# Simple Interest

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.

## What is simple interest?

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 P using simple interest over 2 years.

When calculating 5% of the amount P, we divide P by 100 (to find 1%) and then multiply this value by 5 to achieve 5% of P.

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 multiply the amount by 0.05.
This is why we usually write the value of r as a decimal.

The new value of P × 0.05 is the amount of interest for 1 interval so the amount of interest over 2 intervals would be double this amount or P × 0.05 × 2 or P × 0.10 because the same amount of interest is added for every interval.

This means that we multiply the amount of interest by the number of intervals, t, to find the value of the interest only.

We now need to add the principal amount, P, to this value to find the final amount, A.
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 P by 1.10 is an increase of 5% each year for 2 years on the value of P.

## How to calculate simple interest

In order to calculate simple interest:

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

## Simple interest examples

### Example 1: simple interest – percentage increase

£1500 is invested for 4 years at an annual interest rate of 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

2Substitute the values into the formula.

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

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

3Solve the equation.

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

### Example 2: simple interest – percentage decrease

A car is bought for £10,000 and loses 9% of its value per annum, 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:

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

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

### Example 3: simple interest – different time scale

£7600 is borrowed for 2 years on a credit card. The cost of borrowing is a 1% interest payment per month simple interest for the life of the loan. What is the total cost to pay off after this time?

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

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

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

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

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

### Example 4: simple interest – different percentages

A house is currently valued at £175,000. For the first 3 years, the value of the house increases by the rate of interest of 0.2% per annum. For the following 4 years, the value of the house decreases in value by 0.18% per annum. Using simple interest, calculate the value of the house after these 7 years.

Here we use the formula

$A=P\left(1+r_{1} t_{1}+r_{2} t_{2}\right)$
$P=175,000 \\ r_{1}=0.002 \\ t_{1}=3 \\ r_{2}=-0.0018 \\ t_{2}=4$

Substituting these values into the simple interest formula A=P(1+r1t1+r2t2), we get:

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

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

The US national debt reached $27.75 trillion in the year 2020. If the average amount of debt increased by 1% per year for the next 5 years simple interest, by how much would the debt increase in this time? Write your answer in standard form. Here we can use the formula A=P(rt) as we are calculating the interest only, with: • P = 27.75 trillion = 27,750,000,000,000 • r = 0.01 • t = 5 Substituting these values into the simple interest formula A=P(rt), we get: $\\A=27,750,000,000,000(0.01\times{5})\\$ $\\A=27,750,000,000,000(0.05)\\ \\A=27,750,000,000,000\times{0.05}\\ \\A=1,387,500,000,000\\ \\A=£1.3875\times{10^{12}}\\$ ### Example 6: simple interest – best buy The Chocolatier sells a chocolate bar for £2.40. The Choc Shop sells the same item for £1.99. Both shops have a new offer. Which shop is the cheapest to buy the chocolate bar from? The Chocolatier Save 33% on all chocolate bars The Choc Shop All items 20% off State the formula and the value of each variable. Here we use the formula A=P(1+rt) twice as we are comparing each shop: The chocolatier P = £2.40 r = -0.33 t = 1 The Choc Shop P = £1.99 r = -0.20 t = 1 Substituting these values into the simple interest formula A=P(1+rt), we get: The Chocolatier $\\A=2.41(1-0.33\times{1})\\$ The Choc Shop $\\A=1.99(1-0.2\times{1})\\$ The cheapest shop is The Chocolatier by 1p. (Remember to state your conclusion). The Chocolatier A = 2.4(0.67) A = 1.584 A = £1.58 The Choc Shop A = 1.99(0.8) A = 1.592 A = £1.59 ### Common misconceptions • 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: “£7600 is invested for 2 years at 1% per month simple interest. What is the value of the investment after this time?”, a common error is to incorrectly write the value of t as 2 rather than 24. • Using the incorrect value for the percentage change. A common error is to use the actual percentage as the value for r (therefore not dividing the percentage by 100). E.g. When using simple interest to increase £100 by 2% for 5 years, this incorrect calculation could be made: $\\A=100(1+2\times{5})\\ \\A=100\times{11}\\ \\A= £{1100}\\$ 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 r is negative. Simple interest is part of our series of lessons to support revision on simple interest and compound interest. You may find it helpful to start with the main simple interest and compound interest lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include: ### Practice simple interest questions 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) ### Simple interest GCSE questions 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)

300 x (1+0.03 x 5)

(1)

£345

(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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