CAT 2023 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like **Speed-time-distance**, **Mixture and Alligations**. This year, there was a surprise. The questions from **Geometry** were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 4 : Any non-zero real numbers \(x, y\) such that \(y \neq 3\) and \(\frac{x}{y} \lt \frac{x+3}{y-3}\), will satisfy the condition

- If \(y>10\), then \(-x>y\)
- If \(x \lt 0\), then \(-x \lt y\)
- If \(y \lt 0\), then \(-x \lt y\)
- \(\frac{x}{y} \lt \frac{y}{x}\)

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Case I)

If both x and y are both positive.

Then, \( \frac { x } { y } < \frac { x + 3 } { y - 3 } \) is always true, since We are increasing the numerator and decreasing the denominator.

Case II)

If both x and y are both negative.

\( \frac { x } { y } = \frac { | x | } { | y | } \)

\( \frac { x + 3 } { y - 3 } = \frac { - ( | x | - 3 ) } { - ( | y | + 3 ) } = \frac { | x | - 3 } { | y | + 3 } \)

\( \frac { | x | - 3 } { | y | + 3 } < \frac { | x | } { | y | } \), since we are decreasing the numerator and increasing the denominator.

So, \( \frac { x } { y } > \frac { x + 3 } { y - 3 } \) therefore the given condition \( \frac { x } { y } < \frac { x + 3 } { y - 3 } \) always fails in this case.

Case III)

x is positive and y is negative.

\( \frac { x } { y } = - \frac { x } { | y | } \)

\( \frac { x + 3 } { y - 3 } = - \frac { x + 3 } { | y | + 3 } \)

for \( \frac { x } { y } < \frac { x + 3 } { y - 3 } \),

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

\( \frac { x } { | y | } > \frac { x + 3 } { | y | - 3 } \) . This is always true since we are increasing the numerator and decreasing the denominator.

Case IV)

x is negative and y is positive.

\( \frac { x } { y } = - \frac { | x | } { y } \)

\( \frac { x + 3 } { y - 3 } = - \frac { | x | - 3 } { y - 3 } \)

for \( \frac { x } { y } < \frac { x + 3 } { y - 3 } \)

\( - \left( \frac { x } { y } \right) < - \left( \frac { x + 3 } { y - 3 } \right) \)

\( \frac { | x | } { 4 } < \frac { | x | - 3 } { y - 3 } \)

let \( 3 = k | x | \) , then \( 3 = k \frac { | x | } { y } \times y \)

Observe that k is always positive.

\( \frac { | x | } { 4 } < \frac { | x | - 3 } { y - 3 } \)

\( \frac { | x | } { y } < \frac { | x | ( 1 - k ) } { y \left( 1 - k \frac { | x | } { y } \right) } \)

This is only true when \( ( 1 - k ) > \left( 1 - k \frac { | x | } { y } \right) \)

\( - k > - k \frac { | x | } { y } \)

\( 1 < \frac { | x | } { y } \)

\( y < - x \)

So, the given condition holds good when both x & y are positive or x is positive but y is negative or x is negative, y is positive and \( y < - x \)

Since y is negative in the third option, \( - x < y \), implies that x > |y|, that is x is positive.

We know that when y is negative and x is positive the condition always holds good.

The question is **" Any non-zero real numbers \(x, y\) such that \(y \neq 3\) and \(\frac{x}{y} \lt \frac{x+3}{y-3}\), will satisfy the condition " **

Choice C is the correct answer.

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