Last visit was: 25 Apr 2024, 00:54 It is currently 25 Apr 2024, 00:54

Close
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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Manager
Manager
Joined: 18 Aug 2009
Posts: 214
Own Kudos [?]: 1622 [0]
Given Kudos: 9
Send PM
User avatar
Manager
Manager
Joined: 18 Aug 2009
Posts: 214
Own Kudos [?]: 1622 [0]
Given Kudos: 9
Send PM
avatar
Intern
Intern
Joined: 09 Nov 2009
Posts: 1
Own Kudos [?]: [0]
Given Kudos: 0
Send PM
User avatar
Manager
Manager
Joined: 13 Oct 2009
Affiliations: PMP
Posts: 153
Own Kudos [?]: 250 [0]
Given Kudos: 38
 Q48  V32
Send PM
Re: m13 #13 The unlucky 13! [#permalink]
gbansal745 wrote:
S1 is insufficient
s2 is sufficient because
(x-y) is either even or odd integer, but multiplication with even (x+y) will always be even. So it is sufficient to tell us the given statement result.


Answer is B


Ans. shud be C as we need to know that x,y are integers.

Consider x=13/4, y = 11/4, then x+y=6 and x-y=1/2 and (x-y) (x+y) = 3 (odd)

Archived Topic
Hi there,
Archived GMAT Club Tests question - no more replies possible.
Where to now? Try our up-to-date Free Adaptive GMAT Club Tests for the latest questions.
Still interested? Check out the "Best Topics" block above for better discussion and related questions.
Thank you for understanding, and happy exploring!
GMAT Club Bot
Re: m13 #13 The unlucky 13! [#permalink]
Moderator:
Math Expert
92901 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne