Last visit was: 19 Jul 2025, 19:19 It is currently 19 Jul 2025, 19:19
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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
Back to Forum
Create Topic
User avatar
jabhatta2
User avatar
Joined: 15 Dec 2016
Last visit: 21 Apr 2023
Posts: 1,304
Own Kudos:
283
 [35]
Given Kudos: 188
Posts: 1,304
Kudos: 283
 [35]
If x and y are integers, and x + y < 0, is x — y > 0 08 Apr 2018, 09:41
1
Kudos
Add Kudos
34
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
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:

37% (02:29) correct 63% (02:28) wrong based on 373 sessions

History

Date
Time
Result
Not Attempted Yet
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
742,802
 [7]
Given Kudos: 98,235
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: 102,627
Kudos: 742,802
 [7]
If x and y are integers, and x + y < 0, is x — y > 0 08 Apr 2018, 12:13
4
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.
General Discussion
User avatar
jabhatta2
User avatar
Joined: 15 Dec 2016
Last visit: 21 Apr 2023
Posts: 1,304
Own Kudos:
283
Given Kudos: 188
Posts: 1,304
Kudos: 283
If x and y are integers, and x + y < 0, is x — y > 0 08 Apr 2018, 11:50
Kudos
Add Kudos
Bookmarks
Bookmark this Post
hmm if interested in testing cases -- here is the link to the OA.....you should be able to find it using test cases ...but it takes too long imo

Has to be a simpler way

https://www.manhattanprep.com/gmat/blog ... flowchart/
User avatar
jabhatta2
User avatar
Joined: 15 Dec 2016
Last visit: 21 Apr 2023
Posts: 1,304
Own Kudos:
283
Given Kudos: 188
Posts: 1,304
Kudos: 283
If x and y are integers, and x + y < 0, is x — y > 0 Updated on: 08 Apr 2018, 12:33
Originally posted by jabhatta2 on 08 Apr 2018, 12:30.
Last edited by jabhatta2 on 08 Apr 2018, 12:33, edited 1 time in total.
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.

Wow -- this is really well done ..

Bunuel - do you think its easier to

a) test cases in this case or
b) do you prefer arithmetic

Do you think you could have done your method in 3 mins time pressure ?
User avatar
jabhatta2
User avatar
Joined: 15 Dec 2016
Last visit: 21 Apr 2023
Posts: 1,304
Own Kudos:
283
Given Kudos: 188
Posts: 1,304
Kudos: 283
If x and y are integers, and x + y < 0, is x — y > 0 08 Apr 2018, 12:32
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.

Hi Bunuel - a question on the highlighted portion (bullet b and c)

How is it bullet B tells us x = -1 whereas bullet c tells us x = 1

Dont bullet b and bullet C almost contradict each other in a way ?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
742,802
Given Kudos: 98,235
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: 102,627
Kudos: 742,802
If x and y are integers, and x + y < 0, is x — y > 0 08 Apr 2018, 12:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
[email protected]
Bunuel
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.

Wow -- this is really well done ..

Bunuel - do you think its easier to

a) test cases in this case or
b) do you prefer your method in terms -- Do you think you could have done your method in 3 mins

I'd done this way but it's really subjective.
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Jul 2025
Posts: 102,627
Own Kudos:
742,802
Given Kudos: 98,235
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: 102,627
Kudos: 742,802
If x and y are integers, and x + y < 0, is x — y > 0 08 Apr 2018, 12:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
[email protected]
Bunuel
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

If x and y are integers, and x + y < 0, is x — y > 0?

Notice that the question asks whether x > y.


(1) |x| + |y| > |x|. Cancel |x| on both sides: |y| > 0. Since an absolute value of a number is always more than or equal to 0, then this statement is simply telling us that y ≠ 0. Not sufficient.


(2) x^y = 1. This could be true in three cases:

a. y = 0 and x = any non-zero integer;
b. x = -1 and y = even.
c. x = 1 and y = any integer.

We can have cases satisfying the above and x + y < 0 and giving different answers to the question whether x > y. For example, y = 0 and x = -1 or y = -2 and x = -1. Not sufficient.

(1)+(2) Since (1) rules out y ≠ 0, then we are left with cases b and c from (2):

b. x = -1 and y = even.. This and x + y < 0 to be true y should be negative even number: -2, -4, ... In all these cases x > y.
c. x = 1 and y = any integer. This and x + y < 0 to be true y should be less than -1: -2, -3, -4, ... In all these cases x > y.

In all possible cases we got an YES answer to the question. Sufficient.


Answer: C.

Hope it's clear.

Hi Bunuel - a question on the highlighted portion (bullet b and c)

How is it bullet B tells us x = -1 whereas bullet c tells us x = 1

Dont bullet b and bullet C almost contradict each other in a way ?

Those are different cases for which x^y = 1:
b. x = -1 and y = even. For example, x= -1 and y = 2 --> x^y = 1.
c. x = 1 and y = any integer. For example, x= 1 and y = 3 --> x^y = 1.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Jul 2025
Posts: 16,115
Own Kudos:
74,408
 [1]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,115
Kudos: 74,408
 [1]
If x and y are integers, and x + y < 0, is x — y > 0 10 Apr 2018, 05:54
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

You can imagine the numbers on the number line to solve it.

x and y are integers.

x + y < 0 implies at least one of x and y is negative and if there is one negative number and one positive, the negative one has higher absolute value. I imagine them placed on the number line.

Question: Is x - y > 0
This just means: "Is x to the right of y on the number line?"

1) |x| + |y| > |x|
If absolute value of the two is greater than the absolute value of x alone, it just means that y is not 0.
Not sufficient

2) x^y = 1
This can be done in 2 main ways: Either make y = 0 or make x = 1 (with y as any integer) / -1 (with y even)
If y is 0, x must be negative and answer would be "No"
If x = 1, y must be negative and answer would be "Yes"
Not sufficient

Using both statements, x = 1 (with y as any integer) / -1 (with y even)
If x is 1, answer is "Yes"
If x = -1, y is even. So y is one of -2, -4, -6 etc in which case answer is "Yes". y cannot be positive because the negative value must have higher absolute value.
Sufficient

Answer (C)
User avatar
jabhatta2
User avatar
Joined: 15 Dec 2016
Last visit: 21 Apr 2023
Posts: 1,304
Own Kudos:
283
Given Kudos: 188
Posts: 1,304
Kudos: 283
If x and y are integers, and x + y < 0, is x — y > 0 28 Jan 2020, 17:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasKarishma
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

You can imagine the numbers on the number line to solve it.

x and y are integers.

x + y < 0 implies at least one of x and y is negative and if there is one negative number and one positive, the negative one has higher absolute value. I imagine them placed on the number line.

Question: Is x - y > 0
This just means: "Is x to the right of y on the number line?"

1) |x| + |y| > |x|
If absolute value of the two is greater than the absolute value of x alone, it just means that y is not 0.
Not sufficient

2) x^y = 1
This can be done in 2 main ways: Either make y = 0 or make x = 1 (with y as any integer) / -1 (with y even)
If y is 0, x must be negative and answer would be "No"
If x = 1, y must be negative and answer would be "Yes"
Not sufficient

Using both statements, x = 1 (with y as any integer) / -1 (with y even)
If x is 1, answer is "Yes"
If x = -1, y is even. So y is one of -2, -4, -6 etc in which case answer is "Yes". y cannot be positive because the negative value must have higher absolute value.
Sufficient

Answer (C)

Hi karishma --

on the first highlight in red -- any reason why the following cases were not considered

Case 1) not considered) X and Y are both negative
Case 1, Scenario 1 when both variables are negative) |Y| > |x|
Case 1, Scenario 2 when both variables are negative) |x| > |Y|



Case 2) not considered) X and Y can be zero
Case 2, Scenario 1, x = 0 and Y is negative
Case 2, Scenario 2 y = 0 and X is negative
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Jul 2025
Posts: 16,115
Own Kudos:
74,408
 [1]
Given Kudos: 475
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,115
Kudos: 74,408
 [1]
If x and y are integers, and x + y < 0, is x — y > 0 29 Jan 2020, 02:13
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
[email protected]
VeritasKarishma
[email protected]
If x and y are integers, and x + y < 0, is x — y > 0?

1) |x| + |y| > |x|
2) x^y = 1

The OA solution suggested testing cases ....Any suggestions on doing this via logic or critical thinking ?

You can imagine the numbers on the number line to solve it.

x and y are integers.

x + y < 0 implies at least one of x and y is negative and if there is one negative number and one positive, the negative one has higher absolute value. I imagine them placed on the number line.

Question: Is x - y > 0
This just means: "Is x to the right of y on the number line?"

1) |x| + |y| > |x|
If absolute value of the two is greater than the absolute value of x alone, it just means that y is not 0.
Not sufficient

2) x^y = 1
This can be done in 2 main ways: Either make y = 0 or make x = 1 (with y as any integer) / -1 (with y even)
If y is 0, x must be negative and answer would be "No"
If x = 1, y must be negative and answer would be "Yes"
Not sufficient

Using both statements, x = 1 (with y as any integer) / -1 (with y even)
If x is 1, answer is "Yes"
If x = -1, y is even. So y is one of -2, -4, -6 etc in which case answer is "Yes". y cannot be positive because the negative value must have higher absolute value.
Sufficient

Answer (C)

Hi karishma --

on the first highlight in red -- any reason why the following cases were not considered

Case 1) not considered) X and Y are both negative
Case 1, Scenario 1 when both variables are negative) |Y| > |x|
Case 1, Scenario 2 when both variables are negative) |x| > |Y|



Case 2) not considered) X and Y can be zero
Case 2, Scenario 1, x = 0 and Y is negative
Case 2, Scenario 2 y = 0 and X is negative

As per the highlighted part: x + y < 0 implies at least one of x and y is negative

When we say AT LEAST one of x and y is negative, it includes the case in which both are negative. It also includes the case in which one of them is 0 and the other negative. Also, it includes the case in which one of them is positive and the other negative such that the negative one has higher absolute value.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 37,452
Own Kudos:
1,013
Posts: 37,452
Kudos: 1,013
If x and y are integers, and x + y < 0, is x — y > 0 25 Nov 2022, 02:34
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Bunuel
Math Expert
102627 posts
kingbucky
455 posts