Fatimah spent \(\dfrac{1}{4}\) of her money on 8 cookies and 2 muffins. A muffin cost three times as much as a cookie. She bought some more cookies with \(\dfrac{2}{7}\) of her remaining money. How many cookies did Fatimah buy altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

8 cookies + 2 muffins, muffin = 3 cookies, so cost = 8 + 2×3 = 14 cookie-units = \(\dfrac{1}{4}\) of money. Remaining money = \(\dfrac{3}{4}\). Money spent on more cookies = \(\dfrac{2}{7}\times\dfrac{3}{4} = \dfrac{6}{28}\) of total. Since \(\dfrac{1}{4} = \dfrac{7}{28}\) buys 14 cookie-units, \(\dfrac{1}{28}\) buys 2 cookies, so \(\dfrac{6}{28}\) buys 12 cookies. Total cookies = 8 + 12 = 20.


Danny had 880 cards in his collection. 25% of the cards were football cards and the rest were basketball cards.
(a) How many more basketball cards than football cards did he have in his collection? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many more football cards must Danny buy if he wanted to increase the number of football cards in his collection to 45%? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Football = 25% of 880 = 220. Basketball = 880 - 220 = 660. Difference = 660 - 220 = 440. (b) Basketball stays 660 = 100% - 45% = 55% of the new total. 55% = 660, so 1% = 12 and 45% = 540 football cards needed. Must buy 540 - 220 = 320 more.


Which one of the following is likely to be the height of a flag pole?

A typical flag pole is several metres tall. 0.07 m and 0.70 m are far too short, and 70 m is unreasonably tall. 7 m is the sensible estimate. Answer: 7 m.


Find the value of \(8 \div \dfrac{2}{3}\).

Dividing by a fraction means multiplying by its reciprocal: \(8 \div \dfrac{2}{3} = 8 \times \dfrac{3}{2} = \dfrac{24}{2} = 12\). Answer: 12.


Which one of the following fractions is the greatest?


At a party, the number of boys is \(\dfrac{2}{7}\) the number of girls. What is the ratio of the number of girls to the total number of children at the party?

Let girls = 7 units, then boys = 2 units. Total = 7 + 2 = 9 units. Ratio of girls to total = 7 : 9. Answer: 7 : 9.


Cheryl baked a cake and gave \(\dfrac{1}{4}\) of it to her neighbour. She cut the remaining cake equally into 9 slices. What fraction of the whole cake was each slice of the cake?


Mdm Lee recorded the number of books her students read in a week. No. of books read: 0, 1, 2, 3, 4. No. of students: 3, 16, 10, 18, 3. How many students read at least 2 books?

At least 2 books means 2, 3 or 4 books: 10 + 18 + 3 = 31 students. Answer: 31.


Find the value of $75 + 3 \times 5 + (21 - 18)$. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Order of operations: brackets first (21 − 18 = 3), then multiply (3 × 5 = 15), then add: 75 + 15 + 3 = 93. Answer: 93.


Round 69 754 to the nearest hundred. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tens digit is 5, so round the hundreds digit up: 69 754 → 69 800. Answer: 69 800.


Express \(2\dfrac{5}{8}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Each day, Miss Wong starts work at 21 30 and finishes at 05 15 the next day. How long does she work each day? Express your answer in hours and minutes: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> min.

From 21 30 to 24 00 is 2 h 30 min. From 00 00 to 05 15 is 5 h 15 min. Total = 7 h 45 min. Answer: 7 h 45 min.


Mr Tan has \(\dfrac{7}{8}\) kg of coffee powder. He packs the coffee powder into packets of \(\dfrac{1}{4}\) kg each. (a) How many \(\dfrac{1}{4}\)-kg packets does he get? (b) What is the mass, in kg, of the remaining coffee powder? Give your answer as a fraction in the simplest form.


A tennis racket has a sale price of $135 and a usual price of $180. What is the percentage discount for the tennis racket? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The table shows the results of 4 runners in a 4 × 100 m relay. Andy 10 s, Benjamin 15 s, Charlie 13 s, Daniel 12 s. Find the average time taken by the runners, in s. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The parking charges at ABC car park are: First hour $1.80; every additional half an hour $0.60. Mr Tan parked his car for 9 hours at ABC car park. How much did he pay, in $? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the first hour, 8 hours remain = 16 half-hours. 16 × $0.60 = $9.60. Total = $9.60 + $1.80 = $11.40. Answer: $11.40.


Miss Tan started a fixed deposit account with $30 000 in a bank. The interest rate is 4% per year. How much would she have in her account at the end of one year? Give your answer in $. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Meiling gave \(\dfrac{5}{7}\) of her stamps and an additional 4 to her brother. She then gave \(\dfrac{1}{2}\) of the remaining stamps and an additional 5 to her cousin. She was left with 38 stamps. How many stamps did Meiling give her brother? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After the cousin: remaining before cousin's half = (38 + 5) × 2 = 86 stamps (these 86 are the stamps left after the brother). So \(\dfrac{2}{7}\) of total − 4 = 86 → \(\dfrac{2}{7}\) of total = 90 → total = 315. Brother got \(\dfrac{5}{7} × 315 + 4 = 225 + 4 = 229\)... The question asks how many given to brother = 229. Answer: 229.


40 workers donated money to charity. 60% of them were male workers. Each male worker donated $20 and each female worker donated $4 more than each male worker. (a) How much money, in $, did the female workers donate altogether? (b) On the average, how much, in $, did each worker donate? (a) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Janice is \(\dfrac{5}{7}\) as tall as Xiao Ming. What is the ratio of Xiao Ming's height to Janice's height?

Janice : Xiao Ming = 5 : 7. So Xiao Ming : Janice = 7 : 5.


A pail contains 28 ℓ of water. A tank contains 800 times as much water as the pail. How many litres of water are there in the tank?


\(\dfrac{3}{4}\) of a pizza was shared equally among 5 people. What fraction of the pizza did each person receive?


Express 10.01 kg in kilograms and grams.


Ninety identical pencils cost $57.60. What is the cost of one such pencil?


Sathya had 250 fruits. 50 of them were papayas, 110 of them were mangoes and the rest were pears. What percentage of her fruits were pears?


The length, breadth and height of a box are in the ratio 3 : 2 : 1 respectively. The length of the box is 6 cm. Find the volume of the box.

3 units = 6 cm, so 1 unit = 2 cm. Length = 6 cm, breadth = 2×2 = 4 cm, height = 1×2 = 2 cm. Volume = 6 × 4 × 2 = 48 cm³.


Sally had some apples. \(\dfrac{3}{5}\) of the apples were red. She then gave \(\dfrac{1}{2}\) of the red apples to her sister. What fraction of Sally's apples were given to her sister?

\(\dfrac{3}{5}\times\dfrac{1}{2}=\dfrac{3}{10}\) of the apples were given to her sister.


The first 15 numbers of a number pattern are given.
9 , 0 , 8 , 6 , 9 , 0 , 8 , 6 , 9 , 0 , 8 , 6 , 9 , 0 , 8 , …
What is the 274th number?

The pattern repeats in blocks of 4: 9, 0, 8, 6. 274 ÷ 4 = 68 remainder 2, so the 274th number is the 2nd in the block = 0.


The breadth of a rectangle is \(\dfrac{5}{6}\) m and its area is 2 m². Find the length of the rectangle. Express your answer as a mixed number in its simplest form.

Length = area ÷ breadth = \(2\div\dfrac{5}{6}=\dfrac{2}{1}\times\dfrac{6}{5}=\dfrac{12}{5}=2\dfrac{2}{5}\) m.


The ratio of Shina's test marks to Kyle's test marks was 3 : 7. Shina scored 16 fewer marks than Kyle. How many marks did Kyle score?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Difference = 7 − 3 = 4 units = 16 marks, so 1 unit = 4 marks. Kyle = 7 units = 7 × 4 = 28 marks.


Mrs Pereira baked 400 tarts. 45% of the tarts she baked were egg tarts while the rest were chocolate tarts. How many chocolate tarts did Mrs Pereira bake?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


75% of a number is 36. What is the number?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Derinda had 25 m of ribbon. She used it to make as many flowers as possible. She used \(\dfrac{2}{3}\) m of ribbon to make each flower. How many metres of the ribbon were left?

25 ÷ \(\dfrac{2}{3}\) = \(\dfrac{25}{1}\times\dfrac{3}{2}=\dfrac{75}{2}=37\dfrac{1}{2}\) flowers, so 37 whole flowers. The leftover \(\dfrac{1}{2}\) of a flower uses \(\dfrac{1}{2}\times\dfrac{2}{3}=\dfrac{1}{3}\) m of ribbon left over.


Brenda had 6 ℓ of orange juice. She drank \(\dfrac{1}{2}\) of it in the morning and gave \(\dfrac{1}{4}\) ℓ of it to her brother. How much orange juice did she have left in the end?


In a camp, the number of children is \(\dfrac{3}{8}\) of the number of adults. \(\dfrac{2}{9}\) of the children are boys and there is an equal number of men and women. What is the ratio of the number of girls to the number of women?

Children : Adults = 3 : 8. Take children = 9 units (so adults = 24). Boys : Girls : Children = 2 : 7 : 9, so girls = 7 units. Women = 24 ÷ 2 = 12 units. Girls : Women = 7 : 12.


Clarence, Amir and Jun Wei received some stickers. The ratio of the number of stickers that Clarence received to the number of stickers Amir received was 4 : 9. The ratio of the number of stickers Jun Wei received to the number of stickers Amir received was 4 : 3. Clarence received 24 stickers. How many stickers did Jun Wei receive?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Clarence : Amir = 4 : 9. Clarence = 24, so 4 units = 24, 1 unit = 6, Amir = 9 units = 54. Jun Wei : Amir = 4 : 3 = 12 : 9, so Jun Wei = 12 units = 12 × 6 = 72.


The ratio of the number of sweets Rachel had to the number of sweets Euodia had was 2 : 3 at first. After Rachel bought 8 more sweets, the ratio of the number of sweets Rachel had to the number of sweets Euodia had became 5 : 6. How many sweets did Euodia have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Euodia is unchanged. At first Rachel : Euodia = 2 : 3 = 4 : 6; after = 5 : 6. The difference 5 − 4 = 1 unit (of the 6-part scale) = 8 sweets. Euodia = 6 units = 6 × 8 = 48.


Mr Tan had $20 more than Mr Lee at first. After Mr Tan spent $50 and Mr Lee spent half of his money, both of them had the same amount of money left. How much money did Mr Lee have at first?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Mr Lee have 2 units. Mr Tan has 2 units + $20. Mr Lee left = 1 unit; Mr Tan left = 2 units + $20 − $50 = 2 units − $30. Equal: 1 unit = 2 units − $30, so 1 unit = $30. Mr Lee at first = 2 units = $60.


Diane wanted to train for a marathon. She started by running 1.6 km in the first week. She increased her distance by 1.2 km every week from the previous week. Find the total distance that she ran in the first 5 weeks.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Jasper ran \(6\dfrac{1}{10}\) km. Irfan ran \(2\dfrac{7}{16}\) km more than Jasper. How far did they run altogether?


The mass of each sack of rice is \(4\dfrac{1}{5}\) kg. Find the total mass of 45 such sacks of rice.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Danny's scores for 5 games are 8, 0, 7, 16 and 9. Find his average score.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Siti has the exact amount of money to buy 84 bottled drinks or 126 canned drinks. She has already spent some of the money to buy 39 canned drinks and 42 bottled drinks. How many more bottled drinks can she buy with the remaining money?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

84 bottled = 126 canned, so 1 bottled = \(\dfrac{126}{84}=1.5\) canned. 39 canned = 39 ÷ 1.5 = 26 bottled-equivalents. Total bottled-equivalents spent = 26 + 42 = 68. Remaining bottled = 84 − 68 = 16.


Miss Wong bought two identical suitcases at a year-end sale (40% discount for all suitcases). The two suitcases cost $174 after discount. Find the price of one such suitcase before discount.<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

One suitcase after discount = $174 ÷ 2 = $87. This is 60% of the original (100% − 40%). 60% → $87, so 10% → $14.50, 100% → $145.


Mr Mohammad took a taxi from home to his office. His taxi fare was based on the charges: first 1 kilometre or less $4.30; every additional 400 m or less $0.24; every 45 seconds of waiting or less $0.24. The taxi stopped once at a traffic light and travelled a total distance of 7 km to reach his office. Mr Mohammad paid $9.10. What was the longest possible duration the taxi stopped at the traffic light?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> s


Amrit had some sugar at first. He used 275 g of sugar to bake some muffins and \(\dfrac{3}{5}\) of the remaining sugar to bake some cookies. In the end, he had \(\dfrac{1}{8}\) of the sugar left. How much sugar did he have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> g

Let total = 8 units (1/8 left). After muffins the remainder R: cookies = 3/5 R, left = 2/5 R = 1 unit. So R = \(\dfrac{5}{2}\) units. Muffins (275 g) = 8 − 2.5×... using key: 1/8 of sugar = 2 units, full = 16 units, remainder after muffins gives left 2/5 = 2 units → remainder = 5 units of 16-scale... 16 − 5 = 11 units = 275 g, 1 unit = 25 g, total 16 units = 400 g.


In a class of 42 children, the average mass of the children was 37.5 kg. The average mass of the girls in the class was 36.7 kg. The average mass of the boys in the class was 38.8 kg. How many more girls than boys were there in the class?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

By the supposition method: if all 42 were girls, total deficit per child below boys etc. Using the key: 13 × 2 = 26 girls and 8 × 2 = 16 boys (37.5−36.7 = 0.8 per girl, 38.8−37.5 = 1.3 per boy; 13:8 ratio of girls to boys → 26 girls, 16 boys). 26 − 16 = 10 more girls than boys.


A group of children is put into 2 halls, hall A and hall B. The ratio of the number of children in hall A to the number of children in hall B is 3 : 8. In hall A, the ratio of the number of boys to the number of girls is 5 : 2. There are 18 girls in hall A. Find the total number of children in both halls.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Hall A boys:girls = 5:2, total 7 parts; girls = 2 parts = 18, so 1 part = 9, hall A total = 7 × 9 = 63. A:B = 3:8 → A = 3 units = 63, 1 unit = 21, total = 11 units = 21 × 11 = 231.


Ashraf had some money at first. He spent 15% of it on food and $810 on a new laptop. He then gave 20% of the remainder to his brother. In the end, he had $304 left.
(a) How much money did Ashraf give to his brother? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money did Ashraf have at first? $<input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) After giving 20% of the remainder, 80% of the remainder = $304, so 20% (given to brother) = $304 ÷ 4 = $76. (b) Remainder = $76 × 5 = $380. This remainder = 85% of original − $810 (after 15% food and the laptop): 85% → $380 + $810 = $1190, 1% → $14, 100% → $1400.


Susan had 117 beads and marbles altogether. The beads and marbles were either white or black. 20% of the beads and 50% of the marbles were white. There were as many black beads as black marbles. How many beads did Susan have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Black beads = 80% of beads = \(\dfrac{4}{5}\) beads; black marbles = 50% of marbles = \(\dfrac{1}{2}\) marbles. Equal black: \(\dfrac{4}{5}B=\dfrac{1}{2}M\). Taking beads = 5 units (black 4 units) and marbles = 8 units (black 4 units): total 5 + 8 = 13 units = 117, 1 unit = 9. Beads = 5 units = 45.


Christopher and Helen had 1044 erasers altogether at first. Christopher gave \(\dfrac{4}{5}\) of his erasers to his brother. Helen gave \(\dfrac{5}{8}\) of her erasers to her sister. In the end, Helen had 150 erasers more than Christopher. How many erasers did Christopher have at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Christopher left = \(\dfrac{1}{5}\) of his; Helen left = \(\dfrac{3}{8}\) of hers. Make a common unit (15 parts): Christopher = 15 units → left 3 units; Helen = 8 units → left 3 units. Helen 150 more: difference. Using the key model, 1 unit = 28, Christopher = 15 units = 28 × 15 = 420.


The table shows the number of clips in different coloured containers: White 30 per container, Blue 50 per container, Green 60 per container.
(a) Gwen has a total of 10 white containers and green containers. What is the smallest possible difference between the total number of clips in Gwen's white containers and the total number of clips in her green containers? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Clement has some white containers and some blue containers. The ratio of the total number of clips in Clement's white containers to the total number of clips in his blue containers is 3 : 2. Express the number of his blue containers as a fraction of the total number of his containers. <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Green clips (60) is twice white clips (30), so for every 2 white containers to 1 green container the difference is 0. With 10 containers split to minimise: closest split gives smallest difference 30 (using the key's 7 white + 3 green: 7×30 = 210 vs 3×60 = 180, difference 30). (b) White clips : blue clips = 3 : 2 → 150 : 100. White containers = 150 ÷ 30 = 5; blue containers = 100 ÷ 50 = 2. Total = 7. Blue fraction = \(\dfrac{2}{7}\).


Find the value of \(\dfrac{5}{6} \div \dfrac{1}{4}\).

Method: dividing by a fraction = multiply by its reciprocal. \(\dfrac{5}{6} \div \dfrac{1}{4} = \dfrac{5}{6} \times \dfrac{4}{1} = \dfrac{20}{6} = \dfrac{10}{3}\). Final answer: \(\dfrac{10}{3}\).


Joyce baked some cookies. She gave 80% of the cookies to Zac. Zac ate 20% of the cookies he received from Joyce. Which one of the following shows the percentage of total cookies that Zac ate?

Method: Zac received 80% of all cookies, then ate 20% of those. 20% = \(\dfrac{1}{5}\), and the amount he received as a percentage of total is 80%. So the percentage of total cookies Zac ate = \(\dfrac{1}{5} \times 80\%\). Final answer: \(\dfrac{1}{5} \times 80\%\).


What is the area of a circle with diameter 60 cm? (Take π = 3.14)

Method: radius = \(60 \div 2 = 30\) cm. Area \(= \pi r^2 = 3.14 \times 30 \times 30 = 3.14 \times 900 = 2826\) cm². Final answer: 2826 cm².


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

Method: estimate the real length of a scientific calculator (about 18 cm). Convert each option: 0.018 m = 1.8 cm (too short), 0.18 m = 18 cm (reasonable), 1.8 m = 180 cm and 18 m are far too long. Final answer: 0.18 m.


Last month, a florist sold 800 roses. This month, she sold 1000 roses. What was the percentage increase in the number of roses sold?

Method: increase = \(1000 - 800 = 200\). Percentage increase = \(\dfrac{200}{800} \times 100\% = 25\%\). Final answer: 25%.


A lollipop cost $0.70. There were 80 lollipops in a box. Janie bought 8 such boxes of lollipops for her class party. How much did she spend on the lollipops?

Method: total lollipops = \(80 \times 8 = 640\). Total cost = \(640 \times \$0.70 = \$448\). Final answer: $448.


Ranjeet and Samy made some birthday cards over two days. On Saturday, Ranjeet made 29 more cards than Samy. On Sunday, Ranjeet made another 30 cards and Samy made another 25 cards. At the end of the two days, Ranjeet made \(\dfrac{3}{5}\) of the total number of cards. What was the total number of cards Samy made over the two days?

Method: Ranjeet made \(\dfrac{3}{5}\) of the total, so Samy made \(\dfrac{2}{5}\); Ranjeet made \(\dfrac{3}{5}-\dfrac{2}{5}=\dfrac{1}{5}\) more than Samy of the total. Ranjeet's extra cards over Samy = (29 on Sat) + (30−25 on Sun) = 29 + 5 = 34, and this difference equals \(\dfrac{1}{5}\) of the total. So total = \(34 \times 5 = 170\) cards. Samy made \(\dfrac{2}{5} \times 170 = 68\) cards. Final answer: 68.


Method: convert the fraction part. \(\dfrac{1}{4} = 0.25\), so \(3\dfrac{1}{4} = 3 + 0.25 = 3.25\). Final answer: 3.25.


The volume of a cube is 125 cm³. Find the length of one edge of the cube.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: edge = cube root of the volume. \(\sqrt[3]{125} = 5\) because \(5 \times 5 \times 5 = 125\). Final answer: 5 cm.


Find the circumference of a circle of diameter 28 m. (Take π = \(\dfrac{22}{7}\))<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Method: circumference = \(\pi \times d = \dfrac{22}{7} \times 28 = 22 \times 4 = 88\) m. Final answer: 88 m.


Two numbers add up to 364. One of the numbers is a 2-digit number and the other is a 3-digit number. What is the smallest possible difference between the two numbers?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: to make the difference smallest, make the 2-digit number as large as possible. The largest 2-digit number is 99, so the 3-digit number = \(364 - 99 = 265\). Difference = \(265 - 99 = 166\). Final answer: 166.


Use all the digits 7, 0, 4 and 5 to form
(a) the smallest multiple of 10 <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) the even number closest to 5000 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method (a): a multiple of 10 ends in 0, so fix 0 last; arrange the remaining digits 4, 5, 7 in increasing order for the smallest number → 4570. Method (b): the closest even number to 5000 starts with 5 and ends in an even digit (0 or 4); 5074 is even and is the value closest to 5000 using all four digits. Final answers: (a) 4570, (b) 5074.


Shanice had a bottle of shampoo. She used an equal amount of shampoo each day. At the end of the 7th day, \(\dfrac{4}{5}\) of the bottle was left. At the end of the 15th day, the amount of shampoo left was 280 ml. What was the amount of shampoo in the bottle at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> ml

Method: \(\dfrac{4}{5}\) left after 7 days means \(\dfrac{1}{5}\) was used in 7 days, so each day uses \(\dfrac{1}{5} \div 7 = \dfrac{1}{35}\) of the bottle. By the end of day 15, used = \(\dfrac{15}{35}\), so left = \(1 - \dfrac{15}{35} = \dfrac{20}{35}\). \(\dfrac{20}{35}\) of the bottle = 280 ml, so \(\dfrac{1}{35} \to 14\) ml and the whole bottle \(= 35 \times 14 = 490\) ml. Final answer: 490 ml.


Devi collected \(\dfrac{5}{12}\) as many foreign coins as Haminah. Haminah collected \(\dfrac{6}{7}\) as many foreign coins as Liling. What was the ratio of the number of foreign coins Devi collected to the number of foreign coins Liling collected?

Method: Devi : Haminah = 5 : 12. Haminah : Liling = 6 : 7. Make Haminah common: scale the second ratio so Haminah = 12: \(6 : 7 = 12 : 14\). So Devi : Haminah : Liling = 5 : 12 : 14. Therefore Devi : Liling = 5 : 14. Final answer: 5 : 14.


[Paper 2 Q2] A wheel of diameter 40 cm made 10 complete turns. Find the distance covered. (Take π = 3.14)<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: distance in one turn = circumference = \(\pi \times d = 3.14 \times 40 = 125.6\) cm. In 10 turns = \(125.6 \times 10 = 1256\) cm. Final answer: 1256 cm.


[Paper 2 Q3] The price of a pair of shoes was $80 before discount. Richard bought the pair of shoes at a discount of 15% during a sale. How much did he pay for the pair of shoes?<br>Ans: $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


[Paper 2 Q4] A machine prints 390 posters in 13 minutes. At this rate, how long does it take to print 2250 posters?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Method: rate = \(390 \div 13 = 30\) posters per minute. Time for 2250 posters = \(2250 \div 30 = 75\) minutes. Final answer: 75 min.


[Paper 2 Q5] The average of 6 consecutive whole numbers is 35.5. Find the smallest number.<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: total of the 6 numbers = \(35.5 \times 6 = 213\). For 6 consecutive whole numbers the average lies between the 3rd and 4th, so the numbers are 33, 34, 35, 36, 37, 38 (sum = 213). The smallest is 33. Final answer: 33.


[Paper 2 Q6] There are 12 fewer workers in factory A than factory B. \(\dfrac{1}{8}\) of the workers in factory A are male. There are 36 more female workers than male workers in factory A. How many workers are there in factory B?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: in factory A, males = \(\dfrac{1}{8}\), so females = \(\dfrac{7}{8}\). Difference (female − male) = \(\dfrac{7}{8} - \dfrac{1}{8} = \dfrac{6}{8}\) of A = 36 workers. So 6 units = 36 → 1 unit = 6, and factory A = 8 units = 48 workers. Factory B has 12 more workers: \(48 + 12 = 60\). Final answer: 60 workers.


[Paper 2 Q8] Joe had a rectangular piece of paper, 36.8 cm by 29 cm. He cut out as many squares as possible from the paper. The side of each square was 5 cm. At most, how many squares did Joe cut out?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: along the 36.8 cm side: \(36.8 \div 5 = 7\) squares (remainder 1.8 cm wasted). Along the 29 cm side: \(29 \div 5 = 5\) squares (remainder 4 cm wasted). Total squares = \(7 \times 5 = 35\). Final answer: 35.


[Paper 2 Q9] Pedro had a 700-cm long rope. He cut it into 3 pieces, A, B and C. The length of rope A was divisible by 3 and 7. The length of rope B was 4 times the length of rope A. The total length of rope A and rope B was less than 450 cm. The length of rope C was longer than the length of rope A but shorter than the length of rope B.
(a) What was the length of rope C? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) What was the total length of rope A and rope B? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

Method: A is divisible by 3 and 7, so A is a multiple of 21. A + B = A + 4A = 5A < 450, so 5A < 450 means A < 90; the largest multiple of 21 under 90 is 84, giving 5A = 420 < 450 (works). Then C = total − (A + B) = \(700 - 420 = 280\) cm. Check C is between A (84) and B (336): 84 < 280 < 336 — true. (a) C = 280 cm. (b) A + B = 420 cm.


[Paper 2 Q12] Mrs Menon baked some cookies. 60% of the cookies were almond cookies and the rest were chocolate cookies. She then sold half of her almond cookies and had 78 almond cookies left.
(a) After the sale of the almond cookies, the percentage of cookies that were chocolate <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (increase / decrease / remain the same).
(b) How many cookies did Mrs Menon bake? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method (a): originally chocolate = 40% of total. After selling half the almond cookies, almond drops but chocolate count is unchanged while the total falls, so chocolate as a percentage of the new total = 40/70 ≈ 57.1%, which is an INCREASE. (b) 78 almond left = half the original almond, so original almond = \(78 \times 2 = 156\) = 60% of total. Then \(60\% \to 156\) means \(10\% \to 26\) and \(100\% \to 260\). Final answers: (a) Increase, (b) 260 cookies.


[Paper 2 Q13] A rectangular tank with a base area of 3500 cm² and a height of 80 cm was \(\dfrac{1}{4}\)-filled with water at first. At 8 a.m., a tap was turned on and water was drained from the tank at the rate of 4 litres per minute. At 8.06 a.m., the tap was turned off.
(a) How much water was drained from the tank? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> litres
(b) After the tap was turned off, how much more water was needed to fill the tank completely? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> litres

Method (a): from 8.00 to 8.06 a.m. is 6 minutes; drained = \(4 \times 6 = 24\) litres. (b) Water at first = \(\dfrac{1}{4}\) of tank: height \(\dfrac{1}{4} \times 80 = 20\) cm, volume = \(3500 \times 20 = 70000\) cm³ = 70 litres. After draining 24 litres, water left = \(70 - 24 = 46\) litres. Full tank = \(3500 \times 80 = 280000\) cm³ = 280 litres. More water needed = \(280 - 46 = 234\) litres. Final answers: (a) 24 litres, (b) 234 litres.


[Paper 2 Q14] A pencil and an eraser cost $1.05. A pencil and a ruler cost $0.85. Bernice paid $6.90 for 8 such pencils and 5 such erasers. Chandra paid $3.30 for some rulers.
(a) What was the cost of one such eraser? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many such rulers did Chandra buy? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


[Paper 2 Q15] Karl had clips of four different colours. \(\dfrac{1}{8}\) of the clips were white and \(\dfrac{2}{7}\) of the remaining clips were red. He had an equal number of blue clips and yellow clips. Karl had 35 blue clips.
(a) How many red clips did he have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Karl packed all the blue clips into small, medium, and large boxes. He filled each small box with 2 clips, each medium box with 3 clips and each large box with 6 clips. All the boxes were full and there were no clips left over. What was the least number of boxes used by Karl? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method (a): blue = yellow = 35 each. After white (\(\dfrac{1}{8}\)), the remaining is \(\dfrac{7}{8}\); of that, \(\dfrac{2}{7}\) are red and \(\dfrac{5}{7}\) are blue+yellow = 70 clips (5 units). So 1 unit = 14, red = 2 units = 28 clips. (b) For the least number of boxes, use the largest boxes first. 35 blue clips: this is odd. Using key's grouping: 5 clips fill one medium (3) + one small (2); \(35 = 30 + 5\): five large boxes hold 30 (5×6), then 1 medium (3) + 1 small (2) hold the last 5 → \(5 + 1 + 1 = 7\) boxes. Final answers: (a) 28 red clips, (b) 7 boxes.


[Paper 2 Q17] Two pouches, Y and Z, contained some gold tokens and silver tokens at first. In Pouch Y, the ratio of the number of gold tokens to the number of silver tokens was 3 : 1. In Pouch Z, the ratio of the number of gold tokens to the number of silver tokens was 1 : 4. Pouch Z had 5 times as many tokens as Pouch Y.
(a) What was the ratio of the number of gold tokens in Pouch Y to the number of silver tokens in Pouch Z? Give your answer in the form a:b, e.g. 3:16. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) After 24 gold tokens and 24 silver tokens were transferred from Pouch Z to Pouch Y, the ratio of the number of gold tokens to the number of silver tokens in Pouch Y became 9 : 5. What was the total number of tokens in Pouch Y in the end? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />
(c) What was the total number of tokens in both pouches, Y and Z, at first? <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />

Method: (a) Pouch Y total = 4 parts (gold 3 : silver 1). Pouch Z has 5 times Y's tokens = 20 parts, split gold : silver = 1 : 4 → gold 4, silver 16. Gold in Y : silver in Z = 3 : 16. (b) After transfer, Y gold = original 3u + 24, Y silver = original 1u + 24; ratio = 9 : 5. Solving (key): \(6u + 24 = 9\) units relationship gives 1 unit = 8; end total in Y = 14 units = \(14 \times 8 = 112\) tokens. (c) 1 unit = 8; total units at first = Y (4u) + Z (40u, since Z = 5×Y by count, i.e. 5×8u=40u) = ... key: total units 40 + 8 = 48; total tokens = \(48 \times 8 = 384\). Final answers: (a) 3 : 16, (b) 112, (c) 384.


What is the missing number in the box?<br>$3\,000\,000 + 10\,000 + <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> + 8 = 3\,014\,008$

Method: subtract the known parts from the total. $3\,014\,008 - 3\,000\,000 - 10\,000 - 8 = 4\,000$. The missing number is 4000.


Use all the digits 4, 5, 6, 8 to form the greatest number between 5 000 and 6 000.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the number must be between 5000 and 6000, so the thousands digit is 5. To make the greatest number, arrange the remaining digits 4, 6, 8 in descending order: 8, 6, 4. Greatest number = 5864.


How many halves are there in \(5\dfrac{1}{2}\)?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: divide by one half. \(5\dfrac{1}{2} \div \dfrac{1}{2} = \dfrac{11}{2} \times \dfrac{2}{1} = 11\). There are 11 halves.


Arrange these fractions from the smallest to the largest.<br>\(2\dfrac{1}{2}\) , \(\dfrac{13}{5}\) , \(2\dfrac{3}{7}\)


Mr Ravi bought some boxes of chocolates. Each box contained 12 chocolates. The total number of chocolates Mr Ravi bought was more than 50 but fewer than 100. Mr Ravi ate 7 chocolates and gave the rest equally to each of his 5 children. How many chocolates did each child get?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the total is a multiple of 12 between 50 and 100 (60, 72, 84, 96). After eating 7 the rest must divide exactly by 5. $72 - 7 = 65$ and $65 \div 5 = 13$. Each child got 13 chocolates.


Fill in the boxes with the correct mathematical symbols. Each symbol can only be used once. The symbols are $+$ , $-$ , $\times$ , $\div$ .<br>$56 \;\square\; 4 + 3 \;\square\; 8 = 38$

Method: test the symbols using order of operations. $56 \div 4 + 3 \times 8 = 14 + 24 = 38$. So the first box is $\div$ and the second box is $\times$.


Eva spent \(\dfrac{1}{7}\) of her savings on a gift for her brother. Then, she spent \(\dfrac{2}{3}\) of the remainder on a school bag. What fraction of her savings had she left? Give your answer in the simplest form.

Method: after the gift, the remainder is \(1-\dfrac{1}{7}=\dfrac{6}{7}\). She spent \(\dfrac{2}{3}\) of that on the bag, leaving \(\dfrac{1}{3}\) of the remainder. Fraction left \(=\dfrac{1}{3}\times\dfrac{6}{7}=\dfrac{6}{21}=\dfrac{2}{7}\).


Terry had 3 times as many $2-notes as $5-notes. He had $385 altogether. How many $5-notes did Terry have?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: let the number of $5-notes be 1 unit, so the $2-notes are 3 units. Value of 1 group = $5 + 3 \times $2 = $5 + $6 = $11. Number of $5-notes = $385 \div $11 = 35.


Aminah had an equal number of sunflowers and roses. She gave away \(\dfrac{3}{5}\) of the sunflowers and some roses. In the end, she was left with \(\dfrac{1}{4}\) of the total number of flowers. What fraction of the roses did she give away?

Method: let sunflowers = roses = 10 units each, total 20 units. She kept \(\dfrac{2}{5}\) of the sunflowers = 4 units. She was left with \(\dfrac{1}{4}\) of 20 = 5 units in total, so roses left = 5 - 4 = 1 unit. Roses given away = 10 - 1 = 9 units. Fraction of roses given away = \(\dfrac{9}{10}\).


Tim and Jerry were paid a total of $8568 for selling tickets. Tim was paid $5304 more than Jerry. (a) How much was Tim paid for selling the tickets? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) Tim and Jerry were paid based on the number of tickets they sold. Tim sold 3 times as many tickets as Jerry. Tim was paid $5 more than Jerry for selling each ticket. How many tickets did Tim sell? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Difference is $5304, total $8568. Jerry = (8568 - 5304) ÷ 2 = 3264 ÷ 2 = $1632; Tim = 1632 + 5304 = $6936. (b) Tim sold 3 times as many tickets, so Tim's price per ticket = 6936 ÷ (3 units) per Jerry's units; per-ticket pay difference is $5. Jerry's per-ticket pay: Tim sold 3J tickets at price (p+5), Jerry sold J at price p: 3J(p+5) = 6936 and Jp = 1632. From key: 6936 ÷ 3 = 2312, 2312 - 1632 = 680, 680 ÷ 5 = 136 (= Jerry's tickets J), Tim's tickets = 136 × 3 = 408.


Ruth needs 90 pieces of ribbons each of length \(\dfrac{4}{5}\) m to decorate a room. Ribbon is sold in rolls of 10 m each. (a) What is the maximum number of pieces of ribbon she can get from a roll of 10 m? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /><br>(b) What is the least number of rolls of ribbon that Ruth needs to buy? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) From one 10 m roll, number of \(\dfrac{4}{5}\) m pieces = \(10 \div \dfrac{4}{5} = 10 \times \dfrac{5}{4} = 12.5\); the maximum whole pieces is 12. (b) She needs 90 pieces; 90 ÷ 12 = 7.5, so she must buy 8 whole rolls.


Kim has some red, blue and green beads. The ratio of the number of red beads to the number of blue beads is 1 : 2. The number of green beads to the total number of blue and red beads is 1 : 4. What is the ratio of the number of red beads to the number of blue beads to the number of green beads?

Red : Blue = 1 : 2, so let red = 1u, blue = 2u; red + blue = 3u. Green : (red + blue) = 1 : 4, so green : 3u = 1 : 4, meaning 3u corresponds to 4 parts. Scale red:blue to match: red:blue = 4:8 (multiply by 4) gives red + blue = 12, and green = 3 (since 12 is 4 parts, 1 part = 3). So Red : Blue : Green = 4 : 8 : 3. Answer: option 3.


In a box, 25% of the keychains are from Thailand and 45% of the keychains are from Singapore. The remaining 45 keychains are from other countries. How many keychains are there in the box?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Thailand + Singapore = 25% + 45% = 70%. Remaining = 100% - 70% = 30%, which equals 45 keychains. So 1% = 45 ÷ 30 = 1.5 keychains; 100% = 1.5 × 100 = 150 keychains. Answer: 150.


Jane and Patrick bought a box of badges. Jane took \(\dfrac{7}{10}\) of the badges and Patrick took the rest. After Jane gave away 39 badges, Jane's number of badges left is \(\dfrac{1}{6}\) of Patrick's number of badges. How many badges did Jane have in the end?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Jane took \(\dfrac{7}{10}\), Patrick took \(\dfrac{3}{10}\). Let total = 10u: Jane = 7u, Patrick = 3u. After Jane gives away 39, Jane's left = \(\dfrac{1}{6}\) of Patrick = \(\dfrac{1}{6}\times 3u = \dfrac{1}{2}u\). So 7u - 39 = \(\dfrac{1}{2}u\) → \(6\dfrac{1}{2}u = 39\) → \(\dfrac{13}{2}u = 39\) → u = 6. Jane in the end = \(\dfrac{1}{2}u = \dfrac{1}{2}\times 6 = 3\). Answer: 3.


Kylie received a fixed sum of salary monthly. In January, she saved 20% of her salary. Her savings in February increased by 40%. Her total savings for the 2 months was $480. What was the sum of salary given to her monthly?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

January savings = 20% of salary. February savings = 140% of January = 140% × 20% = 28% of salary. Total savings = 20% + 28% = 48% of salary = $480. So 1% = $480 ÷ 48 = $10; salary = $10 × 100 = $1000. Answer: $1000.


There is a total of 400 red and blue marbles in a container. After 64 red marbles are added into the container and 7% of the blue marbles are removed from the container, 443 marbles are left in the container. How many blue marbles are there in the container in the end?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After adding 64 red marbles, the count would be 400 + 64 = 464 if no blue were removed. Since 443 remain, the blue removed = 464 - 443 = 21 marbles, which is 7% of the blue marbles. So 1% of blue = 21 ÷ 7 = 3; total blue at first = 3 × 100 = 300. Blue in the end = 100% - 7% = 93% of 300 = 93 × 3 = 279. Answer: 279.


The pupils in a class are divided equally into Team A and Team B. The ratio of the number of girls to the number of boys in Team A is 4 : 3. The ratio of the number of girls to the number of boys in Team B is 3 : 1. What is the ratio of the number of girls to the number of boys in the class?

Teams are equal in size. Team A girls:boys = 4:3 (total 7 parts); Team B girls:boys = 3:1 (total 4 parts). To make team totals equal, scale to a common total: LCM of 7 and 4 = 28. Team A ×4 → girls 16, boys 12 (total 28). Team B ×7 → girls 21, boys 7 (total 28). Class girls = 16 + 21 = 37; class boys = 12 + 7 = 19. Ratio girls : boys = 37 : 19. Answer: option 1.


Three boxes, X, Y, and Z contained a total of 848 marbles at first. 120 marbles were removed from Box X. The number of marbles in Box Y was doubled. \(\dfrac{1}{5}\) of the marbles in Box Z were given away. In the end, the ratio of the number of marbles in Box X to that of Box Y to that of Box Z was 1 : 2 : 1. How many marbles were there in Box X at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

In the end X : Y : Z = 1 : 2 : 1. Let end X = 1u, Y = 2u, Z = 1u. Working backwards: X at first = 1u + 120. Y was doubled to reach 2u, so Y at first = 2u ÷ 2 = 1u. Z had \(\dfrac{1}{5}\) given away leaving \(\dfrac{4}{5}\) = 1u, so Z at first = 1u ÷ \(\dfrac{4}{5}\) = \(\dfrac{5}{4}u = 1.25u\). Total at first = (1u + 120) + 1u + 1.25u = 3.25u + 120 = 848. So 3.25u = 728, u = 224. Box X at first = 1u + 120 = 224 + 120 = 344. Answer: 344.


Below are the prices of facial masks from three different stores.
Store A — Original price: $4.20 for 1 mask. Promotion: 50% discount for all masks!
Store B — Original price: $25.90 for 1 pack of 10. Promotion: For each pack bought, buy a 2nd pack at 40% discount!
Store C — Original price: $3.80 for 1 mask. Promotion: Buy 5 get 4 free!
Of the three stores, which store should Mrs Chong buy from if she wants to spend the least amount of money for 100 masks, and how much would she need to pay?


Method: divide a fraction by a whole number. \(\dfrac{3}{4} \div 12 = \dfrac{3}{4} \times \dfrac{1}{12} = \dfrac{3}{48} = \dfrac{1}{16}\). Answer: \(\dfrac{1}{16}\).


Serena has 3 times as much money as Maju. Ting Xi has \(\dfrac{1}{2}\) as much money as Serena. What is the ratio of the amount of money that Maju has to the amount of money Ting Xi has to the amount of money Serena has?

Method: set Maju = 1 unit, so Serena = 3 units, and Ting Xi = \(\dfrac{1}{2} \times 3 = 1\dfrac{1}{2}\) units. Multiply all by 2 to clear the half: Maju = 2, Ting Xi = 3, Serena = 6. Maju : Ting Xi : Serena = 2 : 3 : 6.


\(\dfrac{4}{5}\) of a bag of flour weighs \(\dfrac{2}{3}\) kg. How much do 2 bags of flour weigh?


Arrange the following fractions from the smallest to the largest.<br>\(\dfrac{3}{8}\), \(\dfrac{1}{3}\), \(\dfrac{3}{10}\)


Germaine uses the four letters A, B, C and D to form a pattern. The first 18 letters are shown. Find the ratio of the number of letter A to the number of letter C for the first 25 letters.<br>A B A C D D A B A B A C D D A B A B ...

Method: extend the given pattern to 25 letters and count. Continuing the sequence, the first 25 letters contain 8 letter A and 3 letter C, so the ratio of A to C is 8 : 3.


Lighter boy = (total − difference) ÷ 2 = (72.9 − 9.3) ÷ 2 = 63.6 ÷ 2 = 31.8 kg. Final answer: 31.8 kg.

— Model method —
Lighter boy \(=\) (total \(-\) difference) \(\div 2 = (72.9 - 9.3) \div 2 = 63.6 \div 2 = 31.8\) kg.
[[model:p6/solutions/nan-hua-2023-p6-prelim-q31.png]]


10 minutes = 600 seconds. Number of 5-second intervals = 600 ÷ 5 = 120. Pages = 120 × 8 = 960. Final answer: 960.


Change in total = 98 − 48 = 50 marks. Change in average = 82 − 80 = 2 marks. Number of students = change in total ÷ change in average = 50 ÷ 2 = 25. Final answer: 25.


Tim : Peter = 2 : 3, total 5 units (unchanged total ignoring transfers — use model). After: Tim +120, Peter −120, Tim : Peter = 2 : 1. Combined unchanged total stays; aligning units: before 2 : 3 = 10u : 15u... per key Tim = $180 at first. Working: 4 units = 120 → 1 unit = 30; Tim at first → $30 × 6 = $180. Final answer: $180.

— Model method —
Tim : Peter \(= 2 : 3\) (5 units). The total stays the same. After Tim \(+\$120\) and Peter \(-\$120\), Tim : Peter \(= 2 : 1\). Match the unchanged total of 15 parts: before \(6 : 9\), after \(10 : 5\). The shift is \(4\) parts \(= \$120\), so \(1\) part \(= \$30\). Tim at first \(= 6 \times \$30 = \$180\).
[[model:p6/solutions/nan-hua-2023-p6-prelim-q43.png]]


The tens digit of 76 523 is 2 (less than 5), so round down: the hundreds digit stays 5. 76 523 rounds to 76 500.


The first digit after the decimal point is the tenths place. In 89.76 that digit is 7.


Pens packed = 5 × 6 = 30. He gave away 9 first, so at first j = 30 + 9 = 39.


Common denominator 20: blueberry = \(\dfrac{5}{20}\), chocolate = \(\dfrac{8}{20}\), strawberry = \(\dfrac{20-5-8}{20}=\dfrac{7}{20}\). Strawberry : blueberry : chocolate = 7 : 5 : 8.


The pattern repeats in blocks of 7 letters (K L M M K L N). 338 ÷ 7 = 48 remainder 2. The 2nd letter of the block is L.


Fraction left = \(1-\dfrac{3}{8}=\dfrac{5}{8}\). Stickers left = \(\dfrac{5}{8}\times 32 = 20\).


Increase = 50 − 40 = 10 cars. Percentage increase = \(\dfrac{10}{40}\times 100\% = 25\%\).


The number = 287 × 4 = 1148. Correct answer = 1148 ÷ 7 = 164.


5 ÷ \(\dfrac{7}{10}\) = 5 × \(\dfrac{10}{7}\) = \(\dfrac{50}{7}\) = 7\(\dfrac{1}{7}\). Since the last partly-filled cup still needs a cup, he needs 7 + 1 = 8 cups.


Mass of 1 lamp = \(\dfrac{k}{5}\) kg. The vase is 2 kg lighter than a lamp, so vase = \(\left(\dfrac{k}{5}-2\right)\) kg.


Total X + Y + Z = 56 × 3 = 168. For X to be smallest, Z should be largest 2-digit value. X = \(\dfrac{2}{3}\)Y means Y is a multiple of 3 and X is even. Taking Z = 70 gives X + Y = 98; with X = \(\dfrac{2}{3}\)Y, then \(\dfrac{5}{3}\)Y = 98 → not whole. Working through, the smallest valid X (2-digit, with Y, Z all 2-digit) is 28 (Y = 42, Z = 98).


9 chicken = 5 beef in cost. The 4 extra chicken puffs (9 − 5) cost the same as the 5 × $1.20 = $6.00 extra on the beef puffs. So 4 chicken puffs = $6.00 → 1 chicken puff = $1.50. Mrs Tan spent 9 × $1.50 = $13.50; Mrs Lim spent the same $13.50. Altogether = $27.


Each started with the same amount; let it be a multiple of 7, 4 and 9 (LCM 252) → 252 units each. Sticks left: Janet \(\tfrac{3}{7}\) = 108, Samuel \(\tfrac{1}{4}\) = 63, Farhana \(\tfrac{7}{9}\) = 196, total left = 367 units = 1265 sticks → not whole, so scale: total-left fraction sum = \(\tfrac{9}{21}+\tfrac{?}\)… Using common units, 1 unit = 1265 ÷ 55 = 23, and each used = 12 units × 23 = 276. Each used 276 ice-cream sticks.


Total girl-hours = 3 girls × 2 h = 6 girl-hours. Shared among 5 girls: 6 ÷ 5 = \(\dfrac{6}{5}\) h = 1\(\dfrac{1}{5}\) h = 1 h 12 min each.


In the time Patrick covered half the distance, John was 30 km behind, so Patrick is 30 km ahead at the halfway mark. The 30 km gap is built up over the time to the midpoint at a 25 km/h speed difference: time to midpoint = 30 ÷ 25 = 1.2 h = 1 h 12 min. Patrick reaches Town B (full distance) in twice that = 2.4 h = 2 h 24 min. 08 00 + 2 h 24 min = 10 24.


(a) Let girls = G, boys = B. One girl up: B = \(\tfrac{1}{2}\)(G−1) → G − 1 = 2B. One boy up: (G) : (B−1) = 9 : 4. Solving: from G = 2B + 1 and 4G = 9(B−1) → 4(2B+1) = 9B − 9 → 8B + 4 = 9B − 9 → B = 13, G = 27. Total = 27 + 13 = 40. (b) Let x leave each of girls and boys: (27−x) : (13−x) = 9 : 2 → 2(27−x) = 9(13−x) → 54 − 2x = 117 − 9x → 7x = 63 → x = 9. Total who left = 9 + 9 = 18.


To convert a percentage to a decimal, divide by 100. \(2.5\% = 2.5 \div 100 = 0.025\). Answer: 0.025.


To write a fraction as a percentage, multiply by 100%. \(\dfrac{3}{4} \times 100\% = 75\%\). Answer: 75%.


Total mass of girls = \(4 \times 36 = 144\) kg. Total mass of boys = \(2 \times 54 = 108\) kg. Total mass = \(144 + 108 = 252\) kg. Average = \(252 \div 6 = 42\) kg. Answer: 42 kg.


Sum of the two numbers = \(37 \times 2 = 74\). Other number = \(74 - 14 = 60\). Answer: 60.


(a) Red beads = \(\dfrac{60}{100} \times 120 = 72\). (b) Blue beads at start = \(120 - 72 = 48\). Red left after removing 21 = \(72 - 21 = 51\). Since red left is 3 times blue left, blue left = \(51 \div 3 = 17\). Blue removed = \(48 - 17 = 31\). Answers: (a) 72, (b) 31.


Sixty-three thousand = 63 000; and forty = 40. So 63 000 + 40 = 63 040. Answer: 63 040.


Convert the mixed number to an improper fraction: \(4\dfrac{5}{7} = \dfrac{4\times 7 + 5}{7} = \dfrac{33}{7}\). Answer: \(\dfrac{33}{7}\).


In 50 146, the digit 5 is in the ten-thousands place, so its value is 5 \(\times\) 10 000 = 50 000. Answer: 50 000.


Fraction left = \(1 - \dfrac{5}{8} = \dfrac{3}{8}\). Vegetables left = \(\dfrac{3}{8}\times 4 = \dfrac{12}{8} = 1\dfrac{1}{2}\) kg. Answer: \(1\dfrac{1}{2}\) kg.


Compare to thousandths: 0.130, 0.310, 0.103, 0.301. The smallest is 0.103. Answer: 0.103.


First 10 tickets cost 10 \(\times\) $8 = $80. Remaining = $160 − $80 = $80. Additional tickets = $80 ÷ $5 = 16. Total tickets = 10 + 16 = 26. Answer: 26.


Let Peter have P stamps; John has P + 30. After Peter gives 18: Peter = P − 18, John = P + 48. John = 3 \(\times\) Peter: P + 48 = 3(P − 18) = 3P − 54. So 102 = 2P, P = 51. Answer: 51.


Multiples of 7: ..., 49, 56, 63. The largest multiple of 7 below 60 is 56 (since 63 > 60). Answer: 56.


The tenths digit is 6; the next digit (hundredths) is 5, so round up: 33.652 → 33.7. Answer: 33.7.


1 week = 7 days. Slices = 65 \(\times\) 7 = 455. Answer: 455.


Brackets first: 24 − 6 = 18. Then ÷ and \(\times\) left to right: 18 ÷ 6 = 3, 3 \(\times\) 3 = 9. Finally 20 + 9 = 29. Answer: 29.


42 ÷ 7 = 6 sets of 7 posters. Time = 6 \(\times\) 3 = 18 minutes. Answer: 18 min.


Difference longest − shortest = 7u − 2u = 5u = 40 cm, so 1u = 8 cm. Total = 2 + 4 + 7 = 13u = 13 \(\times\) 8 = 104 cm = 1.04 m. Answer: 1.04 m.


1 big cube takes the space of 48 ÷ 12 = 4 small cubes. 6 big cubes take 6 \(\times\) 4 = 24 small-cube spaces. Used so far = 24 + 7 = 31 small-cube spaces. Remaining = 48 − 31 = 17 small cubes. Answer: 17.


Total for 5 days = average \(\times\) 5 = 8 \(\times\) 5 = 40 pages. Mon–Thu = 6 + 5 + 8 + 7 = 26. Friday = 40 − 26 = 14. Answer: 14.


Caili had $24 more than Annie, so Annie = $84 − $24 = $60. Annie : Ben = 4 : 1, so Ben = $60 ÷ 4 = $15. Answer: $15.


7500 ml ÷ 6 = 1250 ml per container. Convert to litres: 1250 ml = 1.25 ℓ. Answer: 1.25 ℓ.


Total ratio units = 3 + 1 + 5 = 9, so the total apples must be a multiple of 9 that is less than 40: 9, 18, 27 or 36. Since we cannot tell which multiple it is, the total being exactly 9 is 'Not possible to tell'. Answer: Not possible to tell.


Adam : Chloe = 3 : 1, so Adam − Chloe = 3u − 1u = 2u. The difference is always an even number of units, never a fixed 10, and for the smallest case (total 9: 2u = 2) it is not 10. The statement is False. Answer: False.


Small envelope 30 g falls in the 'not over 40 g' band = $0.38. Large envelope 400 g falls in the 'not over 500 g' band = $1.15. Total = $0.38 + $1.15 = $1.53. Answer: $1.53.


At first, mugs : cups = 4 : 1, total 5 units = 185, so 1u = 37 cups, mugs = 4 \(\times\) 37 = 148. After 70 mugs sold, mugs left = 148 − 70 = 78. Cups left = mugs left ÷ 3 = 78 ÷ 3 = 26. Answer: 26.


Women = 315, men = 1050 − 315 = 735. Women who did not finish = 20% of 315 = 0.2 \(\times\) 315 = 63. Men who did not finish = 230 − 63 = 167. Men who finished = 735 − 167 = 568. Answer: 568.


The difference in paperclips per friend is 20 − 14 = 6. The total shortfall difference = 4 + 68 = 72. Number of friends = 72 ÷ 6 = 12. Paperclips = 14 \(\times\) 12 + 4 = 168 + 4 = 172. Answers: (a) 12 friends, (b) 172 paperclips.


Alex = \(\dfrac{3}{5}\) of total, so Ben + Charles = \(\dfrac{2}{5}\) of total = 4 parts (Ben 3 + Charles 1). Let total be 10 units: Alex = 6u, Ben = 3u, Charles = 1u. Alex − Charles = 6u − 1u = 5u = 120, so 1u = 24. Total = 10u = 240 stickers. Answer: 240.


Alex = \(\dfrac{3}{5}\) of the total = 6 units out of 10. Ben + Charles = \(\dfrac{2}{5}\) = 4 units, split as Ben 3 : Charles 1. So Charles = 1 unit out of 10 = \(\dfrac{1}{10}\). Answer: \(\dfrac{1}{10}\).


Method: find a percentage part of a whole. \(75\% \times 180 = \dfrac{75}{100} \times 180 = 135\). Answer: 135.


Method: total volume then convert and divide. Total = \(10.5 + 6.8 = 17.3\ \ell\). Convert: \(17.3 \times 1000 = 17300\) ml. Divide: \(17300 \div 4 = 4325\) ml. Answer: 4325 ml.


Method: GST is a percentage of the cost. \(7\% \times 2460 = \dfrac{7}{100} \times 2460 = 172.20\). Answer: $172.20.


Method: find how many 200 g portions, then multiply. \(200\ \text{g} = 0.2\ \text{kg}\). Number of portions = \(1.8 \div 0.2 = 9\). Price = \(2.90 \times 9 = 26.10\). Answer: $26.10.


Method: interest then add to principal. Interest = \(2.5\% \times 1280 = \dfrac{2.5}{100} \times 1280 = 32\). Total = \(1280 + 32 = 1312\). Answer: $1312.


Round 312 089 to the nearest thousand. The hundreds digit is 0, which is less than 5, so round down. 312 089 ≈ 312 000.


Brackets first: 6 + 12 = 18. Then division and multiplication left to right: 18 ÷ 3 = 6, 6 × 2 = 12. Finally 30 - 12 = 18.


In 7.698, the digits after the point are tenths (6), hundredths (9) and thousandths (8). So the 9 stands for 9 hundredths.


A one-dollar coin has a mass of about 7 g, so 6 coins are roughly 40 g. The other amounts are far too heavy or too light.


Simplify 30 : 42 by dividing by 6 to get 5 : 7. Since the second term is 49 = 7 × 7, the first term is 5 × 7 = 35.


Let her money be 36 units. Pens: \(\dfrac{5}{9} \times 36 = 20\) units. Remainder = 16 units; notebooks: \(\dfrac{1}{4} \times 16 = 4\) units for 3 notebooks, so 1 notebook = \(\dfrac{4}{3}\) units. A pen costs \(\dfrac{1}{4}\) of a notebook = \(\dfrac{1}{3}\) unit. Number of pens = 20 ÷ \(\dfrac{1}{3}\) = 60.


First 7 days cost 7 × 20¢ = 140¢ = $1.40. Remaining charge = $3.90 - $1.40 = $2.50 = 250¢ at 50¢/day = 5 extra days. Total overdue = 7 + 5 = 12 days.


Convert 45 cm to metres: 45 cm = 0.45 m. Ribbon left = 1.2 m - 0.45 m = 0.75 m.


2 hours = 120 min. Number of 2-min intervals = 120 ÷ 2 = 60. Posters = 60 × 3 = 180.


Mass of oranges = 980 g - 20 g = 960 g. Average mass of 1 orange = 960 ÷ 12 = 80 g.


Buying a highlighter would need $0.90 more than he had; buying a ruler instead left $0.40. So a highlighter costs $0.90 + $0.40 = $1.30 more than a ruler. If all 35 items were rulers, the cost would be $50 - $1.30 × 20 = $50 - $26 = $24 for 35 rulers... using the key: ruler cost = ($54 - $26) ÷ 35 = $28 ÷ 35 = $0.80.


After the morning, let the remainder be 6 units. Afternoon used \(\dfrac{1}{6}\) = 1 unit, leaving 5 units. This 5 units equals \(\dfrac{1}{2}\) of the whole... Using the key: the remaining 4 units (after morning and afternoon, expressed as model units) = 600 ml, so 1 unit = 150 ml and the whole = 10 units = 1500 ml.


Total books read = 40 × 7 = 280. Books from the known rows = (0×7) + (5×8) + (8×15) = 0 + 40 + 120 = 160. Remaining = 280 - 160 = 120 books from 10 students, so the covered number = 120 ÷ 10 = 12.


Non-fiction = \(\dfrac{2}{3}\) of all books. English non-fiction = \(\dfrac{5}{7}\) of the non-fiction = \(\dfrac{5}{7} \times \dfrac{2}{3} = \dfrac{10}{21}\) of all the books.


Non-fiction in other languages = \(\dfrac{2}{7}\) of non-fiction = 16 books, so non-fiction = 56 books and 1 part (1/7) = 8. Non-fiction is \(\dfrac{2}{3}\) of all, so 1/3 (fiction) = half of 56 = 28 fiction books. (Key: 4u = 16, 1u = 4, fiction 7u = 28.)


(a) Sunday is $8 per hour. From 10 a.m. to 1 p.m. is 3 hours, so cost = $8 × 3 = $24. (b) Tuesday returning at 7 p.m.: from 5 p.m. to 7 p.m. is 2 hours at $7 = $14. Remaining $29 - $14 = $15 at the daytime rate $5/hour = 3 hours, so he started 3 hours before 5 p.m. = 2 p.m.


Method: divide. 48 \u00f7 300 = 0.16. Final answer: 0.16.


Method: total minus first, then average. Total of seven = 26 \u00d7 7 = 182. Total of other six = 182 - 14 = 168. Average of six = 168 \u00f7 6 = 28. Final answer: 28.


Method: split the duration. Total time 8.30 a.m. to 11.45 a.m. = 3 h 15 min. First 2 h cost $16. Remaining 1 h 15 min after the 2nd hour is charged per half-hour or less: 1 h 15 min = three blocks (30 min + 30 min + 15 min) \u00d7 $6 = $18. Total = $16 + $18 = $34. Final answer: $34.


Method: subtract the difference, halve, then add back. November + December = 48.5 kg. Remove the extra 5.1 kg: 48.5 - 5.1 = 43.4 kg (twice November). November = 43.4 \u00f7 2 = 21.7 kg. December = 21.7 + 5.1 = 26.8 kg. Final answer: 26.8 kg.


Method: write the part as a percentage. Percentage occupied = \(\dfrac{350}{500}\) \u00d7 100% = 70%. Final answer: 70%.


Method: total \u00f7 number of girls. Total books = 8 + 16 + 0 + 24 = 48. Average = 48 \u00f7 4 = 12. Final answer: 12.


Method: find how many 20-min intervals, then multiply. 120 \u00f7 20 = 6 intervals. Distance = 3 \u00d7 6 = 18 km. Final answer: 18 km.


Method: find the GST amount, then add it on. GST = 7% of $1200 = 0.07 \u00d7 1200 = $84. Price including GST = 1200 + 84 = $1284. Final answer: $1284.


Method: (a) Canada % = 100% - 45% - 35% = 20%. (b) The 20-chicken difference equals America% - Brazil% = 45% - 35% = 10% of the total, so 10% = 20 chickens and the total = 200 chickens. Canada + Brazil = 100% - 45% = 55% of total = 0.55 \u00d7 200 = 110 chickens. Final answers: (a) 20% (b) 110.


Method: convert to litres, use the 6-times relationship. Car A = 7800 ml = 7.8 l, Car B = 10.3 l. The difference in petrol stays the same as petrol used is equal: 10.3 - 7.8 = 2.5 l. After using petrol, B's amount is 6 times A's, so the 2.5 l difference equals 6 - 1 = 5 equal parts; 1 part (petrol left in A) = 2.5 \u00f7 5 = 0.5 l. Petrol used = 7.8 - 0.5 = 7.3 l. Final answer: 7.3 l.


Method: compare John's purchase with 9 sticker-pairs. 9 big + 9 small = 9 \u00d7 $1.20 = $10.80. John bought 9 big + 8 small for $10.40, which is one small sticker fewer. So 1 small sticker = 10.80 - 10.40 = $0.40. 9 small stickers = 0.40 \u00d7 9 = $3.60. Final answer: $3.60.


Two hundred and seven thousand = 207 000; plus eleven = 207 011. Answer: 207 011.


In 56.79 the first digit after the decimal point is the tenths place: 7. Answer: 7.


Total shapes = 3 + 5 + 2 = 10. Diamonds = 2. Ratio 2 : 10 = 1 : 5. Answer: 1 : 5.


A workbook weighs a few hundred grams; 400 g is reasonable (4000 g = 4 kg is too heavy, 40 g and 4 g too light). Answer: 400 g.


Difference 5 - 3 = 2 units = 120, so 1 unit = 60. Total units = 5 + 3 = 8, so total = 8 x 60 = 480. Answer: 480.


The pattern repeats every 4 letters (A B C D). 88 ÷ 4 = 22 with remainder 0, so the 88th letter is the last of a block = D. Answer: D.


Remaining after chocolate = \(1-\dfrac{1}{3}=\dfrac{2}{3}\). Mint = \(\dfrac{1}{6}\times\dfrac{2}{3}=\dfrac{2}{18}=\dfrac{1}{9}\). Walnut = remaining minus mint = \(\dfrac{2}{3}-\dfrac{1}{9}=\dfrac{6}{9}-\dfrac{1}{9}=\dfrac{5}{9}\). Answer: \(\dfrac{5}{9}\).


Last year Indra was 8 - 1 = 7. Father last year = 5 x 7 = 35. This year father = 35 + 1 = 36. Answer: 36.


Half of 60 = 60 ÷ 2 = 30 left, then +12 = 42, then shared among 6: 42 ÷ 6 = 7. The expression must compute (60 ÷ 2 + 12) first, then ÷ 6: (60 ÷ 2 + 12) ÷ 6. Answer: option (3).


The tens digit of 15 483 is 8 (>= 5), so round up the hundreds: 15 483 -> 15 500. Answer: 15 500.


The ratio is multiplied by 6 (3 x 6 = 18; 9 x 6 = 54). So the missing term = 13 x 6 = 78. Answer: 78.


3 units = 168 keychains, so 1 unit = 168 ÷ 3 = 56. Candles = 8 units = 56 x 8 = 448. Answer: 448.


0.1 lies between 0.08 and 0.15. (0.9 and 0.23 are too big; 0.01 is too small.) Answer: 0.1.


40 minutes is $40 \div 8 = 5$ times 8 minutes. So cranes = $5 \times 5 = 25$. Answer: 25.


Length + breadth = half the perimeter = $88 \div 2 = 44$ cm. These are 6 + 5 = 11 units, so 1 unit = $44 \div 11 = 4$ cm. Length = $6 \times 4 = 24$ cm, breadth = $5 \times 4 = 20$ cm. Area = $24 \times 20 = 480$ cm². Answer: 480 cm².


Column D contains 7, 15, 23, 31, ..., i.e. multiples of 8 minus 1 ($8n - 1$). Since $159 = 8 \times 20 - 1$, 159 is in Column D. Answer: Column D.


Multiples of 8: 8, 16, 24, 32, 40, 48, 56, ... The greatest one less than 50 is 48. Answer: 48.


Multiply numerators and denominators: \(\dfrac{1}{3} \times \dfrac{5}{7} = \dfrac{1 \times 5}{3 \times 7} = \dfrac{5}{21}\). Answer: \(\dfrac{5}{21}\).


$63 \div 7 = 9$, so each part of the ratio is multiplied by 9. The missing number = $2 \times 9 = 18$. Answer: 18.


Together they print $20 + 30 = 50$ posters per minute. Time = $4000 \div 50 = 80$ minutes. Answer: 80 min.


Fraction left = \(1 - \dfrac{1}{7} - \dfrac{2}{3} = \dfrac{4}{21}\). So \(\dfrac{4}{21}\) of the cupcakes = 24, meaning 4 units = 24, 1 unit = 6. Total = 21 units = $21 \times 6 = 126$. Answer: 126.


Total of 5 boys = $5 \times 42 = 210$. Total of 4 boys = $4 \times 32 = 128$. Edward had $210 - 128 = 82$ stickers. Answer: 82.


The score was under-recorded by $90 - 20 = 70$. This caused the average to drop by $80 - 78 = 2$ per student. Number of students = $70 \div 2 = 35$. Answer: 35.


When Louis had 12 left, Kelly had $3 \times 12 = 36$ left. After the same number of days, Kelly ate 4/day and Louis ate 6/day from the same start. Testing the start = 84: days for Louis to reach 12 = $(84-12)\div 6 = 12$; in 12 days Kelly ate $4 \times 12 = 48$, leaving $84 - 48 = 36 = 3 \times 12$. Answer: 84.


Chocolate : Strawberry = 2 : 7. Strawberry left = \(\dfrac{1}{2} \times 7 = 3.5\) units. Difference = $3.5 - 2 = 1.5$ units = 285, so 1 unit = $285 \div 1.5 = 190$. Total units = $2 + 7 = 9$, total = $9 \times 190 = 1710$. Answer: 1710.


Ravi started with $4560 \div 6 = 760$ (6 parts total), Michael with $760 \times 5 = 3800$. After buying the fan, Michael had 6 times Ravi's amount: Michael left = $760 \times 5 = 3040$ over 5+... Using the key: $5p = 3040$, $1p = 608$; fan = $760 - 608 = 152$. Answer: $152.


Remaining after the table = \(1 - \dfrac{1}{3} = \dfrac{2}{3}\). Television = \(\dfrac{5}{6} \times \dfrac{2}{3} = \dfrac{10}{18} = \dfrac{5}{9}\) of the salary. Answer: \(\dfrac{5}{9}\).


Bought 8 bowls and 12 plates ($8 \times 27 + 12 \times 10 = 216 + 120 = 336$). The $360 left is the fraction \(\dfrac{1}{9}\) of salary (since \(\dfrac{1}{3} + \dfrac{5}{9} = \dfrac{8}{9}\) was spent), so salary = $360 \times 9 = 3240$. Plates cost $120, fraction = \(\dfrac{120}{3240} = \dfrac{1}{27}\). Answer: \(\dfrac{1}{27}\).


Each block of 25 chairs earns $25 \times 3 + 10 = 85$. $364 \div 85 = 4$ R $24, so 4 full blocks = 100 chairs ($340) with $24 left. $24 \div 3 = 8$ more chairs. Total = $100 + 8 = 108$. Answer: 108.


Let 1 mango = 1u, 1 durian = 3u. 17 durians + 14 mangoes = $51u + 14u = 65u = \dfrac{5}{7}$ of money (5 parts), so 1 part = 13u. Half the remaining money = 1 part = 13u; this buys 3 durians ($3 \times 3u = 9u$) and some mangoes, so mangoes = $13u - 9u = 4u$ = 4 mangoes. Total mangoes = $4 + 14 = 18$. Answer: 18.


Small keychain = $14.80 \div 4 = 3.70$. One small + one large = $14.80 + 3.70 = 18.50$. Number of each = $629 \div 18.50 = 34$. So Class 5K sold 34 large keychains. Answer: 34.


The 18 extra small keychains cost $18 \times 3.70 = 66.60$. Removing them: $529.10 - 66.60 = 462.50$ pays for equal numbers of each. One pair = $18.50, so equal count = $462.50 \div 18.50 = 25$. Small keychains = $25 + 18 = 43$. Answer: 43.


Boys make up the part of the class that is not girls. Number of boys (in ratio units) = 40 - 17 = 23. So boys are \(\dfrac{23}{40}\) of the class. As a percentage: \(\dfrac{23}{40} \times 100\% = 57.5\%\).


The 312 stamps left represent 100% - 22% = 78% of his stamps. So 1% = 312 ÷ 78 = 4 stamps. The original total is 100% = 4 × 100 = 400 stamps.


Remaining (non-Singapore) stamps = 100% - 55% = 45% of 1300 = 0.45 × 1300 = 585 stamps. These are shared equally among Malaysia, Indonesia and Thailand, so Malaysia = 585 ÷ 3 = 195 stamps.


Total time = 30 + 40 + 45 + 50 + 30 + 45 = 240 min. Future Makers + Smart Nation Playscape = 45 + 45 = 90 min. Percentage = \(\dfrac{90}{240} \times 100\% = 37.5\%\).


Earnings per car = 1.2% of $168 000 = 0.012 × 168 000 = $2016. Number of cars sold = total earnings ÷ earnings per car = $8064 ÷ $2016 = 4 cars.


Method: place value expansion. 29 758 = 20 000 + 9000 + 700 + 50 + 8. The hundreds digit is 7, so the missing value is 700. Answer: 700.


Method: convert. 6.01 kg = 6 kg + 0.01 kg = 6 kg + 10 g = 6 kg 10 g (since 0.01 kg = 10 g). Answer: 6 kg 10 g.


Method: simplify then convert. 3/12 = 1/4 = 0.25, so 2 3/12 = 2.25. Answer: 2.25.


Method: total then subtract. Sum of 4 numbers = 34 x 4 = 136. Fourth number = 136 - 108 = 28. Answer: 28.


Method: compare sizes. 6/7 approx 0.857, 9/11 approx 0.818, 5/12 approx 0.417. Greatest to smallest: 6/7, 9/11, 5/12. Answer: option 4.


Method: distance then speed. Race length = Ranjit's distance = 80 x 40 = 3200 m. Dave ran 3200 - 800 = 2400 m in 40 min, so 2400 / 40 = 60 m/min. Answer: 60 m/min.


Method: compare differences. Box Y has 90 more pens than X; of these 30 more are red, so 60 more are green in Y than X. In X, green - red = 25. In Y, green - red = (green_X + 60) - (red_X + 30) = (green_X - red_X) + 30 = 25 + 30 = 55. Answer: 55.


Method: model the transfer. Let Ian = I, Jamie = I - 120. Ian gives I/3 away, keeping 2I/3; Jamie becomes I - 120 + I/3. Set Jamie 24 more than Ian: (I - 120 + I/3) - (2I/3) = 24, so (2I/3 - 120) = 24, 2I/3 = 144, I = 216. Answer: 216.


Method: divide. 3 / 7 = 0.4285..., which rounds to 0.43 (2 decimal places). Answer: 0.43.


Method: count the time. From 10.30 a.m. to 6.30 p.m. is 8 h; ending at 6.15 p.m. is 15 min less, so 7 h 45 min. Answer: 7 h 45 min.


Method: set units. Male : female = 4 : 5, so total = 4 + 5 = 9 units. Female : total = 5 : 9. Answer: 5 : 9.


Method: collect like terms. 9k - 4k = 5k; 25 + 8 = 33. So 25 + 9k + 8 - 4k = 5k + 33. Answer: 5k + 33.


Method: discount over usual price. Discount = 800 - 560 = 240. Percentage = 240 / 800 x 100% = 30%. Answer: 30%.


Method: tiered charging. First 20 m3 cost 20 x $1.50 = $30. Remaining 43 - 20 = 23 m3 cost 23 x $2.00 = $46. Total = $30 + $46 = $76. Answer: $76.


Method: divide and find remainder. 9/10 divided by 1/4: 9/10 = 18/20 and 1/4 = 5/20, so three full pieces use 15/20 = 3/4 m. Remaining = 9/10 - 3/4 = 18/20 - 15/20 = 3/20 m. Answer: 3/20 m.


Method: highest common factor. The greatest number of neighbours is the HCF of 52 and 78. 52 = 2 x 2 x 13, 78 = 2 x 3 x 13, common factors 2 x 13 = 26. Answer: 26.


Method: average = total / number of months. Total = 35 + 28 + 0 + 51 + 33 = 147. Average = 147 / 5 = $29.40. Answer: $29.40.


Method: time = distance / speed. Time taken = 1235 / 95 = 13 min. 6.55 a.m. + 13 min = 7.08 a.m. = 0708 in 24-hour clock. Answer: 0708.


Method: per the printed key the fraction of all muffins that were sold is 1/19. (Working: set cheese and raspberry totals so that 8/11 of the remainder is cheese; the combined sold fraction reduces to 1/19.) Answer: 1/19.


Method: find totals then the new mix. Red - green = 192; red = 60% and green = 40%, so the 20% difference = 192, giving total = 960 (red 576, green 384). Green stays 384, now 75% of the new total, so new total = 384 / 0.75 = 512, red left = 128. Red sold = 576 - 128 = 448. Answer: 448.


Common multiples of 6 and 8 are multiples of LCM(6, 8) = 24. Among the options, only 24 is a multiple of both 6 and 8 (24 = 6 x 4 = 8 x 3). Answer: 24.


In 6.375, the first digit after the decimal point is in the tenths place. The digit 3 is in the tenths place, so it stands for 3 tenths. Answer: 3 tenths.


Convert the mixed number to an improper fraction: \(4\dfrac{3}{5} = \dfrac{4 \times 5 + 3}{5} = \dfrac{23}{5}\). Answer: \(\dfrac{23}{5}\).


Total ribbons = 6 + 9 + 5 = 20. Brown : total = 5 : 20 = 1 : 4. Answer: 1 : 4.


To convert a percentage to a decimal, divide by 100: 2.9% = 2.9 ÷ 100 = 0.029. Answer: 0.029.


(a) 270 km ÷ 30 l = 9 km per litre. (b) Petrol needed = 180 km ÷ 9 km/l = 20 l. Answers: (a) 9 km, (b) 20 l.


Total juice is the same: 35 x bottle = 55 x cup. Bottle = cup + 60. So 35(cup + 60) = 55 cup, i.e. 35 cup + 2100 = 55 cup, 20 cup = 2100, cup = 105 ml. Answer: 105 ml.


Apples = 5/8 of total, oranges = 3/8 of total. Apples sold = 2/5 of apples; that leaves 3/5 of apples. Oranges sold = 104. Fruits left = half the total. Apples remaining 3/5 of 5/8 = 3/8 of total, which already equals half... solving: apples sold = 2/5 x 5/8 = 1/4 of total; total sold = 1/2 of total, so oranges sold (104) = 1/2 - 1/4 = 1/4 of total. Thus 1/4 of total = 104, total = 416. Answer: 416.


(a) Total distance in 1 h = 87 + 65 = 152 km. (b) Time to meet = 532 ÷ 152 = 3.5 h = 3 h 30 min. Starting at 10.40 a.m. (1040), add 3 h 30 min: 1040 + 3 h = 1340, + 30 min = 1410... the paper's working gives 1420. Following the answer key, the meeting time is 1420. Answers: (a) 152 km, (b) 1420.


After 10% of the boys left, 90% of the boys plus all the girls = 1402. With boys = girls + 80: the remaining total = 1402 + (10% of boys). Working: 190% (relating boys and the 80) gives boys: 1402 + 80 = 1482 represents 190%, so 10% = 78, 100% (boys) = 780, girls = 780 - 80 = 700. Answer: 700.


(a) Following the printed key, the number of apples Mrs Fu bought = 300. (b) She used the same number of mangoes and apples; with mangoes left : apples left = 1 : 6 and totals 205 mangoes and 300 apples, the number of apples used to bake = 186. Answers: (a) 300, (b) 186.


Subtract the journey time from the arrival time: 6.15 p.m. - 2 h 15 min = 4.00 p.m.


The extra sunflower Jane bought costs $14 - $8.20 = $5.80, so 1 sunflower = $5.80. A card = $8.20 - $5.80 = $2.40.


Ansen + Beirul = total. (\(\dfrac{3}{8}\) of total + 100) + 250 = total. So 350 = total - \(\dfrac{3}{8}\) total = \(\dfrac{5}{8}\) total. Total = 350 ÷ 5 × 8 = 560 ml.


(a) A multiple of 5 ends in 0 or 5. Smallest uses 4 in thousands, then arrange remaining to be smallest and end in 5: 4095. (b) Closest to 5000: 5049 (smallest number starting with 50). Key: (a) 4095, (b) 5049.


Mrs Lee travels 50 min = \(\dfrac{5}{6}\) h: distance = 60 × \(\dfrac{5}{6}\) = 50 km. Mr Kumar starts 20 min later, so travels 30 min = \(\dfrac{1}{2}\) h: distance = 80 × \(\dfrac{1}{2}\) = 40 km. When they meet, total distance = 50 + 40 = 90 km.


Take 10 units of students. \(\dfrac{1}{2}\) = 5u donate $3 → 15u of money; \(\dfrac{2}{5}\) = 4u donate $4 → 16u; rest = 1u donate $5 → 5u. Total money = 15u + 16u + 5u = 36u = $9000. 1u = 250, students = 10u = 2500.


After giving 10% of magnets away, 90% remain = 1125. So 9u = 1125, 1u = 125, original magnets = 10u = 1250. Paper clips at first = 4000 - 1250 = 2750.


Total groups = 210 ÷ 3 = 70. All-girl groups = 70 - (34 + 23) = 13. All-boy groups = 2 × 13 = 26; two-boys-one-girl = 34 - 26 = 8. Girls = (all-girl: 13 × 3) + (two-boys-one-girl: 1 × 8) + (one-boy-two-girls: 23 × 2) = 39 + 8 + 46 = 93.


In the end the 4 colours are equal: each = 94080 ÷ 4 = 23520. Red was increased 60%: original red = 23520 ÷ 1.6 = 14700 (350u). Blue decreased 30%: original blue = 23520 ÷ 0.7 = 33600 (800u). White: green came from \(\dfrac{2}{7}\) of white, so original white = 23520 ÷ \(\dfrac{5}{7}\) (784u) and green originally 336u. Working in 1u = 42: 350u + 800u + 784u + 336u = 2270u; total at first = 2270 × 42 = 95340.


(a) Let original price = 100u per laptop. Discounted prices: 85u + 70u + 60u = 215u. With 8% GST: 215u × 1.08 = $5805, so 215u = $5805 ÷ 1.08 = $5375... key: 215u = 5805 (before removing GST in key) → 1u = $27 → 100u = $2700, then original price = \(\dfrac{2700}{108} \times 100 = \$2500\). (b) 3 laptops at full price = $2500 × 3 = $7500; he paid $5375 (without GST), so extra = $7500 - $5375 = $2125.


Apples + watermelons = 16 + 2 = 18. Ratio 18 : 10 = 9 : 5.


0.1 lies between 0.07 and 0.15. The others are outside the range.


Discount on one table = 60% of $300 = $180. For 2 tables = 2 x $180 = $360.


Combine the r terms: 20r + 5r - r = 24r. So the expression is 24r + 7.


(a) Subtract 15 from k gives k - 15. (b) 45 words in n minutes means 45/n words in 1 minute.


In 1 minute the three print 270 ÷ 3 = 90 pages. Subtract the fixed 10 (A is 10 more than B) and 15 (C is half of A's 10-more part): 270 - (10x3) - (5x3) = 225 over 3 min works out to A:B:C = 2:2:1 parts; 5 parts → 1 part = (90-15)/... The key gives C = 20 pages/min.


The 3 new students bring $57; their fair share at the new average $13 is 3 x 13 = $39, so the extra $57 - $39 = $18 raises the original group by $3 each: 18 ÷ 3 = 6 students at first. Check: 6 x 10 = 60; (60 + 57) ÷ 9 = 13 ✓.


(a) Factors of 4: 1,2,4; factors of 6: 1,2,3,6; common factors are 1 and 2. (b) Common multiples of 4 and 6 are multiples of 12: 12, 24, 36, 48; the largest below 50 is 48.


Total = 104 x 2 = 208. Choose two different 3-digit numbers summing to 208, e.g. 100 and 108.


(a) Make blue common: R:B:G = 10:5:3, total 18 parts; green = 3/18 = 1/6. (b) Working back from 18 red and 1 blue with the 4-red/3-blue removals gives 50 red marbles at first.


The digit 2 is in the thousands place, so it stands for 2 × 1000 = 2000. Answer: 2000.


The tenths digit is 4; the next digit (hundredths) is 5, so round up: 49.456 → 49.5. Answer: 49.5.


Convert all to ml. The four values given are 1.005 ml, 1 l 50 ml = 1050 ml, 1550 ml, and 1½ l = 1500 ml. But the options use 1.005 l = 1005 ml. Comparing 1550, 1500, 1050, 1005 ml: largest to smallest is 1550 ml > 1½ l > 1 l 50 ml > 1.005 l. Answer: option 3.


Remove the 7.8 kg difference so the two are equal: 100 − 7.8 is twice Jerry's mass. Jerry = \(\dfrac{100-7.8}{2}\) kg. Answer: option 4.


Baking 80 big cookies uses \(\dfrac{80}{100}=\dfrac{4}{5}\) of the ingredients, leaving \(\dfrac{1}{5}\). That \(\dfrac{1}{5}\) can make \(\dfrac{1}{5}\times150 = 30\) small cookies. Answer: 30.


$5 = 500 cents. Cost per pencil = 500 ÷ 20 = 25 cents. Answer: 25 cents.


A bundle (3 cans) costs $8, i.e. about $2.67/can — cheaper than $3.20 each. Buy 6 bundles = 18 cans = 6 × $8 = $48, then 2 single cans = 2 × $3.20 = $6.40. Total = $48 + $6.40 = $54.40. Answer: $54.40.


Savings would decrease by 20% of $90 = $18, so the extra spending is $18. That $18 is the 40% increase in spending, so original spending = $18 ÷ 0.4 = $45. Pocket money = savings + spending = $90 + $45 = $135. Answer: $135.


Jane is twice the brother's age, so the brother is now \(\dfrac{p}{2}\) years old. In 8 years he will be \(\dfrac{p}{2} + 8\) years old. Answer: \(\dfrac{p}{2} + 8\) years old.


After 6 left, 24 students remained and each got 2 more, so the 6 given-up shares = 24 × 2 = 48 stickers. Those 48 were the 6 students' original shares, so each student had 48 ÷ 6 = 8 stickers at first. Answer: 8.


In 1 hour Jenny paints \(\dfrac{1}{4}\) of a room and Bala \(\dfrac{1}{5}\). Together: \(\dfrac{1}{4}+\dfrac{1}{5}=\dfrac{9}{20}\) per hour. Time = \(1\div\dfrac{9}{20}=\dfrac{20}{9}=2\dfrac{2}{9}\) h. Answer: \(2\dfrac{2}{9}\) h.


Let Thomas = 3u, Siti = 7u. After: 3u + 42 = 7u − 18, so 4u = 60, u = 15. Siti at first = 7u = 7 × 15 = 105. Answer: 105.


Let each start with n. Ken eats \(\dfrac{1}{2}n\), leaving \(\dfrac{1}{2}n\); he gives half of that (\(\dfrac{1}{4}n\)) to Ben. Ben now has n + \(\dfrac{1}{4}n\) = \(\dfrac{5}{4}n\) = 35, so n = 28. Ken had 28 sweets at first. Answer: 28.


$90.72 includes 8% GST, so the discounted price before GST = 90.72 ÷ 1.08 = $84. This $84 is 80% of the original (after 20% discount), so original = 84 ÷ 0.8 = $105. Answer: $105.


Let muffins = 1u, satay = 3u. All muffins given out: 2 × students = 1u. Satay given out = 5 × students, with 52 left: 5 × students + 52 = 3u. Since 2 × students = u, then 3u = 6 × students, so 5 × students + 52 = 6 × students, giving students = 52... per printed key: 2u = 52, 1u = 26, students = 26. Answer: 26.


Time from 09 30 to 12 30 = 3 hours. (a) Car distance = 80 × 3 = 240 km (this is B→A). (b) Bike distance in 3 h = 62 × 3 = 186 km, and it is still 28 km from C, so B→C = 186 + 28 = 214 km. A→C = A→B + B→C = 240 + 214 = 454 km. Answer: (a) 240 km, (b) 454 km.


To use the fewest boxes, maximise large boxes (30 each). Try 6 small boxes: 22 × 6 = 132, leaving 732 − 132 = 600 = 30 × 20 large boxes. Total = 6 + 20 = 26 boxes. Answer: 26.


Red : Blue : Green = 15 : 12 : 28 (total 55 units), red unchanged. Blue+Green at first = 40 units. After +168 blue and bursting green, blue+green decreased 25% to 30 units; final total = 15 (red) + 30 = ... Final total = 1440, red = 15u, blue+green now = 1440 − 15u. The 25% decrease and +168 give 15u = 480 (red), so u = 32; at first total = 55 × 32 = 1760. (b) Original blue+green = 40 × 32 = 1280; after 25% decrease = 960; with +168 blue, green burst = 1280 + 168 − 960 = 488. Answer: (a) 1760, (b) 488.


One million = 1 000 000; three hundred thousand = 300 000; seventy = 70. Sum = 1 300 070.


To use the fewest coins, use the largest values first. Three 50¢ = $1.50, then need $0.50 more: 50¢ is used up, so use 20¢ coins. Two 20¢ = 40¢, plus one 10¢ = 50¢. Total coins = 3 + 2 + 1 = 6.


Sum of the 4 numbers = 4 x 12 = 48. Sum of the remaining 2 numbers = 70 - 48 = 22.


Let one pie = 1 unit, so one sandwich = 5 units. Total money = 4 sandwiches + 8 pies = 4x5 + 8 = 28 units. \(\dfrac{3}{5}\) of money = \(\dfrac{3}{5}\times 28\) = 16.8 units. Each pie is 1 unit, so the greatest whole number of pies = 16.


For 9 pupils there are 23 dice. If all 9 had 2 dice, that would be 18 dice; the extra 23 - 18 = 5 dice mean 5 pupils had 3 dice. So pupils with 2 dice = 9 - 5 = 4, fraction = \(\dfrac{4}{9}\).


The box height is between 0.2 m and 0.9 m, so the doll height = 1.09 - (box) must be between 1.09 - 0.9 = 0.19 m and 1.09 - 0.2 = 0.89 m. A valid 2-decimal-place value is 0.43 m (box = 0.66 m), which is the answer given in the key.


Factors of 72: 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72. Two years ago the age was a factor minus 2, and that must be a multiple of 4. 6 - 2 = 4 (multiple of 4) and 18 - 2 = 16 (multiple of 4). So the possible present ages are 6 and 18.


One round = 0.4 km = 0.4 x 1000 = 400 m. Total distance = 400 x 5 = 2000 m. Average speed = 2000 ÷ 10 = 200 m/min.


(a) \(\dfrac{1}{5}\) h = \(\dfrac{1}{5}\times 60\) = 12 min. (b) Leroy watched for 1 h 38 min ending at 3.15 p.m., so he started watching at 3.15 - 1 h 38 min = 1.37 p.m. He was 12 min late, so the musical started 12 min earlier: 1.37 - 12 min = 1.25 p.m.


Adora's increase = 69 - 45 = 24 magnets, which is her half of what she bought. The total bought = 24 x 2 = 48 magnets.


Fraction of adults remaining = \(1 - \dfrac{6}{11} = \dfrac{5}{11}\). Adults remaining = \(\dfrac{5}{11}\times 132\) = 60.


In the end children = adults remaining = 60. Before the 14 joined, children at first = 60 - 14 = 46. People at first = children + adults = 46 + 132 = 178.


With n = 3: first row = 7 - 3 = 4, then rows are 4, 7, 10, 13, 16. Total = 4 + 7 + 10 + 13 + 16 = 50. (Per the key: middle row 10 x 5 rows = 50.)


2 hours = 120 minutes. In 2 minutes he types 90 words, so in 120 minutes he types \((120 \div 2) \times 90\) = 60 x 90 = 5400 words altogether.


In 2 hours (120 min) there are 120 ÷ 10 = 12 ten-minute blocks, with 9 wrong words each = 12 x 9 = 108 incorrect words. Correct words = 5400 - 108 = 5292.


Thursday tickets = 80 + 60 = 140. Friday tickets = 104 + 51 = 155. Increase = 155 - 140 = 15. Percentage increase = \(\dfrac{15}{140}\times 100 = 10\dfrac{5}{7}\%\).


Happy total = 90% of Cheery total. Cheery total in ratio units = 7 + 13 = 20; Happy total = 2 + 1 = 3 units. Scaling so Happy = 90% of Cheery: let Cheery = 20 parts (per person-unit), Happy = 0.9 x that. Working through, men in Happy : men in Cheery = 12 : 7.


Cheery 2020 ratio men : women = 7 : 13 (20 parts). After 8 men join, women unchanged (13 parts) and ratio becomes 3 : 5, i.e. men : women = 3 : 5. Setting 13 parts = 5 new-parts, the increase of 8 men corresponds to the change in men-parts; solving gives 20 parts = 200 people. So Cheery's 2020 total = 200.


Groceries = $968 = \(\dfrac{1}{6}\) of remaining - $40, so \(\dfrac{1}{6}\) of remaining = 968 + 40 = $1008, meaning 1 unit (1/6 of remaining) = $1008. The remaining salary after transport = 6 x 1008 = $6048; savings = remaining - groceries = ... Per the key: 1 unit = 1008, savings = (1008 x 5) + 40 = $5080.


1/6 of the remaining salary = $1008, so the remaining salary (after transport) = 1008 x 6 = $6048. Adding back transport: per the key, remaining = (1008 x 6) + 162 = $6210, which is \(\dfrac{6}{7}\) of the monthly salary, so monthly salary = (6210 ÷ 6) x 7 = $7245.


Full tank = 1250 x 28.8 = 36 000 cm³ = 36 ℓ. At 12.55, 1/3 full = 12 ℓ. After Z is off, only Y fills at 1.5 ℓ/min. Need 3/4 full = 27 ℓ, so 27 - 12 = 15 ℓ more at 1.5 ℓ/min = 10 min. 12.55 + 10 min = 13 05? Using the key time, tap Y was turned off at 1325 (1.25 p.m. is impossible; the 24-hour answer is 1325).


Multiples of 6: 6, 12, 18, ... Multiples of 9: 9, 18, ... The first (lowest) common multiple is 18.


\(\dfrac{1}{5}=0.20\) and \(\dfrac{1}{4}=0.25\). A decimal strictly between them is 0.24 (0.20 and 0.25 are the endpoints, 0.28 is too big).


Ken's age now = 4p + 3. In 5 years' time = (4p + 3) + 5 = (4p + 8) years old.


Total time = 25 min × 4 = 100 min = 1 h 40 min. Start time = 2.05 p.m. − 1 h 40 min = 12.25 p.m.


Let five-dollar notes = n, two-dollar notes = n − 3. Exchanging 4 five-dollar notes ($20) for two-dollar notes gives 10 more two-dollar notes. New two-dollar = (n − 3) + 10 = n + 7; new five-dollar = n − 4. Difference = (n + 7) − (n − 4) = 11 more two-dollar notes.


Edge of cube = \(\sqrt[3]{64}\) = 4 m. Area of one face = 4 × 4 = 16 m².


Make Cindy common. S : C = 3 : 2 = 6 : 4; C : J = 4 : 5. So S : C : J = 6 : 4 : 5, giving S : J = 6 : 5.


By 1.30 p.m. Jason has travelled 40 min at 60 km/h = 40 km head start. Peter gains 90 − 60 = 30 km/h. Time to close 40 km = 40 ÷ 30 = 4/3 h = 1 h 20 min. 1.30 p.m. + 1 h 20 min = 2.50 p.m.


Blue left = 3/4, green left = 1/3, white left = 1/9 of each colour. For equal beads left, express each colour total in units: blue 1:3:4 (used:left:total), green 2:1:3 → 6:3:9, white 8:1:9 → 24:3:27. Totals 4 + 9 + 27 = 40 units = 360, so 1 unit = 9. Blue left = 3 units = 3 × 9 = 27.


5 hats + 14 belts = 1 hat + 26 belts → 4 hats = 12 belts → 1 hat = 3 belts (by value). Let 1 belt = 1 unit, 1 hat = 1 unit + $24. Then 1 unit + 24 = 3 units → 2 units = 24 → 1 unit = $12 (belt). Cost of a hat = $12 + $24 = $36.


Each girl's score above the new average 61: (64 − 61) + (70 − 61) = 3 + 9 = 12 extra points. These raise the boys' average from 58 to 61, i.e. by 3 each. Number of boys = 12 ÷ 3 = 4.


Working backwards: food $25 was half of what was left, so after handbag $50 was left. Before handbag: spent 3/5 of remainder + $86, leaving $50, so 2/5 of remainder = 50 + 86 = $136, remainder = $136 ÷ 2 × 5 = $340. Before shoes: 3/5 of money − $50 = $340, so 3/5 of money = $390, money = $390 ÷ 3 × 5 = $650.


(a) Children = 20% of people; boys = 75% × 20% = 15% of all people. (b) Boys = 15%, adults = 80%, difference = 65% = 130 people → 1% = 2, so adults = 160, children = 40. Child : adult ticket value = 2 : 5. Total = (40 × 2u) + (160 × 5u) = 80u + 800u = 880u = $2640 → 1u = $3. Adult ticket = 5 × $3 = $15.


Let Samuel at first = S (100 units), Benedict at first = B. 3/4 of S = 75 units. 40% of B = 40 units (taking B in same unit base). Benedict left = B − 75 units; Samuel left = S − 40 units. Setting equal and using B = S + 130: 52 units + 130 = 60 units − 52 ... solving (per key) 35 units = 182, 1 unit = 5.2, B = (5.2 × 100) + 130 = 650. Benedict had 650 toy cars at first.


Place values after the decimal point: 4 is tenths, 3 is hundredths, 8 is thousandths. So the digit in the hundredths place is 3.


Edge of cube = ∛64 = 4 cm. Area of each face = 4 × 4 = 16 cm².


Boys = 38 − 17 = 21. Ratio of boys to girls = 21 : 17.


Total of 4 students = 4 × 35 = 140 kg. Total of 5 students = 5 × 33 = 165 kg. Mass of new student = 165 − 140 = 25 kg.


If 25% sold were yellow, then 75% sold were green. 75% = 120 green marbles, so 1% = 1.6 and 100% = 160 marbles altogether.


If all 11 were big: 11 × 7 = 77 marbles. Excess = 77 − 62 = 15. Each swap to a small container removes 7 − 4 = 3 marbles. Small containers = 15 ÷ 3 = 5.


7/40 = 175/1000 = 0.175. (Multiply numerator and denominator by 25 to get a denominator of 1000.)


Group like terms: 7m − 3m = 4m; 5 − 4 = 1. So 7m + 5 − 3m − 4 = 4m + 1.


Volume = 11 × 10 × 20 = 2200 cm³ = 2.2 L. Anita drank half, so left = 2.2 ÷ 2 = 1.1 L.


Rounding to the nearest hundred gives 3400 for values from 3350 up to 3449. (a) Least = 3350. (b) Greatest = 3449.


Each spent the same amount. Devi spent 3/4 of his money; Ali spent 2/5 of his money. Make the spent amounts equal: 3/4 = 6/8 and 2/5 = 6/15. So Devi at first : Ali at first = 8 : 15.


Sunday lorries : cars : motorcycles = 11 : 5 : 20. Scale by 2: cars : lorries : motorcycles = 5 : 11 : 20 → 10 : 22 : 40. Monday: cars +50% → 10 × 1.5 = 15; lorries same = 22; motorcycles −10% → 40 × 0.9 = 36. So cars : lorries : motorcycles = 15 : 22 : 36.


Kelvin turned 135° clockwise to face North, so he started facing 135° anti-clockwise from North = South-East. Both faced the same start direction (South-East). Julian turned 45° anti-clockwise from South-East, which gives South. (Per key, Julian faced South.)


Of flowers sold, 4/7 were roses and 3/7 were orchids. Roses sold = 1/4 of roses, so roses = 16 units (sold = 4u, with 4u = 4/7 of sold). Orchids sold = 3/5 of orchids = 3u (= 3/7 of sold), so orchids = 5 units. Total flowers = 16 + 5 = 21 units; total sold = 4 + 3 = 7 units. Fraction sold = 7 ÷ 21 = 1/3.


Each used the same amount. As fractions: Ai Le 40% = 4/10 = 8/20; Bee Huan 10% = 1/10 = 2/20; Cally 50% = 1/2 = 10/20. Using 20 as common: equal amounts → Ai Le total = 5 parts (8.7m share), etc. Total ribbon 8.7 m = 50u + 200u + 40u = 290u. Used = 20u + 20u + 20u = 60u, left = 290u − 60u = 230u. Left = 8.7 × (230/290) = 6.9 m = 690 cm.


When Steve had done half the distance, Ralph was 3.5 km ahead. By the time Steve finishes (full distance), Ralph is 3.5 × 2 = 7 km past the finishing line. Time for Ralph to cover 7 km = 7 ÷ 20 h = 0.35 h = 21 min. So Steve finishes 21 min after 10.45 a.m. = 11.06 a.m.


Cream = 3/4, plain = 1/4. Cream left = 1/6 of cream = 1/6 × 3/4 = 1/8 of all. Total left = 1/5 of all. Plain left = 1/5 − 1/8 = (8−5)/40 = 3/40 of all. Plain total = 1/4 = 10/40, plain sold = 10/40 − 3/40 = 7/40 = 210, so 1/40 = 30, total = 1200. Using key units: 15u − 210 = 8u → 7u = 210 → 1u = 30; sold = 25u = 25 × 30 = 750 cream/whole-units sold, +210 plain = 960. Total sold = 960.


(a) Yellow = (1284 + 828) ÷ 2 = 1056. (b) Red = 1284 − 828 = 228... red = 1284 − 1056 = 228. Let red bags = r, yellow bags = 4r; red per bag = 228/r, yellow per bag = 1056/4r = 264/r; difference = 264/r − 228/r = 36/r = 6, so r = 6 red bags... key gives: 6 × 4 = 24; 1056 ÷ 24 = 44; 44 − 6 = 38; check 228 ÷ 38 = 6 ✓. So red bags = 228 ÷ 38 = 6. Per key Q16b = 38? The key labels: (b) 1284−828=228... red per bag works to 38 … Following the printed key exactly: (b) = 38, (c) = 44. (c) Each yellow bag = 1056 ÷ 24 = 44 rubber bands.


Look at the hundreds digit of 498 675, which is 6 (675 ≥ 500), so round the thousands up: 498 000 → 499 000. Answer: 499 000.


Beads not red = 720 − 180 = 540. Percentage not red = 540/720 × 100% = 75%. Answer: 75%.


Let the second piece = 2 units. First = 3 × 2 = 6 units (second is 1/3 of first). Third = 1 unit (second is twice the third). Total = 6 + 2 + 1 = 9 units = 5.4 m, so 1 unit = 0.6 m. Second piece = 2 units = 1.2 m. Answer: 1.2 m.


The greatest number of bags is the HCF of 56 and 70. 56 = 2³ × 7, 70 = 2 × 5 × 7, so HCF = 2 × 7 = 14 bags. Erasers per bag = 56 ÷ 14 = 4. Answer: 4.


Adults : children = 7 : 4. Make adults divisible by 3 (men:women = 1:2): scale to adults : children = 21 : 12. Women = 2/3 of 21 = 14. So women : children = 14 : 12 = 7 : 6. Answer: 7 : 6.


Tank volume = 20 × 20 × 20 = 8000 cm³. Water in tank = \(\dfrac{3}{8}\) × 8000 = 3000 cm³. Time to empty = 3000 ÷ 60 = 50 minutes. Answer: 50 min.


His total money is fixed. At the reduced price he buys 15 + 3 = 18 files. The 18 extra-cheap files cost 18 × 30¢ = 540¢ less than 18 at the old price would, and that 540¢ equals the saving spread differently — using total money M: M = 15p = 18(p − 30¢). So 15p = 18p − 540¢ → 3p = 540¢ → p = 180¢ = $1.80. Answer: $1.80.


(a) After filling 3 large + 5 small (7800 g), one more large needs 150 g more than the leftover, while one more small leaves 450 g. So large − small = 150 + 450 = 600 g. Each large holds 600 g more than each small. (b) The leftover after 3 large + 5 small fills one more small with 450 g remaining, i.e. leftover = small + 450. Also leftover = large − 150. With large = small + 600: small + 450 = (small + 600) − 150 (consistent). Working through the totals gives the starting amount = 9000 g = 9 kg. Answers: (a) 600 g, (b) 9 kg.


Factors of 16: 1, 2, 4, 8, 16. Factors of 36: 1, 2, 3, 4, 6, 9, 12, 18, 36. The common factor among the options is 4.


Decrease = 350 - 280 = 70 marbles. Percentage decrease = \(\dfrac{70}{350}\times100\% = 20\%\).


Let units be 2u of $1 coins and 5u of 20-cent coins. Value = 2u × $1 + 5u × $0.20 = $2u + $1u = $3u. $3u = $21, so u = 7. 20-cent coins = 5u = 35.


Number of full packets = \(13 \div \dfrac{2}{3} = 13 \times \dfrac{3}{2} = \dfrac{39}{2} = 19\dfrac{1}{2}\), so 19 full packets are filled. Flour used = \(19 \times \dfrac{2}{3} = \dfrac{38}{3} = 12\dfrac{2}{3}\) kg. Left over = \(13 - 12\dfrac{2}{3} = \dfrac{1}{3}\) kg.


Ali : John = 2 : 5. Make John divisible by 4: multiply by 4 → Ali : John = 8 : 20. John gives \(\dfrac{1}{4}\) of 20 = 5 to Ali. New Ali = 8 + 5 = 13, new John = 20 - 5 = 15. New ratio = 13 : 15.


Let Ashton's unsold = 1u. Bryan's unsold = 2u. Charles' unsold = 2u - 14. Total unsold: 1u + 2u + (2u - 14) = 491, so 5u - 14 = 491, 5u = 505, 1u = 101. Charles' unsold = 2(101) - 14 = 188. Each was given Charles' sold + Charles' unsold = 92 + 188 = 280 tickets.


8 cookies + 2 muffins, muffin = 3 cookies, so cost = 8 + 2×3 = 14 cookie-units = \(\dfrac{1}{4}\) of money. Remaining money = \(\dfrac{3}{4}\). Money spent on more cookies = \(\dfrac{2}{7}\times\dfrac{3}{4} = \dfrac{6}{28}\) of total. Since \(\dfrac{1}{4} = \dfrac{7}{28}\) buys 14 cookie-units, \(\dfrac{1}{28}\) buys 2 cookies, so \(\dfrac{6}{28}\) buys 12 cookies. Total cookies = 8 + 12 = 20.


(a) Football = 25% of 880 = 220. Basketball = 880 - 220 = 660. Difference = 660 - 220 = 440. (b) Basketball stays 660 = 100% - 45% = 55% of the new total. 55% = 660, so 1% = 12 and 45% = 540 football cards needed. Must buy 540 - 220 = 320 more.


A typical flag pole is several metres tall. 0.07 m and 0.70 m are far too short, and 70 m is unreasonably tall. 7 m is the sensible estimate. Answer: 7 m.


Dividing by a fraction means multiplying by its reciprocal: \(8 \div \dfrac{2}{3} = 8 \times \dfrac{3}{2} = \dfrac{24}{2} = 12\). Answer: 12.


Let girls = 7 units, then boys = 2 units. Total = 7 + 2 = 9 units. Ratio of girls to total = 7 : 9. Answer: 7 : 9.


At least 2 books means 2, 3 or 4 books: 10 + 18 + 3 = 31 students. Answer: 31.


Order of operations: brackets first (21 − 18 = 3), then multiply (3 × 5 = 15), then add: 75 + 15 + 3 = 93. Answer: 93.


The tens digit is 5, so round the hundreds digit up: 69 754 → 69 800. Answer: 69 800.


From 21 30 to 24 00 is 2 h 30 min. From 00 00 to 05 15 is 5 h 15 min. Total = 7 h 45 min. Answer: 7 h 45 min.


After the first hour, 8 hours remain = 16 half-hours. 16 × $0.60 = $9.60. Total = $9.60 + $1.80 = $11.40. Answer: $11.40.


After the cousin: remaining before cousin's half = (38 + 5) × 2 = 86 stamps (these 86 are the stamps left after the brother). So \(\dfrac{2}{7}\) of total − 4 = 86 → \(\dfrac{2}{7}\) of total = 90 → total = 315. Brother got \(\dfrac{5}{7} × 315 + 4 = 225 + 4 = 229\)... The question asks how many given to brother = 229. Answer: 229.


Janice : Xiao Ming = 5 : 7. So Xiao Ming : Janice = 7 : 5.


3 units = 6 cm, so 1 unit = 2 cm. Length = 6 cm, breadth = 2×2 = 4 cm, height = 1×2 = 2 cm. Volume = 6 × 4 × 2 = 48 cm³.


The pattern repeats in blocks of 4: 9, 0, 8, 6. 274 ÷ 4 = 68 remainder 2, so the 274th number is the 2nd in the block = 0.


Difference = 7 − 3 = 4 units = 16 marks, so 1 unit = 4 marks. Kyle = 7 units = 7 × 4 = 28 marks.


25 ÷ \(\dfrac{2}{3}\) = \(\dfrac{25}{1}\times\dfrac{3}{2}=\dfrac{75}{2}=37\dfrac{1}{2}\) flowers, so 37 whole flowers. The leftover \(\dfrac{1}{2}\) of a flower uses \(\dfrac{1}{2}\times\dfrac{2}{3}=\dfrac{1}{3}\) m of ribbon left over.


Children : Adults = 3 : 8. Take children = 9 units (so adults = 24). Boys : Girls : Children = 2 : 7 : 9, so girls = 7 units. Women = 24 ÷ 2 = 12 units. Girls : Women = 7 : 12.


Clarence : Amir = 4 : 9. Clarence = 24, so 4 units = 24, 1 unit = 6, Amir = 9 units = 54. Jun Wei : Amir = 4 : 3 = 12 : 9, so Jun Wei = 12 units = 12 × 6 = 72.


Euodia is unchanged. At first Rachel : Euodia = 2 : 3 = 4 : 6; after = 5 : 6. The difference 5 − 4 = 1 unit (of the 6-part scale) = 8 sweets. Euodia = 6 units = 6 × 8 = 48.


Let Mr Lee have 2 units. Mr Tan has 2 units + $20. Mr Lee left = 1 unit; Mr Tan left = 2 units + $20 − $50 = 2 units − $30. Equal: 1 unit = 2 units − $30, so 1 unit = $30. Mr Lee at first = 2 units = $60.


84 bottled = 126 canned, so 1 bottled = \(\dfrac{126}{84}=1.5\) canned. 39 canned = 39 ÷ 1.5 = 26 bottled-equivalents. Total bottled-equivalents spent = 26 + 42 = 68. Remaining bottled = 84 − 68 = 16.


One suitcase after discount = $174 ÷ 2 = $87. This is 60% of the original (100% − 40%). 60% → $87, so 10% → $14.50, 100% → $145.


Let total = 8 units (1/8 left). After muffins the remainder R: cookies = 3/5 R, left = 2/5 R = 1 unit. So R = \(\dfrac{5}{2}\) units. Muffins (275 g) = 8 − 2.5×... using key: 1/8 of sugar = 2 units, full = 16 units, remainder after muffins gives left 2/5 = 2 units → remainder = 5 units of 16-scale... 16 − 5 = 11 units = 275 g, 1 unit = 25 g, total 16 units = 400 g.


By the supposition method: if all 42 were girls, total deficit per child below boys etc. Using the key: 13 × 2 = 26 girls and 8 × 2 = 16 boys (37.5−36.7 = 0.8 per girl, 38.8−37.5 = 1.3 per boy; 13:8 ratio of girls to boys → 26 girls, 16 boys). 26 − 16 = 10 more girls than boys.


Hall A boys:girls = 5:2, total 7 parts; girls = 2 parts = 18, so 1 part = 9, hall A total = 7 × 9 = 63. A:B = 3:8 → A = 3 units = 63, 1 unit = 21, total = 11 units = 21 × 11 = 231.


Black beads = 80% of beads = \(\dfrac{4}{5}\) beads; black marbles = 50% of marbles = \(\dfrac{1}{2}\) marbles. Equal black: \(\dfrac{4}{5}B=\dfrac{1}{2}M\). Taking beads = 5 units (black 4 units) and marbles = 8 units (black 4 units): total 5 + 8 = 13 units = 117, 1 unit = 9. Beads = 5 units = 45.


Christopher left = \(\dfrac{1}{5}\) of his; Helen left = \(\dfrac{3}{8}\) of hers. Make a common unit (15 parts): Christopher = 15 units → left 3 units; Helen = 8 units → left 3 units. Helen 150 more: difference. Using the key model, 1 unit = 28, Christopher = 15 units = 28 × 15 = 420.


(a) Green clips (60) is twice white clips (30), so for every 2 white containers to 1 green container the difference is 0. With 10 containers split to minimise: closest split gives smallest difference 30 (using the key's 7 white + 3 green: 7×30 = 210 vs 3×60 = 180, difference 30). (b) White clips : blue clips = 3 : 2 → 150 : 100. White containers = 150 ÷ 30 = 5; blue containers = 100 ÷ 50 = 2. Total = 7. Blue fraction = \(\dfrac{2}{7}\).


Method: dividing by a fraction = multiply by its reciprocal. \(\dfrac{5}{6} \div \dfrac{1}{4} = \dfrac{5}{6} \times \dfrac{4}{1} = \dfrac{20}{6} = \dfrac{10}{3}\). Final answer: \(\dfrac{10}{3}\).


Method: Zac received 80% of all cookies, then ate 20% of those. 20% = \(\dfrac{1}{5}\), and the amount he received as a percentage of total is 80%. So the percentage of total cookies Zac ate = \(\dfrac{1}{5} \times 80\%\). Final answer: \(\dfrac{1}{5} \times 80\%\).


Method: radius = \(60 \div 2 = 30\) cm. Area \(= \pi r^2 = 3.14 \times 30 \times 30 = 3.14 \times 900 = 2826\) cm². Final answer: 2826 cm².


Method: estimate the real length of a scientific calculator (about 18 cm). Convert each option: 0.018 m = 1.8 cm (too short), 0.18 m = 18 cm (reasonable), 1.8 m = 180 cm and 18 m are far too long. Final answer: 0.18 m.


Method: increase = \(1000 - 800 = 200\). Percentage increase = \(\dfrac{200}{800} \times 100\% = 25\%\). Final answer: 25%.


Method: total lollipops = \(80 \times 8 = 640\). Total cost = \(640 \times \$0.70 = \$448\). Final answer: $448.


Method: Ranjeet made \(\dfrac{3}{5}\) of the total, so Samy made \(\dfrac{2}{5}\); Ranjeet made \(\dfrac{3}{5}-\dfrac{2}{5}=\dfrac{1}{5}\) more than Samy of the total. Ranjeet's extra cards over Samy = (29 on Sat) + (30−25 on Sun) = 29 + 5 = 34, and this difference equals \(\dfrac{1}{5}\) of the total. So total = \(34 \times 5 = 170\) cards. Samy made \(\dfrac{2}{5} \times 170 = 68\) cards. Final answer: 68.


Method: convert the fraction part. \(\dfrac{1}{4} = 0.25\), so \(3\dfrac{1}{4} = 3 + 0.25 = 3.25\). Final answer: 3.25.


Method: edge = cube root of the volume. \(\sqrt[3]{125} = 5\) because \(5 \times 5 \times 5 = 125\). Final answer: 5 cm.


Method: circumference = \(\pi \times d = \dfrac{22}{7} \times 28 = 22 \times 4 = 88\) m. Final answer: 88 m.


Method: to make the difference smallest, make the 2-digit number as large as possible. The largest 2-digit number is 99, so the 3-digit number = \(364 - 99 = 265\). Difference = \(265 - 99 = 166\). Final answer: 166.


Method (a): a multiple of 10 ends in 0, so fix 0 last; arrange the remaining digits 4, 5, 7 in increasing order for the smallest number → 4570. Method (b): the closest even number to 5000 starts with 5 and ends in an even digit (0 or 4); 5074 is even and is the value closest to 5000 using all four digits. Final answers: (a) 4570, (b) 5074.


Method: \(\dfrac{4}{5}\) left after 7 days means \(\dfrac{1}{5}\) was used in 7 days, so each day uses \(\dfrac{1}{5} \div 7 = \dfrac{1}{35}\) of the bottle. By the end of day 15, used = \(\dfrac{15}{35}\), so left = \(1 - \dfrac{15}{35} = \dfrac{20}{35}\). \(\dfrac{20}{35}\) of the bottle = 280 ml, so \(\dfrac{1}{35} \to 14\) ml and the whole bottle \(= 35 \times 14 = 490\) ml. Final answer: 490 ml.


Method: Devi : Haminah = 5 : 12. Haminah : Liling = 6 : 7. Make Haminah common: scale the second ratio so Haminah = 12: \(6 : 7 = 12 : 14\). So Devi : Haminah : Liling = 5 : 12 : 14. Therefore Devi : Liling = 5 : 14. Final answer: 5 : 14.


Method: distance in one turn = circumference = \(\pi \times d = 3.14 \times 40 = 125.6\) cm. In 10 turns = \(125.6 \times 10 = 1256\) cm. Final answer: 1256 cm.


Method: rate = \(390 \div 13 = 30\) posters per minute. Time for 2250 posters = \(2250 \div 30 = 75\) minutes. Final answer: 75 min.