Question

A sum of 44,200 is divided between John
and Smith, 12 years and 14 years old
respectively, in such a way that if their
portions be invested at 10 percent per annum
compound interest, they will receive equal
amounts on reaching 16 years of age.
(i) What is the share of each out of
44,200 ?
(ii) What will each receive, when 16 years old? ​

Answer 1

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

Let assume that

Share of John = Rs x

Share of Smith = Rs y

So, x + y = 44200 ----------(1)

Case :- 1

John is 12 years old and he attains his share at the age of 16 years. It means his investment is for 4 years.

So, we have

Amount to be invested, p = Rs x

Rate, r = 10 % per annum compounded annually.

Time, n = 4 years.

Now,

We know that

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

$\rm :\longmapsto\:{ Amount=x\bigg(1+\dfrac{10}{100}\bigg)^{4}}$

$\rm :\longmapsto\:{ Amount=x\bigg(1+\dfrac{1}{10}\bigg)^{4}}$

$\rm :\longmapsto\:{ Amount=x\bigg(\dfrac{10 + 1}{10}\bigg)^{4}}$

$\rm :\longmapsto\:{ Amount=x\bigg(\dfrac{11}{10}\bigg)^{4}} - - - - (2)$

Case :- 2

Smith is 14 years old and he attains his share at the age of 16 years. It means his investment is for 2 years.

So, we have

Amount to be invested, p = Rs y

Rate, r = 10 % per annum compounded annually.

Time, n = 2 years.

Now,

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

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

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

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

$\rm :\longmapsto\:{ Amount=y\bigg(\dfrac{11}{10}\bigg)^{2}} - - - - (3)$

According to statement,

It is given that Amount received by John and Smith at the age of 16 years is same.

$\rm :\longmapsto\:{x\bigg(\dfrac{11}{10}\bigg)^{4} = y\bigg(\dfrac{11}{10}\bigg)^{2}}$

On cancelation the common factor, we get

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

$\rm :\longmapsto\:y = \dfrac{121}{100}x$

$\rm :\longmapsto\:44200 - x = \dfrac{121}{100}x$

[ Using equation (1), x + y = 44200 ]

$\rm :\longmapsto\:44200 = \dfrac{121}{100}x + x$

$\rm :\longmapsto\:44200 = \dfrac{121x + 100x}{100}$

$\rm :\longmapsto\:44200 = \dfrac{221x}{100}$

$\rm :\longmapsto\:200 = \dfrac{x}{100}$

$\bf\implies \:x = 20000 - - - - (4)$

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

$\rm :\longmapsto\:20000 + y = 44200$

$\rm :\longmapsto\: y = 44200 - 20000$

$\bf :\longmapsto\: y = 24200$

Now,

From equation (2) we have

$\rm :\longmapsto\:{ Amount=x\bigg(\dfrac{11}{10}\bigg)^{4}}$

On substituting the value of x, we get

$\rm :\longmapsto\:{ Amount=20000 \times \bigg(\dfrac{11}{10}\bigg)^{4}}$

$\rm :\longmapsto\:{ Amount=20000 \times 1.4641}$

$\bf :\longmapsto\:{ Amount=29282}$

Hence,

John share is Rs 20000

Smith Share is Rs 24200

Amount received by both at age of 16 years = Rs 29282

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

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

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

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