Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 May 2016, 21:21

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5773

Kudos [?]: 70821 [7] , given: 9857

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

Show Tags

12 Nov 2009, 19:09
7
KUDOS
Expert's post
10
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

32% (03:07) correct 68% (01:37) wrong based on 837 sessions

HideShow timer Statistics

Is $$xy<0$$?

(1) $$\frac{x^3*y^5}{x*y^2}<0$$

(2) $$|x|-|y|<|x-y|$$
[Reveal] Spoiler: OA

_________________

Last edited by Bunuel on 05 Jul 2013, 15:21, edited 1 time in total.
Manager
Joined: 11 Sep 2009
Posts: 129
Followers: 5

Kudos [?]: 285 [12] , given: 6

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

12 Nov 2009, 19:44
12
KUDOS
1
This post was
BOOKMARKED
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.

VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 242 [2] , given: 31

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

12 Nov 2009, 19:50
2
KUDOS
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
Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5773

Kudos [?]: 70821 [2] , given: 9857

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

07 Mar 2012, 10:19
2
KUDOS
Expert's post
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?

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.
_________________
VP
Joined: 05 Mar 2008
Posts: 1473
Followers: 11

Kudos [?]: 242 [0], given: 31

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

13 Nov 2009, 21:30
good question..keep posting these types
Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5773

Kudos [?]: 70821 [0], given: 9857

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

13 Nov 2009, 22:07
Expert's post
1
This post was
BOOKMARKED
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.

_________________
VP
Status: There is always something new !!
Affiliations: PMI,QAI Global,eXampleCG
Joined: 08 May 2009
Posts: 1353
Followers: 16

Kudos [?]: 203 [0], given: 10

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

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
_________________

Visit -- http://www.sustainable-sphere.com/
Promote Green Business,Sustainable Living and Green Earth !!

Manager
Joined: 26 Dec 2011
Posts: 117
Followers: 1

Kudos [?]: 27 [0], given: 17

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

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?
Manager
Joined: 26 Dec 2011
Posts: 117
Followers: 1

Kudos [?]: 27 [0], given: 17

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

08 Mar 2012, 04:32
Thanks a lot for your prompt response! Its clear to me now!
Manager
Status: Do till 740 :)
Joined: 13 Jun 2011
Posts: 113
Concentration: Strategy, General Management
GMAT 1: 460 Q35 V20
GPA: 3.6
WE: Consulting (Computer Software)
Followers: 1

Kudos [?]: 8 [0], given: 19

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

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.
Math Expert
Joined: 02 Sep 2009
Posts: 33062
Followers: 5773

Kudos [?]: 70821 [0], given: 9857

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

18 Nov 2012, 05:08
Expert's post
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.
_________________
Intern
Joined: 07 May 2011
Posts: 42
GMAT 1: Q V
GMAT 2: Q V
Followers: 0

Kudos [?]: 14 [0], given: 11

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

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).
Intern
Joined: 06 Apr 2011
Posts: 13
Followers: 0

Kudos [?]: 2 [0], given: 292

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

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.
SVP
Joined: 05 Jul 2006
Posts: 1512
Followers: 5

Kudos [?]: 201 [0], given: 39

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

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.

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.
SVP
Joined: 05 Jul 2006
Posts: 1512
Followers: 5

Kudos [?]: 201 [0], given: 39

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

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
Intern
Joined: 05 Feb 2013
Posts: 29
Location: Ukraine
GMAT 1: 680 Q48 V35
Followers: 0

Kudos [?]: 8 [0], given: 3

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

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.

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
SVP
Joined: 05 Jul 2006
Posts: 1512
Followers: 5

Kudos [?]: 201 [0], given: 39

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

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.

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
Senior Manager
Joined: 13 May 2013
Posts: 472
Followers: 2

Kudos [?]: 130 [0], given: 134

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]

Show Tags

30 Jun 2013, 13:18
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)
Manager
Joined: 31 Mar 2013
Posts: 68
Location: United States
Followers: 0

Kudos [?]: 24 [0], given: 109

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

06 Nov 2013, 11:41
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.

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.
Director
Joined: 23 Jan 2013
Posts: 513
Schools: Cambridge'16
Followers: 2

Kudos [?]: 50 [0], given: 37

Re: Hard inequality: Is xy < 0 [#permalink]

Show Tags

07 Nov 2013, 00:13
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.

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] 07 Nov 2013, 00:13

Go to page    1   2    Next  [ 27 posts ]

Similar topics Replies Last post
Similar
Topics:
1 If xy < 0, is x/y > z ? (1) xyz < 0 (2) x > yz 2 17 Sep 2015, 00:53
8 Is xy < 0 ? 8 08 May 2012, 03:25
3 Is x less than y? (1) x-y+1<0 (2) x-y-1<0 7 26 Feb 2011, 16:40
Is x-y+1 greater than x+y-1 ? (1) x > 0 (2) y < 0 2 18 Aug 2010, 00:30
7 Is xy < 0 ? 12 04 Aug 2010, 15:17
Display posts from previous: Sort by