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

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Senior Manager
Joined: 03 Mar 2010
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If x>1 and y>1, is x<y? [#permalink]

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02 May 2011, 05:29
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Difficulty:

75% (hard)

Question Stats:

21% (00:52) correct 79% (01:14) wrong based on 39 sessions

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

(1) $$\frac{x^2}{(xy+x)} <1$$

(2) $$\frac{xy}{(y^2-y)} <1$$
[Reveal] Spoiler: OA

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Kudos [?]: 361 [0], given: 22

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Re: If x>1 and y>1, is x<y? [#permalink]

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02 May 2011, 05:42
I think its B. The answer is YES

Reason :
S2 reduces to x / (y - 1) < 1

or x + 1 < y

I can cross multiply since x and y are both positive

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

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Re: If x>1 and y>1, is x<y? [#permalink]

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02 May 2011, 18:19
(1)

x < y+1

Not sufficient

(2)

x < y - 1

=> x+1 < y

=> y-x > 1

=> x < y

Sufficient

Answer - B

Could you please recheck the OA/OE ?
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Re: If x>1 and y>1, is x<y? [#permalink]

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12 Aug 2011, 01:02
a x=y=2 N x=2,y=3 Y not sufficient.

b x=2,y=4 Y x=1.1,y= 2.2 Y sufficient.

B it is.
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Re: If x>1 and y>1, is x<y? [#permalink]

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21 Jun 2012, 02:23
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Expert's post
pavanpuneet wrote:
Can someone confirm the OA, it is B according to me as well;

For the following question the 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.

Answer: B.
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Re: If x>1 and y>1, is x<y? [#permalink]

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21 Jun 2012, 02:23
pavanpuneet wrote:
Can someone confirm the OA, it is B according to me as well;

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 $$y$$: $$\frac{x}{y}-y<1$$ --> $$x<y^2$$ --> 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 $$y=\sqrt{3}\approx{1.7}$$ (notice that this set of numbers satisfy both statements), then the answer is NO. Not Sufficient.

Answer: E.
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Re: If x>1 and y>1, is x<y?   [#permalink] 21 Jun 2012, 02:23
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# If x>1 and y>1, is x<y?

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