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.
Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.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.
- 2.Map A has a scale of 1 : 25000 and Map B has a scale of 1 : 50000, both showing the same area. A real distance of 10 km is measured on each map. On which map does this distance appear as the longer length, and how long is it on that map, in centimetres?
- 3.At an animal shelter, the ratio of cats to dogs is 2:5. There are 18 more dogs than cats. Work out the number of cats.
- 4.The price of a share falls by 10% on Monday and then rises by 10% on Tuesday. Work out the overall percentage change from Monday's starting price.
- 5.A scale model of a shipping container is built at a scale of 1 : 30, using material with the same density as the real container. The model has a mass of 400 g. Work out the mass of the real container, giving your answer in kilograms.
- 6.A map has a scale of 1 : 25 000. Two towns are 5 km apart in real life. Work out the distance between the towns on the map, in centimetres.
- 7.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.
- 8.The height of water in a tank, h metres, t minutes after a tap is opened is modelled by h = 0.02t² + 0.5. Estimate the instantaneous rate of change of the height at t = 10, using the gradient of the chord joining t = 9 and t = 11.
- 9.y is directly proportional to x². When x = 3, y = 45. Construct the equation connecting x and y, then work out the value of y when x = 5.
- 10.Two mathematically similar garden ponds have surface areas of 12 m² and 27 m². The fencing needed to go around the smaller pond costs £96. Assuming the cost of fencing is proportional to the perimeter of the pond, work out the cost of fencing the larger pond.
- 11.A cyclist's distance travelled, in metres, is recorded against time, in seconds. From t = 3 to t = 8 the distance increases from 12 m to 32 m. A tangent to the distance–time graph at t = 6 has gradient 3. Work out the average speed of the cyclist over the interval from t = 3 to t = 8.
- 12.A map has a scale of 1 : 25 000. A footpath measures 6 cm on the map. Work out the real length of the footpath, in kilometres.
- 13.The total cost, in pounds, of manufacturing a batch of components is plotted against the number of components produced. A tangent to this graph at a particular point has gradient 3.60. What does the value 3.60 represent?
- 14.A concrete mix is made from gravel and cement in the ratio 5 : 1 by mass. The mass of cement is c kg and the total mass of the mix is m kg. Write down a formula for m in terms of c.
- 15.Two mathematically similar picture frames have perimeters 30 cm and 45 cm. The area of the smaller frame is 40 cm². Work out the area of the larger frame.
Answer key
- (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.)
- (c) Map A, where the distance is 40 cm — 10 km = 1,000,000 cm. On Map A: 1000000 ÷ 25000 = 40 cm. On Map B: 1000000 ÷ 50000 = 20 cm. Since 40 cm is longer than 20 cm, the same real distance appears longer on Map A, the map with the smaller scale number. 'Map B, where the distance is 20 cm' has the correct working for Map B but names the wrong map as the one with the longer length. 'Map A, where the distance is 20 cm' correctly identifies Map A but pairs it with Map B's length. 'Map B, where the distance is 40 cm' correctly identifies Map A's length but attaches it to the wrong map.
- (b) 12 — The difference between the parts of the ratio is 5 − 2 = 3 parts, and this is worth 18. Divide to find one part: 18 ÷ 3 = 6. Cats have 2 parts: 2 × 6 = 12. (30 is the number of dogs, using 5 parts instead of 2. 6 is the value of one part — the number of cats is 2 lots of this, not just one. 9 comes from dividing 18 by 2 and stopping there, instead of dividing by the difference in parts, 3, and then multiplying by 2.)
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (c) 10,800 kg — Method: since the model and the real container are similar and made of the same material, mass scales with volume, so the mass scale factor is the length scale factor cubed. Working: 30³ = 27,000, so the real container's mass is 400 × 27,000 = 10,800,000 g, which is 10,800,000 ÷ 1,000 = 10,800 kg. Answer: 10,800 kg. 12 kg comes from using the length scale factor directly, 400 × 30 = 12,000 g, without cubing it. 360 kg comes from squaring the length scale factor instead of cubing it, 400 × 30² = 360,000 g. 10,800,000 kg comes from correctly cubing the scale factor but then forgetting to convert the mass from grams into kilograms.
- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
- (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.
- (d) 0.40 m/min — To estimate an instantaneous rate of change at a point without a diagram, use the gradient of a chord joining two points close to it, one on each side. At t = 9: 0.02 × 81 = 1.62, so h = 1.62 + 0.5 = 2.12. At t = 11: 0.02 × 121 = 2.42, so h = 2.42 + 0.5 = 2.92. The change in height is 2.92 − 2.12 = 0.80 and the change in time is 11 − 9 = 2, so the gradient of the chord is 0.80 ÷ 2 = 0.40. Reporting the change in height, 0.80, on its own is not a rate, because it has not been divided by the 2 minutes over which it happened. Using the chord from t = 0 (where h = 0.5) to t = 11 instead gives 2.92 − 0.5 = 2.42, and 2.42 ÷ 11 = 0.22, which is the average gradient over the whole 11 minutes, not the instantaneous rate at t = 10. Substituting t = 10 into the formula gives 0.02 × 100 + 0.5 = 2.50, which is the height of the water at that moment, not the rate at which the height is rising. The estimated instantaneous rate of change at t = 10 is 0.40 m/min.
- (d) 125 — Since y is directly proportional to x², y = kx² for a constant k. Using x = 3, y = 45: 45 = k × 9, so k = 45 ÷ 9 = 5. The equation is y = 5x². When x = 5: 5² = 25, and 5 × 25 = 125, so y = 125. Reporting 5² = 25 on its own, without multiplying by the constant k, gives only the square of the new x-value, not the value of y. Treating the proportion as if y were proportional to x itself, rather than to x², gives k = 45 ÷ 3 = 15 and then y = 15 × 5 = 75, which is not this relationship. Squaring the new x-value as though squaring meant doubling it instead gives 5 × 10 = 50, not the true square. When x = 5, y = 125.
- (d) £144 — Method: find the length (perimeter) scale factor by taking the square root of the area ratio, then apply it to the cost. Working: 12 : 27 simplifies to 4 : 9, and the square root of each part gives the length ratio 2 : 3, so the scale factor from the smaller to the larger pond is 3 ÷ 2 = 1.5. Cost = £96 × 1.5 = £144. Answer: £144. £216 comes from using the area ratio itself as the cost ratio, £96 × (27 ÷ 12) = £216, without taking the square root. £64 comes from using the length ratio the wrong way round, £96 × (2 ÷ 3) = £64. £111 comes from simply adding the difference in area, 27 − 12 = 15, onto the original cost, £96 + £15 = £111, instead of scaling proportionally.
- (c) 4 m/s — The average rate of change of distance with respect to time over an interval is the change in distance divided by the change in time — the gradient of the chord joining the two endpoints, not the gradient of any tangent inside the interval. From t = 3 to t = 8 the change in time is 8 − 3 = 5 and the change in distance is 32 − 12 = 20, so the average speed is 20 ÷ 5 = 4 m/s. Reporting the change in distance on its own, as 20 m/s, is not a speed: those 20 metres were covered over the whole 5 seconds, not in one second, so the 20 still has to be divided by the 5. The tangent's gradient of 3 m/s is the instantaneous speed at the single moment t = 6, not the average over the whole 5-second interval, so it must not be used here. Adding the change in distance and the change in time instead of dividing gives 20 + 5 = 25, which is not a speed. The average speed of the cyclist over the interval is 4 m/s.
- (a) 1.5 km — Multiply the map length by the scale: 6 × 25 000 = 150 000 cm. Convert to kilometres: 150 000 cm = 1.5 km. Dividing by only 1000 instead of the full conversion when changing units gives 150 km, a hundred times too large. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 6 × 2500 = 15 000 cm = 0.15 km, a hundred times too small. Leaving the answer as 150 000 without converting units at all, and calling it 150 000 km, mistakes centimetres for kilometres completely.
- (d) £3.60 per component — The gradient of a cost-against-components graph has units of pounds per component, since cost is measured in pounds and the horizontal axis counts components. So 3.60 means it costs an extra £3.60 to produce one more component at that point. Calling it '£3.60 total cost' confuses the gradient, a rate, with the y-value on the graph, which is the total cost itself. Giving it as 3.60 components per pound swaps which axis is on top, giving the units of the reciprocal gradient, not the gradient itself. Turning 3.60 into a percentage invents a unit that has no basis in the graph's axes — a gradient here is a number of pounds, not a percentage. Always build the gradient's units from the two axes' own units, in the order y-axis over x-axis.
- (c) m = 6c — Method: the whole is the sum of the parts in the ratio, and the cement is 1 part, so one part weighs c kg. Working: the mix has 5 + 1 = 6 parts, each of mass c kg, so the total mass is 6 × c, giving m = 6c. Answer: m = 6c. The distractors: m = 5c uses the 5 gravel parts as the multiplier and forgets that the cement is in the mix too, so it gives the mass of the gravel and not the total; m = c + 5 comes from reading the ratio as '5 more than' and adding, which treats a number of parts as a mass in kilograms; m = c/6 turns the relationship upside down, as though the total were shared into the cement rather than the cement multiplied up to the total.
- (c) 90 cm² — The perimeter ratio is 30 : 45, which simplifies to 2 : 3, so the larger frame is 1.5 times the perimeter of the smaller one. Areas scale with the square of this length scale factor: 1.5² = 2.25. 40 × 2.25 = 90, so the larger frame has an area of 90 cm². Giving 60 cm² uses the scale factor, 1.5, without squaring it (40 × 1.5 = 60). Giving 135 cm² cubes the scale factor, 1.5³ = 3.375, as if area scaled like a volume (40 × 3.375 = 135). Giving 2.25 cm² is the squared scale factor on its own, without multiplying by the smaller frame's area of 40 cm².
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