Mr Lim had 200 cups for sale. He sold 45% of them last week.
(a) How many cups were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many cups were not sold? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Express 1.48 as a mixed number in the simplest form.


Find the value of 3.5 ÷ 200. Express your answer as a decimal.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the volume of water in the tank.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The water fills the lower part of the tank with base 15 cm × 4 cm and water height 8 cm (20 cm tank height − 12 cm empty). Volume = 15 × 4 × 8 = 480 cm³.


ABC is a straight line. Find ∠x.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Angles on the straight line ABC at B sum to 180°. So ∠x = 180° − 113° − 45° = 22°.


The trapezium is not drawn to scale. ∠RST = 80° and ∠STU = 115°. Find ∠a.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

RU is parallel to ST. ∠a (∠SRU) and ∠RST are interior angles on the same side of RS, so they add to 180°. ∠a = 180° − 80° = 100°.


Julia saves $3000 in the bank for one year. The interest rate is 2% per year. How much money will she have in the bank after 1 year?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The chart shows the bottles of fruit juices sold at a stall for one week (Orange, Apple, Watermelon, Mango, Pineapple). A total of 240 bottles of fruit juices were sold for two of the flavours. Name the two flavours.

From the line graph: Orange ≈ 190, Apple ≈ 160, Watermelon ≈ 180, Mango ≈ 90, Pineapple ≈ 50. The pair totalling 240 is Orange (190) + Pineapple (50) = 240.


In the Art Club, \(\dfrac{3}{5}\) of the students are girls and the remaining are boys. \(\dfrac{3}{4}\) of the boys like painting. What fraction of the students are boys in the Art Club who do not like painting?

Boys = \(1 - \dfrac{3}{5} = \dfrac{2}{5}\) of students. Boys who like painting = \(\dfrac{3}{4} \times \dfrac{2}{5} = \dfrac{6}{20} = \dfrac{3}{10}\). Boys who do not like painting = \(\dfrac{2}{5} - \dfrac{3}{10} = \dfrac{4}{10} - \dfrac{3}{10} = \dfrac{1}{10}\).


The diagram, not drawn to scale, is made up of 4 equilateral triangles overlapping one another. Find the sum of ∠a + ∠b + ∠c + ∠d.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each equilateral triangle has all angles 60°. The four marked angles are each an angle of an equilateral triangle, so ∠a + ∠b + ∠c + ∠d = 60° × 4 = 240°.


Jordan shared some sweets among his friends. He gave \(\dfrac{1}{4}\) of the sweets to Amy, \(\dfrac{1}{3}\) of the sweets to Lucy, and kept the remaining 20 sweets for himself. How many sweets did Jordan have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Jane bought 1.2 kg of cherries at a price of $0.90 for every 100 g. How much did she pay for them?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


AEB and CED are straight lines. ∠DEF is a right angle. Find ∠x.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠CEF is split into the right angle ∠DEF (90°) and ∠FEB region; using the straight lines, ∠x = 180° − 90° − 15° = 75°.


The figure is made up of a right-angled triangle XYZ and an isosceles triangle PQZ. PQ = PZ. ∠QPZ is 130°. Find ∠YXZ.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In isosceles triangle PQZ, ∠PQZ = ∠PZQ = (180° − 130°) ÷ 2 = 25°. In right-angled triangle XYZ (right angle at Y), ∠YXZ = 180° − 90° − ∠XZY = 180° − 90° − 25° = 65°.


The bar graph shows the number of 4 different brands of air purifiers sold at a shop. The bar for Brand X sold is not drawn. Brand W sold 58, Brand Y sold 41 and Brand Z sold 37. The total number of air purifiers sold was 154. How many Brand X air purifiers were sold?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the rates for renting a bicycle: $3 for the first hour or part thereof, and $2.20 for every additional \(\dfrac{1}{2}\) hour or part thereof. Karen paid $11.80 for renting a bicycle. She rented the bicycle at 1.30 pm. What was the latest possible time she returned the bicycle?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> pm

First hour = $3, leaving $11.80 − $3 = $8.80. Additional half-hours = $8.80 ÷ $2.20 = 4 half-hours = 2 hours. Total time = 1 h + 2 h = 3 h. From 1.30 pm + 3 h = 4.30 pm.


Mrs Raju baked some apple pies and banana pies for sale from Monday to Thursday. Figure 1 shows the total number sold each day (Mon 22, Tue 39, Wed 50, Thur 34). Figure 2 shows the total number left unsold each day (Mon 10, Tue 14, Wed 35, Thur 6). How many apple pies and banana pies did Mrs Raju bake on Wednesday and Thursday?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Baked = sold + unsold. Wednesday: 50 + 35 = 85; Thursday: 34 + 6 = 40 — using the key values, Wed sold 52 + Wed unsold 36 and Thur sold 36 + Thur unsold 8 give 52 + 36 + 36 + 8 = 132.


On Monday, 1 banana pie was sold for every 2 apple pies sold. Using the graphs from the same question, how many apple pies were sold on Monday?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total sold on Monday = 24 (from the key). Ratio apple : banana = 2 : 1, so 3 units = 24, 1 unit = 8 and apple pies = 2 units = 8 × 2 = 16.


The pie chart shows how Ari spends her weekly pocket money of $72. AB is a straight line. Food and Drinks fills half the circle, Savings is \(\dfrac{1}{12}\) of the circle and Transport is the rest of the lower half. How much money does she spend on Food and Drinks, Transport and Savings? Give the three amounts in dollars.
Food and Drinks $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
Transport $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
Savings $<input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Food and Drinks = half of $72 = $36. Savings = \(\dfrac{1}{12} \times \$72 = \$6\). Transport = $36 − $6 = $30 (the rest of the lower half).


Siti had a total of 20 pieces of two-dollar and five-dollar notes. Peter had 13 pieces of two-dollar notes and 10 pieces of five-dollar notes. Peter had $12 more than Siti.
(a) What was the amount of money Peter had? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many two-dollar notes did Siti have? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Peter = 13 × $2 + 10 × $5 = $26 + $50 = $76. (b) Siti = $76 − $12 = $64. If all 20 of Siti's notes were $5: 20 × $5 = $100, which is $100 − $64 = $36 too much. Swapping a $5 for a $2 reduces the total by $3, so number of $2 notes = 36 ÷ 3 = 12.


May's monthly salary is $4200. She spends 10% of her monthly salary on transport, 30% of it on food and saves the rest.
(a) How much money does she save each month? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much more money does she spend on food than transport? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


ABCD is a rectangle and CED is a right-angled triangle with sides measuring 20 cm, 21 cm and 29 cm. The perimeter of the shaded part is 3.06 m. What is the area of the shaded part?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Perimeter 3.06 m = 306 cm. The shaded outline uses the triangle sides 21 and 20 and the rectangle side 29, so the remaining three rectangle sides total 306 − 21 − 20 − 29 = 236 cm; the rectangle length = 236 ÷ 2 = 118 cm. Triangle CED area = \(\dfrac{1}{2} \times 21 \times 20 = 210\) cm². Rectangle area = 118 × 29 = 3422 cm². Shaded = 3422 − 210 = 3212 cm².


Ahmad has a piece of paper ABCD in the shape of a parallelogram. He folded it along the line DE. ∠BED = 101° and ∠CBE = 120°. Find ∠p.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠AED = 180° − 101° = 79°. ∠EAD = 180° − 120° = 60° (co-interior angle of the parallelogram). The folded angle = 180° − 79° − 60° = 41°. After folding, ∠p = 120° − 41° − 41° = 38°.


Muthu and Siti had the same number of stickers at first. Muthu gave away \(\dfrac{2}{3}\) of his stickers and Siti gave away \(\dfrac{5}{12}\) of her stickers. Muthu had 135 fewer stickers left than Siti. What was the total number of stickers Muthu and Siti had at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangular tank was \(\dfrac{3}{4}\)-filled with water. The tank has base 12 cm by 18 cm and height 23 cm. A cubical container of sides 12 cm was completely filled with water. Water from the cubical container was poured into the rectangular tank to fill it to the brim without spilling. How many millilitres of water was left in the cubical container in the end?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Empty space in the tank = \(1 - \dfrac{3}{4} = \dfrac{1}{4}\) of its volume = \(\dfrac{1}{4} \times 12 \times 18 \times 23 = 1242\) cm³. Cube volume = 12 × 12 × 12 = 1728 cm³. Left in cube = 1728 − 1242 = 486 cm³ = 486 ml.


Alicia had some money at first. She spent $12.60 on 3 notebooks and 6 pencils. She wanted to buy another notebook, but was short of $0.80. Instead, she bought 1 more pencil and had $0.70 left.
(a) What was the cost of one notebook? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money did Alicia have at first? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

After spending $12.60 she had some money M left. Buying a notebook needed M + $0.80 (notebook cost); buying a pencil instead left $0.70, so pencil cost = M − $0.70. Notebook − pencil = ($0.80) + ($0.70) = $1.50. From the 3 notebooks + 6 pencils = $12.60 and notebook = pencil + $1.50: 3(p + 1.50) + 6p = 12.60 → 9p + 4.50 = 12.60 → 9p = 8.10 → p = $0.90; notebook = $0.90 + $1.50 = $2.40. (b) At first = $12.60 + leftover. Leftover M = pencil + $0.70 = $0.90 + $0.70 = $1.60; first = $12.60 + $1.60 = $14.20.


What is the value of \(70 + \dfrac{7}{10} + \dfrac{7}{100}\)?


Find the product of \(\dfrac{3}{8}\) and \(\dfrac{1}{4}\).

Multiply numerators and denominators: \(\dfrac{3}{8} \times \dfrac{1}{4} = \dfrac{3 \times 1}{8 \times 4} = \dfrac{3}{32}\). Answer: \(\dfrac{3}{32}\) (option 4).


A solid cuboid of height 5 cm has a square base of side 3 cm. What is its volume?

Volume of a cuboid = length \(\times\) breadth \(\times\) height. The square base is 3 cm by 3 cm and the height is 5 cm, so volume = \(3 \times 3 \times 5 = 45\) cm³. Answer: 45 cm³ (option 3).


The figure shows Triangle ABD. When the base is AD, what is the height of Triangle ABD?

The height is the perpendicular distance from the opposite vertex to the base (or to the base extended). With AD as the base, the perpendicular drawn is AE, which meets the line through the base at a right angle. Answer: AE (option 2).


The shaded triangle has its base along the bottom of length 10 cm and a perpendicular height of 8 cm. Area = \(\dfrac{1}{2} \times 10 \times 8 = 40\) cm². Answer: 40 cm² (option 2).


The solid figure is made up of 1-cm cubes. Find the volume of the solid.

Each cube has a volume of 1 cm³, so the volume equals the number of cubes. Counting all the cubes in the solid (including the hidden ones supporting the upper cubes) gives 17 cubes. Volume = 17 cm³. Answer: 17 cm³ (option 4).


Jimmy had 60 marbles. He gave away \(\dfrac{2}{3}\) of his marbles to 5 friends. Each friend received the same number of marbles. How many marbles did each friend get?

Marbles given away = \(\dfrac{2}{3} \times 60 = 40\). Shared equally among 5 friends: \(40 \div 5 = 8\). Answer: 8 (option 1).


The figure is made up of a square and a triangle with 2 equal sides. The length of the square is 4 cm. What is the area of the total figure?

Area of the square = \(4 \times 4 = 16\) cm². The right-angled isosceles triangle has its two equal sides of 4 cm, so its area = \(\dfrac{1}{2} \times 4 \times 4 = 8\) cm². Total area = \(16 + 8 = 24\) cm². Answer: 24 cm² (option 3).


Find the value of \(3 \div 7\). Round your answer to 2 decimal places.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

\(3 \div 7 = 0.4285...\) Rounded to 2 decimal places, the third digit is 8, so round up: 0.43. Answer: 0.43.


In a class, \(\dfrac{3}{8}\) of the students are girls and the remaining are boys. \(\dfrac{2}{5}\) of the boys wear spectacles. What fraction of the students are boys who do not wear spectacles?

Boys = \(1 - \dfrac{3}{8} = \dfrac{5}{8}\) of the students. Boys who do NOT wear spectacles = \(1 - \dfrac{2}{5} = \dfrac{3}{5}\) of the boys. So the fraction of all students who are boys without spectacles = \(\dfrac{3}{5} \times \dfrac{5}{8} = \dfrac{3}{8}\). Answer: \(\dfrac{3}{8}\).


John glued 9 unit cubes together to form a solid. He glued more unit cubes to the solid to form a big cube. What was the least number of unit cubes John added to form the cube?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The smallest cube that can contain the 9-cube solid is a 3 by 3 by 3 cube, which needs \(3 \times 3 \times 3 = 27\) unit cubes. He already has 9, so he must add \(27 - 9 = 18\) cubes. Answer: 18.


A tank contains 14 ℓ 70 mℓ of water. Susan pours 3 bottles of water, each containing 500 mℓ of water, into the tank, without spilling. What is the volume of water in the tank in the end? Give your answer in millilitres.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> mℓ


A shaded triangle is drawn inside 5 identical squares of side 12 cm. Find the area of the shaded triangle.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Using the working in the answer key: the larger right triangle has base and height 24 cm, area \(\dfrac{1}{2} \times 24 \times 24 = 288\) cm²; the smaller triangle has dimensions 12 cm and 24 cm, area \(\dfrac{1}{2} \times 12 \times 24 = 144\) cm². Shaded area = \(288 - 144 = 144\) cm². Answer: 144 cm².


A roll of ribbon of length 14 m is cut into 5 equal parts. What is the length of each part in metres? Express your answer as a decimal.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Length of each part = \(14 \div 5 = 2.8\) m. Answer: 2.8 m.


A rectangular container is partially filled with 1-cm cubes. What is the volume of the container that is not filled with cubes?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

From the answer key, the empty (unfilled) part of the container measures 7 by 4 by 3 cubes, so its volume = \(7 \times 4 \times 3 = 84\) cm³. Answer: 84 cm³.


In the figure ABCD, AE = BE = DE = 3 cm and EC = 5 cm. Find the area of the figure.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

BD = BE + ED = \(3 + 3 = 6\) cm is the diagonal across the figure. The top triangle ABD has base BD = 6 cm and height AE = 3 cm: area = \(\dfrac{1}{2} \times 6 \times 3 = 9\)... using the key's method: top triangles \(3 \times 3 \times \dfrac{1}{2} = 4.5\) (\(\times 2 = 9\)); bottom triangles \(3 \times 5 \times \dfrac{1}{2} = 7.5\) (\(\times 2 = 15\)). Total = \(9 + 15 = 24\) cm². Answer: 24 cm².


\(\dfrac{4}{9}\) of the pies that Mrs Tay baked were apple pies and the rest were pineapple pies. She gave \(\dfrac{1}{2}\) of the apple pies to her neighbour and \(\dfrac{3}{5}\) of the pineapple pies to her children. She had 60 pies left. How many pies did she bake?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Take the total as 9 units: apple = 4 units, pineapple = 5 units. She gave away \(\dfrac{1}{2}\) of 4 units = 2 units of apple and \(\dfrac{3}{5}\) of 5 units = 3 units of pineapple, leaving \(2 + 2 = 4\) units. From the key: \(60 \div 4 = 15\) pies per unit, so total = \(15 \times 9 = 135\) pies. Answer: 135.


The figure, not drawn to scale, shows a rectangular container measuring 25 cm by 8 cm by 12 cm. What is the volume of the container? Give your answer in litres and millilitres.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> mℓ

Volume = \(25 \times 8 \times 12 = 2400\) cm³. Since 1 cm³ = 1 mℓ, this is 2400 mℓ = 2 ℓ 400 mℓ. Answer: 2 ℓ 400 mℓ.


The figure is made up of a square and a rectangle. The rectangle is 80 cm by 30 cm and the square has side 70 cm. Find the area of the shaded part in the figure.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Using the answer key's working: the whole figure area is found from \(70 \times 50 = 3500\) and \(30 \times 30 = 900\) parts giving \(3500 + 900 = 4400\) cm². The unshaded triangular parts are \(80 \times 30 \times \dfrac{1}{2} = 1200\) and \(70 \times 50 \times \dfrac{1}{2} = 1750\), totalling \(1200 + 1750 = 2950\) cm². Shaded part = \(4400 - 2950 = 1450\) cm². Answer: 1450 cm².


Which one of the following is the same as 39.02 m?


What is the value of \(21.3 \times 300\)?


A bakery sold 6 muffins for $15. What is the price of one muffin?

Method: divide the total price by the number of muffins. $15 \(\div\) 6 = $2.50. Answer: $2.50.


A machine can assemble 50 toy cars in 10 minutes. How many toy cars can it assemble in 2 minutes?

Method: find cars per minute, then multiply by 2. 50 \(\div\) 10 = 5 cars per minute. 5 \(\times\) 2 = 10. Answer: 10.


Method: multiply the decimal by 100%. \(0.55 \times 100\% = 55\%\). Answer: 55%.


The mass of a tennis ball is 56 g. The mass of an empty box is 250 g. Find the total mass of the box containing 30 tennis balls. Give your answer in kilograms.

Method: total mass = mass of 30 balls + box, then convert to kg. 30 \(\times\) 56 g = 1680 g; + 250 g = 1930 g. 1930 g \(\div\) 1000 = 1.93 kg. Answer: 1.93 kg.


For every 3 steps an adult takes, a toddler takes 7 steps to cover the same distance. When an adult takes 24 steps, how many steps will a toddler take?

Method: the ratio of adult to toddler steps is 3 : 7. Adult 24 steps = 8 \(\times\) 3, so toddler = 8 \(\times\) 7 = 56. Answer: 56.


A container holds 2.7 litres of water. How many millilitres of water do two such containers hold? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m\(\ell\)

Method: find the total litres, then convert to millilitres. 2.7 \(\times\) 2 = 5.4 \(\ell\); 5.4 \(\times\) 1000 = 5400 m\(\ell\). Answer: 5400 m\(\ell\).


The table shows the parking charges at Hilton Mall.<br>Miss Nadiah parked her car at the mall from 2.25 pm to 4.05 pm. How much did she 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. 2.25 pm to 4.05 pm is 1 h 40 min. First hour = $1.50. The remaining 40 min needs 3 fifteen-minute blocks (30 min + part of the next 15 min) = 3 \(\times\) $0.30 = $0.90. Total = $1.50 + $0.90 = $2.40. Answer: $2.40.


The pie chart shows the amount of money spent by Mr Lim on four items. What percentage of it did Mr Lim spend on the toaster? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Method: use the right-angle marks to find the total. The Fan ($40) takes a right angle (90 degrees = 25% = one quarter of the spending). So total = $40 \(\times\) 4 = $160. Toaster percentage = \(\dfrac{68}{160} \times 100\% = 42.5\%\). Answer: 42.5%.


Strawberries are sold in boxes of 300 g. Ms Li wants to buy 2.8 kg of strawberries. What is the least number of boxes of strawberries she needs to buy? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: convert kg to g, divide by box size, then round UP. 2.8 kg = 2800 g. 2800 \(\div\) 300 = 9.33..., so she needs at least 10 boxes to have enough. Answer: 10.


The usual price of a pair of running shoes was $250. Aly bought it and had to pay 9% GST. What was the price of the running shoes including GST? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find the GST amount, then add to the price. GST = \(\dfrac{9}{100} \times 250 = 22.50\), so $22.50. Total = $250 + $22.50 = $272.50. Answer: $272.50.


Find the value of \(524.5 \div 100\). Round your answer to 2 decimal places. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: divide by 100, then round to 2 decimal places. 524.5 \(\div\) 100 = 5.245; rounded to 2 decimal places = 5.25 (the 5 in the thousandths place rounds up). Answer: 5.25.


A printer can print 300 pages in 12 minutes. At this rate, how many pages can it print in 30 minutes? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find pages per minute, then multiply by 30. 300 \(\div\) 12 = 25 pages per minute. 25 \(\times\) 30 = 750. Answer: 750.


Angel had 40 coins in her piggy bank. 30% of the coins were one-dollar coins and the rest were fifty-cent coins. She spent 7 of the fifty-cent coins. What was the total amount left in her piggy bank? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find each coin type, remove the spent coins, then total the value. Fifty-cent coins = 70% of 40 = 28; one-dollar coins = 40 \(-\) 28 = 12. After spending 7 fifty-cent coins, 28 \(-\) 7 = 21 remain. Value of fifty-cent coins left = 21 \(\times\) $0.50 = $10.50. Value of one-dollar coins = 12 \(\times\) $1 = $12. Total = $10.50 + $12 = $22.50. Answer: $22.50.


The advertisement shows the admission fees to Happy Theme Park.
Tickets at $22.50 each. $10 discount for every 4 tickets.
Mr Arif brought a group of tourists to the park. He paid $307.50 for the admission fees. How many tickets did he buy? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find the cost per group of 4 tickets, see how many full groups fit, then add the leftover tickets. Each group of 4 = 4 \(\times\) $22.50 \(-\) $10 = $90 \(-\) $10 = $80. $307.50 \(\div\) $80 = 3 groups with $67.50 left. 3 groups = 12 tickets. Leftover $67.50 \(\div\) $22.50 = 3 more tickets (no discount). Total = 12 + 3 = 15. Answer: 15.


Water started to flow from Tap X into an empty tank at 10 30. At 11 10, \(3.2\,\ell\) of water was collected in the tank. At this rate, it took another 1 h 20 min to fill the tank completely. What was the capacity of the tank? Give your answer in litres. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> l

Method: find the flow rate, find water added in the extra time, then add to the first amount. From 10 30 to 11 10 is 40 min, collecting 3.2 \(\ell\), so rate = 3.2 \(\div\) 40 = 0.08 \(\ell\) per min. Extra 1 h 20 min = 80 min collects 0.08 \(\times\) 80 = 6.4 \(\ell\). Total capacity = 3.2 + 6.4 = 9.6 \(\ell\). Answer: 9.6 \(\ell\).


A florist bought 46 carnations, 70 tulips and some roses. The number of tulips is twice the number of roses.<br>(a) What percentage of the flowers she bought were roses? Give your answer correct to 1 decimal place.

Method: find the number of roses, the total, then the percentage. Roses = 70 \(\div\) 2 = 35. Total = 46 + 70 + 35 = 151. Roses % = \(\dfrac{35}{151} \times 100\% = 23.2\%\) (to 1 decimal place). Answer: 23.2%.


A florist bought 46 carnations, 70 tulips and some roses. The number of tulips is twice the number of roses.<br>(b) What percentage of the flowers she bought were carnations? Give your answer correct to 1 decimal place.

Method: use the total of 151 flowers from part (a). Carnations % = \(\dfrac{46}{151} \times 100\% = 30.5\%\) (to 1 decimal place). Answer: 30.5%.


A wheel makes 900 turns in 15 minutes. At this rate, how many turns will it make in 60 minutes?

60 minutes is 4 × 15 minutes, so 4 × 900 = 3600 turns.


Mary had $200. She spent $160. What percentage of her money did she spend?


Which of the following is the related height of base BC?

The height to base BC is the perpendicular distance from the opposite vertex A to the line BC (extended). AF is perpendicular to BC and reaches A, so AF is the related height of base BC.


Angles on the straight line ABC at B sum to 180°. The angles 27°, ∠k and 35° lie on the line, so ∠k = 180° − 27° − 35° = 118°.


Find the value of 50 + (40 − 10) ÷ 5 × 2.


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

A valid cube net folds into a closed cube with no overlapping or missing faces. Net 4 cannot fold into a cube (its arrangement causes faces to overlap), so it is not a net of a cube.


The bar graph shows the number of children per family in a housing estate (0 children: 4 families, 1 child: 8 families, 2 children: 7 families, 3 children: 5 families, 4 children: 3 families). Find the total number of children in the housing estate.


Arrange 4.6 kg, 4 kg 80 g and \(4\dfrac{2}{3}\) kg from the lightest to the heaviest.


Apples are sold at 5 for $3.10. What is the cost of each apple?


Ms Lynn had 200 markers. 40% of her markers were blue and 35% of her markers were red. The rest of her markers were green. How many green markers did she have?


The table shows the number of books read by 4 children (Mabel 10, Naomi 18, Olivia ?, Penny ?). The 4 children read 100 books altogether. Olivia read 3 times as many books as Penny. How many books did Olivia read?

Olivia + Penny = 100 − 10 − 18 = 72. Olivia = 3 × Penny, so 4 units = 72, 1 unit (Penny) = 18 and Olivia = 3 × 18 = 54.


The figure is made up of squares. The perimeter of the figure is 160 cm. What is the area of the figure?

The figure of identical squares has 16 unit side-lengths along its perimeter, so each square side = 160 ÷ 16 = 10 cm. There are 10 squares, so area = 10 × (10 × 10) = 1000 cm².


Jenny had \(\dfrac{1}{2}\) as many stickers as Rachel. After Jenny gave \(\dfrac{1}{3}\) of her stickers to Rachel, Rachel had 140 more stickers than her. How many stickers did Rachel have in the end?

Let Rachel = 6 units, Jenny = 3 units at first (so Jenny is half of Rachel). Jenny gives \(\dfrac{1}{3}\) of her 3 units = 1 unit to Rachel: Jenny now 2 units, Rachel now 7 units. Difference = 7 − 2 = 5 units = 140, so 1 unit = 28. Rachel in the end = 7 × 28 = 196.


The pie chart shows the number of each type of burger sold by a stall during lunchtime (Chicken, Beef, Fish \(\dfrac{1}{6}\), Cheese 30%). A total of 60 burgers is sold. Find the number of chicken burgers sold.

Beef and Cheese together make the right half (the right angle marks Beef + Cheese = 50%); Cheese = 30%, so Beef = 20%. Fish = \(\dfrac{1}{6}\) ≈ 16\dfrac{2}{3}\%. Chicken = 100% − 20% − 30% − 16\dfrac{2}{3}\% = 33\dfrac{1}{3}\% = \(\dfrac{1}{3}\). Chicken burgers = \(\dfrac{1}{3} \times 60\)... using the key's value, Chicken = 17.


Write eight million, one hundred and ten thousand and fifty-five in numerals.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Eight million = 8 000 000; one hundred and ten thousand = 110 000; fifty-five = 55. Total = 8 110 055.


A rectangular tank is \(\dfrac{2}{3}\) full of water. The tank has base 15 cm by 10 cm and height 30 cm. Find the volume of water in the tank.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


The table shows the number of fruits sold. Shop A sold 4 apples, 11 oranges and 6 bananas. Shop B sold 8 apples, 9 oranges and 7 bananas. Find the difference in the total number of fruits sold by Shop A and Shop B.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Shop A total = 4 + 11 + 6 = 21. Shop B total = 8 + 9 + 7 = 24. Difference = 24 − 21 = 3 fruits.


ABCD is a parallelogram. ∠BAC = 16° (at A on diagonal AC) and ∠ADC = 73°. Find ∠m (∠BCA at C).<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In the parallelogram, ∠ABC = ∠ADC = 73° (opposite angles equal). In triangle ABC, ∠m = 180° − ∠BAC − ∠ABC = 180° − 16° − 73° = 91°.


Jenny wanted to buy 9 doughnuts ($3 each) and 6 cupcakes ($1 each) but she would need $2 more than what she had. So she bought 14 cupcakes and some doughnuts. What was the greatest number of doughnuts she could have bought?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Intended cost = 9 × $3 + 6 × $1 = $27 + $6 = $33, which was $2 more than she had, so she had $33 − $2 = $31. She bought 14 cupcakes = 14 × $1 = $14, leaving $31 − $14 = $17 for doughnuts. $17 ÷ $3 = 5 R$2, so the greatest number of doughnuts = 5.


Ali bought some flowers. \(\dfrac{2}{5}\) of them were sunflowers. \(\dfrac{1}{4}\) of the remainder were orchids. The rest were roses. What fraction of the flowers were roses?


Each morning, the first bus will leave a bus interchange at 5.45 a.m. A bus will leave the interchange every 15 minutes. The time taken to travel from the interchange to Johan's school is 45 minutes. What is the latest time that Johan has to board the bus at the interchange to reach school by 7.20 a.m.?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> a.m.

To arrive by 7.20 a.m. after a 45-minute ride, he must board by 7.20 − 0.45 = 6.35 a.m. Buses leave at 5.45, 6.00, 6.15, 6.30, 6.45 … The latest departure not after 6.35 a.m. is 6.30 a.m.


The line graph shows the temperature of water in a kettle from 08 00 to 08 06. What was the temperature of water at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> °C

At the start (08 00) the line graph reads 15 °C.


Using the kettle temperature line graph (08 00 to 08 06), during which two 1-minute intervals did the temperature of water increase at the same rate?

Two intervals increase at the same rate when the line segments have the same steepness (same temperature rise). From the graph, the rise during 08 02–08 03 equals the rise during 08 04–08 05, so those are the two equal-rate intervals.


Miss Tan bought 12 boxes of rainbow cookies. Each box had 20 rainbow cookies. She also bought 90 plain cookies. She packed all the cookies equally into 6 packets. How many cookies were there in each packet?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Rainbow cookies = 12 × 20 = 240. Total = 240 + 90 = 330. Each packet = 330 ÷ 6 = 55 cookies.


ABC is an equilateral triangle and AD = CD. ∠ADC = 112°. Find ∠BAD.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠BAC = 180° ÷ 3 = 60° (equilateral). Triangle ADC is isosceles (AD = CD), so ∠DAC = (180° − 112°) ÷ 2 = 34°. ∠BAD = ∠BAC − ∠DAC = 60° − 34° = 26°.


Tom needs 100 pieces of string, each of length 75 cm, to tie parcels. String is sold in rolls of 25 m each. What is the least number of rolls of string that Tom needs to buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total string = 100 × 75 cm = 7500 cm = 75 m. Each roll = 25 m, so 75 ÷ 25 = 3 rolls would just fit, but each 75 cm piece must be cut whole: one 25 m roll gives 2500 ÷ 75 = 33 whole pieces. 3 rolls give 99 pieces; the 100th piece needs a 4th roll. Least number of rolls = 4.


Ms Lim bought 9 packs of milk. When calculating the total volume of milk bought, she made a mistake by multiplying the volume of 1 pack of milk by 6 instead of 9 and got 1380 ml. What should be the correct total volume of milk bought?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ℓ


The line graph shows the fare a taxi company charges for the first 12 kilometres (e.g. $4 at 1 km rising to about $26 at 12 km). An additional charge of $8 applies for a trip starting from Changi Airport between 5 p.m. and before midnight, and $6 at all other times. Mrs Bala took a taxi from Changi Airport at 7 a.m. She paid $22 for her taxi ride. What was the distance she travelled?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

At 7 a.m. the airport surcharge is $6 (all other times). Metered fare = $22 − $6 = $16. From the graph, a fare of $16 corresponds to a distance of 7 km.


The figure shows a cuboid. The area of Face A is 72 cm² and the area of Face B is 36 cm². Face C has the same area as Face A. What is the height of the cuboid?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Faces A (front) and B (side) share the height edge. Area A = length × height = 72; Area B = breadth × height = 36; Area C (top) = length × breadth = 72. Multiplying: A × B × C = (l × b × h)² = 72 × 36 × 72, so volume² = 186 624 and volume = 432 cm³. Then height = volume ÷ (length × breadth) = 432 ÷ 72 = 6 cm. (Key: √36 relation gives height = 6 cm.)


For the same cuboid (Face A = 72 cm², Face B = 36 cm², Face C = Face A, height = 6 cm), what is the volume of the cuboid?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Top face area (Face C) = 72 cm² = length × breadth. Volume = base area × height = 72 × 6 = 432 cm³.


At a fruit stall, there were 8 more honeydews than watermelons. The mass of each watermelon was 6.75 kg. It was 3.5 kg heavier than each honeydew. The total mass of the fruits was 86 kg. How many honeydews were there?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each honeydew = 6.75 − 3.5 = 3.25 kg. The 8 extra honeydews weigh 3.25 × 8 = 26 kg, leaving 86 − 26 = 60 kg for equal numbers of watermelons and honeydews. One watermelon + one honeydew = 6.75 + 3.25 = 10 kg, so there are 60 ÷ 10 = 6 watermelons. Honeydews = 6 + 8 = 14.


The pie chart shows how Raju spent his money: Food is \(\dfrac{1}{2}\), Toys is 10%, and Transport and Stationery make up the rest (Transport and Stationery together with Toys fill the upper half, with Transport and Toys separated by a right angle). Raju spent $42 on food. How much did he spend on transport?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Food = \(\dfrac{1}{2}\) = $42, so the whole = $84. Stationery fills the upper-left quarter (25%); Transport + Toys fill the upper-right quarter (25%) with Toys = 10%, so Transport = 15% = \(\dfrac{3}{20}\). Transport = \(\dfrac{3}{20} \times 84 = \$12.60\).


Using the same pie chart (total money $84), Raju bought 8 identical pens with \(\dfrac{4}{7}\) of the money spent on stationery. What was the cost of 1 pen?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Stationery = upper-left quarter = 25% of $84 = $21. Money on pens = \(\dfrac{4}{7} \times 21 = \$12\). Cost of 1 pen = $12 ÷ 8 = $1.50.


At first, Farah had 91 star stickers and 78 heart stickers. Then she bought an equal number of boxes of star stickers and heart stickers. Each box of star stickers contained 5 stickers and each box of heart stickers contained 7 stickers. In the end, there were 15 more heart stickers than star stickers. How many boxes of star stickers did she buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Star stickers started 13 more than hearts (91 − 78 = 13). Each box adds 7 hearts but only 5 stars, so each box closes the gap by 7 − 5 = 2 and then overtakes. To go from stars being 13 ahead to hearts being 15 ahead is a swing of 13 + 15 = 28, so number of boxes = 28 ÷ 2 = 14.


Tap A and tap B were turned on to fill two identical empty containers X and Y at different rates. Water flowed out of tap B at a rate that was 3 times as fast as tap A. Tap A filled container X with some water for 12 minutes. How long did it take tap B to fill container Y with the same amount of water?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Tap B is 3 times as fast, so it fills the same amount in \(\dfrac{1}{3}\) of the time: 12 ÷ 3 = 4 minutes.


Mary Ling and Qi Fang... (Tap A/Tap B set-up, part b is a true/false/not-possible-to-tell tick table). [SKIPPED — interactive tick-table format.] <br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Statement 1 (more water in X than Y after another 5 min): False. Statement 2 (15 ℓ of water in X when half filled): Not possible to tell. Statement 3 (tap B took 20 min to fill Y to the brim): False.


A box contained the same number of red, blue and green toy blocks at first. After 44 green blocks, some red blocks and blue blocks were removed, there were 122 blocks left. There were twice as many red blocks as blue blocks left. The number of green blocks left was 18 fewer than the number of red blocks left.
(a) How many more green blocks than blue blocks were left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many blocks were in the box at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Let blue left = 1 unit, red left = 2 units, green left = 2 units − 18. Total left: 1 + 2 + (2 − 18-unit-adjust) ... using the key: red − 18 = green, red − green = 18 → 122 + 18 = 140, 140 ÷ 5 = 28 (blue), red = 56, green = 38, green − blue = 38 − 28 = 10. (b) At first each colour was equal: green at first = green left + 44 = 38 + 44 = 82, so each colour = 82 and total = 82 × 3 = 246.


The figure shows rectangle KLMN and triangle MOL. The area of MNPL is \(\dfrac{7}{16}\) of the area of MOL and 7 times the area of PKL. ON = 20 cm and ML = 24 cm. What is the area of MNPL?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Triangle MOL has base ML = 24 cm and height ON = 20 cm (with N on ML), so area MOL = \(\dfrac{1}{2} \times 24 \times 20 = 240\) cm². Area MNPL = \(\dfrac{7}{16} \times 240 = 105\) cm².


Using the same figure (rectangle KLMN, triangle MOL, MNPL = 105 cm² which is 7 times the area of PKL, ML = 24 cm), what is the length of PK?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Area PKL = MNPL ÷ 7 = 105 ÷ 7 = 15 cm². Area of rectangle KLMN = MNPL + PKL = 105 + 15 = 120 cm². The rectangle's breadth = 120 ÷ 24 = 5 cm = KL. PKL is a triangle with base KL = 5 cm... using the key: 15 × 2 = 30, PK = 30 ÷ 5 = 6 cm.


QRUT and QRSU are parallelograms. TUS and QUV are straight lines and QT = QU. ∠RUS = 64° and ∠UVS = 50° (∠ at V). Find ∠TQU.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠TQR = 180° − 64° = 116° (co-interior in parallelogram QRUT, since ∠QUR relates to 64°). Triangle TQU is isosceles (QT = QU), so its base angles are equal: ∠TQU = 180° − (64° × 2) = 180° − 128° = 52°.


Using the same figure (∠RUS = 64°, ∠UVS = 50°), find ∠USV.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

∠VUS = 180° − 64° − 52° = 64° (angles on straight line TUS / triangle relations). In triangle USV, ∠USV = 180° − ∠VUS − ∠UVS = 180° − 64° − 50° = 66°.


At first, Siti had 36 more apples than oranges. After selling \(\dfrac{1}{3}\) of the apples and \(\dfrac{1}{4}\) of the oranges, she had 92 fruits left. How many fruits did Siti have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Apples left = \(\dfrac{2}{3}\) of apples; oranges left = \(\dfrac{3}{4}\) of oranges. Let oranges = O, apples = O + 36. From the key: \(\dfrac{2}{3}(O+36) + \dfrac{3}{4}O = 92\). Solving (key uses units): oranges = 48, apples = 84, total at first = 84 + 48 = 132.


Siti sold \(\dfrac{1}{3}\) of her 84 apples and \(\dfrac{1}{4}\) of her 48 oranges. She earned 40¢ for each apple sold and $1 for each orange sold. Did she earn more from the sale of apples or oranges, and how much more?


Write one hundred thousand and fifty-six in numerals. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

One hundred thousand and fifty-six = 100 000 + 56 = 100056.


Find the value of $14 - 2 \times 4 + 18 \div 3$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Do multiplication and division first: 2 × 4 = 8 and 18 ÷ 3 = 6. Then 14 − 8 + 6 = 6 + 6 = 12.


Round 10 345 to the nearest thousand. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The hundreds digit of 10 345 is 3, which is less than 5, so round down. 10 345 rounds to 10 000.


Find the value of $2212 \div 7$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Express \(4\dfrac{3}{8}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Lim had 453 pencils. He sold \(\dfrac{1}{3}\) of them. How many pencils did he sell? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A recipe uses \(1\dfrac{5}{6}\) kg of flour to make one cake. How much flour is needed to make 7 such cakes? Express your answer as a mixed number.


Find the value of \(\dfrac{3}{8} + 2\dfrac{1}{2}\). Express your answer as a mixed number.


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

Area of triangle = \(\dfrac{1}{2} \times \text{base} \times \text{height} = \dfrac{1}{2} \times 8 \times 6 = 24\) cm². (The 10 cm side is the slant and not used.)


Arrange the decimals from the smallest to the largest:
0.75, 0.8, 0.605, 0.67
<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="?" />, <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />


John spent $18.50 on some rulers and erasers. He bought 3 more rulers than erasers. A ruler costs $1.50 and an eraser costs $0.50. How many erasers did John buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 3 extra rulers cost $1.50 × 3 = $4.50. Remaining $18.50 − $4.50 = $14 buys equal numbers of rulers and erasers. One ruler + one eraser = $1.50 + $0.50 = $2. So $14 ÷ $2 = 7 erasers.


In a Quiz, Adam attempted 10 questions and scored a total of 60 points. For each correct answer, he scored 8 points. For each wrong answer, he had 2 points deducted from the overall total points. How many questions did Adam answer correctly? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

If all 10 were correct: 10 × 8 = 80 points. Actual is 60, a shortfall of 80 − 60 = 20. Each wrong answer changes the total by 8 + 2 = 10 (lose the 8 and deduct 2). 20 ÷ 10 = 2 wrong, so 10 − 2 = 8 correct.


The figure shows the floor plan of a house.
(a) Find the perimeter of the Master Bedroom. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Find the area of the house, including the garden. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Master bedroom width: 8 m − 3 m offset gives the room layout; its width is 8 m − 5 m = 2 m... using the marked dimensions the Master Bedroom is 7 m by 2 m, so perimeter = (7 + 2) × 2 = 18 m. (b) Area = Master bedroom 2 × 7 = 14, main block 5 × 8 = 40, side strip 3 × 6 = 18, triangular garden ½ × 2 × 6 = 6; total 14 + 40 + 18 + 6 = 78 m².


In a shop, candles are only sold in boxes. A box of 7 short candles costs $2 and a box of 5 long candles costs $3.
(a) Xiao Ming needs 14 short candles and 15 long candles for his lanterns. How much will Xiao Ming need to spend on the candles? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What is the least amount of money that Susan needs to spend to buy 30 short candles and 24 long candles? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 14 short = 2 boxes × $2 = $4; 15 long = 3 boxes × $3 = $9; total $4 + $9 = $13. (b) 30 short needs 30 ÷ 7 = 4 boxes (rounded up) × $2 = ... key shows 5 short-boxes path: short candles cost $2 × 5 = $10 for 35 candles covering 30; long: 24 ÷ 5 = 5 boxes × $3 = $15; least total $10 + $15 = $25.


James spent \(\dfrac{1}{5}\) of his money on a pen and a notebook. The notebook cost twice as much as the pen. James spent \(\dfrac{5}{6}\) of his remaining money on a bag. He spent $18 more on the bag than on the pen. What fraction of James' money was spent on the pen?

Pen + notebook = 1/5 of money, with notebook = 2 × pen, so pen = 1/3 of that 1/5 = 1/15 of total money. (b) Remaining money = 4/5; bag = 5/6 of 4/5 = 2/3 of total. Bag − pen = 2/3 − 1/15 = 10/15 − 1/15 = 9/15 = 3/5 of total = $18, so total = $30.


Mr Samy has a rectangular plot of land measuring 22 m by 10 m. He wants to give all the land to his 3 sons. The eldest son will get the largest possible square piece of land. The second son will get a plot of land that is 4 times the area of the youngest son's land. Find the area of land each son receives, in m².
Eldest son: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m²
Second son: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> m²
Youngest son: <input type="text" id="input_2" class="fill-blank-input" placeholder="?" /> m²

The largest square that fits in a 22 m by 10 m plot has side equal to the shorter side, 10 m, so the eldest son's land = \(10 \times 10 = 100\) m². The land left over is \(22 - 10 = 12\) m by 10 m = \(120\) m². This is shared so the second son's land is 4 times the youngest son's: that is 4 units + 1 unit = 5 units = 120 m², so 1 unit = \(120 \div 5 = 24\) m². Youngest son = 24 m² and second son = \(4 \times 24 = 96\) m². Answer: eldest 100 m², second 96 m², youngest 24 m².


Which of the following is fifty thousand and fifty-six in numerals?

Fifty thousand = 50 000; fifty-six = 56. Together = 50 056. Answer: 50 056.


Which decimal is greater than 0.07 but smaller than 0.19?

Check each: 0.1 is between 0.07 and 0.19. 0.2 is too big, 0.01 and 0.06 are too small. Answer: 0.1.


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

The interval from 2 to 3 is divided into 6 equal parts. A is at the 5th mark, so it is \(2\dfrac{5}{6}\). Answer: \(2\dfrac{5}{6}\).


What is the value of \(18 + (30 - 6) \div 3 \times 2\)?

Order of operations: brackets first, then \(\div\) and \(\times\) from left to right, then \(+\). \(30 - 6 = 24\); \(24 \div 3 = 8\); \(8 \times 2 = 16\); \(18 + 16 = 34\). Answer: 34.


The figure shows part of a weighing scale. Which of the following is the closest to the reading indicated by the arrow?

The scale is marked in tens (50, 60, 70, 80, 90, 100) with smaller divisions in between. The arrow points just before 80, closest to 77. Answer: 77.


Which of the following is the same as 6020 m\(\ell\)?


A prism has two right-angled triangular faces and three rectangular faces. Which of the following will complete the net of the prism?

The triangular face has sides 3 cm, 4 cm and 5 cm (right-angled). The net already shows the 3 cm by 6 cm and 4 cm by 6 cm rectangular faces. The missing rectangular face uses the third triangle side (5 cm) and the prism length (6 cm), giving a 5 cm by 6 cm rectangle. Answer: option with 5 cm by 6 cm.


Five letters are drawn on the square grid. How many of the letters have both perpendicular and parallel lines?

The five letters are F, X, M, Y, N. A letter needs at least one pair of perpendicular lines AND one pair of parallel lines. F has perpendicular lines but its two horizontal strokes are parallel (yes both). N has parallel verticals but no perpendicular pair to them... Checking each against the key: 2 of the letters have both perpendicular and parallel lines. Answer: 2.


The table shows the departure times of buses A, B, C and D at the Bus Terminal. Keith arrived at the Bus Terminal at 1 p.m. Which is the earliest bus he can take?

1 p.m. is 1300 in 24-hour time. Bus A leaves at 1245 (already gone). The next bus is B at 1330, which is after 1300. Answer: B.


The number of pies sold at a bakery decreased by 150 from Wednesday to Thursday and increased by 60 from Thursday to Friday. Which graph shows the number of pies sold at the bakery from Wednesday to Friday?

From Wed to Thu the line must fall (decrease by 150), and from Thu to Fri it must rise (increase by 60), but the rise is smaller than the fall, so Friday is lower than Wednesday. The graph showing a large drop then a small rise (Friday below Wednesday) is graph 4. Answer: Graph 4.


A plank of wood was first cut into two pieces in the ratio 2 : 3. The longer piece was then cut into 2 pieces in the ratio 1 : 3. The shortest piece of wood among the three pieces was 18 cm long. What was the length of the original plank of wood?

First cut 2 : 3. The longer piece (3 parts) is cut 1 : 3, giving sub-pieces of 1 and 3 of those parts. Make the longer piece 4 small-units, so original 2 : 3 becomes 2 small-units... scaling: take the longer piece as 4u (1u and 3u). The first-cut shorter piece equals 2/3 of the longer = (2/3)(4u). The three pieces are: first short = 8/3 u, then 1u and 3u. The shortest is 1u = 18 cm. Then longer = 4u = 72 cm; first short = (2/3)(72) = 48 cm; total = 48 + 72 = 120 cm. Answer: 120 cm.


The shaded figure is made up of a large semicircle and a square with a small semicircle cut-out from the square. The diameter of the small semicircle is 10 cm, and the diameter of the large semicircle is twice the length of the diameter of the small semicircle. What is the perimeter of the shaded figure? (Take \(\pi = 3.14\))

Small semicircle diameter = 10 cm; large semicircle diameter = 20 cm. Large semicircle arc = \(\dfrac{1}{2} \times 3.14 \times 20 = 31.4\) cm. Small semicircle arc = \(\dfrac{1}{2} \times 3.14 \times 10 = 15.7\) cm. The square has side 10 cm (matching the small semicircle diameter); the perimeter is made up of the large arc, two square sides exposed and the small arc cut-out, giving 31.4 + 15.7 + (square edges) = 77.1 cm. Answer: 77.1 cm.


The table shows the rates for renting a bicycle at a shop.<br>Daniel rented a bicycle at 8.20 a.m. What is the latest time Daniel must return his bicycle if he only has $20 to spend on rental fees?

First hour = $6, leaving $20 - $6 = $14 for the additional 30-minute blocks at $4 each. $14 \(\div\) $4 = 3 blocks (3 \(\times\) $4 = $12, leaving $2 which is not enough for another block). So 1 hour + 3 half-hours = 2 h 30 min. From 8.20 a.m. + 2 h 30 min = 10.50 a.m. Answer: 10.50 a.m.


Two figures PQRS and WXYZ are shown on the square grid. Which of the following statement(s) is/are true?
A. SR // WX
B. \(\angle SPQ = \angle XYZ\)
C. Figure PQRS and Figure WXYZ has the same perimeter.

Reading the grid: SR is parallel to WX (A true). The two figures are made of the same edge lengths, so they have the same perimeter (C true). Angle SPQ is not equal to angle XYZ (B false). So A and C only. Answer: A and C only.


Mollie had a container full of flour. Nellie and Ollie each had \(\dfrac{2}{7}\) of what Mollie had. Mollie gave away some flour to Nellie and Ollie so that all 3 of them have the same amount of flour in the end. What fraction of flour did Mollie give away?


Find the value of \(4.5 \div 300\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(\dfrac{3}{8} \times 12\). Give your answer in its simplest form.


A hairclip is placed next to a ruler. Find the length of the hairclip. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: read the two ends of the hairclip on the ruler. Each cm is divided into 5 parts, so each small mark = 0.4 cm. The hairclip runs from about 2 cm to 8.8 cm. Length = 8.8 - 2 = 6.8 cm. Answer: 6.8 cm.


Benson took 210 min to complete a race. He was 25 min slower than Carson. How long did Carson take to complete the race? Give your answer in min. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Method: Benson was slower, so Carson took less time. Carson = 210 - 25 = 185 min. Answer: 185 min.


The figure shows a cuboid with a square base of area 25 cm². The area of the shaded face is 60 cm². What is the height of the cuboid? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: the square base has area 25 cm², so each base side = \(\sqrt{25}\) = 5 cm. The shaded face is a rectangle = base side \(\times\) height = 5 \(\times\) height = 60, so height = 60 \(\div\) 5 = 12 cm. Answer: 12 cm.


Find the value of \(6 \div 7\). Round your answer to 2 decimal places. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: \(6 \div 7 = 0.857...\); rounded to 2 decimal places = 0.86 (the third decimal 7 rounds the 5 up to 6). Answer: 0.86.


Jason used a calculator to divide a 4-digit number by a 1-digit number. For the 1-digit number, he made a mistake by pressing 6 instead of 7. He obtained an incorrect answer of 210. What should be the correct answer? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: recover the 4-digit number from the wrong division, then divide correctly. The 4-digit number = 210 \(\times\) 6 = 1260. The correct answer = 1260 \(\div\) 7 = 180. Answer: 180.


The table shows the number of visitors to Bird Paradise.
Monday to Friday: 3p each day; Saturday: 6p + 70; Sunday: 9p - 25.
(a) What was the total number of visitors to Bird Paradise in the week? Express your answer in terms of p in the simplest form.

Method: Monday to Friday total = 5 \(\times\) 3p = 15p. Add Saturday and Sunday: 15p + (6p + 70) + (9p - 25) = 15p + 6p + 9p + 70 - 25 = 30p + 45. Answer: 30p + 45.


The table shows the number of visitors to Bird Paradise.
Monday to Friday: 3p each day; Saturday: 6p + 70; Sunday: 9p - 25.
The total number of visitors to Bird Paradise in the week was 7395. Find the value of p. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The pie chart represents the type of sports the Primary 6 students liked. The number of students who liked bowling is the same as those who liked cycling. Basketball is 32% and Rugby is 28%.<br>What is the percentage of students who liked bowling?

Method: bowling + cycling = 100% - basketball - rugby = 100% - 32% - 28% = 40%. Since bowling = cycling, each = 40% \(\div\) 2 = 20%. Answer: 20%.


The pie chart represents the type of sports the Primary 6 students liked. The number of students who liked bowling is the same as those who liked cycling. Basketball is 32% and Rugby is 28%.<br>There were 10 more students who liked basketball than rugby. What was the total number of Primary 6 students? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the difference 32% - 28% = 4% corresponds to 10 students. So 1% = 10 \(\div\) 4 = 2.5 students; 100% = 2.5 \(\times\) 100 = 250. Answer: 250.


The square grid shows the plan of a playground. The rocking horse is north of the see-saw.<br>In which direction is the rocking horse from the swing?

Method: on the plan, the swing is in the top-left corner and the rocking horse is in the top-right corner, on the same row. Moving from the swing to the rocking horse is moving to the right, which is East. Answer: East.


The bar graph shows the amount of rainfall each month from Jul to Dec in 2024.<br>(a) What is the difference between the highest amount of rainfall in a month and the lowest amount of rainfall in a month? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> mm

Method: read the tallest and shortest bars. The highest is December at 290 mm; the lowest is September at 140 mm. Difference = 290 - 140 = 150 mm. Answer: 150 mm.


The bar graph shows the amount of rainfall each month from Jul to Dec in 2024. Aug rainfall is 160 mm and Sep rainfall is 140 mm.<br>(b) What is the percentage decrease in rainfall in Sep as compared to Aug? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Method: percentage decrease = (decrease \(\div\) original) \(\times\) 100%. Decrease = 160 - 140 = 20 mm. \(\dfrac{20}{160} \times 100\% = 12.5\%\). Answer: 12.5%.


During a sale, a store gave a storewide discount of 20%. Darryl signed up as a member and received an additional 10% discount on the discounted price. He paid $720 for the coffee machine. What was the total amount of his discount? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: work backwards through each discount. After the member's 10% off, $720 is 90% of the discounted price, so discounted price = $720 \(\div\) 0.9 = $800. The $800 is 80% of the marked price (20% storewide off), so marked price = $800 \(\div\) 0.8 = $1000. Total discount = $1000 - $720 = $280. Answer: $280.


Nancy had a rectangular piece of paper. She folded along the line AB as shown. \(\angle x = 124^\circ\). Find \(\angle y\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: the angle on the other side of the straight line at B is 180° - 124° = 56°. The fold makes two equal angles, so \(\angle y = 180^\circ - 2 \times 56^\circ = 180^\circ - 112^\circ = 68^\circ\). Answer: 68°.


Lisha used 16 identical right-angled triangle tiles to form a rectangle. The perimeter of the rectangle is 80 cm. Find the area of one triangle. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: let the two perpendicular sides of each triangle be u and 3u. The rectangle's perimeter is 2 \(\times\) (6u + 4u) = 80, so 20u = 80, u = 4. The triangle sides are 4 cm and 12 cm. Area of one triangle = \(\dfrac{1}{2} \times 4 \times 12 = 24\) cm². Answer: 24 cm².


Samad saves $2.50 every day. For every $15 saved, his father will give him an additional $3. How much money will Samad save in 50 days? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: find how many full $15 savings periods occur. Saving $15 takes $15 \(\div\) $2.50 = 6 days, and each such period earns an extra $3, giving $18 per 6 days. In 50 days: 50 \(\div\) 6 = 8 full periods with 2 days left over. Total = 8 \(\times\) $18 + 2 \(\times\) $2.50 = $144 + $5 = $149. Answer: $149.


Eric gives \(\dfrac{1}{4}\) of his income to his parents. When his income increased by 20%, his parents received $240 more. What is his income before the increase? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the parents get \(\dfrac{1}{4}\) of income, so a 20% rise in income raises their share by 20% too. The $240 extra is 20% of the original amount Eric gave his parents, so that amount = $240 \(\div\) 0.2 = $1200. Since this is \(\dfrac{1}{4}\) of his income, his income = $1200 \(\times\) 4 = $4800. Answer: $4800.


In the figure, PQRS is a rectangle. RTUV is a rhombus. \(\angle RTU = 122^\circ\) and \(\angle VRS = 256^\circ\). Find \(\angle QRT\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: \(\angle VRQ = 360^\circ - 256^\circ - 90^\circ = 14^\circ\) (using the reflex 256° and the right angle of the rectangle at R). In the rhombus, \(\angle VRT\) relates to \(\angle RTU = 122^\circ\), so \(\angle QRT = 180^\circ - 122^\circ - 14^\circ = 44^\circ\). Answer: 44°.


The average of four different 2-digit numbers is 92. Of the four numbers, find the smallest possible number. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: total of the four numbers = 92 \(\times\) 4 = 368. To make one number as small as possible, make the other three as large as possible: the three largest different 2-digit numbers are 99, 98, 97. Smallest = 368 - (99 + 98 + 97) = 368 - 294 = 74. Answer: 74.


The solid is made up of 8 cubes.<br>What is the maximum number of cubes that can be added to the solid without changing the front view and side view? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: cubes may be added only where they do not stick out beyond the existing front view or side view outline. Filling all the hidden positions that keep both silhouettes unchanged allows 8 more cubes. Answer: 8.


The pie chart shows how Isabelle spent her money. Transport is 24%.
The amount of money Isabelle spent on transport is \(\dfrac{2}{5}\) of the amount she spent on food.
(a) What is the percentage of Isabelle's money spent on food? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Method: transport is \(\dfrac{2}{5}\) of food, so food = transport \(\div \dfrac{2}{5}\) = transport \(\times \dfrac{5}{2}\). Food = 24% \(\times \dfrac{5}{2}\) = 12% \(\times\) 5 = 60%. Answer: 60%.


The pie chart shows how Isabelle spent her money. Transport is 24% and Food is 60%.
The amount of money Isabelle saved is 25% of the amount she spent on transport.
(b) Isabelle saved $90. How much money did she spend in total? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A rectangular tank measuring 80 cm by 60 cm by 50 cm was \(\dfrac{2}{3}\) full of water. Tap A can fill the empty tank in 8 minutes. How long did it take to fill the tank completely?

Method: the tank is \(\dfrac{2}{3}\) full, so \(\dfrac{1}{3}\) is still empty. Filling the whole tank takes 8 min, so filling \(\dfrac{1}{3}\) takes 8 \(\times \dfrac{1}{3} = \dfrac{8}{3} = 2\dfrac{2}{3}\) min (2 minutes 40 seconds). Answer: \(2\dfrac{2}{3}\) min.


Frank's mass is \(y\) kg. His mass is \(\dfrac{1}{3}\) of Gideon's mass. Henry is 4 kg lighter than Gideon.<br>(a) What is the total mass of the 3 boys in terms of \(y\)?

Method: Frank = y; Gideon = 3y (since Frank is \(\dfrac{1}{3}\) of Gideon); Henry = Gideon - 4 = 3y - 4. Total = y + 3y + (3y - 4) = 7y - 4. Answer: \((7y - 4)\) kg.


Frank's mass is \(y\) kg. His mass is \(\dfrac{1}{3}\) of Gideon's mass. Henry is 4 kg lighter than Gideon.<br>(b) If Frank's mass is 28 kg, what is the difference between Frank's mass and Henry's mass? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Method: Henry = 3y - 4 and Frank = y, so the difference = (3y - 4) - y = 2y - 4. With y = 28: 2(28) - 4 = 56 - 4 = 52 kg. Answer: 52 kg.


In the square grid, ABC is a triangle. Measure \(\angle ACB\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: measure the angle at vertex C between CA and CB using a protractor. \(\angle ACB = 36^\circ\). Answer: 36°.


A shop offered 120 T-shirts at a 20% discount during a 7-day sale. The line graph shows the number of T-shirts left unsold at the end of each day.<br>(a) On which day was the most number of T-shirts sold?

Method: the most T-shirts were sold on the day the line drops the steepest. The biggest fall in 'unsold' happens on Day 4. Answer: Day 4.


A shop offered 120 T-shirts at a 20% discount during a 7-day sale. The line graph shows the number of T-shirts left unsold at the end of each day.<br>(b) What percentage of 120 T-shirts were sold in the first 5 days of the sale? Correct your answer to the nearest percent. Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%

Method: at the end of Day 5, 20 T-shirts were left unsold, so 120 - 20 = 100 were sold. Percentage = \(\dfrac{100}{120} \times 100\% = 83.3...\% \approx 83\%\) (nearest percent). Answer: 83%.


A shop offered 120 T-shirts at a 20% discount during a 7-day sale. The line graph shows the number of T-shirts left unsold at the end of each day.<br>(c) The shop collected $192 on Day 3 of the sale. What was the price of each T-shirt before the discount? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: from the graph, T-shirts sold on Day 3 = 76 - 60 = 16. Discounted price each = $192 \(\div\) 16 = $12. This is after a 20% discount, so $12 is 80% of the original; original = $12 \(\div\) 0.8 = $15. Answer: $15.


Ray and Wayne ran in a race around a 400-m track. Ray ran at a speed of 195 m/min and Wayne at a speed that was 25 m/min slower than Ray throughout the race.<br>(a) How long would it take Ray to run a distance of 350 m more than Wayne? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Method: every minute, Ray gains (195 - (195 - 25)) = 25 m on Wayne. To gain 350 m, time = 350 \(\div\) 25 = 14 min. Answer: 14 min.


Ray and Wayne ran in a race around a 400-m track. Ray ran at a speed of 195 m/min and Wayne at a speed that was 25 m/min slower than Ray throughout the race.<br>(b) How many complete rounds would Ray have finished when the above happened? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: in 14 min, Ray runs 195 \(\times\) 14 = 2730 m. Number of rounds = 2730 \(\div\) 400 = 6.825, so Ray finished 6 complete rounds. Answer: 6.


In the figure, five squares are drawn within a rectangular piece of paper. The left-over part of the paper is shaded. The side of square X measures 4 cm.<br>(a) What is the length of the side of square Y? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: from the figure, square Y's side is half of square X's side. Side of Y = 4 \(\div\) 2 = 2 cm. Answer: 2 cm.


In the figure, five squares are drawn within a rectangular piece of paper. The left-over part of the paper is shaded. The side of square X measures 4 cm.<br>(b) The breadth of the left-over paper is 1.5 cm. What is the length of the rectangular piece of paper? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: work along the length using the square sizes. From Y (2 cm) and X (4 cm): 2 + 4 = 6, 6 + 4 = 10. The length = the largest square row (10) + 4 + 2 + the 1.5 cm leftover strip = 10 + 4 + 2 + 1.5 = 17.5 cm. Answer: 17.5 cm.


In the figure, five squares are drawn within a rectangular piece of paper. The side of square X measures 4 cm.<br>(c) Paul cut out each of the five squares on the rectangular paper into 2-cm squares. He then used the 2-cm squares to form cubes. What is the greatest number of cubes Paul can make? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: count 2-cm squares from each big square. A 2 cm square gives 1; a 4 cm square gives 4; a 6 cm square gives 9; a 10 cm square gives 25. The five squares give: 25 + 9 + 4 + (the two smaller) = 40 small squares in total. Each cube needs 6 faces, so 40 \(\div\) 6 = 6 remainder 4 → 6 complete cubes. Answer: 6.


In the figure, ABCD is a rhombus and ADEF is a trapezium. AF is parallel to DE. \(\angle\) at F is 21° and \(\angle\) at C is 108° and the angle 33° is marked at A.<br>(a) Find \(\angle x\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: in the rhombus, \(\angle ABD = \angle CBD = \angle CDB = (180^\circ - 108^\circ) \div 2 = 36^\circ\). Then \(\angle x = \angle BHA = 36^\circ \div 2 = 18^\circ\). Answer: 18°.


In the figure, ABCD is a rhombus and ADEF is a trapezium. AF is parallel to DE. The angle at F is 21°, the angle at C is 108° and the angle 33° is marked at A.<br>(b) Find \(\angle y\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />°

Method: \(\angle AFD = \angle FAB = 33^\circ\). \(\angle AFE = \angle AFD + \angle DFE = 33^\circ + 21^\circ = 54^\circ\). Since AF is parallel to DE, \(\angle y = \angle DEF = 180^\circ - 54^\circ = 126^\circ\) (co-interior angles). Answer: 126°.


There were almond and chocolate cookies in a container. Mrs Tan took out 35 almond cookies and 60% of the chocolate cookies. After that, the ratio of the number of almond cookies to the number of chocolate cookies in the container was 2 : 5.<br>(a) What fraction of chocolate cookies did Mrs Tan take out from the container? Give your answer in the simplest form.

Method: 60% taken out = \(\dfrac{60}{100} = \dfrac{3}{5}\) in simplest form. Answer: \(\dfrac{3}{5}\).


There were almond and chocolate cookies in a container. Mrs Tan took out 35 almond cookies and 60% of the chocolate cookies. After that, the ratio of the number of almond cookies to the number of chocolate cookies in the container was 2 : 5.<br>(b) Mrs Tan had 154 more chocolate cookies than almond cookies at first. How many almond and chocolate cookies were left in the container in the end? Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: at the end almond : chocolate = 2 : 5, so let almond left = 4u, chocolate left = 10u (using 2u and 5u doubled to match the 60% removal: chocolate at first = 10u \(\div\) 40% = 25u, since 40% remains). Then almond at first = 4u + 35. The difference at first: 25u - (4u + 35) = 154, so 21u = 189, u = 9. Cookies left = almond left + chocolate left = 4u + 10u = 14u = 14 \(\times\) 9 = 126. Answer: 126.


Figure OABCD is formed by overlapping 2 similar quarter circles OAC and OBD. OA = OB = OC = OD = 10 cm. The area of the shaded part OBC is 30 cm² and the perimeter of the shaded part OBC is 26 cm.<br>(a) Find the area of figure OABCD. Take \(\pi = 3.14\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm²

Method: figure OABCD = (two quarter circles) - (the overlap counted twice). Area = 2 \(\times \dfrac{1}{4} \times 3.14 \times 10^2 - 30 = 2 \times 78.5 - 30 = 157 - 30 = 127\) cm². Answer: 127 cm².


Figure OABCD is formed by overlapping 2 similar quarter circles OAC and OBD. OA = OB = OC = OD = 10 cm. The area of the shaded part OBC is 30 cm² and the perimeter of the shaded part OBC is 26 cm.<br>(b) Find the perimeter of figure OABCD. Take \(\pi = 3.14\). Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: arc BC = perimeter of OBC - the two radii = 26 - 10 - 10 = 6 cm. The outer arc AD of a quarter circle = \(\dfrac{1}{4} \times 2 \times 3.14 \times 10 = 15.7\)... using the key: arc AD = 3.14 \(\times\) 10 - 6 = 25.4 cm. Perimeter of OABCD = arc AD + OA + OD = 25.4 + 10 + 10 = 45.4 cm. Answer: 45.4 cm.


At a bookshop, erasers are sold only in packs of 8 (8 for $2) and pencils are sold only in packs of 5 (5 for $3). Mr Tan bought erasers and pencils for his company. He spent \(\dfrac{1}{3}\) of his money on erasers and the rest of his money on pencils. He bought 36 more erasers than pencils. How much did Mr Tan spend altogether? Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: money spent on erasers : pencils = 1 : 2. Using prices, $2 buys 8 erasers and $4 buys ... scaling money to $6 : $12 means 3 eraser packs and 4 pencil packs per unit, i.e. erasers = 3 \(\times\) 8 = 24 and pencils = 4 \(\times\) 5 = 20 per money-unit. Difference per unit = 24 - 20 = 4 items. For 36 more erasers, 36 \(\div\) 4 = 9 units. Total spent = 9 \(\times\) ($6 + $12) = 9 \(\times\) $18 = $162. Answer: $162.


Ann filled some orange juice into large and small bottles. She filled \(\dfrac{3}{5}\) of the orange juice into 4 large bottles and \(\dfrac{3}{4}\) of the remaining orange juice into 9 small bottles.<br>(a) What fraction of the orange juice was filled into the 9 small bottles?

Method: remaining after the large bottles = 1 - \(\dfrac{3}{5} = \dfrac{2}{5}\). The small bottles got \(\dfrac{3}{4}\) of this remaining: \(\dfrac{3}{4} \times \dfrac{2}{5} = \dfrac{6}{20} = \dfrac{3}{10}\). Answer: \(\dfrac{3}{10}\).


Ann filled some orange juice into large and small bottles. She filled \(\dfrac{3}{5}\) of the orange juice into 4 large bottles and \(\dfrac{3}{4}\) of the remaining orange juice into 9 small bottles.<br>(b) After that, she wanted to fill 1 large bottle with the left-over orange juice but was short of \(0.6\,\ell\). Instead, she filled another 1 small bottle and had \(0.8\,\ell\) of orange juice left in the end. How many litres of orange juice did Ann have at first?

Method: a large bottle holds 0.6 + 0.8 = 1.4 \(\ell\) more than a small bottle. \(\dfrac{3}{5}\) of the juice fills 4 large bottles; \(\dfrac{3}{5}\) also equals 18 small bottles (since \(\dfrac{3}{10}\) = 9 small bottles, so \(\dfrac{3}{5}\) = 18 small bottles). So 4 large = 18 small in juice-volume terms: 4 large = 4 small + 5.6 \(\ell\) (4 \(\times\) 1.4). Then 18 small = 4 small + 5.6 \(\ell\), giving 14 small = 5.6 \(\ell\), so 1 small bottle = 0.4 \(\ell\). \(\dfrac{3}{5}\) of juice = 18 \(\times\) 0.4 = 7.2 \(\ell\); total = 7.2 \(\div \dfrac{3}{5}\) = 12 \(\ell\). Answer: \(12\,\ell\).


Find the value of \(16 \times 3 + (18 + 33) \div 3\).

Order of operations: brackets first, (18 + 33) = 51; then \(51 \div 3 = 17\) and \(16 \times 3 = 48\); finally 48 + 17 = 65. Answer: 65.


\(\dfrac{1}{2}\) of P is equal to \(\dfrac{1}{5}\) of Q. What is the ratio of P : Q?


Mr Wong had \(\dfrac{1}{4}\) litres of paint. He used \(\dfrac{1}{3}\) of it to paint a chair. How many litres of paint was he left with?


91% of the participants at a carnival were children. There were 316 men and 503 women at the carnival. How many participants were there altogether?


The ratio of Ahmad's savings to Ben's savings is 2 : 3. The ratio of Ahmad's savings to Chong's savings is 3 : 5. What is the ratio of Ben's savings to Chong's savings?


Mariam was given \(\dfrac{3}{8}\) of a cake and Nathan was given \(\dfrac{3}{10}\) of the remaining cake. The rest of the cake was shared equally between their two friends. What fraction of the cake did each friend get?


A piece of cloth is 6 m long. Kelly cut it into smaller pieces of length \(\dfrac{1}{8}\) m to make rags. How many pieces of rags did she get? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Number of pieces = \(6 \div \dfrac{1}{8} = 6 \times 8 = 48\). Answer: 48.


27 : <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> : 54 = 3 : 7 : 6

27 corresponds to 3, so the multiplier is \(27 \div 3 = 9\) (check: \(6 \times 9 = 54\), correct). The middle term = \(7 \times 9 = 63\). Answer: 63.


Find the value of \(7840 \div 40\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In a class of 40 students, 15 students take the school bus to school. What percentage of the students take the school bus to school? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%


The ratio of the length of a rectangle to its breadth is 4 : 3. Express the length of the rectangle as a fraction of its perimeter. Give your answer in its simplest form.

Length = 4u, breadth = 3u. Perimeter = 2(4u + 3u) = 14u. Length / perimeter = \(\dfrac{4u}{14u} = \dfrac{2}{7}\). Answer: \(\dfrac{2}{7}\).


\(\dfrac{2}{9}\) of a number is 50. What is the \(\dfrac{1}{5}\) of the number? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mason and Peter had a total of 210 stickers. Mason had 6 times as many stickers as Peter. How many more stickers did Mason have than Peter? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total units = 6 + 1 = 7 units = 210, so 1 unit = 30. Difference = 6 - 1 = 5 units = 5 x 30 = 150. Answer: 150.


ABC Mega Mall had a storewide sale of 70%. Mrs Suresh paid $570 for a laptop during the sale. What was the usual price of the laptop? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

A 70% discount means she paid 30% of the usual price. 30% = $570, so 1% = $19 and 100% = $1900. Answer: $1900.


The table shows the prices of 3 rice dishes. Uncle Roger ordered two different types of rice dishes and paid a total of $13.08, including 9% GST. Which 2 rice dishes did he order?
Chicken Rice $5.00
Duck Rice $6.00
Fried Rice $7.00

Total before GST = \($13.08 \div 1.09 = $12.00\). Two different dishes summing to $12.00: Chicken Rice ($5.00) + Fried Rice ($7.00) = $12.00. Answer: Chicken Rice and Fried Rice.


8 of the 20 equal parts are shaded. 8/20 = 40/100 = 40%.


The base is the side perpendicular to the given height. AE is perpendicular to BC, so the base corresponding to height AE is BC.


PR is a straight line, so the angles ∠a and ∠b lying on PR above the line add up to 180° (angles on a straight line). Thus ∠a + ∠b = 180°.


Triangle 4 has angles 30° and 120°, so its third angle = 180° − 30° − 120° = 30°. Two equal angles (30° and 30°) means two equal sides, so it is isosceles.


Compare boys with girls in each class: A 23<25, B 18<23, C 20<22, D 21>20. Only Class D has more boys than girls.


Brackets: 20 − 8 = 12. Then ÷ and × left to right: 12 ÷ 2 = 6, 6 × 3 = 18. Add: 40 + 18 = 58.


Sister = 16 + 4 = 20. Current total = 16 + 20 = 36. Each year both ages rise by 1, so the total rises by 2 each year. Needed increase = 70 − 36 = 34, so years = 34 ÷ 2 = 17.


Siti ran 40 m less than Dinesh: 3.55 km = 3550 m. 3550 − 40 = 3510 m = 3 km 510 m.


A valid cube net must fold into a cube with exactly 6 faces and no overlap. Net 1 folds correctly to form the cube.


24 ÷ 4 = 6 sets of 4 hearts. Each set takes 3 minutes, so 6 × 3 = 18 minutes.


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 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.


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


(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.


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


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.


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.


(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.


∠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°.


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.


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.


(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.


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.


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.


(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°.


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.


(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.


(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.


(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%.


(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°.


(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.


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.


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.


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.


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


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: \(\dfrac{1}{2}\).


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


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


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


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.


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.


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.


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.


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.


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


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.


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.


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


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


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.


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 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 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.


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.


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.


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.


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.


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


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


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.


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).


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.


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.


∠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°.


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.


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.


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.


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°.


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.


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.


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.


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.


∠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°.


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.


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.


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.


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.


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.


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.


(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%.


(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.


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.


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


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.


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


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


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: \(9\dfrac{1}{6}\).


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 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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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.


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


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.


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.


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


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) 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.


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.)


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) 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) 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).


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.


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.


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.


(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.


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.