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.

good question..keep posting these types

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

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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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.

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).

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.

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

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)

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