BTGmoderatorAT 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 is odd
(2) wz + yx is odd
Any of the options alone is sufficient? Can any expert explain?
Given: w, x, y and z are integers such that w/x and y/z are integers.
Question: Is (w/x + y/z) odd?
Let's take each statement one by one.
(1) wx + yz is odd
=> Sum of two integers is odd. The sum of two integers can be odd if one of the integers is even and the other is odd.
Case 1: Say wx is odd and yz is even
(a) wx is odd
=> Since if the product of two integers is odd, both the integers are odd; thus, w and x are odd.
(a) yz is even
=> Since if the product of two integers is even, at least one of the integers must be even; thus, y and z are both odd.
Example 1: Say w = x = 1, and y = 6 (taking y as even), and z = 3 (taking z as odd)
=> w/x + y/z => 1/1 + 6/3 = 1 + 2 = 3, odd. The answer is Yes.
Example 2: Say w = x = 1, and y = 6 (taking y as even), and z = 2 (taking z as even)
=> w/x + y/z => 1/1 + 6/2 = 1 + 3 = 4, even. The answer is No.
No unique answer. Insufficient.
(2) wz + yx is odd
Let's take the expression to be tested: w/x + y/z
Taking LCM, we get
w/x + y/z = (wz + xy) / xz
From Statement 2, we know that (wz + xy) is odd and the prompt states that (wz + xy) / xz is integer, thus, (wz + xy) is completely divisible by xz.
To get an integer as the result, upon diving an odd integer, the denominator must also be an odd number. Thus, xz is odd.
(wz + xy) / xz => Odd / Odd = Odd. Sufficient.
The correct answer:
B
Hope this helps!
-Jay
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