CAT 2023 Slot 2 — QA Question 3

Bring to one side and combine over a common denominator:

$$\begin{aligned} &\frac{x}{y}-\frac{x+3}{y-3}\lt 0\\ &\Rightarrow\ \frac{x(y-3)-y(x+3)}{y(y-3)}\lt 0 \end{aligned}$$

Simplify the numerator — the $xy$ terms cancel:

$$\begin{aligned} &x(y-3)-y(x+3)=xy-3x-xy-3y=-3(x+y)\\ &\Rightarrow\ \frac{-3(x+y)}{y(y-3)}\lt 0\\ &\Rightarrow\ \frac{x+y}{y(y-3)}\gt 0 \quad\text{(divide by }-3\text{, flip sign)} \end{aligned}$$

Test the correct option, $y\lt 0$: then $y-3\lt 0$, so the product $y(y-3)\gt 0$. For the fraction to stay positive the numerator must be positive:

$$\begin{aligned} &y(y-3)\gt 0\ \Rightarrow\ x+y\gt 0\\ &\Rightarrow\ -x\lt y \end{aligned}$$

This is exactly what option (d) claims, so the condition is consistent.

Why not (a)? If $y\gt 10$ then again $y(y-3)\gt 0\Rightarrow x+y\gt 0\Rightarrow -x\lt y$, which contradicts "$-x\gt y$". Options (b) and (c) need not hold for all valid $x,y$.