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or (wx+yz)/xz = int
1. wx+yz = odd
2 cases:
A)wx = odd & yz = even
B)wx = even & yz = odd
Can be easily shown to insufficient

2. (wz+xy)/xz = int and wz+xy = odd
Odd/(?) = integer
only possible if (?) = odd
and the integer must also be odd, since only odd = odd*odd

SUFF. _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

I think the answer for this should one will be B .

From A wx + yz is odd that means either

1) wx is odd and yz is even

if this condition is true then w/x is definitely odd but y/z can be even or odd e.g say y =100 z =50 so y/z =2 and in second scenario y =100 z=20 so y/z =5

or 2) wx is even and yz is odd in this case also same reason as above

so insufficient

From B

since w/x and y/z are integers therefore (w/x + y/z) should be integer

since (w/x + y/z) is integer that means ((wz + xy)/xz) should be integer

from condition B we know that wz + xy is odd and since ((wz + xy)/xz) is
integer therefore xz has to be odd because odd cannot be divided by even . hence odd/odd = odd

Therefore if wx is even then yz is odd, and if wx is odd yz = even

when wx is even and yz is odd

wx=even therefore w/x should be even as w and x are integer and so is w/x
Similarily yz is odd then y/z is odd.
Hence w/x +y/z is odd.

Same can be proved when wx is odd and yz is even

2) w/x +y/z = (wz+yx)/xz

Now it is given that wz + yx = odd and also w/x and y/z are integers therefore there sum is also an integer. when numerator is odd the number will be odd.

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