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# If xy < 5, is x < 1 ?

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If xy < 5, is x < 1 ? [#permalink]

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07 Mar 2012, 15:11
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65% (hard)

Question Stats:

52% (03:04) correct 48% (01:22) wrong based on 160 sessions

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If xy < 5, is x < 1 ?

(1) |y| > 5
(2) x/y > 0

[Reveal] Spoiler:
I got the answer is C which is correct but it was a guess work. So here is how I did this question:

Statement 1

|y| > 5 i.e. y > 5 and y > -5 or y > 5 and y < -5.

As this statement doesn't say anything about x, its clearly insufficient.

Statement 2

x/y> 0 --> This statement only tells that x and y have the same sign. Therefore insufficient.

[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Dec 2012, 02:49, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is x < 1? [#permalink]

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07 Mar 2012, 15:24
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enigma123 wrote:
If xy < 5, is x < 1 ?

(1) $$|y| > 5$$
(2) $$\frac{x}{y}> 0$$

I got the answer is C which is correct but it was a guess work. So here is how I did this question:

Statement 1

|y| > 5 i.e. y > 5 and y > -5 or y > 5 and y < -5.

As this statement doesn't say anything about x, its clearly insufficient.

Statement 2

x/y> 0 --> This statement only tells that x and y have the same sign. Therefore insufficient.

|y| > 5 means that y<-5 or y>5.

If xy < 5, is x < 1 ?

(1) $$|y| > 5$$ --> if $$y=10$$ and $$x=0$$ then the answer is YES but if $$y=-10$$ and $$x=2$$ then the answer is NO. Not sufficient.

(2) $$\frac{x}{y}>0$$ --> $$x$$ and $$y$$ have the same sign. Still insufficient: if $$y=-2$$ and $$x=-1$$ then the answer is YES but if $$y=2$$ and $$x=2$$ then the answer is NO. Not sufficient.

(1)+(2) From (2) $$x$$ and $$y$$ have the same sign. Now, if from (1) $$y>5$$ then $$0<x<1$$ (in order $$xy<5$$ to hold true) and if from (1) $$y<-5$$ then $$-1<x<0$$ (again in order $$xy<5$$ to hold true). So in both cases $$x<1$$. Sufficient.

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Re: If xy < 5, is x < 1 ? [#permalink]

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19 Jun 2014, 06:29
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Re: If xy < 5, is x < 1 ? [#permalink]

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13 Mar 2016, 08:45
Here Clearly statements 1 and two are not alone sufficient as x>1 and X<1
but combining them => x<1 as they should be of same sign and |y|>5 => if y is negative x is sufficient
if y is positive => x<<1 as Y>5
hence C
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Re: If xy < 5, is x < 1 ?   [#permalink] 13 Mar 2016, 08:45
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