Printable · GCSE Foundation · ages 14-16

Dividing in a ratio worksheet — GCSE Foundation

Fifteen questions on "dividing in a ratio" — DfE statement R5. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

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Answer key: Dividing in a ratio worksheet — GCSE Foundation

Maths · GCSE Foundation · R5 · Dividing in a ratio · 15 questions

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(c) £90 — Method: split the total amount into the number of parts shown by the ratio, then find Chloe's share. Working: the ratio 2:3 has 2 + 3 = 5 parts, so one part is £150 ÷ 5 = £30, and Chloe's share is 3 × £30 = £90. So Chloe receives £90. Distractor £60 is Ben's share, not Chloe's. Distractor £75 comes from splitting the money into two equal halves, ignoring the ratio. Distractor £30 is the value of one part, found correctly but never multiplied by 3.
(c) 40° — Method: the three angles of a triangle always add up to 180°, so split 180° into the number of parts shown by the ratio. Working: the ratio 2:3:4 has 2 + 3 + 4 = 9 parts, so one part is 180 ÷ 9 = 20°, and the smallest angle is 2 × 20 = 40°. So the smallest angle is 40°. Distractor 60° is the middle-sized angle, 3 parts. Distractor 80° is the largest angle, 4 parts. Distractor 20° is the value of one part, found correctly but never multiplied by 2.
(c) 36 — Add the parts of the ratio: 3 + 4 = 7. Divide the total by the number of parts: 84 ÷ 7 = 12, so one part is worth 12. The smaller share has 3 parts: 3 × 12 = 36. (48 is the larger share, using 4 parts instead of 3. 42 comes from splitting 84 in half, treating the ratio as if it were 1:1. 28 comes from dividing 84 by 3 — one of the ratio numbers — instead of dividing by the total number of parts, 7.)
(c) 12 m — Method: the perimeter of a rectangle is twice the sum of the length and the width, so half the perimeter is one length plus one width; split that half using the ratio. Working: half of 64 is 64 ÷ 2 = 32 m, the ratio 5:3 has 5 + 3 = 8 parts, so one part is 32 ÷ 8 = 4 m, and the width is 3 × 4 = 12 m. So the width is 12 m. Distractor 20 m is the length, 5 parts, not the width. Distractor 24 m comes from splitting the whole perimeter, 64 m, into 8 parts and multiplying by 3, forgetting to halve the perimeter first. Distractor 8 m comes from the same slip stopped one step earlier: splitting the whole perimeter into 8 parts, 64 ÷ 8 = 8, and giving that instead of the width.
(b) £20 — Method: add the parts of the ratio, divide the amount by the number of parts to find the value of one part, then take the parts belonging to the person asked about. Working: 1 + 2 = 3 parts, £60 ÷ 3 = £20 for one part, and Isla has 1 part. Answer: £20. The distractors: £40 is the 2-part share, which belongs to her brother; £30 comes from reading 1:2 as one part out of two and halving the £60; £120 comes from multiplying the £60 by 2 instead of dividing it into 3 parts.
(d) 21 — Add the parts: 5 + 3 = 8. Divide the total by the number of parts: 56 ÷ 8 = 7, so one part is worth 7 counters. Blue has 3 parts: 3 × 7 = 21. (35 is the number of red counters, using 5 parts instead of 3. 28 comes from splitting 56 counters in half instead of in the ratio 5:3. 7 is the value of one part — the number of blue counters is 3 lots of this, not just one.)
(c) 450 g — Method: use the amount of butter given to find the value of one part of the ratio, then find the mass of flour, and finally add flour and butter to get the total. Working: 180 g of butter is 2 parts, so one part is 180 ÷ 2 = 90 g. The flour is 3 parts, so 3 × 90 = 270 g, and the total mass is 270 + 180 = 450 g. So the baker can make 450 g of pastry. Distractor 270 g is only the mass of flour, forgetting to add the butter back on. Distractor 300 g comes from treating the 180 g as 3 parts instead of 2, swapping which ratio number matches the butter. Distractor 540 g comes from multiplying 180 by 3 directly instead of first finding the value of one part.
(b) 24 — Method: let one part of the ratio be worth x, write both groups in terms of x, and use the fact that the two groups end up equal. Working: the boys are 3x and the girls are 4x; after the 6 boys join, 3x + 6 = 4x, so x = 6; the girls are 4 parts, so 4 × 6 = 24. Answer: 24 girls. The distractors: 18 is the number of boys before the 6 join, which is 3 × 6; 30 comes from adding the 6 new members to the girls as well as to the boys; 42 is the total number of members in the choir before the 6 boys join, the 18 boys and the girls together.
(b) 5 litres — Method: adding water changes the total volume but adds no salt, so work out the volume of salt, then the total volume that makes that salt 20% of the mixture, then the extra water. Working: 30% of 10 litres is 0.3 × 10 = 3 litres of salt. For the same 3 litres to be 20% of the new mixture, the new total volume is 3 ÷ 0.2 = 15 litres. The water added is the extra volume, 15 − 10 = 5 litres. Answer: 5 litres. The distractors: 3 litres is the volume of salt in the solution, which is the first step and not what the question asks for; 15 litres is the total volume of the new mixture, which counts the 10 litres already in the container as water that was poured in; 2 litres comes from taking 20% of the original 10 litres, applying the new percentage to the old volume instead of to the new one.
(b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
(c) £3,200 — Method: find the value of one part of the ratio from the first investor's amount, then work out the second investor's share before adding both together. Working: £1,200 is 3 parts, so one part is £1,200 ÷ 3 = £400. The second investor's share is 5 × £400 = £2,000, and the total is £1,200 + £2,000 = £3,200. So the total invested is £3,200. Distractor £2,000 is only the second investor's share, without adding the first investor's £1,200. Distractor £2,400 comes from doubling the first investor's amount instead of using the ratio. Distractor £6,000 comes from multiplying £1,200 by 5 directly instead of first finding the value of one part.
(a) £20,000 — Method: add the parts of the ratio, divide the profit by the number of parts to find the value of one part, then multiply by the parts belonging to the partner asked about. Working: 3 + 5 + 4 = 12 parts, £48,000 ÷ 12 = £4,000 for one part, and that partner has 5 parts, so 5 × £4,000 = £20,000. Answer: £20,000. The distractors: £4,000 is the value of one part; £12,000 is the 3-part share; £16,000 is the 4-part share, which is what a candidate reaches by assuming the biggest share must go with the number written last in the ratio.
(c) 40 cm — Method: split the total length into the number of parts shown by the ratio, then find the value of the shorter share. Working: the ratio 4:5 has 4 + 5 = 9 parts, so one part is 90 ÷ 9 = 10 cm, and the shorter piece is 4 × 10 = 40 cm. So the shorter piece is 40 cm. Distractor 50 cm is the length of the LONGER piece, not the shorter one. Distractor 45 cm comes from splitting the ribbon into two equal halves, ignoring the ratio. Distractor 10 cm is the value of one part, found correctly but never multiplied by 4.
(d) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
(a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)

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