Printable · GCSE Higher · ages 14-16
Gradient as a rate of change worksheet — GCSE Higher
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 Higher
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- 1.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?
- 2.An oven is preheating. Its temperature, T °C, is plotted against time, t minutes, on a straight-line graph. The line passes through the points (2, 60) and (6, 140). Work out the gradient of the line.
- 3.A mobile phone tariff is shown on a straight-line graph with the monthly cost, C pounds, on the vertical axis and the amount of data used, g gigabytes, on the horizontal axis. The line passes through (0, 10) and (8, 26). Work out the gradient and say what it represents.
- 4.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.
- 5.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.
- 6.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.
- 7.Two taxi firms show their charges on straight-line graphs, with the cost in pounds on the vertical axis and the distance in miles on the horizontal axis. Firm A's line passes through (0, 4) and (5, 14). Firm B's line passes through (0, 6) and (5, 21). Work out which firm charges more per mile.
- 8.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.
- 9.A straight line passes through the points (1, 20) and (5, 8). Work out the gradient of the line.
- 10.A water butt is being filled from a hosepipe at a constant rate while a small leak drains water out at a constant rate, giving a constant net rate of change. The volume of water in the butt, V litres, is shown on a straight-line graph against time, t minutes. The line passes through the points (5, 20) and (15, 60). The butt is empty at t = 0. Work out how many minutes it takes to reach a volume of 100 litres.
- 11.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.
- 12.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?
- 13.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.
- 14.A straight-line graph shows the number of pages printed, p, plotted against the time in minutes, t, since a printer started a print job. The line passes through the points (2, 30) and (5, 90). Work out the gradient of the line.
- 15.The amount of fuel left in a car's tank, F litres, is plotted against the distance travelled, d miles, and the points lie on a straight line. The line passes through (0, 45) and (150, 15). Work out the gradient of the line and say what it tells you.
Answer key
- (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) 20 — Gradient = change in T ÷ change in t = (140 − 60) ÷ (6 − 2) = 80 ÷ 4 = 20. A student who subtracts in the wrong order gets −20. A student who divides 80 by 2 instead of 4 gets 40. A student who wrongly treats the line as passing through the origin and uses the point (2, 60) on its own gets 60 ÷ 2 = 30.
- (b) 2, the cost in pounds of each extra gigabyte — Method: the gradient is the change in cost divided by the change in data, so it is the cost of each extra gigabyte; the value where the line meets the vertical axis is the charge before any data is used, which is a different quantity. Working: from (0, 10) to (8, 26) the cost rises by 26 − 10 = 16 pounds while the data rises by 8 − 0 = 8 gigabytes, so the gradient is 16 ÷ 8 = 2, meaning each extra gigabyte costs £2. Answer: 2, the cost in pounds of each extra gigabyte. The distractors: '10, the cost in pounds of each extra gigabyte' reads the intercept as the gradient, but 10 is what the tariff costs when no data at all has been used; '3.25, the cost in pounds of each extra gigabyte' comes from 26 ÷ 8, treating the line as though it passed through the origin when it starts at 10; '2, the fixed monthly charge in pounds' has the gradient right but describes the intercept, and the fixed charge on this tariff is £10.
- (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.
- (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.
- (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) Firm B — £3 per mile against Firm A's £2 per mile — Firm A's gradient is (14 − 4) ÷ 5 = 2, so it charges £2 per mile. Firm B's gradient is (21 − 6) ÷ 5 = 3, so it charges £3 per mile. £3 is more than £2, so Firm B charges more per mile. Swapping the two firms' gradients gives the answer with Firm A at £3 and Firm B at £2, which has the labels the wrong way round. Dividing the change in miles by the change in cost, instead of the other way round, gives 5 ÷ 10 = £0.50 for Firm A and 5 ÷ 15 = £0.33 for Firm B and so names Firm A — that is the gradient upside down. And a positive fixed charge does not mean two firms charge the same rate: the rate is found from the gradient, not from whether the intercept is positive.
- (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.
- (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.
- (d) 25 — Gradient = (60 − 20) ÷ (15 − 5) = 40 ÷ 10 = 4 litres per minute. Since the butt is empty at t = 0, V = 4t. Setting V = 100 gives t = 100 ÷ 4 = 25 minutes.
- (d) £2.00 — Cost for Printer A = 200 × £0.04 = £8. Cost for Printer B = 200 × £0.05 = £10. Difference = £10 − £8 = £2.00.
- (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.
- (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) 20 — The gradient is the change in p divided by the change in t: (90 − 30) ÷ (5 − 2) = 60 ÷ 3 = 20. Choosing 0.05 comes from dividing the change in t by the change in p, the wrong way round (3 ÷ 60). Choosing −20 comes from subtracting the coordinates in the wrong order for one part of the calculation, for example (30 − 90) ÷ (5 − 2), giving a negative value. Choosing 18 comes from dividing the second p-coordinate by the second t-coordinate directly (90 ÷ 5) instead of using the change between the two points.
- (c) −0.2, the car uses 0.2 litres of fuel for each mile — Method: the gradient is the change in the vertical value divided by the change in the horizontal value, which on this graph is a number of litres for each mile, and a negative gradient means the vertical quantity is going down. Working: from (0, 45) to (150, 15) the fuel changes by 15 − 45 = −30 litres while the distance changes by 150 − 0 = 150 miles, so the gradient is −30 ÷ 150 = −0.2, which says the tank loses 0.2 litres for every mile driven. Answer: −0.2, the car uses 0.2 litres of fuel for each mile. The distractors: '0.2, the car gains 0.2 litres of fuel for each mile' comes from subtracting the fuel values the other way round, 45 − 15 = 30, which drops the minus sign and reverses what the graph says; '−5, the car uses 5 litres of fuel for each mile' comes from dividing the change in distance by the change in fuel, 150 ÷ (−30), turning the gradient upside down; '−30, the car uses 30 litres of fuel for each mile' is the change in fuel on its own, never divided by the 150 miles travelled.
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