Question

31. A sum of money is invested at a constant rate of compound interest payable annually. The
interests in two successive years are Rs 225 and Rs 240. Find the interest in the third year.​

Answer 1

Answer:

Amount after 1 year = A = P (1 +r/100)^1 = P (1 + r/100)

Interest = P r /100 = 225

=> P r = 22,500 --- equation 1

Amount after 2 years = B = P(1+r/100)^2 = A (1 + r/100)

Interest during 2nd year = B - A = A r /100 = 240

A r = 24 000 -- equation 2

Hence P (1 + r/100) r = 24 000, now substitute Pr from equation 1

(1 + r/100 ) = 24 000 / 22 500

r /100 = 0.0667

r = 6.67 % per annum

P = 22 500 / r = Rs 3, 375

amount after 1 year = Rs 3 600

amount after 2 nd year = Rs 3, 840