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 pump fills a paddling pool at a rate of 130 litres per hour. Work out the rate in litres per minute, to 2 decimal places.
- 2.A kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
- 3.A machine is bought for £8500. Its value depreciates by 6% each year. Work out the value of the machine after 3 years, to the nearest pound.
- 4.A tank is filled at a rate of 3 litres every 2 minutes. Given that 1 litre = 1000 cm³, work out the flow rate in cm³ per minute.
- 5.A scale model of a shipping container is built at a scale of 1 : 30, using material with the same density as the real container. The model has a mass of 400 g. Work out the mass of the real container, giving your answer in kilograms.
- 6.For the pairs x = 6, y = 15 and x = 10, y = 25, which statement is correct?
- 7.A lorry is carrying a load of mass m tonnes. Given that 1 tonne = 1000 kg, write down an expression for the mass of the load in kilograms.
- 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.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.
- 10.A metal block has a mass of 270 g and a volume of 30 cm³. Work out the density of the block, in g/cm³.
- 11.Aisha invests £3200 in Account A, which pays 5% compound interest each year. She also invests £3200 in Account B, which pays 3% simple interest each year. Work out how much more Account A is worth than Account B after 2 years.
- 12.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.
- 13.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.
- 14.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.
- 15.The number of euros, e, received is directly proportional to the number of pounds, p, exchanged. Exchanging £40 gives 46 euros. Work out how many euros are received for £65, giving your answer to the nearest euro.
Answer key
- (a) 2.17 litres per minute — There are 60 minutes in an hour, so to convert litres per hour to litres per minute you divide by 60: 130 ÷ 60 = 2.1666..., which rounds to 2.17 litres per minute. Multiplying by 60 instead of dividing gives 130 × 60 = 7800.00 litres per minute, using the conversion factor the wrong way round. Leaving the rate unchanged, 130.00, ignores that 'per hour' and 'per minute' are different units. Dividing by 50 instead of 60, misremembering the number of minutes in an hour, gives 130 ÷ 50 = 2.60 litres per minute.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (d) £7060 — A 6% decrease each year means the value becomes 100% − 6% = 94% of the previous year's value, and 94% = 0.94, so the multiplier is 0.94. Apply it once for each of the 3 years: £8500 × 0.94 = £7990 after 1 year, £7990 × 0.94 = £7510.60 after 2 years, £7510.60 × 0.94 = £7059.96 after 3 years, which rounds to £7060 to the nearest pound. (£6970 comes from using simple depreciation instead of compound, taking 6% of the original £8500 three times: £8500 − 3 × £510 = £6970. £7990 is the value after only 1 year, forgetting the remaining 2 years. £7511 is the value after only 2 years, £8500 × 0.94² = £7510.60, forgetting the third year.)
- (b) 1500 — The rate is 3 ÷ 2 = 1.5 litres per minute. Converting to cm³: 1.5 × 1000 = 1500 cm³ per minute. Getting 3000 comes from converting 3 litres to cm³ first (3000 cm³) and forgetting to divide by the 2 minutes. Getting 750 comes from dividing by the 2 minutes a second time after converting (1500 ÷ 2). Getting 2000 comes from converting the 2 minutes as if it were litres (2 × 1000) instead of using the correct rate of 1.5 litres per minute.
- (c) 10,800 kg — Method: since the model and the real container are similar and made of the same material, mass scales with volume, so the mass scale factor is the length scale factor cubed. Working: 30³ = 27,000, so the real container's mass is 400 × 27,000 = 10,800,000 g, which is 10,800,000 ÷ 1,000 = 10,800 kg. Answer: 10,800 kg. 12 kg comes from using the length scale factor directly, 400 × 30 = 12,000 g, without cubing it. 360 kg comes from squaring the length scale factor instead of cubing it, 400 × 30² = 360,000 g. 10,800,000 kg comes from correctly cubing the scale factor but then forgetting to convert the mass from grams into kilograms.
- (c) They are in direct proportion, because y ÷ x = 2.5 for both pairs. — Testing direct proportion means checking that y ÷ x is the same for every pair: 15 ÷ 6 = 2.5 and 25 ÷ 10 = 2.5, so the quantities are in direct proportion. Saying they are not in proportion because x + y differs uses addition, which is not the correct test for proportion. Saying they are not in proportion because y − x differs also uses the wrong test — subtraction, not division. Saying they are in proportion because x × y is 90 and 250 uses multiplication, which is the test for inverse proportion, and the two products are not even equal to each other, so this option also contradicts itself.
- (a) 1000m — Method: kilograms are a smaller unit than tonnes, so change tonnes into kilograms by multiplying by 1000. Working: m tonnes = m × 1000 kg = 1000m kg. So the expression is 1000m. Distractor m/1000 comes from dividing by 1000 instead of multiplying, which would make the number of kilograms smaller than the number of tonnes, the wrong way round. Distractor 1000 + m comes from adding the conversion factor instead of multiplying by it. Distractor m − 1000 comes from subtracting the conversion factor instead of multiplying by it.
- (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.
- (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.
- (c) 9 g/cm³ — Density = mass ÷ volume, so 270 ÷ 30 = 9 g/cm³. Working out 270 − 30 = 240 subtracts the volume from the mass instead of dividing. Writing 270 gives the mass on its own, without dividing by the volume at all. Working out 270 × 30 = 8100 multiplies mass and volume together instead of dividing. The density of the block is 9 g/cm³.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (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.
- (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.
- (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).
- (b) 75 — The exchange rate is constant: k = 46 ÷ 40 = 1.15 euros per pound. For £65, the number of euros is 1.15 × 65 = 74.75, which rounds to 75 euros. Getting 74 comes from rounding 74.75 down instead of to the nearest whole number. Getting 57 comes from using the reciprocal rate (40 ÷ 46) instead of 46 ÷ 40. Getting 71 comes from adding the difference between 65 and 40 (25) onto 46 instead of using the proportional rate.
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