Printable · GCSE Foundation · ages 14-16
Direct and inverse proportion worksheet — GCSE Foundation
Fifteen questions on "direct and inverse proportion" — DfE statement R10. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Direct and inverse proportion worksheet — GCSE Foundation
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- (b) £10 — Method: find the cost of 1 kg by dividing, then multiply by the mass wanted — the unitary method for direct proportion. Working: £6 ÷ 3 = £2 per kg, and £2 × 5 = £10. Answer: £10. The distractors: £11 comes from adding 5 to the £6 instead of scaling; £30 comes from multiplying £6 by 5 without first dividing by 3; £3.60 comes from turning the proportion upside down, dividing by 5 and multiplying by 3.
- (c) 3 hours — Method: in inverse proportion the product of the two quantities is constant, and here that product is the distance. Working: 60 × 4 = 240 km, so at 80 km/h the time is 240 ÷ 80 = 3. Answer: 3 hours. The distractors: 5 hours 20 minutes comes from treating the relationship as direct, working out 4 × 80 ÷ 60; 2 hours 40 minutes comes from cutting the time by the fraction the speed rose by — the speed went up by one third, so the time was cut by one third — which is not how inverse proportion works; 4 hours comes from dividing the 240 km by the original speed of 60 km/h again instead of by the new speed.
- (b) €230.00 — Multiply the amount in pounds by the exchange rate: 200 × 1.15 = 230, so £200 = €230.00. Working out 200 + 1.15 = 201.15 treats the exchange rate as an amount to add rather than a multiplier. Working out 200 × 0.15 = 30 finds only the extra amount earned for every pound and forgets to add it back to the original £200. Working out 200 × 11.5 = 2300.00 misplaces the decimal point in the exchange rate, multiplying by 11.5 instead of 1.15. £200 converts to €230.00.
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (b) 12 — Method: work out what one part of the ratio is worth, then multiply that by the number of parts in the other share. Working: the smaller number matches the 2 parts, so 8 ÷ 2 = 4 for one part, and 3 × 4 = 12. Answer: 12. The distractors: 16 comes from multiplying 8 by 2, the ratio part that belongs to the smaller number; 24 comes from multiplying 8 by 3 without first finding the value of one part; 9 comes from adding the difference between the ratio parts, 3 − 2 = 1, to 8.
- (c) 14 litres — Method: find the amount of fuel used per km first, then use it to find the fuel needed for 175 km. Working: 24 ÷ 300 = 0.08 litres per km, and 0.08 × 175 = 14 litres. So 14 litres are needed. Distractor 24 litres comes from assuming the same amount of fuel is used no matter the distance, without scaling. Distractor 21 litres comes from misreading the original distance as 200 km instead of 300 km. Distractor 1.4 litres comes from a decimal-point slip, giving an answer ten times too small.
- (a) £54 — Method: for direct proportion, wage = rate × hours. Working: £9 × 6 = £54. Wrong options: £15 comes from adding the rate and the hours instead of multiplying (£9 + 6); £63 comes from using 7 hours instead of 6; £1.50 comes from dividing the rate by the hours instead of multiplying (£9 ÷ 6).
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (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) 54 — Method: y = kx, so k = y ÷ x. Working: k = 18 ÷ 5 = 3.6. At x = 15: y = 3.6 × 15 = 54. Wrong options: 28 comes from adding the change in x (10) onto y instead of scaling; 6 comes from treating the relationship as inverse proportion (k = 5 × 18 = 90, then y = 90 ÷ 15 = 6); 60 comes from rounding the constant up to 4 instead of using 3.6.
- (a) £2.80 — Method: find the cost of one pencil first, then use it to find the cost of 7 pencils. Working: £1.20 ÷ 3 = £0.40 for one pencil, and £0.40 × 7 = £2.80. So 7 pencils cost £2.80. Distractor £8.40 comes from multiplying £1.20 by 7 without first dividing by 3 to find the cost of one pencil. Distractor £1.20 comes from assuming the price stays the same no matter how many pencils are bought. Distractor £0.40 is the cost of one pencil, found correctly but never multiplied by 7.
- (c) 15 — Method: the number of cakes is in direct proportion to the mass of flour, so find the multiplier between the two masses and apply it to the number of cakes. Working: 6 ÷ 2 = 3, so there is three times as much flour, and 5 × 3 = 15. Answer: 15. The distractors: 10 comes from multiplying the 5 cakes by 2, the mass in the recipe, instead of by the multiplier 3; 20 comes from multiplying by the difference 6 − 2 = 4, treating a proportion problem as a difference problem; 12 comes from rounding 5 ÷ 2 down to 2 cakes per kilogram and working out 6 × 2.
- (a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
- (a) £40.32 — Find the cost per square metre from the rate given: £14.40 ÷ 20 = £0.72 per m². Then multiply by the area to be covered: £0.72 × 56 = £40.32. Working out 14.40 × 20 ÷ 56 ≈ £5.14 uses the ratio the wrong way round, scaling down as if 56 m² needed less paint than 20 m². Stopping at £0.72 only gives the cost per square metre, not the cost for the whole wall. Working out 14.40 + (56 − 20) = £50.40 adds the extra square metres straight onto the cost in pounds, treating square metres and pounds as the same kind of quantity. Covering 56 m² costs £40.32.
- (c) £4.00 — Find the cost of one pen: £6.40 ÷ 8 = £0.80. Then multiply by 5 pens: £0.80 × 5 = £4.00. Dividing £6.40 by 5 and multiplying by 8 gives £10.24 — that uses the ratio the wrong way round, scaling as if 5 pens were more expensive than 8. Stopping at £0.80 only gives the price of one pen. Multiplying the price of one pen by the difference in the number of pens, (8 − 5) × £0.80, gives £2.40 — the cost of the pens NOT bought, not the cost of the 5 pens bought. 5 pens cost £4.00.
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