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Source: — Data Sufficiency |
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by hemanth28 » Sat Jan 05, 2008 8:51 pm
how did you conclude C....both conditions together are not sufficiant for the response.

xy is even =>both x and y even (or) x even and y odd (or)x odd y even
yz is even =>both y and z even (or) y even and z odd (or)y odd z even

if you take the case that
i)(x odd and y even) and (y even and z odd ) = xz is odd
ii)for anyother case xz is even.

so you cant really conclude using both the statements together.
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by magical cook » Sat Jan 05, 2008 8:53 pm
Ahh, I think I mislead the question "xyz" are even... oops
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by hemanth28 » Sat Jan 05, 2008 8:55 pm
if that is the case than i think the answer is C
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by sankruth » Sun Jan 06, 2008 3:45 am
IMO - D

E x E = E

E x O = E

So, If xy is even, then (xy).z is even irrespective of z being even or odd

Similarly if yz is even, the (yz).x is even irrespective of x being even or odd.
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by gmatguy16 » Sun Jan 06, 2008 10:20 am
imo e
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