Gradients and intercepts - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Gradients and intercepts 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

A straight line has equation \(3x + 2y = 12\).

Find the gradient of the line and the coordinates of the point where it crosses the y-axis.

Model answer

\(2y = -3x + 12\)
\(y = -\dfrac{3}{2}x + 6\)
Gradient \(= -\dfrac{3}{2}\)
The line crosses the y-axis at \((0, 6)\).

Explanation

Make \(y\) the subject so the equation is in the form \(y = mx + c\). Remember to divide every term by 2. The gradient is the coefficient of \(x\) and the y-intercept is the constant term.

Mark scheme

1 mark
B1 for gradient \(-\frac{3}{2}\)
oe, e.g. −1.5; correct answer with no working scores full marks
1 mark
B1 for \((0, 6)\)
accept y-intercept 6

Question 2

2 marks

A straight line has gradient \(-4\) and crosses the x-axis at the point \((3, 0)\).

Find the coordinates of the point where the line crosses the y-axis.

Model answer

\(y = -4x + c\)
\(0 = -4 \times 3 + c\)
\(c = 12\)
The line crosses the y-axis at \((0, 12)\).

Explanation

Use the gradient and the known point to find \(c\), the y-intercept. Thinking of the graph: moving 3 units left from \((3, 0)\) to the y-axis, a line with gradient \(-4\) goes up \(3 \times 4 = 12\) units.

Mark scheme

1 mark
M1 for substituting \((3, 0)\) into \(y = -4x + c\), e.g. \(0 = -12 + c\)
or for a correct counting method: 3 units left means 12 up; correct answer with no working scores full marks
1 mark
A1 for \((0, 12)\)
accept y-intercept 12

Question 3

3 marks

A straight line passes through the points \((2, k)\) and \((k, 8)\), where \(k\) is a constant.
The gradient of the line is 2.

Find the value of \(k\).

Model answer

\(\dfrac{8 - k}{k - 2} = 2\)
\(8 - k = 2(k - 2)\)
\(8 - k = 2k - 4\)
\(12 = 3k\)
\(k = 4\)

Explanation

Write the gradient formula using the two points, set it equal to 2 and solve the resulting linear equation. Multiply both sides by \(k - 2\) to clear the fraction. Check: the points are \((2, 4)\) and \((4, 8)\), and \(\frac{8 - 4}{4 - 2} = 2\).

Mark scheme

1 mark
M1 for a correct equation, e.g. \(\dfrac{8 - k}{k - 2} = 2\)
correct answer with no working scores full marks
1 mark
M1 for removing the fraction and collecting terms, e.g. \(8 - k = 2k - 4\)
1 mark
A1 for \(k = 4\)
cao

Question 4

4 marks

A gym charges a joining fee and then a fixed amount each month.
The total cost for 3 months is £95.
The total cost for 8 months is £195.

The total cost, £\(C\), for \(m\) months is given by \(C = am + b\).
Find the values of \(a\) and \(b\), and explain what each one represents.

Model answer

\(a = \dfrac{195 - 95}{8 - 3} = \dfrac{100}{5} = 20\)
\(95 = 20 \times 3 + b\), so \(b = 95 - 60 = 35\)
\(a = 20\): the gym charges £20 for each month.
\(b = 35\): the joining fee is £35.

Explanation

The extra 5 months cost an extra £100, so each month costs £20; this is the gradient of the cost line. Substituting back gives the intercept, which is the cost at 0 months, the joining fee. Check: \(20 \times 8 + 35 = 195\).

Mark scheme

1 mark
M1 for \(\dfrac{195 - 95}{8 - 3}\) or for setting up \(3a + b = 95\) and \(8a + b = 195\)
1 mark
A1 for \(a = 20\)
1 mark
A1 for \(b = 35\)
1 mark
C1 for \(a\) is the monthly fee (£20 per month) and \(b\) is the joining fee (£35)
both interpretations needed; ft their values

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