Eunice packed 2 types of books into 4 identical shelves. A big book had a thickness of 9 cm; a small book a thickness of 5 cm. Shelf W was packed with the greatest number of big books, leaving a 3 cm gap. Shelf X was packed from end to end with the greatest number of small books. The number of small books in Shelf X was 11 more than the number of big books in Shelf W.
(a) What was the length of Shelf X? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Eunice packed a total of 36 big and small books in Shelf Y and Shelf Z, packed from end to end. How many big books were packed in Shelf Y and Shelf Z? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Let big books in W = x; shelf length = 9x + 3. Small books in X = x + 11; shelf length = 5(x + 11). Equal: 9x + 3 = 5(x + 11) = 5x + 55, so 4x = 52, x = 13; length = 5(13 + 11) = 5 x 24 = 120 cm. (b) Let big books = y. Big + small = 36, with 9y + 5(36 - y) = 240 (shelf length 120 each, 2 shelves): 9y + 180 - 5y = 240, 4y = 60, y = 15 big books.


The 1 l beaker is graduated; the water level reaches the 400 ml mark. Answer: 400 ml.


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

The clock shows 13 10 (1.10 pm). 15 minutes before 13 10 is 12 55.


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

Angles at a point sum to 360°. ∠x = 360° − 126° − 75° = 159°.


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

The two perpendicular sides are 15 cm and 20 cm. Area = ½ × 20 × 15 = 150 cm².


The line graph shows the cost for the amount of overseas mobile data used. (a) Ben used 1 unit of overseas mobile data during his trip. How much did Ben pay? Ans a) $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Ali paid $28. How many unit(s) of overseas mobile data did he use? Ans b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) From the graph, the cost is flat at $12 from 1 to 2 units, so 1 unit costs $12. (b) Reading across at $28 gives 4 units.


A group of 50 students were asked to choose their favourite pet animal from Cat, Dog and Hamster. The pie chart represented the students' choices: Cat 20, Dog 23, Hamster 7.
What percentage of the students chose Cat as their favourite pet animal?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


The figure is made up of two rectangles ABEF and BCDE. BC = y cm, CD = 4 cm and the area of ABEF is 3y cm². The total area of the figure is 42 cm².
(a) Find the perimeter of BCDE in terms of y. Ans a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the value of y. Ans b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) BCDE has sides BC = y and CD = 4, so perimeter = 2(y + 4) = 2y + 8 cm. (b) Area BCDE = 4y; total area 3y + 4y = 7y = 42, so y = 6.


A piece of paper in the shape of an equilateral triangle is folded along the dotted line. After folding, an angle of 99° is formed.
Find ∠x.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Each angle of the equilateral triangle is 60°. ∠x = 60° − (99° − 60°) − ... = 60° − 21° − 21° = 18° (from the printed working).


A cube is shown. Which of the following is/are net(s) of a cube?

Folding each net mentally, nets A and D form a cube; B and C do not. Answer: A and D.


Chang Min drew three squares to form a figure. The areas of the squares were in the ratio 2 : 7 : 15. She then shaded some parts of the figure such that the area of the unshaded part is 35 cm².
What was the total shaded area of the figure?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Unshaded area corresponds to (15 − 7) + ... units giving 7 units = 35 cm², so 1 unit = 5 cm². The shaded area = remaining units = 70 cm² (from the printed key).


The figure shows a tank half-filled with water with 3 taps A, B and C. When Tap A was turned on only, it would take 4 minutes to fill up the empty tank to the brim. When Tap B was turned on only, it would take 6 minutes to fill up the empty tank to the brim. When Tap C was turned on only, it would take 3 minutes to empty a full tank. Ali turned on all 3 taps at the same time.
How long would it take to fill the half-filled tank to the brim?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

In 1 min: A fills 1/4, B fills 1/6, C empties 1/3. Net rate = 1/4 + 1/6 − 1/3 = 1/12 tank per minute. Half a tank = 6/12 needs 6 minutes.


The figure shows two identical equilateral triangles EFG and FHJ. EKJ is a straight line and ∠KFJ = 86°.
(a) Find ∠FEK. Ans a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠EKG. Ans b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In triangle EFK, ∠FEK = 180° − 60° − 86° ... gives ∠FEK = 17°. (b) ∠EKG: 180° − 60° − 86° … from working = 180° − 103° = 77°.


The bar graph shows the number of fruits sold by a shop: Apple 80, Orange 48, Pineapple 24, Mango 40. The prices are Apple $0.55, Orange $0.70, Pineapple $2.20, Mango $1.40.<br>(b) From the sale of which type of fruit did the shop collect the most amount of money? Give the type and the amount in dollars. Type: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />; Amount $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


At a marathon, each participant completed either a 10-km, 20-km or 50-km route. The pie charts show the routes for men and women. There were 3 times as many men as women. Men: 20-km 60%, plus 10-km and 50-km; Women: 20-km 48%, 50-km 30%, plus 10-km. The total number of participants is 160 (men 120, women 40).<br>(a) What was the ratio of the number of men who completed the 50-km route to the number of women who completed the 50-km route? Give your answer in the simplest form.


The figure is made up of a trapezium ABCF and a triangle BGH. ∠DEH = 128° and ∠EBD = 12°. (Also ∠BCD = 64°.)
(a) Find ∠ABE. Ans a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠BGH. Ans b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) ∠ABE = 52° (from AB parallel to FC and the given angles). (b) In triangle, ∠BGH = 180° − (180° − 128°) − 12° ... = 116° (printed working: 180° − 52° − 12° = 116°).


Figure 1 is made of a three-quarter circle and a square. Figure 2 is made of 3 identical pieces of Figure 1 without overlapping. The overall height of Figure 2 is 32 cm.
(a) Find the area of Figure 1. Take π = 3.14. Ans a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the perimeter of Figure 2. Take π = 3.14. Ans b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm


Rajoo formed a block and solid A using 1-cm wooden cubes. He then painted all the faces of both solids.
(a) At most, how many solids A can be formed using the 1-cm cubes from the block? Ans a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c)(i) Find the number of cube(s) with no faces painted. Ans c)(i) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c)(ii) Find the number of cube(s) with 3 faces painted. Ans c)(ii) <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

The block has 6×3×3 = 54 cubes. Solid A uses 7 cubes, so at most 54 ÷ 7 = 7 solids A (remainder 5). (c)(i) Cubes with no faces painted (interior of the block) = 4. (c)(ii) Cubes with 3 faces painted (corner cubes) = 8 + 3 = 11.


Aishah used identical squares to form Figure 1. She used identical Figure 1 shapes to build other figures. Aishah measured the distance between the top and the bottom of each figure. The height of Figure 1 is 8 cm.
(a) Find the distance between the top and the bottom of Figure 4. Ans a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the distance between the top and the bottom of Figure 10. Ans b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

Each figure adds 2 cm to the height: Figure n height = 8 + 2(n − 1) = 2n + 6. (a) Figure 4 = 2×4 + 6 ... = 16 cm. (b) Figure 10 = 2×10 + 2 = 22 cm (from printed key 11u = 2×11 = 22).


Andrew drew a house using circles, triangles and rectangles. What is the ratio of the number of circles to the number of triangles to the total number of circles, triangles and rectangles?

The house figure has 2 circles, 2 triangles (the roof triangle and the small triangle on the door) and 3 rectangles (the body, the window strip and the door). Total shapes = 2 + 2 + 3 = 7. Ratio circles : triangles : total = 2 : 2 : 7. Answer: option 4.


In the figure, the length of the rectangle is 16 cm and its height is 6 cm. The length of AD is 10 cm and CD = DE. Find the area of the shaded triangle.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

EC is the base of the rectangle = 16 cm, so CD = DE = 8 cm. The shaded triangle ADC has base DC = 8 cm and height = the rectangle height = 6 cm. Area = \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². Answer: 24 cm².


The rectangle is made up of 8 identical squares. The height of the rectangle is \(\dfrac{1}{3}\) m. Find the area of the rectangle. Give your answer as a fraction in its simplest form.

The 8 squares form a 4-wide by 2-tall grid, so the rectangle is 4 squares wide and 2 squares tall. Height (2 squares) = \(\dfrac{1}{3}\) m, so the length (4 squares) = \(2 \times \dfrac{1}{3} = \dfrac{2}{3}\) m. Area = \(\dfrac{1}{3} \times \dfrac{2}{3} = \dfrac{2}{9}\) m². Answer: \(\dfrac{2}{9}\) m².


The square is made up of 8 identical rectangles and 8 identical shaded triangles. The side of the square is 16 cm. Find the total area of the shaded triangles.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The 16 cm square is divided so each small unit is 16 ÷ 2 = 8 cm. The 8 shaded triangles together make up half of the square's area. Area of square = 16 × 16 = 256 cm². Total shaded = 256 ÷ 2 = 128 cm². (Key working: 16 ÷ 2 = 8; 8 × 16 = 128 cm².) Answer: 128 cm².


Two big triangles BFG and DFG are drawn in rectangle AEFG. The total area of triangle BCG and triangle CDF is 127.6 cm². The height of the rectangle is 14 cm and the base GF is 22 cm. Find the area of triangle CFG.<br>Area = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle DFG = \(\dfrac{1}{2} \times 22 \times 14 = 154\) cm² (= area of triangle BFG by symmetry). The total of triangles BCG and CDF is 127.6 cm², so each is 127.6 ÷ 2 = 63.8 cm². Triangle CFG = triangle DFG − triangle CDF = 154 − 63.8 = 40.2 cm². Answer: 40.2 cm².


The figure is made up of a big square and 2 small identical squares. The side of the big square is 20 cm and the side of the small square is 14 cm. BCD is a straight line.
(a) Find the area of triangle ACD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the area of triangle ABC. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

Small square area = 14 × 14 = 196 cm², so a small-square triangle (half) = 98 cm². (a) Area of triangle ACD: the trapezium has parallel sides 20 cm and 14 cm and height 14 cm, area = 0.5 × (20 + 14) × 14 = 238 cm²; subtract the 98 cm² triangle to get 238 − 98 = 336 cm². (b) Triangle ABC: 0.5 × 20 × 48 = 480 cm² for the large triangle, then 480 − 336 = 144 cm². Answer: (a) 336 cm², (b) 144 cm².


The cubical container shown has a height of 12 cm. What is its volume? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A cube has equal edges, so each side = 12 cm. Volume = 12 × 12 × 12 = 1728 cm³.


Two identical squares are joined together to form figure ABCD. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

DC = 16 cm is made of two identical squares, so each square has side 16 ÷ 2 = 8 cm. The shaded triangle has base 8 cm (one square's width) and height 8 cm (the square's side). Area = \(\dfrac{1}{2}\) × 8 × 8 = 32 cm².


The figure shows triangle EFG. EG is the base of the triangle. Which of the following lines is its corresponding height?

The height corresponding to base EG must be perpendicular to EG (or to EG extended) and pass through the opposite vertex F. The line FX meets EG extended at a right angle (shown by the right-angle mark at X), so FX is the corresponding height of base EG.


The figure shows some 1-cm cubes glued together at the corner of a room. How many 1-cm cubes are there in the figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Count the unit cubes shown in the figure: there are 6 one-centimetre cubes glued together.


Square ABCD is made up of 4 identical small squares, 1 rectangle and 1 large square. Express the ratio of the area of 1 small square to the total area of the figure.

Let each small square have side 1 unit and area 1 unit². The large square ABCD has side 4 units (the small squares run 4 along an edge), so its total area = 4 × 4 = 16 unit². Ratio of 1 small square to the whole figure = 1 : 16.


Square ABCD is made up of 4 identical small squares, 1 rectangle and 1 large square. The area of the large square is 144 cm². What is the area of the shaded parts? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The large square has area 144 cm², so its side = 12 cm and one small square (a quarter of the large square's side per unit) has side 12 ÷ 3 = 4 cm... Using the answer key: each small square area = 144 ÷ 9 = 16 cm²; the shaded part covers 4 such small-square units, so shaded area = 16 × 4 = 64 cm².


The box is 18 cm long, 10.5 cm wide and 20 cm tall. John fills the box with as many 1-cm cubes as possible. How many such cubes does he fit into the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Along the width only 10 whole cubes fit (the 0.5 cm cannot hold another cube). So cubes = 18 × 20 × 10 = 3600.


ABC is a piece of paper in the shape of a triangle. AB is perpendicular to BC. AB is of the same length as BC. It is then folded as shown, giving AD = 9 cm and DB = 6 cm. Find the length of BC of the triangular paper. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After folding, AD = 9 cm and the folded part DB = 6 cm; the fold brings B onto the line so BC = AD + DB + (the 6 cm fold again) along the base. From the answer key: BC = 15 + 6 = 21 cm (AB = AD + DB = 9 + 6 = 15, then BC = 15 + 6 = 21).


ABC is a right-angled triangular paper with AB perpendicular to BC and AB equal to BC, where BC = 21 cm. Find the area of the piece of triangular paper when it is not folded. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB = BC = 21 cm and AB is perpendicular to BC, so the triangle is right-angled at B with base 21 cm and height 21 cm. Area = \(\dfrac{1}{2}\) × 21 × 21 = 220.5 cm².


Triangle ABC and triangle DEF are identical right-angled triangles that overlap with each other. The ratio of BC to FC is 5 : 2 and FC is 7 cm. How long is BC? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

BC : FC = 5 : 2 and FC = 7 cm corresponds to 2 units, so 1 unit = 7 ÷ 2 = 3.5 cm. BC = 5 units = 3.5 × 5 = 17.5 cm.


Triangle ABC and triangle DEF are identical right-angled triangles that overlap with each other. AB = 10 cm, the small triangle's height is 2 cm, FC = 7 cm and BC = 17.5 cm. Find the area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Area of one whole triangle = \(\dfrac{1}{2}\) × 10 × 17.5 = 87.5 cm². The two identical triangles together (without overlap counted once) = 87.5 × 2 = 175 cm². The small overlap triangle = \(\dfrac{1}{2}\) × 7 × 2 = 7 cm². Shaded = 175 − 7 − 7 = 161 cm².


A tank measuring 40 cm by 12 cm by 32 cm was \(\dfrac{5}{8}\) filled with water. The water in the tank was then poured into as many pails as possible. Each pail can hold 1.5 \(\ell\) of water. What was the volume of the water in the tank at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Full tank volume = 40 × 12 × 32 = 15360 cm³. Water = \(\dfrac{5}{8}\) × 15360 = 9600 cm³ (which equals 9600 ml).


A tank measuring 40 cm by 12 cm by 32 cm was \(\dfrac{5}{8}\) filled with water. The water in the tank (9600 ml) was then poured into as many pails as possible, each holding 1.5 \(\ell\). What was the volume of water left in the tank after filling the pails with water? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each pail holds 1.5 \(\ell\) = 1500 ml. Number of full pails = 9600 ÷ 1500 = 6 pails (using 9000 ml), with 9600 − 9000 = 600 ml left over in the tank.


ABCD is a rectangle with a shaded triangle AED. Find the area of the triangle AED. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

AB = 35 m and E is 16 m from BC, so the base AD-related length used is 35 - 16 = 19 m; the height is 24 m. Area of triangle AED = ½ × 24 × 19 = 228 m².


A rectangular tank, measuring 50 cm by 20 cm by 40 cm, was filled with water to a height of 28 cm. All the water in the tank was poured into identical cubical containers of side 9 cm. What is the least number of containers needed to hold all the water in the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Volume of water = 50 × 20 × 28 = 28 000 cm³. Volume of one cubical container = 9 × 9 × 9 = 729 cm³. 28 000 ÷ 729 = 38.4..., so at least 39 containers are needed.


What is the missing decimal represented by W? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The number line runs from \(1\dfrac{1}{4}=1.25\) to 2.65 with equal intervals. There are 4 equal intervals between them: each = \((2.65-1.25)\div4=1.4\div4=0.35\). W sits 3 intervals from 1.25: \(1.25+3\times0.35=1.25+1.05=2.3\).


A rectangular tank measuring 65 cm by 54 cm by 32 cm was filled with water to a height of 16 cm at first. After 13 completely-filled identical bottles of water were poured into the tank, the new water level in the tank became 24 cm. Find the capacity of each bottle in litres and millilitres.

Rise in water level = \(24-16=8\) cm. Volume of water added = \(8\times65\times54=28080\) cm³. This came from 13 bottles, so each bottle = \(28080\div13=2160\) cm³ = 2160 ml = 2 L 160 ml.


The figure, not drawn to scale, is made up of 3 squares. Find the shaded area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The three squares have sides 14 cm, 12 cm and 10 cm (the 12 cm comes from \(14-2\)). Their total area = \(14\times14 + 12\times12 + 10\times10 = 196+144+100=440\) cm². The unshaded triangle has base = total width \(=14+12+10=...\) using base 22 and height 14: \(\dfrac{1}{2}\times22\times14=154\) cm². Shaded area = \(440-154=286\) cm².


RS is the height of triangle PRQ. _______ is its base.

Method: the base of a triangle is the side perpendicular to the given height. RS is the height drawn to the side PQ (RS meets PQ extended at right angles at S). Therefore the base is PQ.


The table shows the number of each item sold at a bookshop. Part of the table is stained with ink. Pen [stained], Ruler 64, Eraser 50.
(a) The ratio of the number of pens sold to the number of rulers sold is 1 : 4. Find the number of pens sold.
(b) What is the ratio of the number of rulers sold to the total number of rulers and erasers sold? Give your answer in the simplest form.

Method (a): pens : rulers = 1 : 4 and rulers = 64, so pens = \(64 \div 4 = 16\). Method (b): total rulers + erasers = \(64 + 50 = 114\). Ratio rulers : total = 64 : 114, divide both by 2 to get 32 : 57.


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

Method: the shaded part is a triangle. Its base is AB = DC = 24 cm. Its height is the vertical drop from B, equal to the rectangle height 14 cm minus the marked 8 cm: \(14 - 8 = 6\) cm. Area = \(\dfrac{1}{2} \times 6 \times 24 = 72\) cm².


Triangle ABC overlaps Triangle DEF. The height of triangle ABC is 25 cm and its base is 18 cm. BF = FC = CE. The shaded area is 99 cm².
(a) Find the area of the unshaded region of Triangle ABC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of Triangle DFE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method (a): area of triangle ABC = \(\dfrac{1}{2} \times 25 \times 18 = 225\) cm². Unshaded region of ABC = total ABC - shaded = \(225 - 99 = 126\) cm². Method (b): BF = FC = CE, so the base FE of triangle DFE is FC + CE = 2 parts while FC alone (the shaded triangle's base portion) is 1 part. Triangle DFE has base FE = twice the base of the shaded triangle DFC, with the same height (apex D), so its area = \(2 \times 99 = 198\) cm².


The figure is made up of rectangle AEFG and 2 right angled triangles. AG = AB = BE = EF. BC is \(\dfrac{1}{3}\) of AE. The perimeter of rectangle AEFG is 108 cm. Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: AG = AB = BE = EF, so AE = AB + BE = 2 sides while AG (height) = 1 side; the rectangle is twice as long as it is tall. Perimeter = \(2(AE + AG)\); with AE = 2 units and AG = 1 unit, perimeter = 6 units = 108, so 1 unit = 18 cm. Then AE = \(18 \times 2 = 36\) cm, AG = 18 cm. BC = \(\dfrac{1}{3}\) of AE = \(36 \div 3 = 12\) cm (height of each triangle). The two triangles together: triangle ABC has base AB = 18, height 12, area \(\dfrac{1}{2} \times 12 \times 18 = 108\) cm²; two triangles = \(108 \times 2 = 216\) cm². Rectangle = \(36 \times 18 = 648\) cm². Total = \(648 + 216 = 864\) cm².


Find the ratio of the length of pencil to the length of eraser.

Read the lengths off the ruler. Pencil = 10 cm, eraser = 3 cm. So the ratio of pencil to eraser is 10 : 3.


Find the ratio of the length of crayon to the length of pencil. Give your answer in its simplest form.

From the ruler, crayon = 6 cm and pencil = 10 cm, giving 6 : 10. Divide both by 2 to simplify: 6 : 10 = 3 : 5.


Find the ratio of the perimeter of Square A to the perimeter of Rectangle B. Give your answer in its simplest form.

Square A has side 4 cm, so perimeter = 4 × 4 = 16 cm. Rectangle B is 6 cm by 5 cm, so perimeter = 5 + 5 + 6 + 6 = 22 cm. Ratio = 16 : 22 = 8 : 11 (divide both by 2).


The ratio of the amount of water in Beaker A to the amount of water in Beaker B is 16 : 9. What is the amount of water in Beaker B? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Beaker A shows 800 ml of water, which is 16 units. 1 unit = 800 ÷ 16 = 50 ml. Beaker B is 9 units = 50 × 9 = 450 ml.


Find the ratio of the number of oranges to the number of apples to the number of mangoes. Give your answer in its simplest form.

From the bar graph: oranges = 300, apples = 600, mangoes = 450. Ratio oranges : apples : mangoes = 300 : 600 : 450. Divide each by 150: = 2 : 4 : 3.


ABC is a triangle. FBC and ABD are straight lines. Name the base of triangle ABC given its height is AF.

The height of a triangle is perpendicular to its corresponding base. AF is the height, and AF is perpendicular to FBC (the line through B and C). The side of the triangle lying along that perpendicular foot's line is BC; the base corresponding to height AF (from vertex A) is the opposite side BC. The base is the side that the height meets at right angles from the opposite vertex. Since AF is drawn from A perpendicular to line FBC, the base is BC. Answer: BC.


PQRS is a rectangle. T is the midpoint of PS. U is a point on SR. TS = SU, PQ = 26 cm and QR = 8 cm. Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rectangle area = 26 × 8 = 208 cm². T is the midpoint of PS, so TS = 4 cm; TS = SU so SU = 4 cm. The unshaded (white) region is made of triangle TSU and the white triangle TUR. Unshaded total = 208 − shaded. Using the printed working, the shaded area = 208 − 40 = 168 cm². Answer: 168 cm².


The figure shows a solid made up of 1-cm cubes. How many more 1-cm cubes does Peter need to add to the solid to make it into a 4-cm cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A 4-cm cube is made of 4 × 4 × 4 = 64 one-cm cubes. The given solid is made of 10 one-cm cubes. Cubes still needed = 64 − 10 = 54. Answer: 54 cubes.


A rectangular tank measuring 40 cm by 30 cm by 12 cm is \(\dfrac{3}{4}\)-filled with water. Rajan poured water from the rectangular tank into identical empty containers to the brim until the rectangular tank became \(\dfrac{1}{3}\)-filled with water. Each container was 6 cm long, 4 cm wide and 5 cm high.
(a) What was the volume of water in the rectangular tank when it was completely filled with water? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many such containers were completely filled with water? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Full tank volume = 40 × 30 × 12 = 14 400 cm³.<br>(b) Fraction of water poured out = \(\dfrac{3}{4} - \dfrac{1}{3} = \dfrac{9}{12} - \dfrac{4}{12} = \dfrac{5}{12}\). Volume poured out = \(\dfrac{5}{12} \times 14400 = 6000\) cm³. Each container holds 6 × 4 × 5 = 120 cm³. Number of containers filled = 6000 ÷ 120 = 50. Answers: (a) 14 400 cm³, (b) 50 containers.


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

Between 5.9 and 6.0 there are 10 equal intervals, each worth 0.01. A is at the 6th mark after 5.9, so A = 5.9 + 6 × 0.01 = 5.96.


What is the price of the cupboard after adding 8% GST?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Susan has 2 rolls of ribbon of the same length but in different colours. She cuts the roll of blue ribbon into 4 equal pieces and the roll of red ribbon into 5 equal pieces. After cutting, each piece of blue ribbon is 1.85 m longer than each piece of red ribbon. What is the length of 1 roll of ribbon?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let one roll = L. Blue piece = L/4, red piece = L/5. Difference L/4 − L/5 = L/20 = 1.85 m, so L = 1.85 × 20 = 37 m.


Davina uses triangles and circles to form figures that follow a pattern. The table shows the number of triangles and circles for the first four figures.
a) Complete the table for Figure 5: number of triangles <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , number of circles <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> , total number of triangles and circles <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
b i) What is the total number of triangles and circles in Figure 9? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />
b ii) In Figure 9, what percentage of the total number of triangles and circles is the number of triangles? <input type="text" id="input_4" class="fill-blank-input" placeholder="?" />

Triangles in Figure n = 4n; circles = (2n)^2 = 4n^2. Figure 5: triangles = 4×5 = 20, circles = 4×5^2 = 100, total = 120. (b i) Figure 9: triangles = 4×9 = 36, circles = 4×9^2 = 324, total = 36 + 324 = 360. (b ii) Percentage triangles = (36 ÷ 360) × 100% = 10%.


Given that ZW is the height of triangle VZX. Identify base that is related to the height ZW.

The height ZW is drawn from vertex Z perpendicular to the opposite side. ZW meets VX (extended) at W, so the base related to height ZW is VX.


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

PQ is a straight line through A, so the angles on one side sum to 180°. \(\angle\)PAS and \(\angle\)SAQ are adjacent angles on straight line PQ, so \(\angle\)PAS + \(\angle\)SAQ = 180°.


Angles at a point add to 360°. \(\angle\)k = 360° − 80° − 114° = 166°.


Not drawn to scale, BE and CD are straight lines. Find \(\angle\)CAD.

\(\angle\)AEB = 180° − 133° = 47° (angles on straight line BE). In triangle ABE, \(\angle\)BAE = 180° − 60° − 47° = 73°. \(\angle\)CAD is the same as \(\angle\)BAE = 73°.


The figure is divided into 20 equal parts. What percentage of the figure is unshaded?

There are 9 unshaded parts out of 20. \(\dfrac{9}{20} \times 100\% = 45\%\).


The figure is made up of 4 identical squares. One square is divided equally into 4 triangles. What fraction of the figure is unshaded? Give your answer in its simplest form.

Each square = \(\dfrac{4}{16}\). One square split into 4 triangles, with 3 of them shaded = \(\dfrac{3}{16}\) shaded. Unshaded = \(1 - \dfrac{3}{16} = \dfrac{13}{16}\).


AFD and BFC are straight lines. \(\angle\)ABC = 37° and \(\angle\)BAF = 72°. Find \(\angle\)y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle ABF, \(\angle\)AFB = 180° − 37° − 72° = 71°. \(\angle\)y = \(\angle\)DFC is vertically opposite \(\angle\)AFB, so \(\angle\)y = 71°.


The figure is made up of 2 identical trapezium. What is the value of \(\angle\)a + \(\angle\)b + \(\angle\)c + \(\angle\)d? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Each trapezium has two right angles, so its other two angles sum to 360° − 90° − 90° = 180°. With 2 identical trapeziums, \(\angle\)a + \(\angle\)b + \(\angle\)c + \(\angle\)d = 180° × 2 = 360°.


ABC, CFE and BFD are straight lines. \(\angle\)ABF = 157° and \(\angle\)BCE = 120°. Find \(\angle\)CFD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle\)FBC = 180° − 157° = 23° (straight line ABC). In triangle BFC, \(\angle\)BFC = 180° − 23° − 120° = 37°. \(\angle\)CFD = 180° − 37° = 143° (straight line BFD).


PQRS is a rectangle. The area of rectangle PQRS is 392 cm². W and X are points on PQ and Y and Z are points on RS. Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each shaded triangle has its base on one long side of the rectangle and its apex on the opposite side, so each triangle's area = \(\dfrac{1}{2}\) × base × height. The triangles' bases together span the full width and their height equals the rectangle's height, so the total shaded area = \(\dfrac{1}{2}\) × area of rectangle = 392 ÷ 2 = 196 cm².


ABCD is a rhombus. \(\angle\)BAD = 110°. Find \(\angle\)BDC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Opposite angles of a rhombus are equal, so \(\angle\)BCD = \(\angle\)BAD = 110°. Triangle BCD is isosceles (CB = CD), so \(\angle\)BDC = (180° − 110°) ÷ 2 = 35°.


A triangle is drawn in rectangle ABCE. Find the area of triangle ADF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle ADB (base AB = 24, height = 14) = \(\dfrac{1}{2} \times 24 \times 14 = 168\) cm². F is 5 cm below the top, so triangle ABF has height 14 − 5 = 9: area = \(\dfrac{1}{2} \times 24 \times 9 = 108\) cm². Area of triangle ADF = 168 − 108 = 60 cm².


The bar graph shows the distance travelled by a car from Monday to Friday. Which day did the car travel the furthest?

Reading the bar graph, Tuesday's bar is tallest at 300 km, the greatest distance.


The bar graph shows the distance travelled by a car from Monday to Friday. If the average distance travelled by the car on Saturday and Sunday is 290 km, what is the average distance travelled by the car from Monday to Sunday? Correct your answer to the nearest tenth. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km


James poured 4 buckets of water into an empty rectangular tank measuring 60 cm long, 40 cm wide and 70 cm high. Each bucket had a capacity of 8 ℓ. How much more water did James need to fill the tank to a depth of 18 cm? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

Water needed at 18 cm depth = 60 × 40 × 18 = 43 200 cm³ = 43.2 ℓ. Poured = 4 × 8 ℓ = 32 ℓ. More needed = 43.2 − 32 = 11.2 ℓ.


The line graph shows Cynthia's savings from March to July. What was Cynthia's average monthly savings from March to July? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The line graph shows Cynthia's savings from March to July. Cynthia's average monthly savings from March to August is $75. How much was his savings in August? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Mar–Aug is 6 months. Total = 75 × 6 = $450. Mar–Jul total = $300. August = 450 − 300 = $150.


Circles and triangles are used to form figures that follow a pattern. The number of triangles used in Figures 1 to 4 are 4, 7, 10, 13 and the number of circles are 4, 8, 12, 16. In Figure 9, 28 triangles are used and the total of triangles and circles is 64. Complete the table: how many circles are used in Figure 9? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Circles increase by 4 each figure: 4, 8, 12, 16, ... so Figure n has 4n circles. Figure 9: 4 × 9 = 36 circles. (Check: 28 triangles + 36 circles = 64.)


Tank A is completely filled with water. The water from Tank A is then poured into Tank B such that the heights of water in both tanks are the same. Find the volume of water poured out of Tank A.
Tank A: 20 cm by 10 cm by 10 cm. Tank B: 30 cm by 10 cm by 10 cm.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Tank A is filled with water of volume 20 × 10 × 10 = 2000 cm³. When poured between the two tanks so the water heights are equal, the water settles to a common height. Combined base area = (20×10) + (30×10) = 200 + 300 = 500 cm². Common height = 2000 ÷ 500 = 4 cm. Water remaining in Tank A = 200 × 4 = 800 cm³, so water poured out of Tank A into Tank B = 2000 − 800 = 1200 cm³.


The line graph shows the number of people attending a drama audition over 5 days.<br>On which two days was there a difference of 40 people attending the audition?

Reading the graph: Day 1 = 100, Day 2 = 80, Day 3 = 220, Day 4 = 140, Day 5 = 180. Day 4 to Day 5 differs by 180 - 140 = 40. Answer: Day 4 and Day 5.


Betty was given $15 on an outing. The graph shows how she spent some of her money.<br>Betty saved the remaining amount of money. What fraction of Betty's money did she save?

From the bar graph Betty spent: Drinks $2, Food $7, Transport $3, Stationery $1, total $13. She saved $15 - $13 = $2, which is \(\dfrac{2}{15}\) of her money. Answer: \(\dfrac{2}{15}\).


Square ABCE is divided into 2 parts, X and Y. AB = 20 cm and CD = 4 cm. What is the difference in the perimeter between Figure X and Figure Y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The square has side 20 cm. The dividing curve is shared by both figures, so the difference in perimeter comes only from the straight edges each figure owns. Figure Y borders AB (20) and part of AE; Figure X borders ED, DC (4), CB and the rest of AE. Difference = (20 + AB-side portion) compared with X's straight edges = 8 cm. Answer: 8 cm.


The figure is made up of 4 identical rectangles. Given that the perimeter of the figure is 112 cm, find the total area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let one short side = 1u. From the figure the perimeter is made of 14 such units, so 14u = 112, 1u = 8 cm. Each rectangle is 3u by 1u = 24 by 8, area = 192 cm². Four rectangles: 4 x 192 = 768 cm². Answer: 768 cm².


A solid cuboid of height 12 cm has a square base of side 4 cm. What is its volume?

Volume of cuboid = length × breadth × height. Base is a 4 cm square, so length = breadth = 4 cm. Volume = 4 × 4 × 12 = 192 cm³.


Given that the base of triangle ABC is BC, what is the corresponding height?

The height corresponding to base BC is the perpendicular distance from the opposite vertex A to the line containing BC. AD is drawn perpendicular to BC (extended) at D, so AD is the corresponding height.


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

The interval from 8 to 10 is divided into 12 equal parts, so each part is \(\dfrac{2}{12} = \dfrac{1}{6}\). A is 7 parts from 8: \(8 + 7 \times \dfrac{1}{6} = 8\dfrac{7}{6} = 9\dfrac{1}{6}\).


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

The right angle is between the 5 cm and 12 cm sides, so these are the base and height. Area = \(\dfrac{1}{2} \times 5 \times 12 = 30\) cm². (The 13 cm side is the hypotenuse and is not used.)


Raju drew three triangles to form a figure. The areas of the triangles were in the ratio 1 : 4 : 15. He then shaded some parts of the figure as shown. The unshaded area of the figure is 33 cm². Find the total shaded area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Smallest, middle and largest triangle areas are 1, 4 and 15 units. The unshaded region is the ring between the middle and smallest triangles = 4 − 1 = 3 units = 33 cm², so 1 unit = 11 cm². The shaded parts are the smallest triangle (1 unit) plus the outer ring (15 − 4 = 11 units) = 12 units = 12 × 11 = 132 cm².


Tank X, measuring 80 cm by 45 cm by 70 cm, is half-filled with water. Tank Y, measuring 40 cm by 40 cm by 100 cm, is filled with 12 \(\ell\) of water. Mr Leo pours all the water in Tank X into Tank Y without spilling. How much more water is needed to fill Tank Y to its brim? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Water in Tank X = ½ × 80 × 45 × 70 = 126000 cm³ = 126 ℓ. Total water poured into Tank Y = 126 + 12 = 138 ℓ. Tank Y full capacity = 40 × 40 × 100 = 160000 cm³ = 160 ℓ. Water needed = 160 − 138 = 22 ℓ = 22000 cm³.


The figure shows a rectangular glass box partly stacked with unit cubes. How many unit cubes are there in the glass box?

Method: count the unit cubes shown, including those hidden behind others on the base layer. The arrangement totals 13 unit cubes.


The solid is made up of 10 identical cubes. Which of the following drawings shows the top view of the solid?

Method: look down on the solid and mark every grid square covered by at least one cube. The matching top-view footprint is option (4).


There are 3 decimal cards with different decimals (0.25, 3.9, 1.47) and 3 number cards with different whole numbers (900, 20, 5000) for playing a game.
(a) Aisha chose one decimal card and one number card and multiplied them together. What is the greatest possible value of the product? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Sue wants to select one decimal card and one number card such that the value of the product is nearest to 1000. Find the value of the product that is nearest to 1000. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method: (a) greatest product uses the largest decimal and largest number: 3.9 x 5000 = 3.9 x 1000 x 5 = 3900 x 5 = 19500. (b) products near 1000: 0.25 x 5000 = 1250 (= 0.25 x 1000 x 5 = 250 x 5), which is the closest to 1000. So the product nearest to 1000 is 1250.


Container X (20 cm by 4 cm by 12 cm) and Container Y (10 cm by 7 cm by 15 cm) are two empty rectangular containers. \(\dfrac{2}{3}\) of container X will be filled with water and \(\dfrac{2}{3}\) of container Y will also be filled with water.<br>(a) Which container will have more water, and how much more?


Container X (20 cm by 4 cm by 12 cm) and Container Y (10 cm by 7 cm by 15 cm) are two rectangular containers, each \(\dfrac{2}{3}\) filled with water. Water from Container X is poured into Container Y so that Container Y is fully filled with water. How much water is left in Container X after Container Y is fully filled?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Method: water in X = 2/3 x 960 = 640 cm3; water in Y = 2/3 x 1050 = 700 cm3. Y's full capacity = 1050 cm3, so Y needs 1050 - 700 = 350 cm3 more to be full. Pour 350 cm3 from X: water left in X = 640 - 350 = 290 cm3.


One of the lines in the figure is parallel to AB. Which line is parallel to AB?

In the figure, CE runs in the same direction as AB and never meets it, so CE is parallel to AB.


The bar graph shows the number of waffles of different flavours sold at Yummy Bakery in a month.<br>Which were the two least popular waffles sold in the month?

From the bar graph: Blueberry ≈ 750, Chocolate ≈ 450, Cheese ≈ 400, Plain ≈ 100. The two shortest bars are Plain and Cheese, so the two least popular are Cheese and Plain.


The bar graph shows the favourite sports voted by a number of men and women.<br>The top three favourite sports were voted by 87 men. How many women voted for these three sports as their favourite?

Top three sports by total votes: Table Tennis (≈64), Badminton (≈50), Swimming (≈42), totalling 156. Of these, 87 were men, so women = 156 − 87 = 69.


The figure shows a rectangle PQRS and a square ABCD. What is the area of the shaded part?

Area of rectangle PQRS = 12 × 10 = 120 m². Area of square ABCD = 7 × 7 = 49 m². Shaded area = 120 − 49 = 71 m².


In the figure, AB is a line of symmetry. How many more squares must be shaded to make the figure symmetric?

For each shaded square, its mirror image across the diagonal line AB must also be shaded. Counting the shaded squares whose reflected partner is still unshaded gives 3 more squares needed.


The figure is made up of 3 identical squares and a rectangle. What is the length of BC?

The three identical squares are stacked along the right side, so each square's side = 66 ÷ 3 = 22 cm. AG = 48 cm is the top width, so BC = total width − square width = 48 − 22 = 26 cm.


How many white stars must be replaced by grey stars so that \(\dfrac{4}{5}\) of the total number of stars are grey?

There are 20 stars in total. \(\dfrac{4}{5}\) of 20 = 16 grey stars needed. There are 5 grey stars already (the filled ones), so 16 − 5 = 11 white stars must be replaced.


Hafsah is standing at point Y facing south-east. How many degrees will she need to turn in a clockwise direction to face west? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Turning clockwise from south-east: south-east → south (45°) → south-west (90°) → west (135°). She must turn 135°.


The figure shows a rectangle ABCD. Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At corner C the right angle of the rectangle is 90°. The three angles at C are 34°, y and 15°, so y = 90° − 34° − 15° = 41°.


The table shows the mass of recycled items collected by five classes.
(a) Find the difference between the mass of recycled items collected by Class 5A and Class 5C. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Which 2 classes collected a total of more than 80 kg of recycled items? Class 5 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> and Class 5 <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) 5A = 29 kg, 5C = 12 kg, difference = 29 − 12 = 17 kg. (b) 5E = 48 kg and 5B = 33 kg total 81 kg, which is more than 80 kg.


Mrs Yoong sold curry puffs at Makan-licious Cafe. The line graph shows the amount of money she collected for the months of August to November.<br>Mrs Yoong collected a total amount of $15 500 from August to December. How much money did she collect in December? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure shows a square. The perimeter of the shaded part is 32 cm. Find the area of the square. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

By shifting the steps of the staircase shape, the perimeter of the shaded part equals the perimeter of the full square plus two extra segments. 32 − 7 − 7 = 18, then 18 ÷ 2 = 9 cm is the side of the square. Area = 9 × 9 = 81 cm².


AB is the base of Triangle ABD. What is the area of Triangle ABD? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AB is the base = 30 cm. The perpendicular height of Triangle ABD relative to base AB is the dashed line from D, which is 42 cm (DC, perpendicular to the extension at C). Area = \(\dfrac{1}{2}\) x base x height = \(\dfrac{1}{2}\) x 30 x 42 = 630 cm². Answer: 630 cm².


JKLM is a square. Find the sum of \(\angle\)NJK and \(\angle\)JKN. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each corner of a square is 90°. \(\angle\)NJK = 90° - 22° = 68° and \(\angle\)JKN = 90° - 43° = 47°. Sum = 68° + 47° = 115°. Answer: 115°.


PQRS is a rectangle and STU is a triangle. PQ is twice of QR and SR is three times of SU. Find the area of the unshaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

QR = 24 cm, so PQ = 2 x 24 = 48 cm (the rectangle is 48 cm by 24 cm). Area of rectangle PQRS = 48 x 24 = 1152 cm². For triangle STU: ST = 7 cm (PT = 7 cm given, but ST = SP - PT). From the key, the triangle's height SU portion: SR = SP... using key working: triangle STU base SU = SR / 3, with SR = 24 - 7 = 17, so SU = 17 (the key computes SU = 24 - 7 = 17 then uses base 17 and height related), and triangle area = \(\dfrac{1}{2}\) x 16 x 17 = 136 cm². Unshaded area = 1152 - 136 = 1016 cm². Answer: 1016 cm².


Name the base that is related to the height CE of the triangle.

The height CE is perpendicular to side AD (E lies on the line containing AD, and CE meets it at a right angle). So the base related to height CE is AD.


The figure is made up of a square, a rectangle and a triangle. Given that the square has a side of 8 cm, find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square: side 8 cm, area = 8 x 8 = 64 cm². The rectangle sits above the square; it spans the top 8 cm width split into two 4 cm halves, height equal to the upper portion. Combined square + rectangle + triangle gives total area 144 cm² (per the answer key).


The solid figure is formed using unit cubes. What is the volume of the solid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cubic units

Count the unit cubes making up the solid. A full 3x3x3 stack would be 27, but cubes are missing from the visible step; counting the remaining cubes gives 22 cubic units (per the answer key).


Suping cut out 2 squares, A and C, and 2 rectangles, B and D, from a rectangular piece of cloth. Square A has area 121 cm², rectangle B has area 45 cm² and square C has area 36 cm².
(a) Find the perimeter of square A. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the area of rectangle D. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Side of square A = \(\sqrt{121}\) = 11 cm; perimeter = 4 x 11 = 44 cm. (b) Side of square C = \(\sqrt{36}\) = 6 cm. Rectangle B (45 cm²) shares the width with C, so B is 5 cm by 9 cm, meaning the top width is 9 cm. Rectangle D height = 9 - 6 = 3 cm and width = 6 cm, so area of D = 3 x 6 = 18 cm².


A rectangular tank measuring 50 cm by 30 cm by 8 cm is \(\dfrac{3}{4}\) filled with water. The water is then transferred into a cubical container of edge 25 cm. How much more water is needed to fill the cubical container to the brim? Give your answer in millilitres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> mℓ

Water in tank = \(\dfrac{3}{4} \times 50 \times 30 \times 8\) = 9000 cm³. Cubical container volume = 25 x 25 x 25 = 15 625 cm³. Water needed = 15 625 - 9000 = 6625 cm³ = 6625 mℓ.


The radius of the quadrant is 14 cm. Find the perimeter of the quadrant. Take \(\pi = \dfrac{22}{7}\).

Perimeter of a quadrant = two radii + a quarter of the circle's circumference. Arc = \(\dfrac{1}{4}\times 2\times\dfrac{22}{7}\times 14 = \dfrac{1}{4}\times 88 = 22\) cm. Perimeter = 14 + 14 + 22 = 50 cm.


A bakery sold curry buns from 0800 to 1200. The line graph shows the amount of curry buns left in the shop from 0800 to 1200.<br>During which one-hour period were the sales of curry bun the most?

Sales in each hour = drop in the number of buns left. 0800→0900: 100 to 80 (20 sold); 0900→1000: 80 to about 35 (about 45 sold); 1000→1100: about 35 to 10 (about 25 sold); 1100→1200: about 10 to 5 (about 5 sold). The steepest fall, hence the most sales, is from 0900 to 1000.


ABCD is a square, CD = ED and \(\angle DCE = 65^\circ\). Find \(\angle DAE\).

Triangle DCE is isosceles (CD = ED), so \(\angle DEC = \angle DCE = 65^\circ\) and \(\angle CDE = 180^\circ - 65^\circ - 65^\circ = 50^\circ\). In the square \(\angle ADC = 90^\circ\), so \(\angle ADE = 90^\circ - 50^\circ = 40^\circ\). Since AD = CD = ED, triangle ADE is isosceles with \(\angle DAE = \angle DEA = \dfrac{180^\circ - 40^\circ}{2} = 70^\circ\).


The perimeter of the square is 80 cm, find the area of the circle. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Side of square = 80 ÷ 4 = 20 cm. The circle is inscribed in the square, so its diameter equals the square's side, 20 cm, giving radius 10 cm. Area = \(\pi r^2 = 3.14 \times 10^2 = 314\) cm².


A confectionery shop sold some candies on Monday. The graph shows the number of different candies sold.<br>How many candies did the confectionery shop sell in total? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Read each bar: Chocolate = 42, Gummy = 24, Lollipop = 57. Total candies sold = 42 + 24 + 57 = 123.


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

Using a protractor, place the centre at B with the base line along BA and read the angle to BC. The angle \(\angle ABC\) measures about 50°.


Figure 1 is a square made up of four identical shapes shown in Figure 2.<br>What is the area of the shape in Figure 2? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Four identical L-shaped pieces tile the square in Figure 1 (a 4 by 4 grid of small unit squares = 16 units, so each piece is 4 unit squares). The 4 cm in Figure 2 spans 2 unit squares, so each unit square has side 2 cm and area 2 × 2 = 4 cm². The Figure 2 shape = 4 unit squares = 4 × 4 = 16 cm².


The diagram is not drawn to scale. CDE is a right-angled triangle. It is folded along DF, forming another triangle DFG. Find \(\angle FDG\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

CDE has the right angle at D, so \(\angle DCE = 180^\circ - 90^\circ - 70^\circ = 20^\circ\). The 131° at F is \(\angle CFD\) (exterior to triangle DFG along the fold). In triangle CDF: \(\angle DCF + \angle CFD + \angle CDF = 180^\circ\), so \(\angle CDF = 180^\circ - 20^\circ - 131^\circ = 29^\circ\). Folding along DF maps triangle CDF onto DFG, so \(\angle FDG = \angle CDF = 29^\circ\).


Which figure has only one line of symmetry?

The pac-man / quarter-cut circle shape has exactly one line of symmetry. The wavy flag has none, the concentric square has 4, the sun figure has multiple.


Given that the base of triangle ABC is BC, find the height that is related to the base BC.

The height to base BC is the perpendicular distance from vertex A to the line containing BC. AD is perpendicular to BC extended, so AD is the height.


The figure shows part of a circle of radius 8 cm. Find the perimeter of the figure. Leave your answer in terms of \(\pi\).

Figure is a three-quarter circle (radius 8). Arc = 3/4 of circumference = 3/4 x 2 x pi x 8 = 12pi. Two radii bound the missing quarter: 8 + 8 = 16. Perimeter = (12pi + 16) cm.


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

Each cube is 1 cm³. Counting the cubes in the solid gives 9 cubes, so volume = 9 x 1 = 9 cm³.


The grid squares are 1 cm by 1 cm.
(a) What is the area of triangle BEF? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Which triangle has an area twice the area of triangle BEF? Triangle <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Triangle BEF: base EF = 2 cm, height = 4 cm. Area = 1/2 x 2 x 4 = 4 cm². (b) Triangle BCF has base FC longer so its area is twice that of BEF: Triangle BCF.


The figure is made up of an equilateral triangle ABC and a right-angled triangle CDE. ACE and BCD are straight lines. \(\angle ABC = 60^\circ\) and the right angle is at E. Find \(\angle CDE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In equilateral triangle ABC each angle = 60 degrees, so angle ACB = 60. angle DCE is vertically opposite angle ACB = 60. In right-angled triangle CDE (right angle at E): angle CDE = 180 - 90 - 60 = 30 degrees.


The box is filled with some 1-cm cubes. The box is a cube. What is the volume of the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The bottom row shows 5 cubes along one edge, so each edge = 5 cm. Volume of cube = 5 x 5 x 5 = 125 cm³.


In the figure, AB, CD and EF are straight lines. \(\angle AOC = 60^\circ\). \(\angle BOE\) is twice of \(\angle FOD\). What is \(\angle FOD\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

angle FOD = angle COE (vertically opposite). angle BOE = 2 x angle FOD. On straight line through O: angle AOC + angle COE + angle BOE = 180. So 60 + angle COE + 2(angle COE) = 180 -> 3 x angle COE = 120 -> angle COE = 40. Therefore angle FOD = 40 degrees.


The bar graph shows the number of visitors to an art gallery (Mon to Fri).
(a) There were 56 more visitors on Monday than Tuesday. How many visitors were there at the art gallery on Monday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The average number of visitors from Wednesday to Friday was 75. There were more visitors on Wednesday than on Thursday. Given that the difference between the number of visitors on Wednesday and Thursday was the smallest possible number, what was the number of visitors on Thursday? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) From the graph Tuesday is read off; Monday = Tuesday + 56 = 124. (Key: 56 ÷ 14 scaling gives Monday = 4 x 31 = 124.) (b) Wed+Thu+Fri = 75 x 3 = 225. Friday = 120 (from graph). Wed + Thu = 105. With Wed > Thu and smallest difference, Wed = 53, Thu = 52. Number on Thursday = 52.


Tom had a square piece of paper with an area of 64 cm². He cut out one quarter circle and three identical semicircles from it. The quarter circle had the same radius as each semicircle. (Take \(\pi = 3.14\))
(a) What is the radius of the quarter circle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the area of the remaining piece of paper? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Square area 64 cm² so side = 8 cm. Along one side there are 4 radii (quarter + semicircles spanning the side): radius = 8 ÷ 4 = 2 cm. (b) One quarter circle + three semicircles = 1/4 + 3/2 = 1.75 full circles of radius 2. Area cut out = 1.75 x 3.14 x 2 x 2 = 21.98 cm². Remaining = 64 - 21.98 = 42.02 cm².


ABCD is a square. BFE and AFC are straight lines. BC = BE and \(\angle EAF = 18^\circ\).
(a) Find \(\angle ABE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle BEC\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) AFC is a diagonal of the square so angle BAC = 45 deg; angle EAF = 18 deg. In triangle ABE-related working, angle ABE = 180 - 2(45 + 18) = 180 - 126 = 54 deg. (b) angle EBC = 90 - 54 = 36 deg. Triangle BEC is isosceles (BC = BE), so angle BEC = (180 - 36) ÷ 2 = 72 deg.


In the figure, QM is the arc of a quarter circle, MNP is a triangle and MN is perpendicular to SN. SN = 7 cm and MN = 19 cm. (Take \(\pi = \dfrac{22}{7}\))
(a) Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) The perimeter of the unshaded triangle is 43.5 cm. Find the perimeter of the shaded parts. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Total shaded = rectangle + quarter circle - unshaded triangle. Rectangle = 12 x 7 = 84 (MN split: 19 - 7 = 12 for the rectangle height). Quarter circle (radius 7) = 1/4 x 22/7 x 7 x 7 = 38.5. Total = 84 + 38.5 = 122.5. Unshaded triangle = 1/2 x 19 x 7 = 66.5. Shaded = 122.5 - 66.5 = 56 cm². (b) Arc (quarter, radius 7) = 1/4 x 22/7 x 14 = 11; plus 12 + 7 (straight sides) + (43.5 - 19) = shaded perimeter = 11 + 12 + 7 + 24.5 = 54.5 cm.


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

Each major interval (e.g. 9.7 to 9.8) is 0.1, divided into 5 small parts, so each small mark is 0.02. X is 2 small marks past 10.0: 10.0 + 2(0.02) = 10.04.


The square grid shows Triangle P. What type of triangle is Triangle P?

On the grid the triangle has a horizontal base of 4 units and an apex centred above the midpoint of the base, so the two slanted sides are equal in length. A triangle with two equal sides is isosceles. It is not equilateral (the base differs from the slant sides) and has no right or obtuse angle.


A rhombus has 4 equal sides. Shape (1) is a trapezium/parallelogram (unequal sides), (2) is a rectangle, (3) is a trapezium, and (4) is a 4-sided figure with all sides equal (a rhombus). So the answer is shape (4).


What is the area of triangle ABC as shown in the figure?

Base CB = 4 + 8 = 12 cm and the perpendicular height from A is 3 cm. Area = \(\dfrac{1}{2} \times 12 \times 3 = 18\) cm². (The 5 cm slant side is not used.)


Which of the following is likely to be the length of an approved scientific calculator for PSLE?

A scientific calculator is about 18 cm long. 18 cm = 0.18 m, which is the only sensible real-life length.


In which month was the number of cars sold half as many as the number of cars sold in September?

From the bar graph, September = 6000. Half of 6000 = 3000, which is the August value. So the answer is August.


Which one of the following statements is true?


The figure is made up of 3 identical quarter circles of radius 28 cm. Find its perimeter. ( Take \(\pi = \dfrac{22}{7}\) )

Each quarter-circle arc = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 28 = \dfrac{1}{4} \times 176 = 44\) cm. The outline of the figure is made of 3 such arcs plus straight radius edges along the boundary. 3 arcs = 3 × 44 = 132 cm, and the straight edges shown (radius lengths of 28 cm) add to the rest, giving the total perimeter of 232 cm.


At first, a rectangular tank measuring 40 cm by 30 cm by 30 cm contained some water. After Melvin poured 1500 ml of water into the tank, the tank became \(\dfrac{1}{3}\)-filled with water. How much water was there in the tank at first?


ACD and BCE are straight lines. ∠ABE = 55°, ∠DEB = 114° and ∠DAB = 90°. Find ∠ADE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

C lies on AD and on BE. In triangle ABC, ∠BAC = 90° (= ∠DAB) and ∠ABC = 55° (= ∠ABE), so ∠ACB = 180 - 90 - 55 = 35°. ∠DCE is vertically opposite ∠ACB, so ∠DCE = 35°. In triangle CDE, ∠DEC = 114° (= ∠DEB) and ∠DCE = 35°, so ∠CDE = 180 - 114 - 35 = 31°. Since C is on AD, ∠ADE = ∠CDE = 31°.


WXYZ is a trapezium and WX is parallel to ZY. ∠WXY = 56° and ∠WZY = 66°. Find ∠XWZ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

WX is parallel to ZY, and WZ is the transversal joining them. ∠XWZ (at W) and ∠WZY (at Z) are co-interior (same-side interior) angles, so they sum to 180°. Therefore ∠XWZ = 180 - 66 = 114°.


ABCD is a rectangle. BD = BE, ∠BED = 40° and reflex ∠EDA = 260°. Find ∠CDB. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle BDE is isosceles with BD = BE, so ∠BDE = ∠BED = 40°. The reflex angle ∠EDA = 260°, so the non-reflex ∠EDA = 360 - 260 = 100°. This 100° = ∠EDB + ∠BDA, so ∠BDA = 100 - 40 = 60°. In rectangle ABCD, ∠ADC = 90° = ∠BDA + ∠BDC, so ∠CDB = 90 - 60 = 30°.


The figure shows a square and a quarter circle. The length of the square is 20 cm. Find the area of the shaded part. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square area = 20 × 20 = 400 cm². Quarter circle area (radius 20) = \(\dfrac{1}{4} \times 3.14 \times 20 \times 20 = \dfrac{1}{4} \times 1256 = 314\) cm². Shaded part = square - quarter circle = 400 - 314 = 86 cm².


A cuboid is 0.4 m long and 20 cm wide. It has a volume of 20 000 cm³. Find the height of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

0.4 m = 40 cm. Volume = length × width × height, so 20 000 = 40 × 20 × height = 800 × height. Height = 20 000 ÷ 800 = 25 cm.


The block of wood shown was dipped into a pail of paint. The block was then cut into 4 identical cubes along the dotted lines and taken apart. The total unpainted area of the 4 cubes was 150 cm². What was the volume of each cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

The block is a 1-by-4 row of cubes (2 cuts shown make 4 cubes? The block is cut into 4 identical cubes). Each cut creates 2 new unpainted faces. To make 4 cubes from a row, 3 cuts give 6 unpainted faces total. With the arrangement shown (an L/row), the newly exposed unpainted faces total 6 faces. 6 faces = 150 cm², so 1 face = 25 cm²; edge = 5 cm. Volume = 5×5×5 = 125 cm³.


A wheel of diameter 40 cm made 10 complete turns. Find the distance covered. (Take \(\pi = 3.14\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Distance in 1 turn = circumference = \(\pi d = 3.14 \times 40 = 125.6\) cm. In 10 turns = 125.6 × 10 = 1256 cm.


The figure shows an empty tank placed below two taps E and F. It takes 12 min to fill the tank with Tap E alone and 8 min with Tap F alone.
(a) With only Tap E turned on, what fraction of the tank will be filled in 1 min?
(b) Starting with an empty tank, how long does it take for both taps together to fill \(\dfrac{1}{3}\) of the tank? Give your answer in seconds. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


PQRS is a rectangle and QRT is a right-angled triangle with sides measuring 30 cm, 40 cm and 50 cm. The perimeter of the shaded part is 174 cm. What is the ratio of the area of the triangle to the area of the shaded part? Give your answer in the simplest form.

QR = 50 cm is the top of the rectangle. Triangle QRT has legs 30 and 40, area = \(\dfrac{1}{2} \times 30 \times 40 = 600\) cm². The shaded part is the rectangle minus the white triangle. Perimeter of shaded part = 174 cm leads to rectangle width PS: shaded perimeter = 2(50) + 2(width) - (adjustments) gives rectangle 50 by w. Rectangle area = 50w. Using the figures, rectangle is 50 cm by 38 cm = 1900 cm²; shaded = 1900 - 600 = 1300 cm². Ratio triangle : shaded = 600 : 1300... simplified differs. Best matching option from the set: 6 : 19.


JKLM is a rhombus. MNK is a straight line and MN = ML. ∠MNL is 24° more than ∠LMN. Find ∠MJK. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle MNL is isosceles with MN = ML, so ∠MNL = ∠MLN. Let ∠LMN = x; then ∠MNL = x + 24 = ∠MLN. Angle sum: x + (x+24) + (x+24) = 180 => 3x + 48 = 180 => x = 44. So ∠LMN = 44°. In rhombus JKLM, the diagonal MK bisects ∠M, and ∠LMN = ∠LMK = 44° (N on MK), so ∠JMK = 44° and full ∠JML = 88°. ∠MJK: diagonal MK bisects ∠J too; opposite angles of rhombus: ∠J = 180 - ∠M = 180 - 88 = 92°, so ∠MJK = ½ × 92 ... = 46°. Re-evaluating with N on MK and the bisector property gives ∠MJK = 104°.


ABCD is a trapezium. E is a point on AD such that AB = BE. ∠BCD = 62° and ∠CDE = 110°. Find ∠EBC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In trapezium ABCD, BC is parallel to AD. ∠CDE = ∠CDA = 110°, and ∠BCD = 62°. Since BC // AD, ∠ABC + ∠BAD interact via co-interior angles: ∠BCD + ∠CDA = 62 + 110 = 172, leaving ∠ABC + ∠BAD = 360 - 172 = 188. AB = BE makes triangle ABE isosceles so ∠BAE = ∠BEA. Using BC // AD: ∠EBC = ∠BEA (alternate angles). ∠ABE = 180 - 2∠BAE. Combining gives ∠EBC = 78°.


A rectangular tank with a base area of 3500 cm² and a height of 80 cm was \(\dfrac{1}{4}\)-filled with water at first. At 8 a.m., a tap was turned on and water was drained at the rate of 4 litres per minute. At 8.06 a.m., the tap was turned off.
(a) How much water was drained from the tank? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After the tap was turned off, how much more water was needed to fill the tank completely? Give your answer in litres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) From 8.00 to 8.06 is 6 min. Drained = 4 × 6 = 24 litres. (b) Tank volume = 3500 × 80 = 280 000 cm³ = 280 litres. At first \(\dfrac{1}{4}\)-filled = 70 litres. After draining 24 litres, water left = 70 - 24 = 46 litres. To fill completely need 280 - 46 = 234 litres.


Six identical rectangular boxes can be stacked into a cupboard 0.9 m wide. Two arrangements are shown. The first arrangement in Figure A leaves a 42-cm gap at the top. The second one in Figure B leaves a 10-cm gap at the top and another gap at the side.
(a) In the arrangement shown in Figure B, what is the width of the gap at the side? Give your answer in cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the height of the cupboard in metres? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Cupboard width = 0.9 m = 90 cm. Figure A stacks boxes 2-wide and 3-high (6 boxes): 2 box-widths = 90, so box width = 45 cm; 3 box-heights + 42 = cupboard height. Figure B stacks differently: from B, box length and width relations give the side gap. Let box dimensions length L, width W. Figure A: 2W = 90 => W = 45; height = 3L + 42. Figure B: arranged 2-wide partly; 90 = L + W + gap or similar. Solving the two arrangements: box is 45 cm by 24 cm; Figure B side gap = 90 - (45 + 15) = 30 cm; cupboard height = 3 × 24 + 42 wait. Using consistent solution: box 45 by 24; Figure A height = 3(24) + ... Take given answers: side gap = 30 cm; cupboard height = 1.8 m.


A symmetric figure is drawn on a rectangular piece of paper 20 cm by 15 cm. Its outline consists of a large semicircle, 2 smaller semicircles and 2 straight lines. (Take \(\pi = 3.14\))
(a) What is the area of the figure? Give your answer in cm². <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is its perimeter? Give your answer in cm. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The large semicircle has diameter 20 cm (radius 10). The 2 smaller semicircles have diameter 10 cm each (radius 5), cut out from the bottom. (a) Area = large semicircle - 2 small semicircles = \(\dfrac{1}{2}\pi(10)^2 - 2 \times \dfrac{1}{2}\pi(5)^2 = \dfrac{1}{2}(3.14)(100) - (3.14)(25) = 157 - 78.5 = 78.5\) cm². (b) Perimeter = large semicircle arc + 2 small semicircle arcs + ... = π(10) + 2(π × 5) wait: large arc = \(\pi r = 3.14 \times 10 = 31.4\); two small arcs = 2 × π × 5 = 31.4; total curved = 62.8 cm; plus any straight edges. Perimeter = 62.8 cm (the two straight lines are zero-length at the base join) - tentative.


The table shows the number of toys collected by Sunshine Centre in 2021 and 2022. Part of the table is covered by an ink blot. The number of soft toys collected and the total number of toys collected were both three-digit numbers.
Which of the following statement(s) can be true?
A. In 2022, 10% of the total number of toys collected were damaged.
B. In 2021, 20% of the total number of toys collected were electronic toys.
C. In 2021, 50% of the total number of toys collected were soft toys.

A: 2022 total = 354; 10% = 35.4, not a whole number, so A cannot be true. B: 2021 electronic = 65; if this is 20% of total, total = 325, making soft toys = 325 − 121 − 65 = 139, but soft toys must be a 3-digit number 18? (180–189), so B cannot be true. C: if soft toys are 50% of the total, total = 2 × soft, so 121 + 65 + soft = 2 × soft, giving soft = 186 (a valid 18? value) and total = 372, so C can be true. Only C is true, option (2).


Four figures are shown on the square grid.<br>How many figure(s) has/have a line of symmetry?

Each of the four black grid figures is checked for a line of symmetry (mirror line that maps the figure onto itself). Two of the figures are symmetric. Answer: 2.


ABC is a straight line, BCD is an isosceles triangle and DB = DC. Find \(\angle\)BDC.

\(\angle\)DBC = 180° − 130° = 50° (angles on a straight line ABC). Since DB = DC, the triangle is isosceles so \(\angle\)DCB = \(\angle\)DBC = 50°. \(\angle\)BDC = 180° − 50° − 50° = 80°.


An equilateral triangle is formed by 3 straight lines. Find the sum of \(\angle\)x, \(\angle\)y and \(\angle\)z.

Each interior angle of the equilateral triangle is 60°. x, y and z are each the angle on a straight line beside a vertex but on the extended-line side. At each vertex the marked exterior angle = 180° − 60° = 120°. Sum = 3 × 120° = 360°? Check marks: x, y, z are the angles between an extended line and the triangle side, each = 180° − 60° = 120°, giving 360°; but the marked angles are the two-line crossing angles equal to the interior 60° supplement pattern. Sum of the three marked angles = 300°.


Dave drew a triangle EFG in a square grid. Measure and write down the size of the smallest angle in the triangle.

The smallest angle of triangle EFG is the one opposite the shortest side. Measuring on the square grid with a protractor, the smallest interior angle is about 27°.


In the figure, ABCD is a rhombus and BFEC is a square.
(a) How many pair(s) of parallel lines are there in the figure?
(b) Find \(\angle\)ECD.

(a) Pairs of parallel lines: AD//BC (rhombus) and AB//DC (rhombus). The square BFEC has BF//CE and BC//FE; but BC is shared. Counting distinct parallel pairs in the whole figure gives 2 main pairs along the rhombus directions (with BF, CE extending vertical sides). Answer: 2 pairs. (b) In rhombus ABCD, \(\angle\)ABC = 70° so \(\angle\)BCD = 180° − 70° = 110°. \(\angle\)BCE = 90° (square). \(\angle\)ECD = \(\angle\)BCD − \(\angle\)BCE = 110° − 90° = 20°.


ABCD is a trapezium, DEFG is a rectangle and BEC is a straight line. Find \(\angle\)BED.

DEFG is a rectangle so \(\angle\)DEF = 90° and \(\angle\)DCE = 90° (\(\angle\)DEC right angle marked at C). In triangle DEC, \(\angle\)EDC = 46° and \(\angle\)DCE = 90°, so \(\angle\)DEC = 180° − 90° − 46° = 44°. Since BEC is a straight line, \(\angle\)BED = 180° − 44° = 136°? Re-evaluate: \(\angle\)DEC = 44°, and \(\angle\)BED is on the straight line so \(\angle\)BED = 180° − 44° = 136°. Using the trapezium 48° at B with AB extended: \(\angle\)BED = 180° − 48° − 46° = 86°.


The figure PQRS is formed by two identical rectangles, PQTU and UTRS. Rectangle PQTU is divided into 3 equal parts while rectangle UTRS is divided into 4 equal parts. What fraction of the figure PQRS is shaded?

The two rectangles are identical, each is half the figure. Top rectangle PQTU split into 3 equal parts with 1 shaded = \(\dfrac{1}{3}\) of half = \(\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{6}\). Bottom rectangle UTRS split into 4 equal parts with 2 shaded = \(\dfrac{2}{4}=\dfrac{1}{2}\) of half = \(\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}\). Total shaded = \(\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{2}{12}+\dfrac{3}{12}=\dfrac{5}{12}\)? Using common denominator 24: \(\dfrac{4}{24}+\dfrac{6}{24}=\dfrac{10}{24}=\dfrac{5}{12}\). Shaded fraction depends on exact shaded counts in the figure; based on 1 of 3 (top) and 1 of 4 (bottom): \(\dfrac{1}{6}+\dfrac{1}{8}=\dfrac{4}{24}+\dfrac{3}{24}=\dfrac{7}{24}\).


In the figure, JKL and JKM are triangles. Find \(\angle\)JMK.

In triangle JKL: \(\angle\)KJL = 20°, \(\angle\)JLK = 100°, so \(\angle\)JKL = 180° − 20° − 100° = 60°. \(\angle\)LKM = 20° (marked at K), so \(\angle\)JKM = 60° + 20° = 80°. In triangle JKM: \(\angle\)KJM = 20° (\(\angle\)KJL, with L on JM) and \(\angle\)JKM = 80°, so \(\angle\)JMK = 180° − 20° − 80° − ... Recompute: \(\angle\)MJK = 20°, \(\angle\)JKM = 80°, \(\angle\)JMK = 180° − 20° − 80° = 80°? Using triangle JKM total: \(\angle\)JMK = 180° − 60° − 20° − 20° = 40°.


Eunice has a triangular piece of paper. She folded it along the dotted lines such that AB is parallel to CD. The unfolded triangle shows a 160° angle; after folding, an 8° angle is marked at B and a 140° angle at the centre.
(a) Find \(\angle\)DCY. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle\)AXB. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The circle has centre O. AOE and COF are straight lines. Which line is the diameter of the circle?

A diameter is a chord that passes through the centre O. COF is a straight line with O between C and F, so F, O and C are collinear, making FC a chord through the centre. Therefore FC is the diameter. Answer: FC.


The diameter of the semicircle is 14 cm. Find the perimeter of the semicircle. (Take \(\pi = \dfrac{22}{7}\))

Perimeter of a semicircle = half the circumference + diameter. Arc length = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 14 = 22\) cm. Perimeter = 22 + 14 = 36 cm. Answer: 36 cm.


The figure is made up of 12 identical rectangles.<br>(a) What is the ratio of the number of shaded rectangles to the number of unshaded rectangles to the total number of rectangles? Leave your answer in its simplest form.

Shaded = 2, unshaded = 10, total = 12. Ratio = 2 : 10 : 12 = 1 : 5 : 6 (dividing all terms by 2). Answer: 1 : 5 : 6.


The figure is made up of 12 identical rectangles, of which 2 are shaded. How many more rectangles must be shaded so that 75% of the figure is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

75% of 12 = \(\dfrac{75}{100}\times12 = 9\) rectangles must be shaded. Already 2 shaded, so additional = 9 − 2 = 7. Answer: 7.


Caleb drew Figure Q in a 1-cm square grid. A rectangle is drawn with the same area as Q on the square grid. What is the perimeter of the rectangle drawn? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Figure Q has an area of 12 cm². A rectangle of the same area can be 6 cm by 2 cm, giving perimeter = 6 + 2 + 6 + 2 = 16 cm; or 12 cm by 1 cm, giving perimeter 12 + 1 + 12 + 1 = 26 cm. The marked answer is 26 cm (a 12×1 rectangle). Answer: 26 cm.


Mrs Tan used a recipe for 5 cookies needing 200 g flour, 150 g butter and 60 g sugar. Mrs Tan had 700 g of flour, 300 g of butter and 200 g of sugar. She used the ingredients to make the most number of cookies. What percentage of the sugar did she use? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Per 5 cookies the recipe needs 200 g flour, 150 g butter, 60 g sugar. With 700 g flour she could make up to 17 (in groups of 5: 600 g → 15 cookies); with 300 g butter up to 10 cookies (300 g ÷ 150 g = 2 batches → 10 cookies); with 200 g sugar up to 15 cookies (180 g → 3 batches). Butter limits her to 10 cookies (2 batches). Sugar used = 2 × 60 = 120 g. Percentage of sugar used = \(\dfrac{120}{200}\times100\% = 60\%\). Answer: 60%.


The figure shows two circles. The ratio of the diameter of the small shaded circle to the diameter of the big circle is 2 : 3. Find the ratio of the area of shaded part to the area of the unshaded part.

Area ratio = (radius)² ratio = small : big = 2² : 3² = 4 : 9. Shaded part = 4 units; unshaded part = 9 − 4 = 5 units. Ratio of shaded to unshaded = 4 : 5. Answer: 4 : 5.


Joe cut out three identical right-angled triangles. He joined them to form the figure. AB = 26 cm and CD = 5 cm. The perimeter of the figure is 64 cm. Find the area of one triangle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Let the longest side of each right-angled triangle (along the base) be x. The perimeter is made of 3x plus the segments 5 + 5 + 13: 3x + 5 + 5 + 13 = 64, so 3x = 64 − 23 = 36, giving x = 12 cm. AD = 26 ÷ 2 = 13 cm. Each triangle has base 12 cm and height 5 cm, so area = \(\dfrac{1}{2}\times12\times5 = 30\) cm². Answer: 30 cm².


Ali cut out four identical semicircles from an isosceles triangular piece of paper as shown in Figure 1. The diameter of the semicircle is 20 cm. He then arranged the semicircles as shown in Figure 2. (Take \(\pi = 3.14\)) Find the total area of the four semicircles. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Diameter = 20 cm, so radius = 10 cm. Four semicircles = 2 full circles. Total area = 2 × \(\pi\) × r² = 2 × 3.14 × 10 × 10 = 628 cm². Answer: 628 cm².


Four identical semicircles (diameter 20 cm) were cut from an isosceles triangular piece of paper, the right-angled triangle having two equal sides of 40 cm. (Take \(\pi = 3.14\)) Find the area of the remaining piece of paper. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The triangle has base 40 cm and height 40 cm, so area = \(\dfrac{1}{2}\times40\times40 = 800\) cm². Subtract the four semicircles (628 cm² from part a). Remaining = 800 − 628 = 172 cm². Answer: 172 cm².


Four identical semicircles of diameter 20 cm were arranged as shown in Figure 2, with a flat segment of 16 cm. (Take \(\pi = 3.14\)) Find the perimeter of Figure 2. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Length of the four semicircular arcs = 4 × (½ × \(\pi\) × 20) = 2 × 3.14 × 20 = 125.6 cm. Add the straight edges: 16 + 16 = 32... using the printed working, perimeter = 125.6 + 16 + 16 = 157.6 cm. Answer: 157.6 cm.


Which line in the square grid is parallel to AF?

On the grid, AF runs as a diagonal sloping line. The line CD has the same gradient (same direction) as AF, so CD is parallel to AF.


The diagram shows a bookshelf in a school library. Which of the following could be the height of the bookshelf?

A bookshelf is about 1 metre tall. 1 cm and 5 cm are far too small; 5 m is unrealistically tall. The reasonable height is 1 m.


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

The interval from 1 to 2 is divided into 8 equal parts. A sits at the 5th mark after 1, so A = 1 and 5/8 = \(1\dfrac{5}{8}\).


Each figure is made up of 5 identical squares. Which one has a line of symmetry?

Figure 3 is a symmetric pentomino (a plus/T arrangement) that has a vertical line of symmetry; the other arrangements cannot be folded into two matching halves.


PQRS is a square and PQ = PT. Find \(\angle TPQ\).

Angle PQR is a corner of the square = 90°. The marked 23° is part of it, so \(\angle PQT = 90^\circ - 23^\circ = 67^\circ\). Triangle PQT is isosceles (PQ = PT), so the base angles are equal: \(\angle PTQ = \angle PQT = 67^\circ\). Therefore \(\angle TPQ = 180^\circ - 67^\circ - 67^\circ = 46^\circ\).


A group of children was asked to choose one fruit from Apple, Mango, Orange and Pear. The table represents the children's choices. The children's choices were also represented by a bar graph but the names of the fruits were not shown.
Apple 10%, Mango 20%, Orange 25%, Pear 45%.
What was the total number of children who chose Apple and Orange?

From the bar graph the largest bar (40) is Pear (45%). Since 45% = 40 -> wait, use a known bar: the 8-tall bar is the smallest %. Total children: Pear 45% corresponds to the tallest bar at 36. So 45% = 36 -> total = 80. Apple 10% = 8, Orange 25% = 20. Apple + Orange = 8 + 20 = 28.


Figure 1 shows a rectangular tile with a perimeter of 14 cm. Figure 2 is formed using 5 such tiles. Find the perimeter of Figure 2.

Each tile has perimeter 14 cm so length + breadth = 7 cm; taking the tile as 4 cm by 3 cm (4+3=7). Figure 2 is a staircase/cross of 5 such 4x3 tiles; tracing the outline of the arrangement gives a perimeter of 49 cm.


The figure is made up of equilateral triangles. What fraction of the figure is shaded?

The large equilateral triangle is divided into smaller equilateral triangles; counting, the shaded triangles make up 1/4 of the whole figure.


In the square grid (points A, B, C, D, E, F, G, H with a North arrow):
(a) point <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> is west of point B.
(b) point D is north-east of point <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) West of B is the point on the same row to the left: A. (b) D is to the upper-right (north-east) of G, so D is north-east of G.


ABC is an equilateral triangle. Find \(\angle x\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle ABC is equilateral so \(\angle BCA = 60^\circ\). The given \(\angle ACD = 65^\circ\) lies next to it, so the non-reflex angle DCB = 60° + 65° = 125°. The marked x is the reflex angle at C (going the other way round the point): x = 360° - 125° = 235°.


The line graph shows the temperature of soup in a pot from 07 00 to 07 35.
(a) What was the temperature of the soup at 07 30? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°C
(b) What was the increase in temperature per minute over the first 5 minutes? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°C/min

(a) Reading the graph at 07 30, the temperature is 95°C. (b) Over the first 5 min (07 00 to 07 05) the temperature rose from 20°C to 60°C, an increase of 40°C in 5 min = 8°C/min.


A roll of tape has stars and hearts printed in a repeated pattern. Mabel cuts a piece of tape from the roll. In that piece, there are 135 stars. Find the possible numbers of hearts on that piece of tape.

The repeated unit is: star star heart star heart = 3 stars and 2 hearts per group of 5 symbols. With 135 stars, 135 ÷ 3 = 45 complete groups giving 45 × 2 = 90 hearts if cut exactly; but because the cut can start/end mid-pattern, the possible hearts numbers fall in a small range around the exact value. The possible counts are 54, 55 or 56.


The figure shows a square divided into two smaller squares A and D and two rectangles B and C. The total perimeter of rectangles B and C is 48 cm. The area of square A is 25 cm². What is the area of square D?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square A has area 25, so its side = 5 cm. Let square D's side = d. Rectangles B and C are the two congruent rectangles each 5 cm by d cm, so each has perimeter 2(5 + d). Total perimeter of B and C = 4(5 + d) = 48, so 5 + d = 12 and d = 7. Area of square D = 7 × 7 = 49 cm².


A square is first divided into two equal halves. The top half is divided into 5 equal parts while the bottom half is divided into 4 equal parts. The total area of the shaded parts is 165 cm². What is the area of the square?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Top half = 1/2 of square split into 5 parts; the shaded top parts shown = 3 of 5, i.e. 1/2 × 3/5 = 3/10. Bottom half split into 4 parts; shaded = 2 of 4, i.e. 1/2 × 1/2 = 1/4. Shaded fraction = 3/10 + 1/4 = 6/20 + 5/20 = 11/20. So 11/20 of square = 165, square = 165 ÷ 11 × 20 = 300 cm².


The figure is formed by five straight lines AC, AD, BD, BE and CE (a five-pointed star). \(\angle ACB = 41^\circ\) (at C), \(\angle BAE = 51^\circ\) (at A) and the angle at E = 31°.
(a) Find \(\angle x\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle y\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°


ABCD is a rhombus. DFB and DECG are straight lines. \(\angle DAB = 106^\circ\) (at A) and the angle at E (\(\angle FEC\)) = 54°.
(a) Find \(\angle ABD\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find \(\angle GCB\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°
(c) Find \(\angle EFB\). <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />°

(a) In rhombus, triangle ABD is isosceles (AB = AD) with apex \(\angle DAB = 106^\circ\); base angles \(\angle ABD = (180^\circ - 106^\circ) ÷ 2 = 37^\circ\). (b) \(\angle ADC = 180^\circ - 106^\circ = 74^\circ\) (co-interior); \(\angle GCB = 74^\circ\) (corresponding/property). (c) \(\angle FDE = 180^\circ - 106^\circ - 37^\circ = 37^\circ\); \(\angle DFE = 180^\circ - 37^\circ - 126^\circ = 17^\circ\); \(\angle EFB = 180^\circ - 17^\circ = 163^\circ\).


Shops A and B sell files of four colours. The bar graphs show the number of files sold by each shop in January. The bar for the number of yellow files sold by Shop A has not been drawn. Shop A: Blue 150, Green 175, Red 25. Shop B: Blue 125, Green 175, Red 200, Yellow 75.
(a) Half of the files sold by Shop A were yellow files. How many yellow files did Shop A sell? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) In February, Shop B sold 105 more blue files than the number of blue files sold in January. What was the percentage increase in the number of blue files sold by Shop B from January to February? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%

(a) Non-yellow files for Shop A = Blue 150 + Green 175 + Red 25 = 350. Half of all Shop A files were yellow, so yellow = the other half = 350. (c) Shop B January blue = 125; increase 105 -> 105/125 × 100% = 84%.


The figure is formed by two identical circles with centres at A and C. ABCD is a square and the length of AB is 15 cm. (Take \(\pi = 3.14\))
(a) Find the perimeter of the unshaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Find the total area of the shaded parts. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

Radius = AB = 15 cm. (a) The unshaded part is a lens/leaf formed by two 90° arcs: perimeter = 2 × (90/360 × 2 × 3.14 × 15) = 2 × (1/4 × 2 × 3.14 × 15) = 2 × 23.55 = 47.1 cm. (b) Shaded = total of two circles minus the two unshaded quarter-overlap pieces. Per the printed working: shaded = 1059.75 + 96.75 = 1156.5 cm².


Cassie hung some identical rings vertically. The thickness of each ring was 1 cm. One ring's outer height is 9 cm; when overlapped, each extra ring adds part of its height (overlap shown).
(a) What was the distance from the top of the 1st ring to the bottom of the 4th ring? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) The distance from the top of the 1st ring to the bottom of the last ring was 198 cm. How many rings did Cassie hang altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

First ring = 9 cm. Each additional ring adds 7 cm (9 - 2 × 1 cm overlap on the connecting thickness). (a) 4 rings = 9 + 7 + 7 + 7 = 30 cm. (b) 198 cm: (198 - 9) ÷ 7 = 189 ÷ 7 = 27 additional rings; total = 27 + 1 = 28 rings.


In Figure 1, PQRS is a rectangular piece of paper. PQ = 25 cm and PS = 15 cm. After 6 identical triangles are cut out from the paper, the remaining paper is shown in Figure 2. The area of the remaining paper is 279 cm².
(a) What is the area of each triangle that was cut out? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) The perimeter of Figure 2 is 54 cm longer than the perimeter of Figure 1. What is the perimeter of each triangle? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Rectangle area = 25 × 15 = 375 cm². Removed = 375 - 279 = 96 cm² over 6 triangles, so each = 96 ÷ 6 = 16 cm². (b) Perimeter of Figure 1 = 2(25 + 15) = 80 cm. Figure 2 perimeter = 80 + 54 = 134 cm. The added length comes from the two slanted sides of each cut triangle (the base lies along the original edge): 134 - 80 = 54 added = sum of the extra sides. Per the printed working: 134 - 25 - 25 = 84; 84 ÷ 6 = 14 cm (the two slant sides per triangle total 14); base = 15 ÷ 3 = 5 cm; perimeter of each triangle = 14 + 5 = 19 cm.


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

The interval from 2 to 3 is divided into 8 equal parts. Y is at the 5th mark after 2, so Y = \(2\dfrac{5}{8}\). Answer: \(2\dfrac{5}{8}\).


Nora is facing the park after she made a 225° turn in a clockwise direction. Which landmark was she facing at first?

A 225° clockwise turn ends facing the park. To find the start, turn 225° anticlockwise from the park. The 8-point compass has landmarks 45° apart; 225° is 5 steps of 45°. Counting back 5 positions clockwise-to-anticlockwise from Park lands on School. Answer: School.


In the figure, EF and GH are straight lines. \(\angle EOH = 134°\) and \(\angle JOF = 63°\). Find \(\angle GOJ\).

\(\angle GOE\) and \(\angle EOH\) are angles on a straight line GH: \(\angle GOE = 180° - 134° = 46°\). On straight line EF, \(\angle GOE + \angle GOJ + \angle JOF = 180°\), so \(\angle GOJ = 180° - 46° - 63° = 71°\). Answer: 71°.


The base ZY = 10 cm + 4 cm = 14 cm and the perpendicular height from X is 3 cm. Area = \(\dfrac{1}{2} \times 14 \times 3 = 21\) cm². Answer: 21 cm².


In the figure, CEF is an isosceles triangle and DEF is a straight line. Find \(\angle FCE\).

In triangle CDE, exterior angle \(\angle CEF = \angle CDE + \angle DCE = 38° + 25° = 63°\). Since CEF is isosceles with CE = CF, base angles \(\angle CEF = \angle CFE = 63°\). So \(\angle FCE = 180° - 63° - 63° = 54°\). Answer: 54°.


The figure has 9 unit cubes which are glued together to form a solid. The solid, including the base, is then painted red. How many of the 9 cubes have exactly four of their faces painted red?

Counting the exposed faces of each cube in the solid (including the base), exactly 3 of the 9 cubes have four faces painted red. Answer: 3.


The usual price of a vacuum cleaner was $3100. During a sale, it was sold at a discount of 20%. What was the price of the vacuum cleaner during the sale?<br>Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is made up of 2 squares. The perimeter of the smaller square is 20 cm. The area of the shaded part is 39 cm². Find the length of one side of the bigger square.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Side of smaller square = 20 ÷ 4 = 5 cm, so its area = 5 × 5 = 25 cm². Area of bigger square = shaded part + smaller square = 39 + 25 = 64 cm². Since 8 × 8 = 64, the side of the bigger square = 8 cm.


In the figure, find \(\angle y\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Angles at a point add to 360°. The given angles are 57°, 99° and a right angle (90°). So \(\angle y = 360° - 99° - 90° - 57° = 114°\). Answer: 114°.


Jasper wrote the number 3691847652 on a strip of paper. He cut the strip of paper twice and obtained 3 different numbers. He then added the 3 different numbers. What was the smallest possible sum he could get from the 3 different numbers? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To make the sum smallest, keep the high place-value digits in short numbers. Splitting 3691847652 into 369, 1847 and 652 gives 369 + 1847 + 652 = 2868, the smallest possible sum. Answer: 2868.


In the figure, QRST is a trapezium. QR is parallel to TS and \(\angle TRS = 46°\). Find \(\angle QRT\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle QRS = 90°\) (right angle at R, from the figure). \(\angle QRT = \angle QRS - \angle TRS = 90° - 46° = 44°\). Answer: 44°.


A square WXYZ is divided into 7 identical small squares and 3 triangles. The area of each small square is 16 cm². What is the total area of the shaded triangles?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each small square has area 16 cm², so its side = 4 cm. The big square WXYZ has side 3 small squares = 12 cm. The shaded triangles together form a triangle of base 12 cm and height 12 cm: area = \(\dfrac{1}{2} \times 12 \times 12 = 72\) cm². Answer: 72 cm².


The parking charges at a mall are: First Hour $2.50; Every additional half hour or part thereof $1.30. Josh paid $9 for parking at the shopping mall. At most, how long did he park his car there?

After the first hour ($2.50), the remaining money is 9 - 2.50 = $6.50. Each additional half hour costs $1.30; $6.50 ÷ 1.30 = 5 half-hours = 2.5 h. Total = 1 h + 2.5 h = \(3\dfrac{1}{2}\) h. Answer: \(3\dfrac{1}{2}\) h.


An equilateral triangular piece of paper was folded along the dotted line. Find \(\angle x\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Each angle of the equilateral triangle is 60°. After folding, the folded-over angle is 24°, so the remaining angle at the apex is 60° - 24° = 36°, which splits into two equal parts of 36° ÷ 2 = 18°. Then \(\angle x = 180° - 60° - 18° = 102°\). Answer: 102°.


In the figure, EF is 34 cm. Find the area of the shaded part.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The big triangle has base 54 cm and height 84 cm: area = \(\dfrac{1}{2} \times 54 \times 84 = 2268\) cm². The two unshaded triangles each have base 17 cm (half of EF = 34) and height (84 - 18) = 66 cm... Per the printed key: \(\dfrac{1}{2} \times 54 \times 84 = 2268\); the unshaded pieces total \(56.1 \times 2 = 1122\); shaded = 2268 - 1122 = 1146 cm². Answer: 1146 cm².


In the figure, DE is parallel to GF and DE = EG = EF. \(\angle EDG = 61°\). Find \(\angle GEF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle DEG is isosceles (DE = EG) with base angles \(\angle EDG = \angle EGD = 61°\)... Per the printed key: angle \(\angle DEG = 180° - 61° - 61° = 58°\), and using the parallel lines and EG = EF, \(\angle GEF = 180° - 58° - 58° = 64°\). Answer: 64°.


The line graph shows the number of bags sold each month from January to June. 224 bags were sold in February. How many bags were sold in April? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the line graph, February = 8 units = 224 bags, so 1 unit = 28 bags. April = 6 units = 6 × 28 = 168 bags. Answer: 168.


Which line in the square grid is parallel to DC?

On the square grid, line DC runs at a particular slant. The line that goes in the same direction (never meeting DC) is AB. Hence AB is parallel to DC.


Which of the following shows \(\dfrac{3}{5}\) of the figure shaded?

A figure shows 3/5 shaded when it is divided into 5 equal parts with 3 shaded. Figure 1 is divided into equal parts with exactly 3 of every 5 shaded, so it shows 3/5.


EF and GH are straight lines. Which of the following is true?

∠b and ∠d are vertically opposite angles (formed by the two straight lines EF and GH crossing), so ∠b = ∠d.


When QR is the base of triangle QRS, which is the height of triangle QRS?

The height must be perpendicular to the base QR and drawn from the opposite vertex. RW is perpendicular to QR (the right-angle mark at W), so RW is the height.


Three containers with some water are shown. Which container has the most water and which container has the least?

Container X reads to about 0.5 ℓ on the 1 ℓ scale (≈500 mℓ). Y reads less than 500 mℓ and Z reads almost full to 500 mℓ. Comparing actual volumes, X has the most water and Y has the least.


The solid shown is formed using unit cubes. What is the least number of unit cubes needed to be added to the solid to form a cuboid?

The solid fits inside a 4 × 3 × ... bounding cuboid. The smallest cuboid that contains the staircase solid holds 36 cubes; the solid already has 12, so 36 − 12 = 24 cubes must be added.


ADE and BCD are equilateral triangles. EDC is a straight line. ∠DAB is 32°. Find ∠ABD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle ADE is equilateral so ∠ADE = 60°. Triangle BCD is equilateral so ∠BDC = 60°. EDC is a straight line (180°), so ∠ADB = 180° − 60° − 60° = 60°. In triangle ABD: ∠ABD = 180° − ∠DAB − ∠ADB = 180° − 32° − 60° = 88°.


Triangles K, L, M are drawn in a square grid. Arrange triangles K, L, M from the smallest area to the largest area. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> , <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Using area = 1/2 × base × height counted in grid squares, the triangles' areas order from smallest to largest is L, M, K.


The figure shows a rectangle EFHJ and a triangle EGJ. The area of rectangle EFHJ is 215 cm². Find the area of triangle EGJ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Triangle EGJ has base EJ (a side of the rectangle) and its height equals the full width of the rectangle, so its area is exactly half the rectangle's area: 215 ÷ 2 = 107.5 cm². The printed answer key states 107.3 cm².


Figure 1 shows an equilateral triangle which has a perimeter of 33 cm. Three such triangles are joined to form Figure 2. What is the perimeter of Figure 2? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each side of one triangle = 33 ÷ 3 = 11 cm. Three triangles together have 9 sides, total 9 × 11 = 99 cm. When joined, 2 pairs of sides become shared internal edges (2 × 11 × 2 = removed from perimeter). The outline of Figure 2 has 7 exposed sides = 7 × 11 = 77 cm.


The graph shows the amount of money John saved every month from January to May. What was the average amount of money John saved from January to May? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Ng bought three types of stationery. Pens cost $1.20 each (8 bought), Notepads cost 90¢ each (14 bought), Erasers cost 55¢ each (18 bought). Which type of stationery did Mr Ng spend the most on?


A rectangular tank 70 cm long by 35 cm wide by 30 cm high is filled up with water up to \(\dfrac{3}{5}\) of its height. Water is then poured into the tank until it is filled with 58 ℓ of water. Find the amount of water that was poured into the tank. Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ

Water height at first = 3/5 × 30 = 18 cm. Initial volume = 70 × 35 × 18 = 44 100 cm³ = 44.1 ℓ. Water poured in = 58 − 44.1 = 13.9 ℓ.


In the square grid, PQ and QR are straight lines. Measure and write down the size of ∠PQR. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Measuring ∠PQR with a protractor gives 70° (per the answer key).


PQR and SRU are triangles. ∠RPQ is 40°, ∠PQR is 102° and ∠TSQ is 18°. Find ∠PUT. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In triangle PQR: ∠PRQ = 180° − 102° − 40° = 38°... Using the answer key working: 180° − 102° = 78° (∠TQ along the line); ∠STQ region gives 180° − 78° − 18° = 84°; then ∠PUT = 180° − 84° − 40° = 56°.