A sum of money doubles itself in $3$ years at CI, when the interest is compounded annually. In how many years will it amount to $16$ times of itself?

The correct option is C. $12$ years

Let the sum is Rs. $P$

$2P=P{(1+\frac{R}{100})}^3$

$2={(1+\frac{R}{100})}^3$

$\Rightarrow 1+\frac{R}{100}=(2{)}^{\frac{1}{3}}$ .........(1)

Suppose the money becomes $16$ times itself in $n$ years.

$16P=P{(1+\frac{R}{100})}^n$

$16={(1+\frac{R}{100})}^n$

$4^2={(1+\frac{R}{100})}^n$

$(2^2{)}^2={(1+\frac{R}{100})}^n$

$\Rightarrow 1+\frac{R}{100}=(2^4{)}^{\frac{1}{n}}$ ........(2)

From (1) and (2), we have

$(2{)}^{\frac{1}{3}}=(2{)}^{\frac{4}{n}}$

$\Rightarrow \frac{1}{3}=\frac{4}{n}$

$\Rightarrow n=12$ years