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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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If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  27 Jan 2012, 12:44
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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

I did B. But it is not OA.
[Reveal] Spoiler: OA

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Last edited by Bunuel on 23 May 2013, 03:57, edited 1 time in total.
Edited the OA.
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Re: If X>1 and Y>1, is X<Y? [#permalink]  27 Jan 2012, 13:12
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Expert's post
Baten80 wrote:
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I did B. But it is not OA.

OA is wrong. Answer is B.

If x>1 and y>1, is x<y?

(1) $$\frac{x^2}{xy+x}<1$$ --> reduce by $$x$$: $$\frac{x}{y+1}<1$$ --> cross multiply, notice that we can safely do that since $$y+1>0$$: $$x<y+1$$ --> if $$x=2$$ and $$y=2$$, then the answer is NO but if $$x=2$$ and $$y=3$$, then the answer is YES. Not sufficient.

(2) $$\frac{xy}{y^2-y}<1$$ --> reduce by $$y$$: $$\frac{x}{y-1}<1$$ --> cross multiply, notice that we can safely do that since $$y-1>0$$: $$x<y-1$$ --> $$x+1<y$$ ($$y$$ is more than $$x$$ plus 1) --> $$y>x$$. Sufficient.

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Re: If X>1 and Y>1, is X<Y? [#permalink]  27 Jan 2012, 13:14
Baten80 wrote:
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I did B. But it is not OA.

In statement 2, the inequality is well written?
Is this the original expresion?
(XY/Y^2) -Y<1

or?
XY/(Y^2-Y)<1

If it is the first scenario, the answer is E.
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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]  27 Jan 2012, 13:24
Expert's post
metallicafan wrote:
Baten80 wrote:
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I did B. But it is not OA.

In statement 2, the inequality is well written?
Is this the original expresion?
(XY/Y^2) -Y<1

or?
XY/(Y^2-Y)<1

If it is the first scenario, the answer is E.

If (2) is $$\frac{xy}{y^2}-y<1$$, then the answer is indeed E.

If x>1 and y>1, is x<y?

(1) $$\frac{x^2}{xy+x}<1$$ --> reduce by $$x$$: $$\frac{x}{y+1}<1$$ --> cross multiply, notice that we can safely do that since $$y+1>0$$: $$x<y+1$$ --> if $$x=2$$ and $$y=2$$, then the answer is NO but if $$x=2$$ and $$y=3$$, then the answer is YES. Not sufficient.

(2) $$\frac{xy}{y^2}-y<1$$ --> reduce by $$y$$: $$\frac{x}{y}-y<1$$ --> $$x<y^2+y$$ --> if $$x=2$$ and $$y=2$$, then the answer is NO but if $$x=2$$ and $$y=3$$, then the answer is YES. Not sufficient.

(1)+(2) If $$x=2$$ and $$y=3$$, then the answer is YES but If $$x=2$$ and $$y=2$$ (notice that this set of numbers satisfy both statements), then the answer is NO. Not Sufficient.

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Re: If X>1 and Y>1, is X<Y? [#permalink]  05 Jul 2013, 01:43
Bunuel wrote:
Baten80 wrote:
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I did B. But it is not OA.

OA is wrong. Answer is B.

If x>1 and y>1, is x<y?

(1) $$\frac{x^2}{xy+x}<1$$ --> reduce by $$x$$: $$\frac{x}{y+1}<1$$ --> cross multiply, notice that we can safely do that since $$y+1>0$$: $$x<y+1$$ --> if $$x=2$$ and $$y=2$$, then the answer is NO but if $$x=2$$ and $$y=3$$, then the answer is YES. Not sufficient.

(2) $$\frac{xy}{y^2-y}<1$$ --> reduce by $$y$$: $$\frac{x}{y-1}<1$$ --> cross multiply, notice that we can safely do that since $$y-1>0$$: $$x<y-1$$ --> $$x+1<y$$ ($$y$$ is more than $$x$$ plus 1) --> $$y>x$$. Sufficient.

Hi Bunuel,

Can you explain statement 2 using numbers ?

Rohan
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Re: If X>1 and Y>1, is X<Y? [#permalink]  05 Jul 2013, 02:01
Expert's post
RohanKhera wrote:
Bunuel wrote:
Baten80 wrote:
If X>1 and Y>1, is X<Y?
(1) X^2/(XY+X)<1
(2) XY/Y^2-Y<1

I did B. But it is not OA.

OA is wrong. Answer is B.

If x>1 and y>1, is x<y?

(1) $$\frac{x^2}{xy+x}<1$$ --> reduce by $$x$$: $$\frac{x}{y+1}<1$$ --> cross multiply, notice that we can safely do that since $$y+1>0$$: $$x<y+1$$ --> if $$x=2$$ and $$y=2$$, then the answer is NO but if $$x=2$$ and $$y=3$$, then the answer is YES. Not sufficient.

(2) $$\frac{xy}{y^2-y}<1$$ --> reduce by $$y$$: $$\frac{x}{y-1}<1$$ --> cross multiply, notice that we can safely do that since $$y-1>0$$: $$x<y-1$$ --> $$x+1<y$$ ($$y$$ is more than $$x$$ plus 1) --> $$y>x$$. Sufficient.

Hi Bunuel,

Can you explain statement 2 using numbers ?

Rohan

Number plugging is not the best approach to prove that a statement IS sufficient. On DS questions when plugging numbers, goal is to prove that the statement is NOT sufficient.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  05 Jul 2013, 02:40
Hi bunuel
In first statement if x<y+1 can't we say that x<y because even though we add 1 to y x is less than y
Can u plz explain

Thanks

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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  05 Jul 2013, 02:45
[quote="sravanigayatri"]Hi bunuel
In first statement if x<y+1 can't we say that x<y because even though we add 1 to y x is less
than y
Can u plz

They both could be equal.

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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  05 Jul 2013, 02:48
Expert's post
sravanigayatri wrote:
Hi bunuel
In first statement if x<y+1 can't we say that x<y because even though we add 1 to y x is less than y
Can u plz explain

Thanks

Posted from GMAT ToolKit

No, unfortunately that's not correct: x<y+1, does not necessarily means that x<y.

Consider x=2 and y=1.5 --> x<y+1 and x>y.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  16 Sep 2014, 18:39
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  16 Sep 2014, 18:56
Bunuel : how does x/y-y<1 reduce to x<y^2?
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink]  16 Sep 2014, 23:12
Expert's post
bankerboy30 wrote:
Bunuel : how does x/y-y<1 reduce to x<y^2?

There was a typo. Should be x < y^2 + y.
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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1   [#permalink] 16 Sep 2014, 23:12
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