Simple interest and compound interest

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GCSE Maths Number

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}{100})^{n}

Where:

$I$ represents the simple interest
$A$ represents the final amount
$P$ represents the original principal amount
$r$ is the interest rate
$n$ represents the number of times the interest rate is applied

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

Step-by-step guide: Simple interest

Step-by-step guide: Compound interest

What are simple and compound interest?

How to work out simple and compound interest

In order to calculate simple or compound interest:

State the formula and the value of each variable.
Substitute the values into the formula.
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}{100})^{n} \$

Here:

$P=100$
$r=2$
$n=3$

Substituting these values into the compound interest formula

$A=P(1+\frac{r}{100})^{n} \$

We get:

$A = 100(1+(\frac{2}{100})^{3}$

$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.

Explain how to calculate simple or compound interest in 3 steps.

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 interest and compound interest worksheet

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

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Simple interest and compound interest worksheet

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

DOWNLOAD FREE

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?

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?

State the formula and the value of each variable.

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

$P=10000$
$r=-0.09$
$t=8$

Substitute the values into the formula.

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

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

Solve the equation.

\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?

State the formula and the value of each variable.

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)

Substitute the values into the formula.

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

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

Solve the equation.

\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?

State the formula and the value of each variable.

Here we use the formula $A=P(1+\frac{r}{100})^{n}$ with:

$P=8500$
$r=0.3$
$n=5$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P(1+\frac{r}{100})^{n},$ we get:

$A=8500(1+\frac{0.3}{100})^{5}$

Solve the equation.

\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?

State the formula and the value of each variable.

Here we use the formula $A=P(1+\frac{r}{100})^{n}$ with:

$P=399$
$r=-22$
$n=2$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P(1+\frac{r}{100})^{n},$ we get:

$A=399(1-\frac{22}{100})^{2}$

Solve the equation.

\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 $0.24\%$ per month over a period of $2.5$ years compound interest. What is the value of the house after this time?

State the formula and the value of each variable.

This time the interest rate in monthly so we need to convert the time into months.

$2.5$ year is the same as $2.5\times 12 = 30$ months.

Here we use the formula

$A=P(1+\frac{r}{100})^{n}$ with:

$P=150000$
$r=0.24$
$n=30$

Substitute the values into the formula.

Substituting these values into the compound interest formula $A=P(1+\frac{r}{100})^{n},$ we get:

$A=150000(1+\frac{0.24}{100})^{30}$

Solve the equation

\begin{aligned} &A=150000(1+0.0024)^{30}\\\\ &A=150000(1.0024)^{30}\\\\ &A=\pounds161184.40 \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 $A=P(1+\frac{r}{100})^{n}$ with:

$P=150000$
$r=0.24$
$n=2.5$

\begin{aligned} &A=150000(1.0024)^{2.5}\\\\ &A=\pounds150901.62 \end{aligned}

Although the answer here looks reasonable, the $0.24\%$ interest is per month not per year, which is what has been calculated.

Using the incorrect value for the percentage change

Not dividing the percentage rate by 100 is a common error. 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

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.

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 $64,400$ .

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\\

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\\

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 $30,800$ .

\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)

Show answer

(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)

Show answer

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)

Show answer

(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)

The next lessons are

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