A person invests ₹ 5000 for three years at a certain rate of interest compounded annually. At the end of two years this sum amounts to ₹ 6272. Calculate :

(i) the rate of interest per annum.

(ii) the amount at the end of the third year.

(i) For 2 years :

A = ₹ 6272

P = ₹ 5000

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$\Rightarrow 6272 = 5000 \times \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{6272}{5000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{784}{625} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{28}{25}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{28}{25} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{28}{25} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{28 - 25}{25} \\[1em] \Rightarrow r = \dfrac{3}{25} \times 100 \\[1em] \Rightarrow r = 12\%.$

Hence, rate of interest per annum = 12%.

(ii) By formula,

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

Substituting values we get :

$\Rightarrow A = 5000 \times \Big(1 + \dfrac{12}{100}\Big)^3 \\[1em] = 5000 \times \Big(\dfrac{112}{100}\Big)^3 \\[1em] = 5000 \times \Big(\dfrac{28}{25}\Big)^3 \\[1em] = 5000 \times \dfrac{21952}{15625} \\[1em] = \dfrac{109760000}{15625} \\[1em] = ₹7024.64$

Hence, amount in 3 years = ₹ 7024.64