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ExamsIBPS POQuantitative Aptitude

In the given question, two equations numbered I and II are given. Solve both equations and mark the appropriate answer. I. $6x^2 + 33x + 42 = 0$ II. $7y^2 + 15y + 8 = 0$

  1. x > y
  2. x < y
  3. x ≥ y
  4. x ≤ y

Correct answer: x < y

Solution

Factorizing gives $6x^2+33x+42=3(2x+7)(x+2)=0$, so $x=-\tfrac{7}{2}$ or $-2$. Also, $7y^2+15y+8=(7y+8)(y+1)=0$, so $y=-\tfrac{8}{7}$ or $-1$. In all possible cases, $x<y$.

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