Let principal sum = Rs. $$P$$ and rate of interest = $$r\%$$ Amount under compound interest = $$P(1+frac{r}{100})^T$$ Thus, after 4 years => $$P(1+frac{r}{100})^4=4P$$ => $$(1+frac{r}{100})^4=4$$ => $$(1+frac{r}{100})=(4)^{frac{1}{4}}$$ -------------(i) Now, Let after $$t$$ years sum becomes 64 times => $$P(1+frac{r}{100})^t=64P$$ => $$(4)^{frac{t}{4}}=(4)^3$$ Comparing the exponents, we get : => $$frac{t}{4}=3$$ => $$t=4 imes3=12$$ years => Ans - (B)