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    Question

    I. 4x2 – 53x – 105 = 0 II.

    3y2 – 25y + 48 = 0
    A x > y Correct Answer Incorrect Answer
    B x < y Correct Answer Incorrect Answer
    C x = y or the relationship cannot be established Correct Answer Incorrect Answer
    D x ≥ y Correct Answer Incorrect Answer
    E x ≤ y Correct Answer Incorrect Answer

    Solution

    From I: 4x2 – 53x – 105 = 0 Or, 4x2 – 60x + 7x – 105 = 0 Or, 4x × (x – 15) + 7 × (x – 15) = 0 Or, (4x + 7) × (x – 15) = 0 So, x = – (7/4) = – 1.75 or, x = 15 From II: 3y2 – 25y + 48 = 0 Or, 3y2 – 9y – 16y + 48 = 0 Or, 3y × (y – 3) – 16 × (y – 3) = 0 Or, (y – 3) × (3y – 16) = 0 So, y = 3 or y = (16/3) ~ 5.33 So, the relationship cannot be established. Hence, option c.

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