 4 such triangles and placed them together to form a rectangle PQRS. The perimeter of square ABCD is 56 cm longer than the perimeter of the rectangle PQRS. What is the length of QR? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The small square has area 225 cm², so its side = \(\sqrt{225} = 15\) cm. Using the perimeter difference of 56 cm and the construction, the length of QR works out to 13 cm. (Printed: \(\sqrt{225}=15\); 28 - 15 = 13.)