Interest & Depreciation

Simple Interest

What is simple interest?

Interest is extra money added every year (or month) to an original amount of money
Simple interest is interest that is the same amount each time
- It can be good: for example, putting £100 into a bank account and the bank rewarding you with simple interest of 10% every year
  - After one year you’d have £110, after two years you’d have £120, …
- It can be bad: for example, owing £100 to a friend and they charge you simple interest of 10% for every year you don’t pay them back
  - After the first year you’d owe them £110, after the second year you’d owe them £120, …
If £ P is your initial amount of money and simple interest is added to it at a rate of R% per year for T years, then the total amount of interest gained, £ I, is given by the formula

$I equals fraction numerator P R T over denominator 100 end fraction$

Remember that this formula calculates the amount of simple interest added over T years, not the total amount of money after T years
- To find the total amount of money after T years, add the interest £ I to the original amount £ P

Exam Tip

Worked example

A bank account offers simple interest of 8% per year. I put £250 into this bank account for 6 years. Find (a) the amount of interest added over 6 years, (b) the total amount in my bank account after 6 years. (a) Substitute P = 250, R = 8 and T = 6 into the formula

to find the simple interest, I

The amount of interest over 6 years is £120 (b) The total amount after 6 years is the original amount, £250, plus the interest from part (a), £120 250 + 120 The total amount in my bank account after 6 years is £370

Compound Interest

What is compound interest?

Compound interest is where interest is paid on the interest from the previous year (or whatever time frame is being used), as well as on the original amount
This is different from simple interest where interest is only paid on the original amount
- Simple interest goes up by the same amount each time whereas compound interest goes up by an increasing amount each time

How do you work with compound interest?

Keep multiplying by the decimal equivalent of the percentage you want (the multiplier, p)
A 25% increase (p = 1 + 0.25) each year for 3 years is the same as multiplying by 1.25 × 1.25 × 1.25
- Using powers, this is the same as × 1.25³
In general, the multiplier p applied n times gives an overall multiplier of pⁿ
If the percentages change varies from year to year, multiply by each one in order
- a 5% increase one year followed by a 45% increase the next year is 1.05 × 1.45
In general, the multiplier p₁ followed by the multiplier p₂ followed by the multiplier p₃… etc gives an overall multiplier of p₁p₂p₃…

Alternative method: A formula for the final (“after”) amount is where…

..P is the original (“before”) amount, r is the % increase, and n is the number of years
Note that is the same value as the multiplier

Exam Tip

Working example

Jasmina invests £1200 in a savings account which pays compound interest at the rate of 2% per year for 7 years.

To the nearest pound, what is her investment worth at the end of the 7 years?

We want an increase of 2% per year, this is equivalent to a multiplier of 1.02, or 102% of the original amount

This multiplier is applied 7 times;

Therefore the final value after 7 years will be

$£ 1200 space cross times space 1.02 to the power of 7 space equals space £ 1378.42$

Round to the nearest pound

$bold £ bold 1378$

Alternative method
Or use the formula for the final amount $P open parentheses 1 plus r over 100 close parentheses to the power of n space end exponent$
Substitute P is 1200, r = 2 and n = 7 into the formula

$1200 open parentheses 1 plus 2 over 100 close parentheses to the power of 7$

£1378 (to the nearest pound)

Depreciation

What is meant by depreciation?

Depreciation is where an item loses value over time
- For example: cars, game consoles, etc
Depreciation is usually calculated as a percentage decrease at the end of each year
- This works the same as compound interest, but with a percentage decrease

How do I calculate a depreciation?

You would calculate the new value after depreciation using the same method as compound interest
- Identify the multiplier, p (1 – “% as a decimal”)
  - 10% depreciation would have a multiplier of p = 1 – 0.1 = 0.9
  - 1% depreciation would have a multiplier of p = 1 – 0.01 = 0.99
- Raise the multiplier to the power of the number of years (or months etc)
  - $p to the power of n$
- Multiply by the starting value
New value is $A cross times p to the power of n$
- A is the starting value
- p is the multiplier for the depreciation
- n is the number of years
Alternative method: A formula for the final (“after”) amount is
- …P is the original (“before”) amount, r is the % decrease, and n is the number of years
If you are asked to find the amount the value has depreciated by:
- Find the difference between the starting value and the new value

Working example

Mercy buys a car for £20 000. Each year its value depreciates by 15%.

Find the value of the car after 3 full years.

Identify the multiplier

100% – 15% = 85%

p = 1 – 0.15 = 0.85

Raise to the power of number of years

0.85 ³

Multiply by the starting value

£20 000 × 0.85 ³

= £12 282.50

Alternative method
Or use the formula for the final amount $P open parentheses 1 minus r over 100 close parentheses to the power of n space end exponent$
Substitute P is 20 000, r = 15 and n = 3 into the formula