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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]

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New post 27 Jan 2012, 13:44
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A
B
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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, 04:57, edited 1 time in total.
Edited the OA.
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Re: If X>1 and Y>1, is X<Y? [#permalink]

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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.

Answer: B.
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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]

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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x^2/(xy+x)<1

(2) xy/(y^2-y)<1


In the original condition, there are 2 variables(x,y) and 1 equation(x>1 and y>1), which should match with the number of equations. so you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
In 1), from x^2/(xy+x)<1, x/(y+1)<1, x<y+1 is possible but you can't figure out x<y, which is not sufficient.
In 2), from xy/(y^2-y)<1, x/(y-1)<1, x<y-1 is possible. Also, in x<y-1<y, it is always x<y, which is yes and sufficient. Therefore, the answer is B.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: If X>1 and Y>1, is X<Y? [#permalink]

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New post 27 Jan 2012, 14:14
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.
Please confirm.
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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]

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New post 27 Jan 2012, 14:24
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.
Please confirm.


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.

Answer: E.
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Re: If X>1 and Y>1, is X<Y? [#permalink]

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New post 05 Jul 2013, 02:43
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.

Answer: B.


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]

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New post 05 Jul 2013, 03:01
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.

Answer: B.


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]

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New post 05 Jul 2013, 03:40
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]

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New post 05 Jul 2013, 03:45
[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]

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New post 05 Jul 2013, 03:48
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

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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]

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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]

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New post 16 Sep 2014, 19:56
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]

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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]

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New post 23 Jan 2016, 10:48
The question basically asks if x-y<0
Statement 1. (x^2)/(x(y+1))<1 Since x and y both>1 then divide by x and multuply by (y+1) to get x<y+1 ==>x-y<1 Hence x-y can be 0.5 for example or can be -2. Not sufficient

Statement 2. (xy)/(y(y-1)) Again since both x and y are positive we can divide by y and multiply by (y-1) to get x<y-1 Hence x-y<-1 and thus <0 Sufficient
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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] 23 Jan 2016, 10:48
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