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

This guide is written for OCR GCSE (9-1) Mathematics (J560), using the specification Version 2.1 (August 2026). It covers section 7.04, Interpreting graphs: 7.04a real-world graphs, 7.04b gradients and 7.04c areas. Anything the specification lists for the Higher tier only is marked “Higher tier only”; the rest is for both tiers.

Related pages: revision notes, practice questions, the OCR GCSE Mathematics hub, the printable checklist and the free 10-minute diagnostics.

What this unit covers

Section	What you must be able to do	Tier
7.04a	Construct and interpret graphs in real-world contexts, such as distance-time, money conversion and temperature conversion	Both tiers
7.04a	Recognise and interpret graphs that show direct and inverse proportion	Both tiers
7.04b	Understand the relationship between gradient and ratio	Both tiers
7.04b	Interpret straight-line gradients as rates of change, for example the gradient of a distance-time graph as a velocity	Both tiers
7.04b	Calculate or estimate gradients of graphs, and interpret them in distance-time, velocity-time and financial contexts	Higher tier only
7.04b	Use average and instantaneous rate of change (gradients of chords and tangents) in numerical, algebraic and graphical contexts	Higher tier only
7.04c	Calculate or estimate areas under graphs, and interpret them in distance-time, velocity-time and financial contexts	Higher tier only

1. Real-world graphs (7.04a)

Every real-world graph is a story. Before you read any values, do three things.

1. Read both axis labels and their units.
2. Find where the graph starts. The starting value tells you what is there at time zero or quantity zero.
3. Look at the shape: straight, curved, flat, rising or falling.

Conversion graphs

A money conversion graph is a straight line through the origin, because zero pounds is zero of any other currency. The gradient is the exchange rate.

Worked example 1. The rate is £1 = 1.17 euros. A conversion graph is drawn with pounds on the horizontal axis and euros on the vertical axis.

Construct: plot (0, 0) and (100, 117), then join them with a ruler.
Gradient = 117 ÷ 100 = 1.17 euros per pound.
£80 to euros: 80 × 1.17 = 93.6 euros.
150 euros to pounds: 150 ÷ 1.17 = 128.205... = £128.21.

To read from the graph, go up from the pounds value to the line, then across. To go the other way, start on the euro axis.

Temperature conversion graphs

Degrees Celsius to degrees Fahrenheit uses F = 1.8C + 32. This line misses the origin: at C = 0 the value of F is 32.

Worked example 2.

Plot (0, 32) and (100, 212).
Gradient = (212 - 32) ÷ (100 - 0) = 180 ÷ 100 = 1.8.
Intercept = 32 (the Fahrenheit reading when Celsius is 0).
25 °C: F = 1.8 × 25 + 32 = 77 °F.
59 °F: 59 = 1.8C + 32, so C = 27 ÷ 1.8 = 15 °C.
Same reading on both scales: 1.8C + 32 = C, so 0.8C = -32, so C = -40.

Doubling the Celsius value does not double the Fahrenheit value, so this is not a proportion graph.

Distance-time graphs

Time runs along the bottom of a distance-time graph, and the distance from the start runs up the side.

Feature	Meaning
Straight line sloping up	Constant speed away from the start
Horizontal line	Stationary
Straight line sloping down	Constant speed back towards the start
Steeper line	Greater speed

Worked example 3. Tariq cycles from home. The graph joins (0, 0), (1.5, 18), (2, 18), (4, 0), where time is in hours and distance in km.

0 to 1.5 h:   speed = 18 ÷ 1.5 = 12 km/h away from home
1.5 to 2 h:   stationary for 0.5 h (a rest)
2 to 4 h:     speed = 18 ÷ 2 = 9 km/h back towards home
Total distance travelled = 18 + 18 = 36 km
Average speed = 36 ÷ 4 = 9 km/h

He ends at home, so his distance from home is 0 at the end. The distance he has travelled is not 0. Do not mix these up.

2. Direct and inverse proportion graphs (7.04a)

Direct proportion. y is directly proportional to x when y = kx. The graph is a straight line through the origin, and the gradient is k. Doubling x doubles y.

Inverse proportion. y is inversely proportional to x when y = k/x, so xy = k always. The graph is a smooth curve that falls as x rises. It never touches either axis.

Worked example 4. A cleaner is paid £14.50 per hour.

Pay P = 14.5h. The graph is a straight line through (0, 0).
Gradient = 14.5 (pounds per hour).
7.5 hours: P = 14.5 × 7.5 = £108.75.

Worked example 5. Eight painters take 6 days to finish a job. Assume all painters work at the same rate.

xy = 8 × 6 = 48, so days d = 48 ÷ p.
Points: (2, 24), (4, 12), (6, 8), (8, 6), (12, 4).
12 painters: d = 48 ÷ 12 = 4 days.
5 painters: d = 48 ÷ 5 = 9.6 days.

How to tell the two apart on a graph: for direct proportion, y ÷ x is the same at every point. For inverse proportion, x × y is the same at every point. A straight line missing the origin is linear but is not direct proportion.

3. Gradient, ratio and rates of change (7.04b)

The gradient of a straight line is (change in y) ÷ (change in x). It is a ratio: it compares the vertical change with the horizontal change. A gradient of 5/2 means that for every 2 units across, the line rises 5 units, a ratio of 5 : 2.

The units of the gradient come from the axes: units of y per unit of x. So every gradient is a rate.

Graph	Gradient means	Example unit
Distance against time	Velocity or speed	m/s, km/h
Money against quantity	Price per item	£ per kg
Balance against months	Rate of saving	£ per month
Euros against pounds	Exchange rate	euros per £

Worked example 6. A savings balance is a straight line from (0, 200) to (8, 520), with months across and pounds up.

Gradient = (520 - 200) ÷ (8 - 0) = 320 ÷ 8 = 40.
Meaning: the balance rises by £40 each month.
Intercept 200: the starting balance was £200.

Velocity-time graphs and acceleration (Higher tier only)

On a velocity-time graph the gradient is the acceleration, in m/s². A negative gradient is a deceleration.

Worked example 7 (Higher tier only). A train’s velocity-time graph joins (0, 0), (10, 15), (40, 15), (55, 0) with time in seconds and velocity in m/s.

0 to 10 s:   acceleration = 15 ÷ 10 = 1.5 m/s²
10 to 40 s:  gradient 0, so constant velocity 15 m/s
40 to 55 s:  gradient = (0 - 15) ÷ 15 = -1, so deceleration 1 m/s²

Average and instantaneous rate of change (Higher tier only)

On a curve the gradient changes from point to point.

• A chord joins two points on the curve. Its gradient is the average rate of change between those points.
• A tangent touches the curve at one point and has the same direction as the curve there. Its gradient is the instantaneous rate of change at that point. In a financial graph, a chord gives an average rate such as profit per month, and a tangent gives the rate at one moment.

Worked example 8 (Higher tier only). A distance-time graph follows s = 2t², with s in metres and t in seconds.

Average velocity from t = 1 to t = 3:
(s(3) - s(1)) ÷ (3 - 1) = (18 - 2) ÷ 2 = 8 m/s.

To estimate the velocity at exactly t = 2, draw a tangent at (2, 8) with a ruler. Suppose it passes through (1, 0) and (3, 16). The gradient is 16 ÷ 2 = 8 m/s.

Chords from t = 2 get closer to the tangent as the second point moves towards (2, 8):

Chord	Gradient
t = 2 to t = 3	10
t = 2 to t = 2.5	9
t = 2 to t = 2.1	8.2

Algebraically, the chord from t = 2 to t = 2 + h has gradient (2(2 + h)² - 8) ÷ h = (8h + 2h²) ÷ h = 8 + 2h. As h gets smaller this approaches 8, which is the instantaneous rate of change.

Worked example 9 (Higher tier only). A speed-time curve has v = 18t ÷ (t + 3). Its table of values is exact:

t (s)	0	3	6	9	12
v (m/s)	0	9	12	13.5	14.4

Average acceleration from t = 3 to t = 9: (13.5 - 9) ÷ 6 = 0.75 m/s².
A tangent drawn at t = 6 passes through (0, 8) and (12, 16).
Gradient = (16 - 8) ÷ 12 = 2/3 m/s² (about 0.67).

Tangent estimates depend on careful drawing. Show the two points you used.

4. Areas under graphs (7.04c, Higher tier only)

The area under a graph is (vertical quantity) × (horizontal quantity), so its units are the product of the axis units.

Graph	Area represents
Velocity against time (m/s × s)	Distance travelled in m
Rate of earning against time (£ per hour × hours)	Total money in £
Distance against time (km × h)	A quantity in km h, with no everyday meaning on its own

The area can still be calculated, with units km h, but it has no everyday meaning, so the gradient is the useful quantity.

Worked example 10 (Higher tier only). Find the distance the train in example 7 travels.

Shape under the graph: a trapezium with parallel sides 30 s (top) and 55 s (bottom), height 15 m/s.
Area = ½ × (30 + 55) × 15 = 637.5 m.
Check by parts: ½ × 10 × 15 = 75, 30 × 15 = 450, ½ × 15 × 15 = 112.5.
75 + 450 + 112.5 = 637.5 m.
Average speed = 637.5 ÷ 55 = 11.6 m/s (3 s.f.).

Worked example 11 (Higher tier only). A stall’s takings rate (£ per hour) is plotted against hours open. The graph joins (0, 0), (2, 60), (5, 60), (8, 20).

0 to 2 h:   triangle ½ × 2 × 60 = 60
2 to 5 h:   rectangle 3 × 60 = 180
5 to 8 h:   trapezium ½ × (60 + 20) × 3 = 120
Total takings = 60 + 180 + 120 = £360.

Estimating the area under a curve (Higher tier only)

Split the area into vertical strips and treat each as a trapezium. Use the table from example 9 with strips of width 3 s.

Area ≈ 3 × (0/2 + 9 + 12 + 13.5 + 14.4/2)
     = 3 × (34.5 + 7.2)
     = 125.1 m.

Here the curve is concave, so the trapezia fall short of it and the estimate is an underestimate. Narrower strips give a better estimate. Say whether your answer is an over- or under-estimate when the question asks.

Common errors

• Reading a conversion graph without checking the scales on each axis.
• Treating a horizontal line on a distance-time graph as “constant speed”. It means stationary.
• Calling any straight line “direct proportion”. It must pass through the origin.
• Forgetting to convert minutes to hours before finding a speed in km/h.
• Drawing a tangent that cuts the curve instead of touching it.
• Using only the top value of a velocity-time graph as the height of a trapezium. Use parallel sides, with the time axis as the base.
• Giving areas or gradients without units.

Next steps

Condense this unit with the revision notes, then test yourself on the practice questions. The free 10-minute diagnostics show which topics need work, and the checklist lets you tick off what you can do.

Official syllabus

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