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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- (a) 2 km — Method: multiply the map distance by the scale to get the real distance in centimetres, then convert centimetres to kilometres using 100 cm = 1 m and 1000 m = 1 km. Working: 4 × 50 000 = 200 000 cm; 200 000 ÷ 100 = 2000 m; 2000 ÷ 1000 = 2. Answer: 2 km. The distractors: 200 km comes from dividing the 200 000 cm by 1000 in a single step, as if a kilometre were 1000 cm rather than the 100 000 cm it is; 20 km comes from converting to metres correctly, 200 000 ÷ 100 = 2000 m, and then dividing those metres by 100 instead of by 1000; 0.2 km comes from dividing by 1000 to reach metres, as if a metre were 1000 cm, and then dividing by 1000 again, so 200 000 is divided by 1 000 000 altogether.
- (d) 1.00 litres — Squash is 2/9 of the mixture, so the squash volume is 4.5 × 2/9 = 1.00 litres. Using the water's fraction, 7/9, instead of squash's gives 4.5 × 7/9 = 3.50 litres — the volume of water, not squash. Dividing 4.5 by 9 but forgetting to multiply by the numerator 2 gives 4.5 ÷ 9 = 0.50 litres, which is only 1/9 of the mixture. Halving the total volume instead of applying the fraction 2/9 gives 4.5 ÷ 2 = 2.25 litres, which assumes the mixture is half squash.
- (a) 100 — The difference between the parts of the ratio is 7 − 3 = 4 parts, and this is worth 40 beads. Divide to find one part: 40 ÷ 4 = 10. The total number of parts is 7 + 3 = 10, so the total number of beads is 10 × 10 = 100. (40 is just the given difference between gold and silver, not the total. 70 is the number of gold beads only, using 7 parts. 30 is the number of silver beads only, using 3 parts.)
- (a) 1 : 2 000 000 — Method: a scale is a ratio between two lengths written in the same unit, reduced so that the map distance is 1. Working: 120 km = 120 × 1000 × 100 = 12 000 000 cm, so the ratio is 6 : 12 000 000, and dividing both parts by 6 gives 1 : 2 000 000. Answer: 1 : 2 000 000. The distractors: 1 : 12 000 000 comes from writing the real length in centimetres without dividing by the 6 cm on the map; 1 : 200 000 comes from taking 120 km as 1 200 000 cm, one conversion step short, before dividing by 6; 1 : 20 000 comes from converting 120 km to 120 000 m and treating those metres as centimetres.
- (c) Average rate, t = 2 to 6, is −2°C/min — First find the temperature at each end of the interval. At t = 2, T = 80 − 6 × 2 + 0.5 × 2² = 80 − 12 + 2 = 70. At t = 6, T = 80 − 6 × 6 + 0.5 × 6² = 80 − 36 + 18 = 62. The average rate of change over the interval is the change in T divided by the change in t: 62 − 70 = −8, then −8 ÷ 4 = −2°C per minute, so the statement about the average rate is correct. The instantaneous rate at t = 6 is not −2: completing the square gives T = 0.5(t − 6)² + 62, so t = 6 is the turning point of the curve, where the tangent is horizontal and the rate is 0°C per minute — the reaction has stopped cooling by then. The instantaneous rate at t = 2 is not −2 either: a short chord centred on t = 2, from t = 1.9 (T = 70.405) to t = 2.1 (T = 69.605), gives −0.8 ÷ 0.2 = −4°C per minute, so the reaction is cooling twice as fast at the start of the interval as the average over it. Saying the temperature falls 2°C in total confuses the RATE, −2°C per minute, with a TOTAL drop, which is 70 − 62 = 8°C over the four minutes. Always check whether a figure is a rate, per minute, or a total change.
- (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.
- (d) 7/5 — The number of unbroken crayons is 60 − 25 = 35. The comparison is with the broken crayons, so the broken crayons are the denominator: 35/25. Both parts divide by 5: 35 ÷ 5 = 7 and 25 ÷ 5 = 5. The fraction is 7/5, which is greater than 1 because there are more unbroken crayons than broken ones.
- (b) −1% — Method: write each change as a multiplier and multiply them. A 10% fall is × 0.9 and a 10% rise is × 1.1. Working: 0.9 × 1.1 = 0.99, so the final price is 99% of the original, which is 1% less. Answer: an overall change of −1%. The distractors: 0% comes from assuming a 10% fall and a 10% rise cancel — they do not, because the rise is 10% of a smaller amount; +1% has the size right but the sign wrong, from reading the multiplier 0.99 as 1% above 1 instead of 1% below it; −2% comes from finding the 1% fall and then counting it once for each of the two changes.
- (d) 125 — Since y is directly proportional to x², y = kx² for a constant k. Using x = 3, y = 45: 45 = k × 9, so k = 45 ÷ 9 = 5. The equation is y = 5x². When x = 5: 5² = 25, and 5 × 25 = 125, so y = 125. Reporting 5² = 25 on its own, without multiplying by the constant k, gives only the square of the new x-value, not the value of y. Treating the proportion as if y were proportional to x itself, rather than to x², gives k = 45 ÷ 3 = 15 and then y = 15 × 5 = 75, which is not this relationship. Squaring the new x-value as though squaring meant doubling it instead gives 5 × 10 = 50, not the true square. When x = 5, y = 125.
- (d) £5.40 — Method: work out the reduced price at each shop separately, then subtract the smaller from the larger. Working: Shop A's reduced price is £45 × 0.8 = £36, and Shop B's reduced price is £34 × 0.9 = £30.60, so the difference is £36 − £30.60 = £5.40. Answer: £5.40. £11.00 comes from comparing the two ORIGINAL prices, £45 − £34, without applying either shop's reduction at all. £1.60 comes from finding Shop A's reduced price correctly, £36, but then subtracting Shop B's original (unreduced) price of £34 instead of its reduced price. £66.60 comes from adding the two reduced prices together, £36 + £30.60, instead of subtracting them.
- (d) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
- (d) 120 g — Mass = density × volume, so 0.8 × 150 = 120 g. Working out 150 ÷ 0.8 = 187.5 divides by the density instead of multiplying, the wrong way round for finding a mass. Working out 150 × 8 = 1200 misplaces the decimal point in the density, treating 0.8 g/cm³ as 8 g/cm³. Working out 150 − 0.8 = 149.2 simply subtracts the density from the volume, which does not give a mass. The piece of wood has a mass of 120 g.
- (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.
- (d) 12 hours — Since time is inversely proportional to the number of installers, T = k/n. Using n = 4, T = 18: 18 = k ÷ 4, so k = 18 × 4 = 72. The equation is T = 72/n. When n = 6: T = 72 ÷ 6 = 12. Using the original number of installers instead of the new one gives T = 72 ÷ 4 = 18, the wrong value substituted. Treating more installers as needing more time, as if T were directly proportional to n, gives k = 18 ÷ 4 = 4.5 and then T = 4.5 × 6 = 27, the opposite relationship to the one described. Stopping at k = 72 and reporting it gives the time the job would take a single installer working alone — the constant still has to be divided by the new number of installers before it answers the question asked. With 6 installers, the job takes 12 hours.
- (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.)
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