Using and rearranging formulae - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 36 exam-style questions on Using and rearranging formulae 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.

36 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

Make \(x\) the subject of the formula \(y = 3x + 5\)

Model answer

\(y = 3x + 5\)
\(y - 5 = 3x\)
\(x = \dfrac{y - 5}{3}\)

Explanation

Undo the operations in reverse order: first subtract 5 from both sides, then divide both sides by 3. The whole of \(y - 5\) must be divided by 3, so it goes over the fraction line.

Mark scheme

1 mark
M1 for \(y - 5 = 3x\)
or for \(\dfrac{y}{3} = x + \dfrac{5}{3}\); correct answer with no working scores full marks
1 mark
A1 for \(x = \dfrac{y - 5}{3}\)
oe, e.g. \(x = \dfrac{y}{3} - \dfrac{5}{3}\); do not accept \(x = y - \dfrac{5}{3}\)

Question 2

2 marks

A circle has an area of \(49\pi\) cm².

Work out the circumference of the circle.
Give your answer in terms of \(\pi\).

Model answer

\(\pi r^2 = 49\pi\)
\(r^2 = 49\), so \(r = 7\) cm
Circumference \(= 2\pi r = 2 \times \pi \times 7 = 14\pi\) cm

Explanation

Use the area formula backwards to find the radius, then use \(C = 2\pi r\). Dividing by \(\pi\) leaves \(r^2 = 49\), so \(r = 7\). A common slip is to use 49 as the radius.

Mark scheme

1 mark
M1 for \(\pi r^2 = 49\pi\) leading to \(r = 7\)
correct answer with no working scores full marks
1 mark
A1 for \(14\pi\) (cm)
cao

Question 3

3 marks

A metal block is a cuboid measuring 2 cm by 3 cm by 5 cm.
The block has a mass of 240 g.

A solid cube is made from the same metal. The mass of the cube is 1728 g.
Work out the length of one side of the cube.

Model answer

Volume of block \(= 2 \times 3 \times 5 = 30\) cm³
Density \(= \dfrac{\text{mass}}{\text{volume}} = \dfrac{240}{30} = 8\) g/cm³
Volume of cube \(= \dfrac{\text{mass}}{\text{density}} = \dfrac{1728}{8} = 216\) cm³
Side \(= \sqrt[3]{216} = 6\) cm

Explanation

The same metal has the same density, so find the density from the block, then rearrange density = mass ÷ volume to get volume = mass ÷ density. Finally take the cube root of the volume to find the side length.

Mark scheme

1 mark
P1 for the density, \(240 \div (2 \times 3 \times 5) = 8\) (g/cm³)
correct answer with no working scores full marks
1 mark
P1 for the volume of the cube, \(1728 \div 8 = 216\) (cm³)
ft their density
1 mark
A1 for 6 (cm)
cao

Question 4

4 marks

A cone has a perpendicular height that is twice its radius.
The volume of the cone is \(18\pi\) cm³.

Volume of a cone \(= \dfrac{1}{3}\pi r^2h\)

Work out the radius of the cone.

Model answer

\(h = 2r\)
\(V = \dfrac{1}{3}\pi r^2 \times 2r = \dfrac{2}{3}\pi r^3\)
\(\dfrac{2}{3}\pi r^3 = 18\pi\)
\(r^3 = 18 \times \dfrac{3}{2} = 27\)
\(r = 3\) cm

Explanation

Replace \(h\) with \(2r\) so the formula contains only one unknown, then solve. Multiplying \(r^2\) by \(r\) gives \(r^3\), so the last step is a cube root. Check: radius 3 and height 6 give \(\dfrac{1}{3} \times 9\pi \times 6 = 18\pi\).

Mark scheme

1 mark
P1 for substituting \(h = 2r\), e.g. \(\dfrac{1}{3}\pi r^2 \times 2r\)
correct answer with no working scores full marks
1 mark
P1 for \(\dfrac{2}{3}\pi r^3 = 18\pi\)
oe
1 mark
P1 for \(r^3 = 27\)
1 mark
A1 for 3 (cm)
cao

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