Printable · GCSE Foundation · ages 14-16
Gradient as a rate of change worksheet — GCSE Foundation
Fifteen questions on "gradient as a rate of change" — DfE statement R14. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Gradient as a rate of change worksheet — GCSE Foundation
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- 1.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.
- 2.A straight line passes through the points (1, 20) and (5, 8). Work out the gradient of the line.
- 3.A plumber's charge, C pounds, for a job lasting h hours is shown on a straight-line graph. The line passes through the points (1, 45) and (3, 85). Work out what the gradient of this line represents, in context.
- 4.A straight-line graph shows the cost, C pounds, of hiring a small van for m miles driven. The line passes through the origin and the point (25, 20). Work out the gradient of the line.
- 5.A candle burns down at a constant rate. Its height, h cm, is plotted against time, t minutes, on a straight-line graph. The gradient of the line is −0.3. Which statement correctly describes what the gradient means?
- 6.A conversion graph between kilometres and miles passes through the origin and the point (50, 31), with the number of kilometres on the horizontal axis and the number of miles on the vertical axis. Work out the number of miles equivalent to 1 kilometre.
- 7.Two printers' running costs are shown on separate straight-line graphs of total cost against number of pages printed. Printer A's graph has gradient £0.04 per page. Printer B's graph has gradient £0.05 per page. Work out how much cheaper it is to print 200 pages on Printer A than on Printer B, based on these running costs alone.
- 8.A straight-line graph shows the cost, C pounds, of buying n litres of paint. The line passes through the origin and through the point (5, 40). Work out the gradient of the line.
- 9.A window cleaner charges a call-out fee plus an amount per window. The total charge, C pounds, is shown on a straight-line graph against the number of windows cleaned, w. The line passes through the points (2, 14) and (6, 26). Work out the gradient of the line.
- 10.Line P passes through the origin and the point (3, 21). Line Q passes through the origin and the point (5, 45). Both lines represent the distance, in metres, run by an athlete against the time, in seconds. Work out the gradient of the steeper line.
- 11.On a straight-line graph the volume of water, V litres, in a tank is plotted on the vertical axis and the time, t minutes, on the horizontal axis. The line passes through (2, 50) and (6, 130). Work out the gradient of the line and give its units.
- 12.Four graphs are described below. Each shows one quantity plotted against another. Write down the description of the graph that shows two quantities in direct proportion.
- 13.A gym charges members a joining fee plus a fixed amount per month. The total amount paid, C pounds, is shown on a straight-line graph against the number of months, m. The line passes through the points (3, 130) and (7, 210). Work out the total amount a member will have paid after 12 months.
- 14.A straight-line graph shows the cost, C pounds, of hiring a bike for h hours. The line passes through the points (1, 12) and (4, 27). Which of these statements about the line is true?
- 15.A hosepipe fills a paddling pool at a constant rate. The volume of water in the pool increases by 15 litres every 3 minutes. Write down the gradient of the graph of volume against time.
Answer key
- (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.
- (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.
- (a) The plumber charges £20 for each extra hour worked — The gradient is the change in C divided by the change in h: (85 − 45) ÷ (3 − 1) = 40 ÷ 2 = £20. On this graph the gradient represents the extra amount charged for each extra hour worked, so the answer is 'The plumber charges £20 for each extra hour worked'. Not dividing by the change in h gives 40, the option that reads 'charges £40 for each extra hour worked' — that is the total change in cost between the two points, not the rate per hour. Reading off the C-value of the first point, 45, gives the option using £45 — that is the cost of a job lasting 1 hour, not the rate. Treating the gradient as a flat total charge regardless of the time taken misunderstands what a straight-line graph through two different h-values shows: the cost does depend on h, so it cannot be a single fixed total.
- (a) 0.8 — The gradient of a line through the origin is the y-coordinate of a point divided by its x-coordinate: 20 ÷ 25 = 0.8. Choosing 1.25 comes from dividing the wrong way round, 25 ÷ 20. Choosing 20 comes from reading off the cost at the point instead of dividing it by the number of miles. Choosing 5 comes from subtracting the two coordinates (25 − 20) instead of dividing them.
- (d) The candle's height decreases by 0.3 cm every minute. — A negative gradient means the quantity on the vertical axis decreases as the quantity on the horizontal axis increases. The size of the gradient, 0.3, gives the amount of decrease per minute.
- (a) 0.62 miles — The gradient of the line is the change in miles divided by the change in kilometres: 31 ÷ 50 = 0.62, so 1 kilometre converts to 0.62 miles. Dividing the wrong way round, 50 ÷ 31 = 1.612..., rounds to 1.61 miles — that finds how many kilometres are in 1 mile, not the reverse. Doubling the gradient, 1.24 miles, comes from using 62 ÷ 50 instead of 31 ÷ 50. Reading off the y-coordinate of the given point without dividing by the x-coordinate gives 31.00 miles, which is the number of miles for 50 kilometres, not for 1 kilometre.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (b) 8 — Method: the gradient of a straight line is the change in the vertical value divided by the change in the horizontal value between two points on the line. Working: from the origin (0, 0) to (5, 40) the vertical change is 40 − 0 = 40 and the horizontal change is 5 − 0 = 5, so the gradient is 40 ÷ 5 = 8, which here means a cost of £8 for each litre. Answer: 8. The distractors: 0.125 comes from dividing the horizontal change by the vertical change, 5 ÷ 40, which gives litres per pound instead of the gradient; 40 comes from reading off the vertical value of the point and calling it the gradient, ignoring the 5 litres it took to reach that cost; 35 comes from subtracting the two coordinates, 40 − 5, instead of dividing them.
- (a) £3.00 per window — The gradient is the change in C divided by the change in w. Change in C = 26 − 14 = 12. Change in w = 6 − 2 = 4. Gradient = 12 ÷ 4 = £3.00 per window. Dividing the change in w by the change in C instead gives 4 ÷ 12 = £0.33 per window. Dividing 12 by the larger w-value only, 12 ÷ 6 = £2.00 per window, comes from not subtracting the smaller w-value first. Adding the w-values instead of subtracting, 12 ÷ (6 + 2) = £1.50 per window, comes from a sign error when finding the change in w.
- (b) 9 m/s — The gradient of line P is 21 ÷ 3 = 7, so P has a rate of 7 m/s. The gradient of line Q is 45 ÷ 5 = 9, so Q has a rate of 9 m/s. Because 9 is greater than 7, line Q is the steeper line, with gradient 9 m/s. Taking line P's gradient instead of Q's gives 7 m/s, the less steep line. Subtracting the two lines' coordinates directly, (45 − 21) ÷ (5 − 3) = 24 ÷ 2 = 12 m/s, mixes points from different lines rather than using one line's own two points. Adding the two gradients, 7 + 9 = 16 m/s, treats 'steeper' as a total rather than a comparison.
- (a) 20 litres per minute — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, and its units are the vertical unit for each one of the horizontal unit. Working: from (2, 50) to (6, 130) the volume changes by 130 − 50 = 80 litres and the time changes by 6 − 2 = 4 minutes, so the gradient is 80 ÷ 4 = 20, measured in litres for each minute. Answer: 20 litres per minute. The distractors: 25 litres per minute comes from using one point on its own, 50 ÷ 2, which assumes the line starts at the origin when the tank already held 50 litres at 2 minutes; 0.05 litres per minute comes from dividing the change in time by the change in volume, 4 ÷ 80, which gives the time for each litre but is then labelled as litres for each minute; 20 minutes for each litre has the right value with the units the wrong way round, and a tank that needed 20 minutes to gain a single litre would be filling far more slowly than this one.
- (c) A straight line through the origin, rising from left to right — Two quantities are in direct proportion when one is a constant multiple of the other, so their graph is a straight line through the origin: when one quantity is 0 the other is 0 as well, and doubling one doubles the other. A straight line crossing the vertical axis at 5 has a fixed amount added on, so when the horizontal quantity is 0 the vertical quantity is 5, not 0 — a straight line on its own is not enough for direct proportion. A curve that falls steeply and then levels off without touching either axis shows inverse proportion: one quantity grows as the other shrinks, and their product stays the same. A horizontal line at a height of 3 shows a quantity that does not change at all as the other one grows, so it is not proportional to it.
- (c) £310 — Gradient = (210 − 130) ÷ (7 − 3) = 80 ÷ 4 = 20, so the monthly rate is £20. Using C = 20m + c with the point (3, 130): 130 = 60 + c, so c = 70. After 12 months: C = 20 × 12 + 70 = 240 + 70 = £310.
- (a) The hourly rate is £5 and the fixed fee is £7 — The gradient is (27 − 12) ÷ (4 − 1) = 15 ÷ 3 = £5, the hourly rate. Using the point (1, 12): 12 = 5 × 1 + fee, so the fee is 12 − 5 = £7. That gives 'The hourly rate is £5 and the fixed fee is £7'. Swapping the two figures gives the statement with £7 as the rate and £5 as the fee, which has them the wrong way round. Taking the C-value of the first point, £12, as the fixed fee ignores that 1 hour of hire is already included in that £12. Using 15, the change in C, as the hourly rate without dividing by the change in h (3 hours) gives the statement claiming a £15 hourly rate.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
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