Question

The difference between the compound interest and simple interest on a certain sum of money at $10\%$ per annum for $2$ years is $\mathrm{Rs}.500$. Find the sum when the interest is compounded annually.

Answer 1

Step 1: Determine the simple and compound interest.

Assume that the principle amount is $P$.

So, the simple interest is:

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{SI}\mathbf{=}\frac{\mathbf{P}\mathbf{\times }\mathbf{R}\mathbf{\times }\mathbf{T}}{\mathbf{100}} \\ \Rightarrow SI=\frac{P\times 10\times 2}{100} \\ \therefore SI=0.2P \end{gathered}$$

Similarly, the compound interest is:

$\mathbf{CI}\mathbf{=}\mathbf{P}{\mathbf{(}\mathbf{1}\mathbf{+}\frac{\mathbf{R}}{\mathbf{100}}\mathbf{)}}^T-\mathit{P}$

$\begin{array}{rcl}& \Rightarrow & CI=P{\left(1+\frac{10}{100}\right)}^2-P\\ & \Rightarrow & CI=1.21P-P\\ \therefore CI& =& 0.21P\end{array}$

So, the simple and compound interests are $0.2P\mathrm{and}0.21P$ respectively.

Step 2: Determine the amount.

The difference in the interests is $\mathrm{Rs}.500$.

The equation for the difference of interests is:

$\begin{array}{rcl}0.21P-0.2P& =& \mathrm{Rs}.500\\ & \Rightarrow & 0.01P=\mathrm{Rs}.500\\ & \Rightarrow & P=\mathrm{Rs}.50000\end{array}$

The sum after the interest is compounded annually is $\begin{array}{rcl}A& =& P{\left(1+\frac{10}{100}\right)}^2\end{array}$.

Determine the amount:

$\begin{array}{rcl}A& =& P{\left(1+\frac{10}{100}\right)}^2\\ & \Rightarrow & A=\mathrm{Rs}.50000\times 1.21\\ \therefore A& =& \mathrm{Rs}.60500\end{array}$

So, the sum is $\mathrm{Rs}.60500$