CAT 2024 Slot 3 · Quant · Question 2

Two linear equations in two unknowns have no solution exactly when the lines are parallel but not identical.

Writing the equations as px-4y=2 and 3x+ky=a, the coefficient test for no solution is \frac{p}{3}=\frac{-4}{k}\ne \frac{2}{a}.

The first equality gives pk=-12, that is kp+12=0.

Note that this immediately rules out the fourth option, which asserts the opposite.

The second part is the inequality \frac{-4}{k}\ne \frac{2}{a}.

Cross-multiplying, -4a\ne 2k, so 2k+4a\ne 0.

Dividing by 2, k+2a\ne 0, that is 2a+k\ne 0.

This is a genuine requirement for no solution, so it is a necessary condition — the first option.

The second and third options involve a and p only, and neither is forced by the parallel condition, so they can be discarded.

Shortcut. The question asks for a necessary condition, not the full criterion. Since the full criterion is kp+12=0 together with 2a+k\ne 0, any option matching either half qualifies — and option 4 states the negation of the first half, leaving option 1 as the only survivor.