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ExamsIBPS POQuantitative Aptitude

In each question, statements I and II are given. Solve both equations and mark the appropriate answer. I. $4x^2 - 17x + 116 = 3x^2 + 4x + \sqrt{64}$ II. $(y + 14)(y - 14) = 60$ What is the correct relation between $x$ and $y$?

  1. If $x=y$ or no relation can be established
  2. If $x>y$
  3. If $x<y$
  4. If $x\ge y$

Correct answer: If $x=y$ or no relation can be established

Solution

From I, $4x^2 - 17x + 116 = 3x^2 + 4x + 8$ gives $x^2 - 21x + 108 = 0$, so $x=9$ or $12$. From II, $y^2 - 196 = 60$, so $y^2=256$ and $y=\pm16$. Since the possible values do not establish a definite relation, the answer is that no relation can be established.

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