# Examples

## Example – two methods

A Lady put £400 in a savings account for 3 years at 5%. What will she have at the end of 3 years?

You have two ways of completing this.

### First method

Let’s not use a formula let’s just use logic

End of 1st year =£400=100%

x=105%

(See percentages for this method)

End of 1st year -

£400=100%

x=105%

400/x=100/105

Therefore

x=(400times105)/100

x=£420

End of 2nd year

£420=100%

x=105%

420/x=100/105

Therefore

x=(420times105)/100

x=£441

End of 3rd year

£441=100%

x=105%

441/x=100/105

Therefore

x=(441times105)/100

x=£463.05

The lady would have £463.05 at the end of the 3rd year

### Second method

Use the formula

A=P(1+r/n)^(nt)

A=£400(1+0.05/1)^(1times3)

A=£400(1+0.05)^3

A=£400(1.05)^3

A=£400times1.157625

A=£463.05

The lady will have £463.05 at the end of 3 years.

### Example - Interest per year

A woman invests £2000 in a savings account and this account pays 6% interest per year. How much will she have in the account after nine years to the nearest pound?

A=P(1+r/n)^(nt)

A=£2000(1+0.06/1)^(1times9)

A=£2000(1+0.06)^9

A=£2000(1.06)^9

A=£2000times1.689

A=£3,379

After nine years the woman will have £3,379 to the nearest pound in the savings account.

### Example - Interest per month

If you invest £1,000 earning 5% compound monthly.

How much will you have after 15 years to the nearest pound?

A=P(1+r/n)^(nt)

A=£1000(1+0.05/12)^(12times15)

See how to calculate powers on next page

A=£1000(1+0.00417)^180

A=£1000(1.00417)^180

A=£1000times2.1137

A=£2,114

After 15 years you will have £2,114 to the nearest pound (of this amount £1,000.00 is your initial investment).

### Example - Depreciation (Negative interest)

A car cost £8,000 but it depreciates by 9% each year. How much is it worth after 4 years?

NOTE:

The annual interest formula works in exactly the same way

A=P(1+r/n)^(nt)

A=£8000(1+(-0.09)/1)^(1times4)

NOTE:

The minus on the interest rate r.

A=£8000(1-0.09)^4

A=£8000(0.91)^4

A=£8000times0.686

A=£5,486.00

After 4 years the car is only worth £5,486.00

### Example - Interest rate quarterly

A lady invests £5,000 in a savings account which compounds the interest quarterly (i.e. 4 times a year), after 5 years how much will the lady have in the account if the interest rate is 8% per year.

A=P(1+r/n)^(nt)

A=£5000(1+0.08/4)^(4times5)

A=£5000(1+0.02)^20

A=£5000(1.02)^20

A=£5000times1.4859

A=£7,429.74

The lady will have £7,429.74 at the end of five years.

NOTE:

If this had been compound interest yearly then the answer would have been £7,346.64

### Example - Not just for money

A bacteria colony multiplies at the rate of 15% per day. If the colony started with 800 bacteria cells. How many cells would be in the bacteria colony at the end of one week?

A=P(1+r/n)^(nt)

Normally t = time is over a year but the period of growth is per day so in this case t = 1 for one day

Therefore

n= once per day too.

A=800(1+0.15/1)^(1times7)

Note the minus on the interest rate r.

A=800(1+0.15)^7

A=800(1.15)^7

A=800times2.660

A=2,128

There will be 2,128 bacteria cells after one week.

### Example – Finding the principal sum

An investment account with a 10% yearly compound interest rate contains £331 at the end of 3 years. What was the initial investment?

A=P(1+r/n)^(nt)

£331=£P(1+0.1/1)^(1times3)

£331=£P(1+0.1)^3

£331=£P(1.1)^3

£331=£Ptimes1.331

£P=(£331)/1.331

£P=£248.69