Question

A sum of 44200 is divided between John and Smith 12 years and 14 years old respectively in such a way that is their portions be invested at 10% per annum compound interest they will receive equal amounts on waiting 16 years of age for What is the share of each out of 44200 what will each receive when 16 years old

Answer 1

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

$\begin{gathered}\begin{gathered}\bf\: Let-\begin{cases} &\sf{share \: of \: John = x} \\ &\sf{share \: of \: Smith = y} \end{cases}\end{gathered}\end{gathered}$

According to statement,

☆ Total share = 44299

$\red{\sf :\longmapsto\:x + y = 44200 - - - - (1)}$

Since,

☆ John is 12 years old, and he received the maturity when he is 16 years old, so it means John investment is for 4 years.

and

☆ Smith is 14 years old, and he received the maturity when he is 16 years old, so it means John investment is for 2 years.

Case :- 1 Investment for John

• Sum invested, p = x

• Time of investment, n = 4 years

• Rate of interest, r = 10 % per annum

We know,

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

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

$\sf :\longmapsto\:A_1 = x {\bigg(1 + \dfrac{10}{100} \bigg) }^{4}$

$\sf :\longmapsto\:A_1 = x {\bigg(1 + \dfrac{1}{10} \bigg) }^{4}$

$\sf :\longmapsto\:A_1 = x {\bigg( \dfrac{10 + 1}{10} \bigg) }^{4}$

$\sf :\longmapsto\:A_1 = x {\bigg( \dfrac{11}{10} \bigg) }^{4} - - - (1)$

Case :- 2 Investment for Smith

• Sum invested, p = y

• Time of investment, n = 2 years

• Rate of interest, r = 10 % per annum

We know,

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

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

$\sf :\longmapsto\:A_2 = y {\bigg(1 + \dfrac{10}{100} \bigg) }^{2}$

$\sf :\longmapsto\:A_2 = y {\bigg(1 + \dfrac{1}{10} \bigg) }^{2}$

$\sf :\longmapsto\:A_2 = y {\bigg( \dfrac{10 + 1}{10} \bigg) }^{2}$

$\sf :\longmapsto\:A_2 = y {\bigg( \dfrac{11}{10} \bigg) }^{2} - - - (2)$

According to statement,

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

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

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

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

$\rm :\longmapsto\:121x = 100y$

$\rm :\longmapsto\:121x = 100(44200 - x) \: \: \: \: \: \: \: \: \{using \: (1) \}$

$\rm :\longmapsto\:121x = 4420000 - 100x$

$\rm :\longmapsto\:121x + 100x= 4420000$

$\rm :\longmapsto\:221x= 4420000$

$\bf\implies \:x = 20000$

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

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

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

$\bf\implies \:y = 24200$

$\begin{gathered}\begin{gathered}\bf\: Thus-\begin{cases} &\sf{share \: of \: John = 20000} \\ &\sf{share \: of \: Smith = 24200} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

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

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

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

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

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

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