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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- 1.3 builders put up a fence in 12 days. All the builders work at the same rate, so the number of days is inversely proportional to the number of builders. Work out how long 9 builders take to put up the same fence.
- 2.A cyclist's speed, in metres per second, is plotted against time, in seconds, since she sets off. Which statement correctly pairs the gradient of a tangent to this graph, and the area between the graph and the time-axis, with what each one represents?
- 3.Grace's mean mark in her maths tests rises from 70 to 84. Work out the percentage increase in her mean mark.
- 4.A straight-line graph passes through the origin and the point (4, 10). The line represents y = kx. Write down the ratio x : y in its simplest form.
- 5.A fixed job of fitting a solar array is shared between installers, and the time taken is inversely proportional to the number of installers working on it at once. With 4 installers the job takes 18 hours. Construct the equation connecting time T and number of installers n, then work out how long it would take with 6 installers.
- 6.Write 2 m : 150 cm : 50 cm as a ratio of whole numbers in its simplest form.
- 7.A scale drawing of a garden uses a scale of 1 : 20. Write down a formula for the real length, L metres, in terms of the length on the drawing, d centimetres, given that 1 metre = 100 centimetres.
- 8.It takes 8 painters 6 days to paint a fence. Working at the same rate, work out how many days 4 painters would take to paint the same fence.
- 9.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?
- 10.An architect builds a scale model of a staircase before construction. The model is mathematically similar to the real staircase, at a scale of 1 : 10. The model covers a floor area of 0.4 m² on the plan. Work out the floor area the real staircase will cover, giving your answer in m².
- 11.5 decorators paint 3 rooms in 6 hours. All the decorators work at the same rate. Work out how long it takes 10 decorators to paint 5 rooms.
- 12.A recipe for 6 muffins needs 150 g of flour. Aisha wants to make 15 muffins using the same recipe. Work out how much flour she needs.
- 13.A fruit squash is made by mixing squash and water. In a jug, 2/9 of the total volume is squash. A caterer makes up 4.5 litres of the squash mixture. Work out how many litres of squash are in the mixture.
- 14.A sprinter's distance from the start line, in metres, is plotted against time, in seconds. A tangent to the graph at t = 2 seconds has gradient 6. A tangent at t = 8 seconds has gradient 9.5. Which statement correctly compares the sprinter's speed at these two times?
- 15.Two mathematically similar jugs have heights 8 cm and 12 cm. The smaller jug holds 200 ml when it is full. Work out how much the larger jug holds when it is full.
Answer key
- (a) 4 days — Method: the fence is a fixed amount of work, so builders × days is constant; find that product and divide it by the new number of builders. Working: 3 × 12 = 36 builder-days of work, so with 9 builders the time is 36 ÷ 9 = 4 days. Answer: 4 days. The distractors: 6 days comes from halving the 12 days because there are more builders, rather than dividing by the factor of 3 by which the workforce has grown; 36 days is the constant product of builders and days, given as a number of days instead of being shared between the builders; 9 days comes from taking 3 days off the 12, treating three extra builders as three fewer days, which is additive rather than proportional.
- (d) Gradient = acceleration; area = distance travelled. — Method: on a speed–time graph, the gradient of the graph at an instant is the rate of change of speed with time, which is acceleration; the area between the graph and the time-axis over an interval is the total distance covered in that interval, because it accumulates speed × time. Working: gradient = acceleration and area = distance travelled is the correct pairing. Swapping the two quantities completely, gradient = distance travelled and area = acceleration, is the reverse of what each actually measures. Keeping gradient = acceleration correct but then also claiming area = acceleration too is wrong because the area is a different physical quantity, distance, not a second way of finding the same rate. Claiming the gradient itself gives the speed confuses the RATE OF CHANGE of the plotted quantity with the plotted quantity itself — the gradient is how fast the speed is changing, not the speed. On any rate graph, the gradient of the graph is always the RATE at that instant, and the area under the graph is always the TOTAL AMOUNT accumulated — keep straight which of the two questions each one answers.
- (a) 20% — Method: percentage increase = increase ÷ original amount × 100. Working: the increase is 84 − 70 = 14 marks, and 14 ÷ 70 = 0.2, so 0.2 × 100 = 20. Answer: an increase of 20%. The distractors: 14% comes from quoting the 14 mark increase as though marks and per cent were the same thing; 17% comes from dividing the 14 by the new mean 84 instead of by the original 70, which gives 17% to the nearest per cent; 120% is the new mean written as a percentage of the old one, which is the whole of the new mean rather than the increase.
- (c) 2 : 5 — The point (4, 10) gives x = 4, y = 10, so x : y = 4 : 10. Dividing both parts by their highest common factor, 2, gives 2 : 5 in simplest form. Inverting the whole ratio gives 5 : 2, which is y : x instead of x : y. Dividing only the x-part by 2 and leaving the y-part as 10 gives 2 : 10, but scaling one part on its own changes the ratio: 2 : 10 is the same as 1 : 5, not 4 : 10. Dividing only the y-part by 2 and leaving the x-part as 4 gives 4 : 5, the same one-sided mistake made on the other part of the ratio.
- (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) 4:3:1 — Convert every part to the same unit: 2 m = 200 cm, so the ratio is 200 : 150 : 50. Dividing all three parts by 50 gives 4 : 3 : 1. Writing 2 : 150 : 50 has not converted 2 m into centimetres, so the units do not match. Writing 3 : 4 : 1 has the first two parts the wrong way round. Writing 4 : 3 : 2 comes from an arithmetic slip on the last part: 50 ÷ 50 = 1, not 2.
- (c) L = d/5 — The scale 1 : 20 means each cm on the drawing represents 20 cm in real life, so the real length in cm is 20d. Converting to metres by dividing by 100: L = 20d/100 = d/5.
- (b) 12 days — This is inverse proportion: fewer painters take longer. Multiply the original numbers to find the total painter-days needed: 8 × 6 = 48 painter-days. Divide by the new number of painters: 48 ÷ 4 = 12 days. Working out 6 × 4 ÷ 8 = 3 days treats it as direct proportion, as if fewer painters needed less time. Stopping at 48 gives the total painter-days, not the number of days. Working out 6 + (8 − 4) = 10 days adds the change in the number of painters straight onto the number of days, treating painters and days as the same kind of quantity. 4 painters take 12 days.
- (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.
- (a) 40 m² — A scale of 1 : 10 is a length scale factor of 10 from model to real. Areas scale with the square of the length scale factor: 10² = 100. 0.4 × 100 = 40, so the real staircase covers 40 m². Giving 4 m² uses the length scale factor, 10, without squaring it (0.4 × 10 = 4). Giving 0.04 m² divides by the scale factor instead of multiplying by its square (0.4 ÷ 10 = 0.04). Giving 400 m² cubes the scale factor, 10³ = 1000, as if area scaled like a volume (0.4 × 1000 = 400).
- (c) 5 hours — Method: measure the job in decorator-hours, which is in the same ratio as the number of rooms, then share the decorator-hours between the decorators available. Working: 5 decorators × 6 hours = 30 decorator-hours for 3 rooms, so one room takes 30 ÷ 3 = 10 decorator-hours; 5 rooms take 5 × 10 = 50 decorator-hours; shared between 10 decorators that is 50 ÷ 10 = 5 hours. Answer: 5 hours. The distractors: 3 hours comes from halving the 6 hours because the number of decorators doubles, while forgetting that there are also more rooms to paint; 10 hours comes from scaling the 6 hours up for the rooms only, 6 × 5 ÷ 3, and leaving the workforce at 5 decorators; 6 hours comes from assuming that doubling the decorators and increasing the rooms cancel each other out, which they do not, because the rooms rise by a factor of 5/3 and the workforce by a factor of 2.
- (a) 375 g — Find the amount of flour needed per muffin: 150 ÷ 6 = 25 g. Multiply by the new number of muffins: 25 × 15 = 375 g. (60 g comes from using the scale factor the wrong way round, 6/15 × 150. 150 g comes from not scaling the recipe at all. 300 g comes from rounding the scale factor, 15 ÷ 6, down to 2 before multiplying.)
- (d) 1.00 litres — Squash is 2/9 of the mixture, so the squash volume is 4.5 × 2/9 = 1.00 litres. Using the water's fraction, 7/9, instead of squash's gives 4.5 × 7/9 = 3.50 litres — the volume of water, not squash. Dividing 4.5 by 9 but forgetting to multiply by the numerator 2 gives 4.5 ÷ 9 = 0.50 litres, which is only 1/9 of the mixture. Halving the total volume instead of applying the fraction 2/9 gives 4.5 ÷ 2 = 2.25 litres, which assumes the mixture is half squash.
- (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.
- (a) 675 ml — How much a jug holds is a volume, and volumes of similar solids scale with the cube of the length scale factor. The length scale factor is 12 ÷ 8 = 1.5, so the volume scale factor is 1.5 × 1.5 × 1.5 = 3.375. The larger jug holds 200 × 3.375 = 675 ml. Multiplying the scale factor by 3 instead of raising it to the power 3 is the mistake to guard against here.
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