On a certain sum of money, the difference between the compound interest for a year, payable half-yearly, and the simple interest for a year is ₹ 180. Find the sum lent out, if the rate of interest in both cases is 10% per annum.

Let sum of money be ₹ x.

For S.I. :

P = ₹ x

R = 10%

T = 1 years

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 10 \times 1}{100} = \dfrac{x}{10}$.

For C.I. :

For first half-year :

P = ₹ x

R = 10%

T = $\dfrac{1}{2}$ year

I = $\dfrac{P \times R \times T}{100} = \dfrac{x \times 10 \times \dfrac{1}{2}}{100} = \dfrac{x}{20}$.

Amount = P + I = $x + \dfrac{x}{20} = \dfrac{21x}{20}$.

For second year :

P = ₹ $\dfrac{21x}{20}$

R = 10%

T = $\dfrac{1}{2}$ year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{21x}{20} \times 10 \times \dfrac{1}{2}}{100} = \dfrac{21x}{400}$.

Amount = P + I = $\dfrac{21x}{20} + \dfrac{21x}{400} = \dfrac{420x + 21x}{400} = \dfrac{441x}{400}$.

C.I. = Final amount - Initial principal

= $\dfrac{441x}{400} - x = \dfrac{41x}{400}$.

Given,

Difference between C.I. and S.I. = ₹ 180

$\therefore \dfrac{41x}{400} - \dfrac{x}{10} = 180 \\[1em] \Rightarrow \dfrac{41x - 40x}{400} = 180 \\[1em] \Rightarrow \dfrac{x}{400} = 180 \\[1em] \Rightarrow x = 72000.$

Hence, sum lent out = ₹ 72000.