Question

A sum was lent for a year, another sum was lent for 2 years and another sum was lent for 3 years. Each sum was lent at 15% p.a compound interest. If each sum amounted to the same value, the ratio of the first, second and third sums is​

Answer 1

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

Let assume that,

• Sum invested for 1 year be Rs x

• Sum invested for 2 years be Rs y

• Sum invested for 3 years be Rs z.

Rate of interest = 15 % per annum compounded annually.

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 given by

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

Now,

☆ Rs x is invested for 1 year at the rate of 15 % per annum compounded annually, therefore

$\rm :\longmapsto\:{ \bf \: A_1 = x {\bigg(1 + \dfrac{15}{100} \bigg) }^{1}}$

$\rm :\longmapsto\:{ \bf \: A_1 = x {\bigg( \dfrac{100 + 15}{100} \bigg) }}$

$\rm :\longmapsto\:{ \bf \: A_1 = x {\bigg( \dfrac{115}{100} \bigg) }}$

$\rm :\longmapsto\:{ \bf \: A_1 = x {\bigg( \dfrac{23}{20} \bigg) }} - - (1)$

Also,

☆ Rs y is invested for 2 year at the rate of 15 % per annum compounded annually, therefore

$\rm :\longmapsto\:{ \bf \: A_2 = y {\bigg(1 + \dfrac{15}{100} \bigg) }^{2}}$

$\rm :\longmapsto\:{ \bf \: A_2 = y {\bigg(\dfrac{100 + 15}{100} \bigg) }^{2}}$

$\rm :\longmapsto\:{ \bf \: A_2 = y {\bigg(\dfrac{115}{100} \bigg) }^{2}}$

$\rm :\longmapsto\:{ \bf \: A_2 = y {\bigg(\dfrac{23}{20} \bigg) }^{2}} - - - (2)$

Again,

☆ Rs z is invested for 3 year at the rate of 15 % per annum compounded annually, therefore

$\rm :\longmapsto\:{ \bf \: A_3 = z {\bigg(1 + \dfrac{15}{100} \bigg) }^{3}}$

$\rm :\longmapsto\:{ \bf \: A_3 = z {\bigg( \dfrac{100 + 15}{100} \bigg) }^{3}}$

$\rm :\longmapsto\:{ \bf \: A_3 = z {\bigg( \dfrac{115}{100} \bigg) }^{3}}$

$\rm :\longmapsto\:{ \bf \: A_3 = z {\bigg( \dfrac{23}{20} \bigg) }^{3}}$

According to statement,

$\bf :\longmapsto\:A_1 = A_2 = A_3$

$\rm :\longmapsto\:{ \bf \: x{\bigg(\dfrac{23}{20} \bigg) } = y {\bigg(\dfrac{23}{20} \bigg) }^{2} = z{\bigg(\dfrac{23}{20} \bigg) }^{3}}$

$\pink{\rm :\longmapsto\:Dividing \: each \: term \: by{\bigg(\dfrac{23}{20} \bigg) }^{3}}$

$\bf :\longmapsto\:\dfrac{x}{{\bigg(\dfrac{23}{20} \bigg) }^{2}} = \dfrac{y}{{\bigg(\dfrac{23}{20} \bigg) }} = z$

$\bf\implies \:x : y : z = {\bigg(\dfrac{23}{20} \bigg) }^{2} : {\bigg(\dfrac{23}{20} \bigg) } : 1$

$\bf\implies \:x : y : z = {\bigg(\dfrac{529}{400} \bigg) } : {\bigg(\dfrac{23}{20} \bigg) } : 1$

$\bf\implies \:x : y : z = 529 : 460: 400$

☆ Additional Information :-

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

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

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

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