# 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 $$frac{x}{y} \lt \frac{x+3}{y-3}$, will satisfy the condition 1. If $y>10$, then $-x>y$ 2. If $x \lt 0$, then $-x \lt y$ 3. If $y \lt 0$, then $-x \lt y$ 4. $\frac{x}{y} \lt \frac{y}{x}$ ### Video Explanation ## Best CAT Coaching in Chennai #### CAT Coaching in Chennai - CAT 2022Limited Seats Available - Register Now! ### Explanatory Answer Case I$ If both x and y are both positive. Then, $$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. ###### Best CAT Online Coaching Try upto 40 hours for free Learn from the best! ###### Prepare for CAT 2024 with 2IIM's Daily Preparation Schedule ###### Know all about CAT Exam Syllabus and what to expect in CAT ###### Already have an Account? ###### Best CAT Coaching in Chennai Attend a Demo Class ###### Best Indore IPM & Rohtak IPM CoachingSignup and sample 9 full classes for free. Register now! ## CAT Questions | CAT Quantitative Aptitude ## CAT Questions | Verbal Ability for CAT ##### Where is 2IIM located? 2IIM Online CAT Coaching A Fermat Education Initiative, 58/16, Indira Gandhi Street, Kaveri Rangan Nagar, Saligramam, Chennai 600 093 ##### How to reach 2IIM? Mobile:$91$ 99626 48484 / 94459 38484
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