A) 30
B) 45
C) 50
D) 60
Correct Answer: B
$ \Rightarrow $ $ 2x=x{{( 1+\frac{r}{100} )}^{15}} $
$ \Rightarrow $ $ 2={{( 1+\frac{r}{100} )}^{15}} $ (i) Let after n yr sum becomes eight times of itself. Again, $ 8x=x{{( 1+\frac{r}{100} )}^{n}} $ … (ii)
$ \Rightarrow $ $ 8={{( 1+\frac{r}{100} )}^{n}} $
$ \Rightarrow $ $ {{(2)}^{3}}={{( 1+\frac{r}{100} )}^{n}} $
$ \Rightarrow $ $ {{( 1+\frac{r}{100} )}^{3\times 15}}={{( 1+\frac{r}{100} )}^{n}} $
$ \therefore $ $ n=45,yr $