New SIDBI Grade A New Batch Starts on Nov 18

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    Question

    In each question two equations numbered (I) and (II) are given. You should solve both the equations and mark appropriate answer.

    I. 3x2 – 16x + 21 = 0

    II. y2 – 13y + 42 = 0

    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 relation between x and y cannot be established Correct Answer Incorrect Answer

    Solution

    I. 3x2 – 16x + 21 = 0 => 3x2 – 9x - 7x + 21 = 0 => 3x (x – 3) – 7 (x – 3) = 0 => x = 3, 7/3 II. y2 – 13y + 42 = 0 => y2 – 6y – 7y + 42 = 0 => y (y – 6) – 7(y – 6) = 0 => y = 7, 6 Hence, x < y Alternate Method: if signs of quadratic equation is - ve and +ve respectively then the roots of equation will be +ve and +ve. So, roots of first equation = x = 7/3, 3 So, roots of second equation = y = 7, 6 After comparing we can conclude that x < y.

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