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Strategies and techniques for approaching featured GMAT topics

# M13-13

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Math Expert
Joined: 02 Sep 2009
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15 Sep 2014, 23:49
2
2
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Difficulty:

55% (hard)

Question Stats:

48% (00:44) correct 52% (00:49) wrong based on 112 sessions

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

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

(2) $$x + y$$ is even

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Joined: 02 Sep 2009
Posts: 51229

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15 Sep 2014, 23:49
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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13 Dec 2014, 01:11
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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Joined: 02 Sep 2009
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13 Dec 2014, 04:50
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, 17:59
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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24 Feb 2016, 22:49
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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21 Nov 2016, 06:09
What is the source of the question. It is the perfect glimpse of what trips us at GMAT.
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21 Nov 2016, 08:09
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, 04:44
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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18 Feb 2018, 12:31
I think this is a high-quality question and I agree with explanation.
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Joined: 23 Mar 2018
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GMAT 1: 750 Q48 V44

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10 Jun 2018, 23:52
Why are we not considering negative values of x and y?
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11 Jun 2018, 04:41
verysecretive3 wrote:
Why are we not considering negative values of x and y?

The solution shows that the answer is C not mentioning negative or positive nature of x or y. So, (1)+(2) is sufficient no matter whether you consider positive or negative variables.
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24 Jun 2018, 05:50
Bunuel wrote:
Is $$(x - y)*(x + y)$$ an even integer?

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

(2) $$x + y$$ is even

I got this wrong since I considered possibility where X and Y can be negative, such as, X=-2 and Y = 4.
Why we are not considering negative number scenarios
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24 Jun 2018, 09:05
Cbirole wrote:
Bunuel wrote:
Is $$(x - y)*(x + y)$$ an even integer?

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

(2) $$x + y$$ is even

I got this wrong since I considered possibility where X and Y can be negative, such as, X=-2 and Y = 4.
Why we are not considering negative number scenarios

(1)+(2) is sufficient no matter whether you consider positive or negative variables.

If x = -2 and y = 4, then (x - y)*(x + y) = -6*2 = -12 = even. The same answer as we got in the solution above.
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Re: M13-13 &nbs [#permalink] 24 Jun 2018, 09:05
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# M13-13

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