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) 4 — The gradient of a tangent at a point equals the instantaneous rate of change of y with respect to x at that point. A straight line's gradient is found from the change in y divided by the change in x between two points on it. Here the tangent passes through (1, 4) and (5, 20), so the change in y is 20 − 4 = 16 and the change in x is 5 − 1 = 4. The gradient is 16 ÷ 4 = 4. Stopping after finding the change in y, 16, without dividing it by the change in x, is not a gradient at all — that rise happened across 4 units of x, not 1. Adding the two changes instead of dividing one by the other gives 16 + 4 = 20, which is not how a gradient is found. Subtracting in the wrong order, (4 − 20) ÷ (5 − 1), gives −16 ÷ 4 = −4, the wrong sign. The instantaneous rate of change of y with respect to x at x = 3 is 4.
- (d) 5:3:2 — German = 100% − 50% − 30% = 20%. The ratio 50 : 30 : 20 simplifies by dividing every part by 10 to give 5 : 3 : 2.
- (a) 3/2 — Find each average speed: car = 180 ÷ 3 = 60 mph; lorry = 160 ÷ 4 = 40 mph. Put the car's speed over the lorry's speed: 60/40. Divide both numbers by their highest common factor, 20: 60÷20 = 3, 40÷20 = 2, giving 3/2. (2/3 comes from writing the speeds the wrong way round. 9/8 comes from comparing the distances travelled, 180/160, without working out the speeds. 3/4 comes from comparing the times taken, 3/4, instead of the speeds.)
- (c) 1 : 25 — The radii are in the ratio 4 : 20, which simplifies to 1 : 5. Areas scale with the square of the length ratio, so the area ratio is 1² : 5² = 1 : 25. Giving 1 : 5 uses the radius ratio without squaring it. Giving 1 : 10 doubles the radius ratio instead of squaring it. Giving 25 : 1 has the areas the right way round for larger to smaller, not smaller to larger.
- (d) 3 : 2 : 5 — The highest common factor of 45, 30 and 75 is 15. Divide each part by 15: 45 ÷ 15 = 3, 30 ÷ 15 = 2 and 75 ÷ 15 = 5, giving 3 : 2 : 5. Giving 9 : 6 : 15 divides by 5, a common factor but not the highest one. Giving 15 : 10 : 25 divides by 3 only, even further from simplest form. Giving 2 : 3 : 5 has the first two parts swapped.
- (c) 9% — Method: a percentage concentration is the ratio of salt to solution written per 100 g, so scale each concentration to the mass it belongs to, add the two masses of salt, then scale the ratio of salt to mixture back to a denominator of 100. Working: 5:100 = x:400 gives 5 ÷ 100 × 400 = 20 g of salt, and 25:100 = y:100 gives 25 g of salt; the mixture holds 20 + 25 = 45 g of salt in 400 + 100 = 500 g of solution; 45:500 = 9:100. Answer: 9%. The distractors: 15% is the mean of 5% and 25%, which would only be right if the two masses were equal, and here one is four times the other; 21% comes from attaching the concentrations to the wrong masses, working out (400 × 25% + 100 × 5%) ÷ 500; 0.9% comes from working out 45 ÷ 500 = 0.09 and then moving the decimal point one place instead of two when writing the decimal as a percentage.
- (a) 375 g — Find the amount of flour needed per muffin: 150 ÷ 6 = 25 g. Multiply by the new number of muffins: 25 × 15 = 375 g. (60 g comes from using the scale factor the wrong way round, 6/15 × 150. 150 g comes from not scaling the recipe at all. 300 g comes from rounding the scale factor, 15 ÷ 6, down to 2 before multiplying.)
- (a) 3/4 — Write January's total over February's total: 360/480. Both numbers share a factor of 120, so dividing top and bottom by 120 gives 3/4. Choosing 4/3 comes from writing February's amount over January's amount, the wrong way round. Choosing 1/4 comes from finding the difference between the two months (480 − 360 = 120) and writing it over February's amount, instead of using January's amount. Choosing 3/7 comes from writing January's amount over the total received across both months (360 out of 840), instead of over February's amount alone.
- (d) Profit is decreasing by £10 per £1 rise in price. — The gradient of a tangent gives the instantaneous rate of change of profit with respect to price, found from the change in profit divided by the change in price between two points on the tangent. Here the tangent passes through (12, 540) and (18, 480), so the change in profit is 480 − 540 = −60 and the change in price is 18 − 12 = 6. The gradient is −60 ÷ 6 = −10. A negative gradient means profit is decreasing as price increases, so profit is decreasing at an instantaneous rate of £10 for every £1 rise in price. Subtracting the profits in the wrong order, 540 − 480 = 60, and dividing by the same change in price, 60 ÷ 6 = 10, gives a positive value and the wrong direction — profit is not increasing at £15. Stopping after finding only the change in profit, 480 − 540 = −60, without dividing by the change in price, is not a rate at all. Reading off the change in price, 6, and calling it the rate gives £6 per £1 rise in price, but 6 is only the width of the price interval — it is not a change in profit at all, and profit falls across that interval, so the direction is wrong too. The instantaneous rate of change of profit with respect to price at £15 is a decrease of £10 per £1 rise in price.
- (b) 15 — Find the multiplier connecting y to x: 10 ÷ 4 = 2.5. Then apply it to the new value of x: 2.5 × 6 = 15. Working out 10 + (6 − 4) = 12 adds the change in x straight onto y instead of scaling proportionally. Working out 10 × 6 = 60 multiplies the given y-value by the new x-value directly, without finding the multiplier first. Writing 10 keeps y the same as before, not realising it must change with x. When x = 6, y = 15.
- (c) 120 km/h — Method: for a fixed distance the average speed multiplied by the time is constant, and that constant is the distance, so divide the distance by the new time. Working: speed × time = 240, so in 2 hours the speed needed is 240 ÷ 2 = 120 km/h. Answer: 120 km/h. The distractors: 80 km/h is the average speed of the original journey, 240 ÷ 3, which answers for the 3-hour timing rather than the 2-hour one; 160 km/h comes from halving the 3 hours to 1.5 hours and working out 240 ÷ 1.5, instead of using the 2 hours the question gives; 480 km/h comes from multiplying the distance by the 2 hours rather than dividing by it.
- (c) 500 g — Method: a concentration of 10% is the ratio 10:100, and the salt and the solution in the beaker must be in that same ratio, so write 10:100 = 50:m and scale. Working: 50 ÷ 10 = 5, so the salt is 5 times the 10 of the ratio; the solution must be 5 times the 100 of the ratio, giving 5 × 100 = 500 g. Answer: 500 g. The distractors: 5 g comes from working out 10% of 50 g, which treats the 50 g as the whole solution when it is the salt inside it; 450 g comes from scaling correctly and then taking the 50 g of salt away, which gives the mass of water rather than the mass of the whole solution; 5000 g comes from dividing by 0.01 instead of 0.1, that is from writing 10% as 0.01.
- (d) 3/5 — Convert both times to minutes: 2 hours 15 minutes = 135 minutes; 3 hours 45 minutes = 225 minutes. Put the train time over the bus time: 135/225. Divide both numbers by their highest common factor, 45: 135÷45 = 3, 225÷45 = 5, giving 3/5. (5/3 comes from writing the times the wrong way round. 2/5 comes from finding the difference, 225 − 135 = 90 minutes, and writing it as a fraction of the bus time, 90/225. 3/8 comes from comparing the train time to the total time for both journeys, 135/360.)
- (b) 5/7 — The enlargement multiplier is 1.4, which as a fraction is 7/5. To reverse an enlargement, use the reciprocal of the multiplier: flip 7/5 to get 5/7. 7/5 comes from using the enlargement multiplier again, instead of reversing it. 3/5 comes from treating the reverse as 'give back the extra amount', working out 1 − (1.4 − 1) = 0.6, instead of using the reciprocal. 5/2 comes from ignoring the whole number in 1.4 and inverting only the decimal part, 0.4, as if it were the whole multiplier.
- (c) 2.5 — x : y = 2 : 5 means that for every matching pair of values, y ÷ x = 5 ÷ 2 = 2.5. So y = 2.5x, and comparing with y = kx gives k = 2.5. Dividing the other way round, 2 ÷ 5 = 0.4, gives x in terms of y — that is the constant for x = 0.4y, not for y = kx. Taking the y-part of the ratio on its own, 5, reads one number off the ratio instead of dividing the y-part by the x-part; 5 would only be right if the x-part were 1. Subtracting the two parts, 5 − 2 = 3, treats the ratio as a difference, but a ratio compares two quantities by multiplication, not by subtraction. The constant is k = 2.5.
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