At what rate percent per annum compound interest will ₹2304 amount to ₹2500 in 2 years?

Let rate of interest = r.

A = $P\Big(1 + \dfrac{r}{100}\Big)^n$

Given, A = ₹2500, P = ₹2304, n = 2.

Putting values in formula we get,

$\Rightarrow 2500 = 2304\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{2500}{2304} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{50}{48}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{50}{48} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \dfrac{50}{48} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{50 - 48}{48} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{2}{48} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{2}{48} \times 100 \\[1em] \Rightarrow r = \dfrac{200}{48} = 4\dfrac{1}{6}\%.$

Hence, rate of interest = $4\dfrac{1}{6}$%.