Printable · GCSE Higher · ages 14-16
Ratio, proportion and rates of change worksheet — GCSE Higher
Fifteen questions across the ratio, proportion and rates of change statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
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Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
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- (b) 13/2 — Put the number of pages in the novel over the number of pages in the comic: 312/48. Divide both numbers by their highest common factor, 24: 312÷24 = 13, 48÷24 = 2, giving 13/2. (2/13 comes from writing the page counts the wrong way round. 11/2 comes from finding the difference in the page counts, 312 − 48 = 264, and writing it as a fraction of the comic's page count, 264/48. 13/15 comes from comparing the novel's page count to the total number of pages in both books, 312/360.)
- (d) 25 minutes — Method: find the time for one kilometre, then multiply by the number of kilometres — the unitary method with a rate. Working: 10 ÷ 2 = 5 minutes per km, and 5 × 5 = 25. Answer: 25 minutes. The distractors: 20 minutes comes from multiplying the 10 minutes by 2, the distance in the given rate, instead of by the scale factor 2.5; 50 minutes comes from multiplying 10 by 5, treating the 10 minutes as the time for a single kilometre; 15 minutes comes from adding the 5 km on to the 10 minutes, adding quantities that are not the same kind.
- (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) It's an average over 20 days, which may miss the day-5 rate. — Method: a chord's gradient is the AVERAGE rate of change across the whole interval it spans; it only closely approximates the INSTANTANEOUS rate of change at a point inside that interval when the rate of change is roughly constant across the interval, which usually means the interval needs to be short. Working: here the chord spans 20 days while the point of interest, t = 5, is only a quarter of the way along it, so if the reservoir's level rose or fell at different rates over that time, the chord's gradient will not be close to the true gradient of the curve at t = 5 — this is the correct reason. Claiming the chord's gradient needs the water level at every day in between is wrong: a chord's gradient needs only the two endpoint values, at t = 0 and t = 20. Claiming a chord can only estimate the rate at its own endpoints is wrong: a chord between two points can be used to estimate the instantaneous rate of change at any point inside the interval, including one that is not an endpoint — that is exactly the technique being used here, and it is the SIZE of the interval that makes the estimate poor, not the fact that t = 5 is an interior point. Claiming the units do not match is wrong: the chord's gradient and the instantaneous rate of change are both measured in metres per day, so the units are the same. A chord is only a good estimate of an instantaneous rate when the interval it spans is short enough that the rate does not change much within it — always check how long the interval is compared with how far it is to the point you actually want.
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (b) Car A, 50 km/h — Method: speed = distance ÷ time for each car, then compare. Working: Car A = 150 ÷ 3 = 50 km/h. Car B = 180 ÷ 4 = 45 km/h. Since 50 > 45, Car A is faster, travelling at 50 km/h. Wrong options: Car B, 45 km/h correctly finds Car B's speed but wrongly names the slower car as faster; Car A, 45 km/h picks the correct car but uses Car B's speed by mistake; Car B, 50 km/h picks the wrong car but uses Car A's correct speed value.
- (c) 1 : 27 — The edge lengths are in the ratio 2 : 6, which simplifies to 1 : 3. Volumes scale with the cube of the length ratio, so the volume ratio is 1³ : 3³ = 1 : 27. Giving 1 : 3 uses the length ratio without cubing it. Giving 1 : 9 squares the length ratio, which is the rule for areas, instead of cubing it, which is the rule for volumes. Giving 27 : 1 has the ratio the right way round for larger to smaller, not smaller to larger as the question asks.
- (b) 4/3 — Work out the number of bananas: 3/4 of 16 = 12. Form the fraction 16/12 (apples over bananas); both numbers share a factor of 4, so 16 ÷ 4 = 4 and 12 ÷ 4 = 3, giving 4/3. 3/4 comes from giving the fraction that was already stated in the question (bananas as a fraction of apples), instead of inverting it. 4/1 comes from reading '3/4 of the apples' as 'the apples reduced by 3/4', giving only 4 bananas, then 16/4 = 4. 4/7 comes from comparing the apples with the total number of pieces of fruit (16/28), instead of with the bananas alone.
- (b) 1:200000 — '1 cm represents 2 km' means 1 cm on the map is 2 km in real life. Convert 2 km into centimetres, the same unit as the 1: 2 km = 2000 m = 200 000 cm. So the scale is 1 : 200 000. Converting only as far as metres gives 1 : 2000 — the conversion to centimetres was never finished. Losing a zero in the conversion gives 1 : 20 000, ten times too small. Writing 1 : 2 without converting units at all compares 1 cm to 2 km directly, which is not a valid ratio since the units do not match.
- (c) 4 km — Since signal strength is inversely proportional to the square of the distance, S = k/d². Using d = 2, S = 20: 2² = 4, so 20 = k ÷ 4, giving k = 20 × 4 = 80. The equation is S = 80/d². When S = 5: d² = 80 ÷ 5 = 16, so d = 4 (taking the positive root, since distance cannot be negative). Stopping at d² = 16 without taking the square root leaves 16, the square of the distance, not the distance itself. Treating the relationship as inversely proportional to distance itself, rather than to its square, gives k = 20 × 2 = 40 and then d = 40 ÷ 5 = 8, a different relationship. Multiplying by S instead of dividing by it when isolating d² gives d² = 80 × 5 = 400 and d = 20, the wrong operation. The distance at which the signal strength is 5 units is 4 km.
- (d) 8/5 — Two masses can only be compared once they are in the same unit. Since 1 kg is 1000 g, the recipe needs 1200 g. The recipe's mass is being written as a fraction of Dan's mass, so 1200 goes on the top and 750 on the bottom, giving 1200/750. The highest common factor of the two is 150: 1200 ÷ 150 = 8 and 750 ÷ 150 = 5. The fraction is 8/5, which is greater than 1 because the recipe needs more flour than Dan has.
- (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) £82.50 — Find the hourly rate: £52.50 ÷ 7 = £7.50 per hour. For 11 hours: 11 × £7.50 = £82.50. £30 comes from working out the pay for only the extra 4 hours (4 × £7.50), and forgetting to include the original £52.50. £99 comes from misremembering the hourly rate as £9 instead of £7.50, then 11 × £9. £56.50 comes from adding the extra number of hours (4) straight onto the pay in pounds (52.5 + 4), confusing hours with pounds.
- (b) 35 N/m² — Pressure = force ÷ area. 84 ÷ 2.4 = 35 N/m². 201.6 N/m² comes from multiplying the force by the area instead of dividing (84 × 2.4). 81.6 N/m² comes from subtracting the area from the force (84 − 2.4) instead of dividing. 0.03 N/m² comes from dividing the area by the force instead of the force by the area (2.4 ÷ 84).
- (d) 7/4 — A part-to-part ratio a : b gives the fraction a/b when the first quantity is written as a fraction of the second, so 7 : 4 gives 7/4. Writing 4/7 puts the parts the wrong way round — blue as a fraction of red, not red as a fraction of blue. Writing 7/11 uses the total number of counters, 7 + 4 = 11, as the denominator instead of the number of blue counters — that is red as a fraction of the whole bag, not red as a fraction of blue. Writing 11/7 has both the wrong denominator and the parts inverted.
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