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.A patient is given a dose of 200 mg of a drug. Each hour, 30% of the drug remaining in the bloodstream is eliminated, and then a further dose of 50 mg is given. Using the recurrence D_{n+1} = 0.7D_n + 50, with D_0 = 200, work out the amount of drug in the bloodstream after 2 hours.
- 2.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 3.In a mix of mortar the ratio of the mass of sand to the mass of cement is 5 : 2. Write the mass of cement as a fraction of the total mass of the mix.
- 4.A graph shows y plotted against x. The graph is a curve that gets closer to both axes but never touches them, and y decreases as x increases. Write down whether this graph could show direct proportion, inverse proportion, or neither.
- 5.The density of a type of solid plastic is 0.9 g/cm³. Work out the mass of 0.5 m³ of the plastic, in kilograms.
- 6.Amelia uses 2 kg of flour to bake 5 cakes. Using the same recipe, work out how many cakes she can bake with 6 kg of flour.
- 7.The volume of fuel remaining in a storage tank, in thousands of litres, is plotted against time, in hours, since a leak was detected. A tangent to the graph at t = 2 hours has gradient −8.75 thousand litres per hour. A tangent at t = 8 hours has gradient −3.5 thousand litres per hour. Work out by what factor the instantaneous rate of change is greater in size at t = 2 hours than at t = 8 hours.
- 8.A tangent to a curve at the point where x = 3 passes through the points (1, 4) and (5, 20). Work out the gradient of the tangent at x = 3, and hence the instantaneous rate of change of y with respect to x at that point.
- 9.y is directly proportional to √x. When x = 4, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 9.
- 10.A cleaning solution is made by mixing disinfectant and water in the ratio 1 : 9. Priti wants to make 6 litres of the solution in total. Work out how much disinfectant she needs.
- 11.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?
- 12.6 identical printers can print a batch of exam papers in 40 minutes, all working at the same rate. Working at the same rate, work out how many minutes 4 of these printers would take to print the same batch.
- 13.A population of fish in a lake is modelled by the recurrence P_{n+1} = 1.1P_n − 30, where P_n is the population after n years, 30 fish are removed by fishing each year, and P_0 = 400. After how many complete years does the population first exceed 460?
- 14.A straight-line graph shows the cost, C pounds, of hiring a bike for h hours. The line passes through the points (1, 12) and (4, 27). Which of these statements about the line is true?
- 15.A fruit punch is made from orange juice, pineapple juice and lemonade in the ratio 5:3:2. A jug holds 3.5 litres of punch in total. Work out the volume of pineapple juice needed.
Answer key
- (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.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (c) 2/7 — Method: each number in a ratio counts parts, so the whole is all the parts added together, and the fraction wanted is the cement parts over that total. Working: the mix has 5 + 2 = 7 parts altogether, and 2 of those parts are cement, so the cement is 2/7 of the mix. Answer: 2/7. The distractors: 2/5 comes from writing the cement parts over the sand parts, comparing one part with the other part instead of with the whole; 5/7 comes from writing the sand parts over the total, which answers about the wrong material; 7/2 comes from writing the total over the cement parts, turning the fraction upside down.
- (a) Inverse proportion — A curve that decreases and never touches either axis is the standard shape for inverse proportion, y = k/x. Direct proportion graphs are straight lines through the origin, which this is not, so it must be inverse proportion rather than neither.
- (b) 450.00 kg — 1 m³ = 100 × 100 × 100 = 1,000,000 cm³, so 0.5 m³ = 500,000 cm³. Mass = density × volume = 0.9 × 500,000 = 450,000 g. Converting to kilograms by dividing by 1000 gives 450,000 ÷ 1000 = 450.00 kg. Skipping the m³-to-cm³ conversion and multiplying 0.9 × 0.5 = 0.45 treats the volume as if it were already 0.5 cm³, giving 0.45 kg. Finding the mass correctly in grams, 450,000 g, but not converting to kilograms leaves 450000.00 kg, out by a factor of 1000. Using the area conversion factor of 10,000, as if converting m² to cm², instead of the volume factor of 1,000,000 gives 0.5 × 10,000 = 5,000 'cm³', and a mass of 0.9 × 5,000 = 4,500 g, which is 4.50 kg.
- (c) 15 — Method: the number of cakes is in direct proportion to the mass of flour, so find the multiplier between the two masses and apply it to the number of cakes. Working: 6 ÷ 2 = 3, so there is three times as much flour, and 5 × 3 = 15. Answer: 15. The distractors: 10 comes from multiplying the 5 cakes by 2, the mass in the recipe, instead of by the multiplier 3; 20 comes from multiplying by the difference 6 − 2 = 4, treating a proportion problem as a difference problem; 12 comes from rounding 5 ÷ 2 down to 2 cakes per kilogram and working out 6 × 2.
- (a) 2.5 — Method: the factor by which the SIZE of one rate is greater than the size of another is found by dividing the larger magnitude by the smaller magnitude, ignoring their signs, so here the two magnitudes to work with are 8.75 and 3.5. Working: 8.75 ÷ 3.5 = 2.5, so the size of the instantaneous rate of change at t = 2 hours was 2.5 times the size of the instantaneous rate of change at t = 8 hours. Subtracting the two magnitudes, 8.75 − 3.5 = 5.25, gives how many thousand litres per hour greater one rate is than the other, not how many times greater — that is a difference, not a factor. Dividing the magnitudes the wrong way round, 3.5 ÷ 8.75 = 0.4, gives the factor by which the rate at t = 8 hours is smaller than at t = 2 hours, the reciprocal of what was asked for. Adding the magnitudes, 8.75 + 3.5 = 12.25, combines the two rates instead of comparing them, and does not answer a 'by what factor' question at all. A question that asks 'by what factor' is always answered by a division, in the order the question states it — check which rate is on top before you divide.
- (c) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (b) 12 — Since y is directly proportional to √x, y = k√x. Using x = 4, y = 8: √4 = 2, so 8 = k × 2, giving k = 8 ÷ 2 = 4. The equation is y = 4√x. When x = 9: √9 = 3, so y = 4 × 3 = 12. Treating the relationship as if y were proportional to x itself, rather than to √x, gives k = 8 ÷ 4 = 2 and then y = 2 × 9 = 18, which is a different relationship. Multiplying k by the new x-value instead of by its square root gives y = 4 × 9 = 36, skipping the square root altogether. Reporting √9 on its own, without multiplying by k, gives only 3, not the value of y. When x = 9, y = 12.
- (a) 0.6 litres — The ratio 1 : 9 means the solution has 1 + 9 = 10 equal parts in total. Each part is 6 ÷ 10 = 0.6 litres, and disinfectant is 1 part, so Priti needs 0.6 litres of disinfectant. Giving 0.667 litres divides by 9, the number of parts of water, instead of the total number of parts, 10 (6 ÷ 9 ≈ 0.667). Giving 6 litres is the total amount of solution, not just the disinfectant's share of it. Giving 5.4 litres works out the water's share (6 × 9 ÷ 10 = 5.4), not the disinfectant's.
- (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) 60 minutes — Method: for inverse proportion, printers × time is constant. Working: 6 × 40 = 240 (the constant). With 4 printers: 240 ÷ 4 = 60 minutes. Wrong options: 26.7 minutes comes from treating the relationship as direct proportion, scaling the time down as printers decrease (40 × 4 ÷ 6); 24 minutes comes from multiplying the two printer counts together instead of using the constant; 40 minutes comes from not adjusting the time at all for the change in printers.
- (b) 5 years — The recurrence P_{n+1} = 1.1P_n − 30 must be applied once per year, checking after each application whether the population has passed 460. Starting from P_0 = 400: 400 × 1.1 − 30 = 410, so P_1 = 410. Then 410 × 1.1 − 30 = 421, so P_2 = 421. Then 421 × 1.1 − 30 = 433.1, so P_3 = 433.1. Then 433.1 × 1.1 − 30 = 446.41, so P_4 = 446.41, which is still below 460. Then 446.41 × 1.1 − 30 = 461.051, so P_5 = 461.051, the first value above 460. The population first exceeds 460 after 5 complete years. Stopping at P_4 = 446.41 and reporting 4 years reports the last year the population was still below 460, not the first year it was above. Counting the starting value P_0 = 400 as a year of growth makes P_5 the sixth number in the list and gives 6 years, but P_0 is the population before any year has passed, so P_5 is reached after 5 years, not 6. Reading "exceed 460" as "exceed the starting population of 400" instead gives P_1 = 410, already above 400, and 1 year — but the threshold named in the question is 460, not the starting value, so always check every value against the number actually stated in the question.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (c) 1.05 litres — Method: find the value of one part of the ratio from the total volume, then find the share for pineapple juice. Working: the ratio 5:3:2 has 5 + 3 + 2 = 10 parts, so one part is 3.5 ÷ 10 = 0.35 litres, and the pineapple juice is 3 × 0.35 = 1.05 litres. So 1.05 litres of pineapple juice is needed. Distractor 1.75 litres is the volume of orange juice, not pineapple juice. Distractor 0.7 litres is the volume of lemonade, not pineapple juice. Distractor 0.35 litres is the value of one part, found correctly but never multiplied by 3.
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