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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

Maths · GCSE Foundation · Ratio, proportion and rates of change · Calculator · 15 questions

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1.The time taken for a train journey is inversely proportional to the average speed of the train. At an average speed of 60 km/h the journey takes 2 hours. Work out the time taken at an average speed of 40 km/h.
(a)2 hours
(b)1.5 hours
(c)3 hours
(d)1 hour 20 minutes
2.Write 3 hours 45 minutes as a decimal number of hours.
(a)3.67
(b)4.15
(c)3.45
(d)3.75
3.A cyclist rides 17.5 km on Saturday and 12.25 km on Sunday. Write the distance ridden on Saturday as a fraction of the distance ridden on Sunday, giving your answer in its simplest form.
(a)10/7
(b)7/10
(c)10/17
(d)3/7
4.A shop sells rope by the metre. 3 m costs £7.50 and 5 m costs £11.00. Does this data show that the cost is directly proportional to the length of rope bought? Choose the correct verdict and reason.
(a)No — the cost per metre differs: £2.50/m vs £2.20/m
(b)Yes — because both amounts increase as length increases
(c)No — because 5 m costs more than 3 m in total
(d)Yes — the cost per metre is £2.50 in both cases
5.The mass of a length of copper pipe is proportional to its length. A pipe of length 4.5 m has a mass of 12.6 kg. Work out the mass of a pipe of the same type of length 9.5 m.
(a)12.6
(b)3.4
(c)17.6
(d)26.6
6.A coach travels the 54 miles from London to Brighton in 1 hour 30 minutes. Work out the average speed of the coach, in mph.
(a)1.8 mph
(b)41.54 mph
(c)81 mph
(d)36 mph
7.In a fruit bowl, the ratio of apples to oranges is 5 : 3. Work out what percentage of the fruit in the bowl is apples.
(a)37.5%
(b)62.5%
(c)160%
(d)50%
8.A smoothie recipe combines 350 cm³ of mango juice with 650 cm³ of orange juice. Work out the total volume of the smoothie in litres.
(a)0.35 litres
(b)1.00 litres
(c)0.65 litres
(d)1000.00 litres
9.A laptop costs £800 when new. Its value decreases by 25% of its value at the start of each year. Work out how much value the laptop loses in the second year.
(a)£150
(b)£450
(c)£350
(d)£200
10.Two quantities x and y are in direct proportion. When x = 8, the value of y is 20. Work out the value of y when x = 14.
(a)26
(b)5.6
(c)28
(d)35
11.A toy manufacturer makes a model aircraft that is mathematically similar to the real aircraft, at a scale of 1 : 48. The wingspan of the model is 15 cm. Work out the wingspan of the real aircraft, giving your answer in metres.
(a)72 m
(b)7.2 m
(c)0.72 m
(d)0.31 m
12.Two broadband providers show their monthly cost on straight-line graphs, with the cost in pounds on the vertical axis and the data used in gigabytes on the horizontal axis. Provider X's line passes through (0, 15) and (100, 35). Provider Y's line passes through (0, 5) and (100, 45). Work out which provider is cheaper for a customer who uses 40 gigabytes in a month.
(a)Provider Y — £45.00 against Provider X's £35.00
(b)Provider Y — £21.00 against Provider X's £23.00
(c)Provider X — £15.00 against Provider Y's £5.00
(d)Provider X — £23.00 against Provider Y's £21.00
13.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.
(a)110 km
(b)0.22 km
(c)220 km
(d)0.11 km
14.y is directly proportional to x. When x = 5, y = 18. Work out the value of y when x = 15.
(a)60
(b)28
(c)54
(d)6
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.
(a)675 ml
(b)300 ml
(c)450 ml
(d)900 ml

Answer key

(c) 3 hours — Method: inverse proportion means speed × time is constant for the journey, so find that constant and divide it by the new speed. Working: 60 × 2 = 120, which is the distance in kilometres; at 40 km/h the time is 120 ÷ 40 = 3 hours. Answer: 3 hours. The distractors: 1.5 hours is the ratio of the speeds, 60 ÷ 40, given as a time instead of being used to scale the original 2 hours; 1 hour 20 minutes comes from treating time as directly proportional to speed, 2 × 40 ÷ 60, which has the slower train arriving sooner; 2 hours comes from finding the constant 120 and then dividing it by the original 60 km/h again, so the time never changes.
(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) 10/7 — Put Saturday's distance over Sunday's distance: 17.5/12.25. Multiply both numbers by 100 to clear the decimals: 1750/1225. Divide both by their highest common factor, 175: 1750÷175 = 10, 1225÷175 = 7, giving 10/7. (7/10 comes from writing the distances the wrong way round. 3/7 comes from finding the difference, 17.5 − 12.25 = 5.25 km, and writing it as a fraction of Sunday's distance, 5.25/12.25. 10/17 comes from comparing Saturday's distance to the total distance ridden, 17.5/29.75.)
(a) No — the cost per metre differs: £2.50/m vs £2.20/m — Method: divide cost by length for each pair and compare the unit rates. Working: £7.50 ÷ 3 = £2.50 per m; £11.00 ÷ 5 = £2.20 per m. The rates are different, so this is NOT direct proportion. Wrong options: 'Yes — both amounts increase' wrongly assumes any increasing relationship is proportional; 'No — because 5 m costs more in total' judges by total cost rather than the rate per metre, which is not valid reasoning on its own; 'Yes — the cost per metre is £2.50 in both cases' miscalculates the second rate (11.00 ÷ 5 is £2.20, not £2.50).
(d) 26.6 — Find the constant multiplier — the mass of each metre of pipe: 12.6 ÷ 4.5 = 2.8, so the mass is always 2.8 times the length. For a length of 9.5 m, the mass is 9.5 × 2.8 = 26.6 kg. 17.6 comes from assuming an additive relationship instead of a multiplicative one — adding the increase in length (9.5 − 4.5 = 5) onto 12.6. 3.4 comes from using the multiplier the wrong way round (4.5 ÷ 12.6, rounded to 1 d.p.), then multiplying by 9.5. 12.6 comes from simply repeating the given mass, without applying the multiplier to the new length.
(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.
(b) 62.5% — Total parts = 5 + 3 = 8. Apples make up 5 parts, so the percentage is 5/8 × 100 = 62.5%. A student who finds the oranges' share instead gets 3/8 × 100 = 37.5%. A student who assumes an even split gets 50%. A student who inverts the fraction gets 8/5 × 100 = 160%.
(b) 1.00 litres — Total volume = 350 + 650 = 1000 cm³. Since 1000 cm³ = 1 litre, the smoothie is 1.00 litre. Using only the orange juice's 650 cm³ and converting that gives 0.65 litres, forgetting the mango juice entirely. Using only the mango juice's 350 cm³ gives 0.35 litres, forgetting the orange juice. Adding both volumes correctly to get 1000 cm³ but not converting to litres leaves the answer as 1000.00, which is the volume in the wrong unit.
(a) £150 — Value after year 1: £800 × 0.75 = £600. Value after year 2: £600 × 0.75 = £450. The loss during the second year alone is £600 − £450 = £150. £450 comes from giving the value remaining after 2 years, not the amount lost during the second year. £200 comes from working out the loss during the first year instead of the second: £800 − £600 = £200. £350 comes from working out the total loss over both years instead of just the second year's loss: £800 − £450 = £350.
(d) 35 — Method: in direct proportion the ratio y : x is the same for every pair, so find the constant and substitute the new value of x. Working: k = 20 ÷ 8 = 2.5, so y = 2.5x; when x = 14, y = 2.5 × 14 = 35. Answer: 35. The distractors: 26 comes from additive thinking — x rises by 6, so 6 is added to y — which would keep the difference constant rather than the ratio; 28 comes from rounding the constant 2.5 down to 2 and working out 2 × 14, which loses the half in the constant; 5.6 comes from using the constant upside down, 8 ÷ 20 = 0.4, and working out 0.4 × 14.
(b) 7.2 m — Multiply the model wingspan by the scale factor: 15 × 48 = 720. This is in centimetres, and 720 cm = 7.2 m, since 1 m = 100 cm. Giving 0.31 m divides by the scale factor instead of multiplying (15 ÷ 48 ≈ 0.31), scaling the model down rather than the real aircraft up. Giving 72 m converts centimetres to metres by dividing by 10 instead of 100. Giving 0.72 m converts by dividing by 1000 instead of 100.
(b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
(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.
(c) 54 — Method: y = kx, so k = y ÷ x. Working: k = 18 ÷ 5 = 3.6. At x = 15: y = 3.6 × 15 = 54. Wrong options: 28 comes from adding the change in x (10) onto y instead of scaling; 6 comes from treating the relationship as inverse proportion (k = 5 × 18 = 90, then y = 90 ÷ 15 = 6); 60 comes from rounding the constant up to 4 instead of using 3.6.
(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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