GCSE Maths Number

Simple and Compound Interest

Compound Interest

Compound Interest

Here is everything you need to know about compound interest for GCSE maths (Edexcel, AQA and OCR). You’ll learn how to calculate compound interest for increasing and decreasing values, and set-up, solve and interpret growth and decay problems.

Look out for the compound interest worksheet and exam questions at the end.

What is compound interest?

Compound interest is calculated on the principal (original) amount and the interest already accumulated on previous periods. 

For example, take the amount of money in a savings account.

If you put £100 in an account with an annual interest rate of 10%, the value of the money in the account will increase by 10%  in year one.

The new amount of money in the account will be £110 or 110% of the original.

In year two the value of the money in the account will increase by 10% again.

The new amount would be £121 or 121% of the original. This because we have increased £110 by 10%.  

Compound interest formula

Compound interest is interest calculated on top of the original amount including any interest accumulated so far. The compound interest formula is:

\[\\A=P(1+\frac{r}{n})^{nt}\\\]

Where:

  • A represents the final amount
  • P represents the original principal amount
  • r is the percentage change (written as a decimal)
  • n represents the number of times the interest rate is applied over time t
  • t represents the time period

E.g.

Let’s calculate a 3% increase on an amount P using compound interest over 4 years.

When calculating 3% of the amount P, we would divide the amount P by 100 (to calculate 1%) and then multiply this value by 3 (to find 3%). This can also be written as:

\[\\\frac{P}{100}\times{3}\\ \\P\times\frac{3}{100}\\ \\P\times{0.03}\\\]

Again, P × 0.03 is the value of the interest only and so we add this value to P to find the new amount after the first interval. This is written as:

\[\\A=P+P\times{0.03}\\ \\A=P(1+0.03)\\ \\A=P\times{1.03}\\\]

This is the value after the first interval.

The second interval is the new amount, multiplied by a further 1.03 and so on for the amount of intervals. This can be expressed using a table.

YearInterest CalculationSimplified Interest Calculation
Year 0PP × 1.030
Year 1P × 1.03P × 1.031
Year 2P × 1.03 × 1.03P × 1.032
Year 3P × 1.03 × 1.03 × 1.03P × 1.033
Year tP × 1.03 × 1.03 × 1.03 × ...P × 1.03t

After t years, the principal amount P is multiplied by the percentage change for t times. This is why the power is expressed as t.

So for this example, multiplying P by 1.034 would calculate an increase of 3% compound interest for 4 years.

Compound interest formula worksheet

Compound interest formula worksheet

Compound interest formula worksheet

Get your free compound interest formula worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE
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Compound interest formula worksheet

Compound interest formula worksheet

Compound interest formula worksheet

Get your free compound interest formula worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE

How to calculate compound interest

In order to calculate compound interest:

  1. State the formula and the value of each variable.
  2. Substitute the values into the formula.
  3. Solve the equation.

Compound interest formula examples

Example 1: compound interest (percentage increase)

£8500 is invested for 5 years at 0.3% per year compound interest. What is the value of the investment after this time?

  1. State the formula and the value of each variable.

Here we use the formula

\[A=P(1+\frac{r}{n})^{nt}\]

with:

  • P = 8500
  • r = 0.003
  • n = 1
  • t = 5

2Substitute the values into the formula.

Substituting these values into the compound interest formula

\[A=P(1+\frac{r}{n})^{nt}\]

we get:

\[\\A=8500(1+\frac{0.003}{1})^{1\times{5}}\\\]

3Solve the equation.

\[\\A=8500(1+0.003)^{5}\\ \\A=8500(1.003)^{5}\\ \\A=8500\times1.01509027\\ \\A=£ {8628.27}\\\]

Example 2: compound interest (percentage decrease)

A vacuum cleaner is bought for £399 and loses 22% of its value per year compound interest. What is the value of the vacuum cleaner after 2 years?

State the formula and the value of each variable.

Here we use the formula

\[A=P(1+\frac{r}{n})^{nt}\]

with:

  • P = 399
  • r = − 0.22
  • n = 1
  • t = 2

Substituting these values into the compound interest formula

\[A=P(1+\frac{r}{n})^{nt}\]

we get:

\[\\A=399(1-\frac{0.22}{1})^{1\times{2}}\\\]

\[\\A=399(1-0.22)^{2}\\ \\A=399(0.78)^{2}\\ \\A=399\times0.6084\\ \\A=£{242.75}\\\]

Example 3: compound interest (compounding periods)

A house is valued at £150 000. On average, the house price increases by 2.4% monthly compounding over 2.5 years. What is the future value of the house after this time?

Here we use the formula

\[A=P(1+\frac{r}{n})^{nt}\]

with:

  • P = 150000
  • r = 0.024
  • n = 12
  • t = 2.5

Substituting these values into the compound interest formula

\[A=P(1+\frac{r}{n})^{nt}\]

we get:

\[\\A=150000(1+\frac{0.024}{12})^{12\times{2.5}}\\\]

Notice that the exponent is now 12 × 2.5 as the number of months over 2.5 years = 12 × 2.5.

\[\\A=150000(1+0.002)^{30}\\ \\A=150000(1.002)^{30}\\ \\A=150000(1.06177)\\ \\A=150000\times1.06177\\ \\A= £ 159265.94\\\]

Example 4: compound interest (writing an answer in standard form)

Bacteria reproduce at a rate of 3% per hour compound interest. If there are 1.2 x 107 bacteria in a petri dish at hour 0, how many bacteria are in the petri dish after 4 hours?

Here we use the formula

\[A=P(1+r)^{t}\]

as the time scale n=1 with:

  • P = 1.2 × 107
  • r = 0.03
  • t = 4

Substituting these values into the compound interest formula

\[A=P(1+r)^{t}\]

we get:

\[\\A=(1.2\times{10}^{7})(1+0.03)^{4}\\\]

\[\\A=(1.2\times{10}^{7})(1.03)^{4}\\ \\A=(1.2\times{10}^{7})(1.1255)\\ \\A=1.35\times{10}^{7}\\\]

Example 5: compound interest (best buy)

Two offers exist for the same car.

Offer AOffer B
6% APR for 3 years3% APR for 5 years

A car has a value of £7999. Which offers a better value over the period of time stated?

State the formula and the value of each variable.

Here we are calculating compound interest with the formula

\[A=P(1+r)^{t}\]

twice as we are comparing each offer:

Offer AOffer B
P = £7999
r = 0.06
t = 3
P = £7999
r = 0.03
t = 5

Substitute the values into the formula.

Substituting these values into the simple interest formula

\[A=P(1+r)^{t}\]

we get:

Offer A
\\A=7999(1+0.06)^{3}\\

Offer B
\\A=7999(1+0.03)^{5}\\

Solve the equation.

Offer A
\\A=7999\times{1.06}^{3}\\
\\A=\pounds{9526.94}\\

Offer B
\\A=7999\times{1.03}^{5}\\
\\A=\pounds{9273.03}\\

 The best offer is Offer B (remember to state your conclusion).

Common misconceptions

  • Applying the incorrect formula to the question

This is a very common mistake where simple interest is calculated instead of using compound interest.

  • Incorrect percentage change due to different time scales

Here is example 3 with this misconception:

A house is valued at £150,000. On average, the house price increases by 2.4% per month over a period of 2.5 years compound interest. What is the value of the house after this time?

Here we use the formula

\[A=P(1+\frac{r}{n})^{nt}\]

with:

  • P = 150000
  • r = 0.024
  • n = 1
  • t = 2.5
\[\\A=150000(1.024)^{2.5}\\ \\A= £159162.65\\\]

Although the answer here looks reasonable, the 2.4% interest is per month not per year, which is what has been calculated.

  • Using the incorrect value for the percentage change

Using the percentage as the value for r (therefore not dividing the percentage by 100). For example when using compound interest to increase £100 by 2% for 5 years, this calculation is made:

\[\\A=100(1+2)^{5}\\ \\A=100\times{243}\\ \\A= £{24300!}\\\]

This is clearly wrong. We need to instead divide 2 by 100 to give 0.02. This gives a sensible final answer of £101.41.

  • Using a positive value for r when it should be negative

If a value is depreciating (going down), the value of r is negative whereas it is incorrectly used as a positive and so the answer will be larger than the original amount. 

Practice compound interest formula questions

1. The amount of water in a pond changes throughout the year. For 6 months of the year, the water level increases by an average of 1\% . For the other 6 months of the year, the water level drops by 0.7\% . If the pond contains 400L at the beginning of the year, how much water is in the pond at the end of the first year?

401.8L
GCSE Quiz False

401.172L
GCSE Quiz False

407.08L
GCSE Quiz True

407.8L
GCSE Quiz False

The pond starts with 400L of water. We are dealing with two 6 month time periods with a different multiplier for each one.

The amount of water is given by 400\times(1+0.01)^{6}\times(1-0.007)^{6} .

2. A bar of soap decreases in weight every time it is used by 3\% compound interest. If the bar of soap weighs 50g at the start of the week, and it is used 45 times, how much does it now weigh?

5g
GCSE Quiz False

12.7g
GCSE Quiz True

47g
GCSE Quiz False

8g
GCSE Quiz False

Using the compound interest formula results in the following calculation: 50\times(1-0.03)^{45}=12.7 \quad (3sf) .

3. The Amazon rainforest covers an area of 5.5 million km^2 in 2021 . If the rate of deforestation is 0.2\% , what is the expected area of the rainforest in 2030 ?

5,401,788 km^2
GCSE Quiz True

1,100,000 km^2
GCSE Quiz False

4,400,000 km^2
GCSE Quiz False

5,390,000 km^2
GCSE Quiz False

Using the compound interest formula results in the following calculation, with the answer rounded to the nearest integer: 5500000\times(1-0.002)^{9}=5401788 .

4. Account A has an annual percentage yield of 2.5\% compound interest. Account B has an interest rate of 0.2\% per month compound interest. If the initial investment is £3000 , how much more money has Account B accumulated than Account A in interest after 40 months?

£230.67
GCSE Quiz False

£18.93
GCSE Quiz True

£249.60
GCSE Quiz False

£480.27
GCSE Quiz False

After 40 months Account A has accumulated £230.67 in interest as 3 years interest has been accrued.

After 40 months Account B has accumulated £249.60 in interest as 40 months interest has been accrued.

The difference is £18.93 .

5. A bucket is leaking water at a rate of 17\% per hour, compound interest. How many hours would it take for the bucket to be under half full?

2 hours

GCSE Quiz False

3 hours

GCSE Quiz False

4 hours

GCSE Quiz True

5 hours

GCSE Quiz False

After 1 hour the bucket is 83\% full.

After 2 hours the bucket is 68.89\% full.

After 3 hours the bucket is 57.18\% full.

After 4 hours the bucket is 47.46\% full.

6. A battery loses its charge at a rate of 5\% every 2 hours, compound interest. If a battery starts fully charged, what percentage charge is left after 8 hours? Write your answer to 2 decimal places.

80.00\%
GCSE Quiz False

81.45\%
GCSE Quiz True

60.00\%
GCSE Quiz False

85.74\%
GCSE Quiz False

We could use the compound interest formula, or look at how the battery changes over time.

After 2 hours the battery is 95\% full.

After 4 hours the battery is 90.25\% full.

After 6 hours the battery is 85.74\% full.

After 8 hours the battery is 81.45\% full.

Compound interest formula GCSE questions

1. (a) An initial deposit of £1400 is invested for 3 years. The interest payments occur annually at 6\% compound interest. Work out the amount of interest earned after this time.

(b) After the first 3 years, the interest rate falls to 2\% . How much would the investment be worth after a further 4 years?

(4 marks)

Show answer

(a)

1400 × (1.06)^{3}

(1)

£1667.42-1400 = £267.42

(1)

(b)

1667.42 × (1.02)^{4}

(1)

£1804.87

(1)

2. (a) A car is bought for £15 000 . The car loses 10\% of its value every year, compound interest. Which calculation works out the total value after 5 years? Circle your answer.

  • 15000 − 10 × 5
  • (15000 10) × 5
  • 1.1^{5} × 15000
  • 15000 × 0.9^{5}

(b) The car is paid for using a credit card. Every month, £1500 of the bill is paid off, but 0.2\% of the remaining bill is added after. Calculate the balance on the credit card after 3 months.

(4 marks)

Show answer

(a)

15000 × 0.9^{5}

(1)

(b)

13500 × 1.002 = £13527  (month 1)

(1)

12027 × 1.002 = £12051.05 (month 2)

(1)

10551.05 × 1.002 = £10572.16 (month 3)

(1)

3. (a) Kieran invests in multiple financial institutions using the stock market. After the first 3 years, he has made an average of 3\% interest per year, compound interest. For the next two years, his stock prices fall by a compound interest rate of 4\% each year. If the original investment was worth £13,000 , calculate the future value of the investment after 5 years.

 

(b) Harry invested in the same stock, with his investment after 5 years having the present value of £7500 . What was his original investment worth?

(8 marks)

Show answer

(a)

13000 × (1.03)^{3}   

(1)

 

£14205.45

(1)

 

14205.45 × (0.96)^{2}

(1)

 

£13091.74

(1)

 

(b)

7500 / (0.96)^{2}

(1)

 

£8138.02

(1)

 

8138.02 / (1.03)^{3}

(1)

 

£7447.44

(1)

Learning checklist

You have now learned how to:

  • Solve problems involving percentage change, including: percentage increase, decrease and original value problems and simple interest in financial mathematics
  • Set up, solve and interpret the answers in growth and decay problems, including compound interest (and work with general iterative processes)

The next lessons are

  • Reverse percentages
  • Simple interest
  • Iteration

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