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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 1912:30 PM EST
-01:30 PM EST
Learn how Keshav, a Chartered Accountant, scored an impressive 705 on GMAT in just 30 days with GMATWhiz's expert guidance. In this video, he shares preparation tips and strategies that worked for him, including the mock, time management, and more
Nov 2001:30 PM EST
-02:30 PM IST
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.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2407:00 PM PST
-08:00 PM PST
Full-length FE mock with insightful analytics, weakness diagnosis, and video explanations!
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
16 Sep 2014, 00:13
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
40% (02:27) correct
60%
(02:18)
wrong
based on 486
sessions
History
Date
Time
Result
Not Attempted Yet
If \(5x = y + 7\), is \(x - y > 0\)?
(1) \(xy = 6\)
(2) \(x\) and \(y\) are consecutive integers
(1) \(xy = 6\)
(2) \(x\) and \(y\) are consecutive integers
16 Sep 2014, 00:13
Kudos
Bookmarks
Official Solution:
If \(5x = y + 7\), is \(x - y > 0\)?
(1) \(xy = 6\).
From the equation given in the stem, we know that \(x = \frac{y+7}{5}\). Substituting this into the original equation, we obtain \(\frac{y+7}{5}*y = 6\), which simplifies to \(y^2+7y-30=0\). Solving this quadratic equation gives \(y = -10\) or \(y=3\).
If \(y = -10\), then \(x = \frac{y+7}{5}=\frac{-3}{5}\), which means that in this case \(x > y\).
If \(y = 3\), then \(x = \frac{y+7}{5}=2\), which means that in this case \(x < y\).
Since we obtain different answers to the question of whether \(x > y\), statement (1) is insufficient.
(2) \(x\) and \(y\) are consecutive integers.
Note that \(x\) and \(y\) being consecutive integers does not necessarily imply that \(x < y\) and \(x + 1 = y\); it might be that \(x > y\) and \(x - 1 = y\).
If \(x - 1 = y\), then substituting into \(5x = y + 7\) gives \(5x = (x-1) + 7\), which gives \(x=\frac{3}{2}\) and \(y=\frac{1}{2}\). This is not valid since in the case \(x\) and \(y\) are not integers as given in the statement.
If \(x + 1 = y\), then substituting into \(5x = y + 7\) gives \(5x = (x+1) + 7\), which gives \(x=2\) and \(y=3\). Therefore, the answer to the question of whether \(x > y\) is NO. Statement (2) is sufficient.
Answer: B
If \(5x = y + 7\), is \(x - y > 0\)?
(1) \(xy = 6\).
From the equation given in the stem, we know that \(x = \frac{y+7}{5}\). Substituting this into the original equation, we obtain \(\frac{y+7}{5}*y = 6\), which simplifies to \(y^2+7y-30=0\). Solving this quadratic equation gives \(y = -10\) or \(y=3\).
If \(y = -10\), then \(x = \frac{y+7}{5}=\frac{-3}{5}\), which means that in this case \(x > y\).
If \(y = 3\), then \(x = \frac{y+7}{5}=2\), which means that in this case \(x < y\).
Since we obtain different answers to the question of whether \(x > y\), statement (1) is insufficient.
(2) \(x\) and \(y\) are consecutive integers.
Note that \(x\) and \(y\) being consecutive integers does not necessarily imply that \(x < y\) and \(x + 1 = y\); it might be that \(x > y\) and \(x - 1 = y\).
If \(x - 1 = y\), then substituting into \(5x = y + 7\) gives \(5x = (x-1) + 7\), which gives \(x=\frac{3}{2}\) and \(y=\frac{1}{2}\). This is not valid since in the case \(x\) and \(y\) are not integers as given in the statement.
If \(x + 1 = y\), then substituting into \(5x = y + 7\) gives \(5x = (x+1) + 7\), which gives \(x=2\) and \(y=3\). Therefore, the answer to the question of whether \(x > y\) is NO. Statement (2) is sufficient.
Answer: B
General Discussion
11 Feb 2019, 16:05
Kudos
Bookmarks
Hi, Thank you for the detailed explanation.
The question is asking : is (x-y)>0?
Let us go straight to the last line of your explanation, when y
=x+1, x-y= -1, which is <0. Therefore (2) alone is not sufficient
Please advise.
Thanks
The question is asking : is (x-y)>0?
Let us go straight to the last line of your explanation, when y
=x+1, x-y= -1, which is <0. Therefore (2) alone is not sufficient
Please advise.
Thanks
