Growth, decay and compound interest - Edexcel GCSE Maths (Non-calculator) Exam Questions

Practise 37 exam-style questions on Growth, decay and compound interest 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.

37 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

Ravi invests £5000 in a savings account for 2 years.
The account pays 2% compound interest per year.

Work out the amount in the account at the end of the 2 years.

Model answer

After 1 year: 5000 × 1.02 = £5100
After 2 years: 5100 × 1.02 = 5100 + 102 = £5202

Explanation

With compound interest, the interest each year is worked out on the amount at the start of that year, including earlier interest. So the second year's interest is 2% of £5100, not of £5000.

Mark scheme

1 mark
M1 for \(5000 \times 1.02^2\) or 5000 × 1.02 (= 5100) then × 1.02
oe; correct answer with no working scores full marks
1 mark
A1 for £5202
cao; £5200 (simple interest) scores M0 A0

Question 2

2 marks

A sequence is defined by \(u_{n+1} = 3u_n - 4\), with \(u_1 = 3\).

Work out the value of \(u_4\).

Model answer

\(u_2 = 3 \times 3 - 4 = 5\)
\(u_3 = 3 \times 5 - 4 = 11\)
\(u_4 = 3 \times 11 - 4 = 29\)

Explanation

An iterative formula gives each term from the one before. Substitute the previous term each time, working step by step from \(u_1\).

Mark scheme

1 mark
M1 for \(u_2 = 5\) and \(u_3 = 11\)
correct answer with no working scores full marks
1 mark
A1 for 29
cao

Question 3

3 marks

The number of bacteria, \(N\), in a sample \(t\) hours after the start of an experiment is given by \(N = 200 \times 3^t\).

Work out the number of bacteria after 3 hours, and find the value of \(t\) when \(N = 16\,200\).

Model answer

When \(t = 3\): \(N = 200 \times 3^3 = 200 \times 27 = 5400\)
When \(N = 16\,200\): \(3^t = 16\,200 \div 200 = 81 = 3^4\)
So \(t = 4\).

Explanation

Substitute into the formula to find \(N\). To find \(t\), divide by 200 first, then recognise 81 as a power of 3. The number of bacteria triples every hour.

Mark scheme

1 mark
B1 for 5400
correct answer with no working scores full marks
1 mark
M1 for \(3^t = 16\,200 \div 200 = 81\)
or trying values: 5400 at t = 3, 16 200 at t = 4
1 mark
A1 for \(t = 4\)
cao

Question 4

4 marks

Zara puts some money into an account that pays \(r\%\) compound interest per year.
After 1 year there is £2600 in the account. After 2 years there is £2704.

Work out the value of \(r\) and the amount of money Zara put into the account.

Model answer

Interest in year 2 = 2704 − 2600 = 104
\(\dfrac{104}{2600} = \dfrac{4}{100}\), so \(r = 4\)
Amount put in = 2600 ÷ 1.04 = 260 000 ÷ 104 = £2500
Check: 2500 × 1.04 = 2600

Explanation

The second year's interest is \(r\%\) of the amount after 1 year, which gives \(r\). Then reverse one year of growth by dividing by the multiplier 1.04.

Mark scheme

1 mark
P1 for \(2704 \div 2600\) (= 1.04), or interest of 104 in the second year as a percentage of 2600
correct answer with no working scores full marks
1 mark
A1 for \(r = 4\)
1 mark
P1 for \(2600 \div 1.04\)
ft their multiplier
1 mark
A1 for £2500
cao

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