GCSE Maths Number

Simple Interest and Compound Interest

# Simple Interest and Compound Interest

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

Look out for the simple & compound interest worksheets and exam questions at the end.

## What are simple interest and compound interest?

Simple and compound interest are two ways of calculating interest:

Simple interest is calculated on the original (principal) amount, whereas compound interest is calculated on the original amount and on the interest already accumulated on it.

The difference between simple and compound interest is that simple interest is calculated using only the original amount whereas compound interest works out the interest on a previous amount as well.

The formula for calculating the simple interest earned on an investment is

A=Prt

And we can calculate the value of the investment, A, after the time period with the formula:

\begin{aligned} A&=P+Prt \\\\ & =P\left( 1+rt \right) \end{aligned}

The formula for calculating compound interest is:

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

Where:

• I represents the simple interest
• 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

Check out the comparative example below to see the similarities and differences between the two forms of interest.

## How to work out simple and compound interest

In order to calculate simple or compound interest:

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

E.g

\bf{£100} is invested for \bf{3} years at \bf{2\%} per year. Find the final value.

Simple interest

A=P(1+rt) \

Here:

• P=100
• r=0.02 (as 2% = 0.02)
• t=3

Substituting these values into the simple interest formula

A=P(1+rt)

We get:

A = 100(1+0.02\times{3})\

A = 100(1+0.06)\

A = 100(1.06)\

A = 100\times{1.06}\

A = £106

Compound interest

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

Here:

• P=100
• r=0.02
• n=1
• t=3

Substituting these values into the compound interest formula

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

We get:

A = 100(1+(\frac{0.02}{1})^{1\times3}

A = 100(1+0.02)^3\

A = 100(1.02)^3\

A = 100 x 1.02^3\

A = £106.12

Simple and compound interest are used widely in real life, especially in financial mathematics for sale prices, borrowing money on a credit card or taking out a loan, the amount of money invested in the stock market and house prices to name a few.

### Depreciation

When the value of an asset reduces in price, it is known as depreciation.

Depreciation can be calculated the same way as compound interest but we must remember that the multiplier being used will be less than 1.

E.g

A car worth £20 \ 000 depreciates by 10\% per year for 2 years.

To find the value after 2 years we calculate using the multiplier of 0.9.

20 \ 000 \times 0.9^{2}=16 \ 200

The car will be worth £16 \ 200.

Step-by-step guide: Depreciation

## Simple and compound interest examples

### Example 1: Simple Interest – Percentage Increase

£1500 is invested for 4 years at 5\% per year simple 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+rt) with:

• P=1500
• r=0.05
• t=4

2 Substitute the values into the formula.

Substituting these values into the simple interest formula A=P (1+rt) , we get:

A=1500(1+0.05\times{4})

3 Solve the equation.

\begin{aligned} &A=1500(1+0.20)\\\\ &A=1500(1.20)\\\\ &A=1500\times{1.2}\\\\ &A=\pounds1800. \end{aligned}

### Example 2: Simple Interest – Percentage Decrease

A car is bought for £10,000 and loses 9\% of its value per year simple interest. What is the value of the car after 8 years?

Here we use the formula A=P (1+rt) with:

• P=10000
• r=-0.09
• t=8

Substituting these values into the simple interest formula A=P (1+rt) , we get:

A=10000(1-0.09\times{8})

\begin{aligned} &A=10000(1-0.72)\\\\ &A=10000(0.28)\\\\ &A=10000\times{0.28}\\\\ &A=\pounds2800. \end{aligned}

### Example 3: Simple Interest – Different Time Scale

£7600 is invested for 2 years at 1\% per month simple interest. What is the value of the investment after this time?

Here we use the formula A=P (1+rt) with:

• P=7600
• r=0.01
• t= 2 \times 12 = 24 (remember there are 12 months in 1 year)

Substituting these values into the simple interest formula A = P (1 + rt), we get:

A=7600(1+0.01\times{24})

\begin{aligned} &A=7600(1+0.24)\\\\ &A=7600(1.24)\\\\ &A=7600\times{1.24}\\\\ &A=\pounds{9424}. \end{aligned}

### Example 4: Compound Interest – Percentage Increase

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

Here we use the formula A=P(1+\frac{r}{n})^{nt} with:

• P=8500
• r=0.003
• n=1
• t=5

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}}

\begin{aligned} &A=8500(1+0.003)^{5}\\\\ &A=8500(1.003)^{5}\\\\ &A=8500\times1.009027\\\\ &A=\pounds{8576.73} \end{aligned}

### Example 5: 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?

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}}

\begin{aligned} &A=399(1-0.22)^{2}\\\\ &A=399(0.78)^{2}\\\\ &A=399\times0.6084\\\\ &A=\pounds{242.75} \end{aligned}

### Example 6: Compound Interest – Different Time Scale

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=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}}

\begin{aligned} &A=150000(1+0.002)^{30}\\\\ &A=150000(1.002)^{30}\\\\ &A=150000(1.06177)\\\\ &A=150000\times1.06177\\\\ &A=\pounds159265.94 \end{aligned}

### Common misconceptions

• Applying the incorrect formula to the question

This is a very common mistake where the simple interest on an amount is calculated instead of using the compound interest formula.

• Incorrect percentage change due to different time scales

Here is example 6 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

\begin{aligned} &A=150000(1.024)^{2.5}\\\\ &A=\pounds159162.65 \end{aligned}

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:

\begin{aligned} &A=100(1+2)^{5}\\\\ &A=100\times{243}\\\\ &A=\pounds{24300} \end{aligned}

• Is the value increasing or decreasing?

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.

### Simple and Compound Interest practice questions

1. The probability of a team winning their next three football matches is 65\% per match. What is the probability that they win all three matches?

4.29\%

27.46\%

35\%

65\%

We can write the percentage as a fraction and then apply the “and” rule from probability, so the probability is given by

\frac{65}{100} \times \frac{65}{100} \times \frac{65}{100}

This can be converted to a percentage afterwards.

2. The population of bees in a hive is expected to increase by 5\% per year simple interest.

The population of bees in the hive is approximately 56,000 .

Work out the population of bees after 3 years.

64,400

58,800

61,600

64,827

5\% of 56,000 is 2800 , so over the 3 years the population increases by 8,400 .

Adding 8,400 to 56,000 , we get 64,400 .

3. A bucket holds 25L of water. A hole in the bucket starts to leak water, losing 12\% every minute compound interest.

Calculate the total amount of water in the bucket after 10 minutes?

13L

22L

6.96L

77.65L

The original amount is 25 , the decrease is 12\% per minute compounded over 10 minutes, so our calculation is

\\A=25(1-0.12)^{10}\\

\\A=6.96\\

4. The height of an apple tree increases at an average rate of 1.6\% per month, compound interest.

The apple tree is currently 1m tall.

How tall will the apple tree be after 2 years?

1.6m

2.6m

1.46m

1.38m

The original amount is 1 , the increase is 1.6\% per month compounded over 24 months, so our calculation is

\\A=1(1+0.016)^{24}\\

\\\\A=1.46\\

5. The population of chickens on an island is estimated to be around 20,000 .

With few natural predators, the chicken population has exploded, increasing the population by 3.6\% per year simple interest.

Estimate how many chickens there are on the island after 15 years.

72,000

30,800

20,720

33,996

3.6\% of 20,000 is 720 , so over the 15 years the population increases by 10,800 .

Adding 10,800 to 20,000 , we get 30,800 .

6. A vintage car was valued at \pounds 650,000 5 years ago. For the first 3 years, the value of the car depreciates by 2\% every year, using compound interest.

After this, the value increases by 5\% per year for the next 2 years, using compound interest.

What is the current value of the car?

\pounds 587,548

\pounds 752,456

\pounds 674,482

\pounds 611,775

The starting amount is 650,000 . To begin with, there is a 2\% decrease per year compounded over 3 years, then a 5\% increase per year compounded over 2 years. This can be combined into one calculation, so

\\V=650000(1-0.02)^{3}(1+0.05)^{2}\\

\\V=674482\\

### Simple and Compound Interest Exam Questions:

1.  Ed invested \pounds 1500 into a savings account. It earned compound interest at an annual interest rate of 0.72\% for 5 years.

He wrote down how much money he would receive after 5 years. Here is his calculation:

1500\times{0.72}\times{5}=£5400

(a) What are the two mistakes that Ed has made?

• ________________________________________________
• ________________________________________________

(b) What amount would Ed have after 5 years?

(4 marks)

(a)

Ed has used simple interest.

(1)

Ed has not converted the percentage correctly ( 72\% interest).

(1)

(b)

1500 \times 1.0072^5

(1)

\pounds 1554.78

(1)

2.  In 2003 , the population of bats in a cave was approximately 20 million.

After 5 years, the population grew by 3\% every year, and then for the next 3 years, the population gradually declined by 2\% each year.

How many more bats are in that cave in 2011 . Write your answer in standard form to 3 significant figures.

(5 marks)

20,000,000 \times 1.03^5

(1)

20,000,000 \times 1.03^5 \times 0.98^3   or  23185481.49 \times 0.98^3

(1)

21821990

(1)

21,821,990-20,000,000=1821990

(1)

1.82 \times 10^6

(1)

3.  (a)  \pounds 30,000 is invested with an annual percentage rate of 0.1\% simple interest rate per month into Account A . How many months will it take for the investment to reach \pounds 31,200 ?

(b) Account B offers a compound interest rate of 1.5\% per year. Which account would provide the most interest after exactly 3 years?

(7 marks)

(a)

30,000 \times 0.001 = £30 (monthly payment)

(1)

31,200-30,000=£1200 interest

(1)

£1200 \ / \ 30 = 40 months

(1)

(b)

30,000 \times 1.015^3

(1)

£31370.35

(1)

30,000 + (30 \times 36) = 31080

(1)

Account B

(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)

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