At what rate per cent per annum will ₹ 6000 amount to ₹ 6615 in 2 years when interest is compounded annually ?

Given,

P = ₹ 6000

A = ₹ 6615

n = 2 years

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 6615 = 6000 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{6615}{6000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{441}{400} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{20} = 5\%.$

Hence, rate of interest = 5%.