Gradients and areas under graphs - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Gradients and areas under graphs for Edexcel GCSE Maths (Non-calculator). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

28 questions
1–5 marks each
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20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

2 marks

The graph shows the distance, \(d\) metres, travelled by a car \(t\) seconds after it starts. \(P\) and \(Q\) are points on the curve, joined by a straight line.

Work out the gradient of the line \(PQ\) and explain what it represents.

A graph of Distance (m) (vertical axis, 0 to 64, marked every 8) against Time (s) (horizontal axis, 0 to 8, marked every 1), on a grid with 2 small squares to each large square. The curve is a U-shaped parabola: it touches the x-axis at (0, 0). Point P is marked with a dot at (2, 4). Point Q is marked with a dot at (6, 36). A dashed straight line segment joins P (2, 4) to Q (6, 36).

Model answer

\(P = (2, 4)\) and \(Q = (6, 36)\).
Gradient of \(PQ = \dfrac{36 - 4}{6 - 2} = \dfrac{32}{4} = 8\)
This is the average speed of the car between 2 and 6 seconds: 8 m/s.

Explanation

The gradient of a chord on a distance-time graph is the distance travelled divided by the time taken, which is the average speed over that time interval. The car is speeding up, so its actual speed is less than 8 m/s at \(P\) and more than 8 m/s at \(Q\).

Mark scheme

1 mark
M1 for \(\dfrac{36 - 4}{6 - 2}\)
1 mark
A1 for 8, the average speed (in m/s) between \(t = 2\) and \(t = 6\)
both the value and the meaning needed

Question 2

2 marks

The diagram shows the line \(y = 6 - x\).
The shaded region is bounded by the line, the x-axis, the y-axis and the line \(x = 4\).

Work out the area of the shaded region.

A coordinate grid with x from 0 to 7 and y from 0 to 7; each grid square is 1 unit across, and the axes cross at the origin O. Four boundary lines are drawn across the grid: y = 6 − x (solid), y = 0 (solid), x = 0 (solid) and x = 4 (solid). The region satisfying y ≤ 6 − x, y ≥ 0, x ≥ 0 and x ≤ 4 is shaded. The line labelled "y = 6 − x" is a straight line sloping downwards from left to right crossing the y-axis at (0, 6) and crossing the x-axis at (6, 0).

Model answer

When \(x = 0\), \(y = 6\); when \(x = 4\), \(y = 2\).
The region is a trapezium with parallel sides 6 and 2 and width 4.
Area \(= \dfrac{1}{2}(6 + 2) \times 4 = 16\)

Explanation

The area under a straight-line graph between two x-values is a trapezium. Use the equation to find the heights at each end, then apply the trapezium formula. This is the same method used for areas under speed-time graphs.

Mark scheme

1 mark
M1 for \(\frac{1}{2}(6 + 2) \times 4\)
or rectangle 2 × 4 plus triangle ½ × 4 × 4; correct answer with no working scores full marks
1 mark
A1 for 16

Question 3

3 marks

\(P\) and \(Q\) are the points on the curve \(y = 2^x\) with x-coordinates 1 and 3.

Work out the gradient of the chord \(PQ\), and use it to write down an estimate of the gradient of the curve at \(x = 2\).

Model answer

\(P\): \(y = 2^1 = 2\); \(Q\): \(y = 2^3 = 8\)
Gradient of \(PQ = \dfrac{8 - 2}{3 - 1} = \dfrac{6}{2} = 3\)
An estimate of the gradient of the curve at \(x = 2\) is 3.

Explanation

A chord across the point gives the average rate of change between \(x = 1\) and \(x = 3\), which is an estimate of the gradient at \(x = 2\). It is only an estimate: the true gradient is the gradient of the tangent at \(x = 2\). Using a shorter chord, closer to \(x = 2\), would give a better estimate.

Mark scheme

1 mark
B1 for \(P(1, 2)\) and \(Q(3, 8)\)
correct answer with no working scores full marks
1 mark
M1 for \(\dfrac{8 - 2}{3 - 1}\)
ft their coordinates
1 mark
A1 for 3

Question 4

4 marks

A train starts from rest and accelerates at a constant rate for 30 seconds until it reaches a speed of \(V\) m/s.
It then travels at \(V\) m/s for 120 seconds.
It then decelerates at a constant rate and comes to rest after a further 50 seconds.
The train travels a total distance of 4 km.

Work out the value of \(V\).

Model answer

The speed-time graph is a trapezium with parallel sides 120 s and \(30 + 120 + 50 = 200\) s, and height \(V\).
Distance \(= \dfrac{1}{2}(120 + 200) \times V = 160V\)
4 km = 4000 m
\(160V = 4000\)
\(V = 25\)

Explanation

Sketch the speed-time graph: up, flat, down. Its area is the distance, written in terms of \(V\). Make sure the distance is in metres to match speed in m/s and time in seconds before forming the equation.

Mark scheme

1 mark
P1 for 4 km = 4000 m
correct answer with no working scores full marks
1 mark
P1 for an expression for the area under the speed-time graph in terms of \(V\), e.g. \(\frac{1}{2}(200 + 120)V\) or \(15V + 120V + 25V\)
1 mark
P1 for \(160V = 4000\)
ft their area expression and units
1 mark
A1 for 25
cao

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