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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] New post 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] New post 27 Jan 2012, 13:12
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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? [#permalink] New post 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.
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] New post 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.
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] New post 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.

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] New post 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.

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] New post 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] New post 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] New post 05 Jul 2013, 02:48
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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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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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Re: If x>1 and y>1, is x<y? (1) x^2/(xy+x)<1 (2) xy/(y^2-y)<1 [#permalink] New post 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] New post 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] New post 16 Sep 2014, 23:12
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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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