Question

The difference between the compound interest and the simple interest on a certain sum for $3$ years at $10\%$ per annum is $Rs.93$. Find the sum.

Answer 1

Let, $P$ be the sum,

Since,

Simple interest $(S.I)=\frac{P\times T\times R}{100}$

$\Rightarrow S.I=\frac{P\times 3\times 10}{100}$

$\Rightarrow S.I=\frac{3P}{10}$ ---(1)

Compound interest $(C.I)=A-P$

and $A=P(1+\frac{R}{100}{)}^n$

$\Rightarrow A=P(1+\frac{10}{100}{)}^3$

$\Rightarrow A=P(1+\frac{1}{10}{)}^3$

$\Rightarrow A=P(\frac{11}{10}{)}^3$

now, $(C.I)=A-P$

$\Rightarrow (C.I)=P(\frac{11}{10}{)}^3-P$---(2)

It is given that, the difference between the compound interest and the simple interest is $Rs.93$.

i.e., $C.P-S.P=Rs.93$

$\Rightarrow P(\frac{11}{10}{)}^3-P-\frac{3P}{10}=Rs.93$

$\Rightarrow P(\frac{1331}{1000}-1-\frac{3}{10})=Rs.93$

$\Rightarrow P\left(\frac{1331-1000-300}{1000}\right)=Rs.93$

$\Rightarrow \frac{31P}{1000}=Rs.93$

$\Rightarrow P=Rs.93\times \frac{1000}{31}$

$\Rightarrow P=Rs.3\times 1000$

$\Rightarrow P=Rs.3000$

Hence, the required sum is $Rs.3000$.