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

This topic has 1 expert reply and 0 member replies

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

Post Sun Sep 24, 2017 2:37 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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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    Post Sun Sep 24, 2017 2:50 am
    Quote:
    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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    Thanked by: ardz24
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