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.
Answer key: Gradient as a rate of change worksheet — GCSE Foundation
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- (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) −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) 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.
- (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) 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.
- (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.
- (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.
- (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.
- (d) The cyclist's speed, in kilometres per hour — Method: on any straight-line graph the gradient is the change in the quantity on the vertical axis for each 1 unit of the quantity on the horizontal axis, so its meaning is read off the two axis labels. Working: the vertical axis is distance in kilometres and the horizontal axis is time in hours, so the gradient counts kilometres for each hour, and distance for each hour is speed. Answer: the cyclist's speed, in kilometres per hour. The distractors: 'the total distance the cyclist travels, in kilometres' reads the gradient as a value taken off the vertical axis, but a gradient is a rate and no end point of the journey has been given; 'the time the cyclist takes, in hours' names the horizontal axis, which is the quantity the gradient divides by rather than the gradient itself; 'the distance the cyclist travels in 18 hours' treats the 18 as a value of t, when 18 is the steepness of the line and not a point on it.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
- (a) Inverse proportion — A curve that decreases and never touches either axis is the standard shape for inverse proportion, y = k/x. Direct proportion graphs are straight lines through the origin, which this is not, so it must be inverse proportion rather than neither.
- (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) 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) 1.3 — The gradient of a distance-time graph gives the speed. Difference in speed = 4.5 − 3.2 = 1.3 km/h. A student who subtracts the speeds the wrong way round, 3.2 − 4.5, gets −1.3. A student who adds the two speeds instead of comparing them gets 7.7. A student who multiplies the two gradients gets 14.4.
- (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.
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