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      Question

      I. 3x² - 22x + 40 = 0 II. 2y² - 19y +

      44 = 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. 3x² - 22x + 40 = 0 Pairs are: -12, -10 So by changing its signs & dividing by a i.e.3, we will get x = 4, 10/3 II. 2y² - 19y + 44 = 0 Pairs are: -11, -8 So by changing its signs & dividing by a i.e.2, we will get y = 11/2, 4 Hence, x ≤ y.

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