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

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If x>1 and y>1, is x<y? [#permalink]

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New post 02 May 2011, 05:29
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A
B
C
D
E

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

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New post 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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New post 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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New post 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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New post 21 Jun 2012, 02:23
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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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New post 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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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What are GMAT Club Tests?
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Kudos [?]: 128753 [0], given: 12182

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