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

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If w, x, y, and z are integers such that w/x and y/z are [#permalink] New post 25 May 2006, 20:26
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D
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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 + xy is odd.
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 [#permalink] New post 25 May 2006, 20:58
B?

w/x+y/z = int

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.
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 [#permalink] New post 25 May 2006, 21:18
This is indeed trick . But i will try .

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

Hence B should be the answer .

Curious to know the answer.
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 [#permalink] New post 25 May 2006, 21:23
Agree with giddi's solution. Its B.
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 [#permalink] New post 25 May 2006, 22:44
D for me too.

1) given wx + yz = 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.

Hence D
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 [#permalink] New post 26 May 2006, 03:34
jaynayak wrote:
D for me too.



wx=even therefore w/x should be even as w and x are integer and so is w/x

Hence w/x +y/z is odd.



how about w = 100 x = 20, so wx is even but w/x = 5 not even ??
  [#permalink] 26 May 2006, 03:34
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