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Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y

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Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink] New post 12 Nov 2009, 18:09
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Last edited by Bunuel on 05 Jul 2013, 14:21, edited 1 time in total.
Added the OA
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Re: Hard inequality: Is xy < 0 [#permalink] New post 12 Nov 2009, 18:44
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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] New post 12 Nov 2009, 18:50
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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] New post 13 Nov 2009, 20:30
good question..keep posting these types
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Re: Hard inequality: Is xy < 0 [#permalink] New post 13 Nov 2009, 21:07
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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.

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Re: Hard inequality: Is xy < 0 [#permalink] New post 05 May 2011, 21:12
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] New post 07 Mar 2012, 09:05
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] New post 07 Mar 2012, 09:19
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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] New post 08 Mar 2012, 03:32
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] New post 17 Nov 2012, 22:37
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] New post 18 Nov 2012, 04:08
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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.

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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink] New post 19 Nov 2012, 18:42
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: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink] New post 18 Feb 2013, 22:54
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] New post 13 May 2013, 11:13
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] New post 15 May 2013, 04:16
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] New post 15 May 2013, 22:55
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] New post 17 May 2013, 15:45
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: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink] New post 30 Jun 2013, 12:18
Is xy<0?

Is either x or y negative?

(1)(x^3*y^5)/(x*y^2)<0

(x^3 - x^1)*(y^5 - y^2) < 0
(x^2)*(y^3) < 0

(x^2) is positive regardless of whether x is positive or negative. If (x^2)*(y^3) < 0 then (y^3) must be negative and because a number raised to an odd power holds its sign, y must be negative. However, we do not know whether x is positive or negative meaning ab could be (a)*(-b) = -ab OR (-a)*(-b) = ab.
INSUFFICIENT

(2) |x|-|y|<|x-y|

|x|-|y|<|x-y|
|-3| - |-2| < |-3 - (-2)| = 3-2 < |-1| ===> 1<1 Invalid
|3| - |-2| < |3-(-2)| = 3-2 < 5 ===> 1<5 VALID (x, -y)
|2| - |3| < |2-3| = -1 < 1 ===> -1<1 VALID (x, y)
|3| - |2| < |3-2| = 3-2 < 1 ===> 1<1 Invalid
|-3| - |2| < |-3-2| = 3-2 < |-5| ===> 1<5 VALID (-x, y)
INSUFFICIENT

1+2)
|x|-|y|<|x-y|

|2| - |-3| < |2-(-3)|
2-3 < 5 ===> 1<5 Sufficient (x = positive, y = negative)
|-2| - |-3| < |-2 - (-3)|
-1 < 1 ===> -1 < 1 Sufficient (x = negative, y = negative)

There are two valid statements where x and y are negative and where x is positive and y is negative meaning we don't know whether ab is negative or positive.
INSUFFICIENT

(E)
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Re: Hard inequality: Is xy < 0 [#permalink] New post 06 Nov 2013, 10:41
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.


Bunuel, I'm finding this confusing. Can you help me out. Your post on this thread is-x-y-x-y-123108.html says that for statement 2 to be true both x and y need to have the same sign.

Now, this question if you look at statement 2, it says that x and y have opposite signs. Hence, it is never possible for |x|-|y|<|x-y| to be true. So statement 2 seems sufficient. Where is my error in reasoning? Thank you.
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Re: Hard inequality: Is xy < 0 [#permalink] New post 06 Nov 2013, 23:13
emailmkarthik 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.


Bunuel, I'm finding this confusing. Can you help me out. Your post on this thread is-x-y-x-y-123108.html says that for statement 2 to be true both x and y need to have the same sign.

Now, this question if you look at statement 2, it says that x and y have opposite signs. Hence, it is never possible for |x|-|y|<|x-y| to be true. So statement 2 seems sufficient. Where is my error in reasoning? Thank you.




I agree. Inequality rule saying that |x|-|y|<|x-y| is true only when x and y are having different signs exists. Does not it? If so, B is correct
Re: Hard inequality: Is xy < 0   [#permalink] 06 Nov 2013, 23:13
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