Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
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- (a) £9.00 — Rate of pay = total pay ÷ number of hours. £54 ÷ 6 = £9.00 per hour. £324.00 comes from multiplying the pay by the hours instead of dividing (£54 × 6). £0.11 comes from dividing the hours by the pay instead of the pay by the hours (6 ÷ 54). £48.00 comes from subtracting the hours from the pay (£54 − 6) instead of dividing.
- (a) 10 — Method: two equal fractions can be rearranged by cross-multiplying, multiplying each numerator by the other denominator. Working: 4 × 5 = 2 × x, so 2x = 20 and x = 20 ÷ 2 = 10. Answer: 10. The distractors: 20 comes from cross-multiplying to 4 × 5 = 20 and stopping there, without dividing by the 2; 8 comes from multiplying the two numerators, 4 × 2; 2.5 comes from working only with the right-hand fraction, 5 ÷ 2, and ignoring the 4.
- (a) 4 days — Method: the fence is a fixed amount of work, so builders × days is constant; find that product and divide it by the new number of builders. Working: 3 × 12 = 36 builder-days of work, so with 9 builders the time is 36 ÷ 9 = 4 days. Answer: 4 days. The distractors: 6 days comes from halving the 12 days because there are more builders, rather than dividing by the factor of 3 by which the workforce has grown; 36 days is the constant product of builders and days, given as a number of days instead of being shared between the builders; 9 days comes from taking 3 days off the 12, treating three extra builders as three fewer days, which is additive rather than proportional.
- (b) 48 m² — Method: lengths are multiplied by the scale factor, but areas are multiplied by its square. Working: 1 cm on the plan stands for 200 cm = 2 m, so 1 cm² on the plan stands for 2 × 2 = 4 m², and 12 × 4 = 48. Answer: 48 m². The distractors: 24 m² comes from scaling the area by the length factor 2 instead of by its square; 2400 m² comes from multiplying the area by the scale 200 as though it were a length; 4800 m² comes from working in centimetres, 12 × 200² = 480 000 cm², and then dividing by 100 instead of by 10 000 to reach square metres.
- (b) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
- (b) 4/5 — Write the mass of the beans over the mass of the soup: 400/500. Divide the top and bottom by 100 to get 4/5. Choosing 5/4 comes from writing the soup's mass over the beans' mass, the wrong way round. Choosing 1/5 comes from finding the difference in mass (500 − 400 = 100) and writing that over the mass of the soup, instead of using the mass of the beans. Choosing 5/9 comes from writing the mass of the soup over the total mass of both tins (500 out of 900), instead of over the mass of the beans.
- (d) 1:12 — Write the ratio model : real bus using the numbers in the question: 5 : 60. Divide both parts by their highest common factor, 5: 5÷5 = 1, 60÷5 = 12, giving 1 : 12. (12:1 comes from writing the ratio the wrong way round, real bus : model. 1:55 comes from dividing the first part correctly, 5÷5 = 1, but subtracting on the second part, 60 − 5 = 55, instead of dividing it by 5 as well. 5:12 comes from dividing only the real length by 5 and leaving the model length unchanged — both parts of a ratio must be divided by the same number.)
- (a) £102 — Method: find the value of one part of the ratio, then work out each group's share before comparing them. Working: the ratio 5:2 has 5 + 2 = 7 parts, so one part is £238 ÷ 7 = £34. Adults donate 5 × £34 = £170 and children donate 2 × £34 = £68, so adults donate £170 − £68 = £102 more than children. So the difference is £102. Distractor £68 is only the children's donation, without finding the difference. Distractor £170 is only the adults' donation, without finding the difference. Distractor £136 comes from doubling the children's donation instead of subtracting it from the adults' donation.
- (d) 5 — Method: rearrange y = kx to make the constant the subject, then substitute the pair of values given. Working: k = y ÷ x, so k = 10 ÷ 2 = 5. Answer: 5. The distractors: 20 comes from multiplying 10 by 2 instead of dividing, which is the rearrangement done the wrong way round; 12 comes from adding the pair, 10 + 2, treating the relationship as y = x + k; 8 comes from working out 10 − 2, the same additive reading with the operation reversed.
- (c) 25 — Method: divide the larger number by its ratio part to find the value of one part, then multiply by the smaller number's ratio part. Working: 40 ÷ 8 = 5 (value of one part). Smaller number = 5 × 5 = 25. Wrong options: 64 comes from dividing by the smaller ratio part instead of the larger (40 ÷ 5 × 8); 45 comes from adding the value of one part onto 40 instead of scaling down (40 + 5); 35 comes from subtracting the value of one part from 40 (40 − 5) instead of multiplying it by the smaller ratio part.
- (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) 3/10 — Work out the empty space: 500 − 350 = 150 litres. Form the fraction 150/500; both numbers share a factor of 50, so 150 ÷ 50 = 3 and 500 ÷ 50 = 10, giving 3/10. 7/10 comes from writing the fraction of the tank that is full (350/500), instead of the empty space. 1/2 comes from miscalculating 500 − 350 as 250 instead of 150. 3/7 comes from comparing the empty space with the water held (150/350), instead of with the tank's total capacity.
- (d) d ÷ t — Average speed = distance ÷ time, so the expression is d ÷ t. Writing t ÷ d inverts the formula, giving the time per kilometre instead of the speed. Writing d × t confuses speed with the formula for distance travelled (distance = speed × time) used the wrong way round. Writing d + t treats the relationship as additive instead of using division.
- (d) 20 — Method: equivalent ratios are linked by a single multiplier, so find it from the part you know and apply it to the other part. Working: 15 ÷ 3 = 5, so the multiplier is 5, and 4 × 5 = 20. Answer: 20. The distractors: 16 comes from adding the difference between the ratio parts, 4 − 3 = 1, to 15, treating the ratio as a difference; 60 comes from multiplying 15 by 4 without first dividing by 3; 11.25 comes from using the ratio the wrong way round, working out 15 × 3 ÷ 4.
- (c) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
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