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Re: Hard inequality: Is xy < 0 [#permalink]
12 Nov 2009, 18:44

4

This post received KUDOS

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)

Re: Hard inequality: Is xy < 0 [#permalink]
12 Nov 2009, 18:50

2

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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

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
07 Mar 2012, 09:19

2

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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?

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.

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
07 Mar 2012, 09:05

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?

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
19 Nov 2012, 18:42

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).

Re: Hard inequality: Is xy < 0 [#permalink]
13 May 2013, 11:13

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.

Answer: E.

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.

Re: Hard inequality: Is xy < 0 [#permalink]
17 May 2013, 15:45

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.

Answer: E.

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

Re: Is xy<0 ? (1) x^3*y^5/x*y^2 <0 (2) |x|-|y|<|x-y [#permalink]
30 Jun 2013, 12:18

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

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

Re: Hard inequality: Is xy < 0 [#permalink]
06 Nov 2013, 10:41

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.

Answer: E.

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.

Re: Hard inequality: Is xy < 0 [#permalink]
06 Nov 2013, 23:13

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.

Answer: E.

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

gmatclubot

Re: Hard inequality: Is xy < 0
[#permalink]
06 Nov 2013, 23:13