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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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New post 07 Mar 2012, 15:11
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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.

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, 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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New post 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.

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]

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

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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]

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New post 01 Jul 2017, 10:54
Bunuel wrote:
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.


May not be the easiest solution, but I find it interesting that these inequalities can be rigorously solved using graphical visualisation.

Given : xy<5 is the area between the hyperbola.

1) |y| > 5: When y>5, x can only be less than 1. When y<-5, x can only be greater than -1. Thus not sufficient.
2) Same sign, so not sufficient.

Combining.
When y>5, x can only be less 1 BUT greater than 0. So x<1? YES
When y<-5, x can only be greater than -1 BUT less than 0. So, x<1? YES

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Re: If xy < 5, is x < 1 ?   [#permalink] 01 Jul 2017, 10:54
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