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.
Sep 1507:00 PM EDT
-09:00 PM EST
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Tues/Thurs Sept. 15, 2026 - Nov. 15, 2027, 7:00pm-9:00pm EST
Sep 2107:00 PM EDT
-10:00 PM EDT
Boost your GMAT score in less than one month in a live online class + 6 months access to TTP OnDemand video courses included! Class starts Mon, Tues, Wed, Thur, Fri Sept. 21, 2026 - Oct. 9, 2027, 7:00pm-10:00pm EST
20 Dec 2010, 10:21
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
64% (01:13) correct
36%
(01:54)
wrong
based on 11
sessions
History
Date
Time
Result
Not Attempted Yet
Is m^n >= 0?
(1) m < 0 and n is an integer .
(2) n < 0 and even.
(1) m < 0 and n is an integer .
(2) n < 0 and even.
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!
20 Dec 2010, 11:48
Kudos
Bookmarks
surendar26
This is not the GMAT type of question.
Is \(m^n\geq{0}\)
(1) m<0 and n is an integer --> if m=-1 and n=2 then the answer will be YES but if m=-1 and n=3 then the answer will be NO. Not sufficient.
(2) n<0 and even (here also must be mentioned that \(m\) does not equal to zero) --> \(n=-2k\), for some positive integer \(k\) --> \(m^n=m^{-2k}=\frac{1}{m^{2k}}\) --> nonzero number \(m\) in the even power (\(2k\)) is positive, so \(\frac{1}{m^{2k}}>0\) (we needed m to be nonzero number because m goes to the denominator in this statement and division by zero is undefined). Sufficient.
Answer: B.
10 Sep 2016, 02:09
Kudos
Bookmarks
Hi Brunel, since in option 2 since its not mentioned that m is not zero, I solved it assuming a possibility that it could be 0 and arrived at an answer of C.
Can you please explain why C should not be the best choice for the question the way its worded ?
Can you please explain why C should not be the best choice for the question the way its worded ?
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!
