Question

A sum lent on compound interest becomes Rs.2420 in 2 years an Rs.2662 in 3years find the sum​ step by step explain

Answer 1

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

Let assume that

Sum 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 2420

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

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

$\rm :\longmapsto\:2420=x\bigg(1+\dfrac{r}{100}\bigg)^{2} - - - (1)$

Case :- 2

Principal, P = Rs x

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

Time, n = 3 years.

Amount = Rs 2662

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

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

$\rm :\longmapsto\:2662=x\bigg(1+\dfrac{r}{100}\bigg)^{3} - - - (2)$

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

$\rm :\longmapsto\:\dfrac{2662}{2420} = \dfrac{x{\bigg(1 + \dfrac{r}{100} \bigg) }^{3}}{x{\bigg(1 + \dfrac{r}{100} \bigg) }^{2}}$

$\rm :\longmapsto\:\dfrac{11}{10} = 1 + \dfrac{r}{100}$

$\rm :\longmapsto\:\dfrac{11}{10} - 1= \dfrac{r}{100}$

$\rm :\longmapsto\:\dfrac{11 - 10}{10} = \dfrac{r}{100}$

$\rm :\longmapsto\:\dfrac{1}{10} = \dfrac{r}{100}$

$\bf\implies \:r = 10 \: \%$

Now, On substituting r = 10 in equation (1), we get

$\rm :\longmapsto\:2420=x\bigg(1+\dfrac{10}{100}\bigg)^{2}$

$\rm :\longmapsto\:2420=x\bigg(1+\dfrac{1}{10}\bigg)^{2}$

$\rm :\longmapsto\:2420=x\bigg(\dfrac{10 + 1}{10}\bigg)^{2}$

$\rm :\longmapsto\:2420=x\bigg(\dfrac{11}{10}\bigg)^{2}$

$\rm :\longmapsto\:2420=x\bigg(\dfrac{121}{100}\bigg)$

$\rm :\longmapsto\:20=x\bigg(\dfrac{1}{100}\bigg)$

$\bf\implies \:x = 2000$

Hence,

• Sum invested is Rs 2000

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

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

Answer 2

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