IMO E

1) xy + xz is even

so xy and xz both even (a)
or both odd(b)

2) y + xz is odd

y odd, xz even(c)

xz odd, y even(d)

combine 1 and 2

combine (a) and (c)

xz =even
so y = odd

combine (a) and (d)

xz = even
and xz =0dd, contradictory, so this combination is not possible


combine (b) and (c)
xz =odd and xz even

contradictory, so this combination is also not possible

combine (b) and (d)

xz odd and y even

so two case arise:

y =odd, when xz = even
y = even when xz = odd

It is given xy+z is an odd integer
so (case I )xy =odd and z = even
(case II) xy =even and z = odd


Case I: xy =odd, Y can be even as well as odd

y= even,so xz has to be odd
But x =even

so case I is not possible


Case II:

xy = even and z = odd

if y = even, so xz = odd

so x has to be odd because z = odd and xz =odd

if y = odd, then xz =even

but since in CASE II z= odd
so x has to be even


thus we are getting 2 different values of x(even as well as odd)

so answer is E

I did not have time to go through the long explanation, esp because your answer was wrong.
Let me post the explanation for the OA is A.
The Q:
if x,y and z are integers and xy+z is an odd integer, is x an even integer?
1) xy + xz is even
2) y + xz is odd

Given: xy + z = O => xy = O - z; where O represents an odd number
=> if z is odd then xy is even; else if z is even xy is odd.

1) xy + xz = E; where E represents an even number
=> O - z + xz = E; since xy = O - z.
=>z(x-1) = E - O = O => z is O and x-1 is O => x is even. So 1 is sufficient

2) y + xz = O => xz = O - y
If y is even; xz is odd => x is odd
If y is odd; xz = even ---- does not say anything about x.
So x is not definite to be a even or odd. So 2 is insufficient.
Thus OA is A