Tricky GMATPrep question - even-odds integers

Well picking numbers is the easiest way out ...

Given xy + z is odd ...

X = 2 y = 2 z = 1 or

x = 3 y = 2 z = 1

In one case x is even and in one odd

Put these numbers in Stmt 1

x(y+z) will be even only when x is even .Hence Knock of BCE left with AD

Stmt 2 says


xz + y is odd

x = 2 z = 3 y = 1 x is even

x= 3 z = 2 y = 1 x is odd

Hence we are getting a yes and no .
Hence insufficient

Hence A

Hope this helps

i rily think the answer shoul be E

xy + z is odd

stmt 1:

xy + xz is even

i.e xy +z + xz - z

i.e xy +z + z(x-1) is even

since xy +z is odd so z(x-1) must be odd for the above stmt to be true

so x- 1 is odd hence x is even SUFF


stmt 2: xz + y is odd

so xz +y + xy + z is even

i.e x(z+y) +(z+y) is even

i.e (z+y)(x+1) is even

now this can be true if z+y is even

so in that case x+1 can be odd i.e x even & vice versa

INSUFF

A

Good job Samir,
This is the most elegant way to solve this question. I can see you have some history on this site....What's your target score?
Thanks again.