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If xy > 0, does (x-1)(y-1) = 1? (1) x + y = xy (2) x = y [#permalink]
 14 Apr 2008, 13:47
Question Stats:
0% (00:00) correct
 0% (00:00) wrong based on 0 sessions
If xy > 0, does (x-1)(y-1) = 1?
(1) x + y = xy
(2) x = y
So I got this one wrong, but would like to find out where I went wrong. I'll give you folks some time to show your approaches.
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Re: OG 11 Quant Review DS 80 [#permalink]
14 Apr 2008, 16:21
The answer I got is A.
(x-1)(y-1) = xy-(x+y)+1. Since from stat 1 x+y = xy, then (x-1)(x+1)=1
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Re: OG 11 Quant Review DS 80 [#permalink]
14 Apr 2008, 17:01
Did same approach mentioned above.
plugged some numeric values for stmt 2
xy - (x+y) +1 = 1 for x = y = 2
while it does not hold for 3.
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Re: OG 11 Quant Review DS 80 [#permalink]
14 Apr 2008, 17:03
Stat1: xy - (x+y) + 1 = xy - xy + 1 = 1 Stat2: (x-1)*(y-1) becomes (x-1)*(x-1) so x (and y) can be any number other than 0 and satisfy the xy>0 statement, but only 2 (and 0, which we just counted out) satisfy (x-1)*(y-1)=1... thus, Stat2 is out. Stat1 [A]
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Re: OG 11 Quant Review DS 80 [#permalink]
14 Apr 2008, 17:29
ok, cool. youre all on the right track so far. but here is my question:
i simplified the stem down to asking if xy-x-y=0
i see from statement 1 that its suffcient.
now, for statement 2, i subbed in x=y to get y^2-2y=0 --> y(y-2)=0
so for that to be true, either y is 0 or y is 2. now we are told that xy>0, so y cannot be 0. Therefore, y must be 2, and (x-1)(y-1) = (2-1)(2-1) = 1*1 = 1
why isnt that approach for statement 2 correct ?
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Re: OG 11 Quant Review DS 80 [#permalink]
14 Apr 2008, 18:33
pmenon,
your approach did confuse me at first.
your question is valid.
the answer is....you are restricting value of x to be 0 or 2 to satisfy the equality of question stem, one they are asking if can be true.
stem 2 says x=y doesnt restrict valu of x in any way..
Not sure above makes sense...there sure is easier explanation
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Re: OG 11 Quant Review DS 80 [#permalink]
15 Apr 2008, 03:22
well according to the question, xy>0, so we have to take into account when both the variables are negative and when both the variables are positive for statement 2:
xy = x + y
if both x and y are positive, then:
x^2 = 2x
x^2 - 2x = 0
x(x-2) = 0
so x could be 0 or 2, but because none of the variables is suppose to be 0, then x is 2
if both x and y are negative, then:
x^2 = -2x
x^2 + 2x = 0
x(x+2) = 0
so x could be 0 or -2, but because none of the variables is suppose to be 0, then x is -2.
so statement 2 is not suff. Also, because we already found statement 1 to be suff., the answer should be A.
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Re: OG 11 Quant Review DS 80 [#permalink]
16 Apr 2008, 22:11
Agree with A - similar method to first poster.
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Re: OG 11 Quant Review DS 80
[#permalink]
16 Apr 2008, 22:11
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