Question

EXERCISE G
1. A man left * 45,909 for his son and daughter who are 12 and 10 years old in such a way
that if their shares be invested at 2% compound interest annually, they shall receive equal
amount on reaching 15 years of age. How did he divide the money?
[Ans. Son * 23,409 and Daughter 3 22,500)
​

Answer 1

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

Total amount to be share = Rs 45, 909.

➢ Let assume that Son share be Rs x

and

➢ Let assume that Daughter share be Rs y.

$\rm :\longmapsto\:x + y = 45909 - - - (1)$

Case :- 1

Son share = Rs x

Time period for which Rs x invested be 3 years.

Rate of interest be 2 % per annum.

We know that,

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

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

$\rm :\longmapsto\:\rm{ Amount_{son}=x\bigg(1+\dfrac{2}{100}\bigg)^{3}}$

$\rm :\longmapsto\:\rm{ Amount_{son}=x\bigg(1+\dfrac{1}{50}\bigg)^{3}}$

$\rm :\longmapsto\:\rm{ Amount_{son}=x\bigg(\dfrac{50 + 1}{50}\bigg)^{3}}$

$\rm :\longmapsto\:\rm{ Amount_{son}=x\bigg(\dfrac{51}{50}\bigg)^{3}} - - - (2)$

Case :- 2

Daughter share = Rs y

Time period for which Rs y invested be 5 years.

Rate of interest be 2 % per annum.

We know that,

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

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

$\rm :\longmapsto\:\rm{ Amount_{daughter}=y\bigg(1+\dfrac{2}{100}\bigg)^{5}}$

$\rm :\longmapsto\:\rm{ Amount_{daughter}=y\bigg(1+\dfrac{1}{50}\bigg)^{5}}$

$\rm :\longmapsto\:\rm{ Amount_{daughter}=y\bigg(\dfrac{50 + 1}{50}\bigg)^{5}}$

$\rm :\longmapsto\:\rm{ Amount_{daughter}=y\bigg(\dfrac{51}{50}\bigg)^{5}} - - - (3)$

Now,

According to statement,

When son and daughter attains the age of 15 years, their respective amounts are equal.

$\rm :\longmapsto\:\rm{ Amount_{daughter}=Amount_{son}}$

$\rm :\longmapsto\:\rm{x\bigg(\dfrac{51}{50}\bigg)^{3} = y\bigg(\dfrac{51}{50}\bigg)^{5}}$

$\rm :\longmapsto\:x = y\bigg(\dfrac{51}{50}\bigg)^{2}$

$\rm :\longmapsto\:x = y\bigg(\dfrac{2601}{2500}\bigg) - - - - (4)$

On substituting the value of x in equation (1), we get

$\rm :\longmapsto\: y\bigg(\dfrac{2601}{2500}\bigg) + y = 45909$

$\rm :\longmapsto\: \bigg(\dfrac{2601y + 2500y}{2500}\bigg) = 45909$

$\rm :\longmapsto\: \bigg(\dfrac{5101y}{2500}\bigg) = 45909$

$\rm :\longmapsto\: \bigg(\dfrac{y}{2500}\bigg) =9$

$\bf\implies \:y = 22500$

On substituting the value of y in equation (1), we get

$\rm :\longmapsto\:x + 22500 = 45909$

$\rm :\longmapsto\:x = 45909 - 22500$

$\bf\implies \:x = 23409$

Hence,

$\red{\begin{gathered}\begin{gathered}\sf\: Amount \: distributed \: as \: -\begin{cases} &\sf{Son = Rs \: 23409} \\ &\sf{Daughter = Rs \: 22500} \end{cases}\end{gathered}\end{gathered}}$

Additional Information :-

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

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

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

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

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

$\rm :\longmapsto\:\rm{ Amount=P\bigg(1+\dfrac{r}{1200}\bigg)^{12n}}$