A man invests ₹1200 for two years at compound interest. After one year the money amounts to ₹1275. Find the interest for the second year correct to the nearest rupee.

Let rate of interest be r% per annum.

Given, ₹1200 amounts to ₹1275 after one year.

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

Substituting values we get,

$\Rightarrow 1275 = 1200\Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{1275}{1200} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1275}{1200} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1275 - 1200}{1200} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{75}{1200} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{7500}{1200} \\[1em] \Rightarrow r = \dfrac{25}{4} \\[1em] \Rightarrow r = 6\dfrac{1}{4}\%.$

Principal for second year = ₹1275.

Interest for second year = $\dfrac{P \times R \times T}{100}$.

Substituting value we get,

$\text{Interest } = \dfrac{₹1275 \times \dfrac{25}{4} \times 1}{100} \\[1em] = \dfrac{₹1275 \times 25}{400} \\[1em] = ₹\dfrac{31875}{400} \\[1em] = ₹79.6875 \approx ₹80.$

Hence, the interest for the second year = ₹80.