Question

A certain sum of money at compound interest becomes 7.396 in two years and
7.950.70 in three years. Find the rate of interest.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let us assume that

Sum of money invested be Rs x.

Rate of interest be r % per annum compounded annually.

Case :- 1

Principal, P = Rs x

Rate of interest, r = r % per annum compounded annually.

Time, n = 2 years.

Amount = Rs 7396

We know,

Amount on a certain sum of money Rs P invested at the rate of r % per annum compounded annually for n years is

$\rm :\longmapsto\:{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}$

So, on substituting the values, we get

$\rm :\longmapsto\:{7396=x\bigg(1+\dfrac{r}{100}\bigg)^{2}} - - - (1)$

Case :- 2

Principal, P = Rs x

Rate of interest, r = r % per annum compounded annually.

Time, n = 3 years.

Amount = Rs 7950. 70

We know,

Amount on a certain sum of money Rs P invested at the rate of r % per annum compounded annually for n years is

$\rm :\longmapsto\:{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}$

On substituting the values, we get

$\rm :\longmapsto\:{7950.70=x\bigg(1+\dfrac{r}{100}\bigg)^{3}} - - - (2)$

Now, On dividing equation (2) by equation (1), we get

$\rm :\longmapsto\:\dfrac{7950.70}{7396} = \dfrac{x{\bigg(1 + \dfrac{r}{100} \bigg) }^{3}}{x{\bigg(1 + \dfrac{r}{100} \bigg) }^{2}}$

$\rm :\longmapsto\:1.075 = 1 + \dfrac{r}{100}$

$\rm :\longmapsto\:1.075 - 1 = \dfrac{r}{100}$

$\rm :\longmapsto\:0.075 = \dfrac{r}{100}$

$\bf\implies \:r = 7.5 \: \%$

Hence,

• Rate of interest is 7.5 % per annum.

Additional Information :-

Amount on a certain sum of money Rs P invested at the rate of r % per annum compounded annually for n years is

$\rm :\longmapsto\:{ Amount=P\bigg(1+\dfrac{r}{100}\bigg)^{n}}$

Amount on a certain sum of money Rs P invested at the rate of r % per annum compounded semi - annually for n years is

$\rm :\longmapsto\:{ Amount=P\bigg(1+\dfrac{r}{200}\bigg)^{2n}}$

Amount on a certain sum of money Rs P invested at the rate of r % per annum compounded quarterly for n years is

$\rm :\longmapsto\:{ Amount=P\bigg(1+\dfrac{r}{400}\bigg)^{4n}}$