At what rate percent will ₹2000 amount to ₹2315.25 in 3 years at compound interest?

Let rate of interest = r.

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

Given, A = ₹2315.25, P = ₹2000, n = 3.

Putting values in formula we get,

$\Rightarrow 2315.25 = 2000\Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{2315.25}{2000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{231525}{200000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em]$

Taking cube root on both sides we get,

$\Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{20} \\[1em] \Rightarrow r = 5\%.$

Hence, rate of interest = 5%.