OCR GCSE Mathematics: Interpreting graphs: real-world graphs, gradients and areas (J560) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 7.04 of the OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026): real-world graphs, gradients and areas under graphs. Questions marked “(Higher tier only)” test content the specification lists for the Higher tier only. All other questions are suitable for both tiers.

Links: study guide, revision notes, OCR GCSE Mathematics hub and checklist.

Questions

1. A graph converts pounds (horizontal axis) to Danish kroner (vertical axis). It is a straight line through (0, 0) and (50, 430).

(a) Write down the gradient and say what it means. [1] (b) Use the gradient to change 645 kroner into pounds. [2]

2. Temperature in kelvin is given by K = C + 273, where C is the temperature in degrees Celsius. The graph of K against C is a straight line.

(a) Find K when C = 36. [1] (b) Write down the gradient and the intercept of the graph. [2] (c) Explain why K is not directly proportional to C. [1]

3. Priya walks along a path. Her distance-time graph (time in hours, distance from home in km) joins (0, 0), (1, 5), (1.5, 5), (2.5, 11).

(a) Find her speed in the first hour. [1] (b) For how many minutes is she stationary? [1] (c) Find her speed in the final section. [1] (d) Find her average speed for the whole walk. [2]

4. The cost, C pounds, of m metres of fabric is shown by a straight line through (0, 0) and (6, 39).

(a) Find the price per metre. [1] (b) Write an equation linking C and m. [1] (c) Find the cost of 9.5 metres. [1] (d) Another graph is a straight line through (0, 5) and (4, 13). Explain why it does not show direct proportion. [1]

5. Fifteen volunteers take 8 hours to pack parcels. The number of volunteers n and time t hours are inversely proportional.

(a) Find the time taken by 24 volunteers. [2] (b) How many volunteers would finish in 3 hours? [2] (c) Describe the shape of the graph of t against n. [1]

6. A ramp rises 0.9 m over a horizontal distance of 7.2 m.

(a) Write the gradient as a ratio vertical : horizontal in the form 1 : n. [1] (b) A ramp with the same gradient covers a horizontal distance of 20 m. How far does it rise? [2]

7. (Higher tier only) A lift-off test records velocity against time. The graph joins (0, 0), (6, 18), (20, 18), (26, 0), with time in seconds and velocity in m/s.

(a) Find the acceleration in the first 6 seconds. [1] (b) Find the total distance travelled. [3] (c) Find the average speed for the 26 seconds, to 3 significant figures. [1]

8. (Higher tier only) A particle’s distance s metres from a point after t seconds is s = t² + 2t.

(a) Complete the values of s for t = 0, 1, 2, 3, 4. [1] (b) Find the average speed between t = 1 and t = 4. [2] (c) Find the gradient of the chord from t = 2 to t = 3. [1] (d) A tangent to the curve at t = 2 passes through (0, -4) and (4, 20). Find its gradient and say what it represents. [2]

9. (Higher tier only) A shop’s takings rate (pounds per hour) is plotted against hours after opening. The graph joins (0, 10), (4, 50), (6, 50).

(a) Find the takings in the first 4 hours. [2] (b) Find the takings from 4 to 6 hours. [1] (c) Find the mean takings per hour over the 6 hours, to the nearest penny. [2] (d) State the unit of the area under this graph. [1]

10. (Higher tier only) A vehicle’s speed v m/s is given by v = 60t ÷ (t + 5) for t seconds. Values are in the table.

t	0	5	10	15	20
v	0	30	40	45	48

(a) Use four trapezia of equal width to estimate the distance travelled in 20 seconds. [3] (b) State whether your answer is an over-estimate or an under-estimate. Give a reason. [1] (c) A tangent at t = 10 passes through (4, 32) and (16, 48). Find the acceleration at t = 10. [2] (d) Find the average acceleration from t = 0 to t = 20. [2]

11. A mobile plan’s monthly cost C pounds is a straight line against data used g gigabytes, through (0, 12) and (10, 17).

(a) Find the gradient and say what it means. [2] (b) What does the intercept represent? [1] (c) Find the cost when 14 gigabytes are used. [1]

12. (Higher tier only) A cyclist’s velocity-time graph joins (0, 0), (8, 10), (28, 10), (36, 4), (40, 0), with time in seconds and velocity in m/s.

(a) Find the acceleration in the first 8 seconds. [1] (b) Find the deceleration between 28 s and 36 s. [2] (c) Find the total distance travelled. [4] (d) Find the average speed over the 40 seconds. [2]

Answers

1. (a) Gradient = 430 ÷ 50 = 8.6, meaning 8.6 kroner per pound. [1] (b) £ = 645 ÷ 8.6 [1] = £75 [1] Examiner insight: Dividing by the gradient to go from kroner back to pounds is the step that is often reversed, and a quick check that 75 × 8.6 gives 645 protects you.

2. (a) K = 36 + 273 = 309 [1] (b) Gradient = 1 [1], intercept = 273 [1] (c) The line misses the origin (K = 273 when C = 0), so doubling C does not double K. [1] Examiner insight: Base your explanation on the line missing the origin. Saying “it is not a straight line” would be wrong because the line is straight.

3. (a) 5 ÷ 1 = 5 km/h [1] (b) 1 to 1.5 hours is 30 minutes [1] (c) (11 - 5) ÷ (2.5 - 1.5) = 6 km/h [1] (d) Total distance = 11 km in 2.5 hours [1]; 11 ÷ 2.5 = 4.4 km/h [1] Examiner insight: Average speed uses the whole time including the rest, so do not average 5 and 6, and show the total distance and time to protect your method.

4. (a) 39 ÷ 6 = £6.50 per metre [1] (b) C = 6.5m [1] (c) 6.5 × 9.5 = £61.75 [1] (d) The line misses the origin: a length of 0 gives a cost of 5. [1] Examiner insight: A money answer should show pence to two decimal places, so write £61.75 and not £61.7.

5. (a) tn = 15 × 8 = 120 [1]; t = 120 ÷ 24 = 5 hours [1] (b) 120 ÷ 3 [1] = 40 volunteers [1] (c) A smooth curve that falls as n increases and does not touch either axis. [1] Examiner insight: Finding the constant product first shows your method clearly, and a bare wrong answer leaves nothing to show what you intended.

6. (a) 0.9 ÷ 7.2 = 0.125 = 1/8, so 1 : 8 [1] (b) 20 ÷ 8 [1] = 2.5 m [1] Examiner insight: Say which quantity is the vertical one, because writing 8 : 1 reverses the ratio.

7. (a) 18 ÷ 6 = 3 m/s² [1] (b) Trapezium: parallel sides 14 s and 26 s, height 18 [1]; ½ × (14 + 26) × 18 [1] = 360 m [1] (c) 360 ÷ 26 = 13.8 m/s [1] Examiner insight: Show how you split the shape. Using only the 18 and the 26 as sides ignores the 14 s top edge and gives a wrong area.

8. (a) 0, 3, 8, 15, 24 [1] (b) (24 - 3) ÷ (4 - 1) [1] = 7 m/s [1] (c) (15 - 8) ÷ 1 = 7 [1] (d) Gradient = 24 ÷ 4 = 6 [1], the speed at t = 2 s (instantaneous velocity, 6 m/s). [1] Examiner insight: A tangent gradient needs units and an interpretation in context, so write “6 m/s at 2 seconds” rather than just “6”.

9. (a) Trapezium: ½ × (10 + 50) × 4 [1] = £120 [1] (b) 50 × 2 = £100 [1] (c) Total = 220 [1]; 220 ÷ 6 = £36.67 [1] (d) Pounds per hour multiplied by hours gives pounds [1] Examiner insight: A mean per hour needs the total over the whole six hours, so divide by 6 rather than by the number of segments.

10. (a) Strip width 5. Area ≈ 5 × (0/2 + 30 + 40 + 45 + 48/2) [1]; = 5 × 139 [1] = 695 m [1] (b) Under-estimate, because the curve is concave, so the trapezia fall short of it. [1] (c) (48 - 32) ÷ (16 - 4) [1] = 4/3 m/s² (1.33) [1] (d) (48 - 0) ÷ 20 [1] = 2.4 m/s² [1] Examiner insight: A reason must refer to the shape of the curve relative to the trapezia, not just state under- or over-estimate.

11. (a) (17 - 12) ÷ 10 = 0.5 [1]; £0.50 per gigabyte. [1] (b) The fixed monthly charge of £12 when no data is used. [1] (c) 12 + 0.5 × 14 = £19 [1] Examiner insight: An interpretation needs both a number and units, so “50p for each extra gigabyte” says more than “0.5”.

12. (a) 10 ÷ 8 = 1.25 m/s² [1] (b) (4 - 10) ÷ 8 = -0.75 [1]; deceleration 0.75 m/s² [1] (c) 0 to 8 s: ½ × 8 × 10 = 40 [1]; 8 to 28 s: 20 × 10 = 200 [1]; 28 to 36 s: ½ × (10 + 4) × 8 = 56 [1]; 36 to 40 s: ½ × 4 × 4 = 8, total 304 m [1] (d) 304 ÷ 40 [1] = 7.6 m/s [1] Examiner insight: Label each piece of the area with its time interval, since one slip in a multi-part total is easier to find and fix than in a single sum.

Where marks are usually lost

• Giving gradients or areas with no units, or the wrong units.
• Turning a decimal-hour time into minutes incorrectly, for example treating 0.5 hours as 5 minutes.
• Finding the average speed from the final position instead of the total distance travelled.
• Reversing a ratio or an exchange rate when converting back.
• Using a tangent with two points too close together, which magnifies reading errors.
• Forgetting the ½ in the area of a triangle or trapezium.
• Describing the area under a velocity-time graph as speed instead of distance.
• Calling any straight line direct proportion, without checking it passes through the origin.
• Not saying whether an area estimate is an over- or under-estimate when asked.

Next steps

• Re-read the revision notes for any question you lost marks on.
• Go back to the study guide for full explanations.
• Open the OCR GCSE Mathematics hub and the printable checklist.
• Try the free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR GCSE (9-1) Mathematics J560 specification, Version 2.1 (August 2026), for first assessment in 2017, published by OCR. These questions follow section 7.04, Interpreting graphs.