A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹225 and ₹240. Find :

(i) the rate of interest.

(ii) the original sum.

(iii) the interest earned in the third year.

Given,

Interest for first year = ₹225

Interest for second year = ₹240.

Difference = ₹15.

Here, ₹15 is the interest on ₹225 for 1 year.

(i) We know that,

$\text{Rate} = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \dfrac{15 \times 100}{225 \times 1} \\[1em] = \dfrac{20}{3} \\[1em] = 6\dfrac{2}{3}\%.$

Hence, rate of interest = $6\dfrac{2}{3}\%.$

(ii) We know that,

$\text{P} = \dfrac{S.I. \times 100}{R \times T} \\[1em] = \dfrac{225 \times 100}{\dfrac{20}{3} \times 1} \\[1em] = \dfrac{22500}{\dfrac{20}{3}} \\[1em] = \dfrac{67500}{20} \\[1em] = ₹3375.$

Hence, the sum = ₹3375.

(iii) Here,

Amount after second year = ₹3375 + ₹225 + ₹240 = ₹3840.

Interest earned in third year = $\dfrac{3840 \times \dfrac{20}{3} \times 1}{100} = \dfrac{3840 \times 20 \times 1}{300}$ = ₹256.

Hence, interest earned in third year = ₹256.