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    Question

    I. 2x² - 11x + 12 = 0 II. 12y² + 29y +

    15 = 0 In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer.
    A if x>y Correct Answer Incorrect Answer
    B if x≥y Correct Answer Incorrect Answer
    C if x Correct Answer Incorrect Answer
    D if x ≤y Correct Answer Incorrect Answer
    E if x=y or the relationship cannot be established Correct Answer Incorrect Answer

    Solution

    I. 2x² - 11x + 12 = 0 2x² - 8x - 3 x + 12 = 0 2 x(x – 4) – 3(x – 4) = 0 (2 x – 3) (x – 4) = 0 ∴ x = 3/2, 4 II. 12y² + 29y + 15 = 0 12y² + 20y + 9 y + 15 = 0 4 y(3 y + 5) + 3(4 y + 5) = 0 (3 y + 5) (4 y + 3) = 0 ∴ y = -3/4, -5/3 Hence, x>y

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