CAT 2023 Question Paper | Quant Slot 2

CAT Previous Year Paper | CAT Quant Questions | Question 4

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

  1. If \(y>10\), then \(-x>y\)
  2. If \(x \lt 0\), then \(-x \lt y\)
  3. If \(y \lt 0\), then \(-x \lt y\)
  4. \(\frac{x}{y} \lt \frac{y}{x}\)

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Explanatory Answer

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 "

Hence, the answer is 'If \(y \lt 0\), then \(-x \lt y\)'

Choice C is the correct answer.

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