Jul 21 07:00 AM PDT  09:00 AM PDT Attend this webinar to learn a structured approach to solve 700+ Number Properties question in less than 2 minutes Jul 26 08:00 AM PDT  09:00 AM PDT The Competition Continues  Game of Timers is a teambased competition based on solving GMAT questions to win epic prizes! Starting July 1st, compete to win prep materials while studying for GMAT! Registration is Open! Ends July 26th Jul 27 07:00 AM PDT  09:00 AM PDT Learn reading strategies that can help even nonvoracious reader to master GMAT RC
Author 
Message 
TAGS:

Hide Tags

Manager
Joined: 15 Dec 2016
Posts: 103

If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
08 Apr 2018, 09:41
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 ?
Official Answer and Stats are available only to registered users. Register/ Login.



Manager
Joined: 15 Dec 2016
Posts: 103

If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
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
Joined: 02 Sep 2009
Posts: 56306

If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
08 Apr 2018, 12:13
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 nonzero 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
Joined: 15 Dec 2016
Posts: 103

If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
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 nonzero 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 ?



Manager
Joined: 15 Dec 2016
Posts: 103

Re: If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
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 nonzero 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
Joined: 02 Sep 2009
Posts: 56306

Re: If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
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 nonzero 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
Joined: 02 Sep 2009
Posts: 56306

Re: If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
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 nonzero 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
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
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 >



NonHuman User
Joined: 09 Sep 2013
Posts: 11719

Re: If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
Show Tags
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.
_________________




Re: If x and y are integers, and x + y < 0, is x — y > 0
[#permalink]
02 Jul 2019, 21:13






