Question
Rs. 4,600 is invested in three parts all at simple
interest of 10% p.a. These three parts are invested randomly for 1 year, 3 years and 5 years such that the interest received from the given three parts after desired period of investment is same. Find the difference between the smallest and the largest part of the investment.Solution
Let the amount invested in three parts be Rs. ‘p’, Rs. ‘q’ and Rs. ‘r’ ATQ; (p × 0.10 × 1) = (q × 0.10 × 3) = (r × 0.10 × 5) Or,
p = 3q = 5r Let p = 3q = 5r = ‘15t’ So,
p = 15t
q = 5t
r = 3t So,
p + q + r = 4600 Or,
15t + 5t + 3t = 4600 Or,
23t = 4600 Or,
t = 200 So, required difference = 15t − 3t = 12t = 12 × 200 = Rs. 2,400
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