CAT 2023 Question Paper | Quant Slot 2

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

Video Explanation

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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