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if X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2) XY/Y^2-Y<1

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VP
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if X>1 and Y>1, is X<Y? (1) X^2/(XY+X)<1 (2) XY/Y^2-Y<1 [#permalink] New post 13 May 2008, 18:33
00:00
A
B
C
D
E

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(N/A)

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0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Q29.
if X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

Please explain your work
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Re: DS:x<y [#permalink] New post 13 May 2008, 18:43
goalsnr wrote:
Q29.
if X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

Please explain your work


i would go with B..

1) basically boils down to x<(y+1)..suppose x=y..then yeah but x=y and not x<y..insuff
2) basically boils down to x<y-1 or x+1<y..this is clearly sufficient..x<y..
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Re: DS:x<y [#permalink] New post 13 May 2008, 19:16
should be B.

Stmt1 reduces to x-y < 1
x=4.5, y=5 ==> x<y
x=5, y=4.5 ===> x>y
Insuff

Stmt2 reduces to y-x> 1
This ensures that y> x
Suff
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Re: DS:x<y [#permalink] New post 13 May 2008, 23:31
I agree with the above explanations - IMO B.
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Re: DS:x<y [#permalink] New post 14 May 2008, 03:49
B for me as well.

Now, I just want to absolutely confirm something with this quesion: for both the statements, we are able to rearrange the given inequality (i.e. cross multiply) only because we are told in the stem that x>1 and y>1, correct ?

If that was not the case, then we would have to consider multiple case of when the equality in the statement held ?
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Re: DS:x<y [#permalink] New post 14 May 2008, 04:04
Answer option B, similar to fresinha's working.
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Re: DS:x<y [#permalink] New post 14 May 2008, 09:39
Thank you All. I got B as well, but the OB stated was E.
The question is from Set 25.
Re: DS:x<y   [#permalink] 14 May 2008, 09:39
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