A certain sum of money amounts to ₹ 7,260 in 2 years and to ₹ 7,986 in 3 years, interest being compounded annually. Find the rate per cent per annum.

Let original sum of money invested be ₹ x and rate of percent be r%.

By formula,

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

Given,

The sum of money, invested at compound interest, amounts to ₹ 7,260 in 2 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 7260 = x \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 7260 = x\Big(1 + \dfrac{r}{100}\Big)^2 ……(1)$

The sum of money, invested at compound interest, amounts to ₹ 7,986 in 3 years.

$\Rightarrow A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] \Rightarrow 7986 = x \times \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow 7986 = x\Big(1 + \dfrac{r}{100}\Big)^3 ……(2)$

Dividing equation (2) by (1), we get :

$\Rightarrow \dfrac{7986}{7260} = \dfrac{x\Big(1 + \dfrac{r}{100}\Big)^3}{x\Big(1 + \dfrac{r}{100}\Big)^2} \\[1em] \Rightarrow \dfrac{1331}{1210} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \Big(1 + \dfrac{r}{100}\Big) = \dfrac{1331}{1210} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1331}{1210} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1331 - 1210}{1210} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{121}{1210} \\[1em] \Rightarrow r = \dfrac{100 \times 121}{1210} = 10\%.$

Hence, rate percent = 10% p.a.