A sum of money was invested for 3 years, interest being compounded annually. The rates for successive years were 10%, 15% and 18% respectively. If the compound interest for the second year amounted to ₹ 4950, find the sum invested.

Let sum invested be ₹ x.

Amount after 1st year :

$A = x \times \Big(1 + \dfrac{10}{100}\Big) \\[1em] = x \times \dfrac{110}{100} \\[1em] = \dfrac{11x}{10}.$

Amount after 2nd year :

$A = x \times \Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{15}{100}\Big) \\[1em] = x \times \dfrac{110}{100} \times \dfrac{115}{100} \\[1em] = \dfrac{253x}{200}.$

C.I. for 2nd year = Amount after 2 years - Amount after 1 year = $\dfrac{253x}{200} - \dfrac{11x}{10}$

Given,

Compound interest for the second year amounted to ₹ 4950.

$\therefore \dfrac{253x}{200} - \dfrac{11x}{10} = 4950 \\[1em] \Rightarrow \dfrac{253x - 220x}{200} = 4950 \\[1em] \Rightarrow \dfrac{33x}{200} = 4950 \\[1em] \Rightarrow x = \dfrac{4950 \times 200}{33} \\[1em] \Rightarrow x = ₹ 30000.$

Hence, sum invested = ₹ 30000.