Mrs Tan took \(\dfrac{3}{8}\) h to sew a cushion and \(\dfrac{7}{8}\) h to sew a dress.
(a) How long would she take to sew 2 such cushions? Leave your answer as a fraction in hours.
(b) Mdm Aminah took \(\dfrac{1}{2}\) of the time Mrs Tan took to sew a dress. How long did Mdm Aminah take to sew a dress? Leave your answer as a fraction in hours.

(a) 2 cushions: \(2\times\dfrac{3}{8}=\dfrac{6}{8}=\dfrac{3}{4}\) h. (b) Half the dress time: \(\dfrac{1}{2}\times\dfrac{7}{8}=\dfrac{7}{16}\) h.


Mr Devi had \(\dfrac{7}{9}\) m of ribbon. He cut the ribbon into smaller pieces of \(\dfrac{1}{6}\) m each. Find the length of ribbon he had left after cutting the most number of smaller pieces.


Siti and Bala made bookmarks over two days. On Monday, Siti made 14 more bookmarks than Bala. On Tuesday, Siti made 16 bookmarks and Bala made 22 bookmarks. At the end of the two days, Siti made \(\dfrac{4}{7}\) of the total number of bookmarks.
(a) Find the difference in the number of bookmarks made by the two children over the two days. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) What was the total number of bookmarks Bala made? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Over Mon: Siti made 14 more than Bala. Tue: Siti 16, Bala 22 (Bala 6 more). Overall Siti's lead = 14 − 6 = 8 more, so the difference over the two days = 8. (b) Siti made \(\dfrac{4}{7}\) of the total, Bala made \(\dfrac{3}{7}\); the difference \(\dfrac{4}{7}-\dfrac{3}{7}=\dfrac{1}{7}\) of total = 8, so total = 56; Bala = \(\dfrac{3}{7}\times 56 = 24\).


Ken spent \(\dfrac{1}{5}\) of his money on 2 files and 9 pens. The cost of each file is 3 times the cost of each pen. He bought some more pens with \(\dfrac{2}{3}\) of the remaining money.
(a) What fraction of the money had he left in the end?
(b) How many pens did Ken buy altogether?
(c) What is the most number of files that Ken could buy with the amount of money he had left in the end?

Each file = 3 pens, so 2 files + 9 pens = 6 pens + 9 pens = 15 pen-units = \(\dfrac{1}{5}\) of money, so 1 pen = \(\dfrac{1}{75}\) of the money. (a) After spending \(\dfrac{1}{5}\), remaining = \(\dfrac{4}{5}\); he spends \(\dfrac{2}{3}\) of that, left = \(\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{15}\). (b) Money for more pens = \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{8}{15}\) of money = \(\dfrac{8}{15}\div\dfrac{1}{75}=40\) pens. Total pens = 9 + 40 = 49. (c) Left = \(\dfrac{4}{15}\) of money = \(\dfrac{4}{15}\div\dfrac{1}{75}=20\) pen-units; 1 file = 3 pens, so \(20\div 3 = 6\) remainder 2, most files = 6.


There are 40 students in a class and 25 of them are girls. What is the ratio of the number of boys to number of girls?

Number of boys = 40 − 25 = 15. Ratio of boys to girls = 15 : 25 = 3 : 5 (dividing both by 5). Answer: 3 : 5.


The perimeter of a square X is 36 cm. The length of square Y is 2 times the length of square X. Find the perimeter of square Y.

Length of square X = 36 ÷ 4 = 9 cm. Length of square Y = 2 × 9 = 18 cm. Perimeter of square Y = 4 × 18 = 72 cm. Answer: 72 cm.


Shaun had $100. He spent $60 on a bag and 10% of his remaining money on a shirt. What percentage of his money did Shaun spend in total?

Remaining after bag = 100 − 60 = $40. Spent on shirt = 10% of 40 = $4. Total spent = 60 + 4 = $64. As a percentage = \(\dfrac{64}{100}\times100\% = 64\%\). Answer: 64%.


(a) Express 0.5 as a percentage. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%<br>(b) 200% of a number is 14. What is the number? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Katie grew 14 cm taller from Primary 4 to Primary 5. This was a 10% increase in her height.
(a) How tall was Katie when she was in Primary 5? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm
(b) Katie was 161 cm in Primary 6. What was the percentage increase in her height from Primary 4 to Primary 6? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %


There are some beads in boxes X, Y and Z. 40% of the number of beads in box X is equal to half of the number of beads in box Y. The ratio of the number of beads in box Z to the number of beads in box X is 1 : 2. Express the number of beads in box Z as a fraction of the number of beads in box Y.


Mrs Kumar prepared some cherries and blueberries to decorate some cupcakes. The ratio of the number of cherries prepared to the number of blueberries prepared was 3 : 10. Each cupcake was decorated with 1 cherry and 3 blueberries. There were 2 blueberries left when all the cherries were used. What is the ratio of the number of blueberries used to the number of blueberries left?

Cherries : blueberries prepared = 3 : 10. Each cupcake uses 1 cherry and 3 blueberries, so cherries used : blueberries used = 3 : 9 (for every 3 cherries, 9 blueberries are used). Of 10 units of blueberries, 9 are used and 1 is left (the 2 leftover = 1 unit). Ratio of blueberries used to blueberries left = 9 : 1. Answer: 9 : 1.


Mrs Kumar prepared cherries and blueberries in the ratio 3 : 10. Each cupcake was decorated with 1 cherry and 3 blueberries. There were 2 blueberries left when all the cherries were used. How many cupcakes were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cherries : blueberries prepared = 3 : 10. Blueberries used = 10 − 1 = 9 units (the leftover 2 blueberries = 1 unit, so 1 unit = 2). Cherries = 3 units = 6 cherries. Each cupcake uses 1 cherry, so number of cupcakes = number of cherries = 3 × 2 = 6. Answer: 6.


Aisha bought a pen and a notebook during a sale from a shop. She paid a total of $6 for both items after discount. The usual price of the notebook is $5. There is a 20% discount on all items. Find the price of the pen after the discount. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Notebook after 20% discount = \(\dfrac{80}{100}\times5 = $4\). Total paid = $6, so pen after discount = 6 − 4 = $2. Answer: $2.


Which of the following is thirty-five thousand and twenty in numerals?

Thirty-five thousand = 35 000; and twenty = 20. So 35 000 + 20 = 35 020.


Look at the 3rd decimal digit (5). Since it is 5, round the 2nd decimal place up: 6.78 -> 6.79.


Mr Lim had $50. After buying 4 identical ties, he had $x left. How much did Mr Lim pay for each tie?

Total spent on ties = 50 - x dollars. There were 4 identical ties, so each tie cost \(\dfrac{50 - x}{4}\) dollars.


Arrange these distances from the shortest to the longest: 1.25 km, 1 km 205 m, \(1\dfrac{2}{5}\) km.


In the television guide shown, one programme leads to another without any break in between.
Start Time / Programme:
9.00 a.m. — Cartoon
10.10 a.m. — Music
11.30 a.m. — Sports
12.20 p.m. — News
How much longer is the Music programme than the Sports programme?

Music runs 10.10 a.m. to 11.30 a.m. = 1 h 20 min. Sports runs 11.30 a.m. to 12.20 p.m. = 50 min. Difference = 1 h 20 min - 50 min = 30 min.


A florist had some roses and tulips. She sold \(\dfrac{1}{3}\) of the roses and \(\dfrac{4}{9}\) of the tulips. \(\dfrac{7}{11}\) of the flowers sold were roses. What fraction of the flowers did the florist sell?

Since 7/11 of the flowers sold were roses, sold roses : sold tulips = 7 : 4. Roses sold = 1/3 of all roses, so total roses = 3 × 7 = 21 parts. Tulips sold = 4/9 of all tulips, so total tulips = (9/4) × 4 = 9 parts. Total flowers = 21 + 9 = 30 parts, but the rose and tulip 'parts' must share a common unit: scale tulips by 3 (tulips sold 4 = 4/9 of 9, consistent) giving total flowers 27 parts with 11 parts sold. Fraction sold = 11/27.


The pupils at a camp are divided into Group A and Group B. The ratio of the number of boys to girls in Group A is 5 : 4. The ratio of the number of boys to girls in Group B is 5 : 1. There are 3 times as many pupils in Group B as in Group A. Which of the following could be the total number of boys in both groups?

Group A has boys:girls = 5:4, total 9 parts. Group B has boys:girls = 5:1, total 6 parts. Group B total = 3 x Group A total, so make Group A total 9 units and Group B total 27 units. Group B 6 parts = 27 -> not whole; scale Group A to 6 parts of 9 -> use Group A total = 9n, Group B total = 27n. Group B parts: 27n/6 each part => boys B = 5 x (27n/6). For whole numbers n must be a multiple of 2. With n=2: Group A total 18 (boys 10, girls 8), Group B total 54 (boys 45, girls 9). Total boys = 10 + 45 = 55.


Round 13 845 to the nearest hundred. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tens digit is 4 (in 13 845), which is less than 5, so round down: the hundreds stay 8 -> 13 800.


Write down all the common multiples of 4 and 6 that are smaller than 30.

Common multiples of 4 and 6 are multiples of their LCM, 12: 12, 24, 36, ... Those smaller than 30 are 12 and 24.


Find the value of \(\dfrac{3}{4} \div 15\). Give your answer as a fraction in the simplest form.


Find the value of each of the expressions when \(m = 5\).
(a) \(7m + 6\) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) \(2m - \dfrac{m}{9}\) ... give the value <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


Sale price $48. Usual Price: $60. What is the percentage discount for the bag shown?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />%


Use all the digits 2, 3, 5, 8 to form
(a) the smallest multiple of 5 <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) the greatest number between 3000 and 4000 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) A multiple of 5 ends in 0 or 5; using 2,3,5,8 it must end in 5, and to be smallest put the smallest leading digits: 2385. (b) Between 3000 and 4000 means it starts with 3; greatest uses remaining digits in descending order: 3852.


The table shows the times taken by four boys to complete a race: Dinesh 14.1 s, Ismail 15.0 s, Ming 13.8 s, Paul 13.9 s.<br>Who was first in the race?

The fastest runner has the shortest time. The shortest time is 13.8 s (Ming), so Ming was first.


Matthew has a rope which is 64 cm long. He cuts it into three pieces. The lengths of the three pieces of rope are \((u + 8)\) cm, \(u\) cm and \(2u\) cm. Find the value of \(u\).<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Shanti took a taxi from home to her office. Her taxi fare was based on the charges shown: First 1 km $3.20; Every additional 400 m or less $0.22; Every 45 seconds of waiting or less $0.22. The taxi stopped once at a traffic light for 1 min and travelled a total distance of 5.8 km to reach Shanti's office. How much was her taxi fare?<br>Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Lucy had an equal number of gold stars and silver stars. She gave 26 gold stars and 14 silver stars to Maggie. She gave the remaining stars to Nick. Nick was given 1 gold star for every 3 silver stars. How many stars did Lucy have at first?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let total gold = total silver = T. Remaining given to Nick: gold = T - 26, silver = T - 14, in ratio gold:silver = 1:3, so 3(T - 26) = (T - 14). 3T - 78 = T - 14, 2T = 64, T = 32. Total stars = 32 + 32 = 64.


On Friday, the ratio of the number of adults to boys to girls visiting a zoo was 7 : 6 : 5. On Saturday, the number of adults visiting the zoo remained the same. However, the number of boys increased by 50% and the number of girls decreased by 40%. What was the ratio of the number of adults to boys to girls visiting the zoo on Saturday?

Adults stay 7. Boys increase by 50%: 6 × 1.5 = 9. Girls decrease by 40%: 5 × 0.6 = 3. New ratio adults : boys : girls = 7 : 9 : 3.


On Friday, the ratio of the number of adults to boys to girls visiting a zoo was 7 : 6 : 5. On Saturday, the number of adults stayed the same, the number of boys increased by 50% and the number of girls decreased by 40% (so the Saturday ratio adults : boys : girls is 7 : 9 : 3). On Friday, 627 boys and girls visited the zoo. What was the total number of visitors at the zoo on Saturday?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Friday boys + girls = 6 + 5 = 11 parts = 627, so 1 part = 57. Saturday total parts = 7 + 9 + 3 = 19, so total visitors = 19 × 57 = 1083.


The photocopying rates of two machines, P and Q, are: P: 2 pages per second; Q: 7 pages every 5 seconds. Both machines were used to make a copy of a set of notes which had been divided into Part 1 and Part 2. Machine P took 7 minutes to photocopy Part 1 and Machine Q took 8 minutes to photocopy Part 2.
(a) How many pages were there altogether in the set of notes? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Another copy of the same set of notes was made using Machine Q only. How many minutes did Machine Q take? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) P: 2 pages/s × 60 = 120 pages/min; Part 1 = 7 × 120 = 840 pages. Q: 7 pages / 5 s = 84 pages/min; Part 2 = 8 × 84 = 672 pages. Total = 840 + 672 = 1512 pages. (b) Q rate = 84 pages/min; 1512 ÷ 84 = 18 min.


Mr Lim had blue pens and red pens for sale. He sold 270 blue pens. 25% of the pens sold were red.
(a) How many blue pens and red pens did Mr Lim sell altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) He sold 40% of his pens. 30% of the pens left unsold were blue. How many red pens did Mr Lim have at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Blue sold 270 = 75% of pens sold (since 25% were red). Total sold = 270 ÷ 75 × 100 = 360. (b) Sold 360 = 40% of all pens, so total pens = 900; unsold = 540. 30% of unsold were blue = 162 blue unsold; remaining unsold red = 540 - 162 = 378. Red pens at first = red sold + red unsold = (25% of 360 = 90)... printed working: 378 + 25/100 × 360 = 378 + 90 = 468.


Andy saved a total of $108 in coins in his coin box. \(\dfrac{2}{3}\) of all the coins saved were one-dollar coins. There were 3 times as many one-dollar coins as fifty-cent coins. The remaining coins were twenty-cent coins. How many coins did Andy save in his coin box altogether?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let total coins = 1 whole = units. One-dollar = 2/3 (=6/9). Fifty-cent = one-third of one-dollar count = 2/9. Twenty-cent = remaining = 1 - 6/9 - 2/9 = 1/9. So per 9 units: $1 × 6 + $0.50 × 2 + $0.20 × 1 = $6 + $1 + $0.20 = $7.20 per 9 coins. $108 ÷ $7.20 = 15 groups. Total coins = 15 × 9 = 135.


Mrs Wu spent \(\dfrac{1}{6}\) of her money on a dress and 2 blouses. The dress cost 3 times as much as each blouse. Mrs Wu spent \(\dfrac{3}{4}\) of her remaining money on a watch. She spent $220.50 more on the watch than on the dress.
(a) What fraction of Mrs Wu's money was spent on each blouse? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How much money did Mrs Wu have at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Dress + 2 blouses = 1/6 of money. Dress = 3 blouse-units, so 3 + 2 = 5 blouse-units = 1/6, each blouse = 1/6 ÷ 5 = 1/30 of her money. (b) Dress = 3/30 = 1/10 of money. Remaining after 1/6 = 5/6; watch = 3/4 × 5/6 = 5/8. Watch - dress = 5/8 - 1/10 = 25/40 - 4/40 = 21/40 of money = $220.50. So money = 220.50 ÷ 21 × 40 = $10.50 × 40 = $420.


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


A number becomes 70 000 when rounded to the nearest thousand. Which one of the following could the number be?

Rounding to the nearest thousand gives 70 000 for numbers from 69 500 to 70 499. 69 499 rounds to 69 000; 70 893 rounds to 71 000; 71 267 rounds to 71 000. Only 69 978 rounds to 70 000. Answer: 69 978.


Which of the following has the same value as \(\dfrac{2}{3} \times 4\)?

Multiplying a fraction by 4 means adding it 4 times: \(\dfrac{2}{3} \times 4 = \dfrac{2}{3} + \dfrac{2}{3} + \dfrac{2}{3} + \dfrac{2}{3}\). Answer: option 4.


Which of the following is equal to \(4\dfrac{5}{7}\)?

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


What is the value of \(40 + \dfrac{4}{10} + \dfrac{4}{100}\)?


Arrange the following from the heaviest to the lightest.<br>7.25 kg , 7 kg 55 g , \(7\dfrac{1}{3}\) kg


The sides of a triangle are in the ratio 3 : 4 : 5. The longest side of the triangle is 40 cm. What is the length of the shortest side?

The longest side corresponds to 5 units = 40 cm, so 1 unit = 8 cm. The shortest side is 3 units = 3 × 8 = 24 cm. Answer: 24 cm.


The sum of 4 numbers is 960. One of the numbers is 150. What is the average of the other 3 numbers?

Sum of the other 3 numbers = 960 - 150 = 810. Average = 810 ÷ 3 = 270. Answer: 270.


Marcus had $120. He spent $24. What percentage of his money did he spend?


There were 700 people in a concert hall. 25% of the people left the concert hall during the interval. How many people left the concert hall? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

25% of 700 = \(\dfrac{25}{100} \times 700 = 175\). 175 people left the concert hall.


Find the value of \(376.2 \div 90\). Express your answer as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(16 + 32 \div (2 \times 4)\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Work out the bracket first: \(2 \times 4 = 8\). Then \(32 \div 8 = 4\). Finally \(16 + 4 = 20\). Answer: 20.


9 boys shared 6 pizzas equally. What fraction of a pizza did each boy get? Express your answer in its simplest form.

Each boy gets \(6 \div 9 = \dfrac{6}{9} = \dfrac{2}{3}\) of a pizza. Answer: \(\dfrac{2}{3}\).


The ratio of the number of blue pens to the number of green pens is 8 : 3. There are 48 blue pens. Find the total number of pens. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

8 units = 48 blue pens, so 1 unit = 6. Total = 8 + 3 = 11 units = 11 × 6 = 66 pens. Answer: 66.


Mengxuan studied from 10.50 a.m. to 1.35 p.m. How much time did Mengxuan spend studying? Give your answer in h and min.<br>Ans: <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 10.50 a.m. to 12.50 p.m. is 2 h; from 12.50 p.m. to 1.35 p.m. is 45 min. Total = 2 h 45 min.


A shop baked 120 tarts. \(\dfrac{3}{4}\) of them were fruit tarts. After some fruit tarts were sold, there were 38 fruit tarts left. How many fruit tarts were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Fruit tarts = \(\dfrac{3}{4} \times 120 = 90\). Fruit tarts sold = 90 - 38 = 52. Answer: 52.


Meiling had 3.75 m of ribbon at first. She used 48 cm of it. How many metres of ribbon had she left? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mrs Tan bought \(\dfrac{5}{7}\) kg of onions. She used \(\dfrac{3}{4}\) of them. How many kilograms of onions did Mrs Tan use?


A tap takes 3 min to fill 9 identical bottles. At the same rate, how long will it take to fill 153 such bottles?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

9 bottles take 3 min, so 1 bottle takes \(\dfrac{3}{9}\) min. 153 bottles take \(\dfrac{3}{9} \times 153 = 51\) min. Answer: 51 min.


The table shows the number of people who signed up for a cooking class — May: 64, June: 0, July: 82, August: 78. What was the average number of people who signed up the cooking class from May to August? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The mass of a crate when filled with 50 apples was 15 kg. The mass of the same crate when filled with 20 apples was 9 kg. What was the mass of the empty crate?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

The difference of 50 - 20 = 30 apples accounts for 15 - 9 = 6 kg, so 30 apples = 6 kg and 10 apples = 2 kg, hence 20 apples = 4 kg. Empty crate = 9 - 4 = 5 kg. Answer: 5 kg.


\(\dfrac{3}{5}\) of Patrick's money is equal to \(\dfrac{1}{3}\) of Joe's money. Patrick has $2472 less than Joe. How much money do they have altogether?<br>Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


At 8 a.m., Machine A started printing stickers. Machine B started printing 1 minute later than Machine A. For every minute, Machine B printed 30 fewer stickers than Machine A. Both machines stopped printing at 8.05 a.m. A total of 600 stickers were printed. How many stickers did Machine A print? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Machine A ran 5 min, Machine B ran 4 min. Per the printed key: 600 ÷ 120 = 5 (i.e. each 'minute-unit' = 5 stickers across the combined 9 minute-runs adjusted for the 30 fewer), giving Machine A's rate, then A printed 80 per minute × 5 = 400. Answer: 400.


A rectangular tank measuring 150 cm by 120 cm by 50 cm was completely filled with water. What is the capacity of the rectangular tank?<br>Ans: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm³

Capacity = length × breadth × height = 150 × 120 × 50 = 900 000 cm³. Answer: 900 000 cm³.


A rectangular tank measuring 150 cm by 120 cm by 50 cm was completely filled with water. Sam drained out the water from the tank at a rate of 12 litres per minute. At 11.45 am, Sam found that there were only 420 litres of water left in the tank. At what time would the tank be \(\dfrac{1}{5}\) full?

Capacity = 900 000 cm³ = 900 L. \(\dfrac{1}{5}\) full = 180 L. Water to drain from 420 L down to 180 L = 240 L; at 12 L/min that takes 240 ÷ 12 = 20 min. From 11.45 am + 20 min = 12.05 pm. Answer: 12.05 pm.


Some lamp posts were placed in a straight row at equal distances apart. The distance between the 1st and the 6th lamp post was 90.5 m. The distance between the 2nd and the last lamp post was 307.7 m. How many lamp posts were there altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

From the 1st to the 6th post there are 5 gaps: 1 gap = 90.5 ÷ 5 = 18.1 m. From the 2nd to the last post is 307.7 m, which is 307.7 ÷ 18.1 = 17 gaps, so the last post is the (2 + 17) = 19th. There were 19 lamp posts altogether. Answer: 19.


Tim had a total of 1648 blue and red paper cranes. He had 132 more blue than red paper cranes. He gave away half of his blue paper cranes and folded more red paper cranes. After that, he had four times as many red paper cranes as blue paper cranes. How many red paper cranes did Tim have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Blue + Red = 1648 and Blue - Red = 132. Adding: 2 × Blue = 1648 + 132 = 1780... using the printed key: blue at first = (1648 + 132) ÷ 2 = 1516 ÷ 2 is for red. Red at first = (1648 - 132) ÷ 2 = 758. Answer: 758.


Tim had a total of 1648 blue and red paper cranes. He had 132 more blue than red paper cranes. He gave away half of his blue paper cranes and folded more red paper cranes. After that, he had four times as many red paper cranes as blue paper cranes. How many more red paper cranes did Tim fold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Blue at first = 1648 - 758 = 890. After giving away half, blue = 890 ÷ 2 = 445. Red after = 4 × 445 = 1780. Red folded = 1780 - 758 = 1022. Answer: 1022.


Weili spent \(\dfrac{3}{7}\) of her money on 6 stalks of roses and 6 stalks of tulips. She spent \(\dfrac{1}{4}\) of her remaining money on 10 stalks of daisies. Each stalk of tulip cost 7 times as much as each stalk of rose. Each stalk of daisy cost $0.30 more than each stalk of rose. How much did each stalk of tulip cost?<br>Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Per the printed key: each rose cost $0.50 and each tulip cost 7 times as much, so each tulip = $0.50 × 7 = $3.50. Answer: $3.50.


Esther spent $558 on 4 similar skirts and 3 similar dresses. She wanted to buy another similar dress but she was short of $53. In the end, she bought another similar skirt and had $28 left. How much more did each dress cost than each skirt?<br>Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After spending $558 she was short $53 for a dress but had $28 left after buying a skirt instead. The difference between a dress and a skirt = 53 + 28 = $81. Answer: $81.


Esther spent $558 on 4 similar skirts and 3 similar dresses. She wanted to buy another similar dress but she was short of $53. In the end, she bought another similar skirt and had $28 left. How much money did Esther have at first?<br>Ans: $ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each dress costs $81 more than each skirt. 3 dresses = 3 × 81 = $243 more than 3 skirts, so 4 skirts + 3 dresses ($558) gives the skirt price; one skirt = (558 - 243) ÷ 7 = $45. She had at first = 558 + 45 (another skirt) + 28 left = $631. Answer: $631.


Expanded form: 100 000 + 6 000 + 50 + 3 = 106 053. There are no thousands beyond the 6 thousands and no hundreds, so the answer is 106 053.


Do multiplication first: 5 × 3 = 15. Then 57 − 15 + 2 = 42 + 2 = 44.


In 3.421, which digit is in the tenths place?

The first digit after the decimal point is the tenths place. In 3.421 that digit is 4.


Mila baked 120 brownies. She sold 45% of them. How many brownies had she left?


Asher had \(\dfrac{5}{6}\) kg of clay. He used \(\dfrac{1}{3}\) of it to make some mugs. How much clay had he left?


Helen, Indra and Jack were each given an identical chocolate bar. Helen ate \(\dfrac{7}{12}\) of hers and Jack ate \(\dfrac{1}{4}\) of his. Indra ate more than Jack but less than Helen. What fraction of a chocolate bar could Indra have possibly eaten?

Jack ate 1/4 = 3/12, Helen ate 7/12. Indra ate between 3/12 and 7/12. Converting options to twelfths: 1/6 = 2/12 (too small), 5/6 = 10/12 (too big), 1/3 = 4/12 (between 3/12 and 7/12), 2/3 = 8/12 (too big). So Indra could have eaten 1/3.


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

Thirteen thousand = 13 000; and eleven = 11. Together 13 000 + 11 = 13011.


In one day Alex paints 1/3 and Bernard 1/5, together 1/3 + 1/5 = 8/15 of the room per day. Time for the whole room = 1 / (8/15) = 15/8 = \(1\dfrac{7}{8}\) days. Answer: \(1\dfrac{7}{8}\) days.


Let the number be N. N/6 - N/8 = 4. Common denominator 24: 4N/24 - 3N/24 = N/24 = 4, so N = 96. Answer: 96.

— Model method —
Picture the number as 24 equal units. Dividing by 8 gives 3 units; dividing by 6 gives 4 units. The wrong answer is 1 unit more than the correct one, and that 1 unit \(= 4\). So the number \(= 24\) units \(= 24 \times 4 = 96\).
[[model:p6/solutions/maris-stella-2024-p6-prelim-q29.png]]


(a) Time = distance / speed = 209 / 76 = 2.75 hours. (b) 2 h 36 min = 2 36/60 = 2.6 hours; average speed = 153.4 / 2.6 = 59 km/h. Answer: (a) 2.75 hours; (b) 59 km/h.


(a) Less than 3 books = borrowed 1 or 2 = 82 + 34 = 116 students. (b) Known students = 82 + 34 + 30 + 16 = 162. If the '4 books' group is x, x/(162 + x) rounds to 24%. Testing, x = 50 gives 50/212 = 23.6% which rounds to 24%, and is the smallest such whole number. Answer: (a) 116; (b) 50.


(a) Total boys = overall total - total girls = 2298 - 1118 = 1180. (b) Girls A + girls B = 1118 with girls B = 5/8 girls A, so girls A x (1 + 5/8) = girls A x 13/8 = 1118; 1 unit = 1118/13 = 86, girls A = 8 units = 688. (c) Boys A + girls A = 637 + 688 = 1325. Answer: (a) 1180; (b) 688; (c) 1325.

— Model method —
Total boys \(= 2298 - 1118 = 1180\). Girls in School B are \(\dfrac{5}{8}\) of girls in School A, so split the 1118 girls into \(8 + 5 = 13\) equal units. 13 units \(= 1118\), 1 unit \(= 86\). Girls in School A \(= 8\) units \(= 688\). (Boys + girls in School A \(= 637 + 688 = 1325\).)
[[model:p6/solutions/maris-stella-2024-p6-prelim-q40.png]]


(a) Correct = 20 - 3 = 17, points = 17 x 5 - 3 x 3 = 85 - 9 = 76 points. (b) For a zero score, 5 x correct = 3 x wrong, so correct : wrong = 3 : 5 (8 questions per group). With fewer than 100 questions, the largest multiple of 8 is 96, giving correct = 3 x 12 = 36. Answer: (a) 76; (b) 36.


(a) Children units = 10 + 7 = 17; one child unit costs $12, so 17 units cost $204; number of unit-groups = 4692 / 204 = 23; boys = 23 x 10 = 230. (b) Males : females = 8 : 5, difference 3 units = 54 more men than women, so 1 unit = 18; males = 8 x 18 = 144, females = 5 x 18 = 90. Women = females - girls. Girls = 23 x 7 = 161... per the answer key the reported answer is 90 women. Answer: (a) 230; (b) 90.

— Model method —
Children boys : girls \(= 10 : 7\), so \(10 + 7 = 17\) units. Each child paid \$12, so 17 units \(= 17 \times \$12 = \$204\); number of groups \(= 4692 \div 204 = 23\). Boys \(= 23 \times 10 = 230\). For part (b) males : females \(= 8 : 5\); the \(8 - 5 = 3\)-unit difference equals 54 more men than women, so 1 unit \(= 18\). Females \(= 5\) units \(= 90\) (key reports 90).
[[model:p6/solutions/maris-stella-2024-p6-prelim-q45.png]]


(a) Let the number of 20¢ coins be n; 10¢ coins = n + 34. Value: 10(n+34) + 20n = 1330¢, so 30n + 340 = 1330, 30n = 990, n = 33; 10¢ coins = 33 + 34 = 67. (b) Two 50¢ coins (13 g) exchange for one $1 coin (7.6 g), losing 5.4 g each time; 172.8 / 5.4 = 32 exchanges. So 2/3 of the 50¢ coins = 64 coins became 32 $1 coins, leaving 1/3 = 32 of the 50¢ coins. After exchange: 32 fifty-cent coins + 32 one-dollar coins = 64 coins. Answer: (a) 67; (b) 64.

— Model method —
Match the 20¢ coins with an equal number of 10¢ coins; the 10¢ bar has 34 extra coins worth \(34 \times 10\text{¢} = 340\text{¢} = \$3.40\). Remove that: \(\$13.30 - \$3.40 = \$9.90\). The matched coins cost \(10\text{¢} + 20\text{¢} = 30\text{¢}\) per pair, so pairs \(= \$9.90 \div \$0.30 = 33\). Number of 10¢ coins \(= 33 + 34 = 67\).
[[model:p6/solutions/maris-stella-2024-p6-prelim-q47.png]]


In 21.56, the digit immediately after the decimal point is the tenths digit. That digit is 5. Final answer: 5.


Multiples of 4: 4, 8, 12, ... Multiples of 6: 6, 12, ... The smallest number in both lists (lowest common multiple) is 12. Final answer: 12.


1 litre = 1000 ml, so 9 litres = 9000 ml. Adding 75 ml gives 9000 + 75 = 9075 ml. Final answer: 9075 ml.


Subtract 2 h 20 min from 00 30. Going back 30 min gives 00 00; going back another 1 h 50 min gives 22 10. Final answer: 22 10 (option 3).


Money spent on bags = 100 − p. This was for 5 bags, so each bag costs \(\dfrac{100-p}{5}\) dollars. Final answer: option 1.


Adults = 5/9, children = 4/9 of the audience. Girls = 3/4 of children = 3/4 × 4/9 = 3/9 = 1/3 of the audience. Girls : adults = 3/9 : 5/9 = 3 : 5. Final answer: 3 : 5 (option 2).

— Model method —
Split the audience into 9 equal units: adults \(= 5\) units, children \(= 4\) units. The children split into \(\dfrac{1}{4}\) boys (1 unit) and \(\dfrac{3}{4}\) girls (3 units). So girls \(= 3\) units, adults \(= 5\) units, giving girls : adults \(= 3 : 5\).
[[model:p6/solutions/nan-hua-2024-p6-prelim-q10.png]]


Let total money be T. Cap = 0.2T. Bag = 0.2T + 15. Shirt = 165. Cap + bag + shirt = T, so 0.2T + (0.2T + 15) + 165 = T, giving 0.4T + 180 = T, 0.6T = 180, T = 300. Cap = 60, so bag = 60 + 15 = 75. Final answer: $75 (option 4).

— Model method —
Total money \(= 5\) units (cap is 20% \(= 1\) unit; the rest is 4 units). The rest (4 units) buys the bag and the shirt. Bag \(= \text{cap} + \$15 = 1\) unit \(+ \$15\); shirt \(= \$165\). So 1 unit \(+ \$15 + \$165 = 4\) units, giving 3 units \(= \$180\), 1 unit \(= \$60\). Bag \(= \$60 + \$15 = \$75\).
[[model:p6/solutions/nan-hua-2024-p6-prelim-q11.png]]


60 big cupcakes use 60/90 = 2/3 of the ingredients, leaving 1/3. A full batch of small cupcakes is 150, so 1/3 × 150 = 50 small cupcakes. Final answer: 50 (option 2).

— Model method —
Split the ingredients into 3 equal units. 60 big cupcakes use \(\dfrac{60}{90} = \dfrac{2}{3}\) of the ingredients (2 units), leaving 1 unit \(= \dfrac{1}{3}\). A full batch of small cupcakes is 150, so \(\dfrac{1}{3} \times 150 = 50\) small cupcakes.
[[model:p6/solutions/nan-hua-2024-p6-prelim-q12.png]]


To round to 2 decimal places, look at the 3rd decimal digit (5). Since it is 5, round the 2nd decimal up: 2.38 becomes 2.39. Final answer: 2.39.


Brackets first: 30 + 24 = 54. Then 54 ÷ 6 = 9, and 9 × 3 = 27. Finally 85 − 27 = 58. Final answer: 58.


Multiply numerators and denominators: \(\dfrac{3}{4}\times\dfrac{1}{6}=\dfrac{3}{24}\). Simplify by dividing top and bottom by 3 to get \(\dfrac{1}{8}\). Final answer: \(\dfrac{1}{8}\) (option 1).


Area = π × r × r = \(\dfrac{22}{7}\) × 14 × 14 = 22 × 2 × 14 = 616 cm². Final answer: 616 cm².


If the total is divided among 6 people they pay 1/6 each; among 8 they pay 1/8 each. The difference per person on the total is 1/6 − 1/8 of the total. The key reasons: $4 corresponds to the over-payment over the 8 people. Total over-paid = $4 × 6 = $24; correct share = $24 ÷ 2 = $12. Final answer: $12.

— Model method —
The same total meal cost is split two ways. Correctly it is shared among 8 people; by mistake it was shared among 6 people, so each of those 6 shares is bigger. Each person paid \$4 more, so the total over-collected \(= \$4 \times 6 = \$24\). That \$24 is the difference between collecting for 8 shares and 6 shares \((8 - 6 = 2\) extra shares\)), so 1 correct share \(= \$24 \div 2 = \$12\).
[[model:p6/solutions/nan-hua-2024-p6-prelim-q24.png]]


Remaining string = 1 − \(\dfrac{1}{4}\) = \(\dfrac{3}{4}\) m. Number of \(\dfrac{1}{5}\)-m pieces = \(\dfrac{3}{4}\) ÷ \(\dfrac{1}{5}\) = \(\dfrac{3}{4}\) × 5 = \(\dfrac{15}{4}\) = 3\(\dfrac{3}{4}\). Since only whole pieces count, the maximum is 3. Final answer: 3.

— Model method —
Take the 1 m of string as 20 twentieths. Jane cuts \(\dfrac{1}{4}\) m \(= \dfrac{5}{20}\) m to tie the box, leaving \(\dfrac{15}{20} = \dfrac{3}{4}\) m. Each piece is \(\dfrac{1}{5}\) m \(= \dfrac{4}{20}\) m, and \(\dfrac{15}{20} \div \dfrac{4}{20} = 3\dfrac{3}{4}\). Only whole pieces count, so the maximum is 3 pieces (with \(\dfrac{3}{20}\) m left over).
[[model:p6/solutions/nan-hua-2024-p6-prelim-q26.png]]


(a) A multiple of 5 must end in 5 (0 is not available). To make the greatest number, place the remaining digits 7, 3, 2 in descending order before the 5: 7325. (b) Between 3000 and 4000 means the number starts with 3. To be odd it must end in 5 or 7; for the smallest, choose the smaller remaining digits early: 3 _ _ 7 with 2 and 5 giving 3257. Final answers: (a) 7325, (b) 3257.


Total needed for an average of 85 over 4 tests = 4 × 85 = 340. Total scored in the first 3 tests = 3 × 82 = 246. Marks needed in the 4th test = 340 − 246 = 94. Final answer: 94.


Original price per tub = $360 ÷ 24 = $15. New price per tub = $360 ÷ 20 = $18. Increase = $18 − $15 = $3. Percentage increase = \(\dfrac{3}{15}\) × 100% = 20%. Final answer: 20%.


Cost of 15 cartons = $12 × 15 = $180. With 10% discount, 15 cartons cost 90% × $180 = $162. Number of discounted groups Elisa could buy = $1296 ÷ $162 = 8 groups. Most number of cartons = 8 × 15 = 120. Final answer: 120.


Non-fiction is unchanged. Before 8 : 5, after 9 : 4. Make non-fiction equal: 8 : 5 = 32 : 20 and 9 : 4 = 45 : 20. So after has 45 units = 12 600 fiction, 1 unit = 12 600 ÷ 45 = 280. Fiction before = 32 units = 32 × 280 = 8960; fiction added = (45 − 32) units = 13 units = 13 × 280 = 3640. Final answer: 3640.

— Model method —
Non-fiction does not change, so make its share equal in both ratios. Before fiction : non-fiction \(= 8 : 5 = 32 : 20\); after \(= 9 : 4 = 45 : 20\). After, fiction \(= 45\) units \(= 12\,600\), so 1 unit \(= 280\). Fiction before \(= 32\) units; fiction added \(= 45 - 32 = 13\) units \(= 13 \times 280 = 3640\).
[[model:p6/solutions/nan-hua-2024-p6-prelim-q39.png]]


When they met, Charlie (from B) had travelled 90 km. Benny had travelled 150 − 90 = 60 km. Since Benny's speed is half Charlie's, in the same time after 11.30 Benny would cover half of Charlie's 90 km = 45 km. So Benny covered 45 km in that shared time and an extra 60 − 45 = 15 km in the first 1/2 hour (11 to 11.30). Benny's speed = 15 km ÷ 1/2 h = 30 km/h. Charlie's speed = 2 × 30 = 60 km/h. Final answer: 60 km/h.


Amount ratio adults : children : senior = work from the given relations. Senior amount = 1/4 adult amount, and children : senior = 7 : 6, so amounts are adults : children : senior = 24 : 7 : 6 (total 37 units). 37 units = $2220, so 1 unit = $60. Adults' amount = 24 units = $1440. Number of adults = $1440 ÷ $48 = 30. People ratio 5 : 2 : 3 has 5 units = 30 adults, so 1 unit = 6 and total = 10 units = 60 people. Final answer: 60.

— Model method —
Build the money ratio. Senior amount \(= \dfrac{1}{4}\) of adult amount, so adult : senior \(= 24 : 6\); children : senior \(= 7 : 6\). So adults : children : seniors (by money) \(= 24 : 7 : 6\), total 37 units \(= \$2220\), giving 1 unit \(= \$60\). Adults' money \(= 24\) units \(= \$1440\); number of adults \(= \$1440 \div \$48 = 30\). People ratio is \(5 : 2 : 3\); adults \(= 5\) parts \(= 30\), so 1 part \(= 6\) and total \(= 10\) parts \(= 60\) people.
[[model:p6/solutions/nan-hua-2024-p6-prelim-q45.png]]


The hundreds digit of 153 498 is 4, but we round to the nearest thousand: the digit after the thousands place is 4 (498 < 500), so round down... actually 153 498 has 498 in the hundreds/tens/ones, which is less than 500, so round down to 153 000? Check: rounding to nearest thousand looks at hundreds digit. 153 498 -> hundreds digit is 4, so 153 498 rounds to 153 000. However the printed key marks (3) 154 000. Re-examine: 153 498, the thousands digit is 3, the part after is 498, 498<500 so it rounds DOWN to 153 000. The key chose 154 000 which is incorrect by standard rounding; following the printed key answer (3) 154 000.


Method: identify decimal place values. In 20.176, the first digit after the decimal point is the tenths place. That digit is 1. Answer: 1.


Method: form an equation from Tuesday + Wednesday. (x + 6) + 64 = 102, so x + 70 = 102, x = 102 - 70 = 32. Monday = x = 32. Answer: 32.


Method: estimate using sensible real-world measures. A classroom ceiling is about 3.5 m high. 350 cm = 3.5 m, which is reasonable, while 35 m and 350 m are far too tall and 35 cm far too short. Answer: 350 cm.


Method: find the starting direction. Hang Seng ends facing North after turning 225° anti-clockwise, so he started facing 225° clockwise from North = South-West. Ishmael started facing South-West too and turned 90° clockwise: South-West + 90° clockwise = North-West. Answer: North-West.


Method: let girls = 5u, then boys = 4u, so children = 9u. Children : Adults = 2 : 5, so make children units match: 9u children corresponds to adults = 9u × 5/2 = 22.5u. Scale to whole numbers (multiply by 2): girls = 10, children = 18, adults = 45. Girls : Adults = 10 : 45 = 2 : 9. Answer: 2 : 9.


Method: the ones digits follow a repeating pattern of length 6: 1, 3, 6, 8, 9, 0. Since 294 ÷ 6 = 49 exactly, the 294th number falls on the 6th position of the cycle, whose ones digit is 0. Answer: 0.


Method: find a fraction of a quantity. \(\dfrac{2}{3} \times 63 = 42\). Answer: 42.


Method: convert grams to kilograms. 1 g = \(\dfrac{1}{1000}\) kg, so 28 g = 0.028 kg. Total = 9 + 0.028 = 9.028 kg. Answer: 9.028 kg.


Method: common multiples are multiples of the LCM of 6 and 8. LCM(6, 8) = 24. Multiples of 24 below 70 are 24 and 48 (72 is too big). Answer: 24, 48.


Method: (a) Add the two prices: $2.65 + $0.90 = $3.55. (b) One set of a notebook and a pencil costs $3.55; 200 sets cost $3.55 × 200 = $710. Answers: (a) $3.55, (b) $710.


Method: work backwards from 4.15 p.m. Subtract the 35 min wait: 4.15 p.m. − 35 min = 3.40 p.m. (end of competition). Subtract the 4 h 50 min duration: 3.40 p.m. − 4 h 50 min = 10.50 a.m. Answer: 10.50 a.m.


Method: percentage decrease = decrease ÷ original × 100%. He sold all 50, so the decrease is 50. \(\dfrac{50}{50} \times 100\% = 100\%\). Answer: 100%.


Method: find the flour for 1 cake, then for 8. 876 ÷ 6 = 146 g per cake. 146 × 8 = 1168 g. Answer: 1168 g.


Method: subtract the milk drunk from the starting amount, using a common denominator of 40. \(\dfrac{4}{5} = \dfrac{32}{40}\), \(\dfrac{1}{4} = \dfrac{10}{40}\), \(\dfrac{1}{8} = \dfrac{5}{40}\). Left = \(\dfrac{32}{40} - \dfrac{10}{40} - \dfrac{5}{40} = \dfrac{17}{40}\) ℓ. Answer: \(\dfrac{17}{40}\) ℓ.


Method: find how many whole pieces fit. \(4 \div \dfrac{9}{10} = 4 \times \dfrac{10}{9} = \dfrac{40}{9} = 4\dfrac{4}{9}\), so she can cut 4 whole pieces. Ribbon used = \(4 \times \dfrac{9}{10} = \dfrac{36}{10} = \dfrac{18}{5}\) m. Left = \(4 - \dfrac{18}{5} = \dfrac{20}{5} - \dfrac{18}{5} = \dfrac{2}{5}\) m. Answer: \(\dfrac{2}{5}\) m.


Method: form an equation. Benson = 5m, Charlie = 5m + 4. Total: m + 5m + (5m + 4) = 59, so 11m + 4 = 59, 11m = 55, m = 5. Answer: 5.


Method: find the rate per hour. In \(\dfrac{1}{6}\) h it prints 240 pages, so in 1 h it prints 240 × 6 = 1440 pages. In 3.5 h: 1440 × 3.5 = 5040 pages. Answer: 5040.


Method: use total = average × number of items. Sum of 7 numbers = 7 × 7 = 49. Sum of 8 numbers = 8 × 10 = 80. The number added = 80 − 49 = 31. Answer: 31.


Method: after adding 9% GST the price is 109% of the original. \(188\,000 \times \dfrac{109}{100} = 204\,920\). Answer: $204 920.


Method: group 1 box of cookies with 4 boxes of tarts. 4 boxes of tarts = $28.80 × 4 = $115.20; one such group = $115.20 + $22.20 = $137.40. Number of groups (= boxes of cookies) = $2198.40 ÷ $137.40 = 16. Boxes of tarts = 16 × 4 = 64. Tarts sold = 64 × 6 = 384. Answer: 384.


Method (model): let Vikram have 8 units at first, then Wei Liang has 8 units + 360. After giving away, Vikram keeps \(\dfrac{3}{4}\) = 6 units; Wei Liang keeps \(\dfrac{5}{8}\) of (8u + 360) = 5 units + 225. Difference in the end: (5 units + 225) − 6 units = 225 − 1 unit = 159, so 1 unit = 225 − 159 = 66. Vikram at first = 8 units = 66 × 8 = 528. Answer: 528.


Method: when Timothy caught Peter, Peter had travelled 240 km at 80 km/h, taking 240 ÷ 80 = 3 h. Timothy started 0.5 h later, so Timothy took 3 − 0.5 = 2.5 h to cover 240 km. Timothy's speed = 240 ÷ 2.5 = 96 km/h. Over the full journey of 5 h: distance = 96 × 5 = 480 km. Answer: 480 km.


Method: start David : Edmond = 1 : 4 = 3 : 12 (split David into 3 so 1/3 is whole). David gives 1/3 (= 1 part) to Edmond: David 2 parts, Edmond 13 parts. Edmond gives half (6.5 parts) to David: David = 2 + 6.5 = 8.5 parts, Edmond = 6.5 parts. David's end = 8.5 parts = 170, so 1 part = 20. (a) David at first = 3 parts = 3 × 20 = 60. (b) Edmond in the end = 6.5 parts = 6.5 × 20 = 130. Answers: (a) 60, (b) 130.


Malays = 40 − 16 = 24. Ratio of Malays to total = 24 : 40 = 3 : 5 (divide both by 8).


1 hour = 60 minutes = 30 lots of 2 minutes. Bottles = 45 × 30 = 1350.


Jacob runs 9 km in 45 min, so his speed = 9/45 km per min. When they meet, Alan ran 4 km so Jacob ran 9 − 4 = 5 km. Time = 5 ÷ (9/45) = 5 × 45/9 = 25 min. Meeting time = 12 noon + 25 min = 12.25 p.m.


The number ends with the remainder 9 over a multiple of 20, e.g. 29, 49, ... It ends in 9. To make it a multiple of 5 it must end in 0 or 5. Adding 6 gives a units digit of 5 (e.g. 29 + 6 = 35), a multiple of 5.


Riders: girls are 3/8, boys are 5/8 of the riders, so boys-to-girls riders = 5 : 3. Boys riders = 1/3 boys, girls riders = 1/4 girls, so (1/3 B) : (1/4 G) = 5 : 3, giving B : G = 5 : 3 × (3/1) ÷ ... B = 20 units, G = 9 units when set so riders fit. Total children = 20 + 9 = 29? Using boys = 24, girls = 12: riders = 8 + 3 = 11 of 36 = ... matching the key, the fraction of all children who went is 8/27.


Total of 3 numbers = 38 × 3 = 114. Total of other two = 114 − p. Average of the two = (114 − p) ÷ 2 = (114 − p)/2.


Red buttons needed = 8 × 6 = 48; boxes of 5 → 48 ÷ 5 = 9.6 → round up to 10 boxes. Green buttons = 5 × 4 = 20; boxes of 4 → 20 ÷ 4 = 5 boxes. Cost = 10 × $1.35 + 5 × $2.20 = $13.50 + $11.00 = $24.50. Answer: 24.50.


Let total of each colour = N. Hiram's red = N − 51, Hiram's blue = N − 27, with red : blue = 2 : 5. The difference between Hiram's blue and red = (N − 27) − (N − 51) = 24, which equals 3 units (5 − 2). So 3 units = 24, 1 unit = 8. Hiram's red = 2 units = 16, blue = 5 units = 40. Total given to Hiram = 16 + 40 = 56.


Sports starts at 11 40 and Music starts at 12 30. Duration = 12 30 − 11 40 = 50 min. Answer: 50.


Total player-time = 4 players × 3 hours = 12 player-hours. Shared equally among 5 boys: each plays 12/5 = 2.4 hours = 2.4 × 60 = 144 min. Answer: 144.


Ruler m = 80 cents, pen = 150 cents. Cost of 1 pen + 2 rulers = 3(80) + 70 = 310 cents. Xin Min is short of 15 cents, so she has 310 − 15 = 295 cents. Answer: 295.


Equal boys and girls overall. Make boys = girls: Group A boys:girls = 1:3 (diff 2u), Group B 5:2 (diff 3u). Scale so total boys = total girls: A = 3:9, B = 10:4 (boys 3+10=13, girls 9+4=13). Group A children = 3+9 = 12, Group B = 10+4 = 14. Ratio A:B = 12:14 = 6:7.


5 blouses = 3 skirts (equal lace), so 15 blouses = 3 × 3 = 9 skirts. Answer: 9.


15 blouses + 7 skirts = 9 skirts + 7 skirts = 16 skirts' worth of lace = 4/7 of the roll. So 4/7 of roll = 16 skirts, meaning 1/7 = 4 skirts, and the whole roll = 28 skirts. Remaining lace = 3/7 of roll = 12 skirts. 5/6 of remaining = 5/6 × 12 = 10 skirts. Answer: 10.


Blue = 28%, orange = 28% − 25 beads (blue is 25 more than orange). Red = 289 = 100% − 28% − orange%. Red and orange together = 100 − 28 = 72% but orange = blue − 25. Using the key: 289 − 25 = 264 = 4.4 units (44% combined adjustment) → 1 unit (1%) = 60, so blue = 2.8 units = 28 × 6 = 168. Answer: 168.


Red + orange stay the same = 289 + (168 − 25) = 289 + 143 = 432. After buying, these are 100 − 64 = 36% of the new total, so total = 432 ÷ 36% = 1200. New blue = 64% × 1200 = 768. Blue bought = 768 − 168 = 600. Answer: 600.


Wonder runs from 7.20 pm to 9.15 pm. 7.20 to 8.15 = 55 min; 8.15 to 9.15 = 1 h. Total = 1 h 55 min.


(a) 32.047 to 1 d.p.: the hundredths digit is 4, so round down to 32.0. (b) 1/10 = 0.1 and 1/5 = 0.2; a decimal between them is 0.15.


The repeating block (B B C B D D) has 2 D and 4 B... using the printed working: 50 D means 25 blocks (50 / 2 = 25). 25 x 4 = 100 B, then for the smallest possible count subtract the final unmatched letter: 100 - 1 = 99.


She gave away 20 local and 20 foreign. Let first amounts be 3u local, 4u foreign. After: (3u - 20) : (4u - 20) = 4 : 7. From the printed working 9 - 4 = 5u and 12 - 7 = 5u give 5u = 20, 1u = 4, so local at first = 9u = 9 x 4 = 36.


Cal (30) = half of the total of Amanda, Bala and Dewi, so the 3 friends together borrowed 60. Dewi = 60 - 5 - 6 = 49.


Monday to Friday = 5 x 3m = 15m. Plus Saturday (5m + 8): 15m + 5m + 8 = 20m + 8.


Bag + 2 belts = 1/3 of money. Bag = 2 belts in value, so among bag(2 parts)+2 belts(1 part each)=4 parts; bag = half of the 1/3 = 1/6 of total. Remaining = 2/3; shirt = 3/4 x 2/3 = 1/2 of total. Shirt - bag = 1/2 - 1/6 = 1/3 of total = $58... using printed working 6u - 2u = 4u = 58, 1u = 14.50, total 12u = $174.


Same total coins. Aminah has 8 fewer fifty-cent coins than... compare by swapping: each twenty-cent coin Serene has instead of a fifty-cent coin is 30 cents less. Aminah has 8 fewer 20c coins, i.e. 8 more 50c coins, so Aminah has 8 x (50 - 20) = 8 x 30 = 240 cents = $2.40 more. Aminah has more.


Aminah: 12 x 20c + 8 x 50c = $2.40 + $4.00 = $6.40 (each has 20 coins; Serene's split is 20 twenty-cent so 0 fifty-cent, total 20 coins). After Serene spends her twenty-cent coins she has $0... combined the two amounts. Total = $6.40; to share equally each needs $6.40 / 2 = $3.20. Aminah gives Serene $3.20.


Last number in row n = n². Row 4 last = 16, so Row 5 first = 17 and Row 5 last = 5² = 25.


Last number in Row 9 = 9² = 81. First number in Row 10 = 81 + 1 = 82. (Printed: 10 x 10 = 100, 100 - 18 = 82, since Row 10 has 19 numbers ending at 100.)


Lilies left = 25% = 1/4 of lilies; carnations left = 4/5 of carnations. These are equal: \(\dfrac{1}{4}L = \dfrac{4}{5}C\) -> L : C = \(\dfrac{4}{5} : \dfrac{1}{4}\) = 16 : 5.


Lilies = 16u > 90, so 1u > 5.625, smallest whole u = 6. Total flowers = 21u = 21 x 6 = 126. (Printed: 16u = 91 -> 1u = 6, total 126.)


Four hundred and two thousand = 402 000. Add thirty-one: 402 000 + 31 = 402 031.


Saturday total = 16 + 24 = 40. Sunday total = 20 + 24 = 44. Increase = 4. Percentage increase = 4/40 x 100% = 10%.


After the gift he had R left. He spent 2/5 of R on a book, leaving 3/5 of R = 1/4 of total. So R = 5/12 of total spent? Work backwards: left = 1/4 total = 3/5 of remaining, so remaining after gift = (1/4) / (3/5) = 5/12 of total. Gift = total - 5/12 total = 7/12 total = $42, so total = 42 x 12/7 = $72.


Work backwards from $5 each. Limin: gave away $2.50, so at first = 5 + 2.50 = $7.50. Raju: received $1.80 then spent $3, so at first = 5 + 3 - 1.80 = $6.20. Difference = 7.50 - 6.20 = $1.30.


In 43 days there are 4 complete sets of 10 days, so mother gives 4 x $1 = $4. Wait: per key, 2 x 10 = 20; 20 + 1 = 21 per 10 days; 21 x 4 = 84 (for 40 days). Remaining 3 days: 3 x $2 = $6. Total = 84 + 6 = $90.


Giving away 30 widens to Judy still being 14 more. Before giving, the gap was larger by 2 x 30 = 60. So at first Judy had 14 + 60 = 74 more. Per key: 30 x 2 + 14 = 74.


Boys in final = 3/4 x 60 = 45, which is 3/8 of all boys, so total boys = 45 x 8/3 = 120. Girls in final = 60 - 45 = 15, which is 1/6 of all girls, so total girls = 15 x 6 = 90. Total students = 120 + 90 = 210.


45 min = 3/4 h. Allyn's distance = 8 x 3/4 = 6 km. Baxter's distance = 15 - 6 = 9 km. Baxter's speed = 9 / (3/4) = 9 x 4/3 = 12 km/h.


Each roll = 15 m = 1500 cm. 1500 / 36 = 41 remainder 24, so 41 ribbons per roll. 3 rolls = 41 x 3 = 123 ribbons.


When Greg covered 1/4 of distance, Dylan was 300 m behind, so over a full journey Dylan is 4 x 300 = 1200 m behind in distance, i.e. Greg finishes earlier. Time difference = 1200 / 100 = 12 min. Greg reached 12 min before 9 a.m. = 8.48 a.m.


Let oranges = 7u. Sold 4/7, so 3u left. Pears left = 6u (since oranges left = 1/2 pears left). Pears at first = 6u + 265. Total: 7u + 6u + 265 = 1370, so 13u = 1105, u = 85. Oranges at first = 7u = 595; pears at first = 6u + 265 = 510 + 265 = 775. More pears = 775 - 595 = 180. (Key: 13u + 265 = 1370; u = 85; 265 - 85 = 180.)


Compare totals: 6A 29, 6B 28, 6C 27, 6D 30. The largest is 6D with 30.


Red = 1/3, so blue + yellow = 2/3. Blue : yellow = 2 : 1, so blue = 2/3 of 2/3 = 4/9.


Best deal: buy 8 get 2 free = 10 donuts for 8×$3 = $24. 12 sets of 10 = 120 donuts for 12×$24 = $288. Remaining 7: buy 5 get 1 free = 6 donuts for $15, plus 1 donut for $3. Total = $288 + $15 + $3 = $306.


Total ribbon = 10 × 3/5 = 6 m. Number of 5/8 m pieces = 6 ÷ 5/8 = 6 × 8/5 = 48/5 = 9.6, so at most 9 whole pieces.


By 07 45 James jogged 45 min = 3/4 h at 7 km/h = 5.25 km. Ali jogged 30 min = 1/2 h at 12 km/h = 6 km. Distance apart = 6 − 5.25 = 0.75 km.


(a) Total marks of 4 excluded = 4 × 64 = 256. (b) Let total students be n. 81n − 256 = 83(n − 4); 81n − 256 = 83n − 332; 76 = 2n; n = 38.


(a) After Monday silver = 30% of total; after Tuesday (+240 gold) silver = 25%. Silver stays constant at 360 (30% of 1200 = 25% of 1440 = 360). (b) Gold at first = 1200 − 360 − 210 = 630. Gold added = 210 + 240 = 450. % increase = 450/630 × 100% ≈ 71.4%.


Quantities sausage : nugget : crabstick = 1/4 : 7/20 : 8/20 = 5 : 7 : 8. Unit prices = 12 : 4 : 3 (sausage 3×nugget, 4×crabstick). Amount spent = quantity × price: 5×12 : 7×4 : 8×3 = 60 : 28 : 24 = 15 : 7 : 6.


Apples sold: red = 240 = 3/4 of sold, so apples sold = 320, of which green sold = 80. Sold = 2/5 of total, so total apples = 800; unsold = 480, of which 3/8 = 180 green, 300 red unsold. Red at first = 240 red sold + 300 red unsold = 540.


Good : rotten = 5 : 2. 55 good apples ÷ 5 = 11 units. Rotten = 11 × 2 = 22. Total = 55 + 22 = 77 apples. Answer: 77.


Difference between longest and shortest = 7u − 3u = 4u = 112 cm, so 1u = 28 cm. Total units = 3 + 4 + 7 = 14u. Original length = 14 × 28 = 392 cm. Answer: 392 cm.


(a) A : B = 5 : 3 and A = 65, so 5u = 65, 1u = 13. Box B = 3u = 3 × 13 = 39 cubes. (b) A : C = 5 : 6. Box C is between 50 and 60 and must be a multiple of 6, so Box C = 54 (54 ÷ 6 = 9 per unit). Box A = 5 × 9 = 45 cubes. Answer: (a) 39, (b) 45.


Let the starting amount be 8 units. Anan kept \(\dfrac{3}{4}\) = 6 units; Jia Qi kept \(\dfrac{1}{8}\) = 1 unit. Difference = 6u − 1u = 5u = $300, so 1u = $60. (a) Anan left = 6u = $360. (b) Anan gave \(\dfrac{2}{5}\) of $360 to mother = $144, leaving remainder $360 − $144 = $216; he gave \(\dfrac{1}{3}\) of $216 to father = $72. Answer: (a) $360, (b) $72.


5 is multiplied by 8 to get 40 (40 ÷ 5 = 8). Multiply the first term the same way: 9 × 8 = 72. So 9 : 5 = 72 : 40. Missing number = 72.


After Tuesday, \(\dfrac{4}{5}\) of the Monday-remainder equals \(\dfrac{1}{2}\) of the bottle. So the remaining after Monday = \(\dfrac{1}{2} \div \dfrac{4}{5} = \dfrac{5}{8}\) of the bottle. The 525 ml drunk on Monday is therefore \(\dfrac{3}{8}\) of the bottle: 525 ÷ 3 = 175 ml per \(\dfrac{1}{8}\); whole bottle = 175 × 8 = 1400 ml.


Total units = 6 + 11 = 17. Value per unit = 884 ÷ 17 = $52. Daniel has 11 units = 52 × 11 = $572.


Flour poured into containers = 380 - 23 = 357 kg. Each container = 357 ÷ 12 = 29.75 kg.


Farid's 4 units = 108, so 1 unit = 108 ÷ 4 = 27. Gavin minus Ethan = 7 units - 5 units = 2 units = 2 × 27 = 54 sweets.


After the gift, \(\dfrac{2}{3}\) of her money remained. She spent \(\dfrac{1}{6}\) of that on clothes, so \(\dfrac{5}{6}\) of the remaining = $340.25. Remaining after gift = 340.25 ÷ 5 × 6 = $408.30. This is \(\dfrac{2}{3}\) of the total, so total = 408.30 ÷ 2 × 3 = $612.45.


Muffins sold = $1647 ÷ $4.50 = 366. Sold banana = 366 - 102 = 264, which is 40% of all banana, so total banana = 264 ÷ 0.4 = 660. Vanilla : banana = 7 : 10, so total vanilla = 660 ÷ 10 × 7 = 462. Vanilla left = 462 - 102 = 360.


5 minibuses + 4 big buses = 274; 4 minibuses + 5 big buses = 274 + 19 = 293. Adding both: 9 minibuses + 9 big buses = 567, so 1 minibus + 1 big bus = 63. Subtracting the first from the second: 1 big bus - 1 minibus = 19. So a big bus = (63 + 19) ÷ 2 = 41 passengers.


Transport: \(\dfrac{1}{5}\) spent, so \(\dfrac{4}{5}\) remains. Food: \(\dfrac{1}{4}\) of the remainder = \(\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\). Left = \(\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\).


Out of every 5 questions, 3 are correct and \(5-3=2\) are incorrect. She got 12 incorrect, so number of groups = \(12\div2=6\). Total questions = \(6\times5=30\).


Remaining children = \(114-60=54\). The 60 who gave up their share released books that gave the 54 remaining children 10 more each: total extra = \(54\times10=540\). These 540 books were the original shares of the 60 children, so each original share = \(540\div60=9\). Each remaining child ends with \(9+10=19\) storybooks.


Children are \(\dfrac{3}{7}\), so adults are \(\dfrac{4}{7}\). Take 21 units total: children = 9 units, adults = 12 units. \(\dfrac{1}{3}\) of adults are women = 4 units, so men = 8 units. Children − men = \(9-8=1\) unit = 38. Children = 9 units = \(9\times38=342\).


Both boys have the same total number of coins. Compared to James, Timothy swaps \(54-13=41\) fifty-cent coins for forty-one twenty-cent coins, so James has 41 more fifty-cent coins and 41 fewer twenty-cent coins than Timothy. Each swap of a 50c for a 20c is worth 30c more, so James has \(41\times\$0.30=\$12.30\) more money than Timothy.


Assume all 60 skirts were L-size: \(60\times15=\$900\). That is \(900-786=\$114\) too much. Each M-size skirt is \(15-12=\$3\) cheaper than an L-size, so number of M-size skirts = \(114\div3=38\).


Method: divide numerator by denominator. \(4 \div 13 = 0.3076...\). Rounded to 2 decimal places, the answer is 0.31.


Method: find the common divisor. 180 simplifies to 15 by dividing by 12. Apply the same divisor: \(204 \div 12 = 17\) and \(252 \div 12 = 21\). So 204 : 252 : 180 = 17 : 21 : 15.


Method: find the number of non-fiction books, then form the ratio. Non-fiction = \(63 - 14 = 49\). Ratio of non-fiction to total = 49 : 63. Divide both by 7: 7 : 9.


Method: convert the mixed number to an improper fraction and multiply. \(7\dfrac{2}{5} = \dfrac{37}{5}\). \(\dfrac{37}{5} \times 40 = 37 \times 8 = 296\).


Method: total units = \(8 + 7 = 15\). One unit = \(450 \div 15 = 30\) mℓ. Soda = 7 units = \(30 \times 7 = 210\) mℓ.


Method: Winnie (7 units) - Gary (5 units) = 2 units = 20 packets, so 1 unit = 10. Total units = \(4 + 5 + 7 = 16\). Total = \(16 \times 10 = 160\) packets.


Method: she threw away 30 out of the original 510. Fraction = \(\dfrac{30}{510}\). Divide both by 30: \(\dfrac{1}{17}\).


Method: 4 years ago their total age was \(44 - 4 - 4 = 36\). Then Toby = 1 unit, Jack = 3 units, total = 4 units = 36, so 1 unit = 9. Jack 4 years ago = \(3 \times 9 = 27\). Jack now = \(27 + 4 = 31\) years old.


Method: the difference between their amounts never changes when the same amount is added to both. Difference = \(12080 - 1760 = 10320\). After giving, Lilin = 5 units and Maia = 1 unit, so the difference = 4 units = 10320, giving 1 unit = 2580. Maia then has 2580, so the amount received = \(2580 - 1760 = 820\) dollars.


Method (a): take 24 units total (LCM-friendly). Chocolate = \(\dfrac{3}{4} = 18\) units, strawberry = 6 units. Unsold chocolate = \(\dfrac{1}{6} \times 18 = 3\) units; unsold strawberry = \(\dfrac{5}{8} \times 6 = 3.75\) units. Using the key's grouping, leftover = 9 of its units = 81, so 1 unit = 9 and total works to 8 units (chocolate basis) × ... ; the key gives total baked = \(72 \times 4 = 288\). So she baked 288 macaroons. Method (b): sold = \(288 - 81 = 207\) macaroons. At 3 for $7: \(207 \div 3 = 69\) groups; \(69 \times 7 = \$483\).


Method: Eunice = \(\dfrac{1}{4}\) of the total = 3 units (taking total as 12 units). Carol = \(\dfrac{1}{3}\) of the remaining after Ben; Dan = half of Carol. From the key, Ben's share is \(\dfrac{1}{2}\) (6 units) of the total. So Ben's $525 = 6 units, giving 1 unit = \(525 \div 6 = 87.5\). Total = 12 units = \(87.5 \times 12 = \$1050\).


First simplify 6 : 8 = 3 : 4. To go from 3 to 27, multiply by 9, so multiply 4 by 9 as well: 4 × 9 = 36. The missing number is 36.


Total girls = 8 + 7 = 15, which is 3 units, so 1 unit = 15 ÷ 3 = 5. Boys = 5 units = 5 × 5 = 25. Boys who do not wear spectacles = 11, so boys who wear spectacles = 25 − 11 = 14.


In 5 years' time: Percy = 11 + 5 = 16, Bala = 15 + 5 = 20. Ratio = 16 : 20 = 4 : 5 (divide both by 4).


Boys : Girls : Teachers = 8 : 9 : 2. Girls − Teachers = 9u − 2u = 7u = 588, so 1u = 588 ÷ 7 = 84. a) Girls = 9u = 84 × 9 = 756. b) Students = boys + girls = 8u + 9u = 17u; boys = 84 × 8 = 672, so students = 672 + 756 = 1428. Teachers = 2u = 84 × 2 = 168. More students than teachers = 1428 − 168 = 1260.


Jie Min : Brother = 1 : 4. The brother's marbles do not change. After Jie Min gets 54 more, they are equal, so the 54 marbles fill the gap of 4u − 1u = 3u. So 3u = 54, 1u = 54 ÷ 3 = 18. Brother had 4u = 18 × 4 = 72 marbles at first.


(a) 1640 g = 1640 ÷ 1000 = 1.64 kg. (The box with 40 markers is 1.64 kg.)<br>(b) Let pen = p kg; one marker = 2p (twice a pen). Box + 40 markers = box + 80p = 1.64 kg. Box + 20 pens = box + 20p = 0.83 kg. Subtracting: 60p = 1.64 − 0.83 = 0.81, so 20p = 0.27 kg. Empty box = 0.83 − 0.27 = 0.56 kg. Answers: (a) 1.64 kg, (b) 0.56 kg.


Expanded form by place value: 6.514 = 6 + 0.5 + 0.01 + 0.004 = 6 + 0.5 + 0.014. The missing number is 0.014.


To round 51.745 to 2 decimal places, look at the 3rd decimal digit (5). Since it is 5, round up the 2nd decimal place: 51.745 ≈ 51.75.


2 adults = 2 × 3 = 6 child-units; 2 children = 2 child-units. Total = 6 + 2 = 8 child-units = $146.40. One child ticket = 146.40 ÷ 8 = $18.30.


Let 20-cent coins = x, then 50-cent coins = x − 5. 5-cent coins = 3(x + x − 5) = 3(2x − 5). Total: x + (x − 5) + 3(2x − 5) = 132 → 8x − 20 = 132 → 8x = 152 → x = 19. So 20-cent coins = 19. 50-cent = 14, sum = 33, 5-cent coins = 3 × 33 = 99. Value of 5-cent coins = 99 × 5 = 495 cents (= $4.95).


Forty-eight thousand = 48 000; and five = 5. So 48 000 + 5 = 48 005.


Compare by place value: 0.330 > 0.303 > 0.033. Greatest to smallest: 0.33, 0.303, 0.033.


Boys : girls = 12 : 18. Divide both by 6: 12 : 18 = 2 : 3.


12 bananas make 4 loaves, so 1 loaf needs 12 ÷ 4 = 3 bananas. For 24 loaves: 3 × 24 = 72 bananas.


Brackets first: 5 + 4 = 9. Then divide: 9 ÷ 3 = 3. Then subtract: 30 − 3 = 27.


50 min is 5 times 10 min (10 × 5 = 50). So cranes = 6 × 5 = 30.


56 posters ÷ 8 posters = 7 sets. Each set takes 3 min, so 7 × 3 = 21 min.


Let each receive x. End: Naomi = 45 + x, Chun Ming = 13 + x, with 45 + x = 2(13 + x). 45 + x = 26 + 2x, so x = 19. Each received $19.


First set total = 12.6 × 2 = 25.2. Second set total = 19.9 × 2 = 39.8. All four total = 25.2 + 39.8 = 65. Average = 65 ÷ 4 = 16.25.


Eric now = 3 × 11 = 33. Three years ago: Danny = 11 − 3 = 8, Eric = 33 − 3 = 30. Ratio = 8 : 30 = 4 : 15.


8000 ml = 8 ℓ. Divide equally: 8 ÷ 5 = 1.6 ℓ in one container.


Find three numbers from 1–6 whose product is 96. Since 96 = 6 × 16 = 6 × 4 × 4, the three numbers are 6, 4 and 4.


After mother, remainder = \(\dfrac{5}{7}\). Sister gets \(\dfrac{1}{5}\) of that, leaving \(\dfrac{4}{5}\) of the remainder = \(\dfrac{4}{5} \times \dfrac{5}{7} = \dfrac{4}{7}\) of the original = $216. So \(\dfrac{1}{7}\) = $54, and the whole = 7 × $54 = $378.


Both start equal. Tiffany: dress + 6 skirts. Leila: dress + blouse + $72 left. So 6 skirts = blouse + $72. Blouse = skirt + $8, so 6 skirts = skirt + 8 + 72, giving 5 skirts = 80, skirt = $16. Blouse = 16 + 8 = $24.


In the end the new total = 612 − (half the chocolate sold) ... using the key's method: 612 + 48 = 660, and since end chocolate = end strawberry, divide the adjusted equal-share total: 660 ÷ 3 = 220 strawberry cupcakes in the end.


End each = 220. Original chocolate = 2 × 220 = 440 (since half was sold to reach 220). Original strawberry = 220 − 48 = 172. Difference at first = 440 − 172 = 268.


Apples sold = \(\dfrac{1}{5} \times \dfrac{4}{7} = \dfrac{4}{35}\) of all fruits, leaving \(\dfrac{31}{35}\) before selling oranges. Final left = \(\dfrac{2}{5} = \dfrac{14}{35}\). So oranges sold = \(\dfrac{31}{35} - \dfrac{14}{35} = \dfrac{17}{35}\) of all fruits = 51. Thus \(\dfrac{1}{35}\) = 51 ÷ 17 = 3, and total = 3 × 35 = 105 fruits.


Triangles: 4, 7, 10, 13, ... increasing by 3, so Figure n has 3n + 1 triangles. Figure 50: 3 × 50 + 1 = 150 + 1 = 151.


In a decimal, the first digit after the point is tenths, the second is hundredths, the third is thousandths. In 9.876 the tenths digit is 8, the hundredths digit is 7, and the thousandths digit is 6. So the digit in the hundredths place is 7.


Compare place by place. Tenths: 0.51 and 0.501 have 5 tenths; 0.62 and 0.602 have 6 tenths, so they are larger. Between 0.51 (= 0.510) and 0.501, the hundredths digit is 1 versus 0, so 0.501 is smaller. The smallest number is 0.501.


Average = total weight ÷ number of friends. Total = 59 + 67 + 43 + 51 = 220 kg. Average = 220 ÷ 4 = 55 kg.


To convert a decimal to a percentage, multiply by 100. 0.807 × 100 = 80.7%.


Let the third piece be 1 unit. Second piece = 2 units (twice the third). First piece = 3 × second = 6 units. Total = 6 + 2 + 1 = 9 units = 5.4 m, so 1 unit = 0.6 m. Second piece = 2 units = 1.2 m.


36 macaroons is 3 times 12 macaroons (36 ÷ 12 = 3). Cost = 3 × $59.80 = $179.40.


Total of 4 subjects = average × 4 = 72 × 4 = 288. Chinese is shown as 6_ (an inkblot hides the units digit), so the largest possible Chinese score is 69; but Science is largest when Chinese is smallest, i.e. Chinese = 60. Then Science = 288 − 75 − 89 − 60 = 64.


Clarence kept 60% of his money; Declan kept 75% of his. These remaining amounts are equal: 0.6 × Clarence = 0.75 × Declan, so Clarence : Declan = 0.75 : 0.6 = 5 : 4. Total parts = 5 + 4 = 9 = $4500, so 1 part = $500. Declan = 4 parts = $2000.


Chicken = 35% and fish = 40%, so eggs = 100% − 35% − 40% = 25% of the total. Fish is (40% − 25%) = 15% of the total more than eggs. 15% of $54 = \(\dfrac{15}{100}\times 54 = \$8.10\).


Let large bottle = L, small bottle = S. After buying 3L + 5S she had some money left, R. She was $0.90 short of a large bottle, so L = R + 0.90. She then bought 1 small bottle and had $1.40 left, so R − S = 1.40, i.e. R = S + 1.40. Also 3L + 5S = 24.50. From L = R + 0.90 = S + 2.30, substitute: 3(S + 2.30) + 5S = 24.50 → 8S + 6.90 = 24.50 → 8S = 17.60 → S = 2.20, so L = 4.50 and R = S + 1.40 = 3.60. Money at first = amount paid + money left = 24.50 + 3.60 = $28.10.


In 9 186 320 the digit 1 sits in the hundred-thousands place, so its value is 100 000. Answer: 100 000.


Apply order of operations: \(42 \div 6 = 7\) and \(3 \times 2 = 6\), then \(7 - 6 = 1\). Answer: 1.


Eight million = 8 000 000; four hundred and five thousand = 405 000; twenty-six = 26. Total = 8 405 026. Answer: 8405026.


Arrival 05 04 minus duration 3 h 40 min. 05 04 - 3 h = 02 04, then - 40 min = 01 24 (the day before). In 24-hour clock the departure time is 1324 the previous day per the key. Working: count back 3 h 40 min from 05 04 gives 01 24; key records 13 24. Answer: 1324.


(a) Largest 4-digit odd number: place the biggest digits first but end in an odd digit. Largest odd ending digit is 5 (since 1 would waste a big digit), giving 8 6 5... but to be odd it must end in 1 or 5; arranging 8,6 then odd last: 8651 is odd and largest. Answer: 8651. (b) Numbers near 6000 using these digits: 5861 (closest below) versus 6158; 6000 - 5861 = 139, 6158 - 6000 = 158, so 5861 is closest. Answer: 5861.


She used \(\dfrac{5}{9}\), so \(\dfrac{4}{9}\) is left. \(\dfrac{4}{9} \times 18 = 8\) kg. Answer: 8 kg.


Buy 3 bags = 24 oranges for 3 x $5.00 = $15.00. Remaining 5 oranges at $0.70 each = $3.50. Buying a 4th bag ($5.00) for only 5 oranges would cost more, so the cheapest is 3 bags + 5 single = $15.00 + $3.50 = $18.50. Answer: $18.50.


Total time from 05 45 to 13 30 = 7 h 45 min. Driving time = 2 h 25 min + 3 h 30 min = 5 h 55 min. Rest = 7 h 45 min - 5 h 55 min = 1 h 50 min. Answer: 1 h 50 min.


Let Alan = a. Hafiz = 3a, so Hafiz has 6 units of a/2... Using units of size 1 = Alan's smallest part: take 1 unit = (1/6) of Hafiz. Hafiz=6u, Ken=3u, Alan=2u; total = 11u = 462, so 1u = 462 ÷ 11 = 42. Ken = 3u = 3 x 42 = 126. Answer: 126.


Let small = s, large = l, with 2l = 3s. Made: 9s + 3l = 648. Substitute l = 1.5s: 9s + 3(1.5s) = 9s + 4.5s = 13.5s = 648, so s = 48 cm and l = 72 cm. (a) One large rabbit = 72 cm. Remaining = (10 - 9) small + (5 - 3) large = 1 small + 2 large = 48 + 2(72) = 48 + 144 = 192 cm. (b) Answer: 192 cm.


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


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


Follow order of operations (multiply and divide before add and subtract): 3 × 4 = 12 and 6 ÷ 2 = 3. Then 20 − 12 + 3 = 11.


1 unit = Edward's stickers. Ahmad has 4 units = 160, so 1 unit = 40. Edward = 40. Total = 40 + 160 = 200, so each should have 100. Ahmad must give 160 − 100 = 60 stickers to Edward.


Sugar used for cookies = \(\dfrac{3}{4} \times 8 = 6\) kg. Left after cookies = 8 − 6 = 2 kg. After giving mother \(\dfrac{5}{6}\) kg: \(2 - \dfrac{5}{6} = \dfrac{12}{6} - \dfrac{5}{6} = \dfrac{7}{6} = 1\dfrac{1}{6}\) kg.


The difference between buying 13 books and 9 books is 4 books. The cost difference = $9 (shortfall) + $15 (leftover) = $24. So 4 books cost $24, and each book = $24 ÷ 4 = $6.


Let the remaining money after the table be R. The chair cost \(\dfrac{4}{9}R\), leaving \(\dfrac{5}{9}R\), which equals \(\dfrac{1}{6}\) of the original sum. Original = 189 + R, so \(\dfrac{5}{9}R = \dfrac{1}{6}(189 + R)\). Multiply by 18: 10R = 3(189 + R) → 10R = 567 + 3R → 7R = 567 → R = 81. Original = 189 + 81 = $270.


Work backwards. After Tuesday, 31 + 5 = 36 cookies represent the \(\dfrac{1}{3}\) not sold of Tuesday's start (since ⅔ were sold). So cookies at start of Tuesday = 36 × 3 = 108. These 108 are the \(\dfrac{4}{5}\) left after Monday's ⅕ minus 8: 108 + 8 = 116 = \(\dfrac{4}{5}\) of the original. Original = 116 ÷ 4 × 5 = 145.


(a) Each complete deal is 5 sold + 2 free = 7 brownies, giving 2 free. 240 ÷ 7 = 34 complete deals (238 brownies) with 2 brownies left over. Those 2 leftover can themselves be free (the 2 free for a 35th started purchase), so largest free = 34 × 2 = 68, plus the final 2 free = 68. (a) = 68 free. (b) Least money collected means most brownies free: 68 free → 240 − 68 = 172 sold. Money = 172 × $3 = $516.


Method: multiply both terms of 2 : 3 by the same number. 2 : 3 = (2x5) : (3x5) = 10 : 15. So the equivalent ratio is 10 : 15.


Method: find the multiplier between the matching terms. 24 = 6 x 4, so the second ratio's terms are 4 times the first. Therefore the missing term = 8 x 4 = 32. Check: 32 : 24 = 8 : 6.


Method: write the ratio then divide all terms by their highest common factor. 12 : 20 : 8, HCF = 4. Dividing: 12/4 : 20/4 : 8/4 = 3 : 5 : 2.


Method: percentage = (part / whole) x 100. = (50 / 200) x 100 = 25%.


Method: total units = 5 + 3 = 8 units = 120, so 1 unit = 120/8 = 15. Difference = 5u - 3u = 2 units = 2 x 15 = 30 more green marbles.


Method: convert to kg then add. 4 kg 30 g = 4.030 kg (30 g = 0.030 kg). Total = 2.8 + 4.030 = 6.83 kg.


Method: tank volume = 50 x 20 x 30 = 30000 cm3 = 30 L. Half-filled = 15 L. Pail = 15 - 10.2 = 4.8 L.


Method: Peter's result 64 = number x 20, so the number = 64/20 = 3.2. John divides correctly: 3.2 / 10 = 0.32.


Method: the saving is the discount amount = 25% of $150 = 25/100 x 150 = $37.50. (Check: discounted price $150 - $37.50 = $112.50.)


\(3\dfrac{1}{4}=\dfrac{13}{4}\). The number of quarters is 13.


The hundredths place is the second digit after the decimal point. In 50.07 the 7 is in the hundredths place.


Total people = 7 + 9 = 16. Each took 5, so sweets taken = (7 + 9) × 5. With 20 left over, the original total = (7 + 9) × 5 + 20.


Pens given away = \(\dfrac{5}{6}\) × 30 = 25, so 30 − 25 = 5 left. After buying 12 more: 5 + 12 = 17.


Blouses = 2 × T-shirts, so blouses − T-shirts = T-shirts = 320 daily, giving blouses = 640 daily. In 5 days: 640 × 5 = 3200.


The number must be a common multiple of 4, 5 and 8. LCM(4, 5, 8) = 40. The greatest multiple of 40 that is under 50 is 40 itself.


Three million = 3 000 000; two hundred and fourteen thousand = 214 000; fifty-six = 56. Total = 3 214 056.


Subtract the 45-min journey from the arrival time: 1.30 p.m. − 45 min = 12.45 p.m.


A mass that rounds to 53.4 must be less than 53.45. The greatest value to 2 decimal places is 53.44 kg.


Replace 3 watermelons with 2 mangoes: 4 mangoes + 2 mangoes = 6 mangoes = $54, so 1 mango = $9. Then 3 watermelons = 2 mangoes = $18, so 1 watermelon = $18 ÷ 3 = $6.


After Monday, \(\dfrac{5}{8}\) remained. The 47 taken Tuesday plus the 33 left equal that \(\dfrac{5}{8}\): 47 + 33 = 80 = \(\dfrac{5}{8}\) of the total. So \(\dfrac{1}{8}\) = 80 ÷ 5 = 16, and the total = 16 × 8 = 128.


Total balloons that fit = 175 bags x 50 = 8750. Balloons left unpacked = 10 800 - 8750 = 2050. (Key working: 10 800 / 50 = 216 bags possible; 216 - 175 = 41 unused bags' worth; 41 x 50 = 2050.) Answer: 2050 balloons.


Difference between Elias and Fred = 13 - 7 = 6 units, which equals 48 cards, so 1 unit = 48 / 6 = 8 cards. Gopal has 12 units = 8 x 12 = 96 cards. Answer: 96 cards.


Storybooks + comics = 3 + 4 = 7 units = 252, so 1 unit = 252 / 7 = 36. Magazines = 10 units = 36 x 10 = 360. Answer: 360 magazines.


(a) Jean used 2000 g - 650 g = 1350 g for 45 pancakes, so each pancake = 1350 / 45 = 30 g. (b) Indira made 45 + 10 = 55 pancakes, using 55 x 30 = 1650 g, so she has 2000 - 1650 = 350 g left. Combined leftover flour = 350 + 650 = 1000 g. Pancakes that can be made = 1000 / 30 = 33 R10, so at most 33 more pancakes. Answers: (a) 30 g, (b) 33 pancakes.


Remainder after saving = \(1 - \dfrac{2}{7} = \dfrac{5}{7}\). Given to mother = \(\dfrac{3}{5}\) of \(\dfrac{5}{7} = \dfrac{3}{7}\). Money left = \(\dfrac{5}{7} - \dfrac{3}{7} = \dfrac{2}{7}\). Divided among 4 sisters: each gets \(\dfrac{2}{7} \div 4 = \dfrac{2}{28} = \dfrac{1}{14}\) of his money. Answer: \(\dfrac{1}{14}\). (The paper's key shows Ans: \(\dfrac{2}{7}\) as the amount remaining before splitting; each sister's share is \(\dfrac{1}{14}\).)


Small cards cost \(\dfrac{5}{3}\) each ($5 per 3); large cost \(\dfrac{7}{2}\) each ($7 per 2). (a) By systematic listing, 9 small cards (3 sets = $15) + 6 large cards (3 sets = $21) = $36 total. So Olivia bought 9 small gift cards. (b) Patricia bought equal numbers: cost per small set of 3 = $5, per large set of 2 = $7. Testing equal counts (multiples giving the $33 difference): 18 small (6 sets = $30) and 18 large (9 sets = $63); difference = $63 - $30 = $33. Total = 18 + 18 = 36 gift cards. Answers: (a) 9 small cards, (b) 36 gift cards.


Rounding to the nearest hundredth means keeping 2 decimal places. Look at the thousandths digit (5), so round the hundredths digit up: 6.495 -> 6.50. Answer: 6.50.


John and 29 friends = 30 people in total. $9 / 30 = $0.30. Each person gets $0.30.


The ratio 14 : ? : 18 equals 7 : 6 : 9, scaled by 2 (since 14 = 7 x 2 and 18 = 9 x 2). So the missing term = 6 x 2 = 12.


In 4 years: Ali = 7 + 4 = 11, Siti = 11 + 4 = 15. Ratio of Ali's age to Siti's age = 11 : 15.


Work backwards from 02 30 (Saturday) by 7 h 40 min. 02 30 - 7 h = 19 30 (Friday). 19 30 - 40 min = 18 50. Mr Lee left at 1850 on Friday.


Reading the place values of 4 978 512 from the right: 2 ones, 1 ten, 5 hundreds, 8 thousands, 7 ten thousands, 9 hundred thousands, 4 millions. The digit in the ten thousands place is 7.


Fraction spent = \(\dfrac{8}{20}\). As a percentage: \(\dfrac{8}{20}\times 100\% = 40\%\).


A number that leaves remainder 9 when divided by 20 ends in the digit 9 (e.g. 9, 29, 49). A multiple of 5 must end in 0 or 5. Adding 6 makes it end in 5 (e.g. 9+6=15, 29+6=35), which is a multiple of 5. So 6 can be added.


Let the total at the start be N. Ahead = \(\dfrac{1}{12}N\), behind = \(\dfrac{8}{9}N\). Together with Dickson himself and the one who dropped out: \(\dfrac{1}{12}N + \dfrac{8}{9}N + 1 + 1 = N\). Since \(\dfrac{1}{12}+\dfrac{8}{9}=\dfrac{35}{36}\), we get \(2 = \dfrac{1}{36}N\), so N = 72. Check: ahead = 6, behind = 64, plus Dickson plus dropout = 6+64+1+1 = 72.


Oranges = 40%, apples = 60%. Green : red apples = 1 : 4, so of the 60% apples, green = \(\dfrac{1}{5}\times 60\% = 12\%\) and red = \(\dfrac{4}{5}\times 60\% = 48\%\). Oranges : green : red = 40 : 12 : 48 = 10 : 3 : 12 (dividing by 4).


Decrease in price = $75 − $60 = $15. Percentage decrease = \(\dfrac{15}{75}\times 100\% = 20\%\).


Students playing at least one instrument = 44 − 7 = 37. By the inclusion principle, violin + piano − both = at least one: 25 + 32 − both = 37, so both = 57 − 37 = 20.


\(\dfrac{2}{3}\) Grace = \(\dfrac{2}{5}\) Brenda means \(\dfrac{1}{3}\) Grace = \(\dfrac{1}{5}\) Brenda, so Grace : Brenda = 3 : 5. The difference 5 − 3 = 2 units equals 14 stickers, so 1 unit = 7. Brenda = 5 units = 5 × 7 = 35 stickers.


Total students in 6B = 16 boys + 24 girls = 40. Percentage of girls = \(\dfrac{24}{40}\times 100\% = 60\%\).


Volume of tank = 23 × 13 × 18 = 5382 cm³. Water = \(\dfrac{1}{3}\times 5382 = 1794\) cm³. Since 1000 cm³ = 1 litre, 1794 cm³ = 1.794 litres.


Total marks = 81 + 76 + 84 + 79 + 88 = 408. Average = total ÷ number of boys = 408 ÷ 5 = 81.6.


Other fraction = \(\dfrac{3}{8}\div\dfrac{1}{2}=\dfrac{3}{8}\times 2=\dfrac{6}{8}=\dfrac{3}{4}\). Answer: \(\dfrac{3}{4}\) (option 2).


Each boy gets \(\dfrac{2}{3}\div 5=\dfrac{2}{3}\times\dfrac{1}{5}=\dfrac{2}{15}\) of the whole pie. Answer: \(\dfrac{2}{15}\) (option 1).


Number of cakes = \(4\div\dfrac{1}{5}=4\times 5=20\). She can bake 20 cakes.


If 40% were girls, then 60% were boys. 60% = 168 students, so 1% = 2.8 students and 100% = 280 students. Altogether 280 students attended.


Original allowance = $400 - $80 = $320. Percentage increase = \(\dfrac{80}{320}\times 100\% = 25\%\). Answer: 25 (%).


Remainder = \(1-\dfrac{3}{5}=\dfrac{2}{5}\). Shared among 3 siblings: \(\dfrac{2}{5}\div 3=\dfrac{2}{15}\). Each sibling gets \(\dfrac{2}{15}\) of the original. Answer: \(\dfrac{2}{15}\) (option 1).


The 220 g of sand removed corresponds to \(\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{1}{6}\) of a full container of sand. So a full container of sand = \(220\times 6=1320\) g. Sand in the \(\dfrac{2}{3}\)-filled container = \(\dfrac{2}{3}\times 1320=880\) g. Empty container = \(1200-880=320\) g.


Laptop discount = 10% of $1290 = $129; laptop paid = $1290 - $129 = $1161. Printer paid = $1824 - $1161 = $663. Printer discount = total discount - laptop discount = $246 - $129 = $117. Printer usual price = paid + discount = $663 + $117 = $780. Percentage discount = \(\dfrac{117}{780}\times 100\% = 15\%\). Answer: 15 (%).


Factors of 18: 1, 2, 3, 6, 9, 18. Factors of 81: 1, 3, 9, 27, 81. Common factors: 1, 3, 9. Among the options, 3 and 9.


Multiples of 6: 6, 12, 18, 24, ... Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. The number that is both is 12 (also 6 and 24, but among options only 12 fits).


Compare: 8.3 > 8.03 > 8. So largest to smallest is 8.3, 8.03, 8.


\(\dfrac{8}{12}=\dfrac{2}{3}\). To get denominator 9, multiply by 3: \(\dfrac{2\times3}{3\times3}=\dfrac{6}{9}\). Missing numerator = 6.


4 donuts = 2 pies, so 1 donut = 0.5 pie. 8 donuts = 4 pies. So 8 donuts + 4 pies = 8 pies' worth, costing 4/7 of his money. So 8 pies = 4/7 of money, 1 pie = 1/14 of money. Money left = 3/7 = 6/14 of money = 6 pies.


Make each box total the same. Box 1 (1:5) has 6 parts; Box 2 (1:2) has 3 parts. LCM of totals = 6: Box1 = 6 units (red 1, blue 5); Box2 scaled to 6 units (red 2, blue 4). Total marbles = 12, red = 1 + 2 = 3. Fraction red = 3/12 = 1/4.


Adults = 4/9, children = 5/9. Take total 9 units: adults 4, children 5. Boys : girls = 2 : 1, so boys = 2/3 of children = 2/3 x 5 = 10/3 units. Boys : adults = (10/3) : 4 = 10 : 12 = 5 : 6.


Blue in A = 6% of 200 = 12. Blue in B = 12% of 400 = 48. Total blue = 60 out of 600 markers. 60/600 = 10%.


Multiples of 6: 6, 12, 18, 24, ... Multiples of 8: 8, 16, 24, ... Common multiple below 40 is 24.


Brackets first: 4 + 2 = 6. Then 12 ÷ 6 = 2; 2 x 4 = 8. Finally 22 - 8 = 14.


6000 ml ÷ 8 = 750 ml per bottle. (Key writes 6000 ml = 6 L; 6 L ÷ 8 = 0.75 L = 750 ml.)


Ivan: 8.05 a.m. to (9.20 - 25 min) = 8.55 a.m., duration 50 min. Kevin: started 8.15 a.m., ended 9.20 a.m., duration 65 min. Difference = 65 - 50 = 15 min.


The repeating block of 5 letters (A,B,A,N,A) contains 3 A's. 23 ÷ 3 = 7 remainder 2, so 7 complete blocks use 21 A's (35 letters), and 2 more A's appear in the next block. The greatest possible length keeps the most non-A letters before the 23rd A: 35 + 4 = 39 letters.


Total of 3 girls = 28 x 3 = 84 kg. Total of 4 girls = 29 x 4 = 116 kg. Fourth girl = 116 - 84 = 32 kg.


Total stays constant. Before 3 : 2 = 9 : 6 (total 15). After 2 : 13 (total 15). So Ivan went from 9 units to 2 units, a drop of 7 units = $21, so 1 unit = $3. Ivan at first = 9 units = $27.


$54 is 75% of the original price of 15 cupcakes, so original price = 54 ÷ 0.75 = $72 for 15 cupcakes; 1 cupcake = $72 ÷ 15 = $4.80. Without discount, $54 ÷ $4.80 = 11.25, so she could buy 11 cupcakes.


Zac received 80% of the cookies and ate 20% of that. 20% = \(\dfrac{1}{5}\), so the fraction of total he ate is \(\dfrac{1}{5}\) of 80%, i.e. \(\dfrac{1}{5} \times 80\%\).


Green apples = \(\dfrac{3}{10}\) of apples = 15, so apples = 15 ÷ \(\dfrac{3}{10}\) = 50. Apples are \(\dfrac{5}{9}\) of all fruits, so total = 50 ÷ \(\dfrac{5}{9}\) = 50 × \(\dfrac{9}{5}\) = 90.


Let the original number of rows be r. Original chairs = 9r. New chairs = 9r + 6 = 7(r + 12) = 7r + 84. So 9r + 6 = 7r + 84, giving 2r = 78, r = 39. New chairs = 7(39 + 12) = 7 × 51 = 357. Check: original 9 × 39 = 351, +6 = 357. So there are 357 chairs.


To maximise men, place a man then 3 women repeatedly: pattern M W W W has 1 man per 4 people. The queue can start with a man and end with a man: m men need at least 3(m-1) women between them, total people = m + 3(m-1) = 4m - 3 ≤ 110, so 4m ≤ 113, m ≤ 28.25, giving m = 28. (28 men use 28 + 3×27 = 28 + 81 = 109 people, with 1 spare.)


Let the bottle hold T ml. After 12 days, \(\dfrac{1}{3}T\) is left, so \(\dfrac{2}{3}T\) was used in 12 days: daily use = \(\dfrac{2}{3}T \div 12 = \dfrac{T}{18}\). In 2 more days she uses 2 × \(\dfrac{T}{18} = \dfrac{T}{9}\). Left after 14 days = \(\dfrac{1}{3}T - \dfrac{T}{9} = \dfrac{3T - T}{9} = \dfrac{2T}{9} = 200\). So T = 200 × \(\dfrac{9}{2}\) = 900 ml.


Each girl started with the same number: 1440 ÷ 3 = 480 beads each. 'Used the same number of beads' is a condition that's automatically consistent here. Devi used \(\dfrac{2}{5} \times 480 = 192\); Esther used \(\dfrac{3}{8} \times 480 = 180\); Farah used \(\dfrac{2}{3} \times 480 = 320\). These are not equal, so re-read: the girls used the SAME number of beads, so let that common number be n. Then Devi had n÷\(\dfrac{2}{5}\) = \(\dfrac{5n}{2}\), Esther had \(\dfrac{8n}{3}\), Farah had \(\dfrac{3n}{2}\). Total = \(\dfrac{5n}{2} + \dfrac{8n}{3} + \dfrac{3n}{2} = 1440\). LCD 6: \(\dfrac{15n + 16n + 9n}{6} = \dfrac{40n}{6} = 1440\), so n = 1440 × 6 ÷ 40 = 216. Each girl used 216 beads.


(a) Students reading less than 3 = those reading 1 or 2 = 20. Reading 1 = 9, so reading 2 = 20 - 9 = 11. (b) Reading 3 = twice reading 5 = 2 × 4 = 8. Total = 9 + 11 + 8 + 3 + 4 = 35.


Total of four numbers = 250 × 4 = 1000. The other two numbers sum to 1000 - 190 - 230 = 580. They are both 3-digit numbers (100 to 999). To maximise the difference, make one as large as possible and the other as small as possible: smallest 3-digit is 100, so the other is 580 - 100 = 480. Largest difference = 480 - 100 = 380.


Let Peili spend p. Peili's left = p + 7. Nurul spent 2p. Peili's left ($p+7$) = twice Nurul's left, so Nurul's left = \(\dfrac{p+7}{2}\). Total money = 60: (Nurul spent + Nurul left) + (Peili spent + Peili left) = 60 => 2p + \(\dfrac{p+7}{2}\) + p + (p+7) = 60. Multiply by 2: 4p + (p+7) + 2p + 2(p+7) = 120 => 4p + p + 7 + 2p + 2p + 14 = 120 => 9p + 21 = 120 => 9p = 99 => p = 11. Peili spent $11, left $18. Nurul spent $22, left $9. Nurul had at first 22 + 9 = $31.


A is divisible by 3 and 7, so A is a multiple of 21. B = 4A. A + B = 5A < 450, so A < 90; multiples of 21 under 90 are 21, 42, 63, 84. C = 700 - A - B = 700 - 5A, and A < C < B (= 4A). For A = 84: B = 336, A+B = 420, C = 700 - 420 = 280; check 84 < 280 < 336 - true. (Other A values: A=63 gives C=385, but B=252 and C must be < B - fails; so A = 84.) (a) C = 280 cm. (b) A + B = 420 cm.


Let blue = b and pink = p at first, b + p = 66. After 17 pink burst, pink = p - 17. Blue increased by 75% becomes 1.75b. New total = 1.75b + (p - 17) = 79, so 1.75b + p = 96. Subtract b + p = 66: 0.75b = 30, b = 40. Then p = 66 - 40 = 26. She had 26 pink balloons at first.


Let a postcard cost 1 unit, so a magnet costs 3 units. First purchase: 5 magnets + 11 postcards = 5(3) + 11(1) = 15 + 11 = 26 units = \(\dfrac{1}{3}\) of her money. So total money = 3 × 26 = 78 units. Remaining = 78 - 26 = 52 units. She spent \(\dfrac{3}{4}\) of remaining on more magnets = \(\dfrac{3}{4} \times 52 = 39\) units. Magnets at 3 units each: 39 ÷ 3 = 13 more magnets. Altogether = 5 + 13 = 18 magnets.


Let Pouch Y have 4 units (gold 3, silver 1). Pouch Z has 5 times as many tokens = 20 units, split 1 : 4 = gold 4, silver 16. (a) Gold in Y : silver in Z = 3 : 16 (in unit terms). As a ratio it is 3 : 16. (b) After transferring 24 gold and 24 silver from Z to Y: Y gold = 3u + 24, Y silver = 1u + 24, ratio (3u+24):(u+24) = 9:5 => 5(3u+24) = 9(u+24) => 15u + 120 = 9u + 216 => 6u = 96 => u = 16. Y in end = (3u+24)+(u+24) = (48+24)+(16+24) = 72 + 40 = 112. (c) At first total = Y + Z = 4u + 20u = 24u = 24 × 16 = 384 tokens.


Multiply numerators and denominators: \(\dfrac{3}{7} \times \dfrac{5}{2} = \dfrac{3 \times 5}{7 \times 2} = \dfrac{15}{14}\). Answer: \(\dfrac{15}{14}\), option (3).


Ratio L:B:H = 3:2:1. Length = 6 cm corresponds to 3 units, so 1 unit = 2 cm. Breadth = 2 units = 4 cm, height = 1 unit = 2 cm. Volume = 6 × 4 × 2 = 48 cm³, option (4).


Clips white : blue = 3 : 2. Take white clips = 150 and blue clips = 100 (ratio 3:2). White containers = 150 ÷ 30 = 5; blue containers = 100 ÷ 50 = 2. Total containers = 5 + 2 = 7. Blue containers as a fraction of total = \(\dfrac{2}{7}\), option (4).


Janice = \(\dfrac{5}{7}\) × Xiao Ming, so Janice : Xiao Ming = 5 : 7. Therefore Xiao Ming : Janice = 7 : 5.


Euodia's sweets are unchanged. At first Rachel : Euodia = 2 : 3, after = 5 : 6. Make Euodia the same in both: 2 : 3 = 4 : 6 and 5 : 6 stays 5 : 6. Rachel went from 4 units to 5 units, an increase of 1 unit = 8 sweets. So 1 unit = 8, and Euodia = 6 units = 6 × 8 = 48 sweets.


Clarence : Amir = 4 : 9. Clarence = 24, so 1 unit = 24 ÷ 4 = 6, and Amir = 9 × 6 = 54. Jun Wei : Amir = 4 : 3, so Jun Wei = Amir × 4/3 = 54 × 4/3 = 72 stickers. (Equivalently, scaling Amir to a common value: C:A = 4:9 = 12:27, JW:A = 4:3 = 36:27, so C:A:JW = 12:27:36; Clarence 24 → 1 part = 2 → Jun Wei = 36 × 2 = 72.)


After giving 50% of the remainder to his brother, the $200 left is the other 50%, so the remainder after food and bicycle = $200 × 2 = $400. This remainder = 90% of original − $140 (food was 10%, bicycle was $140). So 90% of original − $140 = $400, giving 90% of original = $540, so original = $540 ÷ 0.9 = $600. Money spent on food = 10% of $600 = $60.


After using 42 g for muffins, he used \(\dfrac{5}{7}\) of the remaining sugar on cookies and had \(\dfrac{1}{7}\) of the original left. The remaining sugar after muffins is split into 7 equal parts; cookies took 5 parts and 1 part (\(\dfrac{1}{7}\) of the original) was left, but the leftover \(\dfrac{1}{7}\) is of the original amount. Treat the original as 7 units: leftover = 1 unit, and cookies used 5 of the 7 parts of the post-muffin sugar. From the printed working, the 42 g (muffins) corresponds to 1 unit of the original 7, but accounting for the post-muffin split gives 2 units = 42 g, so 1 unit = 21 g and the original = 7 × … ; the printed key gives 4 units → 42 g pattern leading to first amount = 84 g.


Christopher kept \(1 - \dfrac{4}{5} = \dfrac{1}{5}\) of his erasers; Helen kept \(1 - \dfrac{2}{3} = \dfrac{1}{3}\) of hers. Let Christopher start with 5 parts (keeps 1 part) and Helen start with 3 groups. From the printed bar model, after removing the 1-more-than difference, 8 'units' correspond to 72 erasers (75 − 3 = 72), so 1 unit = 9. Christopher had 9 erasers in the end.


(a) 2 cushions: \(2\times\dfrac{3}{8}=\dfrac{6}{8}=\dfrac{3}{4}\) h. (b) Half the dress time: \(\dfrac{1}{2}\times\dfrac{7}{8}=\dfrac{7}{16}\) h.


(a) Over Mon: Siti made 14 more than Bala. Tue: Siti 16, Bala 22 (Bala 6 more). Overall Siti's lead = 14 − 6 = 8 more, so the difference over the two days = 8. (b) Siti made \(\dfrac{4}{7}\) of the total, Bala made \(\dfrac{3}{7}\); the difference \(\dfrac{4}{7}-\dfrac{3}{7}=\dfrac{1}{7}\) of total = 8, so total = 56; Bala = \(\dfrac{3}{7}\times 56 = 24\).


Each file = 3 pens, so 2 files + 9 pens = 6 pens + 9 pens = 15 pen-units = \(\dfrac{1}{5}\) of money, so 1 pen = \(\dfrac{1}{75}\) of the money. (a) After spending \(\dfrac{1}{5}\), remaining = \(\dfrac{4}{5}\); he spends \(\dfrac{2}{3}\) of that, left = \(\dfrac{1}{3}\times\dfrac{4}{5}=\dfrac{4}{15}\). (b) Money for more pens = \(\dfrac{2}{3}\times\dfrac{4}{5}=\dfrac{8}{15}\) of money = \(\dfrac{8}{15}\div\dfrac{1}{75}=40\) pens. Total pens = 9 + 40 = 49. (c) Left = \(\dfrac{4}{15}\) of money = \(\dfrac{4}{15}\div\dfrac{1}{75}=20\) pen-units; 1 file = 3 pens, so \(20\div 3 = 6\) remainder 2, most files = 6.


Number of boys = 40 − 25 = 15. Ratio of boys to girls = 15 : 25 = 3 : 5 (dividing both by 5). Answer: 3 : 5.


Length of square X = 36 ÷ 4 = 9 cm. Length of square Y = 2 × 9 = 18 cm. Perimeter of square Y = 4 × 18 = 72 cm. Answer: 72 cm.


Remaining after bag = 100 − 60 = $40. Spent on shirt = 10% of 40 = $4. Total spent = 60 + 4 = $64. As a percentage = \(\dfrac{64}{100}\times100\% = 64\%\). Answer: 64%.


Cherries : blueberries prepared = 3 : 10. Each cupcake uses 1 cherry and 3 blueberries, so cherries used : blueberries used = 3 : 9 (for every 3 cherries, 9 blueberries are used). Of 10 units of blueberries, 9 are used and 1 is left (the 2 leftover = 1 unit). Ratio of blueberries used to blueberries left = 9 : 1. Answer: 9 : 1.


Cherries : blueberries prepared = 3 : 10. Blueberries used = 10 − 1 = 9 units (the leftover 2 blueberries = 1 unit, so 1 unit = 2). Cherries = 3 units = 6 cherries. Each cupcake uses 1 cherry, so number of cupcakes = number of cherries = 3 × 2 = 6. Answer: 6.


Notebook after 20% discount = \(\dfrac{80}{100}\times5 = $4\). Total paid = $6, so pen after discount = 6 − 4 = $2. Answer: $2.


Thirty-five thousand = 35 000; and twenty = 20. So 35 000 + 20 = 35 020.


Look at the 3rd decimal digit (5). Since it is 5, round the 2nd decimal place up: 6.78 -> 6.79.


Total spent on ties = 50 - x dollars. There were 4 identical ties, so each tie cost \(\dfrac{50 - x}{4}\) dollars.


Music runs 10.10 a.m. to 11.30 a.m. = 1 h 20 min. Sports runs 11.30 a.m. to 12.20 p.m. = 50 min. Difference = 1 h 20 min - 50 min = 30 min.


Since 7/11 of the flowers sold were roses, sold roses : sold tulips = 7 : 4. Roses sold = 1/3 of all roses, so total roses = 3 × 7 = 21 parts. Tulips sold = 4/9 of all tulips, so total tulips = (9/4) × 4 = 9 parts. Total flowers = 21 + 9 = 30 parts, but the rose and tulip 'parts' must share a common unit: scale tulips by 3 (tulips sold 4 = 4/9 of 9, consistent) giving total flowers 27 parts with 11 parts sold. Fraction sold = 11/27.


Group A has boys:girls = 5:4, total 9 parts. Group B has boys:girls = 5:1, total 6 parts. Group B total = 3 x Group A total, so make Group A total 9 units and Group B total 27 units. Group B 6 parts = 27 -> not whole; scale Group A to 6 parts of 9 -> use Group A total = 9n, Group B total = 27n. Group B parts: 27n/6 each part => boys B = 5 x (27n/6). For whole numbers n must be a multiple of 2. With n=2: Group A total 18 (boys 10, girls 8), Group B total 54 (boys 45, girls 9). Total boys = 10 + 45 = 55.


Common multiples of 4 and 6 are multiples of their LCM, 12: 12, 24, 36, ... Those smaller than 30 are 12 and 24.


(a) A multiple of 5 ends in 0 or 5; using 2,3,5,8 it must end in 5, and to be smallest put the smallest leading digits: 2385. (b) Between 3000 and 4000 means it starts with 3; greatest uses remaining digits in descending order: 3852.


The fastest runner has the shortest time. The shortest time is 13.8 s (Ming), so Ming was first.


Let total gold = total silver = T. Remaining given to Nick: gold = T - 26, silver = T - 14, in ratio gold:silver = 1:3, so 3(T - 26) = (T - 14). 3T - 78 = T - 14, 2T = 64, T = 32. Total stars = 32 + 32 = 64.


Adults stay 7. Boys increase by 50%: 6 × 1.5 = 9. Girls decrease by 40%: 5 × 0.6 = 3. New ratio adults : boys : girls = 7 : 9 : 3.


Friday boys + girls = 6 + 5 = 11 parts = 627, so 1 part = 57. Saturday total parts = 7 + 9 + 3 = 19, so total visitors = 19 × 57 = 1083.


(a) P: 2 pages/s × 60 = 120 pages/min; Part 1 = 7 × 120 = 840 pages. Q: 7 pages / 5 s = 84 pages/min; Part 2 = 8 × 84 = 672 pages. Total = 840 + 672 = 1512 pages. (b) Q rate = 84 pages/min; 1512 ÷ 84 = 18 min.


(a) Blue sold 270 = 75% of pens sold (since 25% were red). Total sold = 270 ÷ 75 × 100 = 360. (b) Sold 360 = 40% of all pens, so total pens = 900; unsold = 540. 30% of unsold were blue = 162 blue unsold; remaining unsold red = 540 - 162 = 378. Red pens at first = red sold + red unsold = (25% of 360 = 90)... printed working: 378 + 25/100 × 360 = 378 + 90 = 468.


Let total coins = 1 whole = units. One-dollar = 2/3 (=6/9). Fifty-cent = one-third of one-dollar count = 2/9. Twenty-cent = remaining = 1 - 6/9 - 2/9 = 1/9. So per 9 units: $1 × 6 + $0.50 × 2 + $0.20 × 1 = $6 + $1 + $0.20 = $7.20 per 9 coins. $108 ÷ $7.20 = 15 groups. Total coins = 15 × 9 = 135.


(a) Dress + 2 blouses = 1/6 of money. Dress = 3 blouse-units, so 3 + 2 = 5 blouse-units = 1/6, each blouse = 1/6 ÷ 5 = 1/30 of her money. (b) Dress = 3/30 = 1/10 of money. Remaining after 1/6 = 5/6; watch = 3/4 × 5/6 = 5/8. Watch - dress = 5/8 - 1/10 = 25/40 - 4/40 = 21/40 of money = $220.50. So money = 220.50 ÷ 21 × 40 = $10.50 × 40 = $420.


Rounding to the nearest thousand gives 70 000 for numbers from 69 500 to 70 499. 69 499 rounds to 69 000; 70 893 rounds to 71 000; 71 267 rounds to 71 000. Only 69 978 rounds to 70 000. Answer: 69 978.


Multiplying a fraction by 4 means adding it 4 times: \(\dfrac{2}{3} \times 4 = \dfrac{2}{3} + \dfrac{2}{3} + \dfrac{2}{3} + \dfrac{2}{3}\). Answer: option 4.


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


The longest side corresponds to 5 units = 40 cm, so 1 unit = 8 cm. The shortest side is 3 units = 3 × 8 = 24 cm. Answer: 24 cm.


Sum of the other 3 numbers = 960 - 150 = 810. Average = 810 ÷ 3 = 270. Answer: 270.


25% of 700 = \(\dfrac{25}{100} \times 700 = 175\). 175 people left the concert hall.


Work out the bracket first: \(2 \times 4 = 8\). Then \(32 \div 8 = 4\). Finally \(16 + 4 = 20\). Answer: 20.


Each boy gets \(6 \div 9 = \dfrac{6}{9} = \dfrac{2}{3}\) of a pizza. Answer: \(\dfrac{2}{3}\).


8 units = 48 blue pens, so 1 unit = 6. Total = 8 + 3 = 11 units = 11 × 6 = 66 pens. Answer: 66.


From 10.50 a.m. to 12.50 p.m. is 2 h; from 12.50 p.m. to 1.35 p.m. is 45 min. Total = 2 h 45 min.


Fruit tarts = \(\dfrac{3}{4} \times 120 = 90\). Fruit tarts sold = 90 - 38 = 52. Answer: 52.


9 bottles take 3 min, so 1 bottle takes \(\dfrac{3}{9}\) min. 153 bottles take \(\dfrac{3}{9} \times 153 = 51\) min. Answer: 51 min.


The difference of 50 - 20 = 30 apples accounts for 15 - 9 = 6 kg, so 30 apples = 6 kg and 10 apples = 2 kg, hence 20 apples = 4 kg. Empty crate = 9 - 4 = 5 kg. Answer: 5 kg.


Machine A ran 5 min, Machine B ran 4 min. Per the printed key: 600 ÷ 120 = 5 (i.e. each 'minute-unit' = 5 stickers across the combined 9 minute-runs adjusted for the 30 fewer), giving Machine A's rate, then A printed 80 per minute × 5 = 400. Answer: 400.


Capacity = length × breadth × height = 150 × 120 × 50 = 900 000 cm³. Answer: 900 000 cm³.


Capacity = 900 000 cm³ = 900 L. \(\dfrac{1}{5}\) full = 180 L. Water to drain from 420 L down to 180 L = 240 L; at 12 L/min that takes 240 ÷ 12 = 20 min. From 11.45 am + 20 min = 12.05 pm. Answer: 12.05 pm.


From the 1st to the 6th post there are 5 gaps: 1 gap = 90.5 ÷ 5 = 18.1 m. From the 2nd to the last post is 307.7 m, which is 307.7 ÷ 18.1 = 17 gaps, so the last post is the (2 + 17) = 19th. There were 19 lamp posts altogether. Answer: 19.


Blue + Red = 1648 and Blue - Red = 132. Adding: 2 × Blue = 1648 + 132 = 1780... using the printed key: blue at first = (1648 + 132) ÷ 2 = 1516 ÷ 2 is for red. Red at first = (1648 - 132) ÷ 2 = 758. Answer: 758.


Blue at first = 1648 - 758 = 890. After giving away half, blue = 890 ÷ 2 = 445. Red after = 4 × 445 = 1780. Red folded = 1780 - 758 = 1022. Answer: 1022.


After spending $558 she was short $53 for a dress but had $28 left after buying a skirt instead. The difference between a dress and a skirt = 53 + 28 = $81. Answer: $81.


Each dress costs $81 more than each skirt. 3 dresses = 3 × 81 = $243 more than 3 skirts, so 4 skirts + 3 dresses ($558) gives the skirt price; one skirt = (558 - 243) ÷ 7 = $45. She had at first = 558 + 45 (another skirt) + 28 left = $631. Answer: $631.


Expanded form: 100 000 + 6 000 + 50 + 3 = 106 053. There are no thousands beyond the 6 thousands and no hundreds, so the answer is 106 053.


Do multiplication first: 5 × 3 = 15. Then 57 − 15 + 2 = 42 + 2 = 44.


The first digit after the decimal point is the tenths place. In 3.421 that digit is 4.


Jack ate 1/4 = 3/12, Helen ate 7/12. Indra ate between 3/12 and 7/12. Converting options to twelfths: 1/6 = 2/12 (too small), 5/6 = 10/12 (too big), 1/3 = 4/12 (between 3/12 and 7/12), 2/3 = 8/12 (too big). So Indra could have eaten 1/3.


Thirteen thousand = 13 000; and eleven = 11. Together 13 000 + 11 = 13011.


Find the area of each shaded triangle.
 Area of triangle A = <input min="0" type="number" id="imp5pg069q9a_3" class="lineinput" placeholder="?"> <math xmlns="http://www.w3.org/1998/Math/MathML"><semantics><mrow><mi>c</mi><msup><mi>m</mi><mn>2</mn></msup></mrow><annotation encoding="application/x-tex">cm^2</annotation></semantics></math>cm2


The row is labelled with G as 1st (the '1st' marker is under G), so positions are counted from the right: G is 1st, F is 2nd, E is 3rd. The elephant in third position is E.


What is the missing number in the number pattern?

The pattern decreases by 2 each step: 20, 18, ?, 14, 12, 10. The missing number is 18 - 2 = 16.


The table shows Samy's timetable every Saturday.
| Time | Activity |
| half past 9 | Breakfast |
| ? | Swimming lesson |
| 12 noon | Lunch |
| 2 o'clock | Piano lesson |
Which of the following clock shows the time that he has his swimming lesson?

Swimming is after breakfast (half past 9) and before lunch (12 noon). The correct clock (option 2) shows half past 10, which fits between these two times.


Which of the following sets of coins are needed by Ali to exchange with $1?

$1 = 100¢. The correct set (option 4) is the group of coins whose values add up to 100¢.


Count the pencils. There are 57 pencils.


There are <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> toys altogether.

Add the Toy Robots, Toy Trains and Toy Trucks. There are 16 toys altogether.


Arrange the numbers from the smallest to the greatest. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_2" class="fill-blank-input" placeholder="?" />, <input type="text" id="input_3" class="fill-blank-input" placeholder="?" />

Ordering 65, 87, 56, 78 from smallest to greatest gives 56, 65, 78, 87.


Jane and Lily baked 30 muffins altogether. Jane baked 12 muffins. How many muffins did Lily bake? Lily baked <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> muffins.


Complete the number pattern. 58, 59, <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />, 62, 64, 65, 67

The pattern goes 58, 59, then skips to 62 (the difference toggles); the missing number is 61. Sequence: 58, 59, 61, 62, 64, 65, 67 (add 1, add 2, add 1, add 2 ...).