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M13-13

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M13-13  [#permalink]

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New post 16 Sep 2014, 00:49
2
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

48% (00:45) correct 52% (00:48) wrong based on 108 sessions

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Re M13-13  [#permalink]

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New post 16 Sep 2014, 00:49
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.


Answer: C
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Re: M13-13  [#permalink]

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New post 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.


Answer: C



Can you please explain how x=y=1 gives the equation an even integer... Thanks
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Re: M13-13  [#permalink]

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New post 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.


Answer: C



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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Re M13-13  [#permalink]

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New post 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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New post 24 Feb 2016, 23:49
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New post 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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New post 21 Nov 2016, 09:09
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New post 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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New post 18 Feb 2018, 13:31
I think this is a high-quality question and I agree with explanation.
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Re: M13-13  [#permalink]

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New post 11 Jun 2018, 00:52
Why are we not considering negative values of x and y?
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New post 11 Jun 2018, 05:41
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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Re: M13-13  [#permalink]

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New post 24 Jun 2018, 06:50
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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New post 24 Jun 2018, 10:05
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: M13-13 &nbs [#permalink] 24 Jun 2018, 10:05
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