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      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.
      A Rs. 2,400 Correct Answer Incorrect Answer
      B Rs. 2,000 Correct Answer Incorrect Answer
      C Rs. 2,200 Correct Answer Incorrect Answer
      D Rs. 1,800 Correct Answer Incorrect Answer
      E None of these Correct Answer Incorrect Answer

      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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