Any non-zero real numbers x, y such that y ≠ 3 and \; x/y

Correct Answer is : If y<0y < 0y<0, and −x<y-x < y−x<y

Correct Answer is : If y<0y < 0y<0, and −x<y-x < y−x<y

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CAT 2023 Slot 2 Solved Paper

Question : 46 of 66

Marks: +1, -0

x, y

y \neq 3

\; \frac{x}{y} < \; \frac{x+3}{y-3}

, Will satisfy the condition.

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

\; \Rightarrow \; \frac{x(y-3)-y(x+3)}{y(y-3)} < 0

\; \Rightarrow \; \frac{x y-3 x-x y-3 y}{y(y-3)} < 0

\; \Rightarrow \; \frac{-3(x+y)}{y(y-3)} < 0

\; \Rightarrow \; \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 B.

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