In what time will ₹1500 yield ₹496.50 as compound interest at 10% per annum 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,

$\Rightarrow 496.50 = 1500\Big[\Big(1 + \dfrac{10}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 496.50 = 1500\Big(\dfrac{110}{100}\Big)^n - 1500 \\[1em] \Rightarrow 496.50 + 1500 = 1500\Big(\dfrac{110}{100}\Big)^n \\[1em] \Rightarrow \dfrac{1996.50}{1500} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \dfrac{199650}{150000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.$

Hence, in 3 years ₹1500 yield ₹496.50 as compound interest at 10% per annum compounded annually.