Question

Find the annual rate of compound interest at which $\text{Rs.}8000$ will become $\text{Rs.}10648$ after $1\text{years}$.

Answer 1

Let the rate of interest be,

$\begin{array}{cc}A& =P{\left(1+\frac{r}{100}\right)}^t\\ 10648& =8000{\left(1+\frac{r}{100}\right)}^1\\ \frac{10648}{8000}& =1+\frac{r}{100}\\ \frac{10648-8000}{8000}& =\frac{r}{100}\\ \frac{r}{100}& =0.331\\ r& =33.1\%\end{array}$

Hence, the rate of interest is equal to $33.1\%$.