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

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.

Answer: E.

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

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?

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.

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

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?

Hi Bunuel
please help me in how to deal with the following expression: |x|−|y|<|x−y| in DS questions in simple way

Please read the entire thread to understand as it has been discussed at is-xy-0-1-x-3-y-5-x-y-2-0-2-x-y-x-y-86780.html#p651477 and is-xy-0-1-x-3-y-5-x-y-2-0-2-x-y-x-y-86780.html#p651854

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 :



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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