A sum of money placed at compound interest doubles itself in 4 years. In how many years will it amount to 8 times?

Correct Answer: 12 years

$$\eqalign{ & {\text{Let}}, \cr & {\text{Principal}} = Rs.\,100 \cr & {\text{Amount}} = Rs.\,200 \cr & {\text{Rate}} = r\% \cr & {\text{Time}} = 4\,{\text{years}} \cr & {\text{Now}}, \cr & A = P \times {\left^n} \cr & 200 = 100 \times {\left^4} \cr & 2 = {\left^4} - - - - - - \left( i \right) \cr & {\text{If}}\,{\text{sum}}\,{\text{become}}\,{\text{8}}\,{\text{times}}\,{\text{in}}\,{\text{the}}\,{\text{time}}\,n\,{\text{years}} \cr & {\text{then,}} \cr & 8 = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {2^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} - - - - - - \left( {ii} \right) \cr & {\text{Using}}\,{\text{eqn}}\,\left( i \right)in\left( {ii} \right),\,{\text{we}}\,{\text{get}} \cr & {\left( {{{\left}^4}} \right)^3} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\left^{12}} = {\left( {1 + \left( {\frac{r}{{100}}} \right)} \right)^n} \cr & {\text{Thus}},\,n = 12\,{\text{years}}. \cr} $$