# 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?

Answer:

You have two ways of completing this.

### First method

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

End of 1^{st} year `=£400=100%`

`x=105%`

(See percentages for this method)

**End of 1 ^{st} year - **

`£400=100%`

`x=105%`

`400/x=100/105`

Therefore

`x=(400times105)/100`

`x=£420`

**End of 2 ^{nd} year**

`£420=100%`

`x=105%`

`420/x=100/105`

Therefore

`x=(420times105)/100`

`x=£441`

**End of 3 ^{rd} year**

`£441=100%`

`x=105%`

`441/x=100/105`

Therefore

`x=(441times105)/100`

`x=£463.05`

Answer:

The lady would have £463.05 at the end of the 3^{rd} 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`

Answer:

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`

Answer:

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`

Answer:

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`

Answer:

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`

Answer:

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`

Answer:

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`

Answer:

The initial investment or starting amount was £248.69