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.
Calculator
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (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.
- (b) Takings rise about £14 per 1°C rise — The gradient here is positive, so as temperature rises, takings rise too: near 22°C, takings increase by about £14 for every 1°C rise in temperature. Reversing this to say takings rise for every 1°C FALL gets the direction of the independent variable backwards — a positive gradient means both quantities move the same way. Saying 'takings are £14 at 22°C' confuses the gradient, a rate of change, with the y-value on the graph, which is the takings itself. Saying takings 'rose £14 in total' from 0°C to 22°C treats the gradient at a single point as if it applied over the whole range from 0°C to 22°C, when it only describes the instant at 22°C. Always keep a rate, a total change and a single reading separate.
- (d) 24.6 km/h — Average speed = distance ÷ time, with time in hours. 47 minutes = 47 ÷ 60 hours. 19.3 ÷ (47 ÷ 60) = 19.3 ÷ 47 × 60 = 24.638…, which rounds to 24.6 km/h (1 d.p.). 0.4 km/h comes from dividing the distance by 47 and treating the result as km/h directly, without converting the minutes to hours at all. 41.1 km/h comes from converting minutes to hours by dividing by 100 instead of 60 (19.3 ÷ 47 × 100). 0.3 km/h comes from dividing the distance by 60 instead of converting the 47 minutes to hours first.
- (d) 10 — First find the height at each end of the chord. At t = 1, h = 30 × 1 − 5 × 1² = 30 − 5 = 25. At t = 3, h = 30 × 3 − 5 × 3² = 90 − 45 = 45. The gradient of the chord estimates the instantaneous rate at the midpoint t = 2: 45 − 25 = 20, then 20 ÷ (3 − 1) = 20 ÷ 2 = 10 m/s. Finding the change in height but forgetting to divide by the change in time gives 20, which is a distance, not a rate. Averaging the two heights instead of finding the difference gives (25 + 45) ÷ 2 = 70 ÷ 2 = 35. Subtracting in the wrong order, 25 − 45 = −20, then −20 ÷ 2 = −10, gives the correct size with the sign flipped — the rocket is rising, not falling, at t = 2 seconds, so a negative rate cannot be right here.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
- (d) 500 ml — Varnish covers a surface, so the amount needed scales with the area scale factor, which is the square of the length scale factor. The length scale factor is 50 ÷ 20 = 2.5, so the area scale factor is 2.5 × 2.5 = 6.25. The varnish needed for the larger statue is 80 × 6.25 = 500 ml. Using 2.5 on its own would scale a length, not a surface.
- (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.
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (c) 2:3 — The white paint is 5 − 2 = 3 litres. The ratio of blue paint to white paint is 2 : 3, which has no common factor, so it is already in simplest form. Getting 2 : 5 compares the blue paint to the total amount of shade instead of to the white paint. Getting 3 : 2 has the two parts the wrong way round. Getting 5 : 3 uses the total amount of shade instead of the blue paint as the first part.
- (d) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
- (a) 1.375 — The multiplier is the new value divided by the original value: 176 000 ÷ 128 000 = 1.375. Choosing 0.375 comes from working out the increase as a fraction of the original (48 000 ÷ 128 000 = 0.375) and stopping there, without adding on the 1 whole that represents the original turnover. Choosing 0.727 comes from dividing the 2023 turnover by the 2024 turnover, the wrong way round (128 000 ÷ 176 000, rounded to 3 decimal places). Choosing 137.5 comes from writing the multiplier as a percentage — the 2024 turnover is 137.5% of the 2023 turnover — instead of converting it to the decimal multiplier 1.375.
- (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) £7.80 — Rate of pay = total pay ÷ number of hours. £58.50 ÷ 7.5 = £7.80 per hour. £438.75 comes from multiplying the pay by the hours instead of dividing (£58.50 × 7.5). £0.13 comes from dividing the hours by the pay instead of the pay by the hours (7.5 ÷ 58.50). £51.00 comes from subtracting the hours from the pay (£58.50 − 7.5) instead of dividing.
- (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.
- (a) 11:00 to 12:00 — Method: the instantaneous rate of change is zero at a turning point, where a rising trend becomes a falling trend; that lies within the first interval whose difference has changed sign from the interval before it. Working: the differences between consecutive readings are +3.4 °C (09:00 to 10:00), +1.7 °C (10:00 to 11:00), −0.3 °C (11:00 to 12:00) and −2.6 °C (12:00 to 13:00); the sign changes from positive to negative within 11:00 to 12:00, since the temperature is still rising up to 11:00 (20.1 °C, the highest recorded value) and has fallen by 12:00, so the instantaneous rate of change was zero somewhere within that interval. Choosing 09:00 to 10:00 picks out the interval with the largest positive difference, +3.4 °C, confusing the fastest rise with no change at all. Choosing 10:00 to 11:00 picks the last interval where the temperature was still rising, one interval too early, without checking that the very next interval turns negative. Choosing 12:00 to 13:00 picks out the interval with the largest-magnitude difference, −2.6 °C, the fastest fall, not where the change is zero. Zero instantaneous rate of change happens at a turning point, where the readings stop rising and start falling — find the FIRST interval whose difference has flipped sign from the one before it, not the biggest change or an interval where the old sign still held.
Build your own mix at the worksheet builder.