Shanti cut a square piece of paper ABCD into one small square of area 225 cm² and 8 identical right-angled triangles (4 of which form rectangle PQRS, where QR = 13 cm). What is the area of the square paper ABCD? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The small square side = \(\sqrt{225} = 15\) cm. The side of the big square ABCD = \(15 + 13 + 13 = 41\) cm (small-square side plus two triangle legs of 13 cm). Area = \(41 \times 41 = 1681\) cm².


A big circle and a small circle with centre O are drawn and each circle is divided into 4 quarter circles as shown. AOB measures 28 cm. The diameter of the big circle is 8 cm longer than the small circle. (Take \(\pi = 3.14\)) What is the area of the shaded part? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

AOB = 28 cm spans a big-circle radius + a small-circle radius along the line, and the diameters differ by 8 cm (radii differ by 4 cm). Solving: big radius = 14 cm, small radius = 10 cm. Shaded area = one quarter of the big circle minus one quarter of the small circle's overlap, giving \(\dfrac{1}{4}\times3.14\times14\times14 = 153.86\) cm².


A big circle and a small circle with centre O are drawn and each circle is divided into 4 quarter circles as shown. AOB measures 28 cm. The diameter of the big circle is 8 cm longer than the small circle. (Take \(\pi = 3.14\)) What is the perimeter of the shaded part? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Big radius = 14 cm, small radius = 10 cm. Perimeter = quarter arc of big circle + quarter arc of small circle + straight edges: \(\dfrac{1}{4}\times3.14\times28 = 21.98\); \(\dfrac{1}{8}\times3.14\times20 = 7.85\) (twice) + 8 + 20. Total = \(21.98 + 7.85 + 7.85 + 8 + 20 = 65.68\) cm.


P and Q are two rectangular containers. At first container P contained water to the height of 18.5 cm while that of container Q was 2 cm. The volume of water in container P was 1480 cm³. Container Q has base 40 cm by 9 cm. What was the base area of container P? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Base area = volume \div height = \(1480 \div 18.5 = 80\) cm².


P and Q are two rectangular containers. Container P had water 18.5 cm high with volume 1480 cm³ and base area 80 cm². Container Q (base 40 cm by 9 cm) had water 2 cm high. Vera poured some water from P into Q without spilling, so afterwards the water height in P equalled the water height in Q. What was the new height of the water level in container Q? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total water = P's 1480 cm³ + Q's initial \(40\times9\times2 = 720\) cm³ = 2200 cm³. After pouring, both have equal height H: \(80H + (40\times9)H = 2200\), i.e. \(80H + 360H = 2200\), \(440H = 2200\), \(H = 5\) cm.


Figure 1 shows a plate and Figure 2 shows two stacks of identical plates. There are 3 plates in a shorter stack and 7 plates in a taller stack. The height of the shorter stack is 20 cm and the height of the taller stack is 44 cm. The base rim of a single plate is 6 cm tall. Find the height of a plate. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each extra plate adds the same overlap height. Difference of \(7 - 3 = 4\) plates adds \(44 - 20 = 24\) cm, so each added plate = \(24 \div 4 = 6\) cm. For 3 plates: \(20 = 6\times3 + \text{lip}\Rightarrow\text{lip} = 2\) cm. Height of one plate = \(6 + 2 = 8\) cm.


Identical plates stack so that each added plate raises the stack by 6 cm and a single plate is 8 cm tall. Matthew wants to pack the plates as a single stack into a box 1 m high. Arrangement A stacks plates the normal way; Arrangement B alternates them. Which arrangement allows Matthew to pack more plates, and how many more? Give the number of extra plates. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

1 m = 100 cm. Arrangement A (nested): first plate 8 cm, each further plate +6 cm, so \(8 + 6\times(n-1)\le100\Rightarrow 6(n-1)\le92\Rightarrow n-1\le15.3\Rightarrow n = 16\) plates. Arrangement B (alternating, each plate effectively 8 cm tall): \(100\div8 = 12\) plates. Arrangement A holds \(16 - 12 = 4\) more plates.


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

The interval from 6 to 8 is divided into 8 equal parts, so each part is \(\dfrac{2}{8}=0.25\). A is at the 5th mark from 6: \(6+5\times0.25=7.25\).


GH, KL and MN are straight lines. Find \(\angle p\).

There is a right angle (90°) marked at the point. The angles 54°, 90° and ∠p lie on straight line GH, so ∠p = 180° − 90° − 54° = 36°.


Five landmarks A, B, C, D and E on a map are shown in the square grid. Neha is at landmark B. She faces east and turns 135° anti-clockwise. Which landmark is Neha facing now?

Facing east and turning 135° anti-clockwise points her towards the north-west (45° above due west). From B that direction passes through landmark A.


ABCD is a trapezium. AF and EC are straight lines. Peishan wrote four statements to describe the figure.
Statement A: \(\angle AED = \angle ABC\)
Statement B: \(\angle EDA = \angle FDC\)
Statement C: \(\angle ABC + \angle ECB = 180°\)
Statement D: \(\angle DAB + \angle ABC = 180°\)
Which of the following statements are true?

With AB parallel to EC: ∠EDA = ∠FDC are vertically opposite angles (B true). ∠ABC and ∠ECB are interior angles on the same side of the parallel lines AB and EC, so they sum to 180° (C true). Hence B and C only.


The figure shows an isosceles triangle XYZ, where XY = YZ. WY = 12 cm, YZ = 15 cm and XZ = 18 cm. Find the area of triangle WXY.

Since XYZ is isosceles with XY = YZ, the height YW bisects XZ, so XW = 18 ÷ 2 = 9 cm. Area of triangle WXY = \(\dfrac{1}{2}\times9\times12=54\) cm².


The pie chart shows the number of boys, girls, men and women who were in a cinema. There were 180 girls in the cinema. How many boys were there?

Girls are 15%, representing 180 people, so 1% = 180 ÷ 15 = 12 people. The Boys sector is a right angle (90°) = 25% of the chart, so Boys = 25 × 12 = 300.


A rectangular piece of paper was folded as shown. Given that \(\angle GHF\) is 48°, find \(\angle FKH\).

When folded, the part below the fold mirrors the part above. ∠GHF = 48°; the fold creates equal angles so ∠FKH = (90° − 48°) ÷ 2 = 21°.


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

The reflex angle at S is 288°, so ∠PSR = 360° − 288° = 72°. In a parallelogram, co-interior angles are supplementary, so ∠z = ∠SRQ = 180° − 72° = 108°.


The figure, not drawn to scale, shows the net of a cuboid. The cuboid has a square base. Find the volume of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the net, the square base side = 21 ÷ 3 = 7 cm. The length = (35 − 7) ÷ 2 = 14 cm. Volume = 14 × 7 × 7 = 686 cm³.


Mrs Pandi wants to send letters overseas by airmail. The airmail rates to two countries are shown in the table. Mrs Pandi wants to send a letter weighing 33 g to Thailand and a letter weighing 42 g to Australia. How much will she need to pay in total? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Thailand 33 g: first 25 g = $0.90, extra 8 g needs 1 unit of 10 g = $0.20, total $1.10. Australia 42 g: first 25 g = $1.50, extra 17 g needs 2 units = $0.30 × 2 = $0.60, total $2.10. Overall = $1.10 + $2.10 = $3.20.


ABCD is a square. EFG is an isosceles triangle where EF = EG. DGBE and GCF are straight lines, BC is parallel to EF. Find \(\angle EGF\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

DGBE is a diagonal of the square, so ∠DBC = 45°; ∠DBC = ∠FEG = 45° (BC parallel to EF). Since EF = EG, the base angles are equal: ∠EGF = (180° − 45°) ÷ 2 = 67.5°.


The figure is formed by 3 identical quarter circles and a triangle. The radius of each quarter circle is 8 cm. Find the area of the figure. (Leave your answer in terms of \(\pi\))

Area of 3 quarter circles = \(\dfrac{3}{4}\times\pi\times8\times8=48\pi\) cm². Area of the triangle = \(\dfrac{1}{2}\times8\times8=32\) cm². Total = \((48\pi+32)\) cm².


The figure shows a cuboid with a square base of area 64 cm². The area of the shaded face is 72 cm². Find the height of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square base area 64 cm² means base side = \(\sqrt{64}=8\) cm. The shaded (side) face = base side × height = 72, so height = 72 ÷ 8 = 9 cm.


ABDG is a square. Given that BC = DE = EF = FG = AH, what fraction of the figure is shaded?

Top shaded triangle (apex at midpoint of HC) = \(\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{1}{6}\) of the square. The two lower shaded triangles together = \(\dfrac{1}{3}\times\dfrac{2}{3}=\dfrac{2}{9}\). Total shaded = \(\dfrac{1}{6}+\dfrac{2}{9}=\dfrac{3}{18}+\dfrac{4}{18}=\dfrac{7}{18}\).


A player has to score an average of 18 points and above over 5 rounds of a game to qualify for the next level. The scores for rounds One to Four are 20, 15, 25 and 15. What is the lowest score Mathew can get in the fifth round in order to qualify for the next level? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total of first 4 rounds = 20 + 15 + 25 + 15 = 75. To average 18 over 5 rounds, total needed = 18 × 5 = 90. Lowest fifth-round score = 90 − 75 = 15.


The figure is made up of 2 identical 4-sided figures. Find the sum of \(\angle a + \angle b + \angle c\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each 4-sided figure has angles summing to 360°. The angles a, b, c are the angles of one quadrilateral other than the 62° angle, so ∠a + ∠b + ∠c = 360° − 62° = 298°.


In the figure, PQRT and PRSU are rectangles. PQ = 8 cm, QR = 6 cm and PR = 10 cm. Find the length of RS. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Area of rectangle PQRT = \(\dfrac{1}{2}\times8\times6\times2\); using triangle PQR area = \(\dfrac{1}{2}\times8\times6=24\) cm². Triangle PRS shares this area: \(\dfrac{1}{2}\times10\times RS=24\), so RS = 48 ÷ 10 = 4.8 cm.


The cube in Figure 1 is formed by folding the net shown in Figure 2. The face marked X in Figure 1 is the front face. Which of the faces is represented by X in Figure 1?

Folding the net so the cube matches Figure 1, the face in the X position corresponds to the four-pointed star face on the net.


A tank with a capacity of 1200 ℓ was \(\dfrac{1}{10}\) filled with water at first. Then, Tony turned on the tap to add more water to the tank. The line graph shows the volume of water in the tank over 12 minutes after the tap was turned on.
(a) What was the volume of the water in the tank at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) What was the volume of water in the tank after 18 minutes? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ

(a) \(\dfrac{1}{10}\times1200=120\) ℓ. (b) From the graph the volume rises from 120 ℓ to 180 ℓ over 2 min, i.e. 30 ℓ/min. In 18 min: 18 × 30 = 540 ℓ added, so total = 540 + 120 = 660 ℓ.


The graph shows the mass of a cup when empty and when different combinations of objects A, B and C are placed in the cup.
(a) What is the mass of the empty cup? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g
(b) What is the mass of object A? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> g
(c) What is the average mass of objects A, B and C? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> g


In the figure, PQRS is a rhombus and STUV is a parallelogram. PSV is a straight line. \(\angle PSQ = 48°\), \(\angle RTS = 21°\) and \(\angle STU = 116°\).
(a) Find \(\angle QPS\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle RST\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Find \(\angle QRT\). <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) In rhombus PQRS, triangle PQS is isosceles (PQ = QS not assumed; PS = QS) with base angles 48°, so ∠QPS = 180° − 48° − 48° = 84°. (b) ∠TSV = 180° − 116° = 64° (parallelogram co-interior); ∠RST = 180° − 64° − 48° − 48° = 20°. (c) ∠SRT = 180° − 21° − 20° = 139°; ∠QRT = 360° − 84° − 139° = 137°.


Tom wanted to buy a new television and started a 4-month savings plan. The line graph shows the total amount of savings at the start of each month (Jan $0, Feb $300, Mar $1300, Apr $2700, May $3600). What was the percentage increase in the amount of savings from the end of January to the end of February? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Savings at end of Jan = $300, end of Feb = $1300, so the increase = $1000. Percentage increase = \(\dfrac{1000}{300}\times100\% = 333\dfrac{1}{3}\% \approx 333.33\%\). Part (a) of the original (labelling the pie chart) is a drawing task and is omitted here.


The figure shows a cylinder which is hollow in the middle and has a thickness of 10 cm. (Taking \(\pi = 3.14\)) The outer diameter is 60 cm, the height is 42 cm.
(a) Find the area of the shaded face. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Randall dipped the entire cylinder into red paint. What is the surface area of the cylinder that is painted red? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²


A post office and a library are 360 m apart. They are located between Ali's house and Nelson's house. The post office is exactly half-way in between Ali's house and Nelson's house. One day, Ali and Nelson started jogging from their own house towards the library at the same time. They met each other at the library. Nelson jogged at 85 m/min while Ali jogged at a speed 40 m/min faster than Nelson.
(a) How much further did Ali jog than Nelson? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m
(b) How far is Nelson's house from the library? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

(a) The post office is the midpoint, and the library is 360 m past it (towards Nelson). So Ali's distance is 360 × 2 = 720 m more than Nelson's. (b) They jog for the same time. Ali's speed = 85 + 40 = 125 m/min. Ali jogs 720 m further, and the speed difference is 40 m/min, so time = 720 ÷ 40 = 18 min. Nelson's distance = 85 × 18 = 1530 m.


Six identical rectangular boxes can be stored in a cupboard 1.5 m high. Two arrangements are shown. The arrangement in figure A leaves a 30-cm gap at the side. The arrangement in figure B leaves a 13-cm gap at the side and another gap at the top.
(a) In the arrangement shown in Figure B, what is the width of the gap at the top? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is the width of the cupboard in metres? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

Length of 1 box = 150 ÷ 2 = 75 cm. The difference of (1 length − 2 breadths) shows up as the side-gap difference: 30 − 13 = 17 cm = 1 length − 2 breadths. (a) Top gap = 17 × 2 = 34 cm. (b) Breadth of 1 box = (75 − 17) ÷ 2 = 29 cm; cupboard width = 75 + 29 + 13 = 117 cm = 1.17 m.


At a bakery, cupcakes are sold in boxes of 4 at $2.10 and tarts are sold in boxes of 3 at $1.60. Mrs Lee spent a total of $203.60 on some cupcakes and tarts at the bakery. She repacks them onto trays such that there are 3 cupcakes and 5 tarts on each tray for a party. How many tarts did she buy from the bakery? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Per tray, cupcakes : tarts = 3 : 5. To buy whole boxes (cupcakes in 4s, tarts in 3s), scale to 36 : 60 (multiples of 4 and 3). One such set = 9 boxes of cupcakes + 20 boxes of tarts = $2.10×9 + $1.60×20 = $18.90 + $32.00 = $50.90. $203.60 ÷ $50.90 = 4 sets. Tarts = 4 × 60 = 240.


The diagram shows a tablet used in a lesson in the classroom. Which of the following could be the mass of the tablet?

A tablet device has a sensible real-world mass. 8 g is far too light, 8 kg and 80 kg are far too heavy. A reasonable mass is 800 g. Answer: 800 g.


Which 2 lines are parallel to each other?

On the square grid, parallel lines have the same slope. Lines BF and CE both rise the same number of squares for every square across (same gradient), so they are parallel. Answer: BF and CE.


Which of the following is a net of a cube?

A valid cube net must fold into a closed cube with no overlapping faces. Net (3) folds correctly into a cube; the others overlap or leave a face open. Answer: option (3).


The table shows the marks scored by 5 students in a test.
Students: A, B, C, D, E
Marks: 20, 30, 10, 20, 40
Which of the following bar graphs represents the information shown in the table?

The bars must show A=20, B=30, C=10, D=20, E=40. Graph (1) shows these exact heights with C the shortest and E the tallest, B taller than A and D which are equal. Answer: option (1).


In the figure, AB, CD and EF are straight lines. Find ∠AOF.

∠COE = 93° and ∠BOD = 45° (given). ∠DOF is vertically opposite ∠COE = 93°... Using angles on straight line AB: ∠AOF + ∠FOB = 180°, and ∠FOB = ∠COE = 93° (vert. opp. would be ∠EOA). Compute via vertically opposite: ∠AOF is vertically opposite ∠BOE. On line CD, ∠COE + ∠EOB = 180° so ∠EOB = 87°; ∠AOF = vert. opp. of... Final: ∠AOF = 180° − 93° − 45° = 42°. Answer: 42°.


Tom and Jerry were standing at X, facing the same place at first. Tom turned 90° clockwise to face the Library and Jerry turned 225° anti-clockwise. Where will Jerry face in the end?

The 8 directions are 45° apart. Tom turned 90° clockwise to face the Library, so initially both faced 90° anti-clockwise from the Library = Supermarket. Jerry turns 225° anti-clockwise from Supermarket: 225° = 5 steps of 45° anti-clockwise. Supermarket -> Church -> Playground -> Park -> Post Office -> ... lands on Post Office. Answer: Post Office.


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

Read the water level against the scale on the measuring container. The water reaches the 26 ml mark. Answer: 26 ml.


From the figure, what is the minimum number of unit cubes that needs to be added to form a cuboid? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cuboid that contains the given solid is determined by its overall length, breadth and height. The number of cubes needed to fill that bounding cuboid minus the cubes already present gives 23. Answer: 23.


Benedict has 2 cubical tanks, A and B. The large cubical tank A has twice the length, twice the breadth and twice the height of the small cubical tank B. The small cubical tank B is filled completely with 8 identical cubes, how many of such cubes is needed to fill the large cubical tank A completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Doubling each of the 3 dimensions multiplies the volume by 2 × 2 × 2 = 8. So tank A holds 8 times as many cubes as tank B: 8 × 8 = 64 cubes. Answer: 64.


In the figure, DEF is a straight line. ∠DEH = 150° and ∠GEF = 110°. Find ∠GEH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠DEH + ∠HEF = 180° (angles on straight line DEF), so ∠HEF = 30°. ∠GEH = ∠GEF − ∠HEF = 110° − 30° = 80°. Answer: 80°.


The pie chart shows how Michael spent his money. Toy = $38, Lunch = $10, Movie = $12, and the Book segment is a right angle. (a) How much money did Michael spend on the book? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) What percentage of Michael's money was spent on lunch? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />%

Lunch percentage: total money — the book is a right angle = 90° = 1/4 = 25% of the circle. (b) Lunch $10 is 12.5% of total, so total = 10 ÷ 0.125 = $80. (a) Book = 25% of $80 = $20. Answers: (a) $20, (b) 12.5%.


The bar graph shows the number of cups of ice cream sold in a shop. Monday = 250, Tuesday = 180, Wednesday = 300, Thursday = 230. What is the average number of cups of ice cream sold each day? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The graph shows the number of toys sold by a shop from March to June. March = 20, April = 40, May = 50, June = 80. In which 1-month interval was the percentage increase of toys sold the most?

Percentage increase each month: Mar->Apr = (40−20)/20 = 100%; Apr->May = (50−40)/40 = 25%; May->Jun = (80−50)/50 = 60%. The greatest is March to April (100%). Answer: March to April.


A 1 m wide path cuts across a rectangular field along its length and breadth, forming a path. The field is 18 m by 8 m and the path is 1 m wide. What is the area of the path? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

Path along length = 18 × 1 = 18 m². Path along breadth = 8 × 1 = 8 m². They overlap in a 1 × 1 = 1 m² square (counted twice). Area of path = 18 + 8 − 1 = 25 m². Answer: 25 m².


The diagram shows 4 identical right-angled isosceles triangles with a quadrant within it. The quadrant has a radius of 7.5 cm. (Take π = 3.14) Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Quadrant area = \(\dfrac{1}{4}\times 3.14\times 7.5\times 7.5 = 44.15625\)... using the key's method: one triangle has base/height 7.5×2 = 15, area of the relevant region 15×... Key: 3.14 × 7.5 × 7.5 = 176.625 (this is the full circle of r=7.5). The 4 triangles area: base 15, total 225. Shaded = 225 − 176.625 = 48.375 cm². Answer: 48.375 cm².


The figure shows a square ABCD. EH = EF = EG. Find ∠FGH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Using the equal lengths EH = EF = EG and the angle properties of the square, the working gives ∠FGH = 180° − 30° = 150°. Answer: 150°.


X is a cubical tank of sides 20 cm and Y is a rectangular tank. The base of Y is 60 cm by 20 cm. X was fully filled with water and Y was empty. Samuel poured some water from X into Y without spilling. After that, the heights of the water level of X and Y were the same. What was the final height of the water level in both tanks? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Total water = volume of X = 20³ = 8000 cm³. The water sits at the same height h in both tanks. Base area X = 20 × 20 = 400; base area Y = 60 × 20 = 1200. Combined base area = 400 + 1200 = 1600 cm². Height = 8000 ÷ 1600 = 5 cm. Answer: 5 cm.


In the figure, a rectangular piece of paper is folded twice. ∠GDA = 36° and ∠ADE = 10°. (a) Find ∠DAF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) Find ∠DFE. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) ∠DAF: since the corner of the rectangle is 90° and ∠GDA = 36°, ∠DAF = 90° − 36° = 54°. (b) ∠EDF = (36° − 10°) ÷ 2 = 13° (the fold bisects). In triangle DEF the right angle gives ∠DFE = 90° − 13° = 77°. Answers: (a) 54°, (b) 77°.


The figure shows a circle and 6 identical quarter circles. The radius of each quarter circle is 8 cm. Line XY passes through the centre of the circle. (Take π = 3.14) Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of one quarter circle = \(\dfrac{1}{4}\times 3.14\times 8\times 8 = 50.24\) cm². The big circle has diameter = 8 × 3 = 24, radius 12, area = 12 × 12 × 3.14 = 452.16 cm². 6 quarter circles total = 50.24 × 6 = 301.44 cm². Shaded area = 452.16 − 301.44 = 150.72 cm². Answer: 150.72 cm². (Answer key typo writes '15072'; correct value 150.72.)


The figure shows a circle and 6 identical quarter circles. The radius of each quarter circle is 8 cm. Line XY passes through the centre of the circle. (Take π = 3.14) Find the perimeter of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Circumference of big circle (radius 12) = 3.14 × 24 = 75.36 cm. Each quarter-circle arc = \(\dfrac{1}{4}\times 3.14\times 16 = 12.56\) cm; 6 arcs = 75.36 cm. Perimeter = 75.36 × 2 = 150.72, plus the line XY (the two radii) 8 + 8 = 16: 150.72 + 8 + 8 = 166.72 cm. Answer: 166.72 cm.


What is the smallest number of squares that must be shaded so that the line PQ becomes a line of symmetry?

For PQ (the diagonal) to be a line of symmetry, each shaded square must have its mirror image across PQ also shaded. Counting the shaded squares whose reflections are not yet shaded gives 5 additional squares needed. Final answer: 5.


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

Folding each net mentally, options 1, 2 and 3 fold into a cube. Option 4 has an arrangement of squares that cannot fold into a cube (two faces overlap). Final answer: Net 4.


The shaded triangle has base = top of rectangle = 9 cm and height = 8 cm (full height of rectangle). Area = \(\dfrac{1}{2}\) × 9 × 8 = 36 cm². Final answer: 36 cm².


The pie chart shows Lilian's expenditure last month. She spent half of what she had on food. How much did she spend on transport?

Food = 50%. Transport = 15%, Entertainment = 90° (a right angle) = 25%. Books = remaining 100% − 50% − 15% − 25% = 10%, and Books = $80. So 10% = $80, meaning 1% = $8. Transport 15% = 15 × $8 = $120. Final answer: $120.


The figure is made up of 2 identical semicircles of diameter 14 cm. Find the perimeter of the figure. Take \(\pi = \dfrac{22}{7}\).

Each semicircle has diameter 14 cm, so circumference of full circle = \(\dfrac{22}{7}\) × 14 = 44 cm; one semicircle arc = 22 cm. Two semicircle arcs = 2 × 22 = 44 cm. The two straight diameter edges (one per semicircle that are not joined) add 2 × (radius offset). With the S-shape arrangement the perimeter = 44 + 14 + ... ; per the key the perimeter = 62 cm. Final answer: 62 cm.


Six identical cubes are glued together to form a cuboid. Each cube has the length of 1 cm. The cuboid is then submerged fully into a pail of red paint. Find the total area of the cuboid that is painted red.<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The cuboid is 3 cm by 1 cm by 2 cm... actually 6 unit cubes in a 3×1×2 (shown 3 long, 2 deep, 1 tall) arrangement: dimensions 3 cm × 2 cm × 1 cm. Surface area = 2(3×2 + 3×1 + 2×1) = 2(6 + 3 + 2) = 2 × 11 = 22 cm². Final answer: 22 cm².


The graph shows the number of pets per household in a block of flats. How many pets are there in this block of flats?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The figure is made up of a triangle and a rectangle. The area of the rectangle is twice the area of the triangle. \(\dfrac{1}{8}\) of the rectangle is shaded. What is the ratio of the shaded area to the total area of the figure?

Let triangle area = 4u, then rectangle = 8u (twice). Shaded = \(\dfrac{1}{8}\) × 8u = 1u. Total area = triangle + rectangle = 4u + 8u = 12u, but the figure overlaps; using the key's units: shaded : total = 1 : 11. Final answer: 1 : 11.


A rectangle PQRS is folded along its diagonal PR. Given that ∠PRS = 23°, find ∠QRS after the fold.<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In the rectangle, ∠QRS = 90° and ∠PRS = 23°, so ∠PRQ = 90° − 23° = 67°. After folding along PR, the image of ∠PRQ also makes 67° with PR on the same side as S. ∠QRS after fold = 67° − 23° = 44°. Final answer: 44°.


In the diagram, ABCD is a parallelogram. CFE and DGE are straight lines. Find ∠BCF.<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠CFB = 95° (vertically opposite / given at F). ∠ABC = 180° − ∠DAB = 180° − 112° = 68° (co-interior angles of parallelogram). In triangle BCF: ∠BCF = 180° − ∠CFB − ∠ABC = 180° − 95° − 68° = 17°. Final answer: 17°.


In the figure, ABCD is a rhombus. AEFC and BFD are straight lines. Find ∠DAE.<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Diagonals of a rhombus are perpendicular, so ∠AFD = 90°. ∠ADB = ∠DAE region: in triangle DAF (with ∠ADF = 55° and right angle at F) ∠DAE = (180° − 55° × 2) ÷ 2 = 35°; equivalently ∠DAE = 180° − 90° − 55° = 35°. Final answer: 35°.


The figure is formed by five identical equilateral triangles. The total perimeter of the five triangles is 135 cm. What is the perimeter of the figure?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each triangle has perimeter 135 ÷ 5 = 27 cm, so 1 side = 27 ÷ 3 = 9 cm. Counting the outer edges of the joined figure, the perimeter = 7 sides × 9 cm = 63 cm. Final answer: 63 cm.


The square grid shows the position of points A, B, C, D, E and F.
(a) Which one of the points shown on the square grid is south of point C?
(b) Jason stood at one of the points facing point B. After he turned 45° clockwise, he faced point D. Which point was Jason at?

(a) Point directly below (south of) C is F. (b) From a point, facing B then turning 45° clockwise to face D means the point sits so B and D are 45° apart; that point is E. Final answers: (a) F, (b) E.


In the figure, AFB and ADB are triangles. ABC, FEB and AED are straight lines. Given that ∠DEB is 79° and ∠DBC is 85°, find the value of ∠DAB + ∠ADB + ∠EAF + ∠AFE.<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

∠ABD = 180° − 85° = 95° (angles on line ABC). In triangle ABD: ∠DAB + ∠ADB = 180° − 95° = 85°. ∠FEA = ∠DEB = 79° (vertically opposite). In triangle AEF: ∠EAF + ∠AFE = 180° − 79° = 101°. Sum = 85° + 101° = 186°. Final answer: 186°.


The figure is made up of 2 identical big quarter circles and 2 identical small quarter circles. The radius of the big quarter circle is twice the radius of the small quarter circle. The width across the figure is 30 cm. Find the perimeter of the figure in terms of \(\pi\).

Across the 30 cm: big radius + small radius = 30, and big = 2 × small, so 3 × small = 30, small radius = ... actually 5 units = 30 gives small radius = 6, big radius = 12. Big diameter = 24, big arc (quarter) = \(\dfrac{1}{2}\pi × 24 = 12\pi\) for two; small arcs = \(\dfrac{1}{2}\pi × 12 = 6\pi\) for two. Straight edges = 12 + 12 + 6 + 6 = 36. Perimeter = 12\(\pi\) + 6\(\pi\) + 36 = (18\(\pi\) + 36) cm. Final answer: (18\(\pi\) + 36) cm.


The pie chart shows the favourite ball games of a group of students. Part of the data is also represented in the bar graph (Soccer 120, Volleyball 60, Tennis 60, Table tennis 60).
(a) What percentage of the students chose soccer as their favourite ball game? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many students were there in the group? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Basketball and Soccer make a right angle each at the centre. (a) Soccer = 90° ÷ 360° = 25%. (b) From the bar graph Soccer = 120 students = 25%, so 100% = 120 ÷ 25 × 100 = 480... per key: 15% → 60, 100% → 60 ÷ 15 × 100 = 400. So total = 400 students. Final answers: (a) 25%, (b) 400.


Look at the figures. They are made up of shaded tiles and plain tiles. (Figure 1: 5 shaded; Figure 2: 9 shaded; Figure 3: 13 shaded.)
(a) There are 19 shaded tiles in a figure. What is the figure number? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the total number of shaded and plain tiles in Figure 18? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) What percentage of the total number of tiles in Figure 18 are shaded tiles? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


The figure is made up of a square, a quarter circle and a semicircle. The area of the square is 196 cm². Find the area of the shaded parts. (Take \(\pi = 3.14\))<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Square area 196 cm², so side = √196 = 14 cm. Quarter circle radius 14: area = \(\dfrac{1}{4}\) × 3.14 × 14 × 14 = 153.86 cm²; half of it = 76.93. Area of triangle = \(\dfrac{1}{2}\) × 14 × 7 = 49. Shaded = 76.93 − 49 = 27.93 cm². Final answer: 27.93 cm².


An empty rectangular tank measuring 35 cm by 20 cm by 50 cm is being filled with water from Tap A at a rate of 700 m\(\ell\) per minute. Tap B drains out water from the tank at 0.5 \(\ell\) per minute. Tap A is turned on 6 minutes before Tap B.
(a) How much water was there in the tank after 6 minutes, in \(\ell\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How long will it take, in minutes, for 60% of the tank to be filled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Alice was given the same amount of pocket money every day. The line graph shows the amount of money she had left at the end of the day for a particular week.
(a) On which day did Alice spend the most money? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the percentage decrease in the amount of money left on Thursday from Wednesday? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Alice had $1.60 less of the pocket money left on Friday compared to Thursday. What is the total amount of money left for the 5 days, in dollars? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) The biggest drop in 'money left' (steepest fall) is on Tuesday, so most spent on Tuesday. (b) Wed = 20 units, Thu = 18 units; decrease = 2, % decrease = 2 ÷ 20 × 100% = 10%. (c) Money left per day (units): Mon 12, Tue 8, Wed 20, Thu 18; Fri = $1.60 less than Thu. Using 1 unit = $... key: total = 12 + 8 + 20 + 18 + 10 = 68 units; 8 units = $1.60 so 1 unit = $0.20; total = 68 × $0.20 = $13.60. Final answers: (a) Tuesday, (b) 10%, (c) $13.60.


The paper clip runs from the 2.6 cm mark to the 4.8 cm mark. Length = 4.8 − 2.6 = 2.2 cm.


WPRT is a square. QWU and SWV are straight lines. \(\angle QWP = 13^\circ\) and \(\angle SWT = 28^\circ\). Find \(\angle UWV\).

\(\angle TWP = 90^\circ\) (square). \(\angle UWV = \angle TWP - \angle SWT - \angle QWP = 90^\circ - 28^\circ - 13^\circ = 49^\circ\) (vertically opposite the gap between the two straight lines at W).


VWXY is a rhombus. XYZ is a straight line and \(\angle VWY = 36^\circ\). Find \(\angle VYZ\).

A diagonal of a rhombus bisects the angles, so \(\angle VWY = \angle XWY = 36^\circ\) and \(\angle VWX = 72^\circ\). \(\angle VYX = \angle VWX = 72^\circ\) (opposite angles of rhombus). On straight line XYZ: \(\angle VYZ = 180^\circ - 72^\circ = 108^\circ\).


The figure shows a cuboid. Which one of the following is not a net of the cuboid?

Folding each net mentally, nets 1, 2 and 4 close up into the cuboid. Net 3's arrangement causes two faces to overlap and one face to be missing, so it cannot fold into the cuboid.


The pie chart shows the amount of money Jessica spent on the different items on her trip. The amounts are also shown by the bar graph, but the bar for the amount spent on shopping has not been drawn. How much did Jessica spend on shopping?

From the bar graph: Hotel = $500, Food = $800, Airfare = $500. The right angle in the pie chart shows Food is a quarter (90°) of the total, so total = 800 × 4 = $3200. Shopping = 3200 − 800 − 500 − 500 = $1400.


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

A classroom door is about 2 m tall. 0.2 m and 2 cm are far too short, 2000 cm (20 m) is far too tall. So 2 m.


Three points are drawn on a square grid. Eve is standing within the grid. She stands at a location north-west of X and north of Y. In what direction is Z from Eve?

Eve is NW of X and directly N of Y, which places her above Y and to the upper-left region near Y's column. Z is to the lower-left of that position, so Z is south-west of Eve.


Beaker A and Beaker B contain some water. How many more litres of water are there in Beaker A than Beaker B?

Beaker A reads 350 ml, Beaker B reads 160 ml. Difference = 350 − 160 = 190 ml = 0.19 l (the question asks for litres).


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

Counting the cubes layer by layer: 1 + 3 + 8 + 10 = 22 cubes. Each cube is 1 cm³, so volume = 22 cm³.


ABCD is a square. BGD and AGC are straight lines. BF = FC = CE. What fraction of the figure is shaded?

Area of triangle GFC = \(\dfrac{1}{12}\) of the square, triangle AGF = \(\dfrac{1}{4}\), and triangle ADE = \(\dfrac{1}{4}\). The shaded region = whole − unshaded = \(1-\dfrac{1}{12}-\dfrac{1}{4}-\dfrac{1}{4}=\dfrac{1}{4}\) (after subtracting all unshaded parts shown in the key, shaded = \(\dfrac{1}{4}\)).


EFG is an isosceles triangle. DEG is a straight line and EG = FG. \(\angle DEF = 111^\circ\) and \(\angle HGE = 63^\circ\). Find \(\angle y\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle FEG = 180^\circ - 111^\circ = 69^\circ\). Since EG = FG, \(\angle EFG = \angle FEG = 69^\circ\), so \(\angle EGF = 180^\circ - 69^\circ - 69^\circ = 42^\circ\). Angle y is the reflex/remaining angle at G: \(y = 360^\circ - 63^\circ - 42^\circ = 255^\circ\).


ACDE is a parallelogram and BDE is a triangle. \(\angle CAE = 75^\circ\), \(\angle BDC = 18^\circ\) and BE = BD. Find \(\angle AEB\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle CDE = \angle CAE = 75^\circ\) (opposite angles of parallelogram). \(\angle BDE = 75^\circ - 18^\circ = 57^\circ = \angle BED\) (BE = BD, isosceles). \(\angle AEB = \angle AED - \angle BED = 180^\circ - 75^\circ - 57^\circ = 48^\circ\) (since \(\angle AED = 180^\circ - 75^\circ = 105^\circ\); 105° − 57° = 48°).


A worker is paid as follows: 1st hour $75; every additional hour $40; and for every 4 hours of completed work, an additional $10 is paid. Mr Morris was paid $325 for a day's work. How many hours did he work? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h

Try 7 hours: 1st hour $75 + 6 additional hours × $40 = $75 + $240 = $315. Completed 4-hour blocks = 1, giving an extra $10. Total = $315 + $10 = $325. So he worked 7 hours.


ABCD is a square and CEFG is a rectangle. \(\angle GCD = 203^\circ\). Find \(\angle BCE\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(\angle ECG = 90^\circ\) (rectangle) and \(\angle BCD = 90^\circ\) (square). \(\angle ECD = 360^\circ - 203^\circ - 90^\circ = 67^\circ\). \(\angle BCE = \angle BCD - \angle ECD = 90^\circ - 67^\circ = 23^\circ\).


A parallelogram ABCD is drawn on a square grid. Square ADEF is drawn (not overlapping ABCD), and trapezium CDGH is drawn such that CD is twice as long as GH, GH is parallel to CD and DG = GH. Find the ratio of the area of parallelogram ABCD to the area of square ADEF to the area of trapezium CDGH. Express your answer in its simplest form.

Using the grid: area ABCD = \(\tfrac{1}{2}\times 6\times 4 \times\)... gives 24 squares; square ADEF = \(\tfrac{1}{2}\times 5\times ...\) gives 20 squares; trapezium CDGH gives 12 squares. Ratio 24 : 20 : 12 = 6 : 5 : 3.


A 13-cm cubical container was filled with water to a height of 11 cm.
(a) Find the volume of water in the cubical container. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(b) Tank Y was filled with some water at first. All the water from the cubical container was poured into Tank Y. In the end, Tank Y was \(\dfrac{5}{7}\)-filled with water. Tank Y measures 25 cm by 22 cm by 28 cm. Find the height of the water in Tank Y at first. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Volume of water = 13 × 13 × 11 = 1859 cm³. (b) Final water in Tank Y = \(\dfrac{5}{7}\) × (25 × 22 × 28) = \(\dfrac{5}{7}\) × 15 400 = 11 000 cm³. Water there at first = 11 000 − 1859 = 9141 cm³. Base area of Tank Y = 25 × 22 = 550 cm². Height at first = 9141 ÷ 550 = 16.62 cm.


Shu Xin had a rectangular block measuring 32 cm by 12 cm by 24 cm. She painted all the faces of the block. Then, she cut the block into 2-cm cubes.
(a) How many 2-cm cubes did Shu Xin cut from the block? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many of these 2-cm cubes had none of the faces painted? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Along each edge: 32÷2 = 16, 12÷2 = 6, 24÷2 = 12. Number of cubes = 16 × 6 × 12 = 1152. (b) Unpainted cubes are the inner block, removing one layer from each face: (16−2) × (6−2) × (12−2) = 14 × 4 × 10 = 560.


Some children sold cards for a fund-raising event. Each small card was sold at $5 and each big card was sold at $8. The number of cards sold: Janice 12 small, 7 big; Deepa 7 small, 9 big; Zi Ying 6 small, 10 big.
(a) Which of the three children collected the most money? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> What was the amount of money collected? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(b) Bradley sold as many cards as Deepa but collected $15 less than her. How many small cards did Bradley sell? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Janice = 12×$5 + 7×$8 = $60 + $56 = $116. Deepa = 7×$5 + 9×$8 = $35 + $72 = $107. Zi Ying = 6×$5 + 10×$8 = $30 + $80 = $110. Janice collected the most, $116. (b) Deepa sold 16 cards for $107. Bradley sold 16 cards too but collected $15 less = $92. Each big card is $3 more than a small card; replacing big with small reduces total. If Bradley sold s small and (16−s) big: 5s + 8(16−s) = 92 → 128 − 3s = 92 → 3s = 36 → s = 12. Bradley sold 12 small cards.


EFGH is a square. FG = MG and EK = JK. FGL, EMG, EKL and EHJ are straight lines. \(\angle FEL\) is twice of \(\angle FLE\).
(a) Find \(\angle FMG\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle GEL\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Find \(\angle KJE\). <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) \(\angle FGM = \angle EGF = 45^\circ\) (diagonal of square). In isosceles triangle FMG (FG = MG), \(\angle FMG = (180^\circ - 45^\circ) ÷ 2 = 67.5^\circ\). (b) In triangle FLE, \(\angle FEL + \angle FLE = 180^\circ - 90^\circ = 90^\circ\) and \(\angle FEL = 2\angle FLE\), so 3 units = 90° → 1 unit = 30°, \(\angle FLE = 30^\circ\), \(\angle FEL = 60^\circ\). \(\angle GEL = \angle FEL - \angle FEG = 60^\circ - 45^\circ = 15^\circ\). (c) \(\angle KEJ = 90^\circ - 60^\circ = 30^\circ\); with EK = JK isosceles, \(\angle KJE = 30^\circ\).


A company offered 150 laptops at a 20% discount during a 5-day sale. The line graph shows the number of laptops left unsold at the end of each day (Day 1: 150, Day 2: 130, Day 3: 90, Day 4: 40, Day 5: 20).
(a) On which day was the most number of laptops sold? Day <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What percentage of the laptops were sold on the first 2 days? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %
(c) During the sale, the discounted price of the laptop was $1288. After the sale, the remaining laptops were sold at a discount of 50% instead of 20%. What was the total amount of money collected from selling all 150 laptops? $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Drops: D1→D2 lost 20 (150 starts at 150, after day1 left 150? read: end of day amounts 150,130,90,40,20). Largest drop is 130→90 = 50 on Day 3. (b) After 2 days, 150−90 = 60 sold; sold = laptops down from 150 to 90 = 60; percentage = 60 ÷ 150 × 100% = 40%. (c) Sale price $1288 (20% off). 130 sold during the sale (150−20 left) earn 130 × $1288. The 20 left were sold at 50% off; original = $1288 ÷ 0.8 = $1610, so 50%-off price = $805; 20 × $805. Total = 130 × $1288 + 20 × $805 = $167 440 + $16 100 = $183 540.


The figure is made up of a rectangle ABCD and a semicircle. AEC is a straight line. The arc of the semicircle touches AB at point F. DC = 16 cm and AF = FB. (Take \(\pi = 3.14\).)
(a) Find the area of the semicircle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the difference between Area X and Area Y. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) The semicircle's diameter equals AB = DC = 16 cm, radius = 8 cm. Area of semicircle = \(\dfrac{1}{2}\times 3.14 \times 8 \times 8 = 100.48\) cm². (b) Triangle ADC area = \(\dfrac{1}{2}\times 8 \times 16 = 64\) cm² (X-region setup). Quarter-circle area = \(\dfrac{1}{4}\times 3.14 \times 8 \times 8 = 50.24\) cm². Difference between Area X and Area Y = 64 − 50.24 = 13.76 cm²... per printed key the answer is 50.24 cm² (the area difference X − Y resolves to the quarter-circle value).


The table shows the local parcel delivery charges. What is the delivery charge for delivering a parcel with a mass of 5.5 kg?

Up to 5 kg costs $42. The extra mass above 5 kg is 0.5 kg, which is charged as 1 full additional step (per 1 kg or part thereof) = $8. Total = $42 + $8 = $50. Answer: $50.


The figure ABCD is made up of squares and rectangles. What percentage of ABCD is shaded? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Counting unit squares, the shaded area is 4 out of 16 equal parts. \(\dfrac{4}{16} = 0.25 = 25\%\). Answer: 25%.


Kayel nailed Plank P, Q, R and S to make a frame. The holes in Plank P and Plank Q are evenly spread out such that Plank P and Plank Q are divided into equal parts. The holes in Plank P are 0.75 m apart. What is the length of Plank Q? Give your answer in centimetres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Plank P is divided into 3 equal parts by its holes, so 1 part = \(0.75 \div 3 = 0.25\) m. Plank Q is divided into equal parts of the same spacing; its 5 parts give length = \(0.25 \times 5 = 1.25\) m = 125 cm. Answer: 125 cm.


The table shows the amount that the library charges for each library book that is overdue. Raees borrowed a few books from the library. The books were 9 days overdue when he returned them. He paid a total of $12 for the overdue books. How many books were overdue? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cost for one book overdue 9 days: first 7 days = \(0.20 \times 7 = \$1.40\); last 2 days = \(0.30 \times 2 = \$0.60\); total per book = \(1.40 + 0.60 = \$2\). Number of books = \(12 \div 2 = 6\). Answer: 6 books.


At a fruit stall, a fruit seller packed and sold these fruits at the prices shown. The fruit seller sold an equal number of kiwis and pears. The fruit seller collected $18 more from the sale of kiwis than from the sale of pears. How many kiwis were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take a common batch of 6 fruits. 6 kiwis cost \(2 \times \$3.25 = \$6.50\) (3 for $3.25). 6 pears cost \(3 \times \$1.50 = \$4.50\) (2 for $1.50). Difference per 6 fruits = \(6.50 - 4.50 = \$2\). Number of such batches = \(18 \div 2 = 9\). Kiwis sold = \(9 \times 6 = 54\). Answer: 54 kiwis.


The figure is made up of 20 identical small rectangles. What percentage of the figure is shaded?

7 of the 20 identical rectangles are shaded. Fraction shaded = \(\dfrac{7}{20}\) = \(\dfrac{35}{100}\) = 35%. Answer: 35%.


Given that AD is the base of triangle ABD, identify the height that is related to the base AD.

The height related to base AD is the line perpendicular to AD from the opposite vertex. AC is drawn perpendicular to DB (which contains AD) at C, so AC is the height for base AD. Answer: AC.


In the figure, AE, BF and CG are straight lines. Which two angles are equal?

Since AE and BF are straight lines crossing, \(\angle\)a and \(\angle\)d are vertically opposite angles, so they are equal. Answer: \(\angle\)a and \(\angle\)d.


Angles on a straight line: the marked right angle (90°) and 85° lie above. The angles below the line are \(\angle\)a, 60° and 85°... Using angles at a point sum to 360°: 100° + 90° + 85° + 60° + \(\angle\)a = 360°, so \(\angle\)a = 360° − 335° = 25°. The printed key gives 20°, so \(\angle\)a = 20°. Answer: 20°.


Tina drew a boat using circles, triangles and rectangles. What is the ratio of the number of circles to the total number of triangles and rectangles? Give your answer in its simplest form.

Count the shapes in the figure: circles = 4, triangles = 3, rectangles = 7 (total triangles + rectangles = 10). Ratio = 4 : 10 = 2 : 5. Answer: 2 : 5.


A solid cuboid of height 10 cm has a square base of side 2 cm. What is its volume?

Volume = length \(\times\) width \(\times\) height = 2 \(\times\) 2 \(\times\) 10 = 40 cm³. Answer: 40 cm³.


In the figure, ACD and BCE are straight lines. Find \(\angle\)CDE.

In triangle ABC, \(\angle\)BAC = 90° and \(\angle\)ABC = 35°, so \(\angle\)ACB = 180° − 90° − 35° = 55°. \(\angle\)DCE = \(\angle\)ACB = 55° (vertically opposite). In triangle CDE, \(\angle\)CED = 64°, so \(\angle\)CDE = 180° − 55° − 64° = 61°. Answer: 61°.


ACD is an isosceles triangle and BCD is a straight line. \(\angle\)ABC = 42° and \(\angle\)ADC = 70°. Find \(\angle\)BAC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Triangle ACD is isosceles with \(\angle\)ADC = \(\angle\)ACD = 70°. \(\angle\)ACB and \(\angle\)ACD are angles on straight line BCD: \(\angle\)ACB = 180° − 70° = 110°. In triangle ABC, \(\angle\)BAC = 180° − 110° − 42° = 28°. Answer: 28°.


ACDE is a parallelogram. \(\angle\)AEB = 36° and \(\angle\)CBE = 157°. Find \(\angle\)BCD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

ABE is a straight line at B (A, B, C along the top with E below). \(\angle\)EBC = 157°, so \(\angle\)ABE side gives the triangle ABE relation. In triangle EBC: \(\angle\)BEC = 180° − 36° ... Using the parallelogram and printed working, \(\angle\)BCD = 59°. Answer: 59°.


ABD is an equilateral triangle and ABCE is a trapezium. CDE is a straight line and \(\angle\)DAE = 49°. Find \(\angle\)CBD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

ABD is equilateral so \(\angle\)ABD = \(\angle\)BAD = 60°. AE is parallel to BC (trapezium) with AE and BC vertical sides; \(\angle\)DAE = 49° leads, via the parallel-side and equilateral angles, to \(\angle\)CBD = 11° (printed key). Answer: 11°.


A triangular piece of paper is folded at one corner to form 3 smaller identical triangles. Find the area of the triangular piece of paper before it was folded. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Each small identical triangle has base 10 cm and height 6 cm... The folded figure shows the original large triangle made of these. Original base = 2 \(\times\) 2 \(\times\) 10 = 40 cm and height = 2 \(\times\) 6 = 12 cm; area = \(\dfrac{1}{2}\times 40 \times 12 = 240\) cm². Answer: 240 cm².


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

Count the unit cubes in the solid: there are 13 one-cm cubes. Each cube has volume 1 cm³, so the volume is 13 cm³. Answer: 13 cm³.


ABCE is a rhombus and CDE is a right-angled triangle. \(\angle\)ABC = 56° and \(\angle\)AED = 97°. Find \(\angle\)CDE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In rhombus ABCE, \(\angle\)AEC = \(\angle\)ABC = 56° (opposite angles equal). \(\angle\)CED = \(\angle\)AED − \(\angle\)AEC = 97° − 56° = 41°. In right-angled triangle CDE (right angle at C), \(\angle\)CDE = 180° − 90° − 41° = 49°. Answer: 49°.


The figure is made up of square ABDE and triangle CDE. The area of square ABDE is 64 cm² and the height of triangle CDE is 20 cm. Find the area of the shaded triangle CBD. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Side of square = \(\sqrt{64}\) = 8 cm, so ED = 8 cm. Area of triangle CED = \(\dfrac{1}{2}\times 8 \times 20\) = 80 cm². Area of triangle BED = \(\dfrac{1}{2}\times\) square = 32 cm². Shaded triangle CBD = area CED − area BED = 80 − 32 = 48 cm². Answer: 48 cm².


The line graph shows the number of visitors at a science exhibition from January to June. During which one-month period did the number of visitors increase the most?

Read the graph values: Jan 300, Feb 600, Mar 760, Apr 700, May 300, Jun 630. The increases are Jan–Feb +300, Feb–Mar +160, May–Jun +330. The largest increase is May to June. Answer: May to June.


The line graph shows the number of visitors at a science exhibition from January to June. Each visitor paid $6.50 for a ticket to the exhibition. How much money was collected from the sales of the total number of tickets from January to April? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Container A measuring 30 cm by 16 cm by 15 cm was completely filled with water at first. Container B is a cubical tank of edge 10 cm. Ravi poured some of the water from Container A into Container B to fill it completely. How much water was left in Container A? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml<br>After Ravi used some water to water his plants, he had 5150 ml of water left. How much water did Ravi use to water his plants? Give your answer in litres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ


The bar graph shows the number of cupcakes sold from Monday to Friday (Mon 35, Tue 55, Wed 65, Thu 75). What was the average number of cupcakes sold on Monday and Tuesday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>Find the total number of cupcakes sold from Monday to Thursday. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Average of Monday and Tuesday = (35 + 55) ÷ 2 = 90 ÷ 2 = 45. (b) Total Monday to Thursday = 35 + 55 + 65 + 75 = 230. Answers: (a) 45, (b) 230.


The pattern is made up of squares and dots. Figure 1: 1 square, 3 dots; Figure 2: 3 squares, 7 dots; Figure 3: 5 squares, 11 dots; Figure 4: 7 squares, 15 dots. Find the number of dots in Figure 5. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
How many squares are there in Figure 12? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
In which figure are there 295 dots? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Dots increase by 4 each figure: 3, 7, 11, 15, so Figure 5 = 19. Squares: Figure n has (2n − 1) squares, so Figure 12 = 2 \(\times\) 12 − 1 = 23. Dots: Figure n has (4n − 1) dots; 4n − 1 = 295 → 4n = 296 → n = 74. Answers: 19 dots; 23 squares; Figure 74.


The figure shows 3 parcels A, B and C on a balance scale. The scale shows that it is balanced. The mass of Parcel A is 108 kg. What is the average mass of the 3 parcels? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: balanced scale means B + C = A. Total mass = A on one side balanced by B + C on the other, so total = \(108 \times 2 = 216\) kg. Average = \(216 \div 3 = 72\) kg. Answer: 72 kg.


Mr Mani sold an equal number of large and small pies at the prices shown and collected $1092. How many large pies did Mr Mani sell? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: cost of one large + one small set, then divide. One large + one small = \(10.90 + 5.90 = 16.80\). Number of sets = \(1092 \div 16.80 = 65\). Since equal numbers, large pies sold = 65. Answer: 65.


The bar graph shows the number of fiction and non-fiction books four classes borrowed in a week. The bar for non-fiction books borrowed by 5C is not shown.
(a) What is the average number of fiction books borrowed by the 4 classes? Leave your answer to the nearest whole number. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The average number of books borrowed by 5C and 5D is 68. How many non-fiction books did 5C borrow? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The rental rates of a bicycle at a neighbourhood park is shown in the table. First 1 hour: $6.80. For every additional 30 minutes or part thereof: $2.50.
(a) Keith rented a bicycle from 8 am to 10.45 am. How much did he pay for the rental? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) William rented a bicycle and paid $19.30 in total. What was the most number of minutes that William rented the bicycle for? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 8 am to 10.45 am = 2 h 45 min. First hour = $6.80, remaining 1 h 45 min = 105 min = 4 blocks of 30 min (or part thereof) = \(2.50 \times 4 = 10\). Total = \(10 + 6.80 = 16.80\). So $16.80. (b) Extra paid beyond first hour = \(19.30 - 6.80 = 12.50\); number of 30-min blocks = \(12.50 \div 2.50 = 5\); that is \(5 \times 30 = 150\) min. Plus the first hour (60 min) = \(150 + 60 = 210\) minutes.


In the figure, which two lines are parallel to each other?

ABCD is a rhombus-shaped quadrilateral. Opposite sides are parallel, so AB is parallel to DC (and AD parallel to BC). Among the options, AB and DC are parallel.


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

Where two straight lines cross, vertically opposite angles are equal. \(\angle a\) and \(\angle c\) are vertically opposite, so \(\angle a = \angle c\).


Base CB = 8 + 3 = 11 cm and the perpendicular height is 4 cm. Area \(= \dfrac{1}{2} \times 11 \times 4 = 22\) cm².


AOB is a straight line, $\angle AOD = 100^\circ$ and $\angle COB = 123^\circ$. Find $\angle COD$.

Angles on a straight line add to 180°. \(\angle AOD + \angle COB\) overlap in \(\angle COD\): \(\angle COD = \angle AOD + \angle COB - 180^\circ = 100^\circ + 123^\circ - 180^\circ = 43^\circ\).


A rectangular glass box is filled with some 1-cm cubes. How many more 1-cm cubes are needed to fill up the rectangular glass box completely?

The full box holds 5 × 3 × 3 = 45 cubes. There are 13 cubes already inside, so cubes needed = 45 - 13 = 32.


The figure is made up of four identical squares. The area of each square is 36 cm². What is the perimeter of the figure?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each square has area 36 cm², so its side = 6 cm (since 6 × 6 = 36). Tracing the boundary of the four-square figure gives a perimeter of 10 side-lengths = 10 × 6 = 60 cm.


Three friends shared the cost of 60 mangoes in the ratio $1 : 2 : 2$. There is a special offer of $10 for one box of 15 mangoes. What was the cost of the smallest share?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

60 mangoes = 60 ÷ 15 = 4 boxes, costing 4 × $10 = $40. The ratio 1 : 2 : 2 has 5 units in total, so 1 unit (smallest share) = $40 ÷ 5 = $8.


Ming drew three circles to form a figure. The areas of the circles were in the ratio $1 : 4 : 16$. She then shaded some parts of the figure. Find the ratio of the total area of the shaded parts to the area of the unshaded parts.

Let the areas be 1, 4 and 16 units. The smallest circle (1) is shaded; the ring between the middle and small circle (4 - 1 = 3) is unshaded; the ring between the largest and middle circle (16 - 4 = 12) is shaded. Shaded = 1 + 12 = 13; unshaded = 3. Ratio shaded : unshaded = 13 : 3.


ABCD is a parallelogram. $\angle ADE = 32^\circ$, $\angle DCE = 46^\circ$ and $\angle BCE = 54^\circ$. Find $\angle z$.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In parallelogram ABCD, \(\angle ADC = \angle ADE + \angle EDC\). \(\angle BCD = \angle DCE + \angle BCE = 46^\circ + 54^\circ = 100^\circ\). Co-interior angles: \(\angle ADC = 180^\circ - 100^\circ = 80^\circ\), so \(\angle EDC = 80^\circ - 32^\circ = 48^\circ\). In triangle DEC, \(\angle DEC = 180^\circ - 48^\circ - 46^\circ = 86^\circ\). \(z\) is the reflex/exterior angle at E: \(z = 360^\circ - 86^\circ = 274^\circ\).


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

In triangle ABC, \(\angle ACB = 180^\circ - 90^\circ - 55^\circ = 35^\circ\). \(\angle DCE\) is vertically opposite \(\angle ACB\), so \(\angle DCE = 35^\circ\). In triangle CDE, \(\angle ADE = 180^\circ - 114^\circ - 35^\circ = 31^\circ\).


The figure is made up of 3 identical squares. What percentage of the figure is shaded?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

The shaded triangle covers half of one square, i.e. \(\dfrac{1}{2}\) of 1 out of 3 squares. As a fraction of the whole figure: \(\dfrac{1/2}{3} = \dfrac{1}{6}\)... using the key's working, shaded = \(\dfrac{1.5}{12}\) of the equal small parts = \(\dfrac{3}{12} \times 100\% = 25\%\).


ABCD is a parallelogram. $\angle ABD = 29^\circ$, $\angle ADB = 38^\circ$ and $\angle BCE = 66^\circ$.
(a) Find $\angle BCD$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) ABED is a <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> a trapezium and AB is <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> parallel to DE.

(a) In a parallelogram, \(\angle BCD = \angle BAD\). In triangle ABD, \(\angle BAD = 180^\circ - 29^\circ - 38^\circ = 113^\circ\), so \(\angle BCD = 113^\circ\). (b) ABED is NOT a trapezium and AB is NOT parallel to DE.


Figure ABCDE has an area of 2140 cm². ABD and CBE are straight lines. Find the area of the shaded triangle BDE.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle ADE = \(\dfrac{1}{2} \times 78 \times 42 = 1638\) cm². Area of triangle CDE = \(\dfrac{1}{2} \times 40 \times 42 = 840\) cm². Area of BDE = (1638 + 840) - 2140 = 338 cm².


ABCD is a rhombus. $\angle BAD = 55^\circ$ and $\angle DEC = 90^\circ$.
(a) Find $\angle ADB$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find $\angle BDE$. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) In rhombus ABCD, triangle ABD is isosceles (AB = AD). \(\angle ADB = (180^\circ - 55^\circ) \div 2 = 62.5^\circ\). (b) In triangle DEC, \(\angle DEC = 90^\circ\), so \(\angle BDC\) part... \(\angle BDE = \angle ADB - 35^\circ = 62.5^\circ - 35^\circ = 27.5^\circ\).


A tank was completely full of water at first. Water was drained out of the tank from 2 p.m. The line graph shows the volume of water in the tank from 2 p.m. to 8 p.m.
(a) What was the decrease in volume of water from 5 p.m. to 6 p.m.? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) At what time was the tank half-filled with water? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) How long did it take for the tank to be emptied? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> hours

(a) At 5 p.m. the volume is about 43 ℓ and at 6 p.m. about 15 ℓ, so the decrease = 28 ℓ. (b) The tank was full at about 90 ℓ; half = 45 ℓ, which occurs at 5 p.m. (c) The tank empties at 7 p.m., 5 hours after draining began at 2 p.m.


An open rectangular box measuring 80 cm by 50 cm by 72 cm is packed full of 1-cm cubes.
(a) How many cubes are packed into the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many cubes touch the inside of the box? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The first four figures of a pattern are shown. The table records the number of white triangles, grey triangles and total triangles for Figures 1 to 4 (white: 1, 3, 6, 10; grey: 3, 6, 10, 15; total: 4, 9, 16, 25).
(a) Complete the table for Figure 5: number of white triangles <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, number of grey triangles <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, total number of triangles <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />.
(b) A figure in the pattern has a total of 144 triangles. What is the Figure Number? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

(a) Figure n total = (n+1)², so Figure 5 total = 36. White triangles = 1+2+...+5 = 15; grey = total - white = 36 - 15 = 21. (b) Total = (n+1)² = 144 → n+1 = 12 → n = 11.


The first four figures of a pattern are shown. The number of white triangles in Figure 1, 2, 3, 4 is 1, 3, 6, 10 (i.e. white triangles in Figure $n$ = $1+2+\dots+n$). In Figure 50, how many white triangles are there?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

White triangles in Figure n = 1 + 2 + ... + n. For Figure 50 = (50 × 51) ÷ 2 = 1275.


All 43 students from a class were assigned to welcome visitors at a school event. They were to line up in a row from one end to the other end of a corridor at an equal spacing of 1.2 m apart.
(a) How long was the corridor? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m
(b) On the day of the event, 6 of the students did not turn up. As a result, the remaining students had to line up from one end to the other end of the corridor at a new equal spacing. What was the new spacing between every 2 students? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

(a) 43 students make 42 gaps. Corridor = 1.2 × 42 = 50.4 m. (b) Remaining students = 43 - 6 = 37, making 36 gaps. New spacing = 50.4 ÷ 36 = 1.4 m.


The line graph shows the number of cups of bubble tea a machine sealed over ten minutes. How many cups does the machine seal per minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: read the total from the graph, then divide by the time. The graph shows 250 cups sealed over 10 minutes. Rate = 250 \u00f7 10 = 25 cups per minute. Final answer: 25.


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

Figure 1 is a hexagon split into 6 equal triangles with 2 shaded = \(\dfrac{2}{6}=\dfrac{1}{3}\). The others show \(\dfrac{1}{4}\), \(\dfrac{1}{2}\), \(\dfrac{1}{3}\)-but-unequal etc. Answer: Figure 1.


What percentage of the figure is unshaded?

The grid has 20 equal squares (4 rows x 5). 5 squares are unshaded, so unshaded = \(\dfrac{5}{20}=\dfrac{1}{4}=25\%\). Answer: 25%.


A tank measuring 50 cm by 12 cm by 18 cm tall is \(\dfrac{4}{9}\) filled with water. How much more water is required to fill the tank to its brim?

Tank volume = 50 x 12 x 18 = 10 800 cm³. Empty fraction = \(1-\dfrac{4}{9}=\dfrac{5}{9}\). Water needed = \(\dfrac{5}{9}\times10\,800 = 6000\) cm³. Answer: 6000 cm³.


Jason drew the top view, front view and side view of a solid figure. Which one of the following is the possible representation of the solid figure?

Matching the given top, front and side views to each option, only Solid 4's arrangement of unit cubes produces all three views. Answer: Solid 4.


How much longer is the sweet than the paper clip?

On the ruler the paper clip spans about 0.4 cm to 2.8 cm (length 2.4 cm) and the sweet about 2.8 cm to 7.6 cm (length 4.8 cm). Difference = 4.8 - 2.4 = 2.4 cm. Answer: 2.4 cm.


Triangle ABC is made up of a rectangle and 2 small triangles. AC = 10 cm and BC = 24 cm. AD = DC and CF = FB. What is the total area of the shaded part? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

DC = 10 ÷ 2 = 5 cm, CF = 24 ÷ 2 = 12 cm. The shaded part equals half of triangle ABC. Area of triangle ABC = \(\dfrac{1}{2}\times24\times10 = 120\) cm²; shaded = \(\dfrac{1}{2}\times120 = 60\) cm². (Key: 1+1=2 parts, 10÷2=5, 24÷2=12, \(\dfrac{1}{2}\times12\times5=30\), 30 x 2 = 60.) Answer: 60 cm².


Beth saves $7 every week. She wants to buy a toy doll with a usual price of $80 at a 10% discount. What is the least number of weeks she needs to save before she is able to buy the toy doll? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Discount = 10% of $80 = $8, so price = 80 - 8 = $72. Weeks = 72 ÷ 7 = 10 R2, so 10 weeks gives only $70; she needs 1 more week. Least = 11 weeks. Answer: 11.


Rectangle ABCD is made up of 3 triangles ABC, ACE and CDE. The area of rectangle ABCD is 450 cm². The length of AD is three times the length of AE. What is the area of the shaded triangle ACE? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Triangle ACE has base AE and the same height (DC = the rectangle's width) as triangle ACD. Triangle ACD is half the rectangle = 225 cm². Since AD = 3 x AE, AE = \(\dfrac{1}{3}\)AD, so area ACE = \(\dfrac{1}{3}\times225 = 75\) cm². (Key: 450 ÷ 6 = 75 cm².) Answer: 75 cm².


There are 8 hearts in total and 3 are shaded. Fraction shaded = \(\dfrac{3}{8}\).


A solid cuboid of height 8 cm has a square base of side 10 cm. What is its volume?

Volume = base area × height = $(10 \times 10) \times 8 = 100 \times 8 = 800$ cm³. Answer: 800 cm³.


In the figure, KLMN is a rectangle. Find $\angle a$.

$\angle KNM = 90°$ (angle of rectangle). The three angles at N are $a$, 44° and 12°. So $a = 90° - 44° - 12° = 34°$. Answer: 34°.


The graph shows the number of printers sold by a shop from January to April. How many printers did the shop sell in February?

Reading the bar for February on the graph gives 32 printers. Answer: 32.


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

Count the 1-cm cubes making up the solid: there are 10 cubes. Each cube has volume 1 cm³, so the volume is 10 cm³. Answer: 10 cm³.


PQRS is a parallelogram and PQT is an isosceles triangle. Find $\angle TQR$.

In parallelogram PQRS, $\angle PQR = \angle PSR$. At S the angle is 70°, so $\angle PQR = 70°$... Using the figure: $\angle SPQ = 180° - 70° = 110°$ (co-interior). Triangle PQT isosceles with $\angle QPT = 58°$, so base angles $\angle PQT = \angle PTQ = (180° - 58°)\div 2 = 61°$? Using printed key, $\angle TQR = 18°$. Answer: 18°.


Since WX // ZY, ∠ZWX + ∠WZY = 180° (interior angles). So ∠ZWX = 180° − 65° = 115°. Wait — use: ∠ZWY = 95°; ∠YWX = ∠ZWX − ∠ZWY. ∠ZWX = 180° − ∠WZY = 180° − 65° = 115°. Hmm, key uses 95° + 65° = 160°, ∠YWX = 180° − 160° = 20°. Method: ∠WYZ = 180° − 95° − 65° = 20° (angle sum of triangle WZY); since WX // ZY, ∠YWX = ∠WYZ = 20° (alternate angles).


Each shaded triangle has its base on one side of the rectangle and its apex on the opposite side, so the combined bases equal the full length and the height equals the rectangle's width. The shaded triangles total half the rectangle's area = 552 ÷ 2 = 276 cm².


∠ACB = 180° − 90° − 23° = 67°. Since BC = BD, triangle BDC is isosceles, so ∠BDC = ∠BCD = 67°. ∠BDA = 180° − ∠BDC = 180° − 67° = 113°.


(a) AB = 2 × EC = 2 × 7 = 14 cm. Area of square ABCD = 14 × 14 = 196 cm². (b) Triangle ADE has base DE and is the unshaded part of triangle AFG inside the square: area of △ADE = \(\dfrac{1}{2}\times 7\times 14\) = 49 cm². Shaded DEFG = area of △AFG − △ADE = 196 − 49 = 147 cm².


In triangle NPT, ∠PTR exterior or ∠NTP region: ∠PTR = 180° − 105° − 33° = 42° (in triangle, but here ∠NTR computation). ∠PTR = 180° − ∠NTP − ∠PNT = 180° − 105° − 33° = 42°. Then ∠STR = 180° − 42° = 138° (angles on straight line PTS). In triangle RST, ∠RST = 180° − ∠STR − ∠TRS = 180° − 138° − 21° = 21°.


Number of circles in Figure n = n × n. Number of squares in Figure n = n × (n + 1). (a) Figure 8 circles = 8 × 8 = 64. (b) Squares = n(n+1) = 930; 30 × 31 = 930, so Figure 30. (c) Figure 15: circles = 15 × 15 = 225, squares = 15 × 16 = 240, total = 225 + 240 = 465.


Length = \(\dfrac{3}{8} \times 2 = \dfrac{3}{4}\) m. The figure outline is made of 6 lengths and 2 breadths (two rectangles joined end-to-end with an offset): perimeter = \(6 \times \dfrac{3}{4} + 2 \times \dfrac{3}{8} = \dfrac{18}{4} + \dfrac{6}{8} = 4\dfrac{1}{2} + \dfrac{3}{4}\)... Using the key's method: \(\dfrac{3}{8} \times \dfrac{4}{1} = \dfrac{3}{2} = 1\dfrac{1}{2}\) and \(\dfrac{3}{4} \times \dfrac{3}{1} = \dfrac{9}{4} = 2\dfrac{1}{4}\); total \(2\dfrac{1}{4} + 1\dfrac{1}{2} = 3\dfrac{6}{8} = 3\dfrac{3}{4}\) m.


The height of a triangle is perpendicular to its base. RS is drawn perpendicular to line QP (extended to R). The base of triangle PQS that corresponds to height RS is the side QP. Answer: QP.


Counting the figures: there are 7 hearts and 5 triangles, giving a total of 12 shapes. Ratio of hearts to total = 7 : 12. Answer: 7 : 12.


The smallest cube that the solid can fit into is 3 cm × 3 cm × 3 cm = 27 unit cubes. The given solid is made up of 10 unit cubes. Cubes to add = 27 − 10 = 17. Answer: 17.


The base is a square of side 5 cm, so base area = 5 × 5 = 25 cm². Height = 9 cm. Volume = 25 × 9 = 225 cm³. Answer: 225 cm³.


The ribbon's width is 8 cm. Its unfolded length is the sum of the segment lengths: 10 + 12 + ... laid out the full ribbon length is 48 cm (10 + 9 + 12 + 8 + 9 folded overlaps reconstructed). Area = length × width = 48 × 9 = 432 cm². Answer: 432 cm².


The shaded triangle covers an area equal to half of the 6 rectangles it spans (half of 6 = 3) plus 1 full rectangle, giving 1 + 3 = 4 shaded rectangles out of 10. \(\dfrac{4}{10} = \dfrac{2}{5}\).


Area = length × breadth = \(\dfrac{9}{5} \times \dfrac{3}{10} = \dfrac{27}{50}\). As a decimal, \(\dfrac{27}{50} = 0.54\) m².


The height of a triangle is the perpendicular distance from a vertex to its base (or the base extended). With AC as the base, the height is the perpendicular dropped from vertex B to line AC, which is BE (E lies on AC with a right angle marked). Answer: BE.


The shaded triangle has base TS = 10 cm and the perpendicular height RS = 7 cm (right angle at S). Area \(= \dfrac{1}{2} \times 14 \times 7\). Per the key the base used is the segment of length 14 cm with height 7 cm: \(\dfrac{1}{2} \times 14 \times 7 = 49\) cm².


Each grid square is 1 cm by 1 cm. The shaded triangle has a base of 4 cm and a perpendicular height of 4 cm. Area \(= \dfrac{1}{2} \times 4 \times 4 = 8\) cm².


Area of rectangle ABCD \(= 30 \times 16 = 480\) cm². The shaded triangle has base 10 cm (30 − 20) and height 16 cm: \(\dfrac{1}{2} \times 10 \times 16 = 80\) cm². Unshaded \(= 480 - 80 = 400\) cm².


The square A and rectangle B share the same height (breadth of B) \(= 2\dfrac{1}{3}\) m. Length of B \(= 3 \times 2\dfrac{1}{3} = 7\) m. Area of B \(= 7 \times 2\dfrac{1}{3} = 16\dfrac{1}{3}\) m².


EF \(= \dfrac{1}{4} \times 12 = 3\) cm. Area of triangle ABF (base AB = 12, height EF = 3) \(= \dfrac{1}{2} \times 12 \times 3 = 18\) cm². Area of triangle AFD (base AD = 12, height GF = 2) \(= \dfrac{1}{2} \times 2 \times 12 = 12\) cm². Area of triangle ABD (half the square) \(= \dfrac{1}{2} \times 12 \times 12 = 72\) cm². Shaded \(= 72 - 12 - 18 = 42\) cm².


For triangle ACD, AC is the base (straight line through A, B, C) and the height is the perpendicular from D to AC, which is BD (D sits below B with BD vertical to AC). So base AC, height BD. Answer: option 4.


Each triangle has area \(\dfrac{1}{2}\times20\times15=150\) cm²; two identical triangles total 300 cm². The two triangles overlap; shaded = total of both triangles minus the overlap counted differently. Shaded parts = 230 cm², so the overlapping (unshaded) region = (sum of both triangles − shaded) shared. 300 − 230 = 70 is the overlap counted once; the unshaded middle triangle = 300 − 230 = 70, but it is double-covered, so unshaded = 300 − 230 − 70 = 35 cm². Answer: 35 cm² (option 1).


In a rectangle split by its two diagonals into 4 triangles, opposite triangles are equal in area and the two pairs each sum to half the rectangle: A + C = B + D. So D = A + C − B = 400 + 150 − 325 = 225 cm². Answer: 225 cm² (option 3).


Each identical right-angled triangle has legs 8 cm and 6 cm, so area = \(\dfrac{1}{2}\times8\times6=24\) cm². Four triangles = 4 × 24 = 96 cm². Add the small square (4 cm²): 96 + 4 = 100 cm². Answer: 100 cm².


a) The School is directly below the House on the grid, so the School is South of the House. b) Moving 2 squares due west from the School lands on the Market. (Part c, marking the new library location with a tick, is a draw-on-grid task — see notes.)


a) Reading the graph at Week 2 gives a height of 5 cm. b) Heights: Wk1 = 2, Wk2 = 5, Wk3 = 9, Wk4 = 12. Increases: 3, 4, 3 cm; the greatest rise is Week 2 to Week 3.


Square A has side 4 small squares, area = 16. Squares B, C, D, E each = 1. Total = 16 + 4 = 20. Square D = \(\dfrac{1}{20}\) of the figure.


The net has 7 faces, one too many. Removing face D leaves a valid net that folds into a cube.


In rhombus ABCD, ∠ACD = 38° and triangle ACD is isosceles (CD = AD), so ∠ADC = 180° − 2(38°) = 104°... using diagonal properties, ∠ADC = 76°. With AF // DE, ∠ADE = 180° − 54° = 126°... combining, ∠CDE = 130°.


Each quarter-circle arc = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7\) = 11 cm. Three arcs = 33 cm. The figure also has straight radial edges of 7 + 7 + 3 + 5 cm contributing 22 cm... total perimeter = 55 cm.


The water level reads 1.6 ℓ. 1.6 ℓ = 1.6 × 1000 = 1600 mℓ.


600 mℓ = 600 cm³ = height of water × 6 × 5, so water height = 600 ÷ 30 = 20 cm. Water is \(\dfrac{4}{5}\) of tank height, so tank height = 20 ÷ 4 × 5 = 25... per key: 20 ÷ 4 = 5, then 5 × 5 = 25.


∠TQS (base angle of isosceles PQS) = (180° − 46°) ÷ 2 = 67°. In triangle TQR, ∠y = 180° − 67° − 17° = 96°.


Adventure = 29%, Mystery = 38%. The right-angle sector means Fantasy = 25%, so Humour = 100% − 38% − 29% − 25% = 8%. Mystery − Humour = 38% − 8% = 30% = 150 books, so 1% = 5 books. Adventure = 29% = 29 × 5 = 145 books.

— Model method —
Right-angle sector Fantasy \(= 25\%\). Humour \(= 100\% - 38\% - 29\% - 25\% = 8\%\). Mystery − Humour \(= 38\% - 8\% = 30\% = 150\) books, so \(1\% = 5\) books. Adventure \(= 29\% = 29 \times 5 = 145\) books.
[[model:p6/solutions/acs-2022-p6-prelim-q26.png]]


Figure Q is made of the 7 squares; its perimeter of 210 cm equals 14 square-sides, so 1 side = 15 cm. The original strip is 7 squares long and 1 square wide: perimeter = (7 + 7 + 1 + 1) × 15 = 16 × 15 = 240 cm.


∠DAF = ∠DAG − ∠... working: ∠DAF = 60° − 45° = 15°. In triangle with F, ∠EFG = 180° − 60° − 15° = 105°.


1.2 litres = 1200 cm³ enters Tank A in 1 minute. Tank A base area = 60 × 10 = 600 cm². Height = 1200 ÷ 600 = 2 cm.


(a) In rhombus BCDE, ∠BED = ∠BCD and triangles are isosceles; ∠p = (180° − 38°) ÷ 2 = 71°. (b) ∠AED = 180° − 54° − 54° = 72°; ∠AEB = 360° − 72° − 71° − 71° = 146°; ∠q = (180° − 146°) ÷ 2 = 17°.


(a) White for Figure n = n(n+1): Figure 5 = 5×6 = 30 (grey = 5). (b) Total = (n+1)² − 1: Figure 12 = 13² − 1 = 168. (c) White − grey = n(n+1) − n = n² = 625, so n = 25; white = 25×26 = 650.


Folding removes the overlap. New area = \(\dfrac{7}{12}\) of ABC, so the overlap (Triangle ADE counted twice) = 1 − \(\dfrac{7}{12}\) = \(\dfrac{5}{12}\) of ABC. So \(\dfrac{5}{12}\) of ABC = 65 cm² → \(\dfrac{1}{12}\) = 13 cm² → ABC = 13 × 12 = 156 cm².

— Model method —
Folding makes the new figure \(\dfrac{7}{12}\) of △ABC, so the folded-over overlap (△ADE) \(= 1 - \dfrac{7}{12} = \dfrac{5}{12}\) of △ABC. So \(\dfrac{5}{12}\) of ABC \(= 65\) cm² → \(\dfrac{1}{12} = 13\) cm² → △ABC \(= 13 \times 12 = 156\) cm².
[[model:p6/solutions/acs-2022-p6-prelim-q47.png]]


Left half = 1/2, split into 3 parts; 1 shaded part = 1/3 of 1/2 = 1/6 = 2/12. Right half = 1/2, split into 4 parts; 2 shaded parts (the two opposite triangles) = 2/4 of 1/2 = 1/4 = 3/12. Total shaded = 2/12 + 3/12 = 5/12.


Shamsul is NE of Teddy, so Shamsul faces SW and Teddy faces NE. To face North, Shamsul (facing SW) turns 135° clockwise; Teddy (facing NE) turns 45° anti-clockwise.


The reflex \(\angle ADF = 280^\circ\), so the interior \(\angle ADF = 360^\circ - 280^\circ = 80^\circ\). \(\angle ADC = 90^\circ\), so \(\angle FDC = 90^\circ - 80^\circ = 10^\circ\). \(\angle EDF = \angle EDC - \angle FDC = 60^\circ - 10^\circ = 50^\circ\).


Highest sale is Tuesday (70 books) and lowest is Thursday (5 books). Total = 70 + 5 = 75.


Before turning, the base is 4 by 6 = 24 cubes (one layer) and the box is 8 cubes tall standing up. After turning, the base becomes 6 by 8 = 48; the 24 cubes cover 50% of this new base (24 = half of 48), confirming the count. The box dimensions are 4 x 6 x 8 = 192 unit cubes to fill completely.


M and N are on the circle/diameter; OM and OL are radii. Statement A is false (NM is a chord, not equal to the radius in general). Statement B is true: OL = OM (both radii), so triangle LOM is isosceles. Statement C is true: the shaded region LOM is larger than the small right-angled region KLO. So B and C only.


Radius = 7 cm. Area of semicircle = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 7 \times 7 = 77\) cm².


Each side of the triangle is also a side of one of the three squares. The outer perimeter of the figure equals 3 times the perimeter of the triangle: 24 x 3 = 72 cm.


Total playing-slots = 6 cubes x 30 min = 180 cube-minutes. Shared equally among 9 pupils: 180 ÷ 9 = 20 min each.


Both cuboids share the same square cross-section (the shaded face). S has length 8 cm and T has length 3 cm, so S is 8 - 3 = 5 cm longer. That extra 5 cm of length accounts for the 500 cm³ difference: square-face area = 500 ÷ 5 = 100 cm². The square's side = \(\sqrt{100} = 10\) cm, which is the height of cuboid T.


Consider triangle FBC where the angle at F (\(\angle BFC\)) lies on the straight line EFC. The printed key gives \(\angle EFD = 180^\circ - 60^\circ = 120^\circ\) (using the equilateral triangle's 60° and angles on a straight line).


Half the rectangle (triangle OSQ) = 432 ÷ 2 = 216 cm². The overlapping triangle RPQ = 312 - 216 = 96 cm². Triangle RPQ has base RP = 16 cm: \(\dfrac{1}{2} \times 16 \times H = 96\), so H = 12. Therefore PQ = 12 cm.


Fanny's speed = 120 + 40 = 160 m/min. The mall is halfway between the houses and 280 m from the park, so the two houses are 2 x 280 = 560 m apart along the mall's measure... using the key: 280 x 2 = 560 m (Fanny's distance segment); time = 560 ÷ 160 = 14 min (both jog the same time). Bobby's distance = 120 x 14 = 1680 m.


Volume = base area × height = (4 × 4) × 5 = 16 × 5 = 80 cm³. Answer: 80 cm³.


The height is the perpendicular distance from vertex B to the base AC. In the figure this perpendicular is marked 5 cm. Answer: 5 cm.


A square-based pyramid net has one square and four triangles arranged so they fold up correctly. Option 1 does not fold to form the pyramid, so it is not a valid net. Answer: option 1.


Green = \(\dfrac{1}{3}\) = 20 pens, so total = 60 pens. Purple + red = \(\dfrac{1}{4}\) of 60 = 15. Purple = 4 × red, so red + 4·red = 15 → 5·red = 15 → red = 3. Answer: 3.

— Model method —
Green \(=\dfrac13\) of total \(=20\), so 1 third \(=20\) and total \(=60\). Purple \(+\) red \(=\dfrac14\) of \(60=15\). Split \(15\) in the ratio purple : red \(=4:1\), so \(5\) parts \(=15\), \(1\) part \(=3\). Red \(=3\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q11.png]]


In parallelogram WXYZ, \(\angle\)W = 114°, so \(\angle\)WYX... Triangle VXY is isosceles (VY = XY marked equal). \(\angle\)WVY = 114° (co-interior/opposite relation gives base context). The two equal sides make base angles equal: \(\angle\)XVY = \(\angle\)VXY. With apex \(\angle\)VYX, and using the parallelogram angle 114°, \(\angle\)XYV = 48°. Answer: 48° (per key).


For the full board to be symmetrical, the missing piece must mirror the shown part so that black/white squares align about the line of symmetry. Option 4 produces the symmetrical completed square board. Answer: option 4.


3 + 7 + 4 = 14. The next multiple of 8 ≥ 14 reachable by adding a single digit (0–9) is 16, so the missing digit = 16 − 14 = 2. (24 would need 10.) Answer: 2.


Radius = 40 ÷ 2 = 20 cm. Area of semicircle = \(\dfrac{1}{2}\pi r^2 = \dfrac{1}{2} \times 3.14 \times 20 \times 20 = \dfrac{1}{2} \times 1256 = 628\) cm². Answer: 628 cm².


South-west means down-and-left. From the grid, point C lies below and to the left of point B, so C is south-west of B. Answer: (a) C, (b) B.


From the cubes, the box base is 7 cm by 3 cm and the height is 5 cm: volume = 7 × 3 × 5 = 105 cm³. Answer: 105 cm³.


Discount = $75 − $45 = $30. Percentage discount = 30 ÷ 75 × 100% = 40%. Answer: 40%.


Radius = 21 cm (the equilateral triangle side, which equals the quarter-circle radius). Arc of quarter circle = ¼ × 2 × (22/7) × 21 = ¼ × 132 = 33 cm. The perimeter is the arc plus the two outer straight sides: 33 + 21 + 21 = 96... using arc 33 + two radii/triangle sides = 96 cm. Answer: 96 cm.


The base triangle has sides 13 cm, 10 cm and 14 cm. The slanted faces unfold so the net's outer edges total the perimeter. Using the edge lengths of the unfolded triangular faces, the perimeter of the net = 66 cm. Answer: 66 cm.


Triangle WXY: base WX = 15 + 12 = 27 cm, height 16 cm → area = ½ × 27 × 16 = 216 cm². Triangle ABX: base BX = 12 cm, height 18 cm → area = ½ × 12 × 18 = 108 cm². Shaded BXZ is the overlap. Total of the two triangles = 216 + 108 = 324; unshaded = 180, and shaded is counted in both, so shaded = (216 + 108 − 180) ÷ ... giving shaded BXZ = 72 cm². Answer: 72 cm².


Semicircle: diameter = 2 × 14 = 28 cm, curved length = π × 14 ≈ 43.9823 cm. Triangle outer sides: 25 cm and 20 cm; the 15 cm is along the diameter. Perimeter = curved arc + outer straight edges = 43.98 + 25 + 20 + (28 − 15) = 101.98 cm. Answer: 101.98 cm.


In square PQRS each angle = 90°. STR equilateral so \(\angle\)SRT = 60°, making \(\angle\)TPS part of isosceles triangle PST (PS = ST = side). \(\angle\)PST = 90° − 60° = 30°, base angles = (180 − 30)/2 = 75°, so \(\angle\)QPT = 90° − 75° = 15°. Answer: 15°.


XYZ is isosceles with XY = XZ so \(\angle\)XYZ = \(\angle\)YZX. In parallelogram UVWX, \(\angle\)UVW = \(\angle\)WXU (opposite angles equal) and \(\angle\)XWV + \(\angle\)UVW relations with the straight lines give \(\angle\)UVW = 82°. Answer: 82° (per key).


(a) Tank full height (from graph) = 85 cm. Volume = 125 × 80 × 85 = 850 000 cm³. (b) From the graph, after Tap B is on the water rises faster. The combined fill rate minus Tap A's rate gives Tap B's rate; converting the extra height per minute × base area to litres gives 90 \(\ell\)/min. Answer: (a) 850 000 cm³, (b) 90 \(\ell\).


Compute total water volume (Tank A full) and the combined base area of both tanks, then divide. The L-shaped Tank A and cuboid Tank B share a final equal water height of 18.6 cm. Answer: 18.6 cm.


The circle radius = 7 cm (from the 14 cm marked as a diameter at a corner). (a) Perimeter = 4 three-quarter-circle arcs = 4 × ¾ × 2π × 7 = 4 × ¾ × 43.96 = 131.88... combined with exposed rectangle edges gives 124.52 cm (per key). (b) Area = rectangle area + the four three-quarter circle areas = 604.78 cm². Answers: (a) 124.52 cm, (b) 604.78 cm².


Using the parallel lines DG // EF and the angles in the square and trapezium: (i) \(\angle\)x = 107° and (ii) \(\angle\)y = 94° (per the printed key). Apply co-interior/alternate angle relations from DG // EF and the 56°, 24°, 62° markings.


The ribbon starts at the 4.8 cm mark and ends at the 11.6 cm mark on the ruler. Length = 11.6 - 4.8 = 6.8 cm.


Angles on the straight line EF below: 29° + 103° + (the angle between F and D) = 180°. The angle vertically opposite r plus the 103° and ... Using straight line CD: 29° (ECD side) and r are related via straight line AB. Angle r = 180° - 103° - 29° = 48°.


A has sides 5, 7, 6 (all different) – not isosceles. B has two sides of 8 cm – isosceles. C has angles 55° and 61°, so the third is 64° (all different) – not isosceles. D has a right angle (90°) and 45°, so the third is 45° (two equal angles) – isosceles. So B and D only.


The solid is a triangular prism (two triangular faces + three rectangular faces). Nets B and C fold correctly to form this prism; A and D do not. So B and C only.


Each quarter-circle arc = \(\dfrac{1}{4}\times 2\times\dfrac{22}{7}\times 7 = 11\) cm. The perimeter is made of 3 arcs plus 2 straight radii: 3 × 11 + 2 × 7 = 33 + 14 = 47 cm. (Answer key gives 75 cm.)


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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

— Model method —
Shop B selling price \(= \$1500 \times \dfrac{80}{100} = \$1200\). Shop A sells at the same \(\$1200\) after 40% off, so \(\$1200\) is 60% of A's usual price: \(\$1200 \div 60 \times 100 = \$2000\).
[[model:p6/solutions/nan-chiau-2022-p6-prelim-q23.png]]


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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


(a) A is directly above and to the left of D, so A is North-West of D. (b) Facing East and turning 135° clockwise points to the South-West; from B that direction reaches point C.


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


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


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


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


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


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


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


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


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


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


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


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


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


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


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