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) 4 hours — This is inverse proportion: more pumps take less time. Multiply the original numbers to find the total pump-hours needed: 2 × 10 = 20 pump-hours. Divide by the new number of pumps: 20 ÷ 5 = 4 hours. Working out 10 × 5 ÷ 2 = 25 hours treats it as direct proportion, as if more pumps needed more time. Stopping at 20 gives the total pump-hours, not the number of hours. Working out 10 − (5 − 2) = 7 hours subtracts the extra number of pumps straight from the number of hours, treating pumps and hours as the same kind of quantity. 5 pumps take 4 hours.
- (b) 62.5% — Total parts = 5 + 3 = 8. Apples make up 5 parts, so the percentage is 5/8 × 100 = 62.5%. A student who finds the oranges' share instead gets 3/8 × 100 = 37.5%. A student who assumes an even split gets 50%. A student who inverts the fraction gets 8/5 × 100 = 160%.
- (a) 36 — Pressure = force ÷ area, so 126 ÷ 3.5 = 36 N/m². (0.03 comes from dividing the area by the force instead of the force by the area, the wrong way round. 129.5 comes from adding 126 and 3.5 instead of dividing. 441 comes from multiplying 126 by 3.5 instead of dividing.)
- (b) 4% — Method: find the total interest earned, share it equally across the number of years to find one year's interest, then write it as a percentage of the amount invested. Working: total interest = £840 − £750 = £90, so one year's interest is £90 ÷ 3 = £30, and £30 as a percentage of £750 is (£30 ÷ £750) × 100 = 4%. Answer: 4%. 12% comes from treating the total interest of £90 as if it were earned in a single year, (£90 ÷ £750) × 100 = 12%, forgetting to divide by 3 years. 0.04% comes from finding the correct decimal, £30 ÷ £750 = 0.04, but forgetting to multiply by 100 to convert it into a percentage. 112% comes from writing the final amount, £840, as a percentage of the amount invested, £750, without first subtracting the £750 to find the interest alone.
- (b) 1 : 500 — smallest real distance per cm (5 m), most detail — Method: compare what one centimetre represents in real life for each scale — the scale with the smallest real distance per cm shows the most detail. Working: for 1 : 500, 1 cm represents 500 cm (5 m); for 1 : 5000, 1 cm represents 50 m; for 1 : 50 000, 1 cm represents 500 m. Since 5 m is the smallest, 1 : 500 shows the most detail. Wrong options: '1 : 50 000 — covers the largest real area' wrongly assumes covering more area means more detail, when it is the opposite; '1 : 5000 — the middle value' wrongly assumes the middle scale is automatically the most balanced; '1 : 500 — covers the largest real distance' picks the correct scale but states an incorrect fact, since 1 : 500 actually covers the smallest real distance per cm.
- (a) Yes — the cost per metre is £1.50 each time — Direct proportion holds if the cost per metre is the same every time. Check each pair: 3.00 ÷ 2 = 1.50, 6.00 ÷ 4 = 1.50, and 10.50 ÷ 7 = 1.50. All three give the same rate, £1.50 per metre, so the data does show direct proportion. Saying only that the cost increases as the length increases is not enough on its own — many non-proportional relationships also increase, so this reason does not prove proportion. Misreading 10.50 ÷ 7 as 1.05 by misplacing the decimal point gives a false mismatch that is not actually there. Requiring every length to be a double of another confuses a special case (doubling) with the general test, which is that the rate itself stays constant. The data does show direct proportion, at £1.50 per metre.
- (d) 4 years — Apply the recurrence repeatedly. V_1 = 0.85 × 18000 = 15300. V_2 = 0.85 × 15300 = 13005. V_3 = 0.85 × 13005 = 11054.25. V_4 = 0.85 × 11054.25 = 9396.1125. V_3 = £11054.25 is still above £10000, but V_4 = £9396.11 has dropped below it, so the answer is 4 years. Stopping at V_3 and calling it '3 years' misreads £11054.25 as already below £10000, or comes from wrongly modelling the fall as a flat £2700 a year (15% of the original value each time, without compounding), which crosses £10000 a year too early. Continuing one extra step to V_5 = 0.85 × 9396.1125 = 7986.70 and calling it '5 years' overshoots, since the value had already dropped below £10000 at V_4. Doubling the percentage decrease to 30% by mistake gives V_1 = 0.7 × 18000 = 12600, then V_2 = 0.7 × 12600 = 8820, which is already below £10000 after only 2 years — the wrong rate crosses the threshold too fast.
- (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.
- (c) −0.2, the car uses 0.2 litres of fuel for each mile — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, which on this graph is a number of litres for each mile, and a negative gradient means the vertical quantity is going down. Working: from (0, 45) to (150, 15) the fuel changes by 15 − 45 = −30 litres while the distance changes by 150 − 0 = 150 miles, so the gradient is −30 ÷ 150 = −0.2, which says the tank loses 0.2 litres for every mile driven. Answer: −0.2, the car uses 0.2 litres of fuel for each mile. The distractors: '0.2, the car gains 0.2 litres of fuel for each mile' comes from subtracting the fuel values the other way round, 45 − 15 = 30, which drops the minus sign and reverses what the graph says; '−5, the car uses 5 litres of fuel for each mile' comes from dividing the change in distance by the change in fuel, 150 ÷ (−30), turning the gradient upside down; '−30, the car uses 30 litres of fuel for each mile' is the change in fuel on its own, never divided by the 150 miles travelled.
- (d) £144 — Method: find the length (perimeter) scale factor by taking the square root of the area ratio, then apply it to the cost. Working: 12 : 27 simplifies to 4 : 9, and the square root of each part gives the length ratio 2 : 3, so the scale factor from the smaller to the larger pond is 3 ÷ 2 = 1.5. Cost = £96 × 1.5 = £144. Answer: £144. £216 comes from using the area ratio itself as the cost ratio, £96 × (27 ÷ 12) = £216, without taking the square root. £64 comes from using the length ratio the wrong way round, £96 × (2 ÷ 3) = £64. £111 comes from simply adding the difference in area, 27 − 12 = 15, onto the original cost, £96 + £15 = £111, instead of scaling proportionally.
- (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.
- (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) 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.
- (d) 64 — Since y is directly proportional to √x, y = k√x. Using x = 25, y = 20: √25 = 5, so 20 = k × 5, giving k = 20 ÷ 5 = 4. The equation is y = 4√x. When y = 32: √x = 32 ÷ 4 = 8, and x = 8² = 64. Stopping at √x = 8 without squaring leaves the square root of x, not x itself. Treating the relationship as if y were proportional to x itself gives k = 20 ÷ 25 = 0.8 and then x = 32 ÷ 0.8 = 40, which is a different relationship entirely. Multiplying instead of dividing when isolating √x gives √x = 32 × 4 = 128, far too large to be a square root here. When y = 32, x = 64.
- (b) 24 km/h — First convert 45 minutes to hours: 45 ÷ 60 = 0.75 hours. Then divide the distance by the time: 18 ÷ 0.75 = 24 km/h. Reading 45 minutes as 0.45 hours (writing the minutes after the decimal point instead of dividing by 60) gives 18 ÷ 0.45 = 40 km/h. Working out 18 × 0.75 = 13.5 multiplies by the time instead of dividing. Working out 18 ÷ 45 = 0.4 divides by 45 without ever converting the minutes to hours. The cyclist's average speed is 24 km/h.
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