CAT 2023

Slot 2

QA

Question & Solution

It is given that $\frac{x}{y}<\ \frac{\ x+3}{y-3}$, which can be written as $\frac{x}{y}-\frac{\ x+3}{y-3}<0$

=> $\ \frac{\ x\left(y-3\right)-y\left(x+3\right)}{y\left(y-3\right)}<0$

=> $\ \frac{\ xy-3x-xy-3y}{y\left(y-3\right)}<0$

=> $\ \frac{\ -3\left(x+y\right)}{y\left(y-3\right)}<0$

=> $\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$

From this inequality, we can say that, when $y<0=>y(y-3)>0$. Now to satisfy the given equation $\ \frac{\ 3\left(x+y\right)}{y\left(y-3\right)}>0$, 

$(x+y)$ must be greater than zero Hence, $x>0$ and $\left|x\right|\ >\left|y\right|$

Therefore, the magnitude of $x$ is greater than the magnitude of $y$. 

Hence, $x>y$, and $\left|x\right|\ >\ \left|y\right|$ =>$-x\ <\ y$ (Since the magnitude of $x$ is greater than the magnitude of $y$.)

The correct option is B.