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

Practise 28 exam-style questions on Substituting into formulae and expressions for Edexcel GCSE Maths (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 power, P watts, used by a resistor is given by the formula
\(P = I^2R\)
where I is the current in amps and R is the resistance in ohms.

Work out the value of P when I = 0.35 and R = 220

Model answer

\(P = I^2R = 0.35^2 \times 220\)
\(= 0.1225 \times 220\)
\(= 26.95\) watts

Explanation

\(I^2R\) means square I and then multiply by R. Only the current is squared. A common mistake is to work out \((0.35 \times 220)^2 = 5929\), which squares R as well.

Mark scheme

1 mark
M1 for \(0.35^2 \times 220\) or \(0.1225 \times 220\)
oe
1 mark
A1 for 26.95
cao; correct answer with no working scores full marks

Question 2

2 marks

A cyclist is slowing down.
The distance, s metres, she travels in t seconds is given by the formula
\(s = ut + \frac{1}{2}at^2\)

Work out the value of s when u = 12.5, a = −2.4 and t = 3.2
Give your answer correct to 3 significant figures.

Model answer

\(ut = 12.5 \times 3.2 = 40\)
\(\frac{1}{2}at^2 = \frac{1}{2} \times (-2.4) \times 3.2^2 = -1.2 \times 10.24 = -12.288\)
\(s = 40 - 12.288 = 27.712\)
s = 27.7 (3 s.f.)

Explanation

Work out each term separately. In \(\frac{1}{2}at^2\) only t is squared, and because a is negative this term is negative. Common mistakes are to square at (giving \(\frac{1}{2} \times (-7.68)^2\)) or to treat −2.4 as positive and get 52.288.

Mark scheme

1 mark
M1 for \(12.5 \times 3.2 + \frac{1}{2} \times (-2.4) \times 3.2^2\) or \(40 - 12.288\)
oe
1 mark
A1 for 27.7
awrt 27.7 (27.712); correct answer with no working scores full marks

Question 3

3 marks

The formula
\(A = P\left(1 + \dfrac{r}{100}\right)^n\)
gives the amount, £A, in a savings account after n years when £P is invested at r% compound interest per year.

Gemma invests £3250 at 2.35% per year compound interest for 7 years.
Work out how much interest Gemma earns.
Give your answer to the nearest penny.

Model answer

\(A = 3250 \times \left(1 + \dfrac{2.35}{100}\right)^7 = 3250 \times 1.0235^7\)
\(= 3250 \times 1.17656... = 3823.827...\)
Interest = 3823.827... − 3250 = 573.827...
Gemma earns £573.83

Explanation

Substitute P = 3250, r = 2.35 and n = 7: the multiplier is 1.0235, raised to the power 7. The formula gives the total amount in the account, so subtract the £3250 invested to find the interest. A common mistake is to use simple interest (3250 × 0.0235 × 7 = £534.63).

Mark scheme

1 mark
M1 for \(3250 \times 1.0235^7\) or \(3250\left(1 + \dfrac{2.35}{100}\right)^7\)
oe; may be implied by 3823.8...
1 mark
M1 for subtracting 3250 from their amount
e.g. 3823.82... − 3250
1 mark
A1 for £573.83
accept 573.82 to 573.83; correct answer with no working scores full marks

Question 4

4 marks

Mia has £8000 to invest for 5 years.

Bank A pays 3.6% per year compound interest, added at the end of each year.
Bank B pays 0.3% per month compound interest, added at the end of each month.

The formula
\(A = P\left(1 + \dfrac{r}{100}\right)^n\)
gives the value, £A, of an investment of £P after n periods at r% compound interest per period.

Which bank will give Mia more money after 5 years, and by how much?
You must show your working.

Model answer

Bank A: \(8000 \times 1.036^5 = 9547.480...\) = £9547.48
Bank B: 5 years = 60 months, \(8000 \times 1.003^{60} = 9575.158...\) = £9575.16
9575.16 − 9547.48 = 27.68
Bank B gives Mia more money, by £27.68.

Explanation

For Bank B the interest is added every month, so there are 5 × 12 = 60 periods at 0.3% each. Although 0.3% a month is the same as 3.6% a year in simple terms, adding interest more often means interest is earned on interest sooner, so Bank B gives slightly more. A common mistake is to use \(1.003^5\).

Mark scheme

1 mark
M1 for \(8000 \times 1.036^5\)
oe; may be implied by 9547.48...
1 mark
M1 for \(8000 \times 1.003^{60}\)
oe; the power must be 60
1 mark
A1 for £9547.48 and £9575.16
accept 9547.4 to 9547.5 and 9575.1 to 9575.2
1 mark
C1 for Bank B, by £27.68
accept £27.6 to £27.7; conclusion must follow correct figures

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