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M16 Q18

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1 KUDOS received
CEO
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Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
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Kudos [?]: 956 [1] , given: 235

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M16 Q18 [#permalink] New post 04 Apr 2010, 12:49
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I think the ans should be E, as y =0 and x as odd integer case is not considered.

Guys input your explanations pls.
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CEO
CEO
User avatar
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2793
Location: Malaysia
Concentration: Technology, Entrepreneurship
Schools: ISB '15 (M)
GMAT 1: 670 Q49 V31
GMAT 2: 710 Q50 V35
Followers: 180

Kudos [?]: 956 [0], given: 235

Reviews Badge
Re: M16 Q18 [#permalink] New post 05 Apr 2010, 13:06
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Re: M16 Q18 [#permalink] New post 05 Apr 2010, 16:24
Expert's post
gurpreetsingh wrote:
I think the ans should be E, as y =0 and x as odd integer case is not considered.

Guys input your explanations pls.


In case x=odd and y=0 --> 2*x*5*y=0=even

Question: 2*x*5*y=even. As there is 2 as a multiple, then this expression will be even if 5xy=integer. Basically we are asked is 5xy=integer true?

Note that x and y may not be integers for 2*x*5*y to be even (example x=\frac{7}{9} and y=\frac{9}{7}) BUT if they are integers then 2*x*5*y is even.


(1) 2+x+5+y=even --> 7+x+y=even --> x+y=odd. Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=1.7 answer NO)

(2) x-y=odd. Not sufficient. (x=1 and y=2 answer YES BUT x=1.3 and y=0.3 answer NO)

(1)+(2) Sum (1) and (2) (x+y)+(x-y)=odd_1+odd_2 --> 2x=even --> x=integer --> y=integer --> Both x and y are integers. Hence sufficient.

Answer: C.
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Re: M16 Q18   [#permalink] 05 Apr 2010, 16:24
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