Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.A process is modelled by the recurrence P_{n+1} = 0.6P_n + 40. As n increases, P_n approaches a long-run value L, which satisfies L = 0.6L + 40. Solve this equation to find L.
- 2.The length of ribbon A is 5/12 of the length of ribbon B. Write the length of ribbon B as a fraction of the length of ribbon A.
- 3.The number of members of a running club increases from 45 to 54. Work out the percentage increase.
- 4.Two quantities x and y are inversely proportional. When x = 2, the value of y is 15. Work out the value of y when x = 5.
- 5.Two cars travel at constant speeds. Car A travels 150 km in 3 hours. Car B travels 180 km in 4 hours. Which car is faster, and what is its speed?
- 6.A conversion graph between kilometres and miles passes through the origin and the point (50, 31), with the number of kilometres on the horizontal axis and the number of miles on the vertical axis. Work out the number of miles equivalent to 1 kilometre.
- 7.A tank is filled at a rate of 3 litres every 2 minutes. Given that 1 litre = 1000 cm³, work out the flow rate in cm³ per minute.
- 8.A delivery driver travels 45 km in the first 30 minutes of a journey, and then a further 75 km in the next 1 hour. Work out her average speed for the whole journey, in km/h.
- 9.A shop sells identical pens. The total cost, C pence, of n pens is given by C = 12n. Write the ratio n : C in its simplest form.
- 10.The price of a cycling helmet rises from £80 to £116. Work out the percentage increase.
- 11.The height of a candle, in cm, is measured as it burns: t = 0 min, height = 20.0; t = 3 min, height = 18.7; t = 5 min, height = 17.5; t = 6 min, height = 16.6; t = 7 min, height = 15.4; t = 9 min, height = 13.4. Which pair of readings gives the best estimate of the instantaneous rate of change of the height at t = 6 minutes, and why?
- 12.It takes 2 identical pumps 10 hours to empty a flooded basement. Working at the same rate, work out how many hours 5 of these pumps would take to empty the same basement.
- 13.In one batch of shortbread, a bakery uses 240 g of butter and 160 g of sugar. Another batch, made in the same ratio, uses 90 g of sugar. Work out how much butter is needed for the second batch.
- 14.A jumper costs £45 at Shop A, where it is reduced by 20%. The same jumper costs £34 at Shop B, where a further 10% reduction is then applied. Work out the difference between the two reduced prices.
- 15.Write 400 g : 1.5 kg as a ratio in its simplest form.
Answer key
- (d) 100 — Rearrange L = 0.6L + 40 by collecting the L terms on one side: L − 0.6L = 40, which gives 0.4L = 40, then L = 40 ÷ 0.4 = 100. Subtracting the other way round, 0.6L − L = 40, gives −0.4L = 40, then L = 40 ÷ (−0.4) = −100 — a sign error that flips the answer negative even though a long-run value here must be positive. Ignoring the 0.6L term completely and solving L = 40 directly gives 40, which throws away the recurrence's own multiplier. Dividing 40 by 0.6 instead of by the correct coefficient 0.4 gives 40 ÷ 0.6 ≈ 66.7, a slip that comes from dividing by the coefficient of L on the RIGHT of the original equation rather than by what is left once the L terms are collected on one side. Always collect the L terms first, then divide by whatever coefficient of L remains.
- (a) 12/5 — If A is 5/12 of B, then B is the reciprocal of that fraction times A: flip 5/12 to get 12/5, so B is 12/5 of A. 5/12 comes from keeping the same fraction without flipping it, treating the relationship as if it works the same way in both directions. 7/12 comes from computing 1 − 5/12 = 7/12, which is not how a fraction reverses. 12/7 comes from subtracting 5 from 12 to get 7, and writing 12 over that, instead of swapping the numerator and denominator of 5/12.
- (c) 20% — Method: percentage increase = (increase ÷ original) × 100. Working: the increase is 54 − 45 = 9, and 9 ÷ 45 = 0.2, so the percentage increase is 0.2 × 100 = 20. Answer: 20%. The distractors: 9% comes from writing the actual increase as a percentage; 16.7% comes from dividing by the new value 54 instead of the original 45; 120% is the multiplier 1.2 written as a change rather than the change itself.
- (a) 6 — Method: for inverse proportion the product xy is the same for every pair, so find that product and use it to work back to the missing value. Working: xy = 2 × 15 = 30, so when x = 5 the equation 5y = 30 gives y = 30 ÷ 5 = 6. Answer: 6. The distractors: 37.5 comes from treating the pair as direct proportion and scaling y up with x, 15 × 5 ÷ 2, although in inverse proportion y falls as x rises; 30 is the constant product itself, given as a value of y rather than used to find one; 12 comes from additive thinking — x rises by 3, so 3 is taken off y — which would make the two quantities differ by a constant instead of multiplying to one.
- (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.
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (b) 1500 — The rate is 3 ÷ 2 = 1.5 litres per minute. Converting to cm³: 1.5 × 1000 = 1500 cm³ per minute. Getting 3000 comes from converting 3 litres to cm³ first (3000 cm³) and forgetting to divide by the 2 minutes. Getting 750 comes from dividing by the 2 minutes a second time after converting (1500 ÷ 2). Getting 2000 comes from converting the 2 minutes as if it were litres (2 × 1000) instead of using the correct rate of 1.5 litres per minute.
- (a) 80 km/h — Average speed = total distance ÷ total time. Total distance = 45 + 75 = 120 km. Total time = 30 minutes + 1 hour = 1.5 hours. 120 ÷ 1.5 = 80 km/h. 60 km/h comes from treating the 30 minutes as a whole hour, giving a total time of 2 hours instead of 1.5 (120 ÷ 2). 82.5 km/h comes from averaging the two separate speeds (45 ÷ 0.5 = 90 km/h and 75 ÷ 1 = 75 km/h, then (90 + 75) ÷ 2) instead of using total distance over total time. 75 km/h comes from using only the second part of the journey (75 km in 1 hour) and ignoring the first part.
- (b) 1:12 — C = 12n means that for every 1 pen there are 12 pence of cost, so n : C = 1 : 12, and the highest common factor of 1 and 12 is 1, so this is already in its simplest form. Writing C : n instead of n : C gives 12 : 1, the ratio the wrong way round. Reading C = 12n as '12 more than n' instead of '12 times n', so n = 1 gives C = 13, gives 1 : 13, from adding instead of multiplying. Choosing n = 12, so C = 12 × 12 = 144, gives the correct pair of values n : C = 12 : 144, but the ratio the right way round left unsimplified — 12 and 144 share a common factor of 12, which has not been cancelled.
- (a) 45% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is £116 − £80 = £36, and 36 ÷ 80 = 0.45, so 0.45 × 100 = 45%. Answer: 45%. The distractors: 36% comes from quoting the £36 increase as though pounds and per cent were the same thing; 31% comes from dividing the £36 increase by the new price £116 instead of by the original £80, which gives 31% to the nearest per cent; 145% is the new price written as a percentage of the original price, which is the whole of the new price rather than the increase.
- (b) t = 5 and t = 7 (closest, evenly spaced) — To estimate the instantaneous rate of change at t = 6, use the chord centred on t = 6 with the closest readings on either side, t = 5 and t = 7. The gradient of this chord is 15.4 − 17.5 = −2.1, then −2.1 ÷ 2 = −1.05 cm per minute. The interval t = 3 to t = 9 is also centred on t = 6 but is wider: 13.4 − 18.7 = −5.3, then −5.3 ÷ 6 ≈ −0.88 cm per minute — this brings in more of the curve's own change in steepness, so it is a worse estimate of the rate at the single instant t = 6. Using t = 6 and t = 7 only gives 15.4 − 16.6 = −1.2, then −1.2 ÷ 1 = −1.2 cm per minute, but this is not centred on t = 6 — it estimates the rate over (6, 7), not at t = 6 itself. Using t = 0 and t = 6 gives 16.6 − 20.0 = −3.4, then −3.4 ÷ 6 ≈ −0.57 cm per minute, the average rate for the whole first six minutes, not the rate at the instant t = 6. Always choose the chord that brackets the point as closely as possible.
- (a) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (c) 135 g — Find the ratio of butter to sugar in the first batch: 240:160, which simplifies to 3:2. For the second batch, sugar = 90 g, so butter = 90 × 3/2 = 135 g. (60 g comes from using the ratio the wrong way round, 90 × 2/3. 170 g comes from subtracting the drop in sugar, 160 − 90 = 70 g, from the original butter amount, 240 − 70, instead of scaling. 240 g comes from not scaling the butter amount at all.)
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (a) 4 : 15 — Convert to the same unit first: 1.5 kg = 1500 g, since 1 kg = 1000 g. This gives the ratio 400 : 1500. Divide both parts by their highest common factor, 100, to get 4 : 15. Giving 40 : 150 divides by 10 only, which is not the highest common factor, so it is not fully simplified. Giving 15 : 4 swaps the order. Giving 4 : 1.5 has not converted 1.5 kg into grams, so the two parts are not measured in the same unit.
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