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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (c) £450 — Method: use the equal ratios 4:5 = 200:x to find Grace's savings, then add the two amounts. Working: Noah's £200 is 4 parts, so one part is £200 ÷ 4 = £50; Grace has 5 parts, so 5 × £50 = £250; altogether £200 + £250 = £450. Answer: £450. The distractors: £250 is Grace's savings on their own, which is the middle step rather than the total the question asks for; £360 comes from reading £200 as the 5 parts instead of the 4, giving one part of £40 and a total of 9 × £40; £400 comes from doubling £200, which treats the two savings as equal and ignores the ratio altogether.
- (c) 5/7 — First find the number of children: 84 − 35 = 49. The question compares the adults with the children, not with everyone on the bus, so the denominator is 49 and the numerator is 35, giving 35/49. Both parts divide by 7: 35 ÷ 7 = 5 and 49 ÷ 7 = 7. In its simplest form the fraction is 5/7.
- (c) 183 mg — Apply the decay, then add the new dose, once for each hour. Hour 1: 0.7 × 200 = 140, then 140 + 50 = 190. Hour 2: 0.7 × 190 = 133, then 133 + 50 = 183, so there is 183 mg after 2 hours. Forgetting the top-up dose and only applying the decay gives 0.7 × 200 = 140, then 0.7 × 140 = 98 — this ignores that a further 50 mg is given every hour. Adding the 50 mg BEFORE the decay is applied, instead of after, gives 0.7 × (200 + 50) = 175, then 0.7 × (175 + 50) = 157.5, which changes how much of the dose is eliminated in the same hour it is given. Multiplying by 0.3, the percentage ELIMINATED, instead of by 0.7, the percentage REMAINING, gives 0.3 × 200 + 50 = 110, then 0.3 × 110 + 50 = 83 — this mixes up the amount that leaves the bloodstream with the amount that stays in it. Always check whether a percentage describes what remains or what is removed before choosing the multiplier.
- (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) 36 cm — A perimeter is built from lengths, so it is multiplied by the length scale factor and not by the square of it. The scale factor from the smaller rectangle to the larger one is 10 ÷ 5 = 2. The larger perimeter is therefore 18 × 2 = 36 cm.
- (c) 45% — Method: adding water changes the total volume but not the amount of fruit juice, so find the juice, find the new total volume, and write the first as a percentage of the second. Working: 3 × 0.6 = 1.8 litres of fruit juice; the new volume is 3 + 1 = 4 litres; 1.8 ÷ 4 = 0.45, which is 45%. Answer: 45%. The distractors: 60% is the strength before the water goes in, and assumes that adding water leaves the strength unchanged; 15% comes from dividing the 60% by the 4 litres of mixture instead of dividing the 1.8 litres of juice by the 4 litres; 75% is the fraction of the new mixture that came out of the original jug, 3 litres out of 4, which ignores that only 60% of that 3 litres was juice.
- (c) 88 km/h — Multiply the speed in mph by the conversion factor: 55 × 1.6 = 88 km/h. Dividing by 1.6 instead of multiplying gives 55 ÷ 1.6 ≈ 34.38 km/h, going the wrong way between the units. Adding the conversion factor instead of multiplying gives 55 + 1.6 = 56.6 km/h, treating the factor as an amount rather than a multiplier. Multiplying by 0.6 instead of 1.6 gives 55 × 0.6 = 33 km/h, using only part of the conversion factor. 55 mph is equal to 88 km/h.
- (b) 2/5 — The ratio red : yellow is 6:15, so write red over yellow: 6/15. Divide both numbers by their highest common factor, 3: 6÷3 = 2, 15÷3 = 5, giving 2/5. (5/2 comes from writing the ratio the wrong way round, yellow over red, 15/6, which simplifies to 5/2. 2/7 comes from comparing the red paint to the total amount of paint, 6 parts out of 21. 5/7 comes from comparing the yellow paint to the total amount of paint, 15 parts out of 21.)
- (c) 24p — Method: change the price to pence, then find what one tenth of 1 kg costs, since 100 g is one tenth of 1 kg. Working: £2.40 = 240p per kg, and 240 ÷ 10 = 24p per 100 g. So the cost is 24p per 100 g. Distractor 2.4p comes from dividing by 100 instead of 10. Distractor 2400p comes from multiplying by 10 instead of dividing. Distractor 240p comes from using the price per kg without scaling it down to 100 g.
- (c) 9% — Method: a percentage concentration is the ratio of salt to solution written per 100 g, so scale each concentration to the mass it belongs to, add the two masses of salt, then scale the ratio of salt to mixture back to a denominator of 100. Working: 5:100 = x:400 gives 5 ÷ 100 × 400 = 20 g of salt, and 25:100 = y:100 gives 25 g of salt; the mixture holds 20 + 25 = 45 g of salt in 400 + 100 = 500 g of solution; 45:500 = 9:100. Answer: 9%. The distractors: 15% is the mean of 5% and 25%, which would only be right if the two masses were equal, and here one is four times the other; 21% comes from attaching the concentrations to the wrong masses, working out (400 × 25% + 100 × 5%) ÷ 500; 0.9% comes from working out 45 ÷ 500 = 0.09 and then moving the decimal point one place instead of two when writing the decimal as a percentage.
- (c) 30 — Method: use y = kx and find k from the given pair of values, then substitute x = 12. Working: k = 20 ÷ 8 = 2.5, so y = 2.5 × 12 = 30. Answer: 30. 24 comes from treating the relationship as additive, adding the increase in x (12 − 8 = 4) straight onto y (20 + 4 = 24), instead of multiplying by k. 14.5 comes from finding k correctly (2.5) but then adding it to x instead of multiplying (12 + 2.5 = 14.5). 4.8 comes from finding k upside down, 8 ÷ 20 = 0.4, and multiplying by x: 12 × 0.4 = 4.8.
- (b) litres per minute — The gradient of a tangent is the change in the quantity on the vertical axis divided by the change in the quantity on the horizontal axis, so its units come from both axes: litres on the vertical axis and minutes on the horizontal axis give litres per minute. Giving the units as minutes for each litre inverts the fraction, giving the units of the RECIPROCAL of the gradient, not the gradient itself. Writing just litres uses only the vertical axis's units and ignores that a gradient is a rate, not an amount. Writing just minutes uses only the horizontal axis's units. A gradient always combines both axes' units as one divided by the other.
- (a) It's an average over 20 days, which may miss the day-5 rate. — Method: a chord's gradient is the AVERAGE rate of change across the whole interval it spans; it only closely approximates the INSTANTANEOUS rate of change at a point inside that interval when the rate of change is roughly constant across the interval, which usually means the interval needs to be short. Working: here the chord spans 20 days while the point of interest, t = 5, is only a quarter of the way along it, so if the reservoir's level rose or fell at different rates over that time, the chord's gradient will not be close to the true gradient of the curve at t = 5 — this is the correct reason. Claiming the chord's gradient needs the water level at every day in between is wrong: a chord's gradient needs only the two endpoint values, at t = 0 and t = 20. Claiming a chord can only estimate the rate at its own endpoints is wrong: a chord between two points can be used to estimate the instantaneous rate of change at any point inside the interval, including one that is not an endpoint — that is exactly the technique being used here, and it is the SIZE of the interval that makes the estimate poor, not the fact that t = 5 is an interior point. Claiming the units do not match is wrong: the chord's gradient and the instantaneous rate of change are both measured in metres per day, so the units are the same. A chord is only a good estimate of an instantaneous rate when the interval it spans is short enough that the rate does not change much within it — always check how long the interval is compared with how far it is to the point you actually want.
- (d) 36 mph — First convert 1 hour 30 minutes to hours: 30 minutes is half an hour, so the time is 1.5 hours. Then divide the distance by the time: 54 ÷ 1.5 = 36 mph. Reading 1 hour 30 minutes as 1.3 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 54 ÷ 1.3 ≈ 41.54 mph. Working out 54 ÷ 30 = 1.8 divides by the number of minutes only, ignoring the hour. Working out 54 × 1.5 = 81 multiplies by the time instead of dividing. The coach's average speed is 36 mph.
- (d) £4 — The gradient of the tangent gives the instantaneous rate of change of cost with respect to the number of passengers, in pounds per passenger. The tangent passes through (16, 184) and (24, 216), so the change in cost is 216 − 184 = 32 and the change in passengers is 24 − 16 = 8. The gradient is 32 ÷ 8 = 4. Stopping after finding the change in cost, without dividing by the change in passengers, leaves 32, not a rate. Adding the two changes instead of dividing gives 32 + 8 = 40, which is not a rate either. Reading off only the change in passengers, 8, is not a rate at all — a rate needs the change in cost as well. The instantaneous rate is £4 per extra passenger.
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