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) £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.
- (d) The line is straight and passes through the origin. — Direct proportion means y = kx for a constant k. This is a straight line, and when x = 0, y = 0, so it passes through the origin. For a positive k it slopes upward from left to right.
- (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.
- (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.
- (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.
- (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.
- (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.
- (b) 8 — The gradient of a line through the origin is the y-value divided by the x-value at any point on the line. Using (10, 80): gradient = 80 ÷ 10 = 8.
- (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.
- (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.
- (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) £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.
- (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.
- (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.
- (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.
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