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.

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

$A=800(1+0.15)^7$

$A=800(1.15)^7$

$A=800times2.660$

$A=2,128$

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