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CAT 2024 Slot 3 — QA Question 10

2 Variable EquationsEasy

For some constant real numbers p, k and a, consider the following system of linear equations in x and y:

px - 4y = 2

3x + ky = a

A necessary condition for the system to have no solution for (x, y), is

Answer & solution

  • 2a + k ≠ 0

  • B

    ap + 6 = 0

  • C

    kp + 12 ≠ 0

  • D

    ap - 6 = 0

Solution

Medium

A 2×22\times 2 linear system has no solution when the two lines are parallel but distinct: the coefficient determinant is zero (parallel), while the constants break the consistency (distinct). Translate both conditions.

1

Parallel condition (determinant =0=0). For px−4y=2px-4y=2 and 3x+ky=a3x+ky=a, the coefficients are proportional when the determinant vanishes.

∣p−43k∣=0⇒ pk−(−4)(3)=0(expand)⇒ pk+12=0\begin{aligned} &\begin{vmatrix}p&-4\\3&k\end{vmatrix}=0\\ &\Rightarrow\ pk-(-4)(3)=0\quad\text{(expand)}\\ &\Rightarrow\ pk+12=0 \end{aligned}
2

Inconsistency condition. No solution requires the constant ratio to differ, i.e. p3=−4k≠2a\dfrac{p}{3}=\dfrac{-4}{k}\ne\dfrac{2}{a}. The inequality part is the extra requirement beyond parallelism.

−4k≠2a⇒ −4a≠2k(cross-multiply)⇒ 2a+k≠0(divide by −2)\begin{aligned} &\frac{-4}{k}\ne\frac{2}{a}\\ &\Rightarrow\ -4a\ne 2k\quad\text{(cross-multiply)}\\ &\Rightarrow\ 2a+k\ne 0\quad\text{(divide by }-2\text{)} \end{aligned}
3

Necessary condition. Among the options, the requirement that must hold for "no solution" (so the lines are not the same line) is the inequality on the constants.

2a+k≠0\begin{aligned} &2a+k\ne 0 \end{aligned}
2a+k≠02a+k\ne 0
Need a hint?

"No solution" = parallel and distinct. Parallel fixes pk+12=0pk+12=0; distinctness forbids the constants from matching, which rearranges to 2a+k≠02a+k\ne 0.

CAT 2024 Slot 3 QA Q10: For some constant real numbers p, k and a, consider the following system of linear equations in x and y: px - — Solution | TheCATExam