The compound interest accrued on an amount of Rs. 22,000 at the end of two years is Rs. 5,596.8. What would be the simple interest accrued on the same amount at the same rate in the same period ?

C.I. = $$Rs. 5,596.8$$

=> $$P [(1 + \frac{R}{100})^T - 1] = 5596.8$$

=> $$22,000 [(1 + \frac{R}{100})^2 - 1] = 5596.8$$

=> $$[(1 + \frac{R}{100})^2 - 1] = \frac{5596.8}{22000} = 0.2544$$

=> $$(1 + \frac{R}{100})^2 = 0.2544 + 1 = 1.2544$$

=> $$1 + \frac{R}{100} = \sqrt{1.2544} = 1.12$$

=> $$\frac{R}{100} = 1.12 - 1 = 0.12$$

=> $$R = 0.12 \times 100 = 12 \%$$

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

= $$\frac{22000 \times 12 \times 2}{100}$$

= $$220 \times 24 = Rs. 5,280$$