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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
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- (b) 9 — Method: divide the known quantity by its ratio number to find the value of one part, then multiply by the other ratio number. Working: 21 ÷ 7 = 3 (value of one part). Red beads = 3 × 3 = 9. Wrong options: 49 comes from dividing by the wrong ratio number (21 ÷ 3 × 7); 24 comes from adding the ratio number for red (3) to 21 instead of scaling; 14 comes from subtracting the ratio number for blue (7) from 21 instead of scaling.
- (c) 5/7 — First find the number of children: 84 − 35 = 49. The question compares the adults with the children, not with everyone on the bus, so the denominator is 49 and the numerator is 35, giving 35/49. Both parts divide by 7: 35 ÷ 7 = 5 and 49 ÷ 7 = 7. In its simplest form the fraction is 5/7.
- (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.)
- (c) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
- (a) 150 g — Method: scale the recipe to find the total sugar needed, then subtract the sugar Sam already has. Working: 200 ÷ 8 × 20 = 500, so 500 g is needed in total; 500 − 350 = 150, so 150 g still to buy. Stopping after finding the total, 500, without subtracting what he has gives 500 g. Scaling the wrong way round, 200 × 8 ÷ 20 = 80, wrongly suggests he already has enough, giving 0 g. Adding the amount he has instead of subtracting it, 500 + 350 = 850, gives 850 g.
- (d) 3/8 — The ratio 3 : 5 has 3 + 5 = 8 parts in total. The first part as a fraction of the whole is 3 out of 8, or 3/8. Writing 3/5 gives the first part compared to the second part, not to the whole. Writing 5/8 gives the second part as a fraction of the whole, not the first. Writing 8/3 has the fraction upside down — the whole must be on the bottom.
- (a) 3/2 — Find each average speed: car = 180 ÷ 3 = 60 mph; lorry = 160 ÷ 4 = 40 mph. Put the car's speed over the lorry's speed: 60/40. Divide both numbers by their highest common factor, 20: 60÷20 = 3, 40÷20 = 2, giving 3/2. (2/3 comes from writing the speeds the wrong way round. 9/8 comes from comparing the distances travelled, 180/160, without working out the speeds. 3/4 comes from comparing the times taken, 3/4, instead of the speeds.)
- (b) 450.00 kg — 1 m³ = 100 × 100 × 100 = 1,000,000 cm³, so 0.5 m³ = 500,000 cm³. Mass = density × volume = 0.9 × 500,000 = 450,000 g. Converting to kilograms by dividing by 1000 gives 450,000 ÷ 1000 = 450.00 kg. Skipping the m³-to-cm³ conversion and multiplying 0.9 × 0.5 = 0.45 treats the volume as if it were already 0.5 cm³, giving 0.45 kg. Finding the mass correctly in grams, 450,000 g, but not converting to kilograms leaves 450000.00 kg, out by a factor of 1000. Using the area conversion factor of 10,000, as if converting m² to cm², instead of the volume factor of 1,000,000 gives 0.5 × 10,000 = 5,000 'cm³', and a mass of 0.9 × 5,000 = 4,500 g, which is 4.50 kg.
- (d) y = 3x/4 — y : x = 3 : 4 means y/x = 3/4. Rearranging to make y the subject gives y = (3/4)x = 3x/4. A student who mixes up which quantity goes on top gets y = 4x/3. A student who treats the ratio numbers as the coefficient and constant of a linear equation instead of a proportional relationship gets y = 3x + 4. A student who mistakes the relationship for inverse proportion gets y = 3/(4x).
- (b) £10 — Method: find the cost of 1 kg by dividing, then multiply by the mass wanted — the unitary method for direct proportion. Working: £6 ÷ 3 = £2 per kg, and £2 × 5 = £10. Answer: £10. The distractors: £11 comes from adding 5 to the £6 instead of scaling; £30 comes from multiplying £6 by 5 without first dividing by 3; £3.60 comes from turning the proportion upside down, dividing by 5 and multiplying by 3.
- (d) 2:3:5 — The highest common factor of 12, 18 and 30 is 6. Divide every part by 6: 12 ÷ 6 = 2, 18 ÷ 6 = 3, 30 ÷ 6 = 5, giving 2 : 3 : 5. Dividing by 2 instead of 6 gives 6 : 9 : 15, which still shares a common factor of 3, so it is not fully simplified. Dividing by 3 instead of 6 gives 4 : 6 : 10, which still shares a common factor of 2, so it is not fully simplified either. Swapping the first two parts gives 3 : 2 : 5, the parts in the wrong order.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
- (a) 135 km — Method: find the distance travelled on one litre, then scale up to 9 litres. Working: 90 ÷ 6 = 15 km per litre, so 15 × 9 = 135 km. Answer: 135 km. 45 km comes from working out the extra distance for the extra 3 litres (15 × 3) but forgetting to add the original 90 km. 60 km comes from using the ratio the wrong way round, 90 × 6 ÷ 9, instead of finding the rate per litre first. 99 km comes from simply adding the number of litres, 9, onto the original distance, 90, instead of scaling the whole journey.
- (c) 2/7 — Method: each number in a ratio counts parts, so the whole is all the parts added together, and the fraction wanted is the cement parts over that total. Working: the mix has 5 + 2 = 7 parts altogether, and 2 of those parts are cement, so the cement is 2/7 of the mix. Answer: 2/7. The distractors: 2/5 comes from writing the cement parts over the sand parts, comparing one part with the other part instead of with the whole; 5/7 comes from writing the sand parts over the total, which answers about the wrong material; 7/2 comes from writing the total over the cement parts, turning the fraction upside down.
- (c) 6.00 m — Method: the ratio of height to shadow length is the same for both objects. Working: road sign height ÷ shadow = 3 ÷ 5 = 0.6. Lamppost height = 0.6 × 10 = 6.00 m. Wrong options: 16.67 m comes from inverting the ratio, using shadow ÷ height instead of height ÷ shadow (10 × 5 ÷ 3); 8.00 m comes from adding the difference between the two shadow lengths to the road sign's height instead of scaling (3 + (10 − 5)); 1.50 m comes from multiplying by the ratio of the two shadow lengths the wrong way round (3 × 5 ÷ 10).
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