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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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Is $$xy<0$$?

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

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

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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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]
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]
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]
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: Hard inequality: Is xy < 0 [#permalink]
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]
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
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Re: Hard inequality: Is xy < 0 [#permalink]
Temurkhon wrote:
Bunuel wrote:
Is $$xy<0$$?

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

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

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.

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

The answer is not B. Consider the following cases to discard: x=-1 and y=-2, for a NO answer and x=1 and y=-2 for an YES answer.

The property you are referring to says that inequality $$|x-y|\geq{|x|-|y|}$$ holds true for any values of x and y. How are you applying it to this problem?
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Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
Hi Bunuel
please help me in how to deal with the following expression: |x|−|y|<|x−y| in DS questions in simple way
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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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hatemnag wrote:
Hi Bunuel
please help me in how to deal with the following expression: |x|−|y|<|x−y| in DS questions in simple way

As for the theory or understanding behind the expression: |x|-|y| < |x-y|, you need to understand what do |x| and |x-y| mean in isolation.

|x| is the distance of 'x' from absolute 0 on the number line.

|x-y| represents the distance of 'x' from 'y' on the number line.

Thus, for S2, you are told that |x|-|y| < |x-y| ---> for such 'complex' absolute value inequality expressions, use numbers to guide you but for algebraic treatment, read along:

|x|-|y| < |x-y| ---> distance of x - distance of y < distance of x from y

Also, note that |any value| = always non-negative ---> |any number| $$\geq$$ 0

Thus, |x| -|y| < a non negative quantity ---> |x| - |y| < 0 (I took 0 as it the simplest and the most straightforward example of a non-negative number).

2 cases on the number line that follow the fact that |x| - |y| < 0 :

Attachment:

5-02-16 8-28-52 AM.jpg [ 32.91 KiB | Viewed 4306 times ]

For case 1, you get xy<0 but for case 2, you get xy> 0 ---> S2 is NOT sufficient as it is giving us 2 different answers to the same question asked.

Hope this helps.
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