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.
May 2110:00 AM EDT
-11:59 PM EDT
Make the most of your break with the most realistic GMAT™ prep. Take up to $700 off select products.
May 2211:00 AM EDT
-11:59 PM EDT
Take 30% off all study plans with code PICNIC - Expires on Monday, May 25th
May 2512:01 AM PDT
-11:59 PM PDT
On the 25th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PST- Jun 10
06:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
28 Apr 2008, 10:02
Kudos
Bookmarks
N
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
(N/A)
Question Stats:
0% (00:00) correct
0%
(00:00)
wrong
based on 0
sessions
History
Date
Time
Result
Not Attempted Yet
OAs [edit] questions are in reverse order, #2 then #1.
Q1 -> E
Q2 -> B
Attachment:
GMP2_2_DS4.jpg
Attachment:
GMP2_2_DS3.jpg
Q1 -> E
Q2 -> B
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!
28 Apr 2008, 11:44
Kudos
Bookmarks
Eliminating A and B is easy. Both the statements give me a range of x in terms of y.
Now lets combine both the equations.
From second eq, y should be negative, because |anything| is always positive.
I prefer to work with positive variables.
So, lets take z=-y, where z is positive
Hence 1st equation becomes |x-3|>-z, which is anyway always true and doesn't give us anything.
2nd eq -> |x-3|<z
Hence -z<x-3<z, which again does not give any definite value of x
Now lets combine both the equations.
From second eq, y should be negative, because |anything| is always positive.
I prefer to work with positive variables.
So, lets take z=-y, where z is positive
Hence 1st equation becomes |x-3|>-z, which is anyway always true and doesn't give us anything.
2nd eq -> |x-3|<z
Hence -z<x-3<z, which again does not give any definite value of x
