### A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?

A. 9 years B. 8 years C. 27 years D. 12 years Answer: Option D
$\begin{array}{l}\text{Let,}\\ \text{P}=\text{Rs.}\ 100\\ \text{A}=\text{Rs.}\ 200\\ \text{Rate}=r\%\\ \text{Time}\ \left(t\right)=4\ \text{years}\\ \text{Now,}\\ \begin{array}{l}A=P\times\left[1+\left(\frac{r}{100}\right)\right]^n\\ 200=100\times\left[1+\left(\frac{r}{100}\right)\right]^4\\ 2=\left[1+\left(\frac{r}{100}\right)\right]^4——(i)\\ \end{array}\end{array}$ If sum become 8 times in time 8 years. Then, $\begin{array}{l}8=\left(1+\left(\frac{r}{100}\right)\right)^n\\ 2^3=\left(1+\left(\frac{r}{100}\right)\right)^n——(ii)\\ \text{Using equetion (i) put in (ii)}\\ \text{we get;}\\ \begin{array}{l}\left(\left[1+\left(\frac{r}{100}\right)\right]^4\right)^3=\left(1+\left(\frac{r}{100}\right)\right)^n\\ \left[1+\left(\frac{r}{100}\right)\right]^{12}=\left(1+\left(\frac{r}{100}\right)\right)^n\\ \text{Thus, n}=12\ \text{years.}\end{array}\end{array}$