Question

A sum of money was lent for $2$ years at $20\%$ compounded annually. If the interest is payable half-yearly instead of yearly, then the interest is $Rs.482$ more. Find the sum.

Answer 1

Since, $A=P(1+\frac{R}{100})$

also, $P=A-C.I$

Let the sum of money be $Rs.x$.

If the interest is compounded annually, then:

$A_1=P_1(1+\frac{R}{100}{)}^n$

$\Rightarrow A_1=x(1+\frac{20}{100}{)}^2$

$\Rightarrow A_1=(\frac{120x}{100}{)}^2$

$\Rightarrow A_1=x(\frac{6}{5}{)}^2$

$\Rightarrow A_1=\frac{36x}{25}$

$\Rightarrow C.I_1=A_1-P$

$\Rightarrow C.I_1=\frac{36x}{25}-x$

$\Rightarrow C.I_1=\frac{36x-25x}{25}$

$\Rightarrow C.I_1=\frac{11x}{25}$ ---(1)

If the interest is compounded half-yearly, then:

$R=\frac{20}{2}=10$, $n=4$ times

$A_2=P_1(1+\frac{R}{100}{)}^n$

$\Rightarrow A_2=x(1+\frac{10}{100}{)}^4$

$\Rightarrow A_2=x(\frac{11}{10}{)}^4$

$\Rightarrow A_2=\frac{14641x}{10000}$

$\Rightarrow C.I_2=A_1-P$

$\Rightarrow C.I_2=\frac{14641x}{10000}-x$

$\Rightarrow C.I_2=\frac{14641x-10000x}{10000}$

$\Rightarrow C.I_2=\frac{4641x}{10000}$ ---(2)

It is given that, if the interest is compounded half-yearly then the interest is $Rs.482$ more.

$\Rightarrow \frac{4641x}{10000}=\frac{11x}{25}+482$

$\Rightarrow 0.4641x=0.44x+482$

$\Rightarrow 0.4641x-0.44x=482$

$\Rightarrow 0.0241x=482$

$\Rightarrow \frac{241x}{10000}=482$

$\Rightarrow x=482\times \frac{10000}{241}$

$\Rightarrow x=2\times 10000$

$\Rightarrow x=20000$

Hence, the required sum is $Rs.20000$