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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Ratio, proportion and rates of change worksheet — GCSE Higher
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- 1.The height of water in a tank, h metres, t minutes after a tap is opened is modelled by h = 0.02t² + 0.5. Estimate the instantaneous rate of change of the height at t = 10, using the gradient of the chord joining t = 9 and t = 11.
- 2.A machine fills bottles at a constant rate. It fills 18 bottles in 3 minutes. Working at the same rate, work out how many bottles the machine fills in 8 minutes.
- 3.A firework rocket's height above the ground, in metres, t seconds after launch, is modelled by h = 30t − 5t². Use a chord between t = 1 second and t = 3 seconds to estimate the instantaneous rate of change of the height at t = 2 seconds, in m/s.
- 4.A straight line passes through the points (1, 20) and (5, 8). Work out the gradient of the line.
- 5.Aisha invests £3200 in Account A, which pays 5% compound interest each year. She also invests £3200 in Account B, which pays 3% simple interest each year. Work out how much more Account A is worth than Account B after 2 years.
- 6.A greenhouse's temperature, in °C, is recorded every hour during the day: 09:00, 15.0; 10:00, 18.4; 11:00, 20.1; 12:00, 19.8; 13:00, 17.2. Work out between which two consecutive readings the instantaneous rate of change of the temperature is most likely to have been zero.
- 7.Two broadband providers show their monthly cost on straight-line graphs, with the cost in pounds on the vertical axis and the data used in gigabytes on the horizontal axis. Provider X's line passes through (0, 15) and (100, 35). Provider Y's line passes through (0, 5) and (100, 45). Work out which provider is cheaper for a customer who uses 40 gigabytes in a month.
- 8.A car is bought for £9000. Its value decreases by 8% each year. Work out its value after 2 years.
- 9.Two mathematically similar hexagonal tiles have areas 18 cm² and 50 cm². Work out the ratio of the side length of the smaller tile to the side length of the larger tile, in simplest form.
- 10.A force of 126 N acts on an area of 3.5 m². Work out the pressure on the area, in N/m².
- 11.y is directly proportional to x. When x = 4, y = 10. Work out the value of y when x = 6.
- 12.A sequence is defined by the iterative rule x_{n+1} = 0.5x_n + 20, with x_0 = 0. By working out x_1, x_2 and x_3, find the value that the sequence is approaching.
- 13.Two mathematically similar logos are printed on a poster. They have areas 20 cm² and 45 cm². Nadia says that the length scale factor from the smaller logo to the larger logo is 45 ÷ 20 = 2.25. Give a reason why Nadia is incorrect, and work out the correct length scale factor.
- 14.A factory machine produces bottles at a constant rate. In 45 minutes it produces 810 bottles. The factory needs 2,160 bottles for an order. Working at the same rate, work out how many minutes it will take to produce the order.
- 15.A printer's ink cartridge level, I millilitres, is plotted against the number of pages printed, p, and the points lie on a straight line. The line passes through (0, 24) and (300, 9). The printer has now printed 300 pages. Work out how many more pages it can print before the cartridge is empty.
Answer key
- (d) 0.40 m/min — To estimate an instantaneous rate of change at a point without a diagram, use the gradient of a chord joining two points close to it, one on each side. At t = 9: 0.02 × 81 = 1.62, so h = 1.62 + 0.5 = 2.12. At t = 11: 0.02 × 121 = 2.42, so h = 2.42 + 0.5 = 2.92. The change in height is 2.92 − 2.12 = 0.80 and the change in time is 11 − 9 = 2, so the gradient of the chord is 0.80 ÷ 2 = 0.40. Reporting the change in height, 0.80, on its own is not a rate, because it has not been divided by the 2 minutes over which it happened. Using the chord from t = 0 (where h = 0.5) to t = 11 instead gives 2.92 − 0.5 = 2.42, and 2.42 ÷ 11 = 0.22, which is the average gradient over the whole 11 minutes, not the instantaneous rate at t = 10. Substituting t = 10 into the formula gives 0.02 × 100 + 0.5 = 2.50, which is the height of the water at that moment, not the rate at which the height is rising. The estimated instantaneous rate of change at t = 10 is 0.40 m/min.
- (d) 48 — Find the rate first: 18 ÷ 3 = 6 bottles per minute. Then apply it to the new time: 6 × 8 = 48 bottles. Working out 18 + (8 − 3) = 23 adds the extra 5 minutes onto the number of bottles instead of scaling proportionally. Working out 18 × 8 = 144 multiplies the given number of bottles by the new number of minutes without finding the rate first. Writing 18 keeps the count the same, not realising it must change with the time. In 8 minutes the machine fills 48 bottles.
- (d) 10 — First find the height at each end of the chord. At t = 1, h = 30 × 1 − 5 × 1² = 30 − 5 = 25. At t = 3, h = 30 × 3 − 5 × 3² = 90 − 45 = 45. The gradient of the chord estimates the instantaneous rate at the midpoint t = 2: 45 − 25 = 20, then 20 ÷ (3 − 1) = 20 ÷ 2 = 10 m/s. Finding the change in height but forgetting to divide by the change in time gives 20, which is a distance, not a rate. Averaging the two heights instead of finding the difference gives (25 + 45) ÷ 2 = 70 ÷ 2 = 35. Subtracting in the wrong order, 25 − 45 = −20, then −20 ÷ 2 = −10, gives the correct size with the sign flipped — the rocket is rising, not falling, at t = 2 seconds, so a negative rate cannot be right here.
- (b) −3 — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, with both changes taken in the same direction along the line. Working: going from (1, 20) to (5, 8) the change in y is 8 − 20 = −12 and the change in x is 5 − 1 = 4, so the gradient is −12 ÷ 4 = −3. Answer: −3, and the negative sign is expected because the line falls from left to right. The distractors: 3 comes from subtracting the smaller y from the larger, 20 − 8 = 12, while still taking the x values from left to right, which loses the minus sign that says the line falls; −12 is the change in y left undivided by the change in x of 4; −1/3 comes from dividing the change in x by the change in y, 4 ÷ (−12), turning the gradient upside down.
- (c) £136 — 5% interest each year means the value becomes 100% + 5% = 105% of the previous year's value, and 105% = 1.05, so the multiplier is 1.05. Account A: £3200 × 1.05 × 1.05 = £3528. Account B (simple interest): £3200 + 2 × (£3200 × 0.03) = £3392. The difference is £3528 − £3392 = £136. (£128 comes from working out Account A with simple interest too, instead of compound: £3200 + 2 × (£3200 × 0.05) = £3520, then £3520 − £3392 = £128. £3528 is the value of Account A on its own, not the difference between the two accounts. £3392 is the value of Account B on its own, not the difference.)
- (a) 11:00 to 12:00 — Method: the instantaneous rate of change is zero at a turning point, where a rising trend becomes a falling trend; that lies within the first interval whose difference has changed sign from the interval before it. Working: the differences between consecutive readings are +3.4 °C (09:00 to 10:00), +1.7 °C (10:00 to 11:00), −0.3 °C (11:00 to 12:00) and −2.6 °C (12:00 to 13:00); the sign changes from positive to negative within 11:00 to 12:00, since the temperature is still rising up to 11:00 (20.1 °C, the highest recorded value) and has fallen by 12:00, so the instantaneous rate of change was zero somewhere within that interval. Choosing 09:00 to 10:00 picks out the interval with the largest positive difference, +3.4 °C, confusing the fastest rise with no change at all. Choosing 10:00 to 11:00 picks the last interval where the temperature was still rising, one interval too early, without checking that the very next interval turns negative. Choosing 12:00 to 13:00 picks out the interval with the largest-magnitude difference, −2.6 °C, the fastest fall, not where the change is zero. Zero instantaneous rate of change happens at a turning point, where the readings stop rising and start falling — find the FIRST interval whose difference has flipped sign from the one before it, not the biggest change or an interval where the old sign still held.
- (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.
- (c) £7617.60 — To decrease by 8% each year, multiply by 0.92 (100% − 8%) twice. £9000 × 0.92 × 0.92 = £7617.60. £7560.00 comes from treating the two 8% decreases as a single flat 16% decrease applied once instead of compounding: £9000 × 0.84 = £7560.00. £8280.00 comes from applying the 8% decrease only once, for 1 year instead of 2: £9000 × 0.92 = £8280.00. £10497.60 comes from multiplying by 1.08 twice, increasing the value instead of decreasing it: £9000 × 1.08 × 1.08 = £10497.60.
- (d) 3 : 5 — Simplify the area ratio: 18 : 50 divides by 2 to give 9 : 25. Areas scale with the square of the length ratio, so take the square root of each part: the square root of 9 is 3, and the square root of 25 is 5, giving a side length ratio of 3 : 5. Giving 5 : 3 has the ratio the right way round for larger to smaller, not smaller to larger. Giving 9 : 25 is the simplified area ratio, without square-rooting it. Giving 18 : 50 is the area ratio before it has even been simplified.
- (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) 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) 40 — Working out successive terms shows where the sequence is heading, but the terms themselves keep changing — the limit is the value where the sequence stops changing, so x_{n+1} = x_n = L there. Substituting into the rule: L = 0.5L + 20. Subtracting 0.5L from both sides: L − 0.5L = 20, so 0.5L = 20, and L = 20 ÷ 0.5 = 40. The individual terms are x_1 = 0.5 × 0 + 20 = 20, x_2 = 0.5 × 20 + 20 = 30 and x_3 = 0.5 × 30 + 20 = 35, getting closer to this value but not equal to it — 35 is only the third term, not the limit. Multiplying by 0.5 instead of dividing at the final step, 20 × 0.5 = 10, undoes the rearrangement rather than completing it, and gives a value smaller than terms the sequence has already passed. Writing the fixed-point equation with the wrong sign, L = 0.5L − 20, gives 0.5L = −20 and L = −40, which cannot be right since every term in the sequence is positive and increasing. The value the sequence is approaching is 40.
- (d) 1.5 — Method: the length scale factor is the square root of the area scale factor, not the area scale factor itself. Working: the area scale factor is 45 ÷ 20 = 2.25, and the square root of 2.25 is 1.5. Answer: 1.5. Nadia's answer, 2.25, is the AREA scale factor — she never took the square root to get back to the length scale factor. 4.5 comes from doubling the area scale factor instead of taking its square root. 0.67 comes from taking the square root in the wrong direction, finding the scale factor from the larger rug to the smaller rug instead of the other way round.
- (b) 120 minutes — Method: find the rate in bottles per minute, then divide the order size by the rate. Working: rate = 810 ÷ 45 = 18 bottles per minute. Time = 2,160 ÷ 18 = 120 minutes. Wrong options: 1,350 minutes comes from subtracting 810 from 2,160 instead of using the rate; 48 minutes comes from dividing the order size by the original time (2,160 ÷ 45) instead of the rate; 108 minutes comes from rounding the rate to 20 bottles per minute before dividing.
- (c) 180 pages — The gradient is (9 − 24) ÷ (300 − 0) = −15 ÷ 300 = −0.05, so the cartridge uses 0.05 ml of ink per page. At p = 300 there are 9 ml left. The extra pages before the cartridge is empty is 9 ÷ 0.05 = 180 pages. Giving 9 as the answer confuses the millilitres of ink remaining with the number of pages remaining — they are different quantities with different units. Multiplying instead of dividing, 9 × 0.05 = 0.45, does not undo the rate correctly. Working out the total number of pages a full cartridge lasts, 24 ÷ 0.05 = 480 pages, answers how many pages the cartridge prints in total from full, not how many more pages it can print from the 300-page point.
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