CAT 2023 Slot 2 · Quant · Question 4

Never cross-multiply when the denominators can change sign — move everything to one side instead.

The condition becomes \frac{x+3}{y-3}-\frac{x}{y}>0.

Over the common denominator y\left( y-3 \right), the numerator is y\left( x+3 \right)-x\left( y-3 \right).

Expanding, xy+3y-xy+3x=3\left( x+y \right).

So the condition is exactly \frac{3\left( x+y \right)}{y\left( y-3 \right)}>0, that is \frac{x+y}{y\left( y-3 \right)}>0.

Now test the option that fixes the sign of y: suppose y<0.

Then y is negative and y-3 is also negative, so their product y\left( y-3 \right) is positive.

With a positive denominator, the fraction is positive only if x+y>0.

Rewriting x+y>0 as x>-y, and multiplying both sides by -1 to flip the inequality, -x<y.

So whenever y<0, the condition forces -x<y — the third option.

The others fail: with y=20 and x=1 the inequality holds but -x>y is false, and with x=-1, y=2 the inequality holds while \frac{x}{y}<\frac{y}{x} does not.

Shortcut. Everything reduces to the single sign statement \frac{x+y}{y\left( y-3 \right)}>0. Fixing y<0 makes the denominator positive at once, so only x+y>0 survives — and that is the third option, rewritten.