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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Answer key: Ratio, proportion and rates of change worksheet — GCSE Foundation
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- (a) 135 minutes — Method: to change hours into minutes, multiply by 60. Working: 2.25 × 60 = 135 minutes. So the film runs for 135 minutes. Distractor 145 minutes comes from reading the '.25' as 25 minutes instead of a quarter of an hour, giving 2 hours 25 minutes. Distractor 225 minutes comes from multiplying by 100 instead of 60. Distractor 150 minutes comes from rounding 2.25 hours to 2.5 hours before converting.
- (a) 5 — Pressure = force ÷ area = 20 ÷ 4 = 5 pascals. Getting 80 comes from multiplying the force and area instead of dividing. Getting 16 comes from subtracting the area from the force (20 − 4 = 16) instead of dividing. Getting 24 comes from adding the force and area (20 + 4 = 24) instead of dividing.
- (c) A straight line through the origin, rising from left to right — Two quantities are in direct proportion when one is a constant multiple of the other, so their graph is a straight line through the origin: when one quantity is 0 the other is 0 as well, and doubling one doubles the other. A straight line crossing the vertical axis at 5 has a fixed amount added on, so when the horizontal quantity is 0 the vertical quantity is 5, not 0 — a straight line on its own is not enough for direct proportion. A curve that falls steeply and then levels off without touching either axis shows inverse proportion: one quantity grows as the other shrinks, and their product stays the same. A horizontal line at a height of 3 shows a quantity that does not change at all as the other one grows, so it is not proportional to it.
- (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.
- (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.
- (a) 4 — Method: for inverse proportion, x × y always stays the same value. Working: when x = 5 and y = 8, the constant is 5 × 8 = 40. When x = 10, y = 40 ÷ 10 = 4. So y = 4. Distractor 16 comes from treating the relationship as direct proportion instead of inverse, working out 8 × 10 ÷ 5. Distractor 3 comes from assuming y decreases by the same amount that x increases, an additive rather than proportional idea. Distractor 0.8 comes from dividing the given y-value, 8, by the new x-value, 10, without first finding the constant.
- (a) 28.8 km/h — Method: first change metres per second into metres per hour, then change metres into kilometres. Working: 8 × 3600 = 28800 metres per hour, then 28800 ÷ 1000 = 28.8 km/h. So the runner's speed is 28.8 km/h. Distractor 28800 km/h comes from stopping after the first step and forgetting to change metres into kilometres. Distractor 2.22 km/h comes from dividing by 3600 instead of multiplying, then multiplying by 1000. Distractor 2.88 km/h comes from using 360 instead of 3600 seconds in an hour, missing a zero.
- (d) £8262 — A fall of 15% is a multiplier of 0.85 and a fall of 10% is a multiplier of 0.9, and each multiplier acts on the value at the start of its own year. After year 1: 12000 × 0.85 = 10200. After year 2: 10200 × 0.9 = 9180. After year 3: 9180 × 0.9 = 8262. The value 3 years after the car was bought is £8262. Adding the percentages to make a single fall of 35% would be wrong, because the later falls are taken from smaller values.
- (a) £85.47 — Value after 2 years: £1200 × 1.035 × 1.035 = £1285.47 (nearest penny). Interest earned = £1285.47 − £1200 = £85.47. £1285.47 is the total value of the account, not the interest earned on top of the original £1200. £84.00 comes from using simple interest instead of compound interest: £1200 × 0.035 × 2 = £84.00. £42.00 comes from working out only the first year's interest and stopping there: £1200 × 0.035 = £42.00.
- (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.
- (c) 250 miles — Find the distance travelled in 1 hour: 150 ÷ 3 = 50 miles. Multiply by 5 hours: 50 × 5 = 250 miles. Giving 300 miles doubles the original distance (150 × 2 = 300) using a scale factor of 2 instead of the correct 5 ÷ 3. Giving 200 miles adds only one extra hour's distance, 50, instead of the two extra hours actually needed (150 + 50 = 200, rather than 150 + 100). Giving 90 miles divides by the scale factor instead of multiplying (150 × 3 ÷ 5 = 90).
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (a) 4/5 — Method: find the June takings first, then write them over the May takings and cancel. Working: the takings fell by £900, so June is £4500 − £900 = £3600; the fraction is 3600/4500, and dividing the numerator and the denominator by 900 gives 4/5. Answer: 4/5 of the May takings. The distractors: 1/5 comes from writing the fall over the May takings, 900/4500, which answers how far the takings dropped rather than what June's takings are compared with May's; 5/4 comes from writing May over June, 4500/3600, reversing the order the question asks for; 4/9 comes from writing June over the two months added together, 3600/8100, a part-to-whole fraction when the comparison asked for is with May alone.
- (c) £2743.60 — Each year the balance is multiplied by 1.03. After the first year: 4000 × 1.03 = 4120. After the second year: 4120 × 1.03 = 4243.60, so that is what Leah takes out. She then spends £1500 of it, which leaves 4243.60 − 1500 = 2743.60. She has £2743.60 left.
- (c) 70 cm — Method: convert the real length to centimetres, then divide by the scale factor. Working: 84 m = 8400 cm. 8400 ÷ 120 = 70 cm. Wrong options: 0.7 cm comes from dividing 84 by 120 without converting metres to centimetres; 1,008,000 cm comes from multiplying instead of dividing (8400 × 120); 7 cm comes from converting 84 m to 840 cm (using ×10 instead of ×100) before dividing.
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