Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
09 Jun 2012, 09:10

1

This post received KUDOS

Expert's post

alphabeta1234 wrote:

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?

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]
14 Jun 2012, 04:21

1

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!

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
14 Jun 2012, 04:31

1

This post received KUDOS

Expert's post

pavanpuneet wrote:

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 \(-1<x<1\).

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
24 Apr 2013, 17:25

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.

Buenel terrific explanation. This system, that you use for solving inequations, is quite fundamental and hard to be wrong. But is it really always the quickest way?? _________________

When you feel like giving up, remember why you held on for so long in the first place.

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
25 Apr 2013, 08:17

Bunuel wrote:

When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.

Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
25 Apr 2013, 23:56

1

This post received KUDOS

Expert's post

oyabu wrote:

Bunuel wrote:

When \(x<0\), then \(|x|=-x\), thus \(\frac{x}{|x|}<x\) becomes \(\frac{x}{-x}<x\) --> \(-1<x\) but since \(x<0\), then \(-1<x<0\).

Hope it's clear.

Thanks. One more follow-up for clarification - Isn't the absolute value of a negative value positive? So if x<0 (x is a negative number), then the absolute value of x should be positive? i.e. |-x|=x?

First of all \(|-x|=|x|\). Next, if \(x<0\), then \(|x|=|negative|=-x=-negative=positive\).

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
23 Jun 2013, 10:37

For number 1 why do we have two values for x<0? In other words why do we have x< 0 and -1<x<0?

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.

Re: If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
11 Jul 2013, 13:43

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

|x|<1 Is -1<x<1

(1) x/|x|< x

x/|x|< x Two cases: x>0, x<0

x>0 x/|x|< x x/x<x 1<x

If x>0 and x>1 then x>1 If x>1 then |x| is NOT less than 1

x<0 x/|x|< x -x/|-x|<x -1<x

-1<x<0 if -1<x<0 then |x| IS less than 1 INSUFFICIENT

(2) |x| > x

If the absolute value of x is greater than x than x must be negative. However, |-x| could be less than 1 or greater than 1 depending on the value of x. INSUFFICIENT

1+2 x must be negative and x from #1 x is either >1 or between -1 and 0. Therefore, we know -1<x<0 which means that |x| is always less than 1. SUFFICIENT

Re: If x is not equal to 0, is |x| less than 1? [#permalink]
06 Aug 2013, 07:42

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 bunel,

hw is that we can have two ranges x<0 & x>0. How we got this ranges?

If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
29 Sep 2014, 06:08

The question asks "is \(-1<x<1\)?" We got that \(x\) is in the range \(-1<x<0\) (red area). Now, as ANY \(x\) from this range (from red area) is indeed in \(-1<x<1\), then we can answer YES to our original question.

Hope it's clear.[/quote]

Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is

That for x=0.5 it is in the range -1<x<1 but it is NOT part of -1<x<0 which we found out. So it is not possible to say that all possible values of -1<x<0 come under -1<x<1 unless ofcourse it is specified that x is an integer.

If x is not equal to 0, is |x| less than 1? (1) x/|x|< x [#permalink]
29 Sep 2014, 06:15

Expert's post

earnit wrote:

Exactly, based on your explanation above, Though the answer is YES it is sufficient to answer but the answer to the question is the range -1<x<1 for x is NO as -1<x<0 is the range and so x could be 0.5 and not be part of the -1<x<0 .

This week went in reviewing all the topics that I have covered in my previous study session. I reviewed all the notes that I have made and started reviewing the Quant...

I was checking my phone all day. I wasn’t sure when I would receive the admission decision from Tepper. I received an acceptance from Goizueta in the early morning...

I started running as a cross country team member since highshcool and what’s really awesome about running is that...you never get bored of it! I participated in...