Answer:
Option E
Explanation:
I. $P\left(1+\frac{R}{100}\right)^{4}$ $=2P$ $\Rightarrow \left(1+\frac{R}{100}\right)^{4}$ $=2$ ---(i)
II. $P\left(1+\frac{R}{100}\right)^{12}$ $=8P$ $\Rightarrow \left(1+\frac{R}{100}\right)^{12}$ $=8$ ---(ii)
III. $P\left(1+\frac{R}{100}\right)^{8}$ $=4P$ $\Rightarrow \left(1+\frac{R}{100}\right)^{8}$ $=4$ ---(iii)
Let the given sum become $16$ times in $n$ years. Then,
$P\left(1+\frac{R}{100}\right)^{n}$ $=16P$
$\Rightarrow \left(1+\frac{R}{100}\right)^{n}$ $=16$ ---(iv)
$\therefore$ Any one of (i), (ii) and (iii) with (iv) will give the value of $n$.
$\therefore$ Correct answer is (E).