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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- (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.
- (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) £32 — First find the gradient: (26 − 14) ÷ (50 − 20) = 12 ÷ 30 = £0.40 per minute. Using the point (20, 14), the charge for 65 minutes is 14 + 0.40 × (65 − 20) = 14 + 18 = £32. Choosing £26 comes from treating the charge as directly proportional to the time, multiplying the gradient by 65 minutes and ignoring the fixed part of the charge (0.40 × 65 = 26). Choosing £40 comes from treating £14 as if it were the charge at 0 minutes, then adding the gradient multiplied by the full 65 minutes (14 + 0.40 × 65 = 40), instead of multiplying by the extra time past 20 minutes. Choosing £33.80 comes from assuming the charge is directly proportional to the minutes already known, scaling up from the point (50, 26) in the ratio 65:50 (65 ÷ 50 × 26 = 33.80).
- (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.
- (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 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.
- (d) 5 — The gradient equals the amount gained divided by the time taken: 15 ÷ 3 = 5 litres per minute.
- (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) −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.
- (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.
- (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) £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.
- (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) £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.
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