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
- (d) 8/5 — Two masses can only be compared once they are in the same unit. Since 1 kg is 1000 g, the recipe needs 1200 g. The recipe's mass is being written as a fraction of Dan's mass, so 1200 goes on the top and 750 on the bottom, giving 1200/750. The highest common factor of the two is 150: 1200 ÷ 150 = 8 and 750 ÷ 150 = 5. The fraction is 8/5, which is greater than 1 because the recipe needs more flour than Dan has.
- (a) 2978 — A rise of 6% is a multiplier of 1.06, applied once for each year. After year 1: 2500 × 1.06 = 2650. After year 2: 2650 × 1.06 = 2809. After year 3: 2809 × 1.06 = 2977.54, which is 2978 to the nearest whole number. Multiplying by 1.18 in one go would be wrong, because the second and third years grow from larger numbers than the first.
- (c) 40 minutes — Method: this is inverse proportion — fewer taps means longer, not shorter — so the number of taps × the time taken stays constant. Working: 6 × 20 = 120, and with 3 taps the time is 120 ÷ 3 = 40 minutes. So 3 taps take 40 minutes. Distractor 10 minutes comes from treating it as direct proportion instead of inverse, working out 20 × 3 ÷ 6. Distractor 30 minutes comes from halving the number of taps and adding half the original time, 20 + 10, instead of doubling the time. Distractor 17 minutes comes from subtracting the number of taps removed, 3, directly from the original time, 20.
- (a) 2.17 litres per minute — There are 60 minutes in an hour, so to convert litres per hour to litres per minute you divide by 60: 130 ÷ 60 = 2.1666..., which rounds to 2.17 litres per minute. Multiplying by 60 instead of dividing gives 130 × 60 = 7800.00 litres per minute, using the conversion factor the wrong way round. Leaving the rate unchanged, 130.00, ignores that 'per hour' and 'per minute' are different units. Dividing by 50 instead of 60, misremembering the number of minutes in an hour, gives 130 ÷ 50 = 2.60 litres per minute.
- (b) 4% — Method: find the total interest earned, share it equally across the number of years to find one year's interest, then write it as a percentage of the amount invested. Working: total interest = £840 − £750 = £90, so one year's interest is £90 ÷ 3 = £30, and £30 as a percentage of £750 is (£30 ÷ £750) × 100 = 4%. Answer: 4%. 12% comes from treating the total interest of £90 as if it were earned in a single year, (£90 ÷ £750) × 100 = 12%, forgetting to divide by 3 years. 0.04% comes from finding the correct decimal, £30 ÷ £750 = 0.04, but forgetting to multiply by 100 to convert it into a percentage. 112% comes from writing the final amount, £840, as a percentage of the amount invested, £750, without first subtracting the £750 to find the interest alone.
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (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).
- (d) Faster at t = 8s — still accelerating — The gradient of a tangent on a distance-time graph is the instantaneous speed, in m/s. At t = 2 seconds the speed is 6 m/s; at t = 8 seconds it is 9.5 m/s, which is faster, so the sprinter is still accelerating between these two times. Saying the sprinter is slower at t = 8s reverses the comparison — 9.5 is greater than 6, not less. Writing 9.5 − 6 = 3.5 and calling this 'metres further covered' turns the difference of two speeds into a distance, which the units do not support: a difference of two speeds is itself a speed, not a distance. Taking 9.5 m/s, the larger of the two instantaneous speeds, as the average speed for the whole race confuses a speed at one instant with an average over the whole distance, which would need the total distance and total time, not two tangent gradients.
- (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) 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.
- (b) No — rate fell by 3.7 thousand/month — Each tangent gradient is the instantaneous growth rate, in thousand subscribers per month. To compare them, subtract the later rate from the earlier one: 4.8 − 1.1 = 3.7. Since 1.1 is less than 4.8, the growth rate has fallen by 3.7 thousand subscribers per month, so the manager is wrong — the app is growing more slowly at 18 months, not faster. Subtracting the other way round and calling the result a rise, 'rate rose by 3.7 thousand/month', gets the direction backwards: the later gradient is the smaller of the two. Adding the two gradients, 4.8 + 1.1 = 5.9, and calling this a combined rate that shows speeding up, is the wrong operation for comparing two rates. Treating the difference 3.7 as a total number of subscribers lost, rather than a rate in thousands per month, confuses a rate with a count. Always subtract the two rates in a sensible order and keep the units in thousands per month.
- (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.
- (b) £16 — Since cost is proportional to the square root of diameter, C = k√d. Using d = 9, C = 12: √9 = 3, so 12 = k × 3, giving k = 12 ÷ 3 = 4. The equation is C = 4√d. When d = 16: √16 = 4, so C = 4 × 4 = 16. Halving the new diameter instead of taking its square root gives 16 ÷ 2 = 8, and then C = 4 × 8 = 32 — halving a number is not the same as taking its square root, as √16 = 4, not 8. Multiplying k by the diameter itself instead of by its square root gives C = 4 × 16 = 64, skipping the square root altogether. Reporting √16 on its own, without multiplying by k, gives only 4, not the cost. The cost of manufacturing a lens of diameter 16 mm is £16.
- (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.)
- (d) Profit is decreasing by £10 per £1 rise in price. — The gradient of a tangent gives the instantaneous rate of change of profit with respect to price, found from the change in profit divided by the change in price between two points on the tangent. Here the tangent passes through (12, 540) and (18, 480), so the change in profit is 480 − 540 = −60 and the change in price is 18 − 12 = 6. The gradient is −60 ÷ 6 = −10. A negative gradient means profit is decreasing as price increases, so profit is decreasing at an instantaneous rate of £10 for every £1 rise in price. Subtracting the profits in the wrong order, 540 − 480 = 60, and dividing by the same change in price, 60 ÷ 6 = 10, gives a positive value and the wrong direction — profit is not increasing at £15. Stopping after finding only the change in profit, 480 − 540 = −60, without dividing by the change in price, is not a rate at all. Reading off the change in price, 6, and calling it the rate gives £6 per £1 rise in price, but 6 is only the width of the price interval — it is not a change in profit at all, and profit falls across that interval, so the direction is wrong too. The instantaneous rate of change of profit with respect to price at £15 is a decrease of £10 per £1 rise in price.
Build your own mix at the worksheet builder.