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If xy < 5, is x < 1 ?

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If xy < 5, is x < 1 ? [#permalink] New post 07 Mar 2012, 14:11
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48% (03:07) correct 52% (01:19) wrong based on 103 sessions
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.

After this I struggled to complete the question, and guessed answer C as correct answer. Can someone please help?
[Reveal] Spoiler: OA

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Last edited by Bunuel on 04 Dec 2012, 01:49, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Is x < 1? [#permalink] New post 07 Mar 2012, 14:24
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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.

After this I struggled to complete the question, and guessed answer C as correct answer. Can someone please help?


|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.

Answer: C.
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Re: If xy < 5, is x < 1 ? [#permalink] New post 19 Jun 2014, 05:29
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Re: If xy < 5, is x < 1 ?   [#permalink] 19 Jun 2014, 05:29
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