NEW HERE? LEARN MORE
closeclose
Last visit was: 25 Apr 2024, 15:15 It is currently 25 Apr 2024, 15:15
Decision Tracker
My Rewards
New posts
New comers' posts

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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel

If x≠0, is |x|<1? (1) x^2/|x| > x

SORT BY:
Date
Kudos
Tags:
Find Similar Topics 
Show Tags
Hide Tags
User avatar
L
BillyZ
Current Student
Joined: 14 Nov 2016
Posts: 1174
Own Kudos [?]: 20713 [26]
Given Kudos: 926
Location: Malaysia
Concentration: General Management, Strategy
Schools: Sloan '23 (D) Tuck '23 (D) Darden '23 (D) Ross '23 (D) Kellogg '23 (D) Wharton '23 (D) Booth '23 (D) Fuqua '24 (A) Harvard '24 (D) Stanford '24 (D)
GMAT 1: 750 Q51 V40 (Online)
GPA: 3.53
Send PM
GMAT ToolKit User Reviews Badge
If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   18 Feb 2017, 05:24
26
Bookmarks
00:00
A
B
C
D
E

Difficulty:

95% (hard)

Question Stats:

34% (02:44) correct 66% (02:30) wrong based on 330 sessions
Hide Show timer Statistics
If \(x ≠ 0\), is \(|x|<1\)?

(1) \(\frac{x^2}{{|x|}} > x\)

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

I have a hard time to answer this question. Could someone please help to explain?
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [8]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   18 Feb 2017, 08:06
2
Kudos
6
Bookmarks
Expert Reply
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.
User avatar
D
ganand
Manager
Manager
Joined: 17 May 2015
Posts: 200
Own Kudos [?]: 3019 [5]
Given Kudos: 85
Send PM
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   18 Feb 2017, 07:50
1
Kudos
4
Bookmarks
ziyuenlau wrote:
If \(x ≠ 0\), is \(|x|<1\)?

(1) \(\frac{x^2}{{|x|}} > x\)

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

I have a hard time to answer this question. Could someone please help to explain?


Hi,

is \(|x|<1\)? => is \(-1 < x < 1\)?

St 1:
Case1: x>0 => |x| = x.
\(\frac{x^2}{{|x|}} = \frac{x^{2}}{x} = x > x\) => No solution.

Case2: x<0 => |x| = -x
\(\frac{x^2}{{|x|}} = \frac{x^{2}}{-x} = -x > x\) => All negative values will satisfy this equation.

Hence, not sufficient.

St 2:
Case 1: x>0 => |x| = x

\(\frac{x}{{|x|}} = \frac{x}{x} = 1 < x\) => solution x>1.

case 2: x<0 => |x| = -x

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

From this statement, we have two sets of solutions. Hence, not sufficient.

By St 1 and st 2, we have
-1< x < 0. => Unique answer. Answer C.

Hope this helps.
General Discussion
User avatar
S
Alexey1989x
Manager
Manager
Joined: 05 Dec 2016
Posts: 194
Own Kudos [?]: 88 [0]
Given Kudos: 49
Concentration: Strategy, Finance
GMAT 1: 620 Q46 V29
Send PM
GMAT ToolKit User
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   28 Feb 2017, 01:41
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.


Hello Bunuel, could you please elaborate more on the (1) condition, I can't grasp how did you get that x is -ve.
Thx
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   28 Feb 2017, 05:52
Expert Reply
Alexey1989x wrote:
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.


Hello Bunuel, could you please elaborate more on the (1) condition, I can't grasp how did you get that x is -ve.
Thx


We have |x| > x.

Now, if x were non-negative, then |x| would be equal to x, for example, if x=2, then |x| = |2| = 2 = x. Only for negative x we can have |x| > x. For example, consider x = -2: |x| = |-2| = 2 > x = 2.
User avatar
L
MahmoudFawzy
Director
Director
Joined: 27 Oct 2018
Status:Manager
Posts: 683
Own Kudos [?]: 1857 [0]
Given Kudos: 200
Location: Egypt
Concentration: Strategy, International Business
Schools: NTU '20 HKUST '21 Japan MBA
GPA: 3.67
WE:Pharmaceuticals (Health Care)
Send PM
GMAT ToolKit User
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   16 Dec 2018, 02:01
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.


Greetings Bunuel

in analyzing statement 2,

you mentioned : "If x < 0, we'll have x/(-x) < x --> -1 < x"

my question: why the inequality is not flipped?
isn't it right to flip the from '<' to '> ?

I think I am confused in this point, please elaborate. :blushing
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   16 Dec 2018, 02:20
Expert Reply
Mahmoudfawzy83 wrote:
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.


Greetings Bunuel

in analyzing statement 2,

you mentioned : "If x < 0, we'll have x/(-x) < x --> -1 < x"

my question: why the inequality is not flipped?
isn't it right to flip the from '<' to '> ?

I think I am confused in this point, please elaborate. :blushing


We are not dividing the entire inequality by x, we are simply reducing x/(-x) by x: x/(-x) = -(x/x) = -1.
avatar
S
GMATaspirant641
Intern
Intern
Joined: 26 Oct 2010
Posts: 12
Own Kudos [?]: 84 [0]
Given Kudos: 14
Send PM
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   25 Jan 2019, 22:18
Bunuel

Please let me know where I am going wrong in my approach with statement 1.

x^2 / |x| > x

we know that the LHS will always be positive.

So, x^2 / |x| > 0 (its not >= 0 since x cannot be 0)

x^2 / |x| > 0

x^2 > 0

|x| > 0

so, x>0 and x<0. Hence, x can be any non zero number.

With this approach, i get answer choice E.

I also understand the approaches mentioned by you and others in the thread. But, I am not able to understand where I am going wrong. Please help.
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   26 Jan 2019, 00:44
Expert Reply
GMATaspirant641 wrote:
Bunuel

Please let me know where I am going wrong in my approach with statement 1.

x^2 / |x| > x

we know that the LHS will always be positive.

So, x^2 / |x| > 0 (its not >= 0 since x cannot be 0)

x^2 / |x| > 0

x^2 > 0

|x| > 0

so, x>0 and x<0. Hence, x can be any non zero number.

With this approach, i get answer choice E.

I also understand the approaches mentioned by you and others in the thread. But, I am not able to understand where I am going wrong. Please help.


Yes, \(\frac{x^2}{|x|} > 0\) is true of all x's except 0. But we don't have \(\frac{x^2}{|x|} > 0\), we have \(\frac{x^2}{|x|} > x\) and as shown above it won't hold true if x is positive. For example, if x = 2, then \(\frac{x^2}{|x|} =2\), so \(\frac{x^2}{|x|} = x\).
User avatar
P
rahul16singh28
Senior Manager
Senior Manager
Joined: 31 Jul 2017
Posts: 435
Own Kudos [?]: 443 [0]
Given Kudos: 752
Location: Malaysia
Schools: INSEAD Jan '19
GPA: 3.95
WE:Consulting (Energy and Utilities)
Send PM
If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   26 Jan 2019, 05:39
hazelnut wrote:
If \(x ≠ 0\), is \(|x|<1\)?

(1) \(\frac{x^2}{{|x|}} > x\)

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

I have a hard time to answer this question. Could someone please help to explain?


Hi chetan2u Bunuel

By adding two equations, we get -

\(\frac{x}{{|x|}}(x-1) > 0\)

Here, x can be -1,2.

Can you please advise where I am going wrong.
User avatar
G
KanishkM
Director
Director
Joined: 09 Mar 2018
Posts: 783
Own Kudos [?]: 453 [0]
Given Kudos: 123
Location: India
Schools: DeGroote'21 Desautels '21 NUS '21
Send PM
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   26 Jan 2019, 07:27
hazelnut wrote:
If \(x ≠ 0\), is \(|x|<1\)?

(1) \(\frac{x^2}{{|x|}} > x\)

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

I have a hard time to answer this question. Could someone please help to explain?


Why didn't i use the values used in A, to negate B , Nevertheless inline is my approach

|x| < 1, will mean that -1<x<1

(1) \(\frac{x^2}{{|x|}} > x\)

when you use -0.5, the value will answer the statement and will answer the question as Yes

but when you use -2, the value will answer the statement and will answer the question as No

4/2 > -2

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

when you use -0.5 , the value will answer the statement and will answer the question as Yes

-0.5 /0.5 < -0.5

But when you use 2, the value will answer the statement and will answer the question as No

Now when we combine both the statements for both the statements to be true together

When you use x =-0.5, equality holds good

C
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   27 Jan 2019, 02:56
Expert Reply
rahul16singh28 wrote:
hazelnut wrote:
If \(x ≠ 0\), is \(|x|<1\)?

(1) \(\frac{x^2}{{|x|}} > x\)

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

I have a hard time to answer this question. Could someone please help to explain?


Hi chetan2u Bunuel

By adding two equations, we get -

\(\frac{x}{{|x|}}(x-1) > 0\)

Here, x can be -1,2.

Can you please advise where I am going wrong.


We cannot add the inequalities the way you did.

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 \(5>1\) --> \(3-5<4-1\).

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 \(2<5\) --> \(3+2<4+5\).

For more check Manipulating Inequalities.
User avatar
S
NinetyFour
Manager
Manager
Joined: 22 Sep 2018
Posts: 191
Own Kudos [?]: 173 [0]
Given Kudos: 78
Send PM
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   08 Feb 2019, 21:17
hazelnut wrote:
If \(x ≠ 0\), is \(|x|<1\)?

(1) \(\frac{x^2}{{|x|}} > x\)

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



Here's my reasoning:

Statement 1:\(x^2 > x{|x|}\)

X can be both a negative integer and a negative decimal. Thus we get a YES and a NO for the stem above.

Statement 2: this tells us \(x < x{|x|}\)
X can be a positive integer or a negative fraction. Once again a YES and a NO for the stem above.

Combining the two the overlapping set is a negative fraction. Hence we can conclude \(|x|<1\)
avatar
B
ChairmanThe
Intern
Intern
Joined: 07 Jan 2020
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 51
Send PM
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   15 May 2020, 03:51
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.





I am having trouble understanding how you simplify the first condition to |x| > x. Can you please elaborate on that?
Thanks
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   15 May 2020, 07:49
Expert Reply
ChairmanThe wrote:
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.





I am having trouble understanding how you simplify the first condition to |x| > x. Can you please elaborate on that?
Thanks


\(\frac{x^2}{|x|}>x\);

\(\frac{|x|*|x|}{|x|}>x\);

\(|x|>x\).
avatar
B
ChairmanThe
Intern
Intern
Joined: 07 Jan 2020
Posts: 3
Own Kudos [?]: 0 [0]
Given Kudos: 51
Send PM
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   15 May 2020, 09:22
Bunuel wrote:
ChairmanThe wrote:
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.





I am having trouble understanding how you simplify the first condition to |x| > x. Can you please elaborate on that?
Thanks


\(\frac{x^2}{|x|}>x\);

\(\frac{|x|*|x|}{|x|}>x\);

\(|x|>x\).




But why are you using |x|∗|x| when you are factoring X^2. Couldn't it be -x *-x and give you the same value for X^2?
Thanks
User avatar
L
Bunuel
Math Expert
Joined: 02 Sep 2009
Posts: 92915
Own Kudos [?]: 619017 [0]
Given Kudos: 81595
Send PM
CAT Tests Forum Quiz Mode
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink] New post   15 May 2020, 09:30
Expert Reply
ChairmanThe wrote:
Bunuel wrote:
ChairmanThe wrote:
Bunuel wrote:
If \(x ≠ 0\), is \(|x|<1\)?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) \(\frac{x^2}{{|x|}} > x\) --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

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

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that \(\frac{x}{{|x|}} < x\) is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Answer: C.





I am having trouble understanding how you simplify the first condition to |x| > x. Can you please elaborate on that?
Thanks


\(\frac{x^2}{|x|}>x\);

\(\frac{|x|*|x|}{|x|}>x\);

\(|x|>x\).




But why are you using |x|∗|x| when you are factoring X^2. Couldn't it be -x *-x and give you the same value for X^2?
Thanks


x^2 = |x|*|x| and if you write it that way you'd be able to simplify \(\frac{x^2}{|x|}>x\) and get \(|x|>x\).
GMAT Club Bot
gmatclubot
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink]
15 May 2020, 09:30
Moderator:
Bunuel
Math Expert
92915 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne
Main navigation
GMAT Resources
Partners
MBA Resources
www.gmac.com|www.mba.com | | GMAT Club Rules | Terms and Conditions