A sum of money is lent at 8% per annum compound interest. If the interest for the second year exceeds that for the first year by ₹ 96, find the sum of money.

Let sum of money be ₹ x.

For first year :

P = ₹ x

T = 1 year

R = 8%

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 8 \times 1}{100} = \dfrac{2x}{25}$.

A = P + I = $x + \dfrac{2x}{25} = \dfrac{25x + 2x}{25} = \dfrac{27x}{25}$.

For 2nd year :

P = ₹ $\dfrac{27x}{25}$

T = 1 year

R = 8%

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{27x}{25} \times 8 \times 1}{100} = \dfrac{216x}{2500}$.

Given,

Interest for 2nd year exceeds interest for first year by ₹ 96

$\therefore \dfrac{216x}{2500} - \dfrac{2x}{25} = 96 \\[1em] \Rightarrow \dfrac{216x - 200x}{2500} = 96 \\[1em] \Rightarrow \dfrac{16x}{2500} = 96 \\[1em] \Rightarrow x = \dfrac{96}{16} \times 2500 \\[1em] \Rightarrow x = 15000.$

Hence, sum of money = ₹ 15000.