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    Question

    I. 20y² - 13y + 2 = 0 II. 6x² - 25x +

    14 = 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 < y 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. 20y² - 13y + 2 = 0 20y² - 8y - 5y + 2 = 0 4y (5y – 2) – 1(5y – 2) = 0 (4y – 1) (5y – 2) = 0 y =2/5, 1/4 II. 6x² - 25x + 14 = 0 6x² - 21x - 4x + 14 = 0 3x (2x – 7) - 2 (2x – 7) = 0 (3x – 2) (2x – 7) = 0 x = 7/2 , 2/3 Hence, x>y

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