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 500,000 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:

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT

1) this only holds true if x is a negative fraction or greater than 1. The question however is asking if |x| is less than 1. The abs val of a number less than -1 will be greater than 1, whereas the abs val of a negative fraction will be less than 1.

x/|x| < x

try -2, -2/|-2| = -1, -1 is greater than -2, so does not hold try -1/2, (-1/2)/|(-1/2)| = -1, -1 is less than -1/2 so this holds. try 1/2, (1/2)/|(1/2)| = 1, 1 is greater than 1/2, so this does not hold try 2, 2/|2| = 1, 1 is less than 2, so this holds

so, -1 < x < 0 and x > 1

the question asks: is |x| < 1?

-1 < x < 0 --> |x| will be a positive fraction, i.e. less than 1 x > 1 --> |x| will be greater than 1

hence insufficient.

2) the abs val of a positive number equals that number so this only holds true for negative numbers (including fractions). The question is asking if |x| is less than 1. Same as in (1), |x| > x

x < 0

try a negative integer, -2, | -2 | = 2, --> |x| will be greater 1

try a negative fraction, -(1/2), | -(1/2) | = 1/2 --> |x| will be less than 1

hence insufficient.

putting both together,

x < 0 and -1< x < 0 and x > 1 this limits x to -1< x < 0, thus |x| will be a positive fraction and will always be less than 1.

therefore the answer is C
_________________

If you like my post, a kudos is always appreciated

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT

1) this only holds true if x is a negative fraction or greater than 1. The question however is asking if |x| is less than 1. The abs val of a number less than -1 will be greater than 1, whereas the abs val of a negative fraction will be less than 1.

x/|x| < x

try -2, -2/|-2| = -1, -1 is greater than -2, so does not hold try -1/2, (-1/2)/|(-1/2)| = -1, -1 is less than -1/2 so this holds. try 1/2, (1/2)/|(1/2)| = 1, 1 is greater than 1/2, so this does not hold try 2, 2/|2| = 1, 1 is less than 2, so this holds

so, -1 < x < 0 and x > 1

the question asks: is |x| < 1?

-1 < x < 0 --> |x| will be a positive fraction, i.e. less than 1 x > 1 --> |x| will be greater than 1

hence insufficient.

2) the abs val of a positive number equals that number so this only holds true for negative numbers (including fractions). The question is asking if |x| is less than 1. Same as in (1), |x| > x

x < 0

try a negative integer, -2, | -2 | = 2, --> |x| will be greater 1

try a negative fraction, -(1/2), | -(1/2) | = 1/2 --> |x| will be less than 1

hence insufficient.

putting both together,

x < 0 and -1< x < 0 and x > 1 this limits x to -1< x < 0, thus |x| will be a positive fraction and will always be less than 1.

therefore the answer is C

I second that... my slip.... i missed the greater part in ST 1...... ! Yes.. it should be C....
_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

1. For Condition 1, x can either be positive integer or -ve fraction. Its true for both i.e if x = 2,3, etc or x = -1/2, -1/3, etc. Hence not sufficient.

2. For condition 2, X has to be -ve fraction or integer. Its true only when x = -1/2, -1/3, etc. or x=-1, -2, etc. Hence B not sufficient.

3.Combine 1 & 2. True only when x is -ve fraction i.e. x=-1/2, -1/3, etc. Hence |x| < 1. Hence C should be the answer.

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

Show Tags

06 Feb 2014, 03:32

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

03 Oct 2015, 11:35

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

13 Nov 2015, 09:21

Bunuel wrote:

jeeteshsingh wrote:

ugimba wrote:

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

(1) \(x/|x| < x\) (2) |x| > x

IMO A...

ST 1: can be written as \(x < x * |x|\) since multiplying by |x| which is always positive would not change the sign.

This gives x as -1<x<0.... therefore sufficient...

ST2: |x| > x gives value of x as x<0... hence could be -2,-3.... Therefore not sufficient...

Cheers! JT

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

Q: is \(-1<x<1\) true?

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

\(x<0\) --> \(\frac{x}{-x} < x\) --> \(x>-1\), but as \(x<0\), then --> \(-1<x<0\);

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

So x can be as in the range {-1,1} as well as out of this range. Not sufficient.

(2) \(|x| > x\) --> \(x<0\). Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is: \(-1<x<0\). Every \(x\) from this range is in the range {-1,1}. Sufficient.

Hi Bunuel, Thanks for this explanation. Bunuel, I am still having a hard time trying to understand the concepts behind the process followed to solve statement 1 in this question.

My reasoning is the following: for example if |x-3|=2, then x-3=2 and x-3=-2. If I follow this same reasoning then for statement one I get the following:

Hi Bunuel, Thanks for this explanation. Bunuel, I am still having a hard time trying to understand the concepts behind the process followed to solve statement 1 in this question.

My reasoning is the following: for example if |x-3|=2, then x-3=2 and x-3=-2. If I follow this same reasoning then for statement one I get the following:

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

x/x<x ---> 1<x x/x<-x ---> 1<-x ---> -1>x

What is the flaw in my reasoning?

Thanks so much for your help!

Let me try to answer your question.

Firstly, |x|<1 ---> -1<x<1 . So the question asks whether x is between -1 and 1 with x\(\neq\)0

Your interpretation of |x-3|=2 is correct by taking the 2 cases. But the way you are approaching statement 1 is not correct. Look below.

Statement 1, x/|x|<x .

Take 2 cases.

Case 1: when x > 0 ---> |x| = x ---> x/|x|<x ---> x>1 . Now you had assumed that x>0 and it gave you x>1. Thus the range for x satisfying this case will thus become x>1. This will give a "no" to "is |x|<1 or is -1<x<1?".

case 2: when x < 0 ---> |x| = -x ---> x/|x|<x ---> x>-1 . Now you had assumed that x<0 and it gave you x>-1. Thus the range for x satisfying this case will thus become -1<x<0. This will give a "yes" to "is |x|<1 or is -1<x<1?".

Thus you get 2 different cases for the same statement, making statement 1 not sufficient.

The issue with your solution is that you are not taking the intersection of the possible values as I have shown in 2 cases above.

|x-3| = 2 --> x-3=2 ONLY when x-3 \(\geq\)0 .

Similarly, |x-3| = 2 --> x-3=-2 ONLY when x-3 < 0.

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

Show Tags

15 Nov 2015, 20:14

Engr2012 wrote:

angierch wrote:

Hi Bunuel, Thanks for this explanation. Bunuel, I am still having a hard time trying to understand the concepts behind the process followed to solve statement 1 in this question.

My reasoning is the following: for example if |x-3|=2, then x-3=2 and x-3=-2. If I follow this same reasoning then for statement one I get the following:

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

x/x<x ---> 1<x x/x<-x ---> 1<-x ---> -1>x

What is the flaw in my reasoning?

Thanks so much for your help!

Let me try to answer your question.

Firstly, |x|<1 ---> -1<x<1 . So the question asks whether x is between -1 and 1 with x\(\neq\)0

Your interpretation of |x-3|=2 is correct by taking the 2 cases. But the way you are approaching statement 1 is not correct. Look below.

Statement 1, x/|x|<x .

Take 2 cases.

Case 1: when x > 0 ---> |x| = x ---> x/|x|<x ---> x>1 . Now you had assumed that x>0 and it gave you x>1. Thus the range for x satisfying this case will thus become x>1. This will give a "no" to "is |x|<1 or is -1<x<1?".

case 2: when x < 0 ---> |x| = -x ---> x/|x|<x ---> x>-1 . Now you had assumed that x<0 and it gave you x>-1. Thus the range for x satisfying this case will thus become -1<x<0. This will give a "yes" to "is |x|<1 or is -1<x<1?".

Thus you get 2 different cases for the same statement, making statement 1 not sufficient.

The issue with your solution is that you are not taking the intersection of the possible values as I have shown in 2 cases above.

|x-3| = 2 --> x-3=2 ONLY when x-3 \(\geq\)0 .

Similarly, |x-3| = 2 --> x-3=-2 ONLY when x-3 < 0.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x/|x| < x (2) |x| > x

If the range of the condition falls into that of the condition in terms of inequalities, the condition is sufficient.

There is 1 variable in the original condition, and there are 2 equations provided by the 2 conditions, so there is high chance (D) will be our answer. For condition 1, if x>0, x/|x|<x --> x/x<x --> 1<x, then 1<x if x<0, x/|x|<x --> x/-x<x --> -1<x, then -1<x<0. Therefore this condition is insufficient because this range does not fall into that of the question. For condition 2, |x|>x --> x<0. This is insufficient for the same reason. Looking at them together, however, -1<x<0 falls into the range of the question, so this is sufficient, and the answer becomes (C).

For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________

There’s something in Pacific North West that you cannot find anywhere else. The atmosphere and scenic nature are next to none, with mountains on one side and ocean on...

This month I got selected by Stanford GSB to be included in “Best & Brightest, Class of 2017” by Poets & Quants. Besides feeling honored for being part of...

Joe Navarro is an ex FBI agent who was a founding member of the FBI’s Behavioural Analysis Program. He was a body language expert who he used his ability to successfully...