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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]

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12 Nov 2009, 19:09
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Is $$xy<0$$?

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

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

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Last edited by Bunuel on 05 Jul 2013, 15:21, edited 1 time in total.
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Re: Hard inequality: Is xy < 0 [#permalink]

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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.

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

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

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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?

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

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13 Nov 2009, 21:30
good question..keep posting these types
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Re: Hard inequality: Is xy < 0 [#permalink]

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13 Nov 2009, 22:07
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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.

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

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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]

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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]

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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]

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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]

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

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

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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]

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

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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]

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

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

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

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

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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

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