Printable · GCSE Foundation · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Foundation
Fifteen questions across the ratio, proportion and rates of change statements at Foundation tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Ratio, proportion and rates of change worksheet — GCSE Foundation
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- 1.A garden plan is enlarged by a scale factor of 3. A flower bed on the original plan is 5 cm long. Work out the length of the flower bed on the enlarged plan, in centimetres.
- 2.A block of wood has a mass of 60 g and a volume of 100 cm³. Work out the density of the block, in g/cm³.
- 3.A cyclist travels 18 km in 45 minutes. Work out the average speed, in km/h.
- 4.A train travels at a constant speed of 90 km/h. Work out the speed in m/s.
- 5.A cyclist rides at an average speed of 24 km/h for 1 hour 15 minutes. Work out the distance travelled, in km.
- 6.Write 3 hours 45 minutes as a decimal number of hours.
- 7.On a straight-line graph the volume of water, V litres, in a tank is plotted on the vertical axis and the time, t minutes, on the horizontal axis. The line passes through (2, 50) and (6, 130). Work out the gradient of the line and give its units.
- 8.A plumber charges a call-out fee plus an hourly rate. The total charge, C pounds, for a job lasting h hours is shown on a straight-line graph. The line passes through the points (2, 70) and (5, 130). Work out the call-out fee, in pounds.
- 9.A company's turnover this year is £180,000. Last year's turnover was £120,000. Write down this year's turnover as a percentage of last year's turnover.
- 10.A dog walker's charge, C pounds, for walking a dog for m minutes is shown on a straight-line graph. The line passes through the points (20, 14) and (50, 26). Work out the charge for a 65-minute walk.
- 11.Ffion is paid £11.20 per hour. She works 6 hours on Monday and 4.5 hours on Tuesday. Work out her total pay for the two days.
- 12.A conversion graph between kilometres and miles passes through the origin and the point (50, 31), with the number of kilometres on the horizontal axis and the number of miles on the vertical axis. Work out the number of miles equivalent to 1 kilometre.
- 13.In a fruit squash mix, the ratio of squash to water is 2 : 9. A shop wants to make up the mix using 3.4 litres of squash. Work out how many litres of water are needed.
- 14.A car is bought for £17,500. Its value decreases by 12% in the first year, and by a further 10% of its reduced value in the second year. Work out the value of the car at the end of the second year, giving your answer to the nearest pound.
- 15.A scale drawing of a park has a scale of 1 : 2500. On the drawing, the distance between the entrance and the lake is 4.4 cm. A jogger runs from the entrance to the lake and then back to the entrance. Work out the total distance the jogger runs, in kilometres.
Answer key
- (d) 15 cm — An enlargement by scale factor 3 means every length is multiplied by 3: 5 × 3 = 15 cm. Dividing by the scale factor instead, 5 ÷ 3 = 1.7 (to 1 d.p.), gives 1.7 cm — dividing shrinks the plan rather than enlarging it. Adding the scale factor instead of multiplying, 5 + 3 = 8, gives 8 cm. Squaring the scale factor, as you would for an area, 5 × 3² = 45, gives 45 cm — length scales by the factor itself, not its square.
- (d) 0.6 g/cm³ — Density = mass ÷ volume. 60 ÷ 100 = 0.6 g/cm³. 1.67 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (100 ÷ 60). 6000 g/cm³ comes from multiplying the mass by the volume instead of dividing (60 × 100). 40 g/cm³ comes from subtracting the mass from the volume (100 − 60) instead of dividing.
- (b) 24 km/h — First convert 45 minutes to hours: 45 ÷ 60 = 0.75 hours. Then divide the distance by the time: 18 ÷ 0.75 = 24 km/h. Reading 45 minutes as 0.45 hours (writing the minutes after the decimal point instead of dividing by 60) gives 18 ÷ 0.45 = 40 km/h. Working out 18 × 0.75 = 13.5 multiplies by the time instead of dividing. Working out 18 ÷ 45 = 0.4 divides by 45 without ever converting the minutes to hours. The cyclist's average speed is 24 km/h.
- (b) 25 m/s — Convert km/h to m/s by multiplying by 1000 (km to m) and dividing by 3600 (hours to seconds): 90 × 1000 ÷ 3600 = 25 m/s. Working out 90 ÷ 60 = 1.5 converts using 60, as if going from hours to minutes rather than to seconds. Working out 90 × 3.6 = 324 multiplies by 3.6 instead of dividing by it, going the wrong way between the units. Working out 90 × 1000 = 90000 converts kilometres to metres but forgets to convert hours to seconds at all. The train's speed is 25 m/s.
- (c) 30 km — First convert 1 hour 15 minutes to hours: 15 minutes is a quarter of an hour, so the time is 1.25 hours. Then multiply by the speed: 24 × 1.25 = 30 km. Reading 15 minutes as 0.15 hours (writing the minutes after the decimal point instead of as a fraction of 60) gives 24 × 1.15 = 27.6 km. Working out 24 × 75 = 1800 multiplies by the number of minutes instead of converting to hours first. Working out 24 ÷ 1.25 = 19.2 divides by the time instead of multiplying. The cyclist travels 30 km.
- (d) 3.75 — 45 minutes is 45 ÷ 60 = 0.75 of an hour, so 3 hours 45 minutes = 3.75 hours. Getting 3.45 comes from writing the minutes directly after the decimal point instead of converting them to a fraction of an hour. Getting 3.67 comes from misreading 45 minutes as 40 minutes and converting 40 ÷ 60 instead. Getting 4.15 comes from rounding 45 minutes up to the next whole hour and adding the remainder as if it were more minutes past that hour.
- (a) 20 litres per minute — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, and its units are the vertical unit for each one of the horizontal unit. Working: from (2, 50) to (6, 130) the volume changes by 130 − 50 = 80 litres and the time changes by 6 − 2 = 4 minutes, so the gradient is 80 ÷ 4 = 20, measured in litres for each minute. Answer: 20 litres per minute. The distractors: 25 litres per minute comes from using one point on its own, 50 ÷ 2, which assumes the line starts at the origin when the tank already held 50 litres at 2 minutes; 0.05 litres per minute comes from dividing the change in time by the change in volume, 4 ÷ 80, which gives the time for each litre but is then labelled as litres for each minute; 20 minutes for each litre has the right value with the units the wrong way round, and a tank that needed 20 minutes to gain a single litre would be filling far more slowly than this one.
- (d) £30.00 — The gradient is (130 − 70) ÷ (5 − 2) = 60 ÷ 3 = £20 per hour. Using the point (2, 70): the cost for 2 hours at £20 per hour is 20 × 2 = £40, so the call-out fee is 70 − 40 = £30. Taking the C-value of the first point as the fee without subtracting the hourly cost gives £70.00 — but that point already includes 2 hours of the hourly rate. Using the gradient itself as the fee, £20.00, confuses the rate per hour with the fixed charge. Subtracting 20 × 3 = 60 instead of 20 × 2 = 40 (using the wrong h-value) gives 70 − 60 = £10.00.
- (c) 150% — Percentage = (180,000 ÷ 120,000) × 100 = 150%.
- (c) £32 — First find the gradient: (26 − 14) ÷ (50 − 20) = 12 ÷ 30 = £0.40 per minute. Using the point (20, 14), the charge for 65 minutes is 14 + 0.40 × (65 − 20) = 14 + 18 = £32. Choosing £26 comes from treating the charge as directly proportional to the time, multiplying the gradient by 65 minutes and ignoring the fixed part of the charge (0.40 × 65 = 26). Choosing £40 comes from treating £14 as if it were the charge at 0 minutes, then adding the gradient multiplied by the full 65 minutes (14 + 0.40 × 65 = 40), instead of multiplying by the extra time past 20 minutes. Choosing £33.80 comes from assuming the charge is directly proportional to the minutes already known, scaling up from the point (50, 26) in the ratio 65:50 (65 ÷ 50 × 26 = 33.80).
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (a) 15.3 litres — Squash : water = 2 : 9, so water is 9 ÷ 2 = 4.5 times the amount of squash. Multiply: 3.4 × 4.5 = 15.3 litres. Using the multiplier upside down — treating squash as 9 ÷ 2 times water, when it is water that is 9 ÷ 2 times squash — and calculating 3.4 × (2 ÷ 9) gives about 0.8 litres (to 1 d.p.); that would be the squash needed for 3.4 litres of water, not the water needed for 3.4 litres of squash. Adding the difference between the ratio parts, 9 − 2 = 7, to the squash amount, 3.4 + 7 = 10.4, mistakes a ratio for a fixed extra amount. Using the total number of parts, 2 + 9 = 11, so the multiplier 11 ÷ 2 = 5.5, gives 3.4 × 5.5 = 18.7 litres — that finds the total mix from the squash amount, not the water alone.
- (b) £13,860 — Method: apply the first year's percentage decrease, then apply the second year's percentage decrease to the new value. Working: after the first year, the car is worth £17,500 × 0.88. Multiplying this result by 0.90 gives the value at the end of the second year, £13,860. Answer: £13,860. £13,650 comes from adding the two percentages together (12% + 10% = 22%) and applying a single 22% decrease, £17,500 × 0.78 = £13,650, instead of applying the decreases one after the other. £15,750 comes from applying only the second year's 10% decrease to the original price, forgetting the first year's decrease entirely, £17,500 × 0.90 = £15,750. £15,400 comes from applying only the first year's 12% decrease and stopping there, forgetting to apply the second year's decrease at all.
- (b) 0.22 km — The real one-way distance is 4.4 × 2500 = 11000 cm. Converting units: 11000 ÷ 100 = 110 m, and 110 ÷ 1000 = 0.11 km. Since the jogger runs there and back, the total distance is 0.11 × 2 = 0.22 km. 0.11 km comes from working out only the one-way distance and forgetting the return journey. 220 km comes from correctly doubling the one-way distance in metres, 110 × 2 = 220, but leaving it mislabelled as kilometres instead of converting metres to kilometres. 110 km comes from working out only the one-way distance in metres, 110, and mislabelling it as kilometres.
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