Let x, y , and z be real numbers satisfying aligned & 4(x^2+y^2+z^2)=a & 4(x-y-z)=3+a aligned Then a equals

CAT 2024 Slot 1 QA Q8 Equations

Question 8 of the Quantitative Ability (QA) section, CAT 2024 Slot 1, with CAT's official answer and a worked solution.

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

Answer: (A) 3 Equations MCQ

Subtract the second equation from the first: $4(x^2 + y^2 + z^2) - 4(x - y - z) = -3$.
Group terms: $4(x^2 - x) + 4(y^2 + y) + 4(z^2 + z) + 3 = 0$.
Complete squares: $4\left(x - \tfrac12\right)^2 + 4\left(y + \tfrac12\right)^2 + 4\left(z + \tfrac12\right)^2 - 3 + 3 = 0$.
A sum of squares is zero only if each is zero: $x = \tfrac12,\ y = -\tfrac12,\ z = -\tfrac12$.
Then $a = 4\left(\tfrac14 + \tfrac14 + \tfrac14\right) = 3$. Check: $4(x - y - z) = 4 \cdot \tfrac32 = 6 = 3 + a$ ✓.

Shortcut. Plug the options into both equations: only $a = 3$ makes the system consistent.