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.
Calculator
Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
MathsUKwww.geekhero.co.uk
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (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.
- (c) 120 km/h — Method: for a fixed distance the average speed multiplied by the time is constant, and that constant is the distance, so divide the distance by the new time. Working: speed × time = 240, so in 2 hours the speed needed is 240 ÷ 2 = 120 km/h. Answer: 120 km/h. The distractors: 80 km/h is the average speed of the original journey, 240 ÷ 3, which answers for the 3-hour timing rather than the 2-hour one; 160 km/h comes from halving the 3 hours to 1.5 hours and working out 240 ÷ 1.5, instead of using the 2 hours the question gives; 480 km/h comes from multiplying the distance by the 2 hours rather than dividing by it.
- (c) C = 1.5n — Method: a fixed ratio between C and n means C is always the same multiple of n, and that multiple is the cost of one bottle. Working: 3.00 ÷ 2 = 1.5, 7.50 ÷ 5 = 1.5 and 12.00 ÷ 8 = 1.5, so every bottle costs £1.50 and C = 1.5n. Answer: C = 1.5n. The distractors: C = n + 1 comes from subtracting on the first row, 3 − 2 = 1, and adding that difference instead of multiplying; it fits the first row and fails the other two, which is why three rows are given; C = 3n reads the £3.00 as the price of one bottle when it is the price of two; C = n/1.5 divides the number of bottles by the price of one bottle, which works out how many bottles a pound buys instead of what n bottles cost.
- (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.
- (b) 1 : 500 — smallest real distance per cm (5 m), most detail — Method: compare what one centimetre represents in real life for each scale — the scale with the smallest real distance per cm shows the most detail. Working: for 1 : 500, 1 cm represents 500 cm (5 m); for 1 : 5000, 1 cm represents 50 m; for 1 : 50 000, 1 cm represents 500 m. Since 5 m is the smallest, 1 : 500 shows the most detail. Wrong options: '1 : 50 000 — covers the largest real area' wrongly assumes covering more area means more detail, when it is the opposite; '1 : 5000 — the middle value' wrongly assumes the middle scale is automatically the most balanced; '1 : 500 — covers the largest real distance' picks the correct scale but states an incorrect fact, since 1 : 500 actually covers the smallest real distance per cm.
- (b) 7980 — After the first year: 8000 × 0.95 = 7600. After the second year: 7600 × 1.05 = 7980. 8000 comes from assuming a 5% decrease followed by a 5% increase returns exactly to the starting number — it does not, because the increase acts on the smaller, already-reduced number. 8400 comes from applying only the second year's 5% increase to the original number: 8000 × 1.05 = 8400. 7600 comes from applying only the first year's 5% decrease and stopping there, without applying the second year's increase.
- (d) d ÷ t — Average speed = distance ÷ time, so the expression is d ÷ t. Writing t ÷ d inverts the formula, giving the time per kilometre instead of the speed. Writing d × t confuses speed with the formula for distance travelled (distance = speed × time) used the wrong way round. Writing d + t treats the relationship as additive instead of using division.
- (d) 3:8 — Convert 2 hours to minutes: 2 hours = 120 minutes. The ratio is 45 : 120. The highest common factor of 45 and 120 is 15. Divide both parts by 15: 45 ÷ 15 = 3 and 120 ÷ 15 = 8, giving 3 : 8. Leaving the hours unconverted gives 45 : 2 — the units on each side are different, so this does not compare like with like. Dividing by 5 instead of 15 gives 9 : 24, which still shares a common factor of 3, so it is not fully simplified. Swapping the order gives 8 : 3, hours to minutes instead of minutes to hours.
- (a) £40.32 — Find the cost per square metre from the rate given: £14.40 ÷ 20 = £0.72 per m². Then multiply by the area to be covered: £0.72 × 56 = £40.32. Working out 14.40 × 20 ÷ 56 ≈ £5.14 uses the ratio the wrong way round, scaling down as if 56 m² needed less paint than 20 m². Stopping at £0.72 only gives the cost per square metre, not the cost for the whole wall. Working out 14.40 + (56 − 20) = £50.40 adds the extra square metres straight onto the cost in pounds, treating square metres and pounds as the same kind of quantity. Covering 56 m² costs £40.32.
- (c) Map A, where the distance is 40 cm — 10 km = 1,000,000 cm. On Map A: 1000000 ÷ 25000 = 40 cm. On Map B: 1000000 ÷ 50000 = 20 cm. Since 40 cm is longer than 20 cm, the same real distance appears longer on Map A, the map with the smaller scale number. 'Map B, where the distance is 20 cm' has the correct working for Map B but names the wrong map as the one with the longer length. 'Map A, where the distance is 20 cm' correctly identifies Map A but pairs it with Map B's length. 'Map B, where the distance is 40 cm' correctly identifies Map A's length but attaches it to the wrong map.
- (a) 5/12 — Work out the distance still to travel: 372 − 217 = 155 miles. Form the fraction 155/372; both numbers share a factor of 31, so 155 ÷ 31 = 5 and 372 ÷ 31 = 12, giving 5/12. 7/12 comes from writing the distance already travelled as the fraction of the journey (217/372 = 7/12), instead of the distance still to travel. 145/372 comes from miscalculating 372 − 217 as 145 instead of 155. 5/7 comes from comparing the remaining distance with the distance already travelled (155/217 = 5/7), instead of with the total journey.
- (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.
- (b) 19.0 g/cm³ — Density = mass ÷ volume. 342.6 ÷ 18 = 19.0333…, which rounds to 19.0 g/cm³ (1 d.p.). 6166.8 g/cm³ comes from multiplying the mass by the volume instead of dividing (342.6 × 18). 324.6 g/cm³ comes from subtracting the volume from the mass (342.6 − 18) instead of dividing. 0.1 g/cm³ comes from dividing the volume by the mass instead of the mass by the volume (18 ÷ 342.6 = 0.0525…, rounded to 1 d.p.).
Build your own mix at the worksheet builder.