Mohit borrowed a certain sum at 5% per annum compound interest and cleared this loan by paying ₹ 12600 at the end of the first year and ₹ 17640 at the end of the second year. Find the sum borrowed.

Let Mohit borrowed ₹ x and ₹ y.

By formula,

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

For ₹ 12600,

P = ₹ x

A = ₹ 12600

r = 5%

n = 1 year

Substituting values in formula we get :

$\Rightarrow 12600 = x \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] \Rightarrow 12600 = x \times \dfrac{105}{100} \\[1em] \Rightarrow x = \dfrac{12600 \times 100}{105} \\[1em] \Rightarrow x = ₹ 12000.$

For ₹ 17640,

P = ₹ y

A = ₹ 17640

r = 5%

n = 2 year

Substituting values in formula we get :

$\Rightarrow 17640 = y \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] \Rightarrow 17640 = y \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow y = \dfrac{17640 \times 20^2}{21^2} \\[1em] \Rightarrow y = ₹ 16000.$

Total money borrowed = ₹ (x + y) = ₹ (12000 + 16000) = ₹ 28000.

Hence, sum borrowed = ₹ 28000.