Last visit was: 04 May 2026, 05:41 It is currently 04 May 2026, 05:41
Home
Messages
Tests
Schools
Events
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 04 May 2026
Posts: 110,051
Own Kudos:
812,820
Given Kudos: 106,014
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 110,051
Kudos: 812,820
GMAT Diagnostic Test Question 43 25 Feb 2012, 12:56
Kudos
Add Kudos
Bookmarks
Bookmark this Post
T740qc
Quote:
Does anyone know how to easily derive from statement (2) the following possibilities: \(x = 1\) and \(y = 0\), and \(x = 2\) and \(y = 2\) ? I don't think it's that obvious. So it's easy to miss one possibility and answer the question differently.

i didnt see an answer to this q. is there an easier way to get to 0,1 or 2? thanks.
Show more

This question is also discussed here: if-x-and-y-are-integers-is-x-y-1-x-y-93713.html#p761563

But I wouldn't worry about it since it needs a revision and will be updated for the next edition.
avatar
steveymc
avatar
Joined: 17 Oct 2012
Last visit: 21 Oct 2012
Posts: 2
Posts: 2
Kudos: 0
GMAT Diagnostic Test Question 43 17 Oct 2012, 09:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer is E,for the first one is not suffice to solve the question cos give if /x/=/y+1/ implies /y+1/ greater than /y/ which is wrong if y is -2.the second one is not suffice to do it so cos if the second statement is right which implies absolute y = x^y-x! which is that /x/ must be greater than x^y-x! from here we know x must be greater or equal to 0,so if x=2,y=5 then the statement is wrong.so if you put them together would still yield the same result,the solution is indeterminant!
avatar
steveymc
avatar
Joined: 17 Oct 2012
Last visit: 21 Oct 2012
Posts: 2
Posts: 2
Kudos: 0
GMAT Diagnostic Test Question 43 17 Oct 2012, 09:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Answer is E,for the first one is not suffice to solve the question cos give if /x/=/y+1/ implies /y+1/ greater than /y/ which is wrong if y is -2.the second one is not suffice to do it so cos if the second statement is right which implies absolute y = x^y-x! which is that /x/ must be greater than x^y-x! from here we know x must be greater or equal to 0,so if x=2,y=5 then the statement is wrong.so if you put them together would still yield the same result,the solution is indeterminant!
User avatar
gmatgeek
User avatar
Joined: 24 Apr 2012
Last visit: 24 Jun 2015
Posts: 33
Own Kudos:
96
Given Kudos: 35
Schools: Kellogg PT '17 Booth PT '17 (M)
GMAT 1: 660 Q48 V33
WE:Information Technology (Healthcare/Pharmaceuticals)
Schools: Kellogg PT '17 Booth PT '17 (M)
GMAT 1: 660 Q48 V33
Posts: 33
Kudos: 96
GMAT Diagnostic Test Question 43 31 Dec 2012, 12:30
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMAT TIGER
Explanation
Rating:
[rating1]yellow/893752[/rating1]
Official Answer: С

Statement (1):
Given that \(|x| = |y+1|\)
If \(x\) and \((y+1)\) both have same signs (either positive or negative), then \(x = y+1\) and \(x>y\), which makes \(x\) and \(y\) consecutive integers.
If \(x\) and \((y+1)\) both do not have the same signs (one is negative and the other is positive), then \(x = -(y+1)\) or \((x + y + 1) = 0\). In that case, either one could be greater in absolute value. S1 is not sufficient by itself.

Statement (2):
If \(x^y = x! + |y|\), \(x\) and \(y\) both have to be positive because \(x! + |y|\) is always a positive integer. If \(y\) is negative, \(x^y\) can never be an integer. So for \(x^y\) to be an integer, \(y\) has to be positive. Let's see what values satisfy the equation:
If \(x = 1\), and \(y = 0\), \(x^y = x! + |y| = 1\).
If \(x = 2\), and \(y = 2\), \(x^y = x! + |y| = 4\).
So \(x\) and \(y\) have at least two possibilities. S2 is not sufficient by itself either.

Statement (1)+Statement (2):
Because of Statement (2), \(x\) and \(y\) have to be positive, in which case, by Statement (1) \(x > y\). Sufficient. Therefore, the answer is C.
Show more

I don't get it..Statement 2 gives you 2 solutions 1,0 and 2,2 so x and y are non-negative .. even combining with statement 2 you get 2 solutions.. so answer is E..
am I missing something ?
User avatar
cjliu49
User avatar
Joined: 01 Jan 2013
Last visit: 02 May 2014
Posts: 28
Own Kudos:
39
Given Kudos: 13
Location: United States
Concentration: Entrepreneurship, Strategy
Schools: Stanford '16 (D) HBS '16 (D) Wharton '16 (D) Haas '16 (WL)
GMAT 1: 770 Q50 V47
WE:Consulting (Consulting)
Schools: Stanford '16 (D) HBS '16 (D) Wharton '16 (D) Haas '16 (WL)
GMAT 1: 770 Q50 V47
Posts: 28
Kudos: 39
GMAT Diagnostic Test Question 43 07 Mar 2013, 18:07
Kudos
Add Kudos
Bookmarks
Bookmark this Post
There's no way this is a 700-level only question, is there? I think I'm a high scorer and the analysis of the correct answer by GMAT Tiger is very sophisticated.

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!
   1   2 
Moderator:
Bunuel
Math Expert
110051 posts