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

These notes condense section 7.04 of the OCR GCSE (9-1) Mathematics (J560) specification, Version 2.1 (August 2026): 7.04a real-world graphs, 7.04b gradients and 7.04c areas. For full explanations and worked examples, read the study guide. Content marked “Higher tier only” is listed for the Higher tier only in the specification.

Links: practice questions, OCR GCSE Mathematics hub, checklist and free 10-minute diagnostics.

Key ideas at a glance

Idea	What to remember
Gradient	(change in y) ÷ (change in x), with units of y per unit of x
Gradient and ratio	Gradient 3/4 means vertical : horizontal = 3 : 4
Direct proportion	y = kx, straight line through the origin, gradient k
Inverse proportion	y = k/x, so xy = k, a falling curve that never meets the axes
Conversion graph	Straight line; through the origin for currency, not for Celsius to Fahrenheit
Distance-time gradient	Speed (velocity if direction matters)
Velocity-time gradient	Acceleration (Higher tier only)
Velocity-time area	Distance (Higher tier only)
Chord gradient	Average rate of change (Higher tier only)
Tangent gradient	Instantaneous rate of change (Higher tier only)

7.04a Real-world graphs

Method: reading a graph

1. Check the axis labels and units.
2. Note the starting value.
3. Read across and up with a ruler, and show the lines if you can.
4. Give the answer with units.

Conversion graphs

• Currency: line through the origin. The gradient is the exchange rate.
• Temperature: F = 1.8C + 32. Gradient 1.8, intercept 32.
• Not a proportion graph if the line misses the origin.

Distance-time features

Feature	Meaning
Up-sloping straight line	Constant speed away
Horizontal	Stationary
Down-sloping straight line	Constant speed back
Steeper	Faster

Average speed = total distance travelled ÷ total time. Returning to the start makes the distance from the start 0, but the distance travelled is not 0.

Proportion graphs

	Direct	Inverse
Equation	y = kx	y = k/x
Shape	Straight line through (0, 0)	Curve falling to the right
Constant	y ÷ x = k	x × y = k
Doubling x	y doubles	y halves

Reminder: 6 workers take 10 days gives k = 60, so 15 workers take 60 ÷ 15 = 4 days.

7.04b Gradients and rates of change

Method: finding a gradient from a straight line

1. Pick two points that are easy to read.
2. Find the change in y and the change in x.
3. Divide, keeping the sign.
4. Add units: y-units per x-unit.

Reminder: a line from (2, 5) to (8, 23) has gradient 18 ÷ 6 = 3.

Higher tier only: velocity-time graphs

• Gradient is acceleration in m/s².
• Negative gradient is deceleration.
• A horizontal line is constant velocity.

Higher tier only: chords and tangents

	Chord	Tangent
What it is	Line joining two points on the curve	Line touching the curve at one point
Gives	Average rate of change	Instantaneous rate of change
How	Gradient between the two points	Draw with a ruler, then read two widely spaced points

Reminder: for s = 3t², the chord from t = 2 to t = 2 + h has gradient 12 + 3h. As h gets smaller, the gradient approaches 12 m/s, the instantaneous rate at t = 2.

7.04c Areas under graphs (Higher tier only)

Method: area under a graph

1. Split the region into triangles, rectangles and trapezia.
2. Trapezium area = ½ × (sum of parallel sides) × width.
3. For a curve, use strips and trapezia to estimate.
4. State the units (axis unit × axis unit) and what the area means.

Graph	Area means
Velocity against time	Distance travelled
Rate of earning (£ per hour) against time	Total money earned

Reminder: velocity rising steadily from 4 m/s to 10 m/s over 6 s gives distance ½ × (4 + 10) × 6 = 42 m.

Over or under: on a concave curve the trapezia fall short of it, so the estimate is too small. On a convex curve the estimate is too big.

Worked reminders

Distance-time. A graph joins (0, 0), (2, 30), (3, 30), (6, 0), with hours across and km up.

0 to 2 h:   30 ÷ 2 = 15 km/h away
2 to 3 h:   horizontal, so stationary for 1 h
3 to 6 h:   30 ÷ 3 = 10 km/h back
Distance travelled = 60 km, so average speed = 60 ÷ 6 = 10 km/h

Inverse proportion. Five machines take 12 hours.

k = 5 × 12 = 60
Eight machines: 60 ÷ 8 = 7.5 hours

Higher tier only: velocity-time. A graph joins (0, 0), (4, 8), (10, 8), (14, 0), with seconds across and m/s up.

Acceleration at the start = 8 ÷ 4 = 2 m/s²
Distance = ½ × (6 + 14) × 8 = 80 m
Check: 16 + 48 + 16 = 80 m

Higher tier only: drawing a tangent. Rest the ruler against the curve at the point so the curve sits evenly on both sides. Draw the line long. Pick two widely spaced points on the line, then divide the rise by the run. Give the answer as an estimate, because a different but sensible tangent will give a slightly different value.

Must-know distinctions

• Speed vs distance from start. Gradient of a distance-time graph: speed. Height of that graph: distance from the start.
• Gradient vs area. Gradient is a rate. Area is an accumulated total.
• Chord vs tangent. Chord means average. Tangent means at one instant.
• Linear vs proportional. Every direct proportion graph is a straight line, but not every straight line is a direct proportion graph.
• Direct vs inverse. Direct: ratio y ÷ x fixed. Inverse: product x × y fixed.

Quick self-test

1. A conversion line goes through (0, 0) and (200, 262) with pounds across and dollars up. State the gradient and use it to convert £45 to dollars.
2. Convert 68 °F to °C using F = 1.8C + 32.
3. y is directly proportional to x, and y = 7 when x = 1. Find y when x = 9.
4. y is inversely proportional to x, and x = 6 when y = 6. Find y when x = 4.
5. A runner covers 15 km in 20 minutes at constant speed. Find the speed in km/h.
6. Find the gradient of the line joining (2, 5) and (8, 23).
7. (Higher tier only) A velocity rises steadily from 4 m/s to 10 m/s over 6 s. Find the acceleration and the distance travelled.
8. (Higher tier only) For s = 3t², find the average rate of change of s between t = 1 and t = 2.
9. (Higher tier only) For s = 3t², show that the chord from t = 2 to t = 2 + h has gradient 12 + 3h, and state what the gradient approaches as h becomes very small.
10. (Higher tier only) A graph shows earnings rate in £ per hour. It rises in a straight line from 0 to 30 over 4 hours. Find the total earned.
11. A ramp has gradient 3/4. How far does it rise over a horizontal distance of 20 m?

Answers

1. Gradient 262 ÷ 200 = 1.31 dollars per pound. £45 gives 45 × 1.31 = $58.95.
2. 68 = 1.8C + 32, so 1.8C = 36 and C = 20 °C.
3. y = 7x, so y = 7 × 9 = 63.
4. xy = 36, so y = 36 ÷ 4 = 9.
5. 20 minutes = 1/3 hour, so speed = 15 ÷ (1/3) = 45 km/h.
6. (23 - 5) ÷ (8 - 2) = 18 ÷ 6 = 3.
7. Acceleration = (10 - 4) ÷ 6 = 1 m/s². Distance = ½ × (4 + 10) × 6 = 42 m.
8. s(2) = 12 and s(1) = 3, so (12 - 3) ÷ (2 - 1) = 9.
9. (3(2 + h)² - 12) ÷ h = (12h + 3h²) ÷ h = 12 + 3h. As h gets very small, the gradient approaches 12.
10. Triangle area = ½ × 4 × 30 = £60.
11. 3/4 × 20 = 15 m.

Where marks are usually lost

• Leaving units off gradients, areas and conversions.
• Reading a scale wrongly when each small square is not 1 unit.
• Writing “constant speed” for a horizontal distance-time line.
• Using 20 minutes as 20 in a km/h calculation instead of 1/3 hour.
• Choosing two points too close together when finding the gradient of a tangent, which magnifies reading errors.
• Counting the base of a trapezium as a parallel side, or forgetting the ½.
• Not showing the two points used to find a gradient, so a small misread cannot be followed.
• Writing “the area is the distance” without checking that the vertical axis is velocity or speed.
• Stating an estimate without saying whether it is an over- or under-estimate when the question asks.

Official syllabus

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