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

Practise 36 exam-style questions on Growth, decay and compound interest 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.

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

Hannah invests £3200 in a savings account for 5 years.
The account pays 2.4% compound interest per year.

Work out the value of Hannah's investment at the end of the 5 years.
Give your answer to the nearest penny.

Model answer

Multiplier = 1 + 0.024 = 1.024
\(3200 \times 1.024^5 = 3602.8797...\)
= £3602.88

Explanation

For compound interest, the interest each year is added on and then earns interest itself, so multiply by 1.024 once for each of the 5 years. A common mistake is to work out simple interest: \(3200 \times 0.024 \times 5 = 384\), giving £3584, which is too small.

Mark scheme

1 mark
M1 for \(3200 \times 1.024^5\) oe
or a complete year-by-year method
1 mark
A1 for £3602.88
accept £3602.87 to £3602.88; correct answer with no working scores full marks

Question 2

3 marks

Leo invests £5000 in an account paying 2.7% compound interest per year.

Work out the least number of whole years it will take for Leo's investment to be worth more than £6000.

Model answer

\(5000 \times 1.027^6 = 5866.68...\) (less than £6000)
\(5000 \times 1.027^7 = 6025.08...\) (more than £6000)
It takes 7 years.

Explanation

Use trial and improvement with the compound interest calculation, trying different numbers of years until the value first goes over £6000. Showing the values for 6 and 7 years proves that 7 is the least number of years. Dividing 1000 by the first year's interest (£135) assumes simple interest and gives the wrong answer.

Mark scheme

1 mark
M1 for \(5000 \times 1.027^n\) for any whole number \(n\) ≥ 2
or a year-by-year method; correct answer with no working scores full marks
1 mark
M1 for \(5000 \times 1.027^6\) (= 5866.68...) and \(5000 \times 1.027^7\) (= 6025.08...)
accept values that round to 5867 and 6025
1 mark
A1 for 7 years
cao; must be supported by working

Question 3

3 marks

Amir invested £2000 in an account paying compound interest at a fixed rate.
After 6 years his investment was worth £2530.64

Work out the annual rate of interest.

Model answer

\(2000 \times m^6 = 2530.64\)
\(m^6 = 2530.64 \div 2000 = 1.26532\)
\(m = \sqrt[6]{1.26532} = 1.0400...\)
The rate of interest is 4% per year.

Explanation

Set up the compound interest equation with an unknown multiplier \(m\), divide by the original amount and take the sixth root to undo the power. The multiplier 1.04 means a 4% increase each year. Dividing the total percentage increase (26.5%) by 6 assumes simple interest and gives the wrong rate.

Mark scheme

1 mark
M1 for \(2530.64 \div 2000\) (= 1.26532)
or \(2000 \times m^6 = 2530.64\); correct answer with no working scores full marks
1 mark
M1 for \(\sqrt[6]{1.26532}\) (= 1.0400...) oe
or trials of \(2000 \times m^6\) with at least two different multipliers
1 mark
A1 for 4%
accept 4.00%

Question 4

4 marks

A patient takes a 200 mg dose of a medicine every 8 hours.
The amount, \(D_n\) mg, of medicine in the body just after the \(n\)th dose is given by
\[ D_{n+1} = 0.35D_n + 200, \quad D_1 = 200 \]
The first few values are 200, 270, 294.5, ...

Show that, however many doses are taken, the amount just after a dose will never be more than 308 mg.

Model answer

The amounts increase: 200, 270, 294.5, 303.075, 306.076..., 307.126..., ...
If the sequence approaches a value \(L\), then \(L = 0.35L + 200\)
\(0.65L = 200\)
\(L = 307.69...\)
If \(D_n < 307.69...\) then \(D_{n+1} < 0.35 \times 307.69... + 200 = 307.69...\), so every amount stays below 307.69...
Since 307.69... < 308, the amount just after a dose is never more than 308 mg.

Explanation

An iterative process like this settles down to a limit where the output equals the input, so solve \(L = 0.35L + 200\). The terms increase towards this value but never reach it, because each new increase is only 35% of the one before. Iterating many times on a calculator is good evidence, but you must say why the values can never pass the limit.

Mark scheme

1 mark
P1 for setting up \(L = 0.35L + 200\) for the value the sequence approaches
or showing at least four further terms, e.g. 303.075, 306.07..., 307.12..., 307.49...
1 mark
P1 for \(0.65L = 200\)
1 mark
A1 for \(L = 307.69...\)
accept 307.7
1 mark
C1 for a complete argument: the amounts increase towards 307.69... and never go beyond it, and 307.69... < 308
e.g. noting that each increase is 0.35 times the previous increase, or that if \(D_n\) < 307.69... then \(D_{n+1}\) < 0.35 × 307.69... + 200 = 307.69...

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