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    Question

    A person named 'P' invested Rs.

    48,000 in an SIP called 'X', which provides compound interest at a rate of 50% per annum, compounded half-yearly, for a duration of 12 months. Afterward, he reinvested the interest earned from SIP 'X' into another SIP, 'Y', which offers compound interest at an annual rate of 'a%', compounded annually, for 3 years. By following this investment strategy, he accumulated a total interest of Rs. 46,656 over 4 years. The expression (1.3a - 2) represents the root of which of the following equations? I) b² - 41b + 378 = 0 II) b² - 51b + 518 = 0 III) b² - 61b + 888 = 0 IV) b² - 71b + 1258 = 0
    A Only I and II Correct Answer Incorrect Answer
    B Only I, II and III Correct Answer Incorrect Answer
    C Only I, III and IV Correct Answer Incorrect Answer
    D Only III Correct Answer Incorrect Answer
    E All I, II, III and IV Correct Answer Incorrect Answer

    Solution

    ATQ, Interest from scheme P = 48000 Ă— (1.252- 1) = 48000 Ă— 0.5625 = 27000 Rs. Interest from scheme Q = 46656 - 27000 = 19656 Rs. 27000 Ă— {(1 + a/100)3- 1} = 19656 (1 + a/100)3- 1 = 0.728 (1 + a/100)3 = 1.728 = 1.23 1 + a/100 = 1.2 a = 20% 1 root of equation = 1.3a - 2 = 1.3 Ă— 20 - 2 = 26 - 2 = 24 Option I, Second root = 41 - 24 = 17, Product of Roots = 24 Ă— 17 = 408 (Wrong) Option II, Second root = 51 - 24 = 27, Product of Roots = 24 Ă— 27 = 648 (Wrong) Option III, Second root = 61 - 24 = 37, Product of Roots = 24 Ă— 37 = 888 (Right) Option IV, Second root = 71 - 24 = 47, Product of Roots = 24 Ă— 47 = 1128 (Wrong) Hence, the equation will be b2 - 61b + 888 = 0

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