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
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- (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.
- (d) 8 — The product of price and number of tickets is constant: k = 4 × 12 = 48. At £6 per ticket, the number of tickets is 48 ÷ 6 = 8. Getting 18 comes from treating price and tickets as directly proportional and working out 12 × 6 ÷ 4 instead of dividing k by the new price. Getting 12 assumes the number of tickets does not change when the price changes. Getting 6 comes from writing down the new price instead of working out the number of tickets.
- (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.)
- (b) Provider Y — £21.00 against Provider X's £23.00 — Provider X's gradient is (35 − 15) ÷ 100 = 0.2, so cost = 15 + 0.2 × 40 = 15 + 8 = £23.00. Provider Y's gradient is (45 − 5) ÷ 100 = 0.4, so cost = 5 + 0.4 × 40 = 5 + 16 = £21.00. £21.00 is less than £23.00, so Provider Y is cheaper: 'Provider Y — £21.00 against Provider X's £23.00'. Getting both costs right but naming Provider X as cheaper compares the two numbers the wrong way round — £23.00 is more than £21.00, not less. Comparing only the fixed fees, £15.00 and £5.00, ignores the cost of the 40 gigabytes actually used. Reading off the costs at 100 gigabytes, £35.00 and £45.00, directly from the graph answers a different usage from the 40 gigabytes the question asks about.
- (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.
- (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) 1 : 4 — Convert to the same unit: 1.2 litres = 1200 ml, since 1 litre = 1000 ml. This gives the ratio 300 : 1200. Divide both parts by their highest common factor, 300, to get 1 : 4. Giving 3 : 12 divides by 100 only, which is a common factor but not the highest one. Giving 1 : 1.2 has not converted 1.2 litres into millilitres, so the two parts are not in the same unit. Giving 4 : 1 swaps the order, comparing water to concentrate instead of concentrate to water.
- (b) 12 — Since y is directly proportional to √x, y = k√x. Using x = 4, y = 8: √4 = 2, so 8 = k × 2, giving k = 8 ÷ 2 = 4. The equation is y = 4√x. When x = 9: √9 = 3, so y = 4 × 3 = 12. Treating the relationship as if y were proportional to x itself, rather than to √x, gives k = 8 ÷ 4 = 2 and then y = 2 × 9 = 18, which is a different relationship. Multiplying k by the new x-value instead of by its square root gives y = 4 × 9 = 36, skipping the square root altogether. Reporting √9 on its own, without multiplying by k, gives only 3, not the value of y. When x = 9, y = 12.
- (c) 135 g — Find the ratio of butter to sugar in the first batch: 240:160, which simplifies to 3:2. For the second batch, sugar = 90 g, so butter = 90 × 3/2 = 135 g. (60 g comes from using the ratio the wrong way round, 90 × 2/3. 170 g comes from subtracting the drop in sugar, 160 − 90 = 70 g, from the original butter amount, 240 − 70, instead of scaling. 240 g comes from not scaling the butter amount at all.)
- (a) 4/5 — Method: find the June takings first, then write them over the May takings and cancel. Working: the takings fell by £900, so June is £4500 − £900 = £3600; the fraction is 3600/4500, and dividing the numerator and the denominator by 900 gives 4/5. Answer: 4/5 of the May takings. The distractors: 1/5 comes from writing the fall over the May takings, 900/4500, which answers how far the takings dropped rather than what June's takings are compared with May's; 5/4 comes from writing May over June, 4500/3600, reversing the order the question asks for; 4/9 comes from writing June over the two months added together, 3600/8100, a part-to-whole fraction when the comparison asked for is with May alone.
- (c) £2717.20 — The recurrence B_{n+1} = 1.02B_n − 200 must be applied once for each month, using the previous month's balance each time. Starting from B_0 = 3000: 3000 × 1.02 = 3060, so B_1 = 3060 − 200 = 2860. Then 2860 × 1.02 = 2917.2, so B_2 = 2917.2 − 200 = 2717.2. Stopping after one month leaves B_1 = £2860.00, not the balance after two months. Applying two months of interest together, 1.02² = 1.0404, and 3000 × 1.0404 = 3121.2, and then subtracting 400 in one go, 3121.2 − 400 = 2721.2, does not reproduce the recurrence, because the second month's interest should be earned on the balance after the first repayment, not on the original £3000. Subtracting £200 twice from B_1 without adding a second month of interest, 2860 − 200 = 2660, drops the interest for the second month altogether. The balance after 2 months is £2717.20.
- (b) £52 — Method: the difference between the two ratio numbers tells you how many parts the £39 difference represents. Working: the difference in parts is 7 − 4 = 3, and this represents £39, so one part is £39 ÷ 3 = £13. Josh's savings are 4 × £13 = £52. So Josh has £52. Distractor £91 is Mia's savings, not Josh's. Distractor £39 comes from using the given £39 difference as the final answer, without scaling it to Josh's number of parts. Distractor £13 is the value of one part, found correctly but never multiplied by 4.
- (b) 2/5 — The ratio red : yellow is 6:15, so write red over yellow: 6/15. Divide both numbers by their highest common factor, 3: 6÷3 = 2, 15÷3 = 5, giving 2/5. (5/2 comes from writing the ratio the wrong way round, yellow over red, 15/6, which simplifies to 5/2. 2/7 comes from comparing the red paint to the total amount of paint, 6 parts out of 21. 5/7 comes from comparing the yellow paint to the total amount of paint, 15 parts out of 21.)
- (d) 3 : 8 — Multiply both parts by 4 to clear the decimal: 0.75 × 4 = 3 and 2 × 4 = 8, giving 3 : 8, which has no common factor other than 1. Giving 75 : 200 multiplies by 100 instead of 4, and has not then been simplified down to 3 : 8. Giving 0.75 : 2 has not been converted into whole numbers at all. Giving 3 : 2 converts the first part correctly but leaves the second part unscaled.
- (b) 1500 — The rate is 3 ÷ 2 = 1.5 litres per minute. Converting to cm³: 1.5 × 1000 = 1500 cm³ per minute. Getting 3000 comes from converting 3 litres to cm³ first (3000 cm³) and forgetting to divide by the 2 minutes. Getting 750 comes from dividing by the 2 minutes a second time after converting (1500 ÷ 2). Getting 2000 comes from converting the 2 minutes as if it were litres (2 × 1000) instead of using the correct rate of 1.5 litres per minute.
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