Substituting into formulae and expressions - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 28 exam-style questions on Substituting into formulae and expressions 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 car is slowing down.
Its velocity, \(v\) m/s, after \(t\) seconds is given by the formula
\[v = u + at\]

Work out the value of \(v\) when \(u = 12\), \(a = -2.5\) and \(t = 4\)

Model answer

\(v = 12 + (-2.5) \times 4 = 12 - 10 = 2\)
\(v = 2\) m/s

Explanation

Multiply before adding: \(at = -2.5 \times 4 = -10\). Then \(12 + (-10) = 2\). The negative acceleration shows the car is slowing down, so the final velocity is less than 12 m/s.

Mark scheme

1 mark
M1 for \(12 + (-2.5) \times 4\)
oe, e.g. \(12 - 10\); correct answer with no working scores full marks
1 mark
A1 for 2
cao

Question 2

2 marks

Einstein's formula connects energy, \(E\) joules, with mass, \(m\) kg:
\[E = mc^2\]

Work out the value of \(E\) when \(m = 2 \times 10^{-3}\) and \(c = 3 \times 10^8\)
Give your answer in standard form.

Model answer

\(c^2 = (3 \times 10^8)^2 = 9 \times 10^{16}\)
\(E = 2 \times 10^{-3} \times 9 \times 10^{16} = 18 \times 10^{13} = 1.8 \times 10^{14}\)

Explanation

Square both parts of \(3 \times 10^8\): \(3^2 = 9\) and \((10^8)^2 = 10^{16}\). Multiply the numbers and add the powers of 10: \(-3 + 16 = 13\). Finally adjust \(18 \times 10^{13}\) so the first number is between 1 and 10.

Mark scheme

1 mark
M1 for \(c^2 = 9 \times 10^{16}\) or \(2 \times 10^{-3} \times 9 \times 10^{16}\)
or \(18 \times 10^{13}\) seen; correct answer with no working scores full marks
1 mark
A1 for \(1.8 \times 10^{14}\)
cao; \(18 \times 10^{13}\) scores M1 A0

Question 3

3 marks

The kinetic energy, \(E\) joules, of an object of mass \(m\) kg moving at \(v\) m/s is given by
\[E = \frac{1}{2}mv^2\]

Ball A has a mass of 0.5 kg and is moving at 12 m/s.
Ball B has a mass of 2 kg and is moving at 6 m/s.

Sam says, "Ball B has more kinetic energy than ball A because its mass is 4 times as big."
Is Sam correct? You must show how you get your answer.

Model answer

Ball A: \(E = \dfrac{1}{2} \times 0.5 \times 12^2 = 0.25 \times 144 = 36\) J
Ball B: \(E = \dfrac{1}{2} \times 2 \times 6^2 = 1 \times 36 = 36\) J
No, Sam is not correct. Both balls have the same kinetic energy, 36 J.

Explanation

The speed is squared in the formula, so ball A's speed being twice as big makes its \(v^2\) four times as big. That exactly balances ball B's mass being four times as big. Substituting the values is the clearest way to test a claim like Sam's.

Mark scheme

1 mark
M1 for substituting correctly for either ball, e.g. \(\dfrac{1}{2} \times 0.5 \times 12^2\) or \(\dfrac{1}{2} \times 2 \times 6^2\)
1 mark
A1 for both energies, 36 J and 36 J
1 mark
C1 for "No" with a correct reason, e.g. both balls have the same kinetic energy
accept a correct explanation that halving the speed divides the energy by 4, which cancels out the mass being 4 times as big, if supported by the values

Question 4

4 marks

The number of diagonals, \(D\), of a polygon with \(n\) sides is given by the formula
\[D = \frac{n(n - 3)}{2}\]

Show that there is no polygon with exactly 30 diagonals.

Model answer

If \(D = 30\), then \(n(n - 3) = 60\).
\(n = 9\): \(D = \dfrac{9 \times 6}{2} = 27\)
\(n = 10\): \(D = \dfrac{10 \times 7}{2} = 35\)
\(D\) increases as \(n\) increases, so a 9-sided polygon has fewer than 30 diagonals and every polygon with 10 or more sides has more than 30.
The number of sides must be a whole number, so no polygon has exactly 30 diagonals.

Explanation

Substitute whole numbers for \(n\) and look for 30. Because the values jump from 27 straight to 35, and keep increasing, 30 is never reached. The key point in the argument is that \(n\) must be a whole number - the number of sides of a polygon.

Mark scheme

1 mark
P1 for forming \(\dfrac{n(n - 3)}{2} = 30\) or \(n(n - 3) = 60\)
or for substituting values of n into the formula
1 mark
P1 for a correct value of D (or n(n − 3)) for n = 9 or n = 10
n = 9 gives D = 27; n = 10 gives D = 35
1 mark
P1 for the values for both n = 9 and n = 10
or for \(n^2 - 3n - 60 = 0\) with a correct reason it has no whole number solution
1 mark
C1 for a complete argument, e.g. 30 lies between 27 and 35 and D increases as n increases, so no whole number n gives 30

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