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i had a slightly slightly different method for statement 1. Do you mind letting me know if the steps I took are OK?

also, what level question is this? My test is coming up soon and I've done all the OG and quant review 2nd edition problems. I wonder if going through these two samurai guides PS/DS would be the best use of my time in the final month and half.

(1). x/|x| < x before even going to x<0 and x>0, i knew that |x| is positive, so I multiplied it across x < x*|x|

case x < 0

divide both sides by x, inequality must flip because x is negative 1 > |x| <-- this says YES to what we asked for

case x > 0

again, divide both sides by x, inequality won't flip 1 < |x| <-- this says NO

insufficient

(2). clearly says x has to be negative, therefore using (1) and (2) it is sufficient.

Though this didn't affect the final answer but the complete solution for (1) would be: If x<0 and 1 > |x| then 1>-x --> -1<x and since x<0 then -1<x<0. If x>0 and 1 < |x| the 1<x.

As for the question I'd say it's >650 level.
_________________

Concentration: General Management, Entrepreneurship

GMAT 1: 750 Q49 V44

GPA: 3.38

WE: Engineering (Computer Software)

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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11 Apr 2012, 16:38

Bunuel wrote:

pinchharmonic wrote:

bunuel,

i had a slightly slightly different method for statement 1. Do you mind letting me know if the steps I took are OK?

also, what level question is this? My test is coming up soon and I've done all the OG and quant review 2nd edition problems. I wonder if going through these two samurai guides PS/DS would be the best use of my time in the final month and half.

(1). x/|x| < x before even going to x<0 and x>0, i knew that |x| is positive, so I multiplied it across x < x*|x|

case x < 0

divide both sides by x, inequality must flip because x is negative 1 > |x| <-- this says YES to what we asked for

case x > 0

again, divide both sides by x, inequality won't flip 1 < |x| <-- this says NO

insufficient

(2). clearly says x has to be negative, therefore using (1) and (2) it is sufficient.

Though this didn't affect the final answer but the complete solution for (1) would be: If x<0 and 1 > |x| then 1>-x --> -1<x and since x<0 then -1<x<0. If x>0 and 1 < |x| the 1<x.

As for the question I'd say it's >650 level.

again thanks Bunuel! that really tied it all together

now if you got to where I got, would you have considered that enough to answer the question? I'm trying to think of an example where not going all the way x by itself w/out abs value would cause some problems and I can't since the derived expression is exactly the same as what was asked.

i had a slightly slightly different method for statement 1. Do you mind letting me know if the steps I took are OK?

also, what level question is this? My test is coming up soon and I've done all the OG and quant review 2nd edition problems. I wonder if going through these two samurai guides PS/DS would be the best use of my time in the final month and half.

(1). x/|x| < x before even going to x<0 and x>0, i knew that |x| is positive, so I multiplied it across x < x*|x|

case x < 0

divide both sides by x, inequality must flip because x is negative 1 > |x| <-- this says YES to what we asked for

case x > 0

again, divide both sides by x, inequality won't flip 1 < |x| <-- this says NO

insufficient

(2). clearly says x has to be negative, therefore using (1) and (2) it is sufficient.

Though this didn't affect the final answer but the complete solution for (1) would be: If x<0 and 1 > |x| then 1>-x --> -1<x and since x<0 then -1<x<0. If x>0 and 1 < |x| the 1<x.

As for the question I'd say it's >650 level.

again thanks Bunuel! that really tied it all together

now if you got to where I got, would you have considered that enough to answer the question? I'm trying to think of an example where not going all the way x by itself w/out abs value would cause some problems and I can't since the derived expression is exactly the same as what was asked.

If the question were is x>-1 then statement (1) would be sufficient, but I don't know what would you answe based on you method.
_________________

Concentration: Entrepreneurship, General Management

GMAT 1: 730 Q49 V42

GPA: 3.44

WE: General Management (Entertainment and Sports)

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Apr 2012, 19:07

Bunuel wrote:

Hussain15 wrote:

If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values: 1 for \(x>0\) --> so we would have: \(1<x\); Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again: \(x<0\)--> \(-x>x\)--> \(x<0\). \(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.

Hi Bunuel,

Sorry to bring this up after the question has been convincingly answered but I have a small doubt:

In the first statement, x/(mod x), while considering the possibility x<0, you write x > x/(-x) and conclude that x>-1. But shouldn't the sign of inequality flip in this case as you are dividing by a negative number? What I mean to say is that shouldn't

x > x/(-x) give us x<-1?

Sorry in advance if I am making an illogical conclusion but if you could clarify, I would appreciate it very much. Thanks.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Apr 2012, 19:17

1

This post received KUDOS

arjuntomar wrote:

Bunuel wrote:

Hussain15 wrote:

If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values: 1 for \(x>0\) --> so we would have: \(1<x\); Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again: \(x<0\)--> \(-x>x\)--> \(x<0\). \(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.

Hi Bunuel,

Sorry to bring this up after the question has been convincingly answered but I have a small doubt:

In the first statement, x/(mod x), while considering the possibility x<0, you write x > x/(-x) and conclude that x>-1. But shouldn't the sign of inequality flip in this case as you are dividing by a negative number? What I mean to say is that shouldn't

x > x/(-x) give us x<-1?

Sorry in advance if I am making an illogical conclusion but if you could clarify, I would appreciate it very much. Thanks.

take an analogy,

x> y/|y|

if y<0 then x> y/(-y) =x> -1

There is no inequality flip because you are transposing negative sign through the inequality you are just preserving the sign to one side.

in the above if you take y=-3 <0

then x> -3/3 = x>-1

there is no sign flip so no worries. Inequality sign only changes when you multiply or divide the both sides of the inequality by a negative number.

Hope this helps..!!
_________________

Practice Practice and practice...!!

If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight.

Concentration: General Management, Entrepreneurship

GMAT 1: 750 Q49 V44

GPA: 3.38

WE: Engineering (Computer Software)

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Apr 2012, 19:22

arjuntomar wrote:

Bunuel wrote:

Hussain15 wrote:

If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values: 1 for \(x>0\) --> so we would have: \(1<x\); Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again: \(x<0\)--> \(-x>x\)--> \(x<0\). \(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.

Hi Bunuel,

Sorry to bring this up after the question has been convincingly answered but I have a small doubt:

In the first statement, x/(mod x), while considering the possibility x<0, you write x > x/(-x) and conclude that x>-1. But shouldn't the sign of inequality flip in this case as you are dividing by a negative number? What I mean to say is that shouldn't

x > x/(-x) give us x<-1?

Sorry in advance if I am making an illogical conclusion but if you could clarify, I would appreciate it very much. Thanks.

arjun,

I know you asked for Bunuel, but I figured I could answer your question. I see what you are thinking, and I guess the rule is a bit misleading the way it is written.

you flip inequalities when you manipulate the equation, see below

1 < 5 (-1) * 1 > (-1) *5 -- multiply both sides by -1, and thus inequality is flipped -1 > -5

you can do the same with dividing by -1, you notice you get the same flipping and answer.

what bunuel did was break the |x| into two different cases. So he wasn't manipulating the equation, but instead subbing in the positive/negative values of x

Concentration: Entrepreneurship, General Management

GMAT 1: 730 Q49 V42

GPA: 3.44

WE: General Management (Entertainment and Sports)

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Apr 2012, 19:37

I think I answered my own question. If we take the case of x<0, then that would make (-x) a positive number, which means that we are dividing a negative number by a positive number and hence, the inequality sign will not flip in this case. Am I right?

Sorry, I am usually strong at inequalities. I guess it's just the stress (exam is in a week's time!). But please do let me know if my reasoning is sound. Thanks so much.

Concentration: Entrepreneurship, General Management

GMAT 1: 730 Q49 V42

GPA: 3.44

WE: General Management (Entertainment and Sports)

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Apr 2012, 19:42

pinchharmonic,

Thanks so much for the prompt reply. I think I just posted the answer to my own question, should have read your explanation first But yes, your reasoning makes perfect sense. I don't why I got confused in the first place, maybe it's because I have done so many questions that my mind's finally giving up. And I thought inequalities was my strength. Thanks so much for the help, let me know if my reasoning also makes sense.

Concentration: General Management, Entrepreneurship

GMAT 1: 750 Q49 V44

GPA: 3.38

WE: Engineering (Computer Software)

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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15 Apr 2012, 23:38

arjuntomar wrote:

pinchharmonic,

Thanks so much for the prompt reply. I think I just posted the answer to my own question, should have read your explanation first But yes, your reasoning makes perfect sense. I don't why I got confused in the first place, maybe it's because I have done so many questions that my mind's finally giving up. And I thought inequalities was my strength. Thanks so much for the help, let me know if my reasoning also makes sense.

hey arjun,

let's say you have this:

1/x < 5

then let's say x is -5

then your inequality is

1/-5 < 5 -1/5 < 5, which makes perfect sense, but notice we never flipped the inequality.

you only flip the inequalities when you perform an action to both the left side and the right side. for example

2x = 4. you divide both sides by 2 and you get x = 2

same with an inequality. only when you perform an action on both left and right sides do you have to consider flipping the inequality. if you're plugging in some value there's no reason to flip the inequality.

i know how you feel. i usually do the best with hard concepts by taking a break from them for 1-2 days (working on something else like verbal). i mean literally not thinking about it at all for 1-2 days. i feel the brain sort of "de-frags" all that information that was so convoluted and confusing. then, when you revisit it, sometimes it becomes so much clearer.

Sorry to bring this up after the question has been convincingly answered but I have a small doubt:

In the first statement, x/(mod x), while considering the possibility x<0, you write x > x/(-x) and conclude that x>-1. But shouldn't the sign of inequality flip in this case as you are dividing by a negative number? What I mean to say is that shouldn't

x > x/(-x) give us x<-1?

Sorry in advance if I am making an illogical conclusion but if you could clarify, I would appreciate it very much. Thanks.

From \(x>\frac{x}{-x}\) we don't dividing the inequality by some negative number, all we do is just reduce fraction. Since \(\frac{x}{-x}=-1\) (the same way as \(\frac{2}{-2}=-1\)) then from \(x>\frac{x}{-x}\) we have that \(x>-1\).

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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17 Apr 2012, 06:08

Eshika wrote:

Bunuel wrote:

udaymathapati wrote:

Hi Bunuel, Thanks for detail explanation. I am finding it difficult only last intersection part. Can you explain it further. My doubut is...If we combine "-1<x<0" or "x>1" these two inequalities, how come range for x fall betweeen -1<x<1 since x>1 is area which will not fit into this equation. Can you explain?

Range from (1): -----(-1)----(0)----(1)---- \(-1<x<0\) or \(x>1\), green area;

Range from (2): -----(-1)----(0)----(1)---- \(x<0\), blue area;

From (1) and (2): ----(-1)----(0)----(1)---- \(-1<x<0\), common range of \(x\) from (1) and (2) (intersection of ranges from (1) and (2)), red area.

Hope it's clear.

Hi Bunuel...I still didnot got it..red area says -1<x<0 but we need that -1<x<1 Please explain

You may think it like this, color part in each row means that it "is a valid" now for 1 & 2 to be valid see which part has both the colored areas

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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09 Jun 2012, 09:52

1

This post was BOOKMARKED

Bunuel wrote:

Hussain15 wrote:

If x is not equal to 0, is |x| less than 1?

(1) x/|x|< x

(2) |x| > x

Will really appreciate if answer is supported by explanation.

\(x\neq{0}\), is \(|x|<1\)? Which means is \(-1<x<1\)? (\(x\neq{0}\))

(1) \(\frac{x}{|x|}< x\) Two cases: A. \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-1<x\). But remember that \(x<0\), so \(-1<x<0\)

B. \(x>0\) --> \(\frac{x}{x}<x\) --> \(1<x\).

Two ranges \(-1<x<0\) or \(x>1\). Which says that \(x\) either in the first range or in the second. Not sufficient to answer whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(3\))

Second approach: look at the fraction \(\frac{x}{|x|}\) it can take only two values: 1 for \(x>0\) --> so we would have: \(1<x\); Or -1 for \(x<0\) --> so we would have: \(-1<x\) and as we considering the range for which \(x<0\) then completer range would be: \(-1<x<0\).

The same two ranges: \(-1<x<0\) or \(x>1\).

(2) \(|x| > x\). Well this basically tells that \(x\) is negative, as if x were positive or zero then \(|x|\) would be equal to \(x\). Only one range: \(x<0\), but still insufficient to say whether \(-1<x<1\). (For instance \(x\) can be \(-0.5\) or \(-10\))

Or two cases again: \(x<0\)--> \(-x>x\)--> \(x<0\). \(x>0\) --> \(x>x\): never correct.

(1)+(2) Intersection of the ranges from (1) and (2) is the range \(-1<x<0\) (\(x<0\) (from 2) and \(-1<x<0\) or \(x>1\) (from 1), hence \(-1<x<0\)). Every \(x\) from this range is definitely in the range \(-1<x<1\). Sufficient.

Answer: C.

Hey Bunuel,

Love your answers. But I want to suggest another method that is giving me some trouble when combining statesment (1) & (2)

1) \(\frac{x}{|x|}< x\) 2) \(x<|x|\)

Since \(0<|x|\) (x cannot equal 0), then we can rewrite statement 2, \(x<|x|\) as \(\frac{x}{|x|}< 1\).

We then subtract statment (1) and (2) as

1) \(\frac{x}{|x|}-\frac{x}{|x|}< x-1\) to get \(0< x-1\) or \(1<x\) showing that x is outside the boundary of \(-1<x<1\) and making the combined statements suffient. But \(1<x\), the derived statement of (1) and (2), contradicts statement (2), which claims that \(0<x\). What am I doing wrong?

Love your answers. But I want to suggest another method that is giving me some trouble when combining statesment (1) & (2)

1) \(\frac{x}{|x|}< x\) 2) \(x<|x|\)

Since \(0<|x|\) (x cannot equal 0), then we can rewrite statement 2, \(x<|x|\) as \(\frac{x}{|x|}< 1\).

We then subtract statment (1) and (2) as

1) \(\frac{x}{|x|}-\frac{x}{|x|}< x-1\) to get \(0< x-1\) or \(1<x\) showing that x is outside the boundary of \(-1<x<1\) and making the combined statements suffient. But \(1<x\), the derived statement of (1) and (2), contradicts statement (2), which claims that \(0<x\). What am I doing wrong?

Thank you!

You can only add inequalities when their signs are in the same direction:

If \(a>b\) and \(c>d\) (signs in same direction: \(>\) and \(>\)) --> \(a+c>b+d\). Example: \(3<4\) and \(2<5\) --> \(3+2<4+5\).

You can only apply subtraction when their signs are in the opposite directions:

If \(a>b\) and \(c<d\) (signs in opposite direction: \(>\) and \(<\)) --> \(a-c>b-d\) (take the sign of the inequality you subtract from). Example: \(3<4\) and \(5>1\) --> \(3-5<4-1\).

So, you cannot subtract \(\frac{x}{|x|}< 1\) from \(\frac{x}{|x|}< x\) since their signs are NOT in the opposite direction.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]

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14 Jun 2012, 05:21

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This post received KUDOS

Hello Bunuel, I tried to solve like this: Basically the question asks whether -1<x<1? Condition 1: x/|x|<x case 1: if x>0; x/x<x ==> 1<x ==> x>1 case 2: if x<0; x/-x<x; since x<0, -x >0 therefore, x<-x2; x+x2 <0 ==> x(1+x)<0, therefore, -1<x<0 {Please confirm if I have derived this right} Given that we do not know the sign we cannot determine, hence condition 1 is insufficient

Condition 2: |x|>x ==> which means x cannot be > then 0. thus, x<0, independently condition 2 is not sufficient;

Combining C1 and C2: we arrive at case 2; but that proves that x does not lie between -1 and 1 but between -1 and 0.

Hence C. Please let me know in case I have done anything conceptually wrong!

Hello Bunuel, I tried to solve like this: Basically the question asks whether -1<x<1? Condition 1: x/|x|<x case 1: if x>0; x/x<x ==> 1<x ==> x>1 case 2: if x<0; x/-x<x; since x<0, -x >0 therefore, x<-x2; x+x2 <0 ==> x(1+x)<0, therefore, -1<x<0 {Please confirm if I have derived this right} Given that we do not know the sign we cannot determine, hence condition 1 is insufficient

Condition 2: |x|>x ==> which means x cannot be > then 0. thus, x<0, independently condition 2 is not sufficient;

Combining C1 and C2: we arrive at case 2; but that proves that x does not lie between -1 and 1 but between -1 and 0.

Hence C. Please let me know in case I have done anything conceptually wrong!

You've done everything right.

Though for case 2 you could do as follows: \(x<0\) --> \(\frac{x}{-x}<x\) --> \(-\frac{x}{x}<x\) --> x is simply reduced: \(-1<x\). Since we consider the range \(x<0\) then \(-1<x<0\).

Also for (1)+(2) we have that \(-1<x<0\). So we can answer yes to the question whether \(-1<x<1\).