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 cartographer is choosing a scale for a map to show a small park in as much detail as possible. Which of these scales would show the MOST detail — 1 : 500, 1 : 5000, or 1 : 50 000? Give the correct scale and reason.
- 2.A printer's ink cartridge level, I millilitres, is plotted against the number of pages printed, p, and the points lie on a straight line. The line passes through (0, 24) and (300, 9). The printer has now printed 300 pages. Work out how many more pages it can print before the cartridge is empty.
- 3.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.
- 4.The cost of manufacturing a spherical container is proportional to the cube of its radius. A container of radius 3 cm costs £54 to manufacture. Construct the equation connecting cost C and radius r, then work out the cost of a container of radius 5 cm.
- 5.The population of a town, P thousand, is plotted against time, t years, since 2020. A tangent to the graph at t = 6 has gradient 2.5. What does this gradient represent?
- 6.Two mathematically similar tins have heights 8 cm and 12 cm. Jayden says that the volume of the larger tin is 1.5 times the volume of the smaller tin. Give a reason why Jayden is incorrect, and work out the correct volume scale factor.
- 7.A shop sells rope by the metre. 3 m costs £7.50 and 5 m costs £11.00. Does this data show that the cost is directly proportional to the length of rope bought? Choose the correct verdict and reason.
- 8.A charity fun run raises money through entry fees and donations. Entry fees raise £1,260, which is 60% of the total amount raised. Work out how much money was raised through donations.
- 9.A lorry is carrying a load of mass m tonnes. Given that 1 tonne = 1000 kg, write down an expression for the mass of the load in kilograms.
- 10.Two mathematically similar triangular flags have areas in the ratio 4 : 25. The height of the smaller flag is 6 cm. Work out the height of the larger flag.
- 11.The strength of a radio signal, in units, is inversely proportional to the square of the distance from the transmitter, in km. At a distance of 2 km the signal strength is 20 units. Construct the equation connecting signal strength S and distance d, then work out the distance at which the signal strength is 5 units.
- 12.The value of a rare coin increases by 12% each year. The coin is currently worth £270. Work out the value of the coin after 2 years, giving your answer to the nearest penny.
- 13.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.
- 14.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.
- 15.A hiker's distance from the start of the trail, in kilometres, is plotted against time, in minutes, since she set off. At t = 45 minutes she stops to rest, so her distance from the start is neither increasing nor decreasing at that instant. Which statement about the tangent to the distance–time graph at t = 45 minutes is correct?
Answer key
- (b) 1 : 500 — smallest real distance per cm (5 m), most detail — Method: compare what one centimetre represents in real life for each scale — the scale with the smallest real distance per cm shows the most detail. Working: for 1 : 500, 1 cm represents 500 cm (5 m); for 1 : 5000, 1 cm represents 50 m; for 1 : 50 000, 1 cm represents 500 m. Since 5 m is the smallest, 1 : 500 shows the most detail. Wrong options: '1 : 50 000 — covers the largest real area' wrongly assumes covering more area means more detail, when it is the opposite; '1 : 5000 — the middle value' wrongly assumes the middle scale is automatically the most balanced; '1 : 500 — covers the largest real distance' picks the correct scale but states an incorrect fact, since 1 : 500 actually covers the smallest real distance per cm.
- (c) 180 pages — The gradient is (9 − 24) ÷ (300 − 0) = −15 ÷ 300 = −0.05, so the cartridge uses 0.05 ml of ink per page. At p = 300 there are 9 ml left. The extra pages before the cartridge is empty is 9 ÷ 0.05 = 180 pages. Giving 9 as the answer confuses the millilitres of ink remaining with the number of pages remaining — they are different quantities with different units. Multiplying instead of dividing, 9 × 0.05 = 0.45, does not undo the rate correctly. Working out the total number of pages a full cartridge lasts, 24 ÷ 0.05 = 480 pages, answers how many pages the cartridge prints in total from full, not how many more pages it can print from the 300-page point.
- (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) £250 — Since cost is proportional to the cube of the radius, C = kr³. Using r = 3, C = 54: 3³ = 27, so 54 = k × 27, giving k = 54 ÷ 27 = 2. The equation is C = 2r³. When r = 5: 5³ = 125, so C = 2 × 125 = 250. Treating the relationship as proportional to r² instead of r³ gives k = 54 ÷ 9 = 6 and then C = 6 × 25 = 150, which models area scaling, not volume scaling. Treating it as proportional to r itself gives k = 54 ÷ 3 = 18 and then C = 18 × 5 = 90. Finding k correctly from the cube but then multiplying it by the radius instead of by the cube of the radius gives 2 × 5 = 10, which applies the right constant to the wrong power of r. The cost of a container of radius 5 cm is £250.
- (d) The population is growing at 2500 people per year. — The gradient of a tangent to a graph gives the instantaneous rate of change of the quantity on the vertical axis with respect to the quantity on the horizontal axis, at that exact point — not the total change and not an average. Here the vertical axis is population in thousands and the horizontal axis is time in years, so the gradient is measured in thousands of people per year. A gradient of 2.5 means the population is growing at an instantaneous rate of 2.5 thousand people per year, and since P is measured in thousands, 2.5 × 1000 = 2500 people per year. This describes the rate of change at that instant, not the total increase over the 6 years and not an average population.
- (a) 3.375 — Method: the volume scale factor between similar shapes is the length scale factor cubed, not the length scale factor itself. Working: the length scale factor is 12 ÷ 8 = 1.5, and 1.5³ = 3.375. Answer: 3.375. Jayden's answer, 1.5, is only the LENGTH scale factor — he never cubed it to get the volume scale factor. 2.25 comes from squaring the length scale factor instead of cubing it, which would give the area scale factor. 4.5 comes from multiplying the length scale factor by 3 (the number of dimensions) instead of cubing it.
- (a) No — the cost per metre differs: £2.50/m vs £2.20/m — Method: divide cost by length for each pair and compare the unit rates. Working: £7.50 ÷ 3 = £2.50 per m; £11.00 ÷ 5 = £2.20 per m. The rates are different, so this is NOT direct proportion. Wrong options: 'Yes — both amounts increase' wrongly assumes any increasing relationship is proportional; 'No — because 5 m costs more in total' judges by total cost rather than the rate per metre, which is not valid reasoning on its own; 'Yes — the cost per metre is £2.50 in both cases' miscalculates the second rate (11.00 ÷ 5 is £2.20, not £2.50).
- (c) £840 — Method: find the total amount raised using the reverse percentage, then subtract the entry fees to find the donations. Working: £1,260 is 60% of the total, so the total is £1,260 ÷ 0.6, and subtracting the entry fees from this total leaves £840 raised through donations. Answer: £840. £2,100 comes from correctly finding the total amount raised but then forgetting to subtract the entry fees, giving the total instead of the donations alone. £504 comes from working out 40% of the entry fees themselves, £1,260 × 0.4 = £504, instead of first finding the total amount raised. £1,890 comes from treating £1,260 as 40% of the total instead of 60%, dividing by 0.4 to get a total of £3,150, and then subtracting the entry fees from that incorrect total.
- (a) 1000m — Method: kilograms are a smaller unit than tonnes, so change tonnes into kilograms by multiplying by 1000. Working: m tonnes = m × 1000 kg = 1000m kg. So the expression is 1000m. Distractor m/1000 comes from dividing by 1000 instead of multiplying, which would make the number of kilograms smaller than the number of tonnes, the wrong way round. Distractor 1000 + m comes from adding the conversion factor instead of multiplying by it. Distractor m − 1000 comes from subtracting the conversion factor instead of multiplying by it.
- (a) 15 cm — Take the square root of each part of the area ratio to find the length ratio: the square root of 4 is 2 and the square root of 25 is 5, giving a length ratio of 2 : 5. Multiply the smaller flag's height by the scale factor 5 ÷ 2 = 2.5: 6 × 2.5 = 15, so the larger flag is 15 cm tall. Giving 37.5 cm uses the area ratio, 25 ÷ 4 = 6.25, directly as the scale factor without square-rooting it first (6 × 6.25 = 37.5). Giving 2.4 cm applies the length ratio the wrong way round, scaling the smaller flag down by 2 ÷ 5 instead of up by 5 ÷ 2 (6 × 0.4 = 2.4). Giving 27 cm adds the difference between the two area-ratio numbers, 25 − 4 = 21, onto the smaller height instead of using it as a scale factor (6 + 21 = 27).
- (c) 4 km — Since signal strength is inversely proportional to the square of the distance, S = k/d². Using d = 2, S = 20: 2² = 4, so 20 = k ÷ 4, giving k = 20 × 4 = 80. The equation is S = 80/d². When S = 5: d² = 80 ÷ 5 = 16, so d = 4 (taking the positive root, since distance cannot be negative). Stopping at d² = 16 without taking the square root leaves 16, the square of the distance, not the distance itself. Treating the relationship as inversely proportional to distance itself, rather than to its square, gives k = 20 × 2 = 40 and then d = 40 ÷ 5 = 8, a different relationship. Multiplying by S instead of dividing by it when isolating d² gives d² = 80 × 5 = 400 and d = 20, the wrong operation. The distance at which the signal strength is 5 units is 4 km.
- (c) £338.69 — To increase by 12% each year, multiply by 1.12 twice. £270 × 1.12 × 1.12 = £338.688, which rounds to £338.69 (nearest penny, since the third decimal place is 8). £334.80 comes from treating the two 12% increases as a single flat 24% increase applied once instead of compounding: £270 × 1.24 = £334.80. £302.40 comes from applying the 12% increase only once, for 1 year instead of 2: £270 × 1.12 = £302.40. £338.68 comes from rounding £338.688 down to the nearest penny instead of up.
- (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.
- (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.
- (c) The tangent is horizontal, so its gradient is 0. — Method: at any point where a distance–time graph is momentarily neither increasing nor decreasing, the tangent to the graph at that point is horizontal, and the gradient of a horizontal line is 0 — this is the instantaneous rate of change at that instant. Working: since the hiker's distance is neither increasing nor decreasing at t = 45 minutes, the tangent there is horizontal, so its gradient is 0. Claiming the tangent is vertical, with an undefined gradient, is the opposite of what the stem says: a vertical tangent would mean the distance was changing infinitely fast at that instant, not that it had stopped changing, and on a distance–time graph it cannot happen at all. Reading the gradient as 45, the time value given in the stem, mistakes a value used to LOCATE the point for the rate of change AT that point. Claiming the gradient cannot be found without also knowing the distance at t = 45 minutes overlooks that 'momentarily stationary' already tells you the rate of change directly, without needing to read any distance value at all. Whenever a stem tells you a quantity is momentarily not changing, that is telling you the instantaneous rate of change directly — it is 0, and no further data is needed to find it.
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