CAT 2023 Slot 2 QA Q4 Inequalities & Modulus

Question 4 of the Quantitative Ability (QA) section, CAT 2023 Slot 2, with CAT's official answer and a worked solution.

Any non-zero real numbers \(x, y\) such that \(y \neq 3\) and \(\frac{x}{y} \lt \frac{x+3}{y-3}\), will satisfy the condition

A
If \(y>10\), then \(-x>y\)
B
If \(x \lt 0\), then \(-x \lt y\)
C
If \(y \lt 0\), then \(-x \lt y\)
D
\(\frac{x}{y} \lt \frac{y}{x}\)

Answer: (C) If y 0 , then -x y Inequalities & Modulus MCQ

Solution

\(\frac{x+3}{y-3} - \frac{x}{y} = \frac{y(x+3) - x(y-3)}{y(y-3)} = \frac{3(x+y)}{y(y-3)} > 0\).
If \(y < 0\), then \(y\) and \(y - 3\) are both negative, so \(y(y-3) > 0\), forcing \(x + y > 0\), i.e. \(-x < y\). That is option C.
A is wrong: for \(y > 10\), \(y(y-3)>0\) gives \(x+y>0\), i.e. \(-x<y\), the opposite of A.
B fails when \(0<y<3\): then \(y(y-3)<0\), so \(x+y<0\), giving \(-x>y\).

Shortcut. Reduce the inequality to \(\frac{x+y}{y(y-3)} > 0\) and test each option's case.