The difference between the compound interest and the simple interest accrued on an amount of ₹ 18,000 in 2 years is ₹ 405. Find the rate of interest per annum.

Let the rate of interest per year = R %

P = ₹ 18,000, R = R %, T = 2 years

S.I. = $\dfrac{P \times R \times T}{100}$

S.I. in 2 years = $\dfrac{18,000 \times R \times 2}{100}$

= ₹ 360R

And, C.I. in 2 years = A - P

= 18,000 $\Big[1 + \dfrac{R}{100}\Big]^2$ - 18,000

Given, C.I. - S.I. = ₹ 405

⇒ 18,000 $\Big[1 + \dfrac{R}{100}\Big]^2$ - 18,000 - 360R = 405

⇒ 18,000 $\Big[1 + \dfrac{R^2}{10000} + \dfrac{2R}{100}\Big]$ - 18,000 - 360R = 405

⇒ $\Big[18,000 + \dfrac{18,000R^2}{10000} + \dfrac{36,000R}{100}\Big]$ - 18,000 - 360R = 405

⇒ $\dfrac{18R^2}{10}$ + 360R - 360R = 405

⇒ $\dfrac{18R^2}{10}$ = 405

⇒ $R^2 = \dfrac{10 \times 405}{18}$

⇒ $R^2 = \dfrac{4050}{18}$

⇒ $R^2 = 225$

⇒ $R = \sqrt{225}$

⇒ R = 15 %

Hence, the rate of interest = 15 %.