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Here we will learn about **simple interest** including how to calculate simple interest for increasing and decreasing values, and how to set up, solve and interpret growth and decay problems.

There are also simple interest worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

**Simple interest **is calculated by finding a percentage of the principal (original) amount and adding it on. It calculates a repeated percentage change of the original amount.

Simple interest is not frequently used in real life for repeated percentage change, however if you are calculating the sale price of an item, the VAT added on top of an item, or the surcharge of borrowing money as a loan, you will most likely use simple interest.

**Simple Interest Formula**

Simple interest can be calculated using the following formula:

\[\\A=P(1+rt)\\\]

Where:

- A represents the final amount.
- P represents the original amount.
- r is the percentage change (written as a decimal).
- t represents the time period (number of intervals).

E.g

Let’s calculate a ** 5% increase** on an amount

When calculating

We can write this as the following:

\[\\\frac{P}{100}\times{5}\\
\\= P\times\frac{5}{100}\\
\\= P\times{0.05}\\\]

So to find ** 5% of an amount**, we

This is why we usually write the value of

The new value of ** P × 0.05 ** is the amount of interest for

This means that we **multiply **the amount of **interest **by the **number of intervals**,

We now need to add the principal amount,

So we now have:

\[\\A=P+P\times{0.10}\\
\\A=P(1+0.10)\\
\\A=P\times1.10\\\]

We have now calculated the percentage change as a multiplier. For this example, multiplying

In order to calculate simple interest:

- State the formula and the value of each variable.
- Substitute the values into the formula.
- Solve the equation.

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

DOWNLOAD FREEGet your free simple interest worksheet of 20+ questions and answers. Includes reasoning and applied questions.

DOWNLOAD FREE**State the formula and the value of each variable**.

Here we use the formula

P = 1500 r = 0.05 t = 4

2**Substitute the values into the formula**.

Substituting these values into the simple interest formula

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

3**Solve the equation**.

\[\\A=1500(1+0.20)\\
\\A=1500(1.20)\\
\\A=1500\times{1.2}\\
\\A= £1800.\]

A car is bought for

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

Here we use the formula

P = 10000 r = – 0.09 t = 8

**Substitute the values into the formula**.

Substituting these values into the simple interest formula

\[\\A=1500(1-0.09\times{8})\\\]

**Solve the equation**.

\[\\A=10000(1-0.72)\\
\\A=10000(0.28)\\
\\A=10000\times{0.28}\\
\\A= £2800.\]

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

Here we use the formula

P = 7600 r = 0.01 t = 2 x12 = 24 (remember there are12 months in1 year)

**Substitute the values into the formula**.

Substituting these values into the simple interest formula

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

**Solve the equation**.

\[\\A=7600(1+0.24)\\
\\A=7600(1.24)\\
\\A=7600\times{1.24}\\
\\A= £{9424}.\]

A house is currently valued at

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

Here we use the formula

\[A=P\left(1+r_{1} t_{1}+r_{2} t_{2}\right)\]

\[P=175,000 \\
r_{1}=0.002 \\
t_{1}=3 \\
r_{2}=-0.0018 \\
t_{2}=4\]

**Substitute the values into the formula**.

Substituting these values into the simple interest formula _{1}t_{1}+r_{2}t_{2})

\[\\A=175,000(1+0.002\times{3}-0.0018\times{4})\\\]

**Solve the equation**.

\[\\A=175,000(1+0.006-0.0072)\\
\\A=175,000(0.9988)\\
\\A=175,000\times{0.9988}\\
\\A= £{174,790}.\]

The US national debt reached

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

Here we can use the formula

P = 27.75 trillion =27,750,000,000,000 r = 0.01 t = 5

**Substitute the values into the formula**.

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

\[\\A=27,750,000,000,000(0.01\times{5})\\\]

**Solve the equation**.

\[\\A=27,750,000,000,000(0.05)\\
\\A=27,750,000,000,000\times{0.05}\\
\\A=1,387,500,000,000\\
\\A=£1.3875\times{10^{12}}\\\]

The Chocolatier sells a chocolate bar for

**The Chocolatier**

Save

**The Choc Shop**

All items

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

Here we use the formula

The chocolatier

The Choc Shop

**Substitute the values into the formula**.

Substituting these values into the simple interest formula

The Chocolatier

\[\\A=2.41(1-0.33\times{1})\\\]

The Choc Shop

\[\\A=1.99(1-0.2\times{1})\\\]

**Solve the equation**.

The cheapest shop is The Chocolatier by

(Remember to state your conclusion).

The Chocolatier

The Choc Shop

**Mixing up simple and compound interest**

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

**Incorrect percentage change due to different time scales**

In example 3: “

**Using the incorrect value for the percentage change**.

A common error is to use the actual percentage as the value for

E.g.

When using simple interest to increase

\[\\A=100(1+2\times{5})\\
\\A=100\times{11}\\
\\A= £{1100}\\\]

r This is incorrect because r should be 0.02 **NOT **2

**Is the value increasing or decreasing?**

If the value is increasing (going up), the value of r is positive.

If a value is depreciating (going down), the value of

1. A technology store has a back to school offer: save 20 % on all full price laptops. Paula buys a laptop that was £689 full price. After 3 years, the value of the purchased laptop has decreased by 4 % per year, simple interest. What is the value of the laptop after these 3 years?

£551.20

£485.06

£529.15

£507.10

The original price was £689 . The sale price was 80 % of this. 80 % of £689 is £551.20 . The value of the laptop has decreased by 4 % of £551.20 per year for three years. We can work out 4 % of 551.20 and multiply by 3 ; this is £66.14 . Subtracting this from the purchase price gives the new value.

2. A saxophone that costs £999 is bought using a finance deal. The total price is divided by the number of months with 7% of the original amount added as an extra charge. The finance deal is spread over 2 years. How much money has been spent on the saxophone after 6 months?

£1068.93

£801.70

£44.54

£267.23

The amount to be paid back is given by 1.07\times999=1068.93

Six months is a quarter of the finance period, so a quarter of the full amount will have been spent on the saxophone, therefore \frac{1}{4}\times1068.93=267.23

3. Two shops have a sale. Shop A advertises a 10 % reduction on all items. Shop B has reduced all items by 8 % with a further 2 % reduction on the sale price of t-shirts. Josie can buy the same t-shirt from each shop that originally costs £16 . Which shop is selling the item cheapest?

Shop A

Shop B

Shop A: After a 10 % reduction, the t-shirt will cost £14.40

Shop B: The 8 % reduction means the t-shirt will be £14.72 , with the additional 2 % reduction making the price £14.43

4. £7342 is invested in a savings account with a 0.4 % simple interest rate per month. What is the value of the investment after 4 years?

£7459.47

£7694.42

£8751.66

£8046.83

The initial amount is 7342 , the simple interest rate is 0.4 % and the number of months is 48 , which results in 7342(1+0.004\times48)

5. A house is valued at £365,500 . The value of the house increases by an average of 0.25 % per year, simple interest. How much is the house worth after 12 years?

£366,413

£376,465

£497,080

£387,430

The initial amount is 365500 , the simple interest rate is 0.25 % and the number of years is 12 , which results in 365500(1+0.0025\times12)

1. (a) £850 is invested for 6 years at 2 % simple interest per year. Work out the total interest?

(b) The interest rate changes to 0.15 % simple interest per month. How does this affect the total amount of money invested over the same time period of 6 years?

**(6 marks)**

Show answer

a) 850 x (1+ 0.02 x 6)

**(1)**

£952

**(1)**

b) 0.15 x 12 x 6 = 10.8 %

**(1)**

850 x (1.108)

**(1)**

£941.80

**(1)**

Decreased by £10.20 over 6 years

**(1)**

2. (a) Freya invests $ 6700 for 2 years. The simple interest rate is 1.2 % per year. Which calculation below works out the total value after 2 years? Circle your answer.

(b) Euan invests $ 6400 for 2 years. The simple interest rate is 3.2 %.

Whose investment is worth more after 2 years and by how much?

You must show your working.

**(4 marks)**

Show answer

a) $ 6700 x 1.024

**(1)**

b) Freya:$ 6700 x 1.024 = £6860.80

**(1)**

Euan: $ 6400 x 1.064 = £6809.60

**(1)**

Freya’s investment is worth more $ 51.20

**(1)**

3. Lauren places £300 into a new bank account with a simple interest rate of 3 %.

How much money would she have after 5 years?

**(2 marks)**

Show answer

300 x (1+0.03 x 5)

**(1)**

£345

**(1)**

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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