The compound interest, calculated yearly, on a certain sum of money for the second year is ₹ 1089 and for the third year it is ₹ 1197.90. Calculate the rate of interest and the sum of money.

Difference between C.I. of two successive years = ₹ 1197.90 - ₹ 1089 = ₹ 108.9

∴ ₹ 108.9 is the interest of one year on ₹ 1089.

By formula,

Rate of interest = $\dfrac{100 \times I}{P \times T} = \dfrac{100 \times 108.9}{1089 \times 1}$ = 10%.

Let sum of money be ₹ x.

For first year :

P = ₹ x

R = 10%

T = 1 year

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

A = P + I = $x + \dfrac{x}{10} = \dfrac{11x}{10}$

For second year :

P = ₹ $\dfrac{11x}{10}$

R = 10%

T = 1 year

I = $\dfrac{P \times R \times T}{100} = \dfrac{\dfrac{11x}{10} \times 10 \times 1}{100} = \dfrac{11x}{100}$.

Given,

C.I. for 2nd year = ₹ 1089

$\therefore \dfrac{11x}{100} = 1089 \\[1em] \Rightarrow x = \dfrac{1089 \times 100}{11} = 9900.$

Hence, rate of interest = 10% and sum of money = ₹ 9900.