Question 4.

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

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

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

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

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

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

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

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

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

The correct option is C.