CAT 2024 Question Paper | Quant Slot 1

CAT Questions

CAT 2024 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 8 :Let $x, y$, and $z$ be real numbers satisfying $$\begin{aligned} & 4\left(x^2+y^2+z^2\right)=a \\ & 4(x-y-z)=3+a \end{aligned}$$ Then $a$ equals

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Video Explanation

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Explanatory Answer

Let $x, y$ and $z$ be real numbers satisfying $$\begin{array}{l} 4\left(x^2+y^2+z^2\right)=a \\ 4(x-y-z)=3+a \end{array}$$ $$\begin{array}{l} x^2+y^2+z^2=a / 4 \\ x-y-z=(3+a) / 4 \end{array}$$ Let $y 1=- y$ and $z 1=- z$ $$x+y 1+z 1=(3+a) / 4$$ We know that $x^2+y^2+z^2 \geq x y+y z+z x$ ( Identity) $$\begin{array}{l} (x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x \\ (x+y+z)^2 \leq 3\left(x^2+y^2+z^2\right) \end{array}$$ Substituting the values, $$\begin{array}{l} \{(3+a) / 4\}^2 \leq 3(a / 4) \\ (3+a)(3+a) \leq 4^* 3 a \\ 9+a^2+6 a \leq 12 a \\ 9+a^2-6 a \leq 0 \\ (a-3)^2 \leq 0 \end{array}$$ The value of a which satisfies the above equation is 3 .

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The question is "Let $x, y$, and $z$ be real numbers satisfying $$\begin{aligned} & 4\left(x^2+y^2+z^2\right)=a \\ & 4(x-y-z)=3+a \end{aligned}$$ Then $a$ equals"

Hence, the answer is '3'

Choice A is the correct answer.

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