If w, x, y and z are integers such that w/x and y/z are inte

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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?

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by GMATGuruNY » Sun Sep 24, 2017 2:50 am
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
Before we evaluate the two statements, we should examine how the question stem can be rephrased.

w/x + y/z = (wz + xy)/xz.
Since w/x and y/z are integers, their sum (w/x + y/z) is an integer.
Thus, (wz + xy)/xz must also be an integer.

The question becomes: Is integer w/x + y/z -- which can be rephrased as (wz + xy)/xz -- odd?

Statement 1: wx + yz = odd.
Let w=1, x=1, y=2 and z=2, so that wx + yz = 1*1 + 2*2 = 5.
Is w/x + y/z odd?
NO, since 1/1 + 2/2 = 2.

Let w=1, x=1, y=6, and z=3, so that wx + yz = 1*1 + 6*3 = 19.
Is w/x + y/z odd?
YES, since 1/1 + 6/3 = 3.
INSUFFICIENT.

Statement 2: wz + xy = odd.
Please note the values highlighted in red:
Just as 10/2=5 is a factor of 10, and 12/3=4 is a factor of 12, so too is (wz + xy)/xz a FACTOR of wz + xy.

Since wz + xy is odd, all of its factors must be odd.
Thus, (wz + xy)/xz must be odd.
SUFFICIENT.

The correct answer is B.
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by rashedais » Sun Nov 04, 2018 8:38 pm
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?
another way to think about this problem is:

Given
w/x is an integer so w=x * a
y/z is an integer so y=z * b
is w/x + y/z odd?
bt putting values
w/x + y/z = (x*a)/x+(z*b)/z= (a+b)= odd?
now the given statements

a. wx + yz = odd

putting these values
(x *a) x+ (z*b)z is odd
so this is a x^2 +bz^2
dont have any info about individual terms so not sufficient

b. wz + xy = odd
(x *a) z+ (z*b)x is odd
so this is xza+xzb = odd
xz(a+b)= odd
now we know that
Odd x Odd = Odd
so (a+b) = odd
hence sufficient

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by Jay@ManhattanReview » Sun Nov 04, 2018 11:04 pm
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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