CAT 2024 Slot 3 QA Q2 Equations

Question 2 of the Quantitative Ability (QA) section, CAT 2024 Slot 3, with CAT's official answer and a worked solution.

For some constant real numbers \(p, k\) and \(a\), consider the following system of linear equations in \(x\) and \(y\) : $$ \begin{aligned} & p x-4 y=2 \\ & 3 x+k y=a \end{aligned} $$ A necessary condition for the system to have no solution for \((x, y)\), is

A
\(2 a+k \neq 0\)
B
\(a p-6=0\)
C
\(a p+6=0\)
D
\(k p+12 \neq 0\)

Answer: (A) 2 a+k ≠ 0 Equations MCQ

Solution

No solution needs \(\frac{p}{3} = \frac{-4}{k} \ne \frac{2}{a}\).
The second part, \(\frac{-4}{k} \ne \frac{2}{a}\), gives \(-4a \ne 2k\), i.e. \(2a + k \ne 0\).
So \(2a + k \ne 0\) is necessary (A). D is wrong since \(kp + 12 = 0\) is required.