Data Sufficiency

Is there anyone out there who can show me how to do this question in an organized fashion....It's easy to get confused with questions like this one....
Thanks

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.