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.
Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.The rent on a flat increases by 10% one year and by a further 10% the following year. Work out the overall percentage increase over the two years.
- 2.A novel has 312 pages. A comic has 48 pages. Write the number of pages in the novel as a fraction of the number of pages in the comic, giving your answer in its simplest form.
- 3.A concrete mix is made from gravel and cement in the ratio 5 : 1 by mass. The mass of cement is c kg and the total mass of the mix is m kg. Write down a formula for m in terms of c.
- 4.A loan of £3000 has interest added at 2% each month, and then a fixed repayment of £200 is made. This is modelled by the recurrence B_{n+1} = 1.02B_n − 200, where B_n is the balance after n months and B_0 = 3000. Work out the balance after 2 months.
- 5.The price of a games console is reduced by 10%. In a later sale the reduced price is reduced by 10% again. Work out the overall percentage decrease.
- 6.A kettle uses 2.1 kWh of energy. A toaster uses 0.9 kWh of energy. Write the energy used by the kettle as a fraction of the energy used by the toaster, giving your answer in its simplest form.
- 7.An online shop's total number of orders received is recorded at the end of each day for five consecutive days: Day 1, 1,284 orders; Day 2, 1,509 orders; Day 3, 1,830 orders; Day 4, 2,296 orders; Day 5, 2,510 orders. Work out between which two consecutive days the average rate of increase in orders was greatest.
- 8.The exchange rate is £1 = €1.15. Convert £200 to euros.
- 9.2.4 kg of cheese costs £36. Work out the cost of 1.5 kg of the same cheese.
- 10.Butter costs £2.40 per kg. Work out the cost of the butter in pence per 100 g.
- 11.y is directly proportional to √x. When x = 4, y = 8. Construct the equation connecting x and y, then work out the value of y when x = 9.
- 12.An amount of money is shared in the ratio 1:2:3. The largest share is £90 more than the smallest share. Work out the total amount that was shared.
- 13.A charity shop and a school share collection-box money in the ratio 5 : 8. The charity shop receives £47.50. Work out how much the school receives.
- 14.A plumber charges a call-out fee plus an hourly rate. The total charge, C pounds, for a job lasting h hours is shown on a straight-line graph. The line passes through the points (2, 70) and (5, 130). Work out the call-out fee, in pounds.
- 15.The depth of water in a tank, in cm, is recorded every 10 seconds: at t = 10, depth = 32; at t = 20, depth = 45; at t = 30, depth = 56. Use the most appropriate chord from these readings to estimate the instantaneous rate of change of depth at t = 20.
Answer key
- (c) 21% — Method: an increase of 10% is a multiplier of 1.1, and two successive increases are found by multiplying the multipliers. Working: 1.1 × 1.1 = 1.21, so the rent is 121% of the original, which is an increase of 21%. Answer: 21%. The distractors: 20% comes from adding the two percentages, which ignores that the second 10% is taken of a larger amount; 121% is the multiplier written as the change rather than the change itself; 11% comes from slipping in the multiplication and getting 1.11 instead of 1.21.
- (b) 13/2 — Put the number of pages in the novel over the number of pages in the comic: 312/48. Divide both numbers by their highest common factor, 24: 312÷24 = 13, 48÷24 = 2, giving 13/2. (2/13 comes from writing the page counts the wrong way round. 11/2 comes from finding the difference in the page counts, 312 − 48 = 264, and writing it as a fraction of the comic's page count, 264/48. 13/15 comes from comparing the novel's page count to the total number of pages in both books, 312/360.)
- (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.
- (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.
- (d) 19% — Method: write each decrease as a multiplier, multiply the multipliers, then compare the result with 100%. Working: a 10% decrease is a multiplier of 0.9, so the two reductions together give 0.9 × 0.9 = 0.81; the final price is 81% of the original, so the price has fallen by 100% − 81% = 19%. Answer: an overall decrease of 19%. The distractors: 20% comes from adding the two reductions, 10% + 10%, which charges the second 10% against the original price instead of against the already reduced price; 21% comes from using the increase multiplier by mistake, since 1.1 × 1.1 = 1.21, and reading that 21% as a decrease; 81% is the percentage of the original price still being paid, not the percentage taken off.
- (d) 7/3 — Put the kettle's energy over the toaster's energy: 2.1/0.9. Multiply both numbers by 10 to clear the decimals: 21/9. Divide both by their highest common factor, 3: 21÷3 = 7, 9÷3 = 3, giving 7/3. (3/7 comes from writing the energy values the wrong way round. 4/3 comes from finding the difference, 2.1 − 0.9 = 1.2 kWh, and writing it as a fraction of the toaster's energy, 1.2/0.9. 7/10 comes from comparing the kettle's energy to the total energy used by both appliances, 2.1/3.0.)
- (b) Day 3 to Day 4 — Method: the average rate of increase between two consecutive days is the difference in the number of orders divided by the number of days between them, which here is just the difference itself, since each gap is one day; comparing all four differences finds which is greatest. Working: the differences are 1,509 − 1,284 = 225 (Day 1 to Day 2), 1,830 − 1,509 = 321 (Day 2 to Day 3), 2,296 − 1,830 = 466 (Day 3 to Day 4), and 2,510 − 2,296 = 214 (Day 4 to Day 5); 466 is the greatest of the four, so the rate of increase was greatest from Day 3 to Day 4. Choosing Day 4 to Day 5 comes from picking the interval that ends on the highest total number of orders, 2,510, confusing the SIZE of the total with the RATE at which it grew. Choosing Day 1 to Day 2 comes from assuming the rate must be greatest at the very start, without working out any of the four differences. Choosing Day 2 to Day 3 comes from comparing only the first two differences, 225 and 321, and stopping there without checking Day 3 to Day 4 or Day 4 to Day 5. Finding the greatest rate of change from a table always means computing every difference between consecutive values and comparing them all — the day with the highest total, or the first pair you check, is not a shortcut.
- (b) €230.00 — Multiply the amount in pounds by the exchange rate: 200 × 1.15 = 230, so £200 = €230.00. Working out 200 + 1.15 = 201.15 treats the exchange rate as an amount to add rather than a multiplier. Working out 200 × 0.15 = 30 finds only the extra amount earned for every pound and forgets to add it back to the original £200. Working out 200 × 11.5 = 2300.00 misplaces the decimal point in the exchange rate, multiplying by 11.5 instead of 1.15. £200 converts to €230.00.
- (c) £22.50 — Method: find the cost of 1 kg, then multiply by the mass wanted. Working: £36 ÷ 2.4 = £15 per kilogram, and 1.5 × £15 = £22.50. Answer: £22.50. The distractors: £15 is the price of 1 kg, which is the first step and not what the question asks for; £54 comes from multiplying the £36 by 1.5 without first reducing it to a price per kilogram; £57.60 comes from using the scale factor upside down, multiplying £36 by 2.4 ÷ 1.5 = 1.6 instead of by 1.5 ÷ 2.4.
- (c) 24p — Method: change the price to pence, then find what one tenth of 1 kg costs, since 100 g is one tenth of 1 kg. Working: £2.40 = 240p per kg, and 240 ÷ 10 = 24p per 100 g. So the cost is 24p per 100 g. Distractor 2.4p comes from dividing by 100 instead of 10. Distractor 2400p comes from multiplying by 10 instead of dividing. Distractor 240p comes from using the price per kg without scaling it down to 100 g.
- (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.
- (d) £270 — Method: the £90 is a difference between two shares, so turn it into a number of parts before finding the value of one part. Working: the largest share is 3 parts and the smallest is 1 part, so the difference is 3 − 1 = 2 parts and 2 parts are worth £90; one part = £90 ÷ 2 = £45; the whole amount is 1 + 2 + 3 = 6 parts, so 6 × £45 = £270. Answer: £270. The distractors: £540 comes from treating the £90 as the value of one part and multiplying it by the 6 parts; £180 comes from finding the £45 correctly but adding only the 1-part and 3-part shares and forgetting the middle share; £135 comes from multiplying £45 by 3 and giving the largest share instead of the total.
- (b) £76.00 — One part of the ratio is £47.50 ÷ 5 = £9.50. The school receives 8 parts, so its share is 9.50 × 8 = £76.00. Dividing £47.50 by 8 instead of 5, treating the charity's amount as if it were 8 parts, gives 47.50 ÷ 8 = 5.9375, then × 5 = £29.69. Adding the charity's amount to the school's amount instead of stopping at the school's own share gives the total collected, 9.50 × 13 = £123.50. Adding one part to the charity's amount instead of multiplying one part by 8 gives 47.50 + 9.50 = £57.00.
- (d) £30.00 — The gradient is (130 − 70) ÷ (5 − 2) = 60 ÷ 3 = £20 per hour. Using the point (2, 70): the cost for 2 hours at £20 per hour is 20 × 2 = £40, so the call-out fee is 70 − 40 = £30. Taking the C-value of the first point as the fee without subtracting the hourly cost gives £70.00 — but that point already includes 2 hours of the hourly rate. Using the gradient itself as the fee, £20.00, confuses the rate per hour with the fixed charge. Subtracting 20 × 3 = 60 instead of 20 × 2 = 40 (using the wrong h-value) gives 70 − 60 = £10.00.
- (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.
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