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:

(2) \(|x| < \frac{1}{x}\) --> since LHS (|x|) is an absolute value which is always non-negative then RHS (1/x), must be positive (as \(|x| < \frac{1}{x}\)), so \(\frac{1}{x}>0\) --> \(x>0\).

Now, if \(x>0\) then \(|x|=x\) and we have: \(x<\frac{1}{x}\) --> since \(x>0\) then we can safely multiply both parts by it: \(x^2<1\) --> \(-1<x<1\), but as \(x>0\), the final range is \(0<x<1\). Sufficient.

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross multiply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).
_________________

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

23 Apr 2012, 19:41

Bunuel wrote:

shikhar wrote:

If x≠0, is |x| <1?

(1) x2<1 (2) |x| < 1/x

Merging similar topics.

shikhar wrote:

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross mulitply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

04 Jul 2013, 04:16

1

This post received KUDOS

boomtangboy wrote:

If x≠0, is |x| < 1 ?

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

Given question stem asks if |x|<1------> Is -1<x<1

from St 1 we have x^2<1 ------> -1<x<1 So Sufficient

from St2 we have

|x|<1/x

Notice that |x| is a positive value and for any Integer value |x|> 1/x -----This implies X is a fraction. In order to satisfy the above equation let us take some fractional value of x and check what happens to the above equation

x= -1/2 so we have 1/2<-2 ------No x=3/4 so we have 3/4< 4/3 ------Yes x=4/3 so we have 4/3 <3/4 ------- no

We see that when fraction is between 0<x<1 then the above equation holds true and hence x is between -1 and 1 Ans should be D....

Bunuel's solution is superb. Saves time
_________________

“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

04 Jul 2013, 11:25

1

This post received KUDOS

If x≠0, is |x| < 1 ?

is x<1 OR is -x<1 x>-1

Is -1<x<1?

(1) x^2 < 1 x^2<1 |x|<1

This tells us exactly what the stem looks for. SUFFICIENT

(2) |x| < 1/x is x < 1/x OR is -x < 1/x is x > -1/x SO -1/x < x < 1/x -1 < x^2 < 1 SUFFICIENT

(I am going to use Bunuel's method only because I think it makes more sense and I believe mine is wrong anyways)

|x|<1/x if |x|<1/x then 1/x MUST be positive as it is greater than an absolute value. If 1/x is positive then x must also be positive and therefore |x|<1/x is actually equal to x<1/x. Because we know that x is positive we can multiply both sides by x to simplify.

(x)* x < 1/x *(x) x^2 < 1 |x|<1 x<1 or x>-1 -1<x<1

This tells us exactly what the stem is looking for SUFFICIENT (D)

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

04 Jul 2013, 12:00

1

This post received KUDOS

WholeLottaLove wrote:

If x≠0, is |x| < 1 ?

(also, could someone explain to me how to solve by the method I originally chose in #2...the one where I get the positive and negative case for |X|)

Thanks!

I.x>0

\(x<\frac{1}{x}\) As x>0, we can safely cross-multiply \(\to x^2<1 \to |x|<1.\)

II.x<0

\(-x<\frac{1}{x}\) multiply both sides by\(x^2\), which is a positive quantity \(\to -x^3<x\)\([x\neq{0}]\)

or \(x(1+x^2)>0 \to\) \((1+x^2)\) and x have same sign and as\((1+x^2)\) is always positive, thus x>0. However, this goes against our assumption. Thus, x is not negative.

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

07 Jul 2013, 17:54

If x≠0, is |x| < 1 ?

(1) x^2 < 1 x<1, x>-1 -1<x<1

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT

(2) |x| < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT

(2) |x| < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

08 Jul 2013, 15:39

It helps a lot. Thanks!

Bunuel wrote:

WholeLottaLove wrote:

If x≠0, is |x| < 1 ?

(1) x^2 < 1 x<1, x>-1 -1<x<1

Valid X is a number that is greater than -1 (i.e. -3/4, -1/2) and less than 1 (i.e. 1/2, 3/4) either way the absolute value of any of those numbers will be less than 1. SUFFICIENT

(2) |x| < 1/x Multiply both sides by x x^2<1 Same solution as above. SUFFICIENT

(D)

Is it valid to multiply the absolute value of x on one side (like in number two) to simplify as long as x is positive?

As, since from |x| < 1/x we concluded that x is positive, then yes we can do that: x < 1/x --> x^2 < 1.

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

21 Jul 2013, 19:38

Bunuel wrote:

shikhar wrote:

If x≠0, is |x| <1?

(1) x2<1 (2) |x| < 1/x

Merging similar topics.

shikhar wrote:

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross mulitply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).

Bunuel,

I am not clear with this part

Also if x<0 then we have -x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: -x^2>1 --> x^2+1<0

I am getting the below:

-x < 1/x ---> when we cross multiply by -ve x, we get (-x*-x ) x2 >( flipping) -1 ( -x/x) ----> X2+1 >0 ??

Pls help...
_________________

"Where are my Kudos" ............ Good Question = kudos

My answer was A... Explanation-> |x|<1 means x should be between -1 and 1.

from (1) x2-1<0 => x between -1 and 1. sufficient

from (2) if x +ve then same as 1 and sufficient if x -ve then |x|=-x -x2<1 or x2 > -1 not sufficient

so and A

Please correct me if wrong

\(x\) cannot be negative. Refer to the solution above.

Also if \(x<0\) then we have \(-x<\frac{1}{x}\) and now if we cross mulitply by negative \(x\) then we should flip the sign: \(-x^2>1\) --> \(x^2+1<0\) which cannot be true for any real value of \(x\) (the sum of two positive value cannot be less than zero).

Bunuel,

I am not clear with this part

Also if x<0 then we have -x<\frac{1}{x} and now if we cross mulitply by negative x then we should flip the sign: -x^2>1 --> x^2+1<0

I am getting the below:

-x < 1/x ---> when we cross multiply by -ve x, we get (-x*-x ) x2 >( flipping) -1 ( -x/x) ----> X2+1 >0 ??

Pls help...

When I say "multiply by negative x", I mean multiply by x, which is negative, so simply by x.
_________________

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

22 May 2014, 22:19

2

This post received KUDOS

1

This post was BOOKMARKED

Hi Mahmud,

|x| <1 can be written as -1<x<1 because 1 is a constant BUT

|x| < \(\frac{1}{x}\) cannot be written as -1/x<x<1/x because 1/x is a variable

Solving,

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

RHS has to be greater than 0 (As LHS can only be +ve or 0) => \(\frac{1}{x}\) > 0 => x>0 (x cannot be -ve or 0 , Because, if x is -ve then \(\frac{1}{x}\) is -ve if x = 0 then\(\frac{1}{x}\) is not defined)

Rgds, Rajat
_________________

If you liked the post, please press the'Kudos' button on the left

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

18 Oct 2014, 12:22

X cannot be 0.

1. X^2 < 1 ==> |X| < 1 since X^2 cannot be negative value so both positive and negative values are possible for X. Sufficient. 2. if X=-1 then 1<-1 not possible X cannot be 0 so X > 0. X^2 < 1 same as st1. Sufficient.

Re: If x≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

31 Oct 2015, 11:07

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≠0, is |x| < 1 ? (1) x^2< 1 (2) |x| < 1/x [#permalink]

Show Tags

10 Sep 2016, 07:34

Hi Bunuel!

A question here.

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

How is St 2 sufficient? You say it yourself that 2) implies that 0<x<1 so that is just part of the range of the question stem ( is -1<x<1 ?) So statement two doesn't cover the negative area of the range. How can it be considered to be sufficient?

Do I make myself clear btw?

The question asks: whether x is between -1 and 1. (2) says that x is between 0 and 1, so the answer to the question is YES.
_________________

Its been long time coming. I have always been passionate about poetry. It’s my way of expressing my feelings and emotions. And i feel a person can convey...

Written by Scottish historian Niall Ferguson , the book is subtitled “A Financial History of the World”. There is also a long documentary of the same name that the...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...