Data Sufficiency

If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1)wx + yz = odd

(2)wz + xy = odd

To find: is w/x + y/z odd

Statement 1: wx + yz = odd

E + O
O + E

For wx Even = w must be Even & x can be ODD or even
For wx Odd = w & x must be ODD

For yz Even = y must be Even & z can be even or Odd
For yz Odd = y & z must be Odd

So, w/x + y/z
==> (w/x) can be Even or Odd -- depending upon "x"
==> (y/z) can be Even or Odd -- depending upon "z"
INSUFFICIENT


Statement 2: wz + xy = odd
w/x + y/z = INT
(wz + xy)/xz = INT
(ODD)/(xz) = INT
that means (xz) must be ODD ==> X & Z must be ODD

Coming back to equation:
w(ODD) + (ODD)y = odd ==> Either w is EVEN or y

Therefore, we have three odd terms and 1 even term and in numerators we have w & y one Odd and other Even
So, EVEN + Odd Or Odd + EVen == ODD
SUFFICIENT
Answer [spoiler]{B}[/spoiler]

For wx Even = w must be Even & x can be ODD or even ....(i have a doubt here)

for wx to be even ... either w or x can be even and then result wx is even ?

abhasjha wrote:If w, x, y, and z are integers such that w/x and y/z are integers, is w/x + y/z odd?

(1)wx + yz = odd

(2)wz + xy = odd

Hi abhasjha,
Given: w/x + y/z = int
(wz + xy)/xz = int
To Find: w/x + y/z = odd => (wz + xy)/xz = odd int ?
Since statement 2 is similar to above equation, lets start with 2 statement.

statement 2 : wz + xy = odd
From given,
Odd/xz = Int => xz = odd.(since, Odd/odd = odd int)
Therefore,
(wz + xy)/xz = odd int

Statement 1: wx + yz = odd
Odd+even = odd or,
even+odd = odd.
lets apply some number,
w=1, x =1,
y=2, z =2
therefore, wx + yz = 1+4 = 5 odd.
apply the same value in the red highlighted equation,
(wz + xy)/xz => (2+2)/2 = 2 which is not odd.
Hence Insufficient.

Answer is B

Regards,
Uva.

abhasjha wrote:For wx Even = w must be Even & x can be ODD or even ....(i have a doubt here)

for wx to be even ... either w or x can be even and then result wx is even ?

We know that w/x is an Integer, that implies the possible scenarios:

=> EVEN/EVEN; where w=EVEN & X=EVEN
=> EVEN/ODD; where w=EVEN & X=ODD
=> ODD/ODD; where w=ODD & X=ODD--------> here, the resultant will be surely ODD

Invalid scenario: ODD/EVEN ---> this is never possible

Now, For wx is EVEN imples:

=> EVEN * EVEN; where w=EVEN & X=EVEN
=> EVEN * ODD; where w=EVEN & X=ODD
=> ODD * EVEN; where w=ODD & X=EVEN -------which is invalid scenario

Hence, "w" must be Even & "x" can be ODD or even