Answer:
Option E
Explanation:
I gives : $C.I$ for $3$ years = Rs. $2522$.
II gives : $(C.I)-(S.I)$ for $2$ years at same rate is Rs. $40$.
$\left[\left(1+\frac{R}{100}\right)^{3}-1\right]$ $=2522$ ---(i)
$\left[\left(1+\frac{R}{100}\right)^{2}-1\right]$ $-\frac{P\times R\times 2}{100}$ $=40$ ---(ii)
On dividing (i) by (ii) we get :
$\frac{\left(1+\frac{R}{100}\right)^{3}-1}{\left(1+\frac{R}{100}\right)^{2}-1-\frac{R}{50}}$ $=\frac{2522}{40}$
$\Rightarrow \frac{\frac{R^{3}}{1000000}+\frac{3R}{100}+\frac{3R^{2}}{10000}}{\frac{R{2}}{10000}}$ $=\frac{1261}{20}$
$\Rightarrow \frac{R}{100}$ $+\frac{300}{R}$ $=\frac{1201}{20}$ $\Rightarrow R^{2}$ $-6005R$ $+30000$ $=0$
$\Rightarrow R^{2}-6000R$ $-5R+30000$ $=0$
$\Rightarrow R(R-6000)$ $-5(R-6000)$ $=0$
$\Rightarrow (R-5)(R-6000)$ $=0$
$\Rightarrow R=5$
$\therefore$ Both I and II are needed to get $R$.
$\therefore$ Correct answer is (E).