CAT 2024 Slot 2 QA Q21 Equations

Question 21 of the Quantitative Ability (QA) section, CAT 2024 Slot 2, with CAT's official answer and a worked solution.

If \(x\) and \(y\) are real numbers such that \(4 x^2+4 y^2-4 x y-6 y+3=0\), then the value of \((4 x+5 y)\) is

TITA: type in the answer. There are no options and no negative marking.

Answer: 7 Equations TITA

Solution

  1. Rewrite: \(4x^2 - 4xy + y^2 + 3y^2 - 6y + 3 = (2x - y)^2 + 3(y-1)^2 = 0\).
  2. A sum of squares is zero only if each is zero: \(y = 1\), \(2x = y \Rightarrow x = \frac12\).
  3. \(4x + 5y = 2 + 5 = 7\).

Same concept in other years

See every question on Rearrange into a sum of squares equal to zero, so each square is zero.

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