# Annual simple interest vs. annual compound interest

100 Euros was invested in a savings account. After 10 years, there is  300 Euros in the account.

Work out how much money would have been in the account after only 5 years if it had been gathering a) annual simple interest. b) annual compound interest. Give each of your answers to the nearest 1 Euro.

c) Will the savings account have more money in it after 15 years if it is gathering annual simple interest or annual compound interest?

The amount in a saving account after $n$ simple interest is given by
$P_n=P_0(1+nI),$
and after $n$ annual compound interest is given by the formula
$P_n=P_0(1+I)^{n},$
where $P_0$ is the initial balance and $I$ is the yearly interest rate.

(a) If it had been gathering annual simple interest. Then
$P_{10}=P_0(1+10 I)=100(1+10I)=300 \Rightarrow 1+10I =3$
$\Rightarrow I=\frac{1}{5}= 20\%.$

Now we compute the amount which have been in the account after only 5 years.

$P_{5}=P_0(1+5 I)=100(1+5 \times \frac{1}{5})=200 \text{Euros}$

(a) If it had been gathering annual interest, then $k=1$. So
$P_{10}=P_0(1+I)^{10}=100(1+I)^{10}=300$
$\Rightarrow (1+I)^{10}=3 \Rightarrow (1+I)=3^{\frac{1}{10}}.$

Now we compute the ammount which have been in the account after only 5 years.
$P_{5}=P_0(1+I)^{5}=100(1+I)^{5}=100(3^{\frac{1}{10}})^{5}=100 \sqrt{3}=173 \text{Euros}.$

(b) If it had been gathering annual interest, then $k=1$. So
$P_{10}=P_0(1+I)^{10}=100(1+I)^{10}=300$
$\Rightarrow (1+I)^{10}=3 \Rightarrow (1+I)=3^{\frac{1}{10}}.$

Now we compute the amount which have been in the account after only 5 years.
$P_{5}=P_0(1+I)^{5}=100(1+I)^{5}=100(3^{\frac{1}{10}})^{5}=100 \sqrt{3}=173 \text{Euros}.$

c) After 15 years, the savings account have more money in it if it is gathering annual compound interest. This is because after 10 years, there is 300 Euros in the account and during years 11-15 the account gathering annual compound interest will be gathering interest on the interests earned during years 11-15.