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# Is xy < 0 ?

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Is xy < 0 ? [#permalink]  04 Aug 2010, 14:17
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Is xy < 0 ?

(1) 1/x < 1/y
(2) x > 0
[Reveal] Spoiler: OA

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Re: Need solution ! [#permalink]  04 Aug 2010, 22:01
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nusmavrik wrote:
Is xy < 0 ?
1. 1/x < 1/y
2. x > 0

Is xy < 0 ?

Notice that the question basically asks whether x and y have the opposite signs.

(1) $$\frac{1}{x}<\frac{1}{y}$$ --> $$x$$ and $$y$$ can have the same sign ($$x=2$$ and $$y=1$$) as well as opposite signs ($$x=-2$$ and $$y=1$$). Not sufficient.

(2) $$x>0$$ --> no info about $$y$$. Not sufficient.

(1)+(2) As from (2) $$x>0$$ then $$\frac{1}{x}<\frac{1}{y}$$ to hold true, $$y$$ must also be positive (left hand side, $$\frac{1}{x}$$ is positive and right hand side, $$\frac{1}{y}$$, to be greater than that, must also be positive). So $$x$$ and $$y$$ have the same sign, they are both positive: $$xy>0$$. Therefore the answer to the question is NO. Sufficient.

Hope it helps.
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Re: Need solution ! [#permalink]  04 Aug 2010, 23:06
Thanks Bunuel ! This question was a hard trap. Thanks for the awesome explanation.
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Re: Is xy < 0 ? [#permalink]  22 Feb 2014, 12:09
Expert's post
Bumping for review and further discussion.
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Re: Is xy < 0 ? [#permalink]  01 Mar 2014, 11:11
Hey Bunuel,
Can you please let me know where I'm going wrong in the following method:
Statement1:
1/x < 1/y
Subtract 1/y from both sides: we get (y-x)/xy <0.
Therefore, either xy<0 or (y-x)<0.
If xy<0 then the answer to our question is YES both are opposite signs.
If y-x<0 -> y<x, then both can have same sign or opposite signs (ie.., y<x<0 OR y<0<x OR 0<y<x)
Therefore, Statement1 is INSUFFICIENT.
Statement2: 0<x so x is Positive but no information about y, So statement 2 is INSUFFICIENT.
(1) and (2) together:
We still have y<0<x or 0<y<x
Therefore, Both statements together are INSUFFICIENT. So E.
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Re: Is xy < 0 ? [#permalink]  02 Mar 2014, 03:55
Expert's post
prsnt11 wrote:
Hey Bunuel,
Can you please let me know where I'm going wrong in the following method:
Statement1:
1/x < 1/y
Subtract 1/y from both sides: we get (y-x)/xy <0.
Therefore, either xy<0 or (y-x)<0.
If xy<0 then the answer to our question is YES both are opposite signs.
If y-x<0 -> y<x, then both can have same sign or opposite signs (ie.., y<x<0 OR y<0<x OR 0<y<x)

Therefore, Statement1 is INSUFFICIENT.
Statement2: 0<x so x is Positive but no information about y, So statement 2 is INSUFFICIENT.
(1) and (2) together:
We still have y<0<x or 0<y<x
Therefore, Both statements together are INSUFFICIENT. So E.

$$\frac{y-x}{xy} <0$$ means that y-x and xy have the opposite sign: +- or -+.

When combined we know that x is positive. Now, if y were negative, then $$xy<0$$, thus $$y-x$$ must be positive, but in this case $$y-x=negative-positive=negative$$, thus this case is not possible, y is NOT negative --> y is positive --> $$xy=positive$$.

Hope it's clear.
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Re: Is xy < 0 ? [#permalink]  02 Mar 2014, 17:08
Thanks Bunuel for this great explanation!
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Re: Is xy < 0 ? [#permalink]  02 Mar 2014, 22:42
nusmavrik wrote:
Is xy < 0 ?

(1) 1/x < 1/y
(2) x > 0

Statement I is insufficient:

x = 2, y = 1 (1/2 < 1/1) (xy > 0) (NO)
x = -1, y = 2 (1/-2 < 1/2) (xy <0) (YES)

Statement II is insufficient:
y can be negative or positive

Combining is sufficient:

(1/x) < (1/y)
If 1/x is positive then 1/y is also positive which means y is also positive. Hence xy is greater than zero
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Re: Is xy < 0 ? [#permalink]  15 Jun 2015, 08:15
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Is xy < 0 ? [#permalink]  16 Jun 2015, 08:09
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Is xy < 0 ?
It's sometimes helpful to think of disproving rather than proving the statement. For xy to be negative, x and y have to be opposite signs.
(1) 1/x < 1/y For this statement consider x and y being both positive or both negative. For example, (1/4) <(1/2) or (-1/2) < (-1/4). Not sufficient.

(2) x > 0 We're given nothing about y so it could be positive as well. Not sufficient.

Multiply both sides of (1/x) < (1/y) by x to give (x/y) > 1 From statement 2 we know that x is positive. For x/y to be greater than one y also has to be positive. Sufficient (no)
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Is xy < 0 ?   [#permalink] 16 Jun 2015, 08:09
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