Mr. Sharma borrowed a certain sum of money at 10% per annum compounded annually. If by paying ₹ 19360 at the end of second year and ₹ 31944 at the end of the third year he clears the debt; find the sum borrowed by him.

Let sum borrowed be ₹ x.

By formula,

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

At the end of 2 years :

$A = x \times \Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{110}{100}\Big)^2 \\[1em] = x \times \Big(\dfrac{11}{10}\Big)^2 \\[1em] = \dfrac{121x}{100}.$

Given,

He pays back ₹ 19360 at the end of second year. So,

Principal for third year = ₹ $\Big(\dfrac{121x}{100} - 19360\Big)$

Amount after 3 year :

$A = \Big(\dfrac{121x}{100} - 19360\Big) \times \Big(1 + \dfrac{10}{100}\Big)^1 \\[1em] = \Big(\dfrac{121x}{100} - 19360\Big) \times \dfrac{110}{100} \\[1em] = \dfrac{1331x}{1000} - 21296.$

Given,

On giving ₹ 31944 at the end of the third year he clears the debt.

$\therefore \dfrac{1331x}{1000} - 21296 = 31944 \\[1em] \Rightarrow \dfrac{1331x}{1000} = 31944 + 21296 \\[1em] \Rightarrow \dfrac{1331x}{1000} = 53240 \\[1em] \Rightarrow x = \dfrac{53240 \times 1000}{1331} \\[1em] \Rightarrow x = ₹ 40000.$

Hence, sum borrowed = ₹ 40000.