Summer is Coming! Join the Game of Timers Competition to Win Epic Prizes. Registration is Open. Game starts Mon July 1st.

It is currently 21 Jul 2019, 05:56

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

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
B
Joined: 15 Dec 2016
Posts: 103
If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 08 Apr 2018, 09:41
1
11
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

31% (02:29) correct 69% (02:24) wrong based on 135 sessions

HideShow timer Statistics


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 ?
Manager
Manager
avatar
B
Joined: 15 Dec 2016
Posts: 103
If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 08 Apr 2018, 11:50
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/
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56306
If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 08 Apr 2018, 12:13
2
1
jabhatta@umail.iu.edu wrote:
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.
_________________
Manager
Manager
avatar
B
Joined: 15 Dec 2016
Posts: 103
If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post Updated on: 08 Apr 2018, 12:33
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
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 ?

Originally posted by jabhatta@umail.iu.edu on 08 Apr 2018, 12:30.
Last edited by jabhatta@umail.iu.edu on 08 Apr 2018, 12:33, edited 1 time in total.
Manager
Manager
avatar
B
Joined: 15 Dec 2016
Posts: 103
Re: If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 08 Apr 2018, 12:32
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
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 ?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56306
Re: If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 08 Apr 2018, 12:34
jabhatta@umail.iu.edu wrote:
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
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.
_________________
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 56306
Re: If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 08 Apr 2018, 12:38
jabhatta@umail.iu.edu wrote:
Bunuel wrote:
jabhatta@umail.iu.edu wrote:
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.
_________________
Veritas Prep GMAT Instructor
User avatar
D
Joined: 16 Oct 2010
Posts: 9449
Location: Pune, India
Re: If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 10 Apr 2018, 05:54
jabhatta@umail.iu.edu wrote:
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)
_________________
Karishma
Veritas Prep GMAT Instructor

Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
Non-Human User
User avatar
Joined: 09 Sep 2013
Posts: 11719
Re: If x and y are integers, and x + y < 0, is x — y > 0  [#permalink]

Show Tags

New post 02 Jul 2019, 21:13
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.
_________________
GMAT Club Bot
Re: If x and y are integers, and x + y < 0, is x — y > 0   [#permalink] 02 Jul 2019, 21:13
Display posts from previous: Sort by

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

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





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