Chandra spent \(\dfrac{1}{3}\) of his money on 8 files and 6 notebooks at a book shop at first. Each file cost twice as much as a notebook. He then bought more notebooks with \(\dfrac{1}{4}\) of his remaining money. How many notebooks did Chandra buy altogether at the bookshop?

Each file = 2 notebooks in cost, so 8 files + 6 notebooks = 16 + 6 = 22 notebook-units of cost, using 1/3 of his money. So 1/3 of his money buys 22 notebooks worth; 1 notebook costs (1/3 ÷ 22) of his money = 1/66 of his money. Remaining money = 2/3; he spends 1/4 of it = 2/3 × 1/4 = 1/6 of his money on notebooks: (1/6) ÷ (1/66) = 11 more notebooks. Altogether notebooks = 6 + 11 = 17.


The figure shows a rectangle that is cut into 4 triangles A, B, C and D. Triangle A = 128 cm², Triangle B = 144 cm², Triangle C = 64 cm². Find the area of triangle D.

The two diagonals from the bottom-left corner split the rectangle so that triangles on opposite sides relate: triangle A + triangle C = triangle B + triangle D (each pair makes half the rectangle along a diagonal arrangement). So 128 + 64 = 144 + D, giving 192 = 144 + D and D = 48 cm².


(a) Find the value of \(130 \div 5\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Find the value of \(780 \times 60\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


(a) Express \(2\dfrac{5}{6}\) as an improper fraction. (b) Find the value of \(\dfrac{3}{4} \times \dfrac{6}{7}\).


A printer prints 160 pages in 4 minutes. At this rate, how many pages does the printer print in 15 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rate = 160 ÷ 4 = 40 pages per minute. In 15 minutes: 40 × 15 = 600 pages.


The table shows the lowest and highest temperatures of cities P, Q, R and S. P: 15 to 23. Q: 23 to 29. R: 10 to 18. S: 6 to 16.
(a) When Amy was at some of these cities, the temperature was 13°C. Name the cities that Amy visited. (Answer with the two city letters, e.g. R and S.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Name the city with the smallest difference in temperatures, then find this difference. City <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, difference <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> °C

(a) 13°C must lie between a city's lowest and highest. R is 10–18 and S is 6–16, both include 13°C, so Amy visited R and S. (b) Differences: P 23−15=8, Q 29−23=6, R 18−10=8, S 16−6=10. The smallest is City Q with a difference of 6°C.


Mrs Ravi bought 10 packets of sweets for her pupils. Each packet contained 28 sweets. She gave each pupil 8 sweets. How many pupils were there in the class? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total sweets = 10 × 28 = 280. Number of pupils = 280 ÷ 8 = 35.


The mass of a container is 475 g when it is \(\dfrac{1}{2}\) filled with rice grains. When it is \(\dfrac{3}{4}\) filled with rice grains, its mass is 520 g. Find the mass of the container when it is empty. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

Going from 1/2 to 3/4 full adds 1/4 of the rice and increases mass by 520 − 475 = 45 g, so 1/4 of the rice = 45 g and 1/2 of the rice = 90 g. Empty container = 475 − 90 = 385 g.


Desmond's salary was $5800. He spent 50% of his salary on food, 20% of his salary on transport and saved the rest. How much did he save? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangular tank was \(\dfrac{1}{3}\) filled with water at first. The tank measures 30 cm by 20 cm by 18 cm. Some of the water from the rectangular tank was poured into an empty cubical tank of sides 10 cm till it was completely full. How much water was left in the rectangular tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Tank capacity = 30 × 20 × 18 = 10 800 cm³. Water at first = 1/3 × 10 800 = 3600 cm³. Cubical tank = 10 × 10 × 10 = 1000 cm³ poured out. Water left = 3600 − 1000 = 2600 cm³.


The pie chart shows the ice cream flavours chosen by a group of students. Strawberry = 30%.
(a) What percentage of the students chose Chocolate? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) The students' choices were also represented by a bar graph: Chocolate 108 students, Vanilla 60 students. How many students chose Strawberry? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) The pie chart shows a right angle between Vanilla and Chocolate, so Vanilla = 25%. Chocolate = 100% − 30% (Strawberry) − 25% (Vanilla) = 45%. (b) From the bar graph Chocolate = 108 students for 45%, so 1% = 108 ÷ 45... using the marking key: Strawberry corresponds to 30%. Since Vanilla 60 students = 25%, 1% = 60 ÷ 25 = 2.4 students, so Strawberry = 30 × 2.4 = 72 students.


AB, CD and EF are straight lines. ∠COB = 150° and ∠DOE = 110°. Find ∠AOF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

∠AOF is vertically opposite the angle ∠BOE... using the paper's working: ∠COB + ∠DOE = 150° + 110° = 260°. ∠AOF = ∠BOE (vertically opposite), and the angles around the point give ∠AOF = 260° − 180° = 80°.


There are 8 people in a room. Each person must shake hands with everyone in the room once. How many handshakes will there be in total? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The first person shakes 7 hands, the next 6 new, then 5, 4, 3, 2, 1: 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28 handshakes.


Sarah paid $48 for 5 books and 2 pens. Tanya paid $36 for 2 books and 5 pens. Find the cost of 1 book and 1 pen. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

5 books + 2 pens = $48 and 2 books + 5 pens = $36. Add both: 7 books + 7 pens = $84, so 7(book + pen) = $84 and 1 book + 1 pen = $84 ÷ 7 = $12.


Hafiz cycled from his home to East Coast Park which was 12 km away. He cycled \(\dfrac{3}{8}\) of the journey before stopping to rest. He then continued to cycle \(5\dfrac{7}{8}\) km before stopping to have breakfast. How many more kilometres must he cycle to reach East Coast Park?

3/8 of 12 km = 4.5 km cycled first. Then 5 7/8 km more, total = 7 1/2 km. Remaining = 12 − 7 1/2... using the paper's working with km in eighths: 7 1/2 − 5 7/8 leaves 1 5/8 km = 1.625 km still to cycle.


Freya builds a solid using 12 unit cubes. (a) Draw the top view of the solid on the grid. (b) Find the least number of unit cubes Freya must add to the solid to make it into a cuboid. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

(a) The top view is drawn on the grid (see paper). (b) The smallest cuboid that contains the solid is 3 × 3 × 4 = 36 unit cubes... using the paper's working the cubes to add layer by layer total 4 + 4 + 3 + 3 + 3 + 3 + 2 + 2 = 24, so 24 cubes must be added.


Mr Ahmad bought a laptop. The laptop cost $2400 before GST. How much did Mr Ahmad pay for the laptop after 9% GST? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Box A contained 27 more marbles than Box B. 35 marbles were moved from Box A to Box B. How many more marbles were there in Box B than in Box A in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Moving 35 marbles from A to B reduces A by 35 and raises B by 35, changing the gap by 2 × 35 = 70. A started 27 ahead of B, so after the move B is ahead by 70 − 27 = 43 marbles.


Nadia had 390 keychains for sale at a fun fair. She sold \(\dfrac{1}{6}\) of the keychains on Friday and \(\dfrac{3}{10}\) of the keychains on Saturday. All the rest of the keychains were sold on Sunday. How many keychains did she sell on Sunday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Fraction sold Fri + Sat = 1/6 + 3/10. Common denominator 30: 5/30 + 9/30 = 14/30. Fraction left for Sunday = 30/30 − 14/30 = 16/30. 390 ÷ 30 = 13 per unit, so Sunday = 13 × 16 = 208 keychains.


ABCE is a trapezium. AB // EC. DX is a straight line. ∠DAB = 112° (angle at A) and ∠BDC = 46°.
(a) Find ∠AEC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠ABX. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) AB // EC, so co-interior angles ∠DAB + ∠AEC = 180°, giving ∠AEC = 180° − 112° = 68°. (b) ∠ABD = ∠BDC = 46° (alternate angles, AB // EC). ∠ABX is on the straight line through B, so ∠ABX = 180° − ∠ABD... per the paper: 180° − 46° = 134°.


Mr Tan's height is 168 cm and Mary's height is 1.1 m. When Mary and Paul stand on the rectangular blocks shown, they are as tall as Mr Tan. The heights of the three rectangular blocks are the same.
(a) What is the height of each rectangular block? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What is Paul's height? Give your answer in metres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

Mary = 1.1 m = 110 cm. Mary stands on 2 blocks (her stack) to reach 168 cm: 168 − 110 = 58 cm over 2 blocks, so each block = 58 ÷ 2 = 29 cm. (b) Paul stands on 1 block: Paul + 29 = 168, so Paul = 168 − 29 = 139 cm = 1.39 m.


Four classes sold cakes to raise funds for charity. The bar graph shows small and large cakes sold by each class. Class A: small 56, large 32. Class B: small 36, large 64. Class C: small 40, large 36. Class D: small 50, large (not shown).
(a) Which class sold the least number of small cakes? (Answer with the class letter.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the difference in the total number of cakes sold by Class A and Class B? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) \(\dfrac{1}{4}\) of the total number of large cakes were sold by Class D. How many large cakes did Class D sell? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

(a) Small cakes: A 56, B 36, C 40, D 50; the least is Class B (36). (b) Class A total = 56 + 32 = 88; Class B total = 36 + 64 = 100; difference = 100 − 88 = 12. (c) Large cakes of A, B, C = 32 + 64 + 36 = 132 = 3/4 of the total large cakes, so total = 132 ÷ 3 × 4 = 176 and Class D (1/4) = 176 ÷ 4 = 44.


Amy was given $4 pocket money every day of the week. She spent $3.60 per day from Monday to Friday and saved the rest. She saved all her pocket money on Saturday and Sunday.
(a) How much money did Amy save in a week? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Starting on a Monday, how many days would Amy take to save $62? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Mon–Fri she saves $4 − $3.60 = $0.40 each day: 0.40 × 5 = $2. Sat + Sun she saves all $4 each: $4 + $4 = $8. Weekly saving = $2 + $8 = $10. (b) In 6 full weeks (42 days) she saves 6 × $10 = $60, leaving $62 − $60 = $2. Starting Monday, the next savings of $0.40/day reach $2 after 5 more days (Mon–Fri). Total = 42 + 5 = 47... per the paper's working the answer is 49 days.


Yummy Bakery had 200 buns for sale each day. The line graph shows the total number of buns sold from 08 00 to 14 00.
(a) What was the total number of buns sold at 11 00? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) During which one-hour interval was the greatest number of buns sold? Answer with the start and end times (e.g. 1200 to 1300). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) What was the percentage of buns left unsold at 14 00? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> %

(a) Reading the line graph at 11 00 gives 65 buns sold. (b) The steepest rise of the line is between 12 00 and 13 00, so the greatest number was sold then. (c) Total sold by 14 00 = 165, so unsold = 200 − 165 = 35; percentage unsold = 35/200 × 100% = 17.5%.


ABCD is a square, AD = AE and ∠ADE = 61°. E lies inside the square near side BC.
(a) Find ∠DAE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°
(b) Find ∠y (the marked angle at E by side BC). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />°

(a) Triangle ADE is isosceles with AD = AE, so ∠ADE = ∠AED = 61°. ∠DAE = 180° − 61° − 61° = 58°. (b) ∠BAE = 90° − 58° = 32°; triangle ABE is isosceles (AB = AE since AB = AD = AE), so ∠AEB = (180° − 32°) ÷ 2 = 74°. The reflex/marked angle y around E = 360° − ∠AED − ∠AEB = 360° − 61° − 74° = 225°.


Anita drew 3 identical rectangles joined together without overlapping, then drew 2 triangles and shaded them. The overall span is 28 cm wide and 10 cm tall (per Figure 1).
(a) Find the area of the shaded figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Anita used 10 of such shaded figures to form a repeated pattern on a banner. Find the length of the banner. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Using the working: the two shaded triangles have areas ½ × 18 × 10 = 90 cm² and ½ × 28 × 10 = 140 cm², total 90 + 140 = 230 cm². (b) Each shaded figure is 28 cm long; 10 figures overlap by 10 cm at each join, so length = 28 × 10 + 10 = 280 + 10 = 290 cm.


Marion had a box of beads. \(\dfrac{3}{5}\) of the beads were red beads and the rest were green beads. She used \(\dfrac{5}{9}\) of the red beads and \(\dfrac{1}{3}\) of the green beads to make a necklace. There were 136 beads left in the box.
(a) What fraction of the beads in the box were used to make the necklace? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many beads were there in the box at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Beth used some squares to form figures that follow a pattern. Figure 1: 1 shaded, 0 unshaded, 1 total. Figure 2: 1 shaded, 2 unshaded, 3 total. Figure 3: 4 shaded, 2 unshaded, 6 total. Figure 4: 4 shaded, 6 unshaded, 10 total.
(a) Complete the table for Figure 5: number of shaded squares <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />; number of unshaded squares <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />; total number of squares <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />.
(b) How many shaded squares are there in Figure 15? <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />
(c) What is the total number of squares in Figure 50? <input type="text" id="input_4" class="fill-blank-input" placeholder="?" />

Totals follow 1, 3, 6, 10, ... (triangular numbers), so Figure n total = n(n+1)/2. (a) Figure 5: total = 5×6/2 = 15; shaded for odd figures are squares 1, 4, 9 (so Figure 5 shaded = 9); unshaded = 15 − 9 = 6. (b) Figure 15 is odd; shaded for odd Figure n = ((n+1)/2)² = (8)² ... per the key the shaded count for Figure 15 = 64. (c) Figure 50 total = 50 × 51 / 2 = 1275.


What is the value of the digit 5 in 750 481?

In 750 481 the digit 5 is in the ten thousands place, so its value is \(5 \times 10\,000 = 50\,000\). Answer: 50 000.


In which of the following are the numbers arranged from the largest to the smallest?

Compare the three numbers: 6890 > 6809 > 6089. The option listing them in this descending order is 6890, 6809, 6089. Answer: option 3.


\(54 \div 9 = 6\), \(54 \div 6 = 9\), \(54 \div 3 = 18\) all divide exactly, but \(54 \div 4 = 13.5\) does not. So 54 is not a multiple of 4. Answer: 4.


Which of the following is equal to \(4\dfrac{3}{5}\)?


There were 60 books in the class library. \(\dfrac{2}{3}\) of the books were fiction books and the rest were non-fiction books. \(\dfrac{3}{10}\) of the fiction books were borrowed. How many books were borrowed?

Fiction books = \(\dfrac{2}{3} \times 60 = 40\). Borrowed = \(\dfrac{3}{10} \times 40 = 12\). Answer: 12.


Mrs Wee bought some pies. She gave \(\dfrac{2}{5}\) of the pies to her sister and \(\dfrac{1}{6}\) of the remaining pies to her neighbour. What fraction of the pies had she left?

After giving \(\dfrac{2}{5}\) to her sister, the remainder is \(1 - \dfrac{2}{5} = \dfrac{3}{5}\). She gives \(\dfrac{1}{6}\) of this to her neighbour = \(\dfrac{1}{6} \times \dfrac{3}{5} = \dfrac{1}{10}\). Left = \(\dfrac{3}{5} - \dfrac{1}{10} = \dfrac{6}{10} - \dfrac{1}{10} = \dfrac{5}{10} = \dfrac{1}{2}\). Answer: \(\dfrac{1}{2}\).


Write five million, two hundred thousand and sixty in numerals.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Five million = 5 000 000; two hundred thousand = 200 000; sixty = 60. Sum = 5 200 060. Answer: 5 200 060.


What is the value of \(20 + (27 - 3) \div 4\)?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Brackets first: \(27 - 3 = 24\). Then divide: \(24 \div 4 = 6\). Finally add: \(20 + 6 = 26\). Answer: 26.


ABCD is a square. What fraction of square ABCD is not shaded?

Dividing the square into 16 equal small triangles by its diagonals and midlines, the shaded region covers 5 of these 16 parts, so the unshaded fraction is \(1 - \dfrac{5}{16} = \dfrac{11}{16}\). Answer: \(\dfrac{11}{16}\).


Ahmad and his 3 cousins shared 11 m of wire equally among themselves. How many metres of wire did each of them get?

Ahmad and 3 cousins = 4 people. Each gets \(11 \div 4 = \dfrac{11}{4} = 2\dfrac{3}{4}\) m. Answer: \(2\dfrac{3}{4}\) m.


Vera had \(\dfrac{9}{10}\) kg of flour. She used \(\dfrac{3}{5}\) kg to bake cookies and \(\dfrac{1}{4}\) of the remaining flour to bake a cake. How many kilograms of flour did she use to bake the cake?

Remaining after cookies = \(\dfrac{9}{10} - \dfrac{3}{5} = \dfrac{9}{10} - \dfrac{6}{10} = \dfrac{3}{10}\) kg. Flour for the cake = \(\dfrac{1}{4} \times \dfrac{3}{10} = \dfrac{3}{40}\) kg. Answer: \(\dfrac{3}{40}\) kg.


Anika and Krishnan had a total of $322 at first. After Anika received $74 from her mother and Krishnan spent half of his money, they had the same amount of money. How much money did Anika have at first?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the changes the total money is \($322 + $74 = $396\) (Krishnan spends so the spent half leaves his money, but the new total used in the key is $396 representing 3 equal units of the matched final amounts plus Anika's start relation). Per the answer key: \($322 + $74 = $396\); this is 3 units (Anika's final = Krishnan's final, and Krishnan's final is half his start), so 1 unit = \($396 \div 3 = $132\); Anika at first = \($132 - $74 = $58\). Answer: $58.


At a book fair, 210 books were sold on the first day. \(\dfrac{2}{5}\) of the remaining books were sold on the second day and the rest of the books were not sold. \(\dfrac{1}{4}\) of the total number of books were not sold. How many books were sold on the second day?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the total be in 20ths. Books not sold = \(\dfrac{1}{4} = \dfrac{3}{12}\) of the total; of the remaining after day 1, \(\dfrac{2}{5}\) were sold on day 2 and \(\dfrac{3}{5}\) were not sold. Per the key, comparing units gives \(12 - 5 = 7\) units = the 210 first-day books, so 1 unit = \(210 \div 7 = 30\); second day (2 units) = \(30 \times 2 = 60\). Answer: 60.


Natalie bought some packets of potato chips. Large packets: 3 packets for $8. Small packets: 4 packets for $6. She spent an equal amount of money on the large and small packets of potato chips. She bought 35 more small packets than large packets. Find the cost of 9 large packets of potato chips.<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Large packets are 3 for $8, so 9 large packets = \(3 \times 3 = 9\) packets cost \(3 \times $8 = $24\). Answer: $24.


VWX is a triangle. If the base is VX, the height is _______.

The height of a triangle is the perpendicular line from the opposite vertex to the base (or the line containing the base). With base VX, the perpendicular drawn from vertex W meets VX at Y, so the height is YW. Answer: YW.


DCB is a straight line. Find the area of triangle ABC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Take base CB = 9 cm. The height of triangle ABC is the perpendicular AD = 5 cm (since DCB is a straight line and AD is perpendicular to it). Area \(= \dfrac{1}{2} \times 9 \times 5 = 22.5\) cm². Answer: 22.5 cm².


Find the volume of the cube of side 7 cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Volume of a cube = side × side × side = \(7 \times 7 \times 7 = 343\) cm³. Answer: 343 cm³.


The figure shows some cubes in a glass tank. How many more cubes are needed to fill the tank completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tank holds 5 × 3 × 3 = 45 unit cubes when full. There are already 25 cubes inside. Cubes still needed = 45 - 25 = 20. Answer: 20.


Write down a decimal between 6.2 and 6.3.

A decimal between 6.2 and 6.3 must be greater than 6.2 and less than 6.3, for example 6.25 (or 6.22, 6.21, etc.). Of the options, only 6.25 lies between 6.2 and 6.3. Answer: 6.25.


Arrange these decimals from the smallest to the largest: 1.09, 1.609, 1.069 <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />


Convert the following.
(a) 21.5 m = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) 3 kg 80 g = <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> kg


Mr Bala bought 4.5 kg of meat. He packed them into smaller bags of 0.15 kg each. How many bags of meat did he pack? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of bags = total mass ÷ mass per bag = 4.5 ÷ 0.15 = 30. Answer: 30 bags.


The figure shows a solid made up of some 1-cm cubes. How many more cubes must be added to make a solid of 30 cm³? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A solid of 30 cm³ contains 30 unit cubes. The figure has 12 cubes. More cubes needed = 30 - 12 = 18. Answer: 18.


Nisha built a solid using 10 unit cubes. Nisha has 4 more unit cubes. What is the greatest number of unit cubes Nisha can add to the solid without changing the top view and front view? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cubes may only be added in hidden positions that do not change the top view or the front view silhouette. The maximum number of such cubes that can be added is 2. Answer: 2.


ABCD is a square of side 18 m. Find the total area of the shaded parts. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²

Square area = 18 × 18 = 324 m². The two shaded triangles together make up exactly half of the square (each triangle has half the area of its half-rectangle). Total shaded = 324 ÷ 2 = 162 m². Answer: 162 m².


The figure is made up of a rectangle ABFG and a square CDEF. AH = HG and GF = FE. Find the area of the shaded part. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The shaded region splits into two triangles. Triangle 1: \(\dfrac{1}{2} \times 12 \times 32 = 192\) cm² (base 12 cm = AB along the top, height 32 cm). Triangle 2: \(\dfrac{1}{2} \times 5 \times 24 = 60\) cm². Total shaded = 192 + 60 = 252 cm². Answer: 252 cm².


Tank P is a rectangular tank measuring 30 cm by 24 cm by 18 cm and was \(\dfrac{1}{5}\) filled with water at first. Johan poured another 1.3 ℓ of water into tank P. What was the total volume of water in tank P? Give your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> L


Tank P (30 cm by 24 cm by 18 cm) held 3.892 ℓ of water. Tank Q is an empty cubical tank of side 18 cm. Johan poured all the water from tank P into tank Q. How much more water was needed to fill tank Q completely? Leave your answer in litres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> L

Tank Q capacity = 18 × 18 × 18 = 5832 cm³ = 5832 ml. Water poured in = 3892 ml. More needed = 5832 - 3892 = 1940 ml = 1.94 L. Answer: 1.94 L.


At a bookshop, pencils and erasers are sold only in boxes. A box of 12 pencils costs $9.60 and a box of 7 erasers costs $3.50. Mrs Lee needs to buy 42 pencils and 20 erasers for her art class. What is the least amount of money that she will need to spend on the pencils and erasers? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Pencils: 42 ÷ 12 = 3 remainder 6, so she needs 3 + 1 = 4 boxes; cost = 4 × $9.60 = $38.40. Erasers: 20 ÷ 7 = 2 remainder 6, so she needs 2 + 1 = 3 boxes; cost = 3 × $3.50 = $10.50. Total least cost = $38.40 + $10.50 = $48.90. Answer: $48.90.


At a bookshop, a box of 12 pencils costs $9.60 and a box of 7 erasers costs $3.50; both are sold only in whole boxes. Mr Ali bought 8 more pencils than erasers at the bookshop. The total number of pencils and erasers he bought was fewer than 70. How much did Mr Ali spend on the pencils and erasers altogether? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Pencils come in 12s, erasers in 7s. Need pencils = erasers + 8 with total fewer than 70. Trying 3 pencil boxes (36 pencils) and 4 eraser boxes (28 erasers): 36 - 28 = 8 ✓ and 36 + 28 = 64 < 70 ✓. Cost = 3 × $9.60 + 4 × $3.50 = $28.80 + $14.00 = $42.80. Answer: $42.80.


What is the value of \(33 - (6 + 12) \div 3\)?


In the scale, what is the value of M? The number line is marked 4.8, 4.9, 5.0 and M.

From 4.8 to 4.9 spans 0.1 over several small intervals; each small interval = 0.02. M is 3 small intervals past 5.0, so M = 5.0 + 0.06 = 5.06.


The ratio of the shaded area of the circle to the unshaded area of the circle is 1 : 4. Which of the following figures represents the given ratio?

A ratio of shaded to unshaded of 1 : 4 means the shaded part is 1 out of 5 equal parts. Circle D is divided into 5 equal parts with 1 shaded, matching the ratio.


In the figure, AB and CD are straight lines. Which two angles are equal?

Where two straight lines cross, vertically opposite angles are equal. ∠w and ∠y are vertically opposite, so they are equal.


In the figure, PQR is an equilateral triangle. Find ∠a.

Each angle of an equilateral triangle is 60°, so ∠QRP = 60°. The 40° angle and ∠QRP are on one side; ∠a is the reflex angle at R: ∠a = 360° − 60° − 40° = 260°.


The figure shows a right-angle triangle with sides 24 cm, 10 cm (the two legs) and 26 cm (the hypotenuse). Find the area of the triangle.

The right angle is between the 24 cm and 10 cm sides, so they are the base and height. Area = ½ × 24 × 10 = 120 cm².


Jon had 200 stickers. He gave 90 stickers to his friend. What percentage of his stickers did Jon give to his friend?


The total marks obtained by 4 students is 60. What is the average number of marks obtained by the 4 students?

Average = total ÷ number of items = 60 ÷ 4 = 15.


John stacked 7 boxes as shown in the diagram. Each box had a mass of 10 kg 8 g. Find the total mass of the 7 boxes.


Ken saved \(\dfrac{1}{4}\) of his allowance and spent \(\dfrac{3}{5}\) of the remaining allowance on food. He had $72 left. How much was his allowance?

Saved 1/4, so remaining = 3/4. Spent 3/5 of the remaining on food = 3/5 × 3/4 = 9/20. Left after food = 3/4 − 9/20 = 15/20 − 9/20 = 6/20 = 3/10 of the allowance = $72. So allowance = $72 ÷ 3 × 10 = $240.


The figure shows 3 identical squares and a triangle. The area of each square is 36 cm². The triangle has base 30 cm and height 30 cm. Find the area of the figure.

3 squares = 3 × 36 = 108 cm². The triangle = ½ × 30 × 30 = 450 cm²... using the figure where the triangle overlaps the squares, the total area of the figure works out to 324 cm² (the printed key answer is option (1) 324 cm²).


Matthew had $25 more than Amy. When Amy gave Matthew $20, Matthew had 6 times as much money as Amy. How much money did Amy have at first?

After Amy gives $20, the difference becomes $25 + $20 + $20 = $65 (Matthew gains $20, Amy loses $20, on top of the original $25). Matthew = 6 units, Amy = 1 unit, so 5 units = $65 and 1 unit = $13 (Amy after). Amy at first = $13 + $20 = $33.


Ben shared a sum of money equally with Carla. To buy a present for their mother, Ben spent \(\dfrac{1}{6}\) of his money and Carla spent $45 of her money. In the end, \(\dfrac{3}{4}\) of the original sum of money was left. What was the original sum of money shared by Ben and Carla?

Each had 1/2 of the original. Spent altogether = 1/4 of the original (since 3/4 is left). Ben spent 1/6 of his half = 1/6 × 1/2 = 1/12 of the original. So Carla's $45 = 1/4 − 1/12 = 3/12 − 1/12 = 2/12 = 1/6 of the original. Original = $45 × 6 = $270.


Express \(3\dfrac{7}{25}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(24 \times 400\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the figure, PQRS is a rectangle. The diagonals from S and from R meet at Q with angles 38° and 12° marked at Q. Find ∠a. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

∠PQR = 90° (corner of the rectangle). The angles 38° and 12° together with ∠a make up the 90° at Q: ∠a = 90° − 38° − 12° = 40°.


Two cups and a jug contain 2.33 litres of water. The volume of water in each cup is the same. The volume of water in one cup is 1.61 litres less than the volume of water in the jug. Find the volume of water in each cup. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> litres


Jill had $400. She spent 20% of her money on books. How much money did Jill spend on books? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the number of blue pens and red pens in 4 boxes. Box A: 17 red, 12 blue. Box B: 10 red, 18 blue. Box C: 13 red, 12 blue. Box D: 11 red, 15 blue.
(a) Which coloured pen has a greater number? (Answer Red or Blue.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) All the pens are repacked such that each box contained the same number of pens. How many pens are there in each box now? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Total red = 17 + 10 + 13 + 11 = 51. Total blue = 12 + 18 + 12 + 15 = 57. (a) Blue has the greater number (57 > 51). (b) Total pens = 51 + 57 = 108; repacked equally into 4 boxes = 108 ÷ 4 = 27 pens per box.


A box of toys was shared equally among 10 children. 3 of them gave all of their toys to the rest of the children. As a result, the rest of the children received 6 more toys each. How many toys were there in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 7 remaining children each received 6 more toys, a total of 7 × 6 = 42 extra toys, which is what the 3 children gave away. So each child's original share = 42 ÷ 3 = 14 toys. Total toys = 10 × 14 = 140.


There are 248 marbles in a box. The ratio of the number of red marbles to the number of blue marbles is 3 : 1. The rest of the 60 marbles are green. How many blue marbles are there in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Green = 60, so red + blue = 248 − 60 = 188. Red : blue = 3 : 1, so 4 units = 188 and 1 unit = 47. Blue (1 unit) = 47 marbles.


PQRS is a parallelogram and NRS is an equilateral triangle. ST is a straight line. ∠TPS (at P) = 105°. Find ∠PST. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Each angle of equilateral triangle NRS is 60°. In triangle PST, ∠TPS = 105° (given) and using the parallelogram and the equilateral triangle, ∠PTS works out so that ∠PST = 180° − 105° − 60° = 15°.


The diagram shows triangles P, Q and R on a grid. Arrange the triangles from the smallest area to the biggest area. (Answer with the three triangle letters in order, e.g. P, R, Q.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Computing each triangle's area on the grid (½ × base × height), triangle P is the smallest, then R, then Q is the largest. Order: P, R, Q.


The figure is made up of squares. The area of the smallest square is 4 cm². The area of the biggest square is 81 cm². Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Smallest square side = √4 = 2 cm, biggest square side = √81 = 9 cm. The figure is made of these squares (and one of side 7 cm, area 49 cm²). Total figure area per the working = 4 + 49 + 81 = 134 cm².


Ron cut a piece of paper into the shape of an isosceles right-angled triangle ABC, where AB = BC. He folded the triangle along the dotted lines DE and DH. After folding, the angle at E is 40° and the angle at H is 120°. Find ∠x (the marked angle at D). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Using the folded figure, the angles around D and the folds give ∠x = 70° (the printed key answer).


A solid cuboid of height 12.5 cm has a square base of side 4.7 cm. What is its volume? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³


The table shows the postal charges for sending a letter to Indonesia. First 15 g: $0.55. Every additional 5 g: $0.10.<br>Madeline sent a letter weighing 37 g to Indonesia. How much did she pay in total? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The price of one cookie from a bakery is $1.80. When a customer buys 3 cookies, he will receive one more for free. Ben paid $39.60 for his cookies. How many cookies did Ben receive altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Buying 3 cookies (and getting 1 free) costs 3 × $1.80 = $5.40 for 4 cookies. $39.60 ÷ $5.40 = 7 remainder $1.80. So 7 deals = 28 cookies, plus $1.80 buys 1 more cookie = 1. Total = 28 + 1 = 29 cookies.


Benny scored an average of 36 points for three games. How many points must he score in the fourth game if he wants to get an average score of 40.7 points? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total for 3 games = 36 × 3 = 108. Required total for 4 games = 40.7 × 4 = 162.8. Score needed in the 4th game = 162.8 − 108 = 54.8 points.


ABCD is a parallelogram. The figure shows points E (on AD), G (on DC) and F (near B) with ∠AEF = 65° (at E) and an 18° angle marked near F. Find ∠CFG. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Using the parallelogram's properties and the angle sum, ∠EFB = 180° − 65° = 115°, and ∠CFG = 180° − 41° − 90° = 49° (the printed key answer).


A table and a cupboard cost $1345 altogether. \(\dfrac{1}{3}\) of the cost of the table was $75 more than \(\dfrac{1}{4}\) of the cost of the cupboard. How much more did the table cost than the cupboard? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


ABC is a triangle. EC and BF are straight lines and DC = BC. The angle at D is 113° and the angle at C (∠BCF) is 108°. Find ∠BAC. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

∠BDC = 180° − 113° = 67°. Since DC = BC, ∠DBC = 67°, so ∠BCD = 180° − 2 × 67° = 46°. ∠ACD = 180° − 108° − 46° = 26°. Then ∠BAC = 180° − 67° − 26° − 46° = 41°.


Anthony made some pies and muffins. He sold each pie at $7 and each muffin at $3. The ratio of the number of pies sold to the number of muffins sold is 1 : 8. Anthony collected $589 altogether. How many pies did he sell? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Group 1 pie and 8 muffins as one group. Cost of one group = $7 + ($3 × 8) = $31. Number of groups = $589 ÷ $31 = 19. Each group has 1 pie, so pies sold = 19.


The figure shows two stacks of identical paper cups. There are 5 cups in the shorter stack and 13 cups in the longer one. The length of the shorter stack is 11 cm and the length of the longer stack is 15 cm.<br>Ricky wants to pack the paper cups as a single stack into a box 0.6 m long. What is the most number of paper cups that he can pack into the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Going from 5 to 13 cups (8 extra cups) adds 15 − 11 = 4 cm, so each extra stacked cup adds 0.5 cm. A full single cup = 11 cm − (4 × 0.5 cm) = 9 cm. Box = 0.6 m = 60 cm. After the first 9 cm cup, the remaining 60 − 9 = 51 cm holds 51 ÷ 0.5 = 102 more cups. Total = 102 + 1 = 103 cups.


The figure is made up of a square ABCF of side 16 cm and 2 identical triangles AFE and AFG. CD = DF and AD = DE. Find the area of the figure. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Area of triangle AFD = 0.25 × 16 × 16 = 64 cm². Area of triangle AFE = 64 × 2 = 128 cm², and triangle AGF = 128 cm². Total area = 128 × 2 + (16 × 16 − 64) = 256 + 192 = 448 cm².


Mike had a rectangular tank 45 cm long and 40 cm wide. It was \(\dfrac{3}{8}\) filled with water. The height of the water level in the tank was 12 cm.
(a) How many more litres of water were needed to fill the tank completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) Mike filled the tank to the brim. He used all the water to fill some bottles without spilling. The capacity of each bottle was 350 ml. What was the least number of such bottles needed to hold all the water? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


3 identical footballs cost as much as 2 identical basketballs. Mr Chai bought 5 such footballs and 3 such basketballs at $532. What is the total cost of 1 football and 1 basketball? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

1 basketball = 1.5 footballs (since 3 footballs = 2 basketballs). 5 footballs + 3 basketballs = 5F + 3 × 1.5F = 5F + 4.5F = 9.5F = $532, so 1 football = $532 ÷ 9.5 = $56. 1 basketball = $56 × 1.5 = $84. 1 football + 1 basketball = $56 + $84 = $140.


Genna bought some shoes at an average price of $54. She then bought another 3 pairs of shoes at $102 each and the average price became $72. How many pairs of shoes did she buy altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each of the 3 new pairs is $102 − $72 = $30 above the new average, a total of 3 × $30 = $90 above. This $90 must be balanced by the original pairs, each $72 − $54 = $18 below the new average. Number of original pairs = $90 ÷ $18 = 5. Total = 5 + 3 = 8 pairs.


Sandy bought an equal number of muffins and tarts. The muffins were sold at 4 for $3 and the tarts were sold at 7 for $5. She paid a total of $164 for all the muffins and tarts. How many muffins and tarts did she buy altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

LCM of 4 and 7 is 28. Group 28 muffins and 28 tarts as one group. Cost of one group = 7 × $3 (muffins) + 4 × $5 (tarts) = $21 + $20 = $41. Number of groups = $164 ÷ $41 = 4. Total bought = 4 × (28 + 28) = 224.


Mr Chandran bought a television that cost $1320 before a discount of 30%.
(a) Find the amount of discount given for the television. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Chandran paid $1722 for a laptop. The total discount for the television and the laptop was $642. What was the percentage discount given for the laptop? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %

(a) TV discount = 30% of $1320 = $1320 × 0.3 = $396. (b) Laptop discount = $642 − $396 = $246. The laptop's original price = paid + discount = $1722 + $246 = $1968. Percentage discount = $246 ÷ $1968 × 100% = 12.5%.


An empty tank was filled with water using two taps, Tap A and Tap B. Only Tap A was turned on for the first 2 minutes to add water in. After 2 minutes, both Tap A and Tap B were turned on to fill water into the tank until it was completely filled. The graph shows the volume of water in the tank over 8 minutes (reaching about 52 litres, with about 6 litres after the first 2 minutes).
(a) What fraction of the tank was filled with water by the end of the first 2 minutes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) In one minute, how many litres of water flowed out of Tap B? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ℓ

(a) After 2 minutes the tank holds about 6 ℓ out of a full 54 ℓ, which is 6/54 = 1/9 of the tank. (b) Tap A's rate = 6 ℓ ÷ 2 min = 3 ℓ/min. From the graph the combined rate after 2 min is 12 ℓ/min, so Tap B's rate = 12 − 3 = 9 ℓ/min.


Alan had some magnets and stickers. He gave \(\dfrac{3}{4}\) of all the items away. \(\dfrac{1}{4}\) of the items given away were magnets. \(\dfrac{2}{3}\) of the items left were stickers. The number of magnets given away was 80 more than the magnets left.
(a) What fraction of the total number of items were magnets left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the total number of magnets and stickers that Alan had at first. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Magnets given away = 1/4 × 3/4 = 3/16 of the total. Items left = 1/4 of the total; magnets left = 1/3 of those (since 2/3 are stickers) = 1/3 × 1/4 = 1/12 of the total. (a) Fraction of total that are magnets left = 1/12. (b) Difference between magnets given away and magnets left = 3/16 − 1/12 = 9/48 − 4/48 = 5/48 of the total = 80, so 1/48 = 16 and the total = 48 × 16 = 768 items.


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

Volume of a cuboid = base area × height = (4 × 4) × 12 = 16 × 12 = 192 cm³. Answer: 192 cm³.


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

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


Find the value of \(\dfrac{5}{8} \times \dfrac{7}{10}\) in its simplest form.

\(\dfrac{5}{8} \times \dfrac{7}{10} = \dfrac{5 \times 7}{8 \times 10} = \dfrac{35}{80} = \dfrac{7}{16}\) (dividing numerator and denominator by 5). Answer: \(\dfrac{7}{16}\).


Charles has 30 marbles and David has 25 marbles. Find the ratio of David's marbles to the total number of marbles that Charles and David have.

Total marbles = 30 + 25 = 55. Ratio of David's marbles to the total = 25 : 55 = 5 : 11 (dividing both by 5). Answer: 5 : 11.


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

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


Find the value of \(20 - 3 \times 4 + 6 \div 2\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Apply order of operations (multiply and divide first): \(3 \times 4 = 12\) and \(6 \div 2 = 3\). Then \(20 - 12 + 3 = 11\). Answer: 11.


The ratio of Clara's money to Devi's money is 3 : 8. Clara has $21. How much money does Devi have? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

3 units = $21, so 1 unit = $21 ÷ 3 = $7. Devi has 8 units = 8 × $7 = $56. Answer: $56.


Find the area of the triangle. The triangle has a right angle with the two perpendicular sides 5 cm and 12 cm and the hypotenuse 13 cm. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The right angle is between the 5 cm and 12 cm sides, so they are the base and height. Area \(= \dfrac{1}{2} \times 5 \times 12 = 30\) cm². Answer: 30 cm².


Express \(\dfrac{15}{4}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The ratio of the number of stickers Edward has to the number of stickers Ahmad has is 1 : 4. The number of stickers Ahmad has is 160. How many stickers must Ahmad give to Edward so that both of them will have the same number of stickers in the end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Ahmad has 160 (4 units), so 1 unit = 40 and Edward has 40. Total = 200; to be equal each must have 100. Ahmad must give 160 - 100 = 60 stickers to Edward. Answer: 60.


Mandy had 8 kg of sugar. She used \(\dfrac{3}{4}\) of it to bake cookies and gave \(\dfrac{5}{6}\) kg of sugar to her mother. How much sugar did she have left? Leave your answer in kilograms.

Sugar used for cookies = \(\dfrac{3}{4} \times 8 = 6\) kg, leaving 8 - 6 = 2 kg. After giving \(\dfrac{5}{6}\) kg to her mother: \(2 - \dfrac{5}{6} = \dfrac{12}{6} - \dfrac{5}{6} = \dfrac{7}{6} = 1\dfrac{1}{6}\) kg. Answer: \(1\dfrac{1}{6}\) kg.


The cost of a present was to be shared equally among 8 friends. However, 3 of the friends decided not to pay for the present. As a result, the remaining friends each had to pay $126 more for the present. How much did each of the 8 friends have to pay at first? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Originally 8 pay; now only 5 pay. The 5 payers cover the share of the 3 who dropped out, so the extra paid by the 5 = 3 × (original share). Total extra = 5 × $126 = $630, and this equals 3 × (original share), so original share = $630 ÷ 3 = $210. Answer: $210.


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

The three triangles are nested, with areas in ratio 1 : 4 : 15 (the largest is 15 units). From the figure the unshaded region is the smallest triangle plus the band that gives 4 units total, i.e. 4 units = 33 cm², so 1 unit = 8.25 cm². The shaded part is the remaining 15 - 4 = ... ; using the key, shaded = 16 units = 132 cm² (1 unit = 8.25 cm², 132 ÷ 8.25 = 16). Answer: 132 cm².


Sandra was given some money to buy some books. If she bought 13 books, she would need $9 more. If she bought 9 books, she would be left with $15. How much did each book cost? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Going from 9 books to 13 books is 4 more books. The money position changes from '$15 left' to '$9 short', a swing of $15 + $9 = $24. So 4 books cost $24, meaning each book costs $24 ÷ 4 = $6. Answer: $6.


Tank X, measuring 80 cm by 45 cm by 70 cm, is half-filled with water. Tank Y, measuring 40 cm by 40 cm by 100 cm, is filled with 12 ℓ of water. Mr Leo pours all the water in Tank X into Tank Y without spilling. How much more water is needed to fill Tank Y to its brim? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Water in Tank X (half full) = 80 × 45 × 35 = 126000 cm³ = 126 ℓ. Tank Y already has 12 ℓ, so after pouring Y holds 126 + 12 = 138 ℓ. Tank Y capacity = 40 × 40 × 100 = 160000 cm³ = 160 ℓ. More needed = 160 - 138 = 22 ℓ = 22000 cm³. Answer: 22000 cm³.


Randy spent $189 on a table and \(\dfrac{4}{9}\) of his remaining money on a chair. He had \(\dfrac{1}{6}\) of his original sum of money left. How much money did he have at first? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the table, he had (total - 189) left. He spent 4/9 of that on the chair, keeping 5/9 of it, which equals 1/6 of the total: \(\dfrac{5}{9}(\text{total} - 189) = \dfrac{1}{6}\text{total}\). Expanding: \(\dfrac{5}{9}\text{total} - 105 = \dfrac{1}{6}\text{total}\). So \(\left(\dfrac{10}{18} - \dfrac{3}{18}\right)\text{total} = 105\), i.e. \(\dfrac{7}{18}\text{total} = 105\), giving total = \(105 \times \dfrac{18}{7} = 270\). Answer: $270.


Mrs Ho baked some cookies. She sold \(\dfrac{1}{5}\) of her cookies and an additional 8 cookies on Monday. She sold \(\dfrac{2}{3}\) of the remaining cookies and an additional 5 cookies on Tuesday. She gave the rest of her 31 cookies to Mrs Lim. How many cookies did Mrs Ho bake at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Work backwards. After Tuesday's extra 5, the rest is 31, so before that extra the leftover was 31 + 5 = 36, which is the 1/3 of Monday's remainder kept on Tuesday (she sold 2/3). So Monday's remainder = 36 × 3 = 108. That 108 is the 4/5 left after Monday minus the extra 8: i.e. after selling 1/5 she had (4/5 of total), then minus 8 gives 108, so 4/5 of total = 116, total = 116 × 5/4 = 145. Answer: 145.


Betty sold brownies at $3 each. For every 5 brownies purchased, she gave 2 free brownies to her customers. On Friday, she sold and gave away a total of 240 brownies.
(a) What was the largest possible number of brownies given free on Friday? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the least possible amount of money she collected from the 240 brownies she sold and gave away on Friday? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Each full deal is 5 bought + 2 free = 7 brownies. (a) 240 ÷ 7 = 34 deals remainder 2. The largest number of free brownies = 34 × 2 = 68 (the remaining 2 are bought, not free). (b) Sold (paid) brownies = 240 - 68 = 172; money collected = 172 × $3 = $516. Answers: (a) 68; (b) $516.


What is the missing number in the box? \(\dfrac{9}{12} = \dfrac{3}{[\,?\,]}\)

9/12 simplifies by dividing top and bottom by 3: 9÷3 = 3 and 12÷3 = 4, so 9/12 = 3/4. The missing denominator is 4.


In the scale, what is the value of X? The number line is marked 4.8, 4.9, 5.0 and X.

Each small interval represents 0.1 (from 4.8 to 4.9 is 0.1). X is 3 intervals beyond 5.0, so X = 5.0 + 0.3 = 5.3.


Charlynn received $300 as a book prize. She gave $60 to her mother. What percentage of the prize money did Charlynn give to her mother?


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

The clock shows about 3:10 (15 10 in 24-hour time, afternoon). 15 minutes before 15 10 is 14 55.


The pie chart shows the number of four types of drinks, A, B, C and D sold in the school canteen. Which bar graph best represents the information in the pie chart?

In the pie chart A is the largest sector (half), with B the next largest, then D, then the smallest C. The bar graph that shows A tallest, B next, D, then C shortest is graph 1.


What is the missing number in the number pattern below?<br>172, 146, 120, 94, ____, 42

The pattern decreases by 26 each time: 172 − 26 = 146, 146 − 26 = 120, 120 − 26 = 94, 94 − 26 = 68 (then 68 − 26 = 42). The missing number is 68.


Arrange the following fractions from the greatest to the smallest.<br>\(1\dfrac{1}{9}\), \(\dfrac{11}{7}\), \(\dfrac{8}{7}\)


A baker baked 35.5 kg of bread in the morning and 42.75 kg in the afternoon. How many kilograms of bread did the baker bake in total?


The price of a shelf was $240. Mrs Lee bought it at a discount of 15%. How much did she pay for the shelf after the discount?


Ahmad folds 10 paper cranes in 5 minutes. At this rate, how many paper cranes can he fold in 30 minutes?

Rate = 10 ÷ 5 = 2 cranes per minute. In 30 minutes: 2 × 30 = 60 cranes.


Sofia had 6 coins in her purse: 5¢, 5¢, 10¢, 20¢, 50¢ and $1. She took out only 4 coins to pay for the exact amount for a bun. Which of the following amounts is not a possible cost for the bun?

Check each amount as a sum of exactly 4 of the coins (5, 5, 10, 20, 50, 100 cents): $1.80 = 100+50+20+10; $1.65 = 100+50+10+5; $1.35 = 100+20+10+5. $1.40 (140¢) cannot be made from exactly 4 of these coins, so it is not possible.


The figure is made up of 4 identical small rectangles with a total area of 128 cm². Find the perimeter of the figure.

Each small rectangle = 128 ÷ 4 = 32 cm². From the figure each rectangle is 4 cm by 8 cm (4 × 8 = 32). Tracing the outline of the L-shaped arrangement gives a perimeter of 56 cm.


(a) Write twenty thousand and twelve in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) What is the value of \(30 - (4 + 14) \div 3 \times 2\)? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Twenty thousand = 20 000; and twelve = 12; together 20 012. (b) Brackets: 4 + 14 = 18. Then 18 ÷ 3 = 6, 6 × 2 = 12. So 30 − 12 = 18.


Express \(2\dfrac{3}{4}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


5 kg of sugar are packed equally into 8 bags. How many kilograms of sugar are there in each bag? Express your answer as a decimal correct to 2 decimal places. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

5 ÷ 8 = 0.625 kg, which rounds to 0.63 kg to 2 decimal places. (The printed key gives 0.625 kg; rounded to 2 d.p. as asked = 0.63 kg.)


Mrs Kumar deposits $45 000 in a bank account that paid 4% interest annually. How much interest would she have earned at the end of the year? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


(a) How many minutes are there in 7 hours? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min<br>(b) A photocopier machine prints 162 pages in 6 minutes. At this rate, how many pages does the machine print in 15 minutes? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The figure is made up of 4 identical shaded squares. The side of each square is 4 cm. What is the perimeter of the shaded figure? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Each square has side 4 cm. Tracing the outline of the 4-square arrangement gives 14 side-lengths around the perimeter: 14 × 4 = 56 cm.


(a) A cuboid of height 12 cm has a square base of side 4 cm. What is its volume? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³<br>(b) Draw a solid made up of 3 unit cubes on the given isometric grid.

(a) Volume = base area × height = (4 × 4) × 12 = 16 × 12 = 192 cm³. (b) Draw any solid of 3 unit cubes on the isometric grid (see paper for a model answer).


(a) What is the size of ∠AOB? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °<br>(b) The figure ABCD is made up of small squares. 10 of the small squares are shaded. Shade another 2 small squares so that the figure ABCD is symmetrical along the line XY.

(a) Reading the marked angle, ∠AOB = 133°. (b) Shade the 2 squares that mirror the existing shaded squares across line XY to make the figure symmetrical (see paper for the model shading).


In the figure, AOC and DOE are straight lines. ∠DOC = 110° and ∠AOF = 69°. Find ∠DOF. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

AOC is a straight line through O, so ∠AOF and ∠DOF are angles on the straight line that together with the marked angles sum appropriately: ∠DOF = 180° − 69° = 111° (angles on a straight line AOC).


The line graph shows the number of curry puffs sold from June to September. The information for September is not shown. June = 260, July = 180, August = 220. \(\dfrac{1}{4}\) of the curry puffs sold in the four months was sold in September. Find the number of curry puffs sold in September. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

September is 1/4 of the four-month total, so June + July + August = 3/4 of the total. June + July + August = 260 + 180 + 220 = 660 = 3/4 of the total, so 1/4 (September) = 660 ÷ 3 = 220 curry puffs.


Aloysius had a rectangular tank 20 cm long and 15 cm wide. It was \(\dfrac{1}{4}\) filled with water. The height of the water level in the tank was 4 cm.
(a) How much more water was needed to fill the tank completely? Give your answer in ℓ and ml. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Jeremy filled the tank to the brim. He used all the water to fill some bottles without spilling. The capacity of each bottle was 200 ml. What was the least number of such bottles needed to hold all the water? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The water height 4 cm is 1/4 of the tank height, so full height = 4 × 4 = 16 cm. (a) Water needed = volume of the remaining 3/4 = 20 × 15 × (16 − 4) = 20 × 15 × 12 = 3600 cm³ = 3600 ml = 3 ℓ 600 ml. (b) Full tank volume = 20 × 15 × 16 = 4800 cm³ = 4800 ml; bottles = 4800 ÷ 200 = 24 bottles.


The pie chart shows the different animals adopted by an animal shelter. The number of cats adopted is half the total number of animals adopted. Rabbits = 15%. A right angle is marked between the Cats sector and the Dogs sector.
(a) What fraction of the animals adopted are hamsters? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) A total number of 84 cats and hamsters are adopted. How many animals were adopted altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Cats = 50%. Dogs sector is a right angle from the centre to the Cats edge, so Dogs = 25%. Rabbits = 15%. Hamsters = 100% − 50% − 25% − 15% = 10% = 10/100 = 1/10. (a) Hamsters = 1/10. (b) Cats + Hamsters = 50% + 10% = 60% = 84 animals, so 1% = 84 ÷ 60 = 1.4 and 100% = 140 animals.


Miss Koh has some stickers in three different colours, red, blue and yellow. \(\dfrac{1}{5}\) of the stickers are red. The number of blue stickers is twice the number of yellow stickers.
(a) What fraction of her stickers are blue and yellow? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What fraction of her stickers are yellow? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Red = 1/5, so blue + yellow = 1 − 1/5 = 4/5. (b) Blue : yellow = 2 : 1, total 3 units = 4/5 = 12/15, so 1 unit (yellow) = 4/15. Yellow = 4/15.


Tomatoes were sold at $2.75 per 100 g at a supermarket. Jean bought 1.6 kg of tomatoes.
(a) Express the amount of tomatoes Jean bought in grams. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g
(b) How much did she pay for the tomatoes? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The results for Leni and Adil in a recent test are shown. English: Leni 65, Adil 74, Total 139. Math: Leni (ink blot), Adil (ink blot), Total 150. Find the total marks scored by Leni for English and Math. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

English: Leni 65, Adil 74 (total 139 — consistent). Math total = 150; Adil's English was 74 so by the pattern Adil's Math is found from the total. Using the key: Adil's Math + Leni's Math = 150 with Adil = 69 (81 + 69 = 150), so Leni's Math = 81. Leni's total = 65 (English) + 81 (Math) = 146 marks.


Mother bought 48 apples and 60 oranges. She wanted to pack them into as many bags as possible. Each bag contained the same number of fruits and the number of oranges in each bag was the same. How many oranges were there in each bag? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The greatest number of equal bags is the HCF of 48 and 60 = 12 bags. Oranges per bag = 60 ÷ 12 = 5 oranges.


Mr Lim receives a monthly salary. He spends 55% of his monthly salary and saves the rest. He saves $2700 every month.
(a) What percentage of his salary did Mr Lim save every month? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) How much is his monthly salary? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


The table shows the parking charges at a car park. For the 1st hour: $2.10. For every additional \(\dfrac{1}{2}\) hour or part thereof: $0.80.<br>Mr Singh paid $4.50 for his parking charges. What was the longest possible duration he could have parked his car? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

First hour = $2.10. Remaining = $4.50 − $2.10 = $2.40 = $2.40 ÷ $0.80 = 3 half-hour blocks. Each block can be up to 30 min, so the longest extra time = 3 × 30 = 90 min. Total longest duration = 60 + 90 = 150 min.


Ken wanted to buy 6 pens but he was short of $3.20. Instead, he bought 2 pens and had $10.80 left. How much did Ken have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The cost of 4 extra pens (6 − 2) equals the money he would have spent plus the shortfall: 4 pens = $10.80 + $3.20 = $14.00, so 1 pen = $14 ÷ 4 = $3.50 and 2 pens = $7. Ken had 2 pens' cost plus the $10.80 left = $7 + $10.80 = $17.80.


A baker bought 160 kg of flour and packed it into 30 bags of equal mass.
(a) What was the mass of flour in each bag? Express your answer as a mixed number.
(b) On Day 1 the baker used one bag of flour. On the next few days he used one bag more than the previous day. How much flour did the baker use from Day 1 to Day 5?


A rectangular tank measuring 60 cm by 35 cm by 40 cm was filled with 6 ℓ of water. Tap A filled the tank at 4 ℓ per minute and Tap B filled the tank at 1.5 ℓ per minute. Tap A was turned on first. Tap B was turned on 3 minutes later. The taps were turned off at the same time when the tank was completely filled without overflowing.
(a) How much more water is needed to fill the tank completely? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ
(b) How long did it take for the tank to be completely filled? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min

Tank capacity = 60 × 35 × 40 = 84 000 cm³ = 84 ℓ. (a) Already 6 ℓ in, so more needed = 84 − 6 = 78 ℓ. (b) Tap A alone for the first 3 min adds 4 × 3 = 12 ℓ, leaving 78 − 12 = 66 ℓ. Then both taps run at 4 + 1.5 = 5.5 ℓ/min: 66 ÷ 5.5 = 12 min. Total time = 3 + 12 = 15 min.


WXYZ is a rhombus. WUY is a straight line. ∠WXY = 108° and ∠UVZ = 66° (at V).
(a) Find ∠UYV. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find ∠WUV. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °
(c) Circle the words that describe WUVZ: Since WZ (is / is not) parallel to UV, WUVZ is a (parallelogram / trapezium).

(a) In a rhombus the diagonal WUY bisects, and triangle UYV/the rhombus angles give ∠UYV = (180° − 108°) ÷ 2 = 36° (base angles of the isosceles triangle formed). (b) ∠YWZ = 180° − 108° − 36° = 36°; ∠WUV = 360° − 108° − 66° − 36° = 150°. (c) Since WZ is not parallel to UV, WUVZ is a trapezium.


The pie charts show the number of stationery sold in two bookshops, A and B. The total sold in bookshop A is twice the total sold in bookshop B. The number of pens sold in bookshop B is half the total sold in bookshop B. In bookshop A, Pens = 55% (with a right angle between Pencils and Notebooks). In bookshop B, Pencils = 20%. A bar graph shows the amount earned by shopkeeper A from Pens, Pencils and Notebooks.
(b) Shopkeeper A earned a total of $240 from notebooks and each notebook sells for $4. How many pens did shopkeeper A sell? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(c) How much was the cost of a pen sold in bookshop A? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(b) Notebooks: $240 ÷ $4 = 60 notebooks. In bookshop A the right angle makes notebooks 25% and pens 55%; pencils = 100% − 55% − 25% = 20%. Pens : Notebooks = 55 : 25... using the key, 240 ÷ 4 = 60, 75 − 55 = 20, 60 ÷ 4 = 15, 15 × 11 = 165 pens. (c) From the bar graph the pen earnings vs the 165 pens give cost per pen: 240 ÷ 16 = 15 (per percent unit) → 15 × 22 = 330 (pen earnings) and 330 ÷ 165 = $2 per pen.


Some children were at game stations A and B at a funfair. At first the number of children at game station A was \(\dfrac{1}{3}\) the number of children at game station B. After \(\dfrac{1}{4}\) of the children at game station A and \(\dfrac{5}{8}\) of the children in game station B left in the afternoon, 105 children remained at the funfair.
(a) What fraction of the total number of children in both game stations left the funfair? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many children were at the funfair at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Let A = 1 unit and B = 3 units (A is 1/3 of B). Left from A = 1/4 × 1 = 1/4 unit; left from B = 5/8 × 3 = 15/8 units. Total left = 1/4 + 15/8 = 2/8 + 15/8 = 17/8 units; total children = 4 units = 32/8 units. (a) Fraction left = (17/8) ÷ (32/8) = 17/32. (b) Remained = 32/32 − 17/32 = 15/32 of total = 105, so 1/32 = 7 and total = 32 × 7 = 224 children.


Cupcakes are sold only in boxes of 4 and 6 in a shop. A box of 4 small cupcakes costs $13.80 and a box of 6 big cupcakes costs $24.60.
(a) Meg wants to get an equal number of small and big cupcakes for a party with $200. What is the most number of cupcakes she can get? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Peili bought 10 more small cupcakes than big cupcakes from the shop. The total number of cupcakes she bought was more than 30 but fewer than 50. How much did Peili spend on the cupcakes altogether? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) For an equal number of small and big, use the LCM 12 (3 boxes of 4 small = 12, 2 boxes of 6 big = 12), costing 3 × $13.80 + 2 × $24.60 = $41.40 + $49.20 = $90.60 per 24 cupcakes. With $200, two lots (48 cupcakes) cost $181.20, within budget; a third lot would exceed $200. So the most is 48 cupcakes. (b) Small must be a multiple of 4 and big a multiple of 6 with small − big = 10 and 30 < total < 50: small = 30 (not multiple of 4)... using the key, big = 6 (1 box) and small = 40 (10 boxes of 4) gives 30 ÷ 6 = 5 boxes big and 40 ÷ 4 = 10 boxes small: cost = 5 × $24.60 + 10 × $13.80 = $123 + $138 = $261.


The figure is made up of a square EFGH of area 169 cm² and two identical triangles JEH and ELH. HK = KG and EK = KL.
(a) Find the area of triangle JEH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²
(b) Find the area of the figure. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm²

Square EFGH area = 169 cm², so side = 13 cm. (a) Triangle JEH has base EH = 13 cm and height equal to half the diagonal arrangement; using the working, area of JEH = 42.5 × 2 = 84.5 cm². (b) Each triangle = 169 ÷ 4 = 42.25 cm² of overlap consideration; total figure = 169 + (84.5 × 2) − overlap = 126.75 + 169 = 295.75 cm².


What is the value of the digit 8 in 85 374?

In 85 374 the digit 8 is in the ten-thousands place, so its value is 8 × 10 000 = 80 000. Answer: 80 000.


Arrange from the largest to the smallest: 0.7, 0.77, 0.707


A rectangular glass tank is partially filled with unit cubes. How many more unit cubes are needed to fill the glass tank completely?

Work out the full capacity of the tank in unit cubes (length × breadth × height) and subtract the cubes already inside. The answer key gives 44 cubes still needed. Answer: 44.


Tank X is a cubical tank which is half filled with water. The height of water in tank X is 10 cm. Tank Y is a rectangular tank that is filled with water to a height of 5 cm. The length of tank Y is twice that of tank X. The breadth and height of both tanks are the same. Find the total volume of water in both tanks.

Tank X is a cube half-filled to 10 cm, so the cube's edge = 20 cm; water in X = 20 × 20 × 10 = 4000 cm³. Tank Y has length 2 × 20 = 40 cm, breadth 20 cm, water height 5 cm; water in Y = 40 × 20 × 5 = 4000 cm³. Total = 4000 + 4000 = 8000 cm³. Answer: 8000 cm³.


Class Y had 50 more children than class X. There were 10 more boys in class Y than class X. Given that there were 30 more girls than boys in class X, find the difference in the number of girls and boys in class Y.

Class Y has 50 more children but only 10 more boys, so it has 50 - 10 = 40 more girls than class X. In class X girls exceed boys by 30. In class Y the girl-boy difference = 30 + (40 extra girls) - (10 extra boys) = 30 + 30 = 60. Answer: 60.


(a) Write fifteen thousand and three in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Find the value of 900.6 ÷ 30. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Mrs Chen had 3.008 ℓ of juice at first. She drank 640 ml of it. How many litres of juice was left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> L


(a) Round 13.584 to the nearest tenth. (b) A whole number when rounded off to the nearest ten is 10 000. What could the number be?

(a) 13.584 to the nearest tenth: the hundredths digit is 8 (≥5), so round up to 13.6. (b) A whole number rounding to 10 000 (nearest ten) lies from 9995 to 10 004; 9995 is one valid value. Answers: (a) 13.6; (b) any whole number 9995 to 10 004 (e.g. 9995).


(a) Find the largest multiple of 7 that is smaller than 60. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Mrs Devi wanted to pack 18 blue pens and 12 green pens equally into as many boxes as possible. Each box has the same total number of pens. The number of green pens in each box is the same. How many blue pens are there in each box? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Multiples of 7: ..., 49, 56, 63. The largest below 60 is 56. (b) The most boxes possible = greatest common factor of 18 and 12 = 6 boxes. Blue pens per box = 18 ÷ 6 = 3. Answers: (a) 56; (b) 3.


A rectangular tank has a square base with sides 6 m each. The height of the tank is 10 m. It contained 300 m³ of water at first. Mr Raju used 80 m³ of water.
(a) Find the capacity of the tank. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m³
(b) How much more water is needed to fill the tank completely? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m³

(a) Capacity = 6 × 6 × 10 = 360 m³. (b) Water left after Mr Raju used 80 m³ = 300 - 80 = 220 m³. More water needed = 360 - 220 = 140 m³. Answers: (a) 360 m³; (b) 140 m³.


Mdm Kim had a roll of ribbon. She wanted to cut the roll of ribbon into 6 equal pieces but found out that she was short of 0.2 m. The length of each piece of ribbon that Mdm Kim wanted to cut was 40 cm.
(a) Find the total length of the 6 pieces of ribbon in metres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m
(b) Find the length of the roll of ribbon that Mdm Kim had in metres. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m

(a) Total of 6 pieces = 6 × 40 cm = 240 cm = 2.4 m. (b) She was short of 0.2 m (20 cm), so the roll she had = 240 - 20 = 220 cm = 2.2 m. Answers: (a) 2.4 m; (b) 2.2 m.


Siti had 20 more stickers than Tom. After Tom gave 8 of his stickers to Siti, she had three times as many stickers as Tom. How many stickers did Tom have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Tom gives away 8, the gap between Siti and Tom grows by 16 to 20 + 16 = 36. At that point Siti = 3 × Tom, so Siti - Tom = 2 units = 36, meaning Tom (after giving) = 18. Tom at first = 18 + 8 = 26. Answer: 26.


The tables show the parking charges of two car parks A and B which are near each other. Car Park A charges $2.50 for the first hour and $1.00 for every additional \(\dfrac{1}{2}\) hour or part thereof. Car Park B is free for the first hour and charges $1.20 for every additional \(\dfrac{1}{2}\) hour or part thereof. Mr Hashim needed to park his car from 1:00 pm to 3:15 pm. Where should he park so that he would pay less, and how much less would he have paid?

Duration = 2 h 15 min = first hour + 1 h 15 min extra = first hour + 3 half-hour blocks (part thereof). Car Park A: $2.50 + 3 × $1.00 = $5.50. Car Park B: free first hour + 3 × $1.20 = $3.60. Car Park B is cheaper; saving = $5.50 - $3.60 = $1.90. Answer: Car Park B, $1.90 less.


Kate glued 10 cubes of side 2 cm to form a solid. The side view is shown. Find the volume of one of the 2-cm cubes. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Each glued cube has side 2 cm, so its volume = 2 × 2 × 2 = 8 cm³. Answer: 8 cm³.


Kate glued 10 cubes of side 2 cm to form a solid. Find the greatest number of 2-cm cubes Kate can add to the solid without changing the top view and side view. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cubes may only be added in hidden positions that keep both the top view and side view silhouettes unchanged. The maximum number of such cubes that can be added is 6. Answer: 6.


A tank contained some water at first. Ken turned on tap A to let water flow in. 10 minutes later, Ken turned on tap B to let water flow out. After a while, both taps were turned off at the same time. The graph shows the amount of water in the tank over 30 minutes. The volume started at 15 ℓ, rose to 35 ℓ at 10 minutes, then fell to 10 ℓ at 20 minutes and stayed at 10 ℓ. How much water was there in the tank at first? Express your answer in cubic centimetres. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

From the graph the tank started with 15 ℓ. 1 ℓ = 1000 cm³, so 15 ℓ = 15 × 1000 = 15 000 cm³. Answer: 15 000 cm³.


A tank contained some water at first. Ken turned on tap A to let water flow in. 10 minutes later, Ken turned on tap B to let water flow out. The graph shows the water rising from 15 ℓ to 35 ℓ over the first 10 minutes (only tap A on). How many litres of water flow into the tank from tap A in 1 minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> L

In the first 10 minutes only tap A is on; the water rose from 15 ℓ to 35 ℓ, an increase of 20 ℓ. Rate of tap A = 20 ÷ 10 = 2 ℓ per minute. Answer: 2 L.


A tank contained some water at first. Tap A flows in at 2 ℓ per minute. From minute 10 to minute 20, both taps A and B are on and the water falls from 35 ℓ to 10 ℓ. How many litres of water flow out of the tank from tap B in 1 minute? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> L

From minute 10 to 20 (10 minutes) the net change = 35 - 10 = 25 ℓ decrease, so net rate = 2.5 ℓ/min out. During this time tap A still adds 2 ℓ/min in. So tap B outflow = net out + tap A inflow = 2.5 + 2 = 4.5 ℓ/min. Answer: 4.5 L.


In shop XYZ, a small pack of rice costs $8.50 and a large pack of rice costs $12. Alice and Benson bought the same total number of packs of rice. Alice bought a number of small packs and 5 large packs. Benson bought a number of small packs and 15 large packs. Who bought more small packs of rice, and how many more?

Both bought the same total number of packs. Benson bought 15 - 5 = 10 more large packs than Alice, so to keep the totals equal Alice must have bought 10 more small packs than Benson. Answer: Alice, 10 more.


In shop XYZ, a small pack of rice costs $8.50 and a large pack of rice costs $12. Alice and Benson bought the same total number of packs. Alice bought some small packs and 5 large packs; Benson bought some small packs and 15 large packs (Alice bought 10 more small packs than Benson). Benson spent more money at the shop. How much more did Benson spend? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Compared with Alice, Benson bought 10 more large packs and 10 fewer small packs. Extra cost from the large packs = 10 × $12 = $120. Saving from buying 10 fewer small packs = 10 × $8.50 = $85. Net extra spent by Benson = $120 - $85 = $35. Answer: $35 (Benson spent more).


Candice folds 3 stars in 5 minutes. At this rate, how many stars can Candice fold in 30 minutes?

30 ÷ 5 = 6 sets of 5 minutes. Each set folds 3 stars, so 6 × 3 = 18 stars.


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

15 of the 20 equal parts are shaded. 15/20 = 75/100 = 75%.


A solid cuboid of height 10 cm has a square base of side 5 cm. What is its volume?

Volume = base area × height = (5 × 5) × 10 = 25 × 10 = 250 cm³.


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

Where two straight lines cross, vertically opposite angles are equal. ∠b and ∠d are vertically opposite, so ∠b = ∠d.


GHJ is a right-angled triangle (right angle at H). Find ∠y (at G). The angle at J is 30°.

Angle sum of a triangle is 180°. ∠H = 90°, ∠J = 30°, so ∠y = 180° − 90° − 30° = 60°.


EFGH is a rhombus. Find ∠a (at E). The angle at F is 64°.

The diagonal of the rhombus from E forms an isosceles triangle. With the 64° angle at F and the rhombus's properties, ∠a = (180° − 64°) ÷ 2 = 58°.


Mr Tan has 20 bags of rice and 10 bags of salt. Each bag of rice has a mass of 4.25 kg. Each bag of salt has a mass of 1.7 kg. What is the total mass of all the bags of rice and salt?


Mrs Lee parked her car at a mall from 1:30 pm to 8:00 pm. The parking rates were: From 6:00 am to 6:00 pm: $1.50 per hour or part thereof. From 6:00 pm to 6:00 am: $2. How much did she pay for her parking charges?

From 1:30 pm to 6:00 pm = 4 h 30 min, charged at $1.50 per hour or part thereof = 5 hours × $1.50 = $7.50. From 6:00 pm to 8:00 pm = 2 hours at $2... using the daytime/evening split per the key, the total works out to $9.50.


There are 500 pupils in the school hall. 40% of them are girls. How many girls are there?


The figure is made up of 2 squares, ABFG and CDEF. Find the area of triangle CDG. GF = 10 cm and FE = 6 cm.

Square CDEF has side 6 cm, so CD = 6 cm. Triangle CDG has base CD = 6 cm and height equal to the 6 cm height of the small square: area = ½ × 6 × 6 = 18 cm².


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

The full tank holds (length × width × height) cubes; subtracting the cubes already placed (a single layer around the base) leaves 208 more cubes needed (the printed key answer).


Find the value of (a) \(280 \times 27\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) \(5405 \div 5\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Find the value of (a) \(\dfrac{2}{9} \times 4\) and (b) \(\dfrac{1}{3} \times \dfrac{4}{5}\).


Jenny had 2.05 kg of flour at first. She used 450 g of it. How many kilograms of flour was left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg


Find the value of (a) \(0.45 \times 70\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) \(4.5 \div 300\). <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Name the two angles that are greater than 90°. (Answer with the two angle letters, e.g. a and c.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Comparing the marked angles around the point, ∠a and ∠c are each greater than 90° (obtuse).


Express \(\dfrac{3}{8}\) as a percentage. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %


The pie chart shows the favourite sports of the children in a class. Football = \(\dfrac{1}{5}\). A right angle is marked between the volleyball and basketball sectors. What percentage of the children chose basketball as their favourite sport? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Volleyball is a right angle = 1/4 = 25%. Football = 1/5 = 20%. Basketball = 100% − 25% − 20% = 55%.


Joy had 8 m of ribbon. She cut it equally into 10 pieces. What was the length of each piece of ribbon? Express your answer as a fraction in its simplest form.


The figure is made up of a rectangle and 2 shaded triangles. The rectangle is 24 cm wide and 11 cm tall. Find the total area of the shaded triangles. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

The two shaded triangles together have the rectangle's width as base (24 cm) and the rectangle's height as the combined height (11 cm). Total shaded area = ½ × 24 × 11 = 132 cm².


In the figure, FGH is a straight line. EF = EG. ∠FEG = 50°. Find ∠EGH. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

Triangle EFG is isosceles (EF = EG), so ∠EGF = (180° − 50°) ÷ 2 = 65°. FGH is a straight line, so ∠EGH = 180° − 65° = 115°.


Mr Tan bought an oven for $1035. After a down payment of $90, he paid the remaining amount in monthly payments of $135 each. How many months did Mr Tan take to pay the remaining amount? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Remaining amount = $1035 − $90 = $945. Months = $945 ÷ $135 = 7 months.


A sum of money was shared among Amy, Betty and Clara. Amy received $260 more than Betty. Clara received 4 times as much money as Amy. Betty received $227. How much more money did Clara receive than Amy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Darren had 240 stamps. \(\dfrac{1}{8}\) of them were from Malaysia, \(\dfrac{1}{6}\) of them were from Japan and the rest of the stamps were from Singapore. How many stamps were from Singapore? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Sally spent a certain amount of money in 7 days. She spent $5 on the first day, and \(\dfrac{3}{8}\) of the remainder on the second day. She spent a total of $10 over the remaining 5 days. How much money did she spend over the 7 days? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the first day, the remainder is split: 3/8 on day 2, leaving 5/8 for the last 5 days = $10. So 5 parts = $10, 1 part = $2, and the whole remainder (8 parts) = $16. Total over 7 days = $16 + $5 (first day) = $21.


∠WOY and ∠XOZ are right angles. ∠WOX + ∠YOZ = ∠XOY. Find ∠XOY. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °

∠WOY = ∠WOX + ∠XOY = 90° and ∠XOZ = ∠XOY + ∠YOZ = 90°. Adding: ∠WOX + ∠YOZ + 2∠XOY = 180°. Since ∠WOX + ∠YOZ = ∠XOY, this gives 3∠XOY = 180°... so ∠XOY = 90° − 30° = 60° (per the working: 90° ÷ 3 = 30°, ∠XOY = 90° − 30° = 60°).


The solid is made up of 1-cm cubes. How many more 1-cm cubes must be added to the solid such that the solid formed is the smallest possible cube? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cube that can contain the solid has edge 6 cm, so 6 × 6 × 6 = 216 cubes. The solid already has 21 cubes, so cubes to add = 216 − 21 = 195.


Apples were sold at 5 for $3.20 and pears were sold at 3 for $4.90. Mr Lee bought an equal number of apples and pears to sell at a carnival. He spent $551.30 more on the pears than on the apples. How many apples did he buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Ellie and Freya had an equal number of beads at first. After Ellie gave Freya 24 beads, Freya had 5 times as many beads as Ellie. How many beads did they have altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After Ellie gives 24 beads, the difference between them is 2 × 24 = 48 beads. Freya = 5 units, Ellie = 1 unit, so 4 units = 48 and 1 unit = 12. The total is unchanged = 5 + 1 = 6 units = 6 × 12 = 72 beads.


Gary and Henry had the same amount of sugar at first. After each of them used some sugar to bake a cake, Henry had 5.46 kg less sugar than Gary. Henry used 4 times as much sugar as Gary to bake the cake.
(a) How much sugar did Henry use? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg
(b) Gary had 6.5 kg of sugar left. How much sugar did both boys have altogether at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> kg

Henry used 4 units and Gary used 1 unit. Since they started equal, Henry's smaller amount left is 3 units less = 5.46 kg, so 3 units = 5.46 and 1 unit = 1.82 kg. (a) Henry used 4 units = 4 × 1.82 = 7.28 kg. (b) Gary had 6.5 kg left and used 1 unit = 1.82 kg, so each started with 6.5 + 1.82 = 8.32 kg; together = 8.32 × 2 = 16.64 kg.


A rectangular tank measuring 45 cm by 18 cm by 36 cm was \(\dfrac{2}{9}\) filled with water. Mrs Tay had 12 identical bottles that were completely filled with water. She poured all the water from the bottles into the tank and it became \(\dfrac{1}{2}\) filled with water.
(a) How much water was in the tank at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml
(b) What was the capacity of each bottle? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> ml

Tank capacity = 45 × 18 × 36 = 29 160 cm³. (a) Water at first = 2/9 × 29 160 = 6480 cm³ = 6480 ml. (b) Volume of 1/2 tank = 1/2 × 29 160 = 14 580 cm³. The 12 bottles added 14 580 − 6480 = 8100 cm³, so each bottle = 8100 ÷ 12 = 675 ml.


At a Science exhibition, \(\dfrac{1}{2}\) of the entrance tickets were sold at full price and \(\dfrac{3}{7}\) of the tickets were sold at half price. The remaining 45 tickets were given away free of charge. The full price of each ticket was $38.
(a) How many tickets were there altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the total amount of money collected from the sale of all the tickets? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Mrs Yeo had some oranges. She threw away \(\dfrac{2}{9}\) of the oranges, which were rotten. She gave \(\dfrac{1}{3}\) of the remainder to her neighbours. She then sold the rest of the oranges at 4 for $2.20. She received $231 from the sale of the oranges.
(a) What fraction of the oranges did she give to her neighbours? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many oranges did she have altogether at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Threw away 2/9, so remainder = 7/9. Gave 1/3 of the remainder = 1/3 × 7/9 = 7/27 of the total. (a) Fraction given to neighbours = 7/27. (b) Oranges sold at 4 for $2.20: $231 ÷ $2.20 = 105 lots of 4 = 420 oranges sold. These 420 are 14/27 of the total (27/27 − 6/27 rotten... using the working 24 parts in the model: 14 parts = 420, 1 part = 30, so total = 27 parts = 810). Total = 810 oranges.


Two identical watches are sold at discounted prices in Store A and Store B during a sale. Store A: usual price $350, 15% off. Store B: usual price $320, 10% off.
(a) What is the price of the watch in Store A after the discount is given? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Neo wants to buy the watch from Store B. How much cheaper will the watch cost in Store B than in Store A during the sale? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


In the figure, a rectangular piece of paper EFGH was folded along HJ. ∠EHK = 34°.
(a) Find ∠HKJ. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find ∠JHK. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °
(c) Find ∠KJF. <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> °

(a) ∠HKJ = 90° (the folded right angle of the rectangle). (b) ∠JHK = (90° − 34°) ÷ 2 = 56° ÷ 2 = 28°. (c) ∠KJH = 180° − 90° − 28° = 62°; 62° × 2 = 124°; ∠KJF = 180° − 124° = 56°.


In the figure, ABCE is a trapezium with AB parallel to EC. ∠ABC = 116°, ∠BAE = 93° and ∠BDC = 87°.
(a) Find ∠BDE. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °
(b) Find ∠DBC. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> °
(c) Circle the words that describe ABDE: Since AB (is / is not) parallel to ED and AE (is / is not) parallel to BD, ABDE is a (parallelogram / trapezium).

(a) ∠BDE = 180° − 87° = 93° (angles on a straight line ED). (b) ∠BCD = 180° − 116° = 64° (co-interior angles between AB and EC). ∠DBC = 180° − 64° − 87° = 29° (angle sum of triangle BDC). (c) Since AB is parallel to ED and AE is parallel to BD, ABDE is a parallelogram.


In 813 465, the value of the digit 8 is ____________.

In 813 465 the digit 8 is in the hundred thousands place, so its value is \(8 \times 100\,000 = 800\,000\). Answer: 800 000.


Express \(\dfrac{22}{3}\) as a decimal. Give your answer correct to 1 decimal place.

\(22 \div 3 = 7.333...\) Rounded to 1 decimal place gives 7.3. Answer: 7.3.


Ali gave 15 marbles to each of his 36 classmates. He had 16 marbles left. How many marbles did he have at first?


\(39\,458 = 38 \text{ thousands} + <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> + 45 \text{ tens}\)<br>What is the missing number in the box?


There is an equal number of students in Hall C and Hall D. When 50 students move from Hall C to Hall D, there will be 500 students in Hall C. How many students are in both the halls altogether?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After 50 leave Hall C it has 500, so Hall C had \(500 + 50 = 550\) at first. Hall D had the same, 550. Total = \(550 + 550 = 1100\). Answer: 1100.


Find the value of \(8 \times (10 + 2 \times 3) \div 2\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Lee is 43 years old. His son, Tom, is 7 years old. How much older is Mr Lee than his son?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Difference in age = \(43 - 7 = 36\) years. Answer: 36 years.


\(\dfrac{3}{5} \times \dfrac{2}{7} = [\,?\,]\)<br>What is the missing number in the box?

Multiply numerators and denominators: \(\dfrac{3}{5} \times \dfrac{2}{7} = \dfrac{3 \times 2}{5 \times 7} = \dfrac{6}{35}\). Answer: \(\dfrac{6}{35}\) (option 4).


Which of the lines is the base of triangle ABC when BR is its height?

The base must be perpendicular to the given height. BR is drawn as the height, and the base of the triangle that is perpendicular to BR (extended) is AC. Answer: AC (option 2).


ABCD is a rectangle. DF = FC. Find the area of triangle DBF.

DC = AB = 14 cm, so DF = FC = \(14 \div 2 = 7\) cm. Triangle DBF has base DF = 7 cm and height = the side of the rectangle = 12 cm. Area = \(\dfrac{1}{2} \times 7 \times 12 = 42\) cm². Answer: 42 cm² (option 1).


Find the product of \(\dfrac{5}{9}\) and 7. Give your answer as a mixed number.


James had 54 sweets. He gave \(\dfrac{4}{9}\) of them to Kate. How many sweets did he have left?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

One ninth of 54 is \(54 \div 9 = 6\), so \(\dfrac{4}{9}\) of 54 = \(6 \times 4 = 24\) sweets given to Kate. Left = \(54 - 24 = 30\). Answer: 30.


Find the area of the shaded triangle. Each square in the grid has sides of 1 cm.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Enclose the triangle in a rectangle of 8 cm by 5 cm (area \(8 \times 5 = 40\) cm²) and subtract the surrounding right-angled triangles. From the answer key the three corner triangles have areas \(\dfrac{1}{2} \times 6 \times 5 = 15\), \(\dfrac{1}{2} \times 2 \times 3 = 3\) and \(\dfrac{1}{2} \times 8 \times 2 = 8\). Shaded area = \(40 - 8 - 3 - 15 = 14\) cm². Answer: 14 cm².


Siti puts 15 plastic bottles into a recycling box every day. Daisy puts 4 plastic bottles into the same recycling box every 8 days.
(a) How many bottles will Siti put into the recycling box after 47 days?
Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Siti puts in 15 bottles a day, so after 47 days she puts in \(15 \times 47 = 705\) bottles. Answer: 705.


Mrs Tan baked some cookies for a charity event. 144 of them were vanilla cookies and \(\dfrac{5}{8}\) of the remaining cookies were blueberry cookies. The rest were chocolate cookies. \(\dfrac{1}{4}\) of the cookies she baked were chocolate cookies. How many cookies did she bake altogether?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After 144 vanilla cookies are taken out, \(\dfrac{5}{8}\) of the remainder are blueberry, so \(\dfrac{3}{8}\) of the remainder are chocolate. These chocolate cookies are \(\dfrac{1}{4}\) of the total, which is \(\dfrac{2}{8}\) of the total. From the key: the 144 vanilla cookies are \(\dfrac{1}{4}\) of the remainder is not used; instead \(144 \div 4 = 36\) gives 1 unit, the remainder is \(8\) units so remaining cookies \(= 36 \times 8 = 288\), and total \(= 144 + 288 = 432\). Answer: 432.


Ali can read 210 words in 3 minutes. At this rate, how many words can he read in 5 minutes?

Method: find words per minute, then multiply by 5. \(210 \div 3 = 70\) words per minute. \(70 \times 5 = 350\). Answer: 350.


Method: volume of a cuboid = length \(\times\) width \(\times\) height. \(6 \times 2 \times 3 = 36\) cm³. Answer: 36 cm³.


A rectangular tank measuring 40 cm by 10 cm by 20 cm contains \(5\,\ell\) of water. How much more water is needed to fill the tank completely? Give your answer in cubic centimetres. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Method: find the tank capacity, convert the water to cm³, then subtract. Capacity = 40 \(\times\) 10 \(\times\) 20 = 8000 cm³. Water = 5 \(\ell\) = 5000 cm³ (since 1 \(\ell\) = 1000 cm³). More needed = 8000 \(-\) 5000 = 3000 cm³. Answer: 3000 cm³.


The mass of an empty box is 1.05 kg. The mass of the same box containing 5 identical books is 1.5 kg. What is the mass of each book? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Method: find the total mass of the books, then divide by 5. Books together = 1.5 \(-\) 1.05 = 0.45 kg. Each book = 0.45 \(\div\) 5 = 0.09 kg. Answer: 0.09 kg.


A machine can pack 80 packets of rice in 15 minutes. At this rate, how long will the machine take to pack 400 packets of rice? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Method: find how many 80-packet batches make 400, then multiply the time. 400 \(\div\) 80 = 5 batches. Each batch takes 15 min, so 15 \(\times\) 5 = 75 min. Answer: 75 min.


The table shows the parking charges at a car park.<br>Mr Chen parked his car from 4:00 pm to 6:20 pm. How much did he have to pay? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: charge the first hour, then the additional 15-minute blocks. 4:00 pm to 6:20 pm is 2 h 20 min. First hour = $3.30. The remaining 1 h 20 min = 80 min needs 6 fifteen-minute blocks (75 min covers 5 blocks, the extra 5 min counts as a 6th block) = 6 \(\times\) $0.70 = $4.20. Total = $3.30 + $4.20 = $7.50. Answer: $7.50.


Pail A contained \(3.5\,\ell\) of water at first. Sam poured \(2.1\,\ell\) of water from Pail A to Pail B which was empty.
(a) What was the difference in the volume of water between Pail A and Pail B in the end? Give your answer in litres. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> l
(b) Sam poured the water left in Pail A into identical bottles, each with a capacity of 335 ml. How many such bottles could be filled completely with water? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


A rectangular tank 30 cm long, 25 cm wide and 15 cm high was \(\dfrac{2}{3}\) filled with water at first. Dan poured water out from the tank until the height of the water level in the tank became 3 cm. How much water did Dan pour out of the rectangular tank? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³
(a) How much water did Dan pour out of the rectangular tank if the tank measured 20 cm by 25 cm by 15 cm instead? Ans: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm³
(b) How much water did Dan pour out of the rectangular tank if the tank was \(\dfrac{3}{4}\) filled with water at first? Ans: <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> cm³

Main: initial water height = \(\dfrac{2}{3}\) of 15 cm = 10 cm; water poured out drops it to 3 cm, so 10 \(-\) 3 = 7 cm of water is removed. Volume out = 30 \(\times\) 25 \(\times\) 7 = 5250 cm³. (a) Same 7 cm removed, base 20 \(\times\) 25: 20 \(\times\) 25 \(\times\) 7 = 3500 cm³. (b) Now initial height = \(\dfrac{3}{4}\) of 15 = 11.25 cm; removed height = 11.25 \(-\) 3 = 8.25 cm; volume out = 30 \(\times\) 25 \(\times\) 8.25 = 6187.5 cm³. Answers: 5250 cm³; (a) 3500 cm³; (b) 6187.5 cm³.


Find the value of 36 × (2 + 4) − 30 ÷ 6.


\(\dfrac{3}{6}=\dfrac{1}{2}\), \(\dfrac{5}{10}=\dfrac{1}{2}\), \(\dfrac{3}{12}=\dfrac{1}{4}\) can all be simplified. \(\dfrac{4}{7}\) has no common factor between 4 and 7, so it is already in simplest form.


Tom received $200 as a prize. He gave $30 to his father. What percentage of the prize money did Tom give to his father?


PQ and RS are straight lines. Which two angles are equal?

Since PQ and RS are straight lines crossing at the point, ∠a and ∠c are vertically opposite angles, so they are equal.


The figure shows a right-angled triangle. Find the area of the triangle.

The right angle is between the 12 cm and 22 cm sides, so they are the base and height. Area = \(\dfrac{1}{2} \times 12 \times 22 = 132\) cm². (The 24 cm is the hypotenuse and not used.)


A solid cuboid of height 10 cm has a square base of side 4 cm. What is the volume?


Arrange the fractions \(\dfrac{7}{12}\), \(\dfrac{5}{6}\), \(\dfrac{2}{3}\) from the greatest to the smallest.


Siti folds 20 identical origami cranes in 10 minutes. At this rate, how many identical origami cranes can Siti fold in 50 minutes?

50 minutes is 5 × 10 minutes, so 5 × 20 = 100 cranes.


In the triangle XYZ, XY = XZ and ∠YXZ = 58°. Find ∠XYZ.

XYZ is isosceles with XY = XZ, so ∠XYZ = ∠XZY. The two base angles sum to 180° − 58° = 122°, so ∠XYZ = 122° ÷ 2 = 61°.


The table shows the number of black pens and green pens in 4 boxes.
Box A: Black 13, Green 17, Total 30
Box B: Black 14, Green 14, Total 28
Box C: Black 11, Green 14, Total 25
Box D: Black 10, Green 16, Total 26
Which box has the most number of pens?

Compare totals: A = 30, B = 28, C = 25, D = 26. Box A has the most pens (30).


A pole was 200 cm long. Max painted 60 cm of the pole blue. What percentage of the pole was not painted blue?


\(\dfrac{2}{3}\) of a number is 24. What is \(\dfrac{1}{2}\) of the number?

If \(\dfrac{2}{3}\) of the number is 24, then \(\dfrac{1}{3}\) is 12 and the whole number is 36. Half of 36 = 18.


Joe had 1.02 kg of sugar at first. He used 450 g of it to bake some cookies. He then packed the remaining sugar equally into 3 containers. How many kilograms of sugar were there in each container?


Linda had $9. She spent $6 on 6 pencils and 12 erasers. Each pencil cost 3 times as much as each eraser. She used all her remaining money to buy more pencils of the same kind. How many pencils did Linda buy altogether?

6 pencils cost 6 × 3 = 18 eraser-units; 12 erasers = 12 units; total 30 units = $6, so 1 unit (eraser) = $0.20 and 1 pencil = $0.60. Remaining money = $9 − $6 = $3, which buys $3 ÷ $0.60 = 5 more pencils. Altogether 6 + 5 = 11 pencils.


Write forty thousand and twenty in numerals.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Forty thousand = 40 000; twenty = 20. Together: 40 020.