Printable · GCSE Foundation · ages 14-16

Proportion as equality of ratios worksheet — GCSE Foundation

Fifteen questions on "proportion as equality of ratios" — DfE statement R7. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.

VersionsA only A + B A + B + Csame questions, different order

AnswersQuestions only With answer key Key only

CalculatorMixed Non-calculator (Paper 1)Calculator (Papers 2 and 3)

New sheet

Answer key: Proportion as equality of ratios worksheet — GCSE Foundation

Maths · GCSE Foundation · R7 · Proportion as equality of ratios · 15 questions

MathsUKwww.geekhero.co.uk

(d) 20 litres — The ratio of concentrate to water is 2 : 5, so water = concentrate × 5 ÷ 2. 8 × 5 ÷ 2 = 20, so Priya needs 20 litres of water. Giving 40 litres multiplies by 5 but forgets to divide by 2 (8 × 5 = 40). Giving 3.2 litres uses the ratio inverted, multiplying by 2 ÷ 5 instead of 5 ÷ 2 (8 × 2 ÷ 5 = 3.2). Giving 11 litres uses additive reasoning instead of multiplicative: it adds the difference between the ratio parts, 5 − 2 = 3, onto the amount of concentrate (8 + 3 = 11), but ratios scale by multiplying, not by adding a fixed amount.
(b) 12 cm — Method: height:width = 3:2 means both lengths are built from parts of the same size, so write 3:2 = 18:x, find one part and multiply by the number of parts in the width. Working: the height is 3 parts and measures 18 cm, so one part is 18 ÷ 3 = 6 cm; the width is 2 parts, so 2 × 6 = 12 cm. Answer: 12 cm. The distractors: 27 cm comes from using the ratio the wrong way round, 18 ÷ 2 × 3, which makes the width longer than the height even though 2 is the smaller part; 6 cm is the value of one part, given as the width instead of being doubled; 17 cm comes from treating the ratio as a difference — 3 − 2 = 1, so 1 cm is taken off the height — but a ratio scales the lengths, it does not subtract from them.
(d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
(c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
(c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
(c) £21 — Method: the two amounts are in the same ratio as 5:3, so write 5:3 = 35:x, find the value of one part and then take the number of parts asked for. Working: Harry's £35 is 5 parts, so one part is £35 ÷ 5 = £7; Isla has 3 parts, so 3 × £7 = £21. Answer: £21. The distractors: £7 is the value of a single part, given as Isla's share instead of being multiplied by the 3 parts she has; £28 comes from finding one part correctly and then working out £35 − £7, subtracting one part from Harry's money instead of taking three parts; £33 comes from treating the ratio additively — the parts differ by 5 − 3 = 2, so £2 is taken off Harry's £35, but a ratio compares by multiplying, not by subtracting.
(b) 1.5 km — Multiply the map length by the scale factor: 6 × 25000 = 150000 cm. Convert to kilometres, using 100 cm = 1 m and 1000 m = 1 km, so 100000 cm = 1 km: 150000 ÷ 100000 = 1.5 km. (1500 km comes from converting only as far as metres, 150000 ÷ 100 = 1500 m, and then writing kilometres on the end. 15 km comes from dividing by 10000 instead of 100000. 0.15 km comes from dividing by 1000000 instead of 100000.)
(a) 42 — Find the value of one part: 18 ÷ 3 = 6. Find the number of dogs: 4 × 6 = 24. Add the cats and the dogs to find the total: 18 + 24 = 42. (24 is the number of dogs only, not the total number of animals. 126 comes from multiplying the number of cats by the total number of parts, 18 × 7, instead of finding one part first. 25 comes from adding the ratio numbers 3 and 4 directly to the number of cats.)
(a) 135 km — Method: find the distance travelled on one litre, then scale up to 9 litres. Working: 90 ÷ 6 = 15 km per litre, so 15 × 9 = 135 km. Answer: 135 km. 45 km comes from working out the extra distance for the extra 3 litres (15 × 3) but forgetting to add the original 90 km. 60 km comes from using the ratio the wrong way round, 90 × 6 ÷ 9, instead of finding the rate per litre first. 99 km comes from simply adding the number of litres, 9, onto the original distance, 90, instead of scaling the whole journey.
(c) Yes — 700 ÷ 7 = 100 ml for the 1 part of concentrate — Method: add the ratio parts to find the total number of parts, divide the total volume by this, then use the ratio to find concentrate's share. Working: 1 + 6 = 7 parts. 700 ÷ 7 = 100 ml per part. Concentrate = 1 part = 100 ml, so Freya is correct. Wrong options: 'divide 700 by 6' uses only one of the ratio numbers instead of the total of 7 parts, giving about 117 ml; '600 ml is concentrate' swaps which ratio number belongs to the concentrate and which belongs to the water; 'half of 700 ml should be concentrate' ignores the ratio altogether and assumes an equal split.
(b) 250 g — Method: adding water changes the total mass but not the mass of salt, so find the salt, hold it fixed, use the new ratio to find the new total mass and subtract the mass already in the beaker. Working: 12:100 = x:500 gives 12 ÷ 100 × 500 = 60 g of salt; that 60 g must be 8% of the new mixture, so 8:100 = 60:y gives y = 60 ÷ 8 × 100 = 750 g; the water added is 750 − 500 = 250 g. Answer: 250 g. The distractors: 750 g is the mass of the diluted solution, given without taking away the 500 g that was in the beaker to start with; 60 g is the mass of salt, the quantity that stays the same, given instead of the mass of water; 20 g comes from treating the fall from 12% to 8% as 4% of the original 500 g, which measures a change in concentration as though it were a mass of water.
(c) 9% — Method: a percentage concentration is the ratio of salt to solution written per 100 g, so scale each concentration to the mass it belongs to, add the two masses of salt, then scale the ratio of salt to mixture back to a denominator of 100. Working: 5:100 = x:400 gives 5 ÷ 100 × 400 = 20 g of salt, and 25:100 = y:100 gives 25 g of salt; the mixture holds 20 + 25 = 45 g of salt in 400 + 100 = 500 g of solution; 45:500 = 9:100. Answer: 9%. The distractors: 15% is the mean of 5% and 25%, which would only be right if the two masses were equal, and here one is four times the other; 21% comes from attaching the concentrations to the wrong masses, working out (400 × 25% + 100 × 5%) ÷ 500; 0.9% comes from working out 45 ÷ 500 = 0.09 and then moving the decimal point one place instead of two when writing the decimal as a percentage.
(d) 75 pages — Method: find the number of pages printed in one minute, then scale up to 10 minutes. Working: 45 ÷ 6 = 7.5 pages per minute, so 7.5 × 10 = 75 pages. Answer: 75 pages. 27 pages comes from using the ratio upside down, 45 × 6 ÷ 10, instead of finding the rate per minute first. 55 pages comes from simply adding the extra minutes, 10, onto the original number of pages, 45. 70 pages comes from rounding the rate down to 7 pages per minute before multiplying by 10, instead of using the exact rate of 7.5.
(c) 6.00 m — Method: the ratio of height to shadow length is the same for both objects. Working: road sign height ÷ shadow = 3 ÷ 5 = 0.6. Lamppost height = 0.6 × 10 = 6.00 m. Wrong options: 16.67 m comes from inverting the ratio, using shadow ÷ height instead of height ÷ shadow (10 × 5 ÷ 3); 8.00 m comes from adding the difference between the two shadow lengths to the road sign's height instead of scaling (3 + (10 − 5)); 1.50 m comes from multiplying by the ratio of the two shadow lengths the wrong way round (3 × 5 ÷ 10).
(a) 0.6 litres — The ratio 1 : 9 means the solution has 1 + 9 = 10 equal parts in total. Each part is 6 ÷ 10 = 0.6 litres, and disinfectant is 1 part, so Priti needs 0.6 litres of disinfectant. Giving 0.667 litres divides by 9, the number of parts of water, instead of the total number of parts, 10 (6 ÷ 9 ≈ 0.667). Giving 6 litres is the total amount of solution, not just the disinfectant's share of it. Giving 5.4 litres works out the water's share (6 × 9 ÷ 10 = 5.4), not the disinfectant's.

Build your own mix at the worksheet builder.