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Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
 12 Nov 2009, 19:09
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Question Stats:
35% (02:42) correct
64% (01:33) wrong based on 40 sessions
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Re: Hard inequality: Is xy < 0 [#permalink]
12 Nov 2009, 19:44
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Statement 1: \frac{x^3*y^5}{x*y^2} < 0
x^2*y^3 < 0
We know that x^2 must be positive, so therefore y < 0. Since we do not know whether x is positive or negative, however, this is insufficient.
Statement 2: |x|-|y| < |x-y|
I'll approach this the way I would as if I was writing it on the GMAT with time constraints, by picking numbers:
(1) Is it possible for xy > 0 AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = 3)
(2) Is it possible for xy < 0 AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = -3)
Therefore, not sufficient.
Evaluating Both Statements
We know from Statement 1 that y < 0. Therefore we can revise the above approach for Statement 2 to specifics:
(1) Is it possible for y < 0, x < 0 (i.e. xy > 0) AND the above criteria to be satisfied? Yes. (i.e. x = -2, y = -3)
(2) Is it possible for y < 0, x > 0 (i.e. xy < 0) AND the above criteria to be satisfied? Yes. (i.e. x = 2, y = -3)
Therefore, insufficient.
The answer is E.
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Re: Hard inequality: Is xy < 0 [#permalink]
12 Nov 2009, 19:50
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Bunuel wrote: Is xy<0?
(1) \frac{x^3*y^5}{x*y^2}<0
(2) |x|-|y|<|x-y| I'm going with e on this one (1) \frac{x^3*y^5}{x*y^2}<0 x^3-1 * y*5-2 < 0 x^2*y^3<0 y < 0 but x<0 or x > 0 so insufficient because it doesn't matter what x is the answer will be negative however plugging back into the original equation gives two different answers (2) |x|-|y|<|x-y|[/quote] if x = 1 and y = -2 |1| - |-2| < |1 + 2| if x>0 and y > 0 then the equations will be equal if x> 0 and y < 0 yes if x<0 and Y < 0 yes and both would make xy<0 a different answer
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Re: Hard inequality: Is xy < 0 [#permalink]
13 Nov 2009, 21:30
good question..keep posting these types
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Re: Hard inequality: Is xy < 0 [#permalink]
13 Nov 2009, 22:07
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Re: Hard inequality: Is xy < 0 [#permalink]
05 May 2011, 22:12
a gives x can > or < 0. y is <0. not sufficient. b x>0,y>0 or x>0 and y< 0 both satisfies. Not sufficient. a+b both x >0 | x<0 and y <0. Not sufficient. Thus E
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
07 Mar 2012, 10:05
For the second case, I tried to solve by squaring on both the sides. So you get, -2|x||y|<-2xy, which is ....|x||y|>xy...only case possible is when one of them is negative... will it then not prove that xy<0. Please let me know where am I wrong?
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
07 Mar 2012, 10:19
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pavanpuneet wrote: For the second case, I tried to solve by squaring on both the sides. So you get, -2|x||y|<-2xy, which is ....|x||y|>xy...only case possible is when one of them is negative... will it then not prove that xy<0. Please let me know where am I wrong? Welcome to GMAT Club. Below is an answer to your question. You can square both sides of an inequality if and only both sides are non-negative. For |x|-|y|<|x-y| we know that |x-y| is non-negative, but |x|-|y| can be negative, as well as positive (non-negative), hence you cannot apply squaring. GENERAL RULE: A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).For example: 2<4 --> we can square both sides and write: 2^2<4^2; 0\leq{x}<{y} --> we can square both sides and write: x^2<y^2; But if either of side is negative then raising to even power doesn't always work. For example: 1>-2 if we square we'll get 1>4 which is not right. So if given that x>y then we can not square both sides and write x^2>y^2 if we are not certain that both x and y are non-negative. B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).For example: -2<-1 --> we can raise both sides to third power and write: -2^3=-8<-1=-1^3 or -5<1 --> -5^2=-125<1=1^3; x<y --> we can raise both sides to third power and write: x^3<y^3. Hope it helps.
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
08 Mar 2012, 04:32
Thanks a lot for your prompt response! Its clear to me now!
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
17 Nov 2012, 23:37
Hi bunuel,
) Holds true when:
A. 0<x<y
B. x<0<y
C. y<x<0
D. y<0<x
pls let me know how this works
when you have |x-y|>|x| -|y|
if we take x=8 y=4
this will be always equal.|x-y| can be never be greater than |x| -|y| if both x and y have same signs
Please let me know how you deduced it.
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
18 Nov 2012, 05:08
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
19 Nov 2012, 19:42
I really like one of the ideas forwarded above.
data point 1 was fairly obvious.
with data point 2, we need only to stare at it and the question xy<0 for a few seconds.
the more you think about it the more it becomes obvious that you would need to pick numbers such that xy<0 can be proven and xy<0 can be debunked.
you could do this by using x>0 and y<0 (to prove xy<0) and using x>0,y>0 (to debunk).
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Re: Hard inequality: Is xy < 0 [#permalink]
19 Nov 2012, 19:44
amit2k9 wrote: a gives x can > or < 0. y is <0. not sufficient.
b x>0,y>0 or x>0 and y< 0 both satisfies. Not sufficient.
a+b both x >0 | x<0 and y <0. Not sufficient.
Thus E This is very well done. Quick, clean reasoning. I love it.
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
18 Feb 2013, 23:54
excellent question bunuel. after 3 mins of trying and testing numbers i frustatingly put C as an answer....  .. Thanks for explanation.
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Re: Hard inequality: Is xy < 0 [#permalink]
13 May 2013, 12:13
Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when:
A. 0<x<y
B. x<0<y
C. y<x<0
D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Hi Bunuel, I want to double check the following with you ... the 4 cases u ve mentioned is detailed version of the following or not for the inequality to hold true 1) xy<0 2) /y/>/x/ and this is because the equality (=) in /y-x/ > = /x/ - /y/ holds true only in 2 cases /x/ = /y/ or /x/>/y/ and xy>0 am i right ?? plz let me know.. thanking you in advance.
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
15 May 2013, 05:16
Bunuel wrote: shankar245 wrote: Hi bunuel,
) Holds true when:
A. 0<x<y
B. x<0<y
C. y<x<0
D. y<0<x
pls let me know how this works
when you have |x-y|>|x| -|y|
if we take x=8 y=4
this will be always equal.|x-y| can be never be greater than |x| -|y| if both x and y have same signs
Please let me know how you deduced it. No, the red part is not correct. Consider x=4 and y=8 OR x=-4 and y=-8. x=4 , y = 8 /4-8/ = 4 , /4/ - /8/ = -4 x=-4, y=-8 /-4-(-8)/ = 4 , /-4/-/-8/ = 4-8 = -4
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Re: Hard inequality: Is xy < 0 [#permalink]
15 May 2013, 23:55
Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when:
A. 0<x<y
B. x<0<y
C. y<x<0
D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Hi Bunuel, Can you explain how you got the following set of inequality? : A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x Thanks
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Re: Hard inequality: Is xy < 0 [#permalink]
17 May 2013, 16:45
Sergiy wrote: Bunuel wrote: Yes, OA is E.
Number plugging as AKProdigy87 did is probably the best way here. But still if needed below is short overview of algebra of this question:
(1) y<0. Not sufficient.
(2) Holds true when:
A. 0<x<y
B. x<0<y
C. y<x<0
D. y<0<x
Not sufficient.
(1)+(2) C. and D. options are left from (2) but still insufficient, as xy may or may not be negative.
Answer: E. Hi Bunuel, Can you explain how you got the following set of inequality? : A. 0<x<y B. x<0<y C. y<x<0 D. y<0<x Thanks /x-y/ >= /x/-/y/ is always true ( try it ) and the equality (equal sign) holds in 2 cases 1) when xy> 0 ( same sign) and /x/>/y/ ( both conditions together) 2) x=y=0 Now , the inequality holds true in any other case (/x-y/ > /x/-/y/) i.e. xy<0 ( different signs) or /x/</y/ if u translate this last bit into inequalities it gives the above 4 scenarios that i reformatted below A. xy>0 but /y/> /x/ b. xy<0 , y>x c. xy<0 and /y/ > /x/ d. xy<0 , x>y hope this helps
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Re: Hard inequality: Is xy < 0
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17 May 2013, 16:45
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