Events & Promotions
| Last visit was: 06 Sep 2026, 10:40 |
It is currently 06 Sep 2026, 10:40 |
|
|
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 0712:01 AM PDT
-11:59 PM PDT
On the 7th, GMAT Club Tests will be Free! Including all the Quizzes, Questions, and Tests. 12 AM - 11:59 PM PDT
Sep 0712:30 AM EDT
-01:30 AM EDT
Tejas studied hard but kept repeating the same mistakes. After 3 attempts at 645, a focused 4-week strategy helped him identify and fix what was holding him back—leading to a 695 with just 12–15 hours of study per week.- Sep 07
10:00 AM PDT
-11:00 AM PDT
A complete walkthrough of GMAT Club’s free 12-week GMAT study plan: what you study each week, how progress and goals are tracked, how the error log works, and how your practice unlocks paid tools at no cost. 
Sep 0201:00 PM EDT
-11:59 PM EDT
Take 30% off all study plans with code COURSEWORK - Expires on Monday 9/7
Sep 0401:30 AM EDT
-02:30 AM EDT
This Labor Day, turn your hard work into real score gains. From 615→715 in 15 days to 705 in 30 days, our students prove what's possible with personalized, expert guidance. Save up to 25% on 1-on-1 tutoring.
Sep 0410:00 AM EDT
-11:59 PM EDT
We’ve worked incredibly hard to build TTP into the best test prep experience possible, and it would mean a lot to us to win Newsweek’s 2026 Readers’ Choice Award for Best Test Prep. If TTP has helped you, we’d be incredibly grateful for your vote.
Sep 0609:30 AM EDT
-12:30 PM EST
Top scores are possible when you enroll in a powerful EA course, taught live online + 6 months access to TTP OnDemand video courses included! Perfect class schedule and easy course access for working professionals. Class starts Sundays Sept. 6, 2026.- Sep 09
07:00 AM PDT
-10:00 AM PDT
Meet AdComs and explore top Master’s programs - MiM, MiF, MSc, MSBA and more. - Application Fee Waivers - Free 1-Week of GMAT Club Tests: - Master's Application Toolkit - Grand Prize Giveaway - Sep 09
10:00 AM PDT
-11:00 AM PDT
Three MBA applications. Three rejections. No interview invites. A year later, Aman was admitted to Cambridge Judge, SMU, and IMD. In this episode of MBA Admit Stories, Aman shares how he rebuilt his MBA application... 
Sep 1008:00 PM PDT
-09:00 PM PDT
Free- full-length test + 15 concept videos + 150+ short videos + study plan!
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 2108:00 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
08 May 2012, 00:21
Kudos
Bookmarks
Is \(|x - y| \gt |x + y|\) ?
1) \(x^2 - y^2 = 9\)
2) \(x - y = 2\)
From S1, can we assume that x has to equal 5 and y has to equal 4 so when you square x and y, their difference would be 9. Thus S1 would be sufficient.
S2 is insufficient.
Thus the answer is A, but the Official Answer is C. Can someone please explain. Thanks.
1) \(x^2 - y^2 = 9\)
2) \(x - y = 2\)
From S1, can we assume that x has to equal 5 and y has to equal 4 so when you square x and y, their difference would be 9. Thus S1 would be sufficient.
S2 is insufficient.
Thus the answer is A, but the Official Answer is C. Can someone please explain. Thanks.
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 below for better discussion and related questions.
Thank you for understanding, and happy exploring!
08 May 2012, 00:30
Kudos
Bookmarks
Samwong
Why would you assume that? Why not x=-5 and y=-4? Or x=3 and y=0? \(x^2 - y^2 = 9\) has infinitely many solutions for x and y, and only one of them is x=5 and y=4.
Is \(|x - y| \gt |x + y|\) ?
(1) \(x^2 - y^2 = 9\) --> \((x-y)(x+y)=9\). Not sufficient.
(2) \(x - y = 2\) --> don't know the value of \(x+y\). Not sufficient.
(1)+(2) Since from (2) \(x - y = 2\) then from (1) \((x-y)(x+y)=2(x+y)=9\) --> \(x+y=4.5\). Sufficient.
Answer: C.
Hope it helps.
08 May 2012, 00:54
Kudos
Bookmarks
Thanks Bunuel for the reply. That make sense now. I did a similar problem on MGMAT CAT. The difference is that on the MGMAT problem there is a constrain of x < y < 0
If x and y are integers such that x < y < 0, what is x – y?
(1) (x + y)(x – y) = 5
(2) xy = 6
Official Explanation:
The problem stem tells you that x and y are both negative integers, and x is less than y. You are asked for the value of x – y. Note that in order to answer this question, you might not need the separate values of x and y; however, if you can find those separate values, then you can solve for x – y or any other “combo.”
(1) SUFFICIENT: This statement is tricky. One approach is to distribute the left side to get the difference of squares:
x2 – y2 = 5
Now, since both x and y are integers, x2 and y2 are integers as well. Let’s look at the sequence of squares, and consider their differences (we can leave out zero, since neither x nor y can be zero):
1 4 9 16 25 36…
Diff = 3 5 7 9 11…
As you can see, the difference between squares grows as the squares themselves get larger. The only difference between two squares that equals 5 is the difference between 4 and 9. Since x and y are both negative, this tells us that x = –3 and y = –2; therefore, x – y = –1.
Alternatively, we could note that both x + y and x – y are themselves integers. Looking at the statement, we have
(x + y)(x – y) = 5
int × int = 5
The only possible integer pairs that give 5 as a product are (5, 1) and (-5, -1), since 5 is prime. Now, because both x and y are negative, the (5, 1) pair won’t work either way (either with x + y = 5 or with x + y = 1). So let’s try (-5, -1):
x + y = –5
x – y = –1
Adding these two equations, we get 2x = -6, or x = -3. Substituting back into x + y = –5, we get y = –2. (If we had assigned x + y = –1 and x – y = –5, we would have gotten y = 2, which doesn’t fit the problem constraints.)
(2) INSUFFICIENT: There are two pairs of integers that fit the constraint x < y < 0 and the statement xy = 6: (-3)(-2) = 6 AND (-6)(-1) = 6.
The correct answer is A.
If x and y are integers such that x < y < 0, what is x – y?
(1) (x + y)(x – y) = 5
(2) xy = 6
Official Explanation:
The problem stem tells you that x and y are both negative integers, and x is less than y. You are asked for the value of x – y. Note that in order to answer this question, you might not need the separate values of x and y; however, if you can find those separate values, then you can solve for x – y or any other “combo.”
(1) SUFFICIENT: This statement is tricky. One approach is to distribute the left side to get the difference of squares:
x2 – y2 = 5
Now, since both x and y are integers, x2 and y2 are integers as well. Let’s look at the sequence of squares, and consider their differences (we can leave out zero, since neither x nor y can be zero):
1 4 9 16 25 36…
Diff = 3 5 7 9 11…
As you can see, the difference between squares grows as the squares themselves get larger. The only difference between two squares that equals 5 is the difference between 4 and 9. Since x and y are both negative, this tells us that x = –3 and y = –2; therefore, x – y = –1.
Alternatively, we could note that both x + y and x – y are themselves integers. Looking at the statement, we have
(x + y)(x – y) = 5
int × int = 5
The only possible integer pairs that give 5 as a product are (5, 1) and (-5, -1), since 5 is prime. Now, because both x and y are negative, the (5, 1) pair won’t work either way (either with x + y = 5 or with x + y = 1). So let’s try (-5, -1):
x + y = –5
x – y = –1
Adding these two equations, we get 2x = -6, or x = -3. Substituting back into x + y = –5, we get y = –2. (If we had assigned x + y = –1 and x – y = –5, we would have gotten y = 2, which doesn’t fit the problem constraints.)
(2) INSUFFICIENT: There are two pairs of integers that fit the constraint x < y < 0 and the statement xy = 6: (-3)(-2) = 6 AND (-6)(-1) = 6.
The correct answer is A.
08 May 2012, 01:02
Kudos
Bookmarks
Samwong
Actually you have more than one constraint:
(i) x and y are integers;
(ii) x<y<0
If x and y are integers such that x<y<0 what is x-y?
(1) (x+y)(x-y)=5. x and y are integers means that both x+y and x-y are integers. So, we have that the product of two integer factors equal to 5. There are only two combination of such factors possible: (1, 5) and (-1, -5). Since given that x and y are both negative then the first case is out, so x-y is either -1 or -5, but it can not be -5, because in this case x+y must be -1 and no sum of two negative integers yields -1. Hence x-y=-1. Sufficient.
(2) xy= 6. If x=-3 and y=-2 then x-y=-1 but if x=-6 and y=-1 then x-y=-5. Not sufficient.
Answer: A.
Hope it helps.
