Find the time (in years) in which ₹12500 will produce ₹3246.40 as compound interest at 8% per annum, interest compounded annually.

Let the time be n years.

$C.I. = P \Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big].$

Substituting values we get,

$\Rightarrow 3246.40 = 12500\Big[\Big(1 + \dfrac{8}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 3246.40 = 12500\Big(\dfrac{108}{100}\Big)^n - 12500 \\[1em] \Rightarrow 3246.40 + 12500 = 12500\Big(\dfrac{108}{100}\Big)^n \\[1em] \Rightarrow \dfrac{15746.40}{12500} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \dfrac{1574640}{1250000} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \dfrac{19683}{15625} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^3 = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.$

Hence, in 3 years ₹12500 yield ₹3246.40 as compound interest at 8% per annum compounded annually.