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Num properties even/odd

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by Patrick_GMATFix » Mon Jul 12, 2010 5:10 am
The key to this is a good rephrase. For a product of 2 integers to be even, at least one of them must be even.

REPHRASE: Is x or y even?

The statements should be evaluated with even/odd in mind. For instance look at (1). 4y is always even. so 5x-4y is 5x-even. For this to have an even result, 5x itself must be even, so x must be even. This answers our rephrase with a definitive YES.

The 2nd statement should be handled the same way. Again, think in terms of even/odd property. The answer is D

This is GMATPrep question 1510. To practice similar questions, set topic='Number Properties' and difficulty='400-500' in the Drill Generator

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by kvcpk » Mon Jul 12, 2010 5:12 am
gmatruler wrote:If x and y are positive integers, is the product xy even?

(1) 5x - 4y is even

(2) 6x + 7y is even
5x-4y is even

if 5x-4y is even, then
5x and 4y both are even.. or 5x and 4y both are odd.
4y cannot be odd. So 5x cannot be odd.
hence 5x has to be even. which means x is even.

if xy has to be even, then either x or y or both needs to be even.

hence SUFF

6x + 7y is even
if 6x + 7y is even, then
6x and 7y both are even.. or 6x and 7y both are odd.
6x cannot be odd.
therefore 7y is even. implies y is even.
if xy has to be even, then either x or y or both needs to be even.

hence SUFF

pick D.
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by Rahul@gurome » Mon Jul 12, 2010 5:14 am
gmatruler wrote:If x and y are positive integers, is the product xy even?

(1) 5x - 4y is even

(2) 6x + 7y is even
(1) 5x - 4y is even implies x should be an even integer, only then 5x - 4y can be even.
Whether y is even or odd, since x is even so (even)(even) = even and (even)(odd) = even
So, xy is even. Hence, (1) is SUFFICIENT.

(2) 6x + 7y is even implies y should be an even integer, only then 6x + 7y can be even.
(Even)(Even) = even and (Odd)(even) = even
Hence, (2) is SUFFICIENT.

[spoiler]The correct answer is (D).[/spoiler]
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by outreach » Mon Jul 12, 2010 9:21 am
i picked sample numbers
got the ans as D
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by gmatruler » Sun Jul 18, 2010 6:29 am
Thanks guys!
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