Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2912:30 AM EDT
-01:30 AM EDT
Learn how Kamakshi achieved a GMAT 675 with an impressive 96th %ile in Data Insights. Discover the unique methods and exam strategies that helped her excel in DI along with other sections for a balanced and high score.
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
Apr 3001:30 AM EDT
-02:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
May 0111:00 AM PDT
-12:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
May 0306:30 AM PDT
-08:30 AM PDT
Verbal trouble on GMAT? Fix it NOW! Join Sunita Singhvi for a focused webinar on actionable strategies to boost your Verbal score and take your performance to the next level.
14 Mar 2017, 02:20
Kudos
Bookmarks
A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
64% (02:10) correct
36%
(02:23)
wrong
based on 25
sessions
History
Date
Time
Result
Not Attempted Yet
If x and y are integers and x + y = 21, is x > 19?
(1) 2y > 19 + y
(2) 3x/y > 51x
(1) 2y > 19 + y
(2) 3x/y > 51x
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
15 Mar 2017, 05:53
Kudos
Bookmarks
x +y = 21
St 1: 2y>19+y or y>19. hence x<2. therefore xis not >19 . ANSWER
St 2: 3x/y >51x
or x/y>17x
or x(1-17y)/y>0
therefore x>0, 0<y<1/17. for this x>19
or x<0 , y<0 or y>1/17. for this we cannot say. INSUFFICIENT
Option A
St 1: 2y>19+y or y>19. hence x<2. therefore xis not >19 . ANSWER
St 2: 3x/y >51x
or x/y>17x
or x(1-17y)/y>0
therefore x>0, 0<y<1/17. for this x>19
or x<0 , y<0 or y>1/17. for this we cannot say. INSUFFICIENT
Option A
Mo2men
Joined: 26 Mar 2013
Last visit: 09 May 2023
Posts: 2,426
Own Kudos:
Given Kudos: 641
Concentration: Operations, Strategy
Schools: Erasmus (II)
Products:
13 Dec 2017, 08:24
Kudos
Bookmarks
Bunuel
Hi Bunuel,
Are you sure that answer is A? How do you prove that Statement 2 is insufficient?
My answer as follows:
2) 3x/y > 51x............x/y>17x
If y>0.... then x>17xy.....x-17xy>0....x(1-17y)>0
As y is positive... (1-17y) is negative........x must be negative in order to have x(1-17y)>0
x is NOT greater than 19
If y 51x
x/y > 17x
In the resulting inequality, x can be positive only if y is also positive.
There is ONLY ONE SOLUTION for x+y = 21 such that x>19 and x and y are both positive:
x=20, y=1.
If this solution does not satisfy x/y > 17x, then it is not possible that x>19.
Plugging x=20 and y=1 into x/y > 17x, we get:
20/1 > 17*20
20 > 340.
Doesn't work.
Thus, it is not possible that x>19, implying that the answer to the question stem is NO.
SUFFICIENT.
Thanks
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Data Sufficiency (DS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
