GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 17 Nov 2019, 23:12

Close

GMAT Club Daily Prep

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.

Close

Request Expert Reply

Confirm Cancel

If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Find Similar Topics 
Manager
Manager
avatar
B
Status: May The Force Be With Me (D-DAY 15 May 2012)
Joined: 06 Jan 2012
Posts: 193
Location: India
Concentration: General Management, Entrepreneurship
Reviews Badge
If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post Updated on: 09 Jun 2019, 22:05
13
84
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

62% (01:42) correct 38% (01:48) wrong based on 1487 sessions

HideShow timer Statistics

If x ≠ 0, is |x| < 1 ?

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

_________________
Giving +1 kudos is a better way of saying 'Thank You'.

Originally posted by boomtangboy on 08 Apr 2012, 09:43.
Last edited by Bunuel on 09 Jun 2019, 22:05, edited 3 times in total.
Renamed the topic and edited the question.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 08 Apr 2012, 10:01
28
28
boomtangboy wrote:
If x≠0, is |x| <1?
(1) x2<1
(2) |x| < 1/x


If \(x\neq{0}\), is \(|x| <1\)?

Is \(|x| <1\)? --> is -\(1<x<1\) (\(x\neq{0}\))?

(1) \(x^2<1\) --> \(-1<x<1\). Sufficient.

(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.

Answer D.
_________________
General Discussion
Manager
Manager
avatar
Joined: 14 Feb 2012
Posts: 90
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 19 Apr 2012, 13:25
1
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
_________________
The Best Way to Keep me ON is to give Me KUDOS !!!
If you Like My posts please Consider giving Kudos

Shikhar
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 19 Apr 2012, 21:44
2
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 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).
_________________
Intern
Intern
avatar
Joined: 01 Mar 2012
Posts: 18
Concentration: Operations, Finance
GMAT 1: 740 Q49 V41
GPA: 3.3
WE: Engineering (Manufacturing)
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 23 Apr 2012, 20:41
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).


Well Explained Bunuel
Director
Director
User avatar
Joined: 25 Apr 2012
Posts: 654
Location: India
GPA: 3.21
WE: Business Development (Other)
Reviews Badge
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 04 Jul 2013, 05:16
1
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.”
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 398
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 04 Jul 2013, 12:25
1
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|)

Thanks!
Verbal Forum Moderator
User avatar
B
Joined: 10 Oct 2012
Posts: 584
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 04 Jul 2013, 13:00
1
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.

Sufficient.
_________________
Senior Manager
Senior Manager
User avatar
Joined: 13 May 2013
Posts: 398
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 07 Jul 2013, 18:54
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?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 07 Jul 2013, 23:18
1
1
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.

Hope it helps.
_________________
Manager
Manager
User avatar
B
Joined: 14 Jan 2013
Posts: 130
Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 21 Jul 2013, 20:38
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... :|
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 21 Jul 2013, 21:44
Mountain14 wrote:
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... :|


When I say "multiply by negative x", I mean multiply by x, which is negative, so simply by x.
_________________
Retired Moderator
User avatar
P
Status: The best is yet to come.....
Joined: 10 Mar 2013
Posts: 481
GMAT ToolKit User
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 22 May 2014, 20:33
If |x| <1 --> -1<x<1, why not |x| < 1/x --> -1/x<x<1/x? If so, how does it lead to x>0?
_________________
Hasan Mahmud
Intern
Intern
User avatar
Joined: 20 May 2014
Posts: 34
Location: India
Schools: IIMC
GMAT 1: 700 Q51 V32
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 22 May 2014, 23:19
2
1
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
Senior Manager
Senior Manager
User avatar
S
Joined: 08 Dec 2015
Posts: 285
GMAT 1: 600 Q44 V27
Reviews Badge
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 10 Sep 2016, 08:34
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?

Do I make myself clear btw?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 59095
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 11 Sep 2016, 03:21
1
iliavko wrote:
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?

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.
_________________
IIMA, IIMC School Moderator
User avatar
V
Joined: 04 Sep 2016
Posts: 1371
Location: India
WE: Engineering (Other)
CAT Tests
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 27 Aug 2017, 09:25
1
Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
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).


why am I getting \(x^2\) > -1 for this step? I multiplied both sides of \(-x<1/x\) with -x
_________________
It's the journey that brings us happiness not the destination.

Feeling stressed, you are not alone!!
CEO
CEO
avatar
S
Joined: 20 Mar 2014
Posts: 2570
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
GMAT ToolKit User Reviews Badge
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 01 Sep 2017, 05:20
1
adkikani wrote:
Bunuel mikemcgarry IanStewart shashankism Engr2012

Quote:
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).


why am I getting \(x^2\) > -1 for this step? I multiplied both sides of \(-x<1/x\) with -x


You are making an algebraic mistake. If you do assume that x<0, |x| = -x; multiplying both sides of the inequality -x<1/x by -x you should get, -x*(-x) < (1/x)*(-x) ---> x^2 < -1 and this is not possible for any real 'x'. Hence, x<0 can not be true.

Additionally, you dont need to flip the signs when you multiply by -x as we are assuming that x<0, making -x >0 and when you multiply an inequality by a positive quantity, you do not need to flip the signs.

Hope this helps.
Math Expert
avatar
V
Joined: 02 Aug 2009
Posts: 8182
If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 23 Nov 2018, 19:33
adkikani wrote:

Project DS Butler: Day 18: Data Sufficiency (DS35)


For DS butler Questions Click Here



If \(x\neq{0}\), is \(|x|\)< 1?

(1) \(x^2\) < 1

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



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

(1) \(x^2\) < 1
\(x^2<1\) means |x|<1, since only fractions with absolute value less than 1 will have their square less than 1.
Sufficient

(2) \(|x|\) < \(1/x\)
Since |x| >0, we can write 1/|x|x>1
Since |x| and 1 are positive, x also must be positive for above to be true.
And since X is also positive, |X|=X, and X<1/X can be written as x^2<1
Therefore X is between 0 and 1
Sufficient

D
_________________
Manager
Manager
avatar
B
Joined: 07 Apr 2018
Posts: 101
Location: United States
Concentration: General Management, Marketing
GMAT 1: 600 Q45 V28
GPA: 3.8
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x  [#permalink]

Show Tags

New post 25 Nov 2018, 16:06
the slow but robust way of doing this is solving it for x>0 and x<0.
Attachments

IMG_20181125_175840.jpg
IMG_20181125_175840.jpg [ 1.96 MiB | Viewed 3969 times ]

GMAT Club Bot
Re: If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x   [#permalink] 25 Nov 2018, 16:06

Go to page    1   2    Next  [ 30 posts ] 

Display posts from previous: Sort by

If x ≠ 0, is |x| < 1 ? (1) x^2 < 1 (2) |x| < 1/x

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  





Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne