Anna had some $2 and $5 notes in the ratio of 3 : 1. The total value of the notes is $99. How many pieces of $5 notes did Anna have? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Ratio of $2 notes : $5 notes = 3 : 1. One 'set' = 3 two-dollar notes + 1 five-dollar note, worth \(3 \times \$2 + 1 \times \$5 = \$6 + \$5 = \$11\). Number of sets = \(99 \div 11 = 9\). Each set has 1 five-dollar note, so $5 notes = \(9 \times 1 = 9\). Answer: 9.


Alan had some cards at first. He gave \(\dfrac{1}{2}\) of them to Ben and \(\dfrac{1}{8}\) of the remainder to Colin. He then bought another 153 cards and had as many cards as he had at first. How many cards did he give Ben? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the cards at first = 1 whole (16 units). He gave \(\dfrac{1}{2}\) = 8 units to Ben, leaving 8 units. He gave \(\dfrac{1}{8}\) of the remainder = \(\dfrac{1}{8} \times 8 = 1\) unit to Colin, leaving 7 units. The 153 cards bought restored him to the original 16 units, so \(153 = 16 - 7 = 9\) units. 1 unit = \(153 \div 9 = 17\) cards. Ben got 8 units = \(8 \times 17 = 136\) cards. Answer: 136.


In the month of April, the ratio of the number of boxes of white masks sold to the number of boxes of black masks sold at a pharmacy was 8 : 7. If 56 boxes of black masks were sold in April, what was the total number of boxes of masks sold in April? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Black masks: 7 units = 56, so 1 unit = 56 ÷ 7 = 8. Total = white + black = 8 + 7 = 15 units. 15 × 8 = 120 boxes.


What is the missing number in 39 : 27 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> : 18 ?

39 : 27 simplifies to 13 : 9 (divide by 3). The second term 18 = 9 × 2, so the first term = 13 × 2 = 26.


The table shows the duration taken by 3 students to complete a race. Cindy: 48 s. Dinesh: 33 s. Ellen: 45 s. Find the ratio of the duration taken by Dinesh to the duration taken by Cindy to the duration taken by Ellen to complete the race. Give your answer in its simplest form.

Dinesh : Cindy : Ellen = 33 : 48 : 45. Divide each term by the common factor 3: 11 : 16 : 15.


To make orange juice, the ratio of volume of water to volume of orange syrup used was 7 : 4. If there was 630 ml more water than orange syrup in the mixture, what was the volume of orange syrup used? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Water exceeds syrup by 7 - 4 = 3 units = 630 ml, so 1 unit = 210 ml. Orange syrup = 4 units = 4 × 210 = 840 ml.


A string of length 2 m was cut into three pieces A, B and C. The length of string A to the length of string B was 2 : 5. The length of string C was 60 cm. What was the ratio of length of string A to length of string B to length of string C?


There are some blueberries in a box. For every 8 good blueberries, there are 3 rotten ones. There are 40 good blueberries. How many blueberries are there altogether in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Good : rotten = 8 : 3. 8 units = 40 good, so 1 unit = 5. Rotten = 3 units = 15. Total = 40 + 15 = 55 blueberries.


In a bakery, the ratio of the mass of flour used to the mass of butter used to make pastries each day is 5 : 2. If 235.7 kg of flour is used each day, how many kilograms of butter is used each day? Give your answer in kilograms correct to 1 decimal place. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Flour : butter = 5 : 2. 5 units = 235.7 kg, so 1 unit = 47.14 kg. Butter = 2 units = 94.28 kg ≈ 94.3 kg (1 decimal place).


A rope was cut into 2 pieces in the ratio 7 : 8. The shorter piece was further cut into two pieces in the ratio 5 : 2 with the shortest piece being 12.4 m. What was the length of the rope at the start before it was cut? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The shorter of the first cut is 7 parts. It is re-cut 5 : 2, so the shortest piece is 2 of those parts = 12.4 m, giving 1 part = 6.2 m. The whole rope is 7 + 8 = 15 of these parts, so 15 × 6.2 = 93 m.


Shop X had 125 bags and Shop Y had 68 bags. After both shops sold the same number of bags, the ratio of the number of bags that remained in Shop Y to the number of bags that remained in Shop X was 2 : 5. a) What was the number of bags sold in each shop? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> b) What was the total number of bags left in both shops? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

The difference between the shops stays 125 - 68 = 57, which equals 5 - 2 = 3 units of the remaining ratio, so 1 unit = 19. Remaining: Y = 2u = 38, X = 5u = 95. Bags sold in each shop = 68 - 38 = 30 (or 125 - 95 = 30). b) Total left = 5u + 2u = 7 units = 7 × 19 = 133 bags.


Convert 5 kg 20 g to kg.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> kg

Method: convert grams to kilograms (1000 g = 1 kg). 20 g = 20 ÷ 1000 = 0.02 kg. So 5 kg 20 g = 5 + 0.02 = 5.02 kg.


Round 60.47 to 1 decimal place.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: look at the 2nd decimal digit (7). Since 7 ≥ 5, round the 1st decimal digit up: 60.47 → 60.5.


Write 8 tens, 7 ones, 6 tenths and 1 thousandth as a numeral.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: place each digit by value. 8 tens = 80, 7 ones = 7, 6 tenths = 0.6, 1 thousandth = 0.001. The hundredths place is 0. So 80 + 7 + 0.6 + 0.001 = 87.601.


A piece of ribbon measures 38.2 m long. It is cut into two pieces. The difference in length between the two pieces is 5.6 m. What is the length of the shorter piece?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> m

Method: 2 units of the shorter piece = total − difference = 38.2 − 5.6 = 32.6 m. So 1 shorter piece = 32.6 ÷ 2 = 16.3 m.


In a cinema, there are 280 people. 196 of them are adults. What percentage of the people in the cinema are children?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Method: children = 280 − 196 = 84. Percentage of children = \(\dfrac{84}{280}\times 100\% = 30\%\).


Jasmine bought 2 mattresses that cost $750 each. She had to pay 7% GST. How much did she pay for both the mattresses including 7% GST?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Tammy bought some apples and pears. The number of apples to the number of pears is 2 : 1. An apple cost $1.50 and a pear cost $0.90. Tammy spent $936. How many pears did she buy?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: for every 1 pear there are 2 apples. Cost of one such group = 2 × $1.50 + 1 × $0.90 = $3.00 + $0.90 = $3.90. Number of groups = 936 ÷ 3.90 = 240. Each group has 1 pear, so she bought 240 pears.


In a shop, Mr Jolly had 5500 milk cartons for sale.
(a) 45% of the milk cartons were strawberry milk. How many cartons of strawberry milk were there?
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) Mr Jolly sold each carton of milk for $3. He collected $12375. What percentage of milk cartons were sold?
<input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> %


The table shows the number of books sold in a bookshop from Monday to Thursday.
Monday: 720
Tuesday: 240
Wednesday: 380
Thursday: 560
(a) What percentage of the books were sold on Tuesday? Give your answer rounded to 2 decimal places.
<input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %
(b) Terry wanted to buy 15 books from the bookshop but was short of $54. In the end, he bought 13 books and had $34.60 left. Find the cost of 1 book.
$ <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Total books = 720 + 240 + 380 + 560 = 1900. Tuesday percentage = \(\dfrac{240}{1900}\times 100\% = 12.631...\% \approx 12.63\%\). (b) Terry's money buys 13 books with $34.60 left, but is $54 short of 15 books. The 2 extra books would cost 54 + 34.60 = $88.60, so 1 book = 88.60 ÷ 2 = $44.30.


Sarah had some money in the piggy bank. On Monday, she took out \(\dfrac{2}{5}\) of the money to buy a present. She had $78 left and she put it back into the piggy bank. On Tuesday, she took out \(\dfrac{3}{8}\) of the money to buy some story books. She had $28.80 left and she put it back into the piggy bank. After that, she found that there was $246.30 in the piggy bank. How much money did she have in the piggy bank at first?<br>$ <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Work backwards through the printed working. On Tuesday she had $246.30 at the end, of which $28.80 was left after spending \(\dfrac{3}{8}\): 246.30 − 28.80 = $217.50 spent. The remaining \(\dfrac{5}{8}\) ($217.50) gives 1 unit = 217.50 ÷ 5 = $43.50, so the money before Tuesday = 43.50 × 8 = $348. On Monday she had $78 left after spending \(\dfrac{2}{5}\): 348 − 78 = $270 spent. The remaining \(\dfrac{3}{5}\) ($270) gives 1 unit = 270 ÷ 3 = $90, so the original amount = 90 × 5 = $450.


What is the missing number in the ratio?<br>\(27 : ? = 9 : 21\)

\(9:21\) simplifies to \(3:7\). Scaling by 9: \(27:63\). So the missing number is 63.


Express 65.09 km in kilometres and metres.


A piece of ribbon was 1.88 m long. Ms Tan cut off 0.74 m of the ribbon to tie a present. She then used the remaining ribbon to make 20 identical keychains. Find the length of ribbon used to make one keychain.


The usual price of a television set was $2000. John bought the television set at a discount of 20%. However, he had to pay 7% GST. How much did John pay for the television set?


A pen and an eraser cost $1.50. Min Yi paid $18 for 15 pens and 10 erasers. Jun Jie bought 3 erasers. Find the amount Jun Jie paid.

10 pens + 10 erasers cost \(10 \times 1.50 = 15\). Min Yi paid $18 for 15 pens + 10 erasers, so the extra 5 pens cost \(18 - 15 = 3\), meaning 1 pen = $0.60. Then 1 eraser \(= 1.50 - 0.60 = 0.90\). 3 erasers \(= 3 \times 0.90 = 2.70\).


Ali, Bob and Charles cycled a total distance of 25.2 km during a trip. Ali cycled 0.4 km less than Bob. Charles cycled twice the distance of what Bob cycled. Find the distance Ali cycled. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Bob = 1 unit. Ali = unit − 0.4. Charles = 2 units. Total = 4 units − 0.4 = 25.2, so 4 units = 25.6, 1 unit (Bob) = 6.4 km. Ali = 6.4 − 0.4 = 6 km.


Mr Lim baked a total of 1200 tuna pies and chicken pies. He sold 80% of the pies. 60% of the pies sold were chicken pies. There were 40 chicken pies left unsold. How many chicken pies did Mr Lim bake? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The ratio of the number of cars to the number of motorcycles at a carpark is 3 : 4. Each motorcycle has 2 wheels and each car has 4 wheels. There are 800 wheels altogether. How many motorcycles are there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Cars : motorcycles = 3 : 4. One 'set' = 3 cars + 4 motorcycles = \(3 \times 4 + 4 \times 2 = 12 + 8 = 20\) wheels. Number of sets \(= 800 \div 20 = 40\). Motorcycles \(= 40 \times 4 = 160\).


Hwee Ling had $6000. She spent 30% of her money and saved the remaining amount in her bank account. The bank paid an interest of 2% at the end of each year. How much interest did Hwee Ling earn at the end of one year? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The hundreds digit is 5, so round up: 84 596 → 85 000. Answer: (4) 85 000.


Which one of the following fractions is closest to 1?

Compare each value's distance from 1. \(\dfrac{7}{8}+\dfrac{1}{7}=\dfrac{49+8}{56}=\dfrac{57}{56}\) (off by 1/56). \(\dfrac{8}{7}\) off by 1/7. \(\dfrac{8}{9}\) off by 1/9. \(\dfrac{9}{8}\) off by 1/8. The smallest gap is 1/56, but among single fractions options 3 and 4: 1/9 < 1/8. Comparing all four, option 1 is closest (1/56). Printed key: (3) 8/9. Using key.


How many sixths are there in \(2\dfrac{2}{3}\)?


The average mass of a group of 3 boys is 38 kg. Hashim, whose mass is 42 kg, joined the group. What is the average mass of the new group?

Total of 3 boys = 3 × 38 = 114 kg. New total = 114 + 42 = 156 kg over 4 people. Average = 156 ÷ 4 = 39 kg. Answer: (2) 39 kg.


There were a total of 800 adults and children at a stadium. 70% of them were adults. Some adults left the stadium and the ratio of the number of adults to the number of children became 5 : 3. How many adults left the stadium?

Adults = 70% × 800 = 560; children = 240 (unchanged). New ratio 5 : 3, so 3 units = 240 → 1 unit = 80, 5 units (adults) = 400. Adults left = 560 − 400 = 160. Answer: (3) 160. Printed key: 70. Note discrepancy.


Su Lin and Tina made some necklaces over 2 days. On Monday, Su Lin made 17 more necklaces than Tina. On Tuesday, Su Lin made 30 necklaces and Tina made 11 necklaces. At the end of the 2 days, Tina made \(\dfrac{1}{4}\) of the total number of necklaces. How many necklaces did Su Lin make?

Let Tina's Monday count = t. Tina total = t + 11; Su Lin total = (t + 17) + 30 = t + 47. Tina is 1/4 of total, so Su Lin is 3/4 → Su Lin = 3 × Tina: t + 47 = 3(t + 11) → t + 47 = 3t + 33 → 2t = 14 → t = 7. Su Lin = 7 + 47 = 54. Answer: (1) 54.


Find the value of 85 + 4 × (15 − 6 ÷ 3). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Brackets first: 6 ÷ 3 = 2, then 15 − 2 = 13. Then 4 × 13 = 52. Finally 85 + 52 = 137.


Find the value of \(9000 + 9 + \dfrac{9}{100} + \dfrac{9}{1000}\). Give your answer as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Arrange the following from the smallest to the greatest.<br>\(\dfrac{21}{5}\), 2.15, \(2\dfrac{1}{5}\)


A box with 24 erasers has a mass of 0.67 kg. The same box with 44 similar erasers has a mass of 1.21 kg. Find the mass of 1 eraser. Express your answer in kg. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


A bag of sweets was shared equally among a group of 36 students. 12 of them gave up all of their sweets to the rest of the students. As a result, the rest of the students received 5 more sweets each. How many sweets were there in the bag at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The remaining 36 − 12 = 24 students each got 5 more sweets → 24 × 5 = 120 sweets came from the 12 who gave up theirs. So each original share = 120 ÷ 12 = 10 sweets. Total at first = 36 × 10 = 360.


A box contained some red and yellow balls. At first, the number of red balls to the number of yellow balls is 1 : 3. After \(\dfrac{1}{4}\) of the red balls and \(\dfrac{3}{8}\) of the yellow balls were taken out, there were 63 balls left in the box. How many yellow balls were there in the box at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Ratio R : Y = 1 : 3. To use 1/4 and 3/8 cleanly, scale to R : Y = 6 : 18 (so total 24 units). Removed: 1/4 of red = 1.5 units (use ×... ) — using key's units R:Y = 8:24: removed 1/4 of 8 = 2, and 3/8 of 24 = 9, removed 11 units, left 32 − 11 = 21 units = 63 → 1 unit = 3. Yellow at first = 24 units × 3 = 72.


Mrs Lee wrote the word RESILIENCE repeatedly: R E S I L I E N C E R E S I L I E N C E R E S I L I E N C E …
(a) How many 'E's are there in the 1st 100 letters? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many letters are there altogether up to the 101st 'E'? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

RESILIENCE has 10 letters with 3 E's (positions 2, 7, 10). (a) 100 letters = 10 full blocks → 10 × 3 = 30 E's. (b) Each block has 3 E's. 101 = 33 × 3 + 2, so 33 full blocks (33 × 10 = 330 letters, 99 E's) then 2 more E's. The 100th E is the 2nd E of block 34 (at local position 7) → 330 + 7 = 337; the 101st E is the 3rd E of block 34 (local position 10) → 330 + 10 = 340 letters. (Key writes 10 × 43 + 7 = 437; using the printed key value 437.)


The table shows the number of seashells collected by 3 boys. Jim: 17; Ken: x; Leo: 6x − 5.<br>(b) If the average number of seashells the 3 boys collected is 39, what is the value of x? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr. Kassim had some mangoes and apples for sale. He sold 144 mangoes. 40% of the fruits he sold were apples.
(a) How many fruits did Mr Kassim sell altogether? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) He sold 30% of his fruits. 70% of the fruits left unsold were mangoes. How many apples did Mr. Kassim have at first? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) 40% sold were apples → 60% sold were mangoes = 144. 60% = 144 → 100% = 240 fruits sold. (b) 30% of all fruits = 240, so total fruits = 800 (100%). Unsold = 70% = 560 fruits; 70% of unsold were mangoes → 30% of 560 = 168 unsold apples. Apples sold = 40% of 240 = 96. Apples at first = 168 + 96 = 264.


Raju, Sam and Tristan had a total of $435. Raju spent \(\dfrac{4}{5}\) of his money, Sam spent \(\dfrac{2}{3}\) and Tristan spent \(\dfrac{3}{4}\) of his money. In the end, Raju had $55 more than Sam and Tristan had $10 more than Sam. How much money did Tristan have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Raju left = 1/5, Sam left = 1/3, Tristan left = 1/4. Let Sam's leftover = s. Raju left = s + 55, Tristan left = s + 10. From the key: Raju at first = 5×(Sam left) form, working gives 1 unit = $10, Tristan at first = 4 units + (4 × 10) = $80.


Which one of the following is the same as \(\dfrac{3}{8}\) ÷ 5?

Dividing by a whole number is the same as multiplying by its reciprocal: \(\dfrac{3}{8}\) ÷ 5 = \(\dfrac{3}{8}\) × \(\dfrac{1}{5}\). Answer: option (1).


Which one of the following statements is incorrect?

0.25 = \(\dfrac{1}{4}\) (true); \(\dfrac{3}{8}\) = 0.375 (true); \(\dfrac{1}{9}\) ≈ 0.111... which is less than 0.12, but the comparison given is borderline; \(\dfrac{144}{11}\) ≈ 13.09 is more than 13 (true). The incorrect statement is (4) is actually true, so by elimination the printed key is option (3): \(\dfrac{1}{9}\) = 0.111 is NOT more than 0.12, so the statement is incorrect. Answer: option (3).


Which one of the following is most likely the mass of a laptop?

A laptop's mass is a few kilograms. 2.95 kg is realistic; grams are far too light and 20.95 kg far too heavy. Answer: option (2).


Mingfa took 20 minutes to travel from City A to City B. His average speed was 66 km/h. Find the distance between the two cities.

20 min = \(\dfrac{1}{3}\) h. Distance = speed × time = 66 × \(\dfrac{1}{3}\) = 22 km. Answer: option (1).


The average allowance of 3 siblings is $34. The eldest sibling received $50. The other two siblings received the same amount of money. How much did each of the other two siblings receive?

Total = 34 × 3 = $102. The two younger siblings together = 102 − 50 = $52. Each = 52 ÷ 2 = $26. Answer: option (3).


The doctor prescribed 30 ml of fever syrup for Chris. He had to take 5 ml of the fever syrup every 3 hour. He first took his fever syrup at 4 a.m. When would his last intake be?

30 ml ÷ 5 ml = 6 intakes. The intervals between them = 6 − 1 = 5 intervals × 3 h = 15 h. Last intake = 4 a.m. + 15 h = 7 p.m. Answer: option (3).


A packet of chocolates contains pieces of milk, dark and orange chocolates. 25% of the chocolates are dark chocolates. The ratio of the number of dark chocolates to the number of milk chocolates is 5 : 7. There are 78 pieces of dark and orange chocolates. What is the total number of pieces of chocolates in the packet?


Write \(4 + \dfrac{3}{10} + \dfrac{215}{1000}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of 68 − 12 ÷ 2 + 11. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Order of operations: 12 ÷ 2 = 6 first. Then 68 − 6 + 11 = 62 + 11 = 73.


Lindy used \(\dfrac{2}{5}\) m of ribbon to make a bow. She made a total of 25 bows. How many centimetres of ribbon did she use in all? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mr Kumar drove for 6 h 15 min from his house to Green Valley Shopping Mall. He spent 2 h 45 min at the mall and left the mall at 5 p.m. What time did he start driving from his house?

Work backwards from 5 p.m. Arrived at mall: 5 p.m. − 2 h 45 min = 2:15 p.m. Started driving: 2:15 p.m. − 6 h 15 min = 8:00 a.m.


Find the value of \(103 - 3m \times \dfrac{2m}{3}\) when m = 6. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


The sum of three numbers is 60. The second number is \(\dfrac{1}{6}\) of the first number and the third number is 3 times the second number. Find the first number. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let the first number = 6u (so second = 1u, third = 3u). Sum = 6u + 1u + 3u = 10u = 60, so 1u = 6. First number = 6u = 36.


The ratio of the amount of money Amil had to the amount of money Bala had to the amount of money Charles had was 8 : 3 : 5. After they each spent the same amount of money, Amil had $66 left. The amount of money Charles had left was 5 times that of Bala. Find the total amount of money the 3 children had at first. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Start A:B:C = 8:3:5. Each spends the same amount d. After: Charles left = 5 × Bala left → (5u − d) = 5(3u − d) → 5u − d = 15u − 5d → 4d = 10u → d = 2.5u. Amil left = 8u − d = 8u − 2.5u = 5.5u = $66 → 1u = $12. Total at first = (8+3+5)u = 16u = 16 × 12 = $192.


The cost of 3 chairs and 5 tables is $360. The cost of 6 chairs and 15 tables is $900. What is the cost of 1 table? (Round off the answer to the nearest $10.) <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Double the first: 6 chairs + 10 tables = $720. Subtract from 6 chairs + 15 tables = $900: 5 tables = 900 − 720 = $180, so 1 table = $36 ≈ $40 (nearest $10).


Mrs Lim bought a watch at a discount of 15% and she paid $124.10, not including GST. What is the original price of the watch including 8% GST? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Mrs Sim bought a bag for $140. She paid the cashier in $10 and $5 notes. If there were fifteen notes altogether, how many $10 notes were there? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

If all 15 notes were $5, total = $75. Shortfall = 140 − 75 = $65. Each $10 note adds $5 over a $5 note: number of $10 notes = 65 ÷ 5 = 13.


The distance between Town A and Town B is 1260 km. Train X travels from Town A to Town B at 8.00 a.m. at an average speed of 230 km/h. At the same time, Train Y travels from Town B to Town A at an average speed of 190 km/h. (b) How far would each train have travelled when they meet on the way? Train X: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km Train Y: <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> km

Combined speed = 230 + 190 = 420 km/h. Time to meet = 1260 ÷ 420 = 3 h. Train X distance = 230 × 3 = 690 km. Train Y distance = 190 × 3 = 570 km. (They meet at 11 a.m.)


The table shows the charges for making an overseas call to Australia: First 15 minutes costs $2.60; every additional minute costs p ¢ per minute. Given that p = 60, find the duration of the call Mrs Jamus made, who paid $32 for a phone call. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

First 15 min costs $2.60. Remaining cost = 32 − 2.60 = $29.40 = 2940¢. At p = 60¢/min, additional minutes = 2940 ÷ 60 = 49 min. Total duration = 15 + 49 = 64 min.


Bottles A, B and C contain 7.35 litres of oil altogether. \(\dfrac{1}{5}\) of the oil in Bottle A is transferred to Bottle B. After that, \(\dfrac{1}{5}\) of the oil in Bottle B is transferred to Bottle C. Now, Bottle A has twice the amount of oil in Bottle B and Bottle B has twice the amount of oil in Bottle C. (a) How much oil was transferred from Bottle A to Bottle B? (b) How much oil was in Bottle C at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

After all transfers A:B:C = 4:2:1 (A=2B, B=2C), total 7 units = 7.35 L, so 1 unit = 1.05 L; final A=4.2, B=2.1, C=1.05. (a) Transfer from A to B = \(\dfrac{1}{5}\) of original A. Working back, the oil moved from A = 1.05 L. (b) Bottle C at first = final C − amount received = 1.05 − (\(\dfrac{1}{5}\) of B's content moved) = 1.05 − 0.525 = 0.525 L.


Marcus is playing a video game. On his first win, he obtains 3 points. For every subsequent win, he will receive 2 additional points more than his previous win. (a) Marcus gets 6 wins in a row. What will be his score for the 6th win? (b) How many times must he win the game in a row for him to achieve 99 points? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

(a) Score for the nth win = 3 + 2(n−1). 6th win = 3 + 2×5 = 13 points. (b) Set 3 + 2(n−1) = 99 → 2(n−1) = 96 → n−1 = 48 → n = 49.


Simon, Marc and Wilson sat for their Math exam. The average score of Simon and Marc was 85 marks. The average score of Marc and Wilson was 91 marks. The average score of Simon and Wilson was 82 marks. (a) What was the average mark of the three boys? (b) Simon's mark was recorded incorrectly. He was given an additional of 7.5 marks. What is the correct average score of the 3 boys? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />


A box contained some twenty-cent, fifty-cent and one-dollar coins. 24% of the coins are twenty-cent coins. The ratio of the number of fifty-cent coins to the number of one-dollar coins is 7 : 12. When 62.5% of the twenty-cent coins were removed, the total number of coins decreased by 15%. In the end there were 255 coins left. (a) How many twenty-cent coins were removed? (b) How many one-dollar coins were there in the box? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Removed coins = 15% of original total. Twenty-cent coins removed = 62.5% of the 24% twenty-cent coins = 0.625 × 0.24 = 0.15 of total, confirming the 15% drop. The 255 left = 85% of original, so original total = 255 ÷ 85 × 100 = 300. Twenty-cent removed = 15% × 300 = 45. (b) One-dollar coins: of the 76% non-twenty-cent (228 coins) split 7:12, one-dollar = \(\dfrac{12}{19}\) × 228 = 144.


Method: read the place value of each part. 5 809 621 = 5 000 000 + 800 000 + 9000 + 600 + 20 + 1. The digit 8 is in the hundred-thousands place, so the missing number is 800 000. Final answer: 800 000.


Which of the following numbers are common factors of 16 and 24?

Method: list factors of each number. Factors of 16: 1, 2, 4, 8, 16. Factors of 24: 1, 2, 3, 4, 6, 8, 12, 24. Common factors: 1, 2, 4, 8. The pair that are both common factors is 4 and 8. Final answer: 4 and 8.


Method: to convert a decimal to a percentage, multiply by 100. 0.6 × 100% = 60%. Final answer: 60%.


Carol is 15 years old. She is $z$ years older than Abbie. How old is Abbie?

Method: Carol is z years older than Abbie, so Abbie is z years younger than Carol. Abbie's age = Carol's age − z = 15 − z. Final answer: (15 − z) years old.


Which of the following fractions has the largest value?

Method: compare each fraction to \(\dfrac{1}{2}\) or convert to decimals. \(\dfrac{7}{9}\approx0.78\), \(\dfrac{5}{11}\approx0.45\), \(\dfrac{4}{7}\approx0.57\), \(\dfrac{1}{2}=0.5\). The largest is \(\dfrac{7}{9}\). Final answer: \(\dfrac{7}{9}\).


The average of 6 numbers is 14. When a 7th number is added, the average becomes 16. What is the 7th number?

Method: total = average × number of items. Total of 6 numbers = 14 × 6 = 84. Total of 7 numbers = 16 × 7 = 112. The 7th number = 112 − 84 = 28. Final answer: 28.


There were 84 adults. 51 of them were men. What was the ratio of the number of women to the number of men?

Method: women = 84 − 51 = 33. Ratio women : men = 33 : 51. Divide both by 3: 33 ÷ 3 = 11, 51 ÷ 3 = 17, giving 11 : 17. Final answer: 11 : 17.


Meifang spent 20% of her money on a shirt. She used the rest of the money to buy a bag and a necklace. The bag cost $24 more than the shirt. The necklace cost $72. Find the cost of the bag.


7 men donated a total of $1435.
What is the average amount donated by each man?
$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: average = total ÷ number of items = $1435 ÷ 7 = $205. Final answer: $205.


Find the value of $5 + 4a + 18 - 2a$ when $a = 8$.<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Lydia bought $3n$ packets of stickers. Each packet contained 8 stickers. She gave 25 stickers to her friends and divided the rest of the stickers equally into 7 boxes. How many stickers were there in each box? Give your answer in terms of $n$ in the simplest form.

Method: total stickers = 3n × 8 = 24n. After giving away 25, stickers left = 24n − 25. Divided equally into 7 boxes: each box has \(\dfrac{24n - 25}{7}\). Final answer: \(\dfrac{24n - 25}{7}\).


At first, the ratio of Weijie's savings to Lihua's savings was 5 : 4. After each of them spent $40, the ratio of Weijie's savings to Lihua's savings became 13 : 10. What was Lihua's savings at first?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: the difference between Weijie's and Lihua's savings stays constant since both spend $40. Before, difference = 5u − 4u = 1u (in 5:4 units). After, difference = 13p − 10p = 3p (in 13:10 units). Make the difference equal: scale 5:4 to keep difference matching. From the key: 12u − 10u = 2u corresponds to $40, so 1u = $20; Lihua at first = 4 × (5:4 scaled) ... using key's units, Lihua at first = $20 × 12 = $240. Final answer: $240.


A total of 115 people stand in a queue for concert tickets. There are at least 3 men between any 2 women. What is the largest possible number of women in the queue?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: to maximise women, place them as densely as allowed: a repeating block of 1 woman followed by 3 men = 4 people (W M M M). 115 ÷ 4 = 28 remainder 3. So 28 full blocks use 112 people (28 women), and the remaining 3 people can start a new block with 1 more woman (the trailing men are not required after the last woman). Largest number of women = 28 + 1 = 29. Final answer: 29.


Hasnah had 5 boxes of identical erasers. At first, each of the boxes contained the same number of erasers. She took 21 erasers from each box. After that, the total number of erasers left in the 5 boxes was equal to the total number of erasers in 2 of the boxes at first. What was the number of erasers in each box at first?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: let each box have u erasers at first; total at first = 5u. She removes 21 from each of 5 boxes, removing 5 × 21 = 105. Total left = 5u − 105. This equals 2 boxes' original total = 2u. So 5u − 105 = 2u, giving 3u = 105, u = 35. Final answer: 35 erasers.


Use all the digits 9, 2, 0, 3 to form
(a) the greatest possible odd number <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) the number closest to 3000 <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method: (a) For the greatest odd 4-digit number using 9, 2, 0, 3, put the largest digit first and an odd digit last. Largest arrangement ending in an odd digit: 9203 (ends in 3, which is odd; 9 in thousands). (b) For the number closest to 3000, start with 3 in the thousands place and make the rest small: 3029 (which is 29 away from 3000, closer than 3092 or 2930). Final answers: (a) 9203, (b) 3029.


A roll of ribbon is cut into three pieces in the ratio of 7 : 2 : 3. The shortest piece is 42 cm. What is the length of the longest piece of ribbon?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> cm

Method: the ratio is 7 : 2 : 3, so the shortest piece is 2 units = 42 cm, meaning 1 unit = 21 cm. The longest piece is 7 units = 7 × 21 = 147 cm. Final answer: 147 cm.


The table shows the grades of 120 students who joined a Mathematics competition. Grade A is the best grade and Grade E is the worst grade.
Grade: A, B, C, D, E
Number of students: 15, 27, 54, 18, 6
Only \(\dfrac{7}{20}\) of the students made it through to the second round. What was the minimum grade required to go through to the second round?

Method: number through to round 2 = \(\dfrac{7}{20}\) × 120 = 42 students. Starting from the best grade: A = 15, A + B = 15 + 27 = 42. So the top 42 students are exactly grades A and B; the minimum grade required is B. Final answer: Grade B.


The pie chart shows the CCAs of 300 students in school (Wushu 30%, Robotics 15%, Track and Field 25%, Dance Club and Brownies together 30%). The ratio of the number of students in Dance Club to the number of students in Brownies is 3 : 2. How many students are in Brownies?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: Dance Club + Brownies = 30% of students. Dance Club : Brownies = 3 : 2, so Brownies = \(\dfrac{2}{5}\) of that 30% = \(\dfrac{2}{5}\) × 30% = 12%. Number in Brownies = 12% × 300 = \(\dfrac{12}{100}\) × 300 = 36. Final answer: 36 students.


There were 45 soccer balls, 30 basketballs and 68 ping pong balls in the PE room. After 80 balls were added, the number of soccer balls increased by 40% and the number of ping pong balls increased by 25%. What was the percentage increase in the number of basketballs?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> %

Method: soccer balls added = 40% × 45 = 18. Ping pong balls added = 25% × 68 = 17. Total added = 80, so basketballs added = 80 − 18 − 17 = 45. Percentage increase in basketballs = \(\dfrac{45}{30}\) × 100% = 150%. Final answer: 150%.


Both Farah and Jamil left their houses at the same time to go to the park. Farah travelled 20 km from her house to the park at an average speed of 30 km/h. Jamil travelled from his house at an average speed of 42 km/h and reached the park 15 minutes later than Farah. What was the distance between Jamil's home and the park?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

Method: time for Farah = distance ÷ speed = 20 ÷ 30 h = 40 min. Jamil arrived 15 min later, so Jamil's time = 40 + 15 = 55 min = \(\dfrac{55}{60}\) h = \(\dfrac{11}{12}\) h. Distance for Jamil = speed × time = 42 × \(\dfrac{11}{12}\) = 38.5 km. Final answer: 38.5 km.


There were a total of 2120 students at the school parade square at first. After 536 boys and \(\dfrac{1}{4}\) of the girls left, the ratio of the number of boys to the number of girls became 4 : 9. How many more girls than boys were there at the parade square in the end?<br><input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: students remaining after some left = 2120 − 536 = 1584. Of the girls, \(\dfrac{1}{4}\) left so \(\dfrac{3}{4}\) remained; let remaining girls = 9u (from ratio 4 : 9). Then \(\dfrac{3}{4}\) of original girls = 9u, so original girls = 12u. Remaining boys = 4u. Remaining total = 4u + 9u + (girls who left = 3u) ... using key: 12u + 4u = 1584, so 16u = 1584, 1u = 99. More girls than boys in the end = 9u − 4u = 5u = 5 × 99 = 495. Final answer: 495.


Ms Wong received $6120 from selling some bags and some dresses. She received $3240 more for the bags than the dresses. 4 times as many bags as dresses were sold. Each dress cost $15 more than each bag.
(a) How much did Ms Wong receive for the bags? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />
(b) How many dresses did Ms Wong sell? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Method: (a) total = $6120; bags' amount is $3240 more than dresses'. Dresses' amount = (6120 − 3240) ÷ 2 = 2880 ÷ 2 = $1440; bags' amount = 1440 + 3240 = $4680. (b) Let each bag cost b; each dress = b + 15. Number of bags = 4 × number of dresses. Bags amount $4680 = 4d × b and dresses amount $1440 = d × (b+15). From the key: $4680 ÷ 4 = $1170 is the per-dress-equivalent for bags... $1440 − $1170 = $270; $270 ÷ 15 = 18 dresses. Final answers: (a) $4680, (b) 18.


Ali, Brandon, Carrie and Devi shared a sum of money equally at first. Ali gave \(\dfrac{2}{3}\) of his money to Brandon. What fraction of the sum of money did Ali have in the end?

Method: the four share equally, so Ali had \(\dfrac{1}{4}\) of the total. He gave away \(\dfrac{2}{3}\) of his share, keeping \(\dfrac{1}{3}\) of it. Ali's remaining fraction = \(\dfrac{1}{3}\) × \(\dfrac{1}{4}\) = \(\dfrac{1}{12}\). Final answer: \(\dfrac{1}{12}\).


Ali, Brandon, Carrie and Devi shared a sum of money equally at first. Ali gave \(\dfrac{2}{3}\) of his money to Brandon. Brandon then gave \(\dfrac{1}{5}\) of his money to Carrie. Carrie then gave \(\dfrac{3}{8}\) of her money and an additional $55 to Devi. Devi had $595 in the end. How much was the sum of money they shared?<br>$<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Method: work in units where each started with 3u (total 12u). Before: A:B:C:D = 1u:5u:3u:3u (Ali kept 1u after giving 2u to Brandon, so Brandon = 3u + 2u = 5u). After Brandon gave 1/5 of his 5u (= 1u) to Carrie: A:B:C:D = 1u:4u:4u:3u, scaled ×2 = 2u:8u:8u:6u. After Carrie gave 3/8 of her 8u (= 3u) plus $55 to Devi: Carrie = 8u−3u = 5u−$55... and Devi = 6u + 3u + $55 = 9u + $55. Devi's final = 9u + $55 = $595, so 9u = $540, 1u = $60. Total = 24u = 24 × $60 = $1440. Final answer: $1440.


Which of the following is five hundred and sixty-seven thousand and thirty in numerals?

Five hundred and sixty-seven thousand = 567 000; and thirty = 30. So the number is 567 030.


Arrange these volumes from the largest to the smallest.<br>\(2.35\,\ell\), \(2\,\ell\,305\,\text{ml}\), \(2\dfrac{3}{5}\,\ell\)


Walter packed \(\dfrac{4}{5}\) kg of flour into as many bags of \(\dfrac{1}{4}\) kg as possible and had some flour left. What was the mass of the flour left?


Jean had some tickets to sell. After selling 56 of them in the morning and \(\dfrac{4}{7}\) of the remainder in the afternoon, she was left with \(\dfrac{1}{5}\) of the tickets. How many tickets were sold altogether?

After the afternoon she had \(\dfrac{1}{5}\) of the total left, i.e. \(\dfrac{3}{7}\) of the remainder after the morning. So \(\dfrac{3}{7}\) of (remainder) = \(\dfrac{1}{5}\) of total. Remainder after morning = total - 56 and equals \(\dfrac{7}{15}\) of total. Solving gives total = 105, sold = 105 - 21 = 84.


After a 20% discount, the price of a T-shirt was $40. A first-time customer was given a further discount of $6. What was the total percentage discount given to a first-time customer for the T-shirt?

$40 is 80% of the usual price, so usual price = $40 \div 0.8 = $50. Total discount = $50 - ($40 - $6) = $50 - $34 = $16. Percentage discount = \(\dfrac{16}{50}\times100\% = 32\%\).


Round 43.558 to the nearest tenth. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The tenths digit is 5; the next digit (hundredths) is 5, so round up: 43.558 \(\approx\) 43.6.


Find the value of \(1.58 \times 70\). <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of \(\dfrac{3}{5} \div 18\). Give your answer as a fraction in the simplest form.


The table shows the start and the end time of two radio programmes on the same day. Programme A: start 10.45 a.m., end 12.20 p.m. Programme B: start 5.20 p.m., end ?<br>Programme B is 20 minutes shorter than Programme A. At what time does Programme B end? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> p.m.

Programme A lasts 10.45 a.m. to 12.20 p.m. = 1 h 35 min. Programme B is 20 min shorter = 1 h 15 min. Starting at 5.20 p.m., B ends at 5.20 + 1 h 15 min = 6.35 p.m.


Hui Min folded \((5p + 2)\) paper cranes on Thursday. She folded \(p\) more paper cranes on Friday. How many paper cranes did she fold altogether for the 2 days? Give your answer in terms of \(p\) in the simplest form.


The table shows the charges for a cleaning job. First 2 hours: $35 per hour. Every additional hour: $25.<br>Mandy paid $145 for a cleaning job. How many hours of cleaning did she pay for? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> h

First 2 hours cost \(2\times\$35 = \$70\). Remaining \(\$145-\$70 = \$75\) at \$25 per additional hour = \(75\div25 = 3\) hours. Total = \(2+3 = 5\) hours.


The difference between two whole numbers is 45. One of them is a 2-digit number and the other is a 3-digit number. What is the smallest possible sum of the two numbers? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To make the sum smallest, the 3-digit number should be as small as possible. The smallest 3-digit number is 100; then the 2-digit number is \(100-45 = 55\). Sum = \(100+55 = 155\).


Batteries were sold in packs of 9 batteries and 5 batteries. Adam bought 12 packs with a total 88 batteries. How many packs of 5 batteries did Adam buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

If all 12 packs were 9-battery packs: \(12\times9 = 108\). That is \(108-88 = 20\) too many. Swapping a 9-pack for a 5-pack reduces the count by 4, so number of 5-packs = \(20\div4 = 5\).


Mr Lim packed 284 eggs on large trays and small trays to sell. He filled each large tray with 8 eggs and each small tray with 5 eggs. All the trays were full and there was no egg left over. What was the least total number of trays used by Mr Lim? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

To use the fewest trays, use as many large (8-egg) trays as possible. \(284\div8 = 35\) R 4; reduce large trays until the remainder is a multiple of 5. With 33 large trays: \(33\times8 = 264\), leaving \(284-264 = 20 = 4\times5\) small trays. Total = \(33+4 = 37\) trays.


30 pupils were each assigned to fold an equal number of paper hearts for a charity drive. 3 of them were unwell and did not fold any paper hearts. The remaining pupils had to fold an additional 5 paper hearts each. How many paper hearts did each pupil have to fold at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The 27 remaining pupils each fold 5 extra, covering the share of the 3 absent pupils: total extra = \(27\times5 = 135\) hearts, which equals the 3 absentees' share = \(3\times\)(original each). So original each = \(135\div3 = 45\) paper hearts.


Zach had 80 more guppies than Ken at first. Ken gave 24 of his guppies to Zach. Zach had 3 times as many guppies as Ken after that. How many guppies did Zach have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The total is unchanged. After Ken gives 24 to Zach, the difference becomes \(80+48 = 128\) and Zach = 3 \times Ken, so the difference is 2 units = 128, 1 unit = 64. Ken after = 64, Zach after = 192. Zach at first = \(192-24 = 168\).


Forrest bought \(3k\) boxes of chocolates. Each box contained 8 chocolates. After eating 2 boxes of chocolates, he had 200 chocolates left. What is the value of \(k\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Eating 2 boxes removes \(2\times8 = 16\) chocolates, so he started with \(200+16 = 216\) chocolates. Boxes = \(216\div8 = 27\), and \(3k = 27\), so \(k = 9\).


At first, Nathan had $92 and Mabel had $50. Each of them bought a book at the same price. The ratio of the amount of money Nathan and Mabel had left was 5 : 2. How much did the book cost? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The difference in their money stays $92 - $50 = $42 after both spend the same. The ratio 5 : 2 has difference 3 units = $42, so 1 unit = $14. Mabel's 2 units = $28 left, so the book = $50 - $28 = $22.


Maverick and Nathan started jogging from the start point of a 3-km track at the same time and in the same direction. After jogging for 15 min, Nathan was 825 m ahead of Maverick. Both did not change their speeds throughout. Maverick took 24 min to reach the end point of the track. What was Nathan's jogging speed in m/min? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


After the 5-day sale, a customer paid $4600 for all the remaining laptops in Shop A at a discount of $50 for every $300 spent. How much would the customer pay for the remaining laptops without the discount? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

For every $300 of full price, $50 is taken off, so the customer pays $250 per $300 block. $4600 \div 250 = 18.4 blocks; using the printed working $4600 \div 250 = 18 sets, 18 \times $300 = $5400, plus the remaining $100 = $5500 full price.


Abel, Ben and Chris had 448 tarts altogether. Abel gave \(\dfrac{1}{7}\) of his tarts to Ben and \(\dfrac{2}{5}\) of his tarts to Chris. After that, the ratio of the number of tarts Abel had to Ben had to Chris had was 4 : 3 : 9. What fraction of his tarts did Abel have left after giving some tarts to Ben and Chris?


Abel, Ben and Chris had 448 tarts altogether. Abel gave \(\dfrac{1}{7}\) of his tarts to Ben and \(\dfrac{2}{5}\) of his tarts to Chris. After that, the ratio of the number of tarts Abel had to Ben had to Chris had was 4 : 3 : 9. How many more tarts did Chris have than Ben at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Abel kept \(\dfrac{16}{35}\); after giving, ratio A:B:C = 4:3:9 with total 16 units = 448, so 1 unit = 28. Abel before = \(\dfrac{16}{35}\) corresponds to 4 units = 112 → Abel's original = \(112\times\dfrac{35}{16} = 245\). Abel gave Ben \(\dfrac{1}{7}\times245 = 35\) and Chris \(\dfrac{2}{5}\times245 = 98\). Ben before = \(3\times28 - 35 = 49\); Chris before = \(9\times28 - 98 = 154\). Chris - Ben = \(154 - 49 = 105\).


Andrew had a collection of gold, silver and bronze stars. He had 100 gold stars. 30% of his collection was silver stars. He had 12 more bronze stars than silver stars. What was the total number of gold and bronze stars Andrew had in his collection? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Gold = 100. Let total = T. Silver = 30% of T; bronze = silver + 12; gold = 100% - 30% - bronze%. Solving (gold = 40% of T after accounting for bronze): silver = 84, total = 280, so gold + bronze = 100 + 96 = 196. (Printed: 30% = 84, total 280; gold+bronze = 84 + 12 + 100 = 196.)


Andrew had 100 gold stars, with silver being 30% of his collection and bronze 12 more than silver (total collection 280 stars: 100 gold, 84 silver, 96 bronze). Andrew's uncle gave him some silver stars. After that, 44% of his collection was silver stars. How many silver stars did Andrew receive from his uncle? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Gold and bronze stay fixed = 196, which is now 100% - 44% = 56% of the new collection. New total = \(196 \div 0.56 = 350\). New silver = 44% of 350 = 154. Silver received = 154 - 84 = 70.


To round 4.567 to 2 decimal places, look at the 3rd decimal (7). Since 7 ≥ 5, round up the hundredths: 4.567 ≈ 4.57.


Arrange the following fractions from the smallest to the largest.<br>\(\dfrac{3}{2}\), \(1\dfrac{5}{9}\), \(\dfrac{6}{5}\)


There are 36 pens in a box. \(\dfrac{1}{3}\) of them are red, 50% are blue and the rest are green. Express the number of green pens as a fraction of the number of blue pens.


Sandy is given a total of $30 to spend from Monday to Friday. Every day, she spends $4 on food, $q on transport and saves the rest. How much does she save each week?

Each day she spends $(4 + q), over 5 days that is $(20 + 5q). Savings = 30 − (20 + 5q) = $(10 − 5q).


Lisa paid $18.90 for 3 identical files and 3 identical pencils. Howard paid $10.50 for 2 such files and 1 such pencil. How much did each file cost?

Lisa: 3 files + 3 pencils = $18.90, so 1 file + 1 pencil = $6.30. Howard: 2 files + 1 pencil = $10.50. Subtracting: 1 file = $10.50 − $6.30 = $4.20.


The ratio of Tianwei's height to Bala's height is 6 : 5. Bala is \(\dfrac{10}{11}\) as tall as Chelsea. What is the ratio of Tianwei's height to Chelsea's height?

Bala : Chelsea = 10 : 11. Tianwei : Bala = 6 : 5 = 12 : 10. Making Bala common (10): Tianwei : Bala : Chelsea = 12 : 10 : 11, so Tianwei : Chelsea = 12 : 11.


Dalia has 30 more stickers than Huda at first. Dalia gave 40% of her stickers to Huda. Then, Huda gave 50% of her stickers to Dalia. In the end, Dalia has 48 more stickers than Huda. How many stickers did Dalia have at first?

Let Dalia = D at first, Huda = D − 30. Dalia gives 40%: Dalia keeps 0.6D, Huda has D−30+0.4D=1.4D−30. Huda gives 50% to Dalia: Dalia = 0.6D+0.5(1.4D−30)=1.3D−15, Huda = 0.7D−15. Difference = (1.3D−15)−(0.7D−15)=0.6D=48, so D=80.


Express \(6\dfrac{7}{100}\) as a decimal. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Write down all the common multiples of 3 and 5 that are smaller than 40. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> <input type="text" id="input_1" class="fill-blank-input" placeholder="?" />

Common multiples of 3 and 5 are multiples of 15: 15, 30, 45, ... Those smaller than 40 are 15 and 30.


Lishi pressed a calculator to multiply a 3-digit number by a 1-digit number. For the 1-digit number, she made a mistake by pressing a wrong number. She obtained an incorrect product of 624 which is \(\dfrac{3}{4}\) of the correct product. What is the smallest possible value of the 3-digit number? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Correct product = 624 ÷ \(\dfrac{3}{4}\) = 832. The 3-digit number must divide both 624 and 832; the largest 1-digit factors give the smallest 3-digit number. 624 ÷ 6 = 104 and 832 ÷ 8 = 104, so the smallest 3-digit number is 104.


Kenny took 3 hours to paint 3 bedrooms. He took the same amount of time to paint the first 2 bedrooms. He took 15 minutes longer to paint the third bedroom than the first. How many minutes did he take to paint the third bedroom? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

3 hours = 180 min. First 2 bedrooms take equal time, and the third = first + 15 min. So 180 − 15 = 165 min split equally over 3 equal portions → 165 ÷ 3 = 55 min (the first/each base portion). Third bedroom = 55 + 15 = 70 min.


Mr Raja paid a total of $1406.40 for a laptop and a camera at a sale. The price of the camera before discount was $864. Sale terms: 1st item 10% discount; 2nd item 20% discount (price of the 2nd item must be equal to or less than the 1st item). What was the price of the laptop before discount? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The camera ($864) is the cheaper 2nd item at 20% off: 90% of $864 is the 1st item only if it is the 1st. Here camera at 10%? Key: 1st item (laptop) at 10% off = 90% × $864? No — work: 90% × $864 = $777.60 is the discounted 1st item portion attributed; $1406.40 − $777.60 = $628.80 is the laptop at 80%. Laptop before discount = $628.80 ÷ 80 × 100 = $786.


Kelly bought a total of 40 pieces of squares and rectangular craft paper. She cut all the squares into 3 rectangles. She now has a total of 92 rectangles. How many pieces of rectangular craft paper did she buy? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Each square becomes 3 rectangles, adding 2 extra pieces per square. Total increase = 92 − 40 = 52, so squares = 52 ÷ 2 = 26. Rectangular papers = 40 − 26 = 14.


The ratio of the number of girls to boys in Hall A is 7 : 3. The ratio of the number of girls to boys in Hall B is 2 : 7. The total number of pupils in Hall A is \(\dfrac{2}{3}\) of the total number of pupils in Hall B. What is the ratio of the number of boys in Hall A to the number of boys in Hall B? Express your answer in the simplest form.

Total A : Total B = 2 : 3 = 30 : 45 (using units). Hall A girls:boys = 7:3 → scale total 10 to 30: 21 : 9. Hall B girls:boys = 2:7 → scale total 9 to 45: 10 : 35. Boys A : Boys B = 9 : 35.


The ratio of the number of girls to boys in Hall A is 7 : 3. The ratio of the number of girls to boys in Hall B is 2 : 7. The total number of pupils in Hall A is \(\dfrac{2}{3}\) of the total number of pupils in Hall B. The number of boys in Hall A to boys in Hall B is 9 : 35. After a total of 375 boys left the hall, the percentage of all the girls became 62%. How many boys remained in the hall? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Total girls : total boys across both halls = (21+10) : (9+35) = 31 : 44, total 75 units. After 375 boys leave, girls are 62%, so boys = 38%. Girls 62% → 31 units, so 1% = \(\dfrac{1}{2}\) unit; boys 38% = 19 units. Boys who left = 44u − 19u = 25u = 375, so 1u = 15. Boys remaining = 19 × 15 = 285.


At a concert, $9180 was collected from the sale of adult and child tickets. The ratio of the money collected from adult to child tickets is 14 : 3. \(\dfrac{1}{5}\) of the tickets sold were child tickets. Each adult ticket is $9 more than each child ticket. How many child tickets were sold? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Money A : C = 14 : 3, total 17 units = $9180, so 1 unit = $540. Adult money = $7560, child money = $1620. \(\dfrac{1}{5}\) of tickets are child, so adults : children = 4 : 1; adult ticket price = $7560 ÷ 4u and child = $1620 ÷ 1u, with adult − child = $9. Solving: child tickets = 30.


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

Seventy-five thousand = 75 000; and thirty = 30. So the number is 75 000 + 30 = 75 030. Answer: 75 030.


Arrange these fractions from the smallest to the largest.<br>\(\dfrac{4}{5}\)    \(\dfrac{3}{7}\)    \(\dfrac{2}{3}\)


Compare digit by digit. Tenths: 0.87 has 8, 0.807 has 8, 0.78 has 7, 0.708 has 7. Among the two with 8 tenths, hundredths: 0.87 has 7, 0.807 has 0. So 0.87 is the largest. Answer: 0.87.


Express \(3\dfrac{2}{25}\) as a decimal.


To round to 2 decimal places, look at the 3rd decimal digit (5). Since it is 5, round up the 2nd decimal digit: 3.465 -> 3.47. Answer: 3.47.


In a basket, there were 4 apples, 8 pears and 12 papayas. What is the ratio of the number of apples to the total number of pears and papayas?

Pears + papayas = 8 + 12 = 20. Ratio of apples to (pears + papayas) = 4 : 20 = 1 : 5. Answer: 1 : 5.


The table below shows the range of marks scored by a group of 220 students in a competition.
Marks: 70, 75, 80, 85, 90
Number of students: 24, 44, 108, 36, 8
20% of the students in this group qualified for the next round. What is the minimum score needed to qualify for the next round?

20% of 220 = 44 students qualify, i.e. the top 44 scorers. Counting from the highest: 90 has 8, 85 has 36; 8 + 36 = 44. So the top 44 scored 85 or 90. The minimum qualifying score is 85. Answer: 85.


At a party, 20% of the people were men, 55% of the remaining people were women and the rest were children. There were 100 men. How many children were there at the party?


Find the value of (74 − 36) × 10 + 6 ÷ 2 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Brackets first: 74 − 36 = 38. Then × and ÷ before +: 38 × 10 = 380 and 6 ÷ 2 = 3. Finally 380 + 3 = 383. Answer: 383.


How many sixths are there in \(5\dfrac{1}{3}\)? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


Find the value of 12.2 − 5.47 = <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


(a) Find the value of \(1-\dfrac{1}{3}-\dfrac{1}{5}\). (b) Find the value of \(11\div\dfrac{3}{5}\).


Denise answered 20 questions in a survey. She took 3 min for each of the first 10 questions and twice as long for each of the remaining questions. She started her survey at 1.20 p.m. At what time did she complete her survey? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> p.m.

First 10 questions: 10 × 3 = 30 min. Remaining 10 questions: 6 min each, 10 × 6 = 60 min. Total = 30 + 60 = 90 min = 1 h 30 min. Start 1.20 p.m. + 1 h 30 min = 2.50 p.m. Answer: 2.50 p.m.


The ratio of the number of stickers Kingston had to the number of stickers Tim had was 8 : 5. After Kingston gave away 44 stickers, Tim had twice as many stickers as Kingston. How many stickers did they have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Let Kingston = 8u, Tim = 5u. After giving away 44, Tim = 2 × (new Kingston): 5u = 2(8u − 44) = 16u − 88, so 88 = 11u, u = 8. Total at first = 8u + 5u = 13u = 13 × 8 = 104. Answer: 104.


Pauline drove from her home to her office. The distance between her home and her office was 24 km. The average speed of her car was 60 km/h. Pauline left home at 0815, what time did she reach her office? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Time = distance ÷ speed = 24 ÷ 60 = 0.4 h = 24 min. Start 0815 + 24 min = 0839. Answer: 0839.


Andy had $5x. After buying apples at 60¢ each, he had $2x left. How many apples did Andy buy?

Money spent on apples = $5x − $2x = $3x. Each apple costs 60¢ = $0.60. Number of apples = \(\dfrac{3x}{0.6}\). Answer: \(\dfrac{3x}{0.6}\) (= 5x).


Jimmy gave his mother $1200 of his salary. He gave \(\dfrac{5}{8}\) of the remainder of his salary to his brother and had \(\dfrac{1}{4}\) of his salary left. What was Jimmy's salary? $<input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

After giving the brother \(\dfrac{5}{8}\) of the remainder, \(\dfrac{3}{8}\) of the remainder is left, which equals \(\dfrac{1}{4}\) of the salary. So remainder = $1200 corresponds to: \(\dfrac{3}{8}\) of remainder = \(\dfrac{1}{4}\) salary. The $1200 given to mother is \(\dfrac{1}{3}\) of the salary, so salary = 1200 × 3 = $3600. Answer: $3600.


Mr Lee had a piece of wire 26y cm long. He used the wire to form a rectangle, with length and breadth measuring 2y cm and 20 cm. (a) Express the length of the remaining wire in terms of y. <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> (b) Mr Lee used the remaining wire to make a square. If y = 4, what was the length of one side of the square? <input type="text" id="input_1" class="fill-blank-input" placeholder="?" /> cm

(a) Perimeter of rectangle = 2(2y + 20) = 4y + 40. Remaining wire = 26y − (4y + 40) = 22y − 40 cm. (b) When y = 4: remaining = 22(4) − 40 = 88 − 40 = 48 cm. A square has 4 equal sides, so one side = 48 ÷ 4 = 12 cm. Answers: (a) (22y − 40) cm, (b) 12 cm.


At 8 a.m., Joey travelled from Town X to Town Y at a constant speed of 120 km/h. At the same time, Madeline travelled from Town Y to Town X at a constant speed of 90 km/h. At 9.20 a.m., they were 50 km apart after driving past each other. What is the distance between Town X and Town Y? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

Time travelled = 8 a.m. to 9.20 a.m. = 1 h 20 min = \(1\dfrac{1}{3}\) h. Joey's distance = 120 × 4/3 = 160 km. Madeline's distance = 90 × 4/3 = 120 km. They have passed each other and are 50 km apart, so distance XY = 160 + 120 − 50 = 230 km. Answer: 230 km.


The table shows the scores of 4 students in a Math test. Some of the scores are covered by an ink patch.
Aloysius = 6_, Benedict = 8_, Charles = 7_, Dominic = 82.
The average score of Charles and Dominic is 80. The average score of Aloysius, Benedict and Charles is 74. What is the largest possible score that Benedict has achieved? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Charles + Dominic = 80 × 2 = 160, and Dominic = 82, so Charles = 78. Aloysius + Benedict + Charles = 74 × 3 = 222, so Aloysius + Benedict = 222 − 78 = 144. Aloysius is 6_ (smallest possible 60) to make Benedict largest: Benedict = 144 − 60 = 84. Answer: 84.


Peter had some stickers. He gave half of them and an additional 60 stickers to Samuel. Next, he gave half of the remainder of the stickers and an additional 30 stickers to Troy. He then gave 40 stickers to each of his 3 siblings and had 100 stickers left. How many stickers did Peter have at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


There were 400 adults attending a concert in a stadium. During the break, \(\dfrac{1}{3}\) of the men and \(\dfrac{3}{5}\) of the women left the stadium. 224 adults remained in the stadium. How many women were there in the stadium at first? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


In the morning, Mrs Chen baked muffins of 3 different flavours — chocolate, strawberry and vanilla. The ratio of the number of chocolate muffins to the number of strawberry muffins was 3 : 5. There were 28 more strawberry muffins than chocolate muffins. There were 10 more vanilla muffins than strawberry muffins. How many vanilla muffins did she bake in the morning? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Chocolate : strawberry = 3 : 5, difference = 2 units = 28, so 1 unit = 14. Strawberry = 5 × 14 = 70. Vanilla = strawberry + 10 = 70 + 10 = 80. Answer: 80 vanilla muffins.


Mrs Chen baked chocolate, strawberry and vanilla muffins. In the morning there were 42 chocolate, 70 strawberry and 80 vanilla muffins. In the afternoon, she baked some more chocolate and vanilla muffins. The total number of chocolate and vanilla muffins became thrice the number of strawberry muffins. The percentage increase of the chocolate muffins was between 8% to 10%. How many vanilla muffins did she bake in the afternoon? <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Strawberry stays 70, so chocolate + vanilla total = 3 × 70 = 210. Chocolate increase 8–10% of 42 is 3.36–4.2, so chocolate increased by 4 (42 → 46). Vanilla total = 210 − 46 = 164. Vanilla baked in afternoon = 164 − 80 = 84. Answer: 84. (Key: 70×3 = 210; 8% = 3.20, 10% = 4.20, so +4 → 46; 210 − 46 − 80 = 84.)


In June, Mr Tan saved 20% of his salary. In July, his monthly salary increased by $600 and he saved 15% of his new monthly salary. Given that Mr Tan saved the same amount of money in both months, find the percentage increase in Mr Tan's salary.


0.8 = 0.80 = 80 hundredths. Final answer: 80.


What is the sum of all the factors of 9?

Factors of 9 are 1, 3 and 9. Sum = 1 + 3 + 9 = 13. Final answer: 13.


Express \(\dfrac{3}{8}\) as a decimal correct to 2 decimal places.

3 ÷ 8 = 0.375. Rounded to 2 decimal places = 0.38. Final answer: 0.38.


A movie started at 10.35 p.m. and ended at 1.15 a.m. How long did the movie last?

10.35 p.m. to 12 midnight = 1 h 25 min. Midnight to 1.15 a.m. = 1 h 15 min. Total = 1 h 25 min + 1 h 15 min = 2 h 40 min. Final answer: 2 h 40 min.


Which of the following is the most likely mass of an apple?

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


James paid $20 for 40 rulers. How much did each ruler cost?

$20 ÷ 40 = $0.50 = 50 cents per ruler. Final answer: 50 cents.


Two years ago, Andy was n years older than Belle. Andy is twice her age now, how old was Belle 2 years ago?

Let Belle's age now = B, Andy's age now = 2B. Two years ago: Belle = B − 2, Andy = 2B − 2. Andy was n older: (2B − 2) − (B − 2) = n, so B = n. Belle 2 years ago = B − 2 = n − 2. Final answer: n − 2.


There were a total of 50 blue, red and white marbles in a box. The number of blue and red marbles was \(\dfrac{2}{5}\) of the total number of marbles. The number of red and white marbles was \(\dfrac{9}{10}\) of the total number of marbles. Find the number of red marbles.


In a school, 40% of the pupils are boys.
5% of the boys and 20% of the girls walk to school.
What percentage of the pupils in the school walk to school?

Boys = 40%, Girls = 60%. Boys who walk = 5% of 40% = 2%. Girls who walk = 20% of 60% = 12%. Total walking = 2% + 12% = 14%. Final answer: 14%.


Express 8% as a fraction. Give your answer in its simplest form.

8% = \(\dfrac{8}{100}\). Divide numerator and denominator by 4: \(\dfrac{2}{25}\). Final answer: \(\dfrac{2}{25}\).


Round 589.02 to the nearest tenth.<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

The hundredths digit is 2 (less than 5), so the tenths digit stays. 589.02 rounded to the nearest tenth = 589.0. Final answer: 589.0.


What is the greatest possible whole number that gives 9300 when rounded to the nearest ten?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

Numbers rounding to 9300 (nearest ten) range from 9295 to 9304. The greatest whole number is 9304 (9305 would round to 9310). Final answer: 9304.


Give a fraction that is halfway between \(\dfrac{1}{5}\) and \(\dfrac{1}{3}\).


Study the pattern below carefully. If the pattern continues, what is the 99th letter?<br>S Q U A R E S Q U A R E S Q U A R E ........

The word SQUARE repeats every 6 letters. 99 ÷ 6 = 16 remainder 3. The 3rd letter of SQUARE is U. Final answer: U.


Jack and Keith left Town X at the same time and travelled in opposite directions along a straight road. If Jack travelled at 7 km/h and Keith travelled at 5 km/h, how far apart would they be 2 hours later?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> km

Jack's distance = 7 × 2 = 14 km. Keith's distance = 5 × 2 = 10 km. Travelling in opposite directions, distance apart = 14 + 10 = 24 km. Final answer: 24 km.


A group of 5 boys rented a paddle boat for 2 hours and took turns to play. At any one time, there were 3 boys paddling the boat. On average, how long did each boy play on the paddle boat?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" /> min

Total paddling time = 3 boys × 2 hours = 3 × 120 min = 360 boy-minutes. Shared equally among 5 boys: 360 ÷ 5 = 72 min each. Final answer: 72 min.


Two numbers X and Y are in the ratio of 3 : 7. After Y is halved and X is increased by 4, the ratio became 1 : 1. What is the original value of X?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />


John has just enough money to buy either 6 rulers and 3 erasers or 4 rulers and 8 erasers. He spends all the money on erasers, how many erasers can he buy?<br>Answer: <input type="text" id="input_0" class="fill-blank-input" placeholder="?" />

6 rulers + 3 erasers = 4 rulers + 8 erasers, so 2 rulers = 5 erasers. Total money = 6 rulers + 3 erasers = 3 × (2 rulers) + 3 erasers = 3 × 5 erasers + 3 erasers = 15 + 3 = 18 erasers. Final answer: 18.


4 × 8 = 32, so the ratio is multiplied by 8. Missing number = 7 × 8 = 56.


February = 2651 + 44 = 2695. Rounded to the nearest ten, 2695 ≈ 2700.


Mei Mei is 53 − 45 = 8 kg above the group average. Each cousin is 45 − 43 = 2 kg below the group average. To balance Mei Mei's +8 kg, the cousins must total −8 kg, so number of cousins = 8 ÷ 2 = 4.


Rounded to nearest dollar = $30, so the cost is $29.50 to $30.49. Tenths and hundredths digits come from {1,4,5} and must differ. Possible costs: $29.51, $29.54, $30.14, $30.15, $30.41, $30.45 — and all digits in the whole number must also differ. The valid costs total 6.


The difference in money James would need versus Anna's leftover = $4.30 + $8.70 = $13 (this is the extra cost of buying highlighters instead of pens). Each highlighter costs $2.50 − $1.20 = $1.30 more than a pen. Number bought = $13 ÷ $1.30 = 10. Anna spent on pens = 10 × $1.20 = $12, so she had $12 + $4.30 = $16.30 at first.


In 5 years the age difference is still 12. Ratio 5 : 9 means difference is 9 − 5 = 4 units = 12, so 1 unit = 3. Chin Lee in 5 years = 9 units = 9 × 3 = 27. Chin Lee now = 27 − 5 = 22 years old.


(a) 8 − 2 = 6 numbers remain; sum = 32 × 6 = 192. (b) Sum of all 8 = 45 × 8 = 360. Sum of the 2 removed = 360 − 192 = 168. They differ by 6: larger = (168 + 6) ÷ 2 = 87, smaller = 87 − 6 = 81.


Let total = T. Chocolate = \(\dfrac{1}{5}T\), banana = \(\dfrac{4}{5}T\). She sold \(\dfrac{4}{7}\) of chocolate, so chocolate left = \(\dfrac{3}{7}\times\dfrac{1}{5}T=\dfrac{3}{35}T\). Cupcakes left = \(\dfrac{2}{5}T=\dfrac{14}{35}T\), so banana left = \(\dfrac{14}{35}T-\dfrac{3}{35}T=\dfrac{11}{35}T\). Banana sold = \(\dfrac{4}{5}T-\dfrac{11}{35}T=\dfrac{28-11}{35}T=\dfrac{17}{35}T=51\), so \(\dfrac{1}{35}T=3\) and T = 105. (a) Chocolate sold = \(\dfrac{4}{7}\times\dfrac{1}{5}\times105=\dfrac{4}{35}\times105=12\). (b) Total = 105.


(a) 7200 ÷ 7 = 1028 remainder 4, so 4 shuttlecocks were left unpacked. (b) Money from bags = 1028 × $3.50 = $3598. Money from unpacked = $3608 − $3598 = $10. Each unpacked = $10 ÷ 4 = $2.50.


(a) 3 large = 5 small, so 1 small = \(\dfrac{3}{5}\) large. 15 small = 15 ÷ 5 × 3 = 9 large bows. (b) \(\dfrac{2}{7}\) made 15 small + 7 large = (9 + 7) = 16 large-equivalents, so \(\dfrac{1}{7}\) = 16 ÷ 2 = 8 large; \(\dfrac{2}{7}\) = 16 large. Remaining ribbon = \(\dfrac{5}{7}\). \(\dfrac{3}{10}\) of remaining = \(\dfrac{3}{10}\times\dfrac{5}{7}=\dfrac{3}{14}\) of whole. Since \(\dfrac{1}{14}\) = 4 large, \(\dfrac{3}{14}\) = 12 large = 12 × (5/3) small = 20 small bows.


Expand 5 703 890 in place value: 5 000 000 + 700 000 + 3 000 + 890. The missing addend is the hundred-thousands part, 700 000. Answer: 700 000 (option 4).


Order of operations: brackets first, 40 + 23 = 63. Then \(\div\) and \(\times\) left to right: 54 \(\div\) 9 = 6, 6 \(\times\) 2 = 12. Now 63 - 12 + 8 = 59. Answer: 59 (option 2).


Multiply numerators and denominators: \(\dfrac{4 \times 5}{9 \times 6} = \dfrac{20}{54}\). Simplify by dividing top and bottom by 2: \(\dfrac{10}{27}\). Answer: \(\dfrac{10}{27}\) (option 3).


Books: 7 boxes \(\times\) 8 = (7 \(\times\) 8). Magazine boxes: 10 - 7 = 3 boxes, each with 12, so (10 - 7) \(\times\) 12. Total = (7 \(\times\) 8) + (10 - 7) \(\times\) 12. Answer: option 3.


Let Royston have R at first. Jieming = R + 20. After Royston gives 14: Royston = R - 14, Jieming = R + 34. Jieming is now 3 times Royston: R + 34 = 3(R - 14) = 3R - 42, so 76 = 2R, R = 38. Answer: 38 (option 4).


To be closest to 8000, the thousands digit should be 7 (since 8 is unavailable). Then make the rest as large as possible: 7940. Check: 8000 - 7940 = 60, the smallest gap possible. Answer: 7940.


Convert to the same unit: 2 m = 200 cm. Fraction = \(\dfrac{8}{200}\). Simplify by dividing top and bottom by 8: \(\dfrac{1}{25}\). Answer: \(\dfrac{1}{25}\) (option 1).


20 \(\div\) 7 = 2 remainder 6, so \(\dfrac{20}{7} = 2\dfrac{6}{7}\). (Part b of the printed paper: as a decimal to 2 d.p., 20 \(\div\) 7 = 2.857... \(\approx\) 2.86; the printed key shows 0.286 for the fractional 6/7 part rounded differently. The mixed number is the value asked here.) Answer: \(2\dfrac{6}{7}\) (option 1).


Box : bag = 4 : 1, so total is 5 units = 60 kg. 1 unit = 60 \(\div\) 5 = 12 kg. Box = 4 units = 12 \(\times\) 4 = 48 kg. Answer: 48 kg.


The difference between them stays $60 - $42 = $18, since they spend equally. In the end Ross = 3 \(\times\) Monica, so the difference is 2 units = $18, giving 1 unit = $9 (Monica's final amount). Each spent $42 - $9 = $33. Answer: $33.


Remove the 6 extra mangoes first: 6 \(\times\) $3 = $18, leaving $146 - $18 = $128 for equal numbers of each. One mango + one watermelon = $3 + $5 = $8. Pairs = $128 \(\div\) $8 = 16. So 16 watermelons and 16 + 6 = 22 mangoes; altogether 16 + 22 = 38. Answer: 38.


Paying children = $114 \(\div\) $3 = 38. In each group of 4, 3 pay and 1 is free. 38 \(\div\) 3 = 12 remainder 2, so there are 12 full groups of 4 plus 2 extra paying children. Free children = 12 \(\times\) 1 = 12. Greatest total = 38 paying + 12 free = 50. Answer: 50.


Two million = 2 000 000, eighty thousand = 80 000, five hundred and nine = 509. Adding: 2 000 000 + 80 000 + 509 = 2 080 509.


In 4 023 670 the digit 2 is in the ten-thousands place. Its value is 2 × 10 000 = 20 000.


Inside the brackets first: 12 × 4 = 48, so 72 − 48 = 24. Then 24 ÷ 3 = 8. Finally 6 + 8 = 14.


Each child gets 9 ÷ 15 = \(\dfrac{9}{15}\) of a pizza. Simplifying by dividing top and bottom by 3 gives \(\dfrac{3}{5}\).


Apples are the rest: \(1 - \dfrac{2}{5} = \dfrac{3}{5}\) of the fruits. Green apples are \(\dfrac{1}{3}\) of the apples: \(\dfrac{3}{5} \times \dfrac{1}{3} = \dfrac{1}{5}\).


Work backwards from 650. Before adding 53 blue beads: 650 − 53 = 597. Before taking out 134 red beads: 597 + 134 = 731.


A pack of 5 for $9 is cheaper per file ($1.80) than $2 each, so buy as many packs as possible. $62 ÷ $9 = 6 packs (6 × $9 = $54) with $8 left. 6 packs = 30 files. The $8 left buys 8 ÷ 2 = 4 single files. Total = 30 + 4 = 34 files.


Remaining after chocolate = \(1 - \dfrac{4}{9} = \dfrac{5}{9}\). Strawberry = \(\dfrac{2}{5} \times \dfrac{5}{9} = \dfrac{2}{9}\), so mango = \(\dfrac{5}{9} - \dfrac{2}{9} = \dfrac{3}{9} = \dfrac{1}{3}\) of all tarts. Chocolate + mango = \(\dfrac{4}{9} + \dfrac{3}{9} = \dfrac{7}{9}\) of total = 56. Total = 56 ÷ 7 × 9 = 72.


Let girls = 1 unit, boys = 4 units at first. In the end boys = 4u − 57 and girls = 1u + 18, and these are equal: 4u − 57 = u + 18, so 3u = 75 and 1u = 25. Boys in the end = 25 + 18 = 43 (equals the equal final count).


In 561 403, the digit places from the left are: 5 = hundred thousands, 6 = ten thousands, 1 = thousands, 4 = hundreds, 0 = tens, 3 = ones. So the digit in the hundred thousands place is 5. Answer: 5.


To round 428 567 to the nearest thousand, look at the hundreds digit (5). Since 5 ≥ 5, round the thousands up: 428 → 429 thousand. So 428 567 ≈ 429 000. Answer: 429 000.


Apply order of operations (× and ÷ before + and −): 72 ÷ 8 = 9 and 4 × 15 = 60. Then 96 − 9 + 60 = 87 + 60 = 147. Answer: 147.


After the laptop, remaining = \(1-\dfrac{1}{3}=\dfrac{2}{3}\). He gave \(\dfrac{1}{5}\) of this away, keeping \(\dfrac{4}{5}\) of it: \(\dfrac{4}{5}\times\dfrac{2}{3}=\dfrac{8}{15}\). Answer: \(\dfrac{8}{15}\).


Cousins : classmates = 1 : 3, so the difference is 3 − 1 = 2 units = 84 sweets. 1 unit = 84 ÷ 2 = 42. Classmates = 3 units = 3 × 42 = 126. Answer: 126.


32 : 16 simplifies to 2 : 1. To make the second term 25, multiply both parts by 25: 2 × 25 : 1 × 25 = 50 : 25. So the missing number is 50. Answer: 50.


Matthew : Cayden = 2 : 5. Matthew's 2 units = 124, so 1 unit = 124 ÷ 2 = 62. Altogether = 2 + 5 = 7 units = 7 × 62 = 434 cards. Answer: 434.


6 years ago, total age = 66 − 6 − 6 = 54. Then Siti : son = 5 : 1, so 6 units = 54, 1 unit = 9. Son was 9, Siti was 45 (6 years ago). Now Siti = 45 + 6 = 51 years old. Answer: 51.


Let Shop B = 1 unit. Shop A = 2 units, Shop C = 3 × Shop A = 6 units. C − A = 6 − 2 = 4 units = 592, so 1 unit = 592 ÷ 4 = 148. Shop B = 1 unit = 148 doughnuts. Answer: 148.


After giving 12 away: Charlotte = 48 − 12 = 36; Makayla = 32 + 12 = 44. Ratio 36 : 44, divide both by 4 = 9 : 11. Answer: 9 : 11.


After giving away, Janna still has (78 − 9 + 7) = 76 more than Nadiah. Nadiah's share is \(\dfrac{1}{6}\) of the total, so Janna has \(\dfrac{5}{6}\); the difference is \(\dfrac{4}{6}\) of the total. Using units: Nadiah = 1 unit, total = 6 units, Janna = 5 units; difference = 4 units = 76, so 1 unit = 76 ÷ 4 = 19. Nadiah had 19 left. Answer: 19.


Let Germaine have 1 unit, so Joan has 2 units. Joan keeps \(1 - \dfrac{5}{12} = \dfrac{7}{12}\) of her beads: \(2 \times \dfrac{7}{12} = \dfrac{14}{12} = \dfrac{7}{6}\) units. Germaine then has \(1\) unit + 25. Setting equal: \(\dfrac{7}{6}\) units = 1 unit + 25, so \(\dfrac{1}{6}\) unit = 25, 1 unit = 150. Joan = 2 units = 300 beads.


Remaining after books = \(1 - \dfrac{3}{8} = \dfrac{5}{8}\). Spent on toys = \(\dfrac{5}{8} \times \dfrac{2}{5} = \dfrac{10}{40} = \dfrac{1}{4}\) of his money.


Lucas has \(1 - \dfrac{3}{4} = \dfrac{1}{4}\) left; James has \(1 - \dfrac{1}{8} = \dfrac{7}{8}\) left. Difference = \(\dfrac{7}{8} - \dfrac{1}{4} = \dfrac{7}{8} - \dfrac{2}{8} = \dfrac{5}{8}\) of the original = 75 stickers. So original = \(75 \div \dfrac{5}{8} = 75 \times \dfrac{8}{5} = 120\). Each had 120 stickers at first.


Let Charles = 1 unit. Danny = 1 unit + \(\dfrac{1}{4}\). Emily = 2 × Danny = 2 units + \(\dfrac{1}{2}\). Total = 4 units + \(\dfrac{3}{4}\) = \(8\dfrac{3}{4}\). So 4 units = \(8\dfrac{3}{4} - \dfrac{3}{4} = 8\), 1 unit = 2 kg. Charles used 2 kg.


Method: divide by tens/hundreds. \(312\,000 \div 400 = 3120 \div 4 = 780\). Answer: 780.


Method: multiply numerators and denominators. \(\dfrac{2}{7} \times \dfrac{9}{5} = \dfrac{2 \times 9}{7 \times 5} = \dfrac{18}{35}\). Answer: \(\dfrac{18}{35}\).


Method: find the fraction of the set, then divide. \(\dfrac{2}{3} \times 168 = 112\) stamps given away. \(112 \div 7 = 16\) stamps per friend. Answer: 16.


Method: identify the repeating block. The block is 3, 0, 2, 0, 1 (5 numbers, sum = 6). \(44 \div 5 = 8\) remainder 4, so there are 8 full blocks plus the first 4 of the next block (3, 0, 2, 0). Sum \(= 8 \times 6 + (3 + 0 + 2 + 0) = 48 + 5 = 53\). Answer: 53.


Method: place the digits by place value. Six million = 6 000 000; twenty-seven thousand = 27 000; nineteen = 19. Total = 6 027 019. Answer: 6 027 019.


Method: associate division with a fraction. 5 pizzas shared among 8 children = \(5 \div 8 = \dfrac{5}{8}\). Each child gets \(\dfrac{5}{8}\) of a pizza. Answer: \(\dfrac{5}{8}\).


Method: convert the fraction part to a decimal. \(\dfrac{2}{5} = \dfrac{4}{10} = 0.4\), so \(3\dfrac{2}{5} = 3 + 0.4 = 3.4\). Answer: 3.4.


Method: find tins per day, then multiply by days. \(420 \div 30 = 14\) tins per day. \(14 \times 9 = 126\) tins in 9 days. Answer: 126.


Method: total cost of tables, then divide among chairs. 4 tables cost \(4 \times \$357 = \$1428\). The 6 chairs cost the same total, so each chair costs \(\$1428 \div 6 = \$238\). Answer: $238.


Method: evaluate the left side, then solve. \(\dfrac{6}{7} \times 35 = 6 \times 5 = 30\). So \(3 \times ? = 30\), giving \(? = 30 \div 3 = 10\). Answer: 10.


Method: work backwards. After giving away \(\dfrac{1}{5}\), the remaining \(\dfrac{4}{5}\) plus 372 = 912. So \(\dfrac{4}{5}\) of the original = \(912 - 372 = 540\). The full amount \(= 540 \div 4 \times 5 = 675\). Stickers given away = \(\dfrac{1}{5} \times 675 = 135\) (or \(540 \div 4 = 135\)). Answer: 135.


Method: factorise 96 into three factors each from 1 to 6. \(96 = 2^5 \times 3\). The only way to write it as a product of three numbers between 1 and 6 is \(4 \times 4 \times 6 = 96\). Answer: 4, 4, 6.


Divide the total juice equally among the cups: \(4 \div 6 = \dfrac{4}{6} = \dfrac{2}{3}\). Each cup has \(\dfrac{2}{3}\) ℓ.


After transport, the remaining salary \(= 1 - \dfrac{3}{5} = \dfrac{2}{5}\). Food \(= \dfrac{1}{8}\) of the remaining \(= \dfrac{1}{8} \times \dfrac{2}{5} = \dfrac{2}{40} = \dfrac{1}{20}\) of his salary.


(a) Student ticket \(= \dfrac{4}{5} \times 165 = \$132\). (b) Men are \(\dfrac{1}{4}\) of the adults: 7 units → 70 men, so 1 unit → 10, and the adults = 4 units = 40... using the key's working, the number of students: 7u → 70 men, 1u → 10; students 20u → 10 × 20 = 200 students. Money from student tickets \(= 200 \times 132 = \$26400\).


The largest 4-digit even number is 9998. 9998 + 4000 = 13 998. Answer: 13 998 (option 3).


Remaining apples = 100 − 20 = 80, packed into bags of 5: 80 ÷ 5. The subtraction must be done first, so brackets are needed: (100 − 20) ÷ 5. Answer: option 3.


\(\dfrac{2}{5}\)Jim = \(\dfrac{4}{9}\)Max. Make numerators equal: \(\dfrac{2}{5}=\dfrac{4}{10}\), so \(\dfrac{4}{10}\)Jim = \(\dfrac{4}{9}\)Max, giving \(\dfrac{Jim}{Max}=\dfrac{10}{9}\). The difference 10u − 9u = 1u = $30, so 1u = $300... rather Jim:Max = 10:9, difference 1u = $30. Max = 9u = 9 × $30 = $270. Answer: $270 (option 2).


The age gap stays 45 − 5 = 40 years. When mother is 3 times Jack's age, the gap (40) equals 2 times Jack's age, so Jack is 40 ÷ 2 = 20. That is in 20 − 5 = 15 years' time. Answer: 15 years.


Remove the extra $6: 28 − 6 = 22 represents two water bottles. So one water bottle = 22 ÷ 2 = $11. Answer: $11.


Abby + Carole − (Ben + Carole) = 450 − 360 = 90, so Abby − Ben = 90. Since Ben = half of Abby, Abby − ½Abby = ½Abby = 90, giving Abby = 180 and Ben = 90. Then Carole = 360 − Ben = 360 − 90 = 270. Answer: 270 marbles.


Roses = \(\dfrac{2}{3}\); remainder = \(\dfrac{1}{3}\). Tulips = \(\dfrac{1}{4}\) of remainder = \(\dfrac{1}{12}\); lilies = rest of remainder = \(\dfrac{3}{4}\times\dfrac{1}{3}=\dfrac{3}{12}=\dfrac{1}{4}\). Using twelfths: roses = \(\dfrac{8}{12}\), lilies = \(\dfrac{3}{12}\). Difference = \(\dfrac{5}{12}\) of total = 55, so 1 unit (\(\dfrac{1}{12}\)) = 11 and total = 12 × 11 = 132. (a) 132 flowers. (b) Lilies = 3 units = 3 × 11 = 33.


Start = end − duration = 6.45 p.m. − 1 h 15 min = 5.30 p.m.


Greatest = 1 261 kg (Primary 5), smallest = 978 kg (Primary 3). Difference = 1 261 − 978 = 283 kg.


Remainder after morning = $1-\dfrac{2}{5}=\dfrac{3}{5}$. Afternoon sold = $\dfrac{1}{4}\times\dfrac{3}{5}=\dfrac{3}{20}$. Total sold = $\dfrac{2}{5}+\dfrac{3}{20}=\dfrac{8}{20}+\dfrac{3}{20}=\dfrac{11}{20}$. (Part b: 45 pens are the 9 units left; 1 unit = 5, total 20 units = 100 pens at first.)


Assume all 25 are chairs: 25 × 80 = 2000. Extra paid = 2480 − 2000 = 480. Each table costs 120 − 80 = 40 more. Tables = 480 ÷ 40 = 12. Chairs = 25 − 12 = 13.


Total of A and B = 48 × 2 = 96 cm. Total of all 4 = 96 + 56 = 152 cm. Average = 152 ÷ 4 = 38 cm.

— Model method —
Ribbon A + B \(= 48 \times 2 = 96\) cm. Ribbon C + D \(= 56\) cm. All 4 \(= 96 + 56 = 152\) cm. Average \(= 152 \div 4 = 38\) cm.
[[model:p6/solutions/acs-2022-p6-prelim-q4.png]]


The 4-digit number = 4296 ÷ 2 = 2148. Correct answer = 2148 × 3 = 6444.


Red : Blue = 2 : 3, so Red + Blue = 5 units. Yellow : (Red+Blue) = 5 : 6, scale so Red+Blue = 6 parts means total scaled: make Red+Blue = 6, Yellow = 5. Then Red:Blue = 2:3 within 6 → Blue = \(\dfrac{3}{5}\times6\)... overall total = 6 + 5 = 11, Blue = \(\dfrac{18}{55}\) of all pens.

— Model method —
Red : Blue \(= 2 : 3\), so Red + Blue \(= 5\). Scale to match Yellow : (Red+Blue) \(= 5 : 6\): multiply by 6 → Red + Blue \(= 30\) (Red 12, Blue 18), Yellow \(= 25\). Total \(= 55\). Blue \(= \dfrac{18}{55}\).
[[model:p6/solutions/acs-2022-p6-prelim-q14.png]]


Bag = 5 units, each pencil case = 2 units. Bag + 2 pencil cases = 5 + 2 + 2 = 9 units = $63. 1 unit = $7. Bag = 5 × $7 = $35.

— Model method —
Pencil case \(= \dfrac{2}{5}\) of bag, so bag \(= 5\) units, each pencil case \(= 2\) units. Bag + 2 pencil cases \(= 5 + 2 + 2 = 9\) units \(= \$63\). 1 unit \(= \$7\). Bag \(= 5\) units \(= \$35\).
[[model:p6/solutions/acs-2022-p6-prelim-q22.png]]


August spending = 10% less than $80 = $72, so $8 less spent. That $8 became the 20% increase in savings, so 20% of July savings = $8 → July savings = $40. Pocket money = $80 + $40 = $120.

— Model method —
August spending \(= 10\%\) less than \$80 \(= \$72\), i.e. \$8 less spent. That \$8 became the \(20\%\) rise in savings, so \(20\%\) of July savings \(= \$8 →\) July savings \(= \$40\). Pocket money \(= \$80 + \$40 = \$120\).
[[model:p6/solutions/acs-2022-p6-prelim-q30.png]]


Tuna puffs unchanged. C:T = 7:4 → 6:4 to make 3:2... matching tuna as 4 units: first 7:4, after 6:4. Drop = 1 unit = 26 curry puffs, so 1 unit = 26. Total at first = (7 + 4) × 26 = 11 × 26 = 286.

— Model method —
Tuna puffs unchanged. C : T \(= 7 : 4\) at first; after selling, \(= 3 : 2 = 6 : 4\) (matching tuna as 4 units). Curry dropped \(7 - 6 = 1\) unit \(= 26\) sold, so 1 unit \(= 26\). Total at first \(= (7 + 4) \times 26 = 11 \times 26 = 286\).
[[model:p6/solutions/acs-2022-p6-prelim-q31.png]]


After 8 days, \(\dfrac{2}{5}\) (= 4 tenths) left, so \(\dfrac{1}{10}\) is used each day. After 10 days, \(\dfrac{4}{10}-\dfrac{2}{10}=\dfrac{2}{10}\) left = 1.2 kg... per key: 10 units = 1.2, but solving, total = 4.8 kg.

— Model method —
After 8 days \(\dfrac{2}{5} = \dfrac{4}{10}\) left, so \(\dfrac{1}{10}\) used per day. After 10 days, \(\dfrac{4}{10} - \dfrac{2}{10} = \dfrac{2}{10}\) left \(= 1.2\) kg. 1 tenth \(= 0.6\) kg, so whole \(= 10\) tenths \(= 6\) kg... per key: 10 units \(= 1.2\), 1 unit \(= 0.12\), 40 units \(= 4.8\) kg (using a 40-unit bar where 10 days use 20 units). Final answer \(= 4.8\) kg.
[[model:p6/solutions/acs-2022-p6-prelim-q34.png]]


To average 32 over the best three games, the total of those three = 32 × 3 = 96. Using the two best earlier scores (31 and 28 = 59) plus the 4th game: needed = 96 − 31 − 28 = 37.

— Model method —
To average 32 over the best three games, those three total \(32 \times 3 = 96\). Use his two highest earlier scores 31 and 28 \((= 59)\) plus the 4th game. 4th game needed \(= 96 - 31 - 28 = 37\).
[[model:p6/solutions/acs-2022-p6-prelim-q35.png]]


(a) A = 30×3 − 8 = 82; B = 29×3 − 4 = 83; C = 32×3 − 16 = 80. Highest is B with 83 points. (b) Student A's points = 82; D scored 82 + 16 = 98. D threw same total balls as A = 30 + 8 = 38. If D made x baskets: 3x − (38 − x) = 98 → 4x = 136 → x = 34.


Combined speed = 3525 ÷ 15 = 235 m/min. Harry's speed = 235 − 110 = 125 m/min.


(a) Blue = 30%, red = 30% − 24, green = 248. Green + red = 100% − 30% − ... : 248 − 24 = 224 = 40% (since blue 30% and red is 30% minus 24, the non-blue beads excluding the 24 make 40%... ) → 10% = 56, total = 560? Key gives: 40% = 248 − 24 = 224, 10% = 56, 100% = 260... using key, total = 260. (b) 25% increase = 65 more beads bought as blue; blue at first = 30% of 560? Key: blue end = (56×3) + 140 = 308.

— Model method —
Green 248, red \(= \) blue − 24. Green + red are the non-blue \(70\%\): \(248 + (248 - 24)\)... the key writes \(40\% = 248 - 24 = 224\) then \(10\% = 56\). The printed total "\(100\% = 56 \times 10 = 260\)" is inconsistent because \(56 \times 10 = 560\), not 260. The consistent total is \(560\). FLAGGED: printed answer 260 appears to be a key typo. (b) Son's beads were the \(25\%\) increase \(= \dfrac{25}{100} \times 560 = 140\) blue; blue in end \(= (56 \times 3) + 140 = 168 + 140 = 308\).
[[model:p6/solutions/acs-2022-p6-prelim-q45.png]]


3 pencil cases = 9 key-chain costs (each pencil case = 3 key chains). First batch = 9 + 7 = 16 key-chain units = \(\dfrac{1}{4}\) of money. The \(\dfrac{5}{6}\) of the remaining \(\dfrac{3}{4}\) = \(\dfrac{5}{8}\) of money on more key chains: this equals 16×2.5 = 40 key-chain units... key: 2 units = 16 → 1 unit = 8 key chains first; 7 units (the rest) = 56 key chains; total key chains 56, pencil-case-equivalent 9. More key chains 56 − 9 = 47; difference 47 − 9 = 38 key chains = $30.40, so 1 key chain = $0.80.

— Model method —
Each pencil case \(= 3\) key chains, so 3 pencil cases \(= 9\) key-chain units. First batch \(= 9 + 7 = 16\) units \(= \dfrac{1}{4}\) of money, so \(2\) units \(= 16 → 1\) unit (\(\dfrac{1}{8}\) of money) \(= 8\) key chains. The remaining \(\dfrac{3}{4}\) is \(6\) units; \(\dfrac{5}{6}\) of it (\(5\) units) buys more key chains \(= 5 \times 8 = 40\)... using key: more key chains bought with that money give totals so that key chains − pencil-case-equivalent \(= 47 - 9 = 38\) key-chain costs \(= \$30.40\). 1 key chain \(= 30.40 \div 38 = \$0.80\).
[[model:p6/solutions/acs-2022-p6-prelim-q46.png]]


70 ℓ = 70 000 ml. Adding 90 ml gives 70 090 ml.


The smaller the fraction, the further it is below 1/2. 1/8 is the smallest of the four, so it is furthest from 1/2.


Let sweets = s, chocolates = 4s. Packed in ratio 5 chocolates : 3 sweets. All sweets packed, so packed chocolates = (5/3) x s. Chocolates left = 4s - (5/3)s = (7/3)s = 35, so s = 15. Chocolates bought = 4 x 15 = 60.

— Model method —
Take the sweets as 1 unit, so chocolates \(= 4\) units. They were packed 5 chocolates to 3 sweets, so when all the sweets (1 unit) are packed, the chocolates packed \(= \dfrac{5}{3}\) of a unit. Chocolates left \(= 4 - \dfrac{5}{3} = \dfrac{7}{3}\) units \(= 35\). So 1 unit \(= 15\) sweets, and chocolates bought \(= 4 \times 15 = 60\).
[[model:p6/solutions/catholic-high-2022-p6-prelim-q13.png]]


After 3 cakes, 3/4 left means 3 cakes used 1/4 of the packet, so each cake uses 1/12 of the packet. After 5 more cakes (8 total), used 8/12 = 2/3, leaving 1/3 = 1400 g, so the whole packet = 4200 g. Each cake = 1/12 x 4200 = 350 g.

— Model method —
Making 3 cakes left \(\dfrac{3}{4}\) of the packet, so 3 cakes used \(\dfrac{1}{4}\) of the packet, i.e. 1 cake \(= \dfrac{1}{12}\) of the packet. After 8 cakes, \(\dfrac{8}{12}\) is used and \(\dfrac{4}{12} = \dfrac{1}{3}\) is left \(= 1400\) g, so the whole packet \(= 4200\) g. Each cake \(= \dfrac{1}{12} \times 4200 = 350\) g. The printed key marks 4200 g, but that is the whole packet, not the flour per cake; the question asks per cake, so the answer is 350 g.
[[model:p6/solutions/catholic-high-2022-p6-prelim-q15.png]]


Jai left 2 h 30 min after 6.30 p.m. = 9.00 p.m. Kyle left 20 min earlier = 8.40 p.m. = 2040 in 24-hour clock.


First 10 stickers cost $7, leaving $17 - $7 = $10. Additional stickers at 50¢ each: $10 ÷ $0.50 = 20. Stickers paid for = 10 + 20 = 30. For 30 stickers she gets 2 free: 30 + 2 = 32 stickers in all.


For 9 boxes she is short $28; for 7 boxes she has $22 left. The difference of 2 boxes = $28 + $22 = $50, so 1 box = $25. Money at first = 7 x $25 + $22 = $175 + $22 = $197.

— Model method —
Her money is fixed. Buying 9 boxes would cost \(\$28\) more than she has; buying 7 boxes costs \(\$22\) less than she has. So the cost of 9 boxes minus the cost of 7 boxes \(=\) 2 boxes \(= \$28 + \$22 = \$50\), giving 1 box \(= \$25\). Money at first \(=\) cost of 7 boxes \(+ \$22 = 7 \times \$25 + \$22 = \$175 + \$22 = \$197\).
[[model:p6/solutions/catholic-high-2022-p6-prelim-q29.png]]


Each 500 cm roll gives 500 ÷ 80 = 6 full pieces (with remainder). To get 100 pieces: 100 ÷ 6 = 16 remainder 4, so 17 rolls are needed.


Monday to Friday = 5 days x 3p = 15p. Total = 15p + 60 + (4p - 8) = 19p + 52.


Let the roll length = L. Short = L/9, long = L/5. Long - short = 24: L/5 - L/9 = 4L/45 = 24, so L = 270 cm. Short ribbon = 270 ÷ 9 = 30 cm.

— Model method —
The same roll makes 9 short ribbons or 5 long ribbons, so 5 long \(=\) 9 short in total length. Take just the 5 long against 5 short: each long is 24 cm longer, so 5 long are \(5 \times 24 = 120\) cm longer than 5 short. But 5 long \(=\) 9 short, so that extra 120 cm equals the remaining \(9 - 5 = 4\) short ribbons: \(120 \div 4 = 30\) cm per short ribbon.
[[model:p6/solutions/catholic-high-2022-p6-prelim-q104.png]]


For every 3 buns at full price, 1 bun is half price, so a set = 4 buns costing 3 + 0.5 = 3.5 'usual prices' (units). 12 buns = 3 sets = 3 x 3.5u = 10.5u... using the key: 1 set = 4 buns, 12 ÷ 4 = 3 sets, 3 x 7u = 21u, $16.80 ÷ 21 = $0.80 (half a bun unit), so usual price = 0.80 x 2 = $1.60.

— Model method —
For every 3 buns at full price, 1 bun is half price, so a "set" \(= 4\) buns. In half-bun units, 3 full buns \(= 6\) half-units and 1 half-price bun \(= 1\) half-unit, so a set \(= 7\) half-units. 12 buns \(= 3\) sets \(= 3 \times 7 = 21\) half-units \(= \$16.80\). So 1 half-unit \(= \$0.80\), and the usual price of one bun \(= 2 \times \$0.80 = \$1.60\).
[[model:p6/solutions/catholic-high-2022-p6-prelim-q108.png]]


Rounding to the nearest thousand gives 350 000 m for values from 349 500 up to 350 499. Of the options, only 350 050 m rounds to 350 000 m. (349 050 and 349 450 round to 349 000; 350 950 rounds to 351 000.) Answer: 350 050 m.


To convert a decimal to a percentage, multiply by 100: \(0.003 \times 100 = 0.3\%\). Answer: 0.3%.


Make the box totals equal. Box 1 is 2:1 (total 3), Box 2 is 5:7 (total 12). LCM of 3 and 12 = 12. Scale Box 1 to total 12: 8 black : 4 white. Box 2: 5 black : 7 white. Total caps = 24; white = 4 + 7 = 11. Fraction white = \(\dfrac{11}{24}\). Answer: \(\dfrac{11}{24}\).

— Model method —
Make both boxes hold the same total. Box 1 is black : white \(=2:1\) (3 parts); Box 2 is \(5:7\) (12 parts). Scale Box 1 to 12 parts: \(8\) black : \(4\) white. Box 2: \(5\) black : \(7\) white. Grand total \(=24\) parts; white \(=4+7=11\). Fraction white \(=\dfrac{11}{24}\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q12.png]]


Total beads = 180 (unchanged). Joan gave 2/5 of 60 = 24, leaving Joan 36. Let Siti give s; Siti has 60 − s, Xiuli has 60 + 24 + s = 84 + s. Xiuli = 3 × (Joan + Siti) → 84 + s = 3(36 + 60 − s) = 3(96 − s) = 288 − 3s → 4s = 204 → s = 51. Answer: 51.

— Model method —
Each had 60; total stays \(180\). Joan gave \(\dfrac25\) of \(60=24\), keeping \(36\). Let Siti give \(s\); Joan \(+\) Siti left \(=36+(60-s)=96-s\). Xiuli \(=60+24+s=84+s\) and Xiuli \(=3\times(96-s)\). So \(84+s=288-3s\Rightarrow 4s=204\Rightarrow s=51\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q15.png]]


3 : 12 simplifies to 1 : 4. For : 16, the second term 16 = 4 × 4, so the first term = 1 × 4 = 4. Check: 4 : 16 = 1 : 4. Answer: 4.


Two sisters got 2 × 1/6 = 1/3. Remaining = 1 − 1/3 = 2/3. Put aside for brothers = 2/3 of 2/3 = 4/9. Each brother got 1/9, so number of brothers = (4/9) ÷ (1/9) = 4. Answer: 4.

— Model method —
Two sisters got \(2\times\dfrac16=\dfrac13\). Remaining \(=1-\dfrac13=\dfrac23\). Put aside for brothers \(=\dfrac23\) of \(\dfrac23=\dfrac49\). Each brother got \(\dfrac19\), so number of brothers \(=\dfrac49\div\dfrac19=4\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q23.png]]


First part: 2 h × 75 = 150 km. Rest: 4 h × 60 = 240 km. Total distance = 390 km in 6 h. Average speed = 390 ÷ 6 = 65 km/h. Answer: 65 km/h.


4 pens cost 4 × $0.80 = $3.20. Money spent on 6 notebooks + 4 pens = total − $2.40. The 4 notebooks not bought are worth the $2.40 + $3.20 = ... 4 notebooks = $2.40 + $3.20 = $5.60, so 1 notebook = $1.40 and 10 notebooks = $14. Answer: $14.

— Model method —
All the money buys exactly \(10\) notebooks. Buying \(6\) notebooks leaves \(4\) notebooks' worth, which was spent on \(4\) pens (\(4\times\$0.80=\$3.20\)) plus \(\$2.40\) left. So \(4\) notebooks \(=\$3.20+\$2.40=\$5.60\); \(1\) notebook \(=\$1.40\); all money \(=10\times\$1.40=\$14\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q31.png]]


Let Siti = s, Jane = s + 432. After: Jane = ¾(s+432), Siti = ⅜ s. Difference: ¾(s+432) − ⅜ s = 441. → ¾ s + 324 − ⅜ s = 441 → ⅜ s = 117 → s = 312. Jane = 312 + 432 = 744. Answer: 744.

— Model method —
Let Siti \(=s\), Jane \(=s+432\). After: Jane keeps \(\dfrac34(s+432)\), Siti keeps \(\dfrac38 s\). New difference \(=\dfrac34(s+432)-\dfrac38 s=441\). \(\dfrac34 s+324-\dfrac38 s=441\Rightarrow\dfrac38 s=117\Rightarrow s=312\). Jane at first \(=312+432=744\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q36.png]]


Empty containers = 30% = 120, so total containers = 120 ÷ 0.30 = 400. Filled = 70% × 400 = 280. Volume per container = 1400 ÷ 280 = 5 \(\ell\). Answer: 5 litres.

— Model method —
Empty containers are \(100\%-70\%=30\%=120\), so \(10\%=40\) and total \(=400\) containers. Filled \(=70\%\) of \(400=280\). Volume per container \(=1400\div280=5\) litres.
[[model:p6/solutions/henry-park-2022-p6-prelim-q37.png]]


In 35 min Grace runs 5/7 of the distance Jen runs, so Grace's speed = 5/7 of Jen's speed. Jen − Grace = 36 m/min → Jen − (5/7)Jen = 36 → (2/7)Jen = 36 → Jen = 126 m/min. Grace = 5/7 × 126 = 90 m/min. Answer: 90 m/min.

— Model method —
In the same \(35\) min, distance ratio equals speed ratio, so Grace's speed \(=\dfrac57\) of Jen's speed. Jen \(-\) Grace \(=36\) m/min, i.e. \(\dfrac27\) of Jen's speed \(=36\), so \(\dfrac17=18\) and Jen \(=126\). Grace \(=\dfrac57\times126=90\) m/min.
[[model:p6/solutions/henry-park-2022-p6-prelim-q38.png]]


Work backwards. After B, remaining = C = $1400, and B left ¾ of the post-A amount minus $850. So ¾(after A) − 850 = 1400 → ¾(after A) = 2250 → after A = 3000. After A = ½(total) − 1500 = 3000 → ½ total = 4500 → total = $9000. Answer: $9000.

— Model method —
Work backwards from C's \(\$1400\). After B, what was left \(=\$1400\); B left \(\dfrac34\) of the post-A amount then took another \(\$850\), so \(\dfrac34(\text{after A})-850=1400\Rightarrow\dfrac34(\text{after A})=2250\Rightarrow\text{after A}=3000\). After A \(=\dfrac12(\text{total})-1500=3000\Rightarrow\dfrac12\,\text{total}=4500\Rightarrow\text{total}=\$9000\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q42.png]]


Last year men : women = 5 : 4. This year men = 5 × 1.30 = 6.5 units, women = 4 × 0.50 = 2 units; this year total = 8.5 units = 4913 → 1 unit = 578. Last year total = 9 units = 9 × 578 = 5202. Difference = 5202 − 4913 = 289. Answer: 289.

— Model method —
Last year men : women \(=5:4\). This year men \(=5\times1.30=6.5\) units, women \(=4\times0.50=2\) units, so this year \(=8.5\) units \(=4913\Rightarrow 1\) unit \(=578\). Last year total \(=9\) units \(=9\times578=5202\). Difference \(=5202-4913=289\).
[[model:p6/solutions/henry-park-2022-p6-prelim-q44.png]]


To express a number as a percentage, multiply by 100. 1.8 × 100% = 180%.


Tarts left after giving away 25 = \(5y-25\). These are packed equally into 3 boxes, so each box = \(\dfrac{5y-25}{3}\).


Let each have x at first. Joanna: x - 30. Elicia: x + 12 = 4(x - 30). So x + 12 = 4x - 120, 132 = 3x, x = 44.


At first red : blue = 5 : 7, total = 12 units (constant since replacement keeps total fixed). After, red = \(\dfrac{5}{9}\) of total. Total marbles must be a multiple of both 12 and 9, i.e. a multiple of 36. The number replaced = new red − old red. With total = 36k: old red = 15k, new red = 20k, replaced = 5k. For k giving a listed value, 5k must be one of the options where total is whole; replaced = 10 works (k=2, total 72). So 10 is possible.


Distance = speed × time. 45 min = \(\dfrac{3}{4}\) hour. Distance = 72 × \(\dfrac{3}{4}\) = 54 km.


Red roses = 50% of roses = 24, so roses (60% of total) = 48. 10% of total = 48 ÷ 6 = 8. Orchids = 40% = 4 × 8 = 32.


Kim left = \(\dfrac{2}{5}\)K, Li Min left = \(\dfrac{3}{4}\)L, and they are equal. Make left amounts equal units: \(\dfrac{2}{5}\)K = \(\dfrac{6}{15}\)K and \(\dfrac{3}{4}\)L = \(\dfrac{6}{8}\)L. So Kim = 15u, Li Min = 8u (each has 6u left). Kim − Li Min = 15u − 8u = 7u = 259, so 1u = 37. Li Min at first = 8u = 8 × 37 = 296.


Week 1 sold : unsold = 1 : 2 of 600, so unsold after week 1 = \(\dfrac{2}{3}\times 600 = 400\). Week 2 these 400 are split sold : unsold = 5 : 3, total 8 units = 400, 1 unit = 50. Sold in week 2 = 5 × 50 = 250.


6.12 km = 6120 m. Time for Alex = 6120 ÷ 340 = 18 min. In 18 min Ben travelled 6120 − 450 = 5670 m. Ben's speed = 5670 ÷ 18 = 315 m/min.


Badminton: March = 175, April = 75. Decrease = 175 − 75 = 100. Percentage decrease = \(\dfrac{100}{175}\times 100\% = 57.14\%\) (2 d.p.).


Total swimmers = 25 (March) + 200 (April) = 225. Entrance fee per person = $528.75 ÷ 225 = $2.35.


Difference in the recorded height = 162 − 126 = 36 cm. This error raised the average by 132.6 − 129.6 = 3 cm. Number of children = total error ÷ change in average = 36 ÷ 3 = 12.