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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- (c) 20 cm — Convert 5 km to centimetres: 5 km = 5000 m = 500 000 cm. Divide by the scale factor: 500 000 ÷ 25 000 = 20, giving 20 cm. Converting only as far as metres, 5000 ÷ 25 000 = 0.2, gives 0.2 cm — the conversion to centimetres was never finished. Dropping a zero in the division gives 2 cm, ten times too small. Misreading the scale as 1 : 2500 instead of 1 : 25 000 gives 500 000 ÷ 2500 = 200 cm, ten times too big.
- (b) 20 km — The scale 2 cm : 5 km means each 1 cm on the map represents 5 ÷ 2 = 2.5 km in real life. The footpath is 8 cm on the map, so its real length is 8 × 2.5 = 20 km. 40 km comes from multiplying 8 by 5 directly, ignoring that the scale's '2 cm' has to be divided out first: 8 × 5 = 40. 3.2 km comes from dividing 8 by 2.5 instead of multiplying: 8 ÷ 2.5 = 3.2. 5 km comes from multiplying 2.5 by the scale's '2' instead of by the footpath's 8 cm: 2.5 × 2 = 5.
- (c) Height rising at 2 m/s at t = 1.5 s — A tangent's gradient on a height-time graph is the instantaneous rate of change of height, in metres per second, so gradient 2 means the ball's height is increasing at 2 m/s at t = 1.5 s. Saying the height 'is 2 m' confuses the gradient, a rate, with the y-value on the graph, which is the ball's height itself. Saying the ball 'travelled 2 m from t = 1 to t = 2' treats the instantaneous gradient at one instant as if it were the total distance risen over a whole one-second interval, which is a different quantity found from two height readings, not from one tangent. Saying the speed 'is 2 m/s²' uses the wrong units — m/s² measures acceleration, the rate of change of speed, not speed itself. Always check that the units quoted match what a height-time graph's gradient can actually give you: metres per second.
- (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) 0.22 km — The real one-way distance is 4.4 × 2500 = 11000 cm. Converting units: 11000 ÷ 100 = 110 m, and 110 ÷ 1000 = 0.11 km. Since the jogger runs there and back, the total distance is 0.11 × 2 = 0.22 km. 0.11 km comes from working out only the one-way distance and forgetting the return journey. 220 km comes from correctly doubling the one-way distance in metres, 110 × 2 = 220, but leaving it mislabelled as kilometres instead of converting metres to kilometres. 110 km comes from working out only the one-way distance in metres, 110, and mislabelling it as kilometres.
- (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).
- (a) 7 — The average rate of change of y with respect to x over an interval is the change in y divided by the change in x between its two endpoints — the gradient of the chord joining them, not a rate at a single point. At x = 1: 2 × 1 = 2, so y = 1 + 2 = 3. At x = 4: 2 × 4 = 8, so y = 16 + 8 = 24. The change in y is 24 − 3 = 21 and the change in x is 4 − 1 = 3, so the average rate of change is 21 ÷ 3 = 7. Reporting the change in y, 21, on its own is not a rate of change, because it has not been divided by the 3 units of x over which it happened. Reporting 3 is not a rate either — 3 is the width of the interval, and also the value of y at x = 1, and neither of those measures how fast y is changing. Subtracting in the wrong order gives −21 ÷ 3 = −7, the wrong sign. The average rate of change of y with respect to x over the interval is 7.
- (b) 4 — Method: find the length scale factor by taking the cube root of the volume scale factor, then square it to get the area scale factor. Working: 8 = 2³, so the length scale factor is 2, and the area scale factor is 2² = 4. Answer: 4. 8 comes from using the volume scale factor itself as if it were the area scale factor. 64 comes from squaring the volume scale factor, 8² = 64, instead of first taking its cube root. 2 comes from correctly finding the length scale factor but then forgetting to square it.
- (c) 1.2 cm/s — To estimate an instantaneous rate of change at a point from a table of readings, use the chord that spans the point symmetrically — equal steps either side — because the over-estimate on one side and the under-estimate on the other largely cancel. Here that is the chord from t = 10 to t = 30. The change in depth is 56 − 32 = 24 and the change in time is 30 − 10 = 20, so the estimate is 24 ÷ 20 = 1.2 cm/s. The one-sided chord from t = 20 to t = 30 gives (56 − 45) ÷ (30 − 20) = 11 ÷ 10 = 1.1 cm/s, which estimates the rate somewhere between t = 20 and t = 30 rather than at t = 20 itself. Dividing the 20-second change in depth by the 10-second gap between consecutive readings gives 24 ÷ 10 = 2.4, mixing the change from one interval with the time from another. Reporting the change in depth, 24, on its own is not a rate at all, because it has not been divided by a time. The best estimate of the instantaneous rate of change of depth at t = 20 is 1.2 cm/s.
- (c) L = d/5 — The scale 1 : 20 means each cm on the drawing represents 20 cm in real life, so the real length in cm is 20d. Converting to metres by dividing by 100: L = 20d/100 = d/5.
- (c) 27.1 cm — The model length is 20.6 ÷ 76 = 0.271052... metres. Converting to centimetres by multiplying by 100 gives 27.1052..., which rounds to 27.1 cm. Forgetting to convert metres to centimetres leaves the answer as 0.271052... metres, which rounds to 0.3 cm if the unit is simply relabelled. Multiplying by 1000 instead of 100 when converting metres to centimetres gives 271.052..., which rounds to 271.1 cm. Multiplying by 76 instead of dividing, 20.6 × 76 = 1565.6, uses the scale factor the wrong way round — that would be the real length if the model were 20.6 units long, not the other way round.
- (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.
- (d) £117.60 — Add the hours worked over the two days: 6 + 4.5 = 10.5 hours. Multiply by the rate of pay: 10.5 × £11.20 = £117.60. (£67.20 is Monday's pay only. £50.40 is Tuesday's pay only. £106.40 comes from mistakenly adding the hours as 6 + 3.5 = 9.5 — misreading Tuesday's 4.5 hours as 3.5 — and then multiplying by £11.20.)
- (a) 1.375 — The multiplier is the new value divided by the original value: 176 000 ÷ 128 000 = 1.375. Choosing 0.375 comes from working out the increase as a fraction of the original (48 000 ÷ 128 000 = 0.375) and stopping there, without adding on the 1 whole that represents the original turnover. Choosing 0.727 comes from dividing the 2023 turnover by the 2024 turnover, the wrong way round (128 000 ÷ 176 000, rounded to 3 decimal places). Choosing 137.5 comes from writing the multiplier as a percentage — the 2024 turnover is 137.5% of the 2023 turnover — instead of converting it to the decimal multiplier 1.375.
- (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.
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