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# M13-13

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Math Expert
Joined: 02 Sep 2009
Posts: 42250

Kudos [?]: 132650 [0], given: 12331

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16 Sep 2014, 00:49
Expert's post
5
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BOOKMARKED
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Difficulty:

55% (hard)

Question Stats:

51% (00:43) correct 49% (00:44) wrong based on 87 sessions

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Is $$(x - y)*(x + y)$$ an even integer?

(1) $$x$$ and $$y$$ are integers

(2) $$x + y$$ is even
[Reveal] Spoiler: OA

_________________

Kudos [?]: 132650 [0], given: 12331

Math Expert
Joined: 02 Sep 2009
Posts: 42250

Kudos [?]: 132650 [0], given: 12331

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16 Sep 2014, 00:49
Expert's post
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BOOKMARKED
Official Solution:

(1) $$x$$ and $$y$$ are integers. If $$x=y=1$$, then the answer is YES but if $$x=1$$ and $$y=0$$, then the answer is NO. Not sufficient.

(2) $$x+y$$ is even. If $$x=y=1$$ then the answer is YES but if $$x=1.8$$ and $$y=0.2$$, then the answer is NO, since in this case $$(x-y)*(x+y)$$ is not an integer at all. Not sufficient.

(1)+(2) Since from (1) $$x$$ and $$y$$ are integers and from (2) $$x+y$$ is even then either $$x$$ and $$y$$ are both even or both odd. In either case $$(x-y)*(x+y)$$ is even. Sufficient.

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Joined: 05 Oct 2013
Posts: 37

Kudos [?]: 7 [0], given: 51

Location: United States
Concentration: General Management, Strategy
WE: Design (Transportation)

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13 Dec 2014, 02:11
Bunuel wrote:
Official Solution:

(1) $$x$$ and $$y$$ are integers. If $$x=y=1$$, then the answer is YES but if $$x=1$$ and $$y=0$$, then the answer is NO. Not sufficient.

(2) $$x+y$$ is even. If $$x=y=1$$ then the answer is YES but if $$x=1.8$$ and $$y=0.2$$, then the answer is NO, since in this case $$(x-y)*(x+y)$$ is not an integer at all. Not sufficient.

(1)+(2) Since from (1) $$x$$ and $$y$$ are integers and from (2) $$x+y$$ is even then either $$x$$ and $$y$$ are both even or both odd. In either case $$(x-y)*(x+y)$$ is even. Sufficient.

Can you please explain how x=y=1 gives the equation an even integer... Thanks

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Math Expert
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Kudos [?]: 132650 [0], given: 12331

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13 Dec 2014, 05:50
Arnav180 wrote:
Bunuel wrote:
Official Solution:

(1) $$x$$ and $$y$$ are integers. If $$x=y=1$$, then the answer is YES but if $$x=1$$ and $$y=0$$, then the answer is NO. Not sufficient.

(2) $$x+y$$ is even. If $$x=y=1$$ then the answer is YES but if $$x=1.8$$ and $$y=0.2$$, then the answer is NO, since in this case $$(x-y)*(x+y)$$ is not an integer at all. Not sufficient.

(1)+(2) Since from (1) $$x$$ and $$y$$ are integers and from (2) $$x+y$$ is even then either $$x$$ and $$y$$ are both even or both odd. In either case $$(x-y)*(x+y)$$ is even. Sufficient.

Can you please explain how x=y=1 gives the equation an even integer... Thanks

If x = y = 1, then (x - y)*(x + y) = 0 = even (recall that 0 is an even integer).
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24 Feb 2016, 18:59
I think this is a poor-quality question and I don't agree with the explanation. If X and Y are both 5 and 5 this doesn't work

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Math Expert
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24 Feb 2016, 23:49
Binglai wrote:
I think this is a poor-quality question and I don't agree with the explanation. If X and Y are both 5 and 5 this doesn't work

I think you should brush up basics. 0 is an even integer.
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Joined: 22 Feb 2016
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Location: India
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GMAT 1: 690 Q42 V47
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21 Nov 2016, 07:09
What is the source of the question. It is the perfect glimpse of what trips us at GMAT.

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Math Expert
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21 Nov 2016, 09:09
AmritaSarkar89 wrote:
What is the source of the question. It is the perfect glimpse of what trips us at GMAT.

Source: GMAT Club's Tests.
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21 Sep 2017, 05:44
This is very good question. My first answer is B as I thought that A and B both must be odd or even and should be enough to answer the question. However, I missed the point that X and Y can be decimal.

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Re: M13-13   [#permalink] 21 Sep 2017, 05:44
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# M13-13

Moderators: Bunuel, chetan2u

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