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    Question

    Rs. 3,600 is invested in three parts all at simple

    interest of 5% p.a. These three parts are invested randomly for 2 years, 3 years and 6 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. 1,200 Correct Answer Incorrect Answer
    B Rs. 1,800 Correct Answer Incorrect Answer
    C Rs. 1,600 Correct Answer Incorrect Answer
    D Rs. 2,400 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 × 0.05 × 2) = (q × 0.05 × 3) = (r × 0.05 × 6) Or,
    2p = 3q = 6r Let 2p = 3q = 6r = ‘6t’ So,
    p = 3t
    q = 2t
    r = t So,
    p + q + r = 3600 Or,
    3t + 2t + t = 3600 Or,
    6t = 3600 Or,
    t = 600 So, required difference = 3t − t = 2t = 2 × 600 = Rs. 1,200

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