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.
Answer key: Ratio, proportion and rates of change worksheet — GCSE Higher
MathsUKwww.geekhero.co.uk
- (b) t = 5 and t = 7 (closest, evenly spaced) — To estimate the instantaneous rate of change at t = 6, use the chord centred on t = 6 with the closest readings on either side, t = 5 and t = 7. The gradient of this chord is 15.4 − 17.5 = −2.1, then −2.1 ÷ 2 = −1.05 cm per minute. The interval t = 3 to t = 9 is also centred on t = 6 but is wider: 13.4 − 18.7 = −5.3, then −5.3 ÷ 6 ≈ −0.88 cm per minute — this brings in more of the curve's own change in steepness, so it is a worse estimate of the rate at the single instant t = 6. Using t = 6 and t = 7 only gives 15.4 − 16.6 = −1.2, then −1.2 ÷ 1 = −1.2 cm per minute, but this is not centred on t = 6 — it estimates the rate over (6, 7), not at t = 6 itself. Using t = 0 and t = 6 gives 16.6 − 20.0 = −3.4, then −3.4 ÷ 6 ≈ −0.57 cm per minute, the average rate for the whole first six minutes, not the rate at the instant t = 6. Always choose the chord that brackets the point as closely as possible.
- (d) 90 cm — Method: scale each dimension by the scale factor, then find the perimeter. Working: model height = 240 ÷ 8 = 30 cm; model width = 120 ÷ 8 = 15 cm. Perimeter = 2 × (30 + 15) = 90 cm. Wrong options: 11.25 cm comes from squaring the scale factor as if finding an area (720 ÷ 64); 510 cm comes from scaling only one dimension and leaving the other at full size; 720 cm comes from finding the real perimeter (2 × (240 + 120)) but forgetting to scale it down at all.
- (a) 8 cm — Method: a scale of 1 : n means the real distance is n times the distance on the map, so to go from the real distance back to the map distance, put both lengths in the same unit and then divide by the scale. Working: 1 km = 100 000 cm, so 8 km = 8 × 100 000 = 800 000 cm; 800 000 ÷ 100 000 = 8. Answer: 8 cm. The distractors: 800000 cm comes from converting the 8 km into centimetres and stopping there, so the division by the scale — the inverse operation the question asks for — is never done; 80 cm comes from taking a metre to be 1000 cm, which turns 8 km into 8 × 1000 × 1000 = 8 000 000 cm and gives 8 000 000 ÷ 100 000 = 80; 0.08 cm comes from taking a kilometre to be 1000 cm, which turns 8 km into 8000 cm and gives 8000 ÷ 100 000 = 0.08.
- (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.
- (c) 450 g — Method: use the amount of butter given to find the value of one part of the ratio, then find the mass of flour, and finally add flour and butter to get the total. Working: 180 g of butter is 2 parts, so one part is 180 ÷ 2 = 90 g. The flour is 3 parts, so 3 × 90 = 270 g, and the total mass is 270 + 180 = 450 g. So the baker can make 450 g of pastry. Distractor 270 g is only the mass of flour, forgetting to add the butter back on. Distractor 300 g comes from treating the 180 g as 3 parts instead of 2, swapping which ratio number matches the butter. Distractor 540 g comes from multiplying 180 by 3 directly instead of first finding the value of one part.
- (d) 1 : 1 — Sugar and butter together make 3 + 5 = 8 parts of the mixture. Comparing flour to this, 8 : 8, simplifies to 1 : 1. Giving 1 : 2 compares flour with the whole mixture (8 + 3 + 5 = 16 parts, giving 8 : 16 = 1 : 2) instead of with the rest of the mixture. Giving 3 : 5 is the ratio of sugar to butter, not of flour to the rest of the mixture. Giving 8 : 3 compares flour only with sugar, leaving butter out altogether.
- (c) 1.2 — Add the parts of the ratio: 6 + 1 = 7. Divide the total amount by the number of parts: 8.4 ÷ 7 = 1.2 litres, which is the value of one part and also the amount of syrup, since syrup is 1 part. (7.2 litres is the amount of water, using 6 parts instead of 1. 1.4 comes from dividing 8.4 by 6 — the water's part of the ratio — instead of dividing by the total number of parts, 7. 0.84 comes from dividing 8.4 by 10 instead of by 7.)
- (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).
- (c) m = 6c — Method: the whole is the sum of the parts in the ratio, and the cement is 1 part, so one part weighs c kg. Working: the mix has 5 + 1 = 6 parts, each of mass c kg, so the total mass is 6 × c, giving m = 6c. Answer: m = 6c. The distractors: m = 5c uses the 5 gravel parts as the multiplier and forgets that the cement is in the mix too, so it gives the mass of the gravel and not the total; m = c + 5 comes from reading the ratio as '5 more than' and adding, which treats a number of parts as a mass in kilograms; m = c/6 turns the relationship upside down, as though the total were shared into the cement rather than the cement multiplied up to the total.
- (d) 132 — Method: find the value of one part of the ratio, use it to find Leo's pages, then add both amounts together. Working: 84 ÷ 7 = 12 (value of one part). Leo's pages = 12 × 4 = 48. Total = 84 + 48 = 132. Wrong options: 48 gives only Leo's pages and forgets to add Mia's; 147 comes from reversing the ratio parts (84 ÷ 4 × 7 = 147) and stopping there; 231 comes from reversing the ratio parts and then adding Mia's pages (84 + 147).
- (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.
- (b) The ratio C : m is not constant because the formula includes a fixed charge of £3 as well as the charge per mile. — For the ratio C : m to stay constant, C must be directly proportional to m, i.e. C = km with no constant term. Because of the +3 fixed charge, C is not directly proportional to m: for example m = 1 gives C = 5.5 (ratio 5.5 : 1), while m = 10 gives C = 28 (ratio 2.8 : 1) — the ratio has changed.
- (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.
- (d) 218 — The rule x_{n+1} = 0.8x_n + 50 must be applied once for each step, using the result of the previous step every time — not the same starting value repeated. Starting from x_0 = 200: 0.8 × 200 = 160, so x_1 = 160 + 50 = 210. Then 0.8 × 210 = 168, so x_2 = 168 + 50 = 218. Stopping after one iteration leaves x_1 = 210, not x_2. Applying only the multiplier twice without adding 50 at each step uses 0.8² = 0.64, and 0.64 × 200 = 128, which drops the 50 completely. Adding 50 twice at the end instead of once per step, 128 + 100 = 228, still does not reproduce the actual recurrence, because the 50 added at the first step is itself multiplied by 0.8 at the second step. After two iterations, x_2 = 218.
- (a) 46.8 km/h — Method: 1 hour = 60 minutes, so a rate given in kilometres per minute is converted to kilometres per hour by multiplying by 60. Working: 0.78 × 60 = 46.8, so the instantaneous rate of change is 46.8 kilometres per hour. Keeping the given value unchanged and only relabelling the unit gives 0.78 km/h, which ignores that the time unit has changed. Dividing by 60 instead of multiplying — as you would when converting to a larger length unit — gives 0.78 ÷ 60 = 0.013 km/h, the wrong direction for a rate measured against a larger time unit. Adding 60 to the given rate instead of multiplying by it gives 0.78 + 60 = 60.78 km/h. Converting a rate always means multiplying or dividing by the conversion factor between the units, never adding it, and the direction depends on whether the new time unit is bigger or smaller than the old one.
Build your own mix at the worksheet builder.