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 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.
- 2.6 identical printers can print a batch of exam papers in 40 minutes, all working at the same rate. Working at the same rate, work out how many minutes 4 of these printers would take to print the same batch.
- 3.y is inversely proportional to x², and x is positive. When x = 2, y = 45. Construct the equation connecting x and y, then work out the value of x when y = 5.
- 4.Aisha invests £3200 in Account A, which pays 5% compound interest each year. She also invests £3200 in Account B, which pays 3% simple interest each year. Work out how much more Account A is worth than Account B after 2 years.
- 5.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
- 6.Maya buys a vintage car for £15000. Its value is expected to increase by 4% each year for 2 years. She then plans to sell the car and use £2000 of the money from the sale to buy a trailer. Work out how much money she will have left after selling the car and buying the trailer.
- 7.A company had 8000 employees. The number of employees decreased by 5% in the first year, and then increased by 5% in the second year. Work out the number of employees at the end of the second year, to the nearest whole number.
- 8.A car uses fuel at a constant rate. It uses 24 litres of fuel to travel 300 km. Assuming the same rate, work out how many litres of fuel are needed for a journey of 175 km.
- 9.Two mathematically similar water bottles have a volume scale factor of 8 from the smaller bottle to the larger bottle. Work out the surface area scale factor from the smaller bottle to the larger bottle.
- 10.A cyclist travels d kilometres in t hours. Write down an expression, in terms of d and t, for the cyclist's average speed in km/h.
- 11.A fixed job of fitting a solar array is shared between installers, and the time taken is inversely proportional to the number of installers working on it at once. With 4 installers the job takes 18 hours. Construct the equation connecting time T and number of installers n, then work out how long it would take with 6 installers.
- 12.A population of fish in a lake is modelled by the recurrence P_{n+1} = 1.1P_n − 30, where P_n is the population after n years, 30 fish are removed by fishing each year, and P_0 = 400. After how many complete years does the population first exceed 460?
- 13.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 14.A kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
- 15.A map has a scale of 2 cm : 5 km. A footpath measures 8 cm on the map. Work out the real length of the footpath, in kilometres.
Answer key
- (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.
- (a) 60 minutes — Method: for inverse proportion, printers × time is constant. Working: 6 × 40 = 240 (the constant). With 4 printers: 240 ÷ 4 = 60 minutes. Wrong options: 26.7 minutes comes from treating the relationship as direct proportion, scaling the time down as printers decrease (40 × 4 ÷ 6); 24 minutes comes from multiplying the two printer counts together instead of using the constant; 40 minutes comes from not adjusting the time at all for the change in printers.
- (a) 6 — Since y is inversely proportional to x², y = k/x². Using x = 2, y = 45: 2² = 4, so 45 = k ÷ 4, giving k = 45 × 4 = 180. The equation is y = 180/x². When y = 5: x² = 180 ÷ 5 = 36, so x = 6, taking the positive root because the question states that x is positive. Treating the relationship as inversely proportional to x itself gives k = 45 × 2 = 90 and then x = 90 ÷ 5 = 18, a different relationship. Multiplying by y instead of dividing by it when isolating x² gives x² = 180 × 5 = 900 and x = 30, the wrong operation. Stopping at x² = 36 without taking the square root leaves 36, not the value of x. When y = 5, x = 6.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
- (a) £14224 — Value after 2 years: £15000 × 1.04 × 1.04 = £16224. Money left after buying the trailer: £16224 − £2000 = £14224. £14200 comes from treating the two 4% increases as a single flat 8% increase applied once instead of compounding: £15000 × 1.08 = £16200, and £16200 − £2000 = £14200. £13600 comes from applying the 4% increase only once, for 1 year instead of 2: £15000 × 1.04 = £15600, and £15600 − £2000 = £13600. £18224 comes from adding the £2000 instead of subtracting it: £16224 + £2000 = £18224.
- (b) 7980 — After the first year: 8000 × 0.95 = 7600. After the second year: 7600 × 1.05 = 7980. 8000 comes from assuming a 5% decrease followed by a 5% increase returns exactly to the starting number — it does not, because the increase acts on the smaller, already-reduced number. 8400 comes from applying only the second year's 5% increase to the original number: 8000 × 1.05 = 8400. 7600 comes from applying only the first year's 5% decrease and stopping there, without applying the second year's increase.
- (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.
- (b) 4 — Method: find the length scale factor by taking the cube root of the volume scale factor, then square it to get the area scale factor. Working: 8 = 2³, so the length scale factor is 2, and the area scale factor is 2² = 4. Answer: 4. 8 comes from using the volume scale factor itself as if it were the area scale factor. 64 comes from squaring the volume scale factor, 8² = 64, instead of first taking its cube root. 2 comes from correctly finding the length scale factor but then forgetting to square it.
- (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) 12 hours — Since time is inversely proportional to the number of installers, T = k/n. Using n = 4, T = 18: 18 = k ÷ 4, so k = 18 × 4 = 72. The equation is T = 72/n. When n = 6: T = 72 ÷ 6 = 12. Using the original number of installers instead of the new one gives T = 72 ÷ 4 = 18, the wrong value substituted. Treating more installers as needing more time, as if T were directly proportional to n, gives k = 18 ÷ 4 = 4.5 and then T = 4.5 × 6 = 27, the opposite relationship to the one described. Stopping at k = 72 and reporting it gives the time the job would take a single installer working alone — the constant still has to be divided by the new number of installers before it answers the question asked. With 6 installers, the job takes 12 hours.
- (b) 5 years — The recurrence P_{n+1} = 1.1P_n − 30 must be applied once per year, checking after each application whether the population has passed 460. Starting from P_0 = 400: 400 × 1.1 − 30 = 410, so P_1 = 410. Then 410 × 1.1 − 30 = 421, so P_2 = 421. Then 421 × 1.1 − 30 = 433.1, so P_3 = 433.1. Then 433.1 × 1.1 − 30 = 446.41, so P_4 = 446.41, which is still below 460. Then 446.41 × 1.1 − 30 = 461.051, so P_5 = 461.051, the first value above 460. The population first exceeds 460 after 5 complete years. Stopping at P_4 = 446.41 and reporting 4 years reports the last year the population was still below 460, not the first year it was above. Counting the starting value P_0 = 400 as a year of growth makes P_5 the sixth number in the list and gives 6 years, but P_0 is the population before any year has passed, so P_5 is reached after 5 years, not 6. Reading "exceed 460" as "exceed the starting population of 400" instead gives P_1 = 410, already above 400, and 1 year — but the threshold named in the question is 460, not the starting value, so always check every value against the number actually stated in the question.
- (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.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
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