Last visit was: 23 Apr 2026, 23:49 It is currently 23 Apr 2026, 23:49
Home
Messages
Tests
Schools
Events
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
Back to Forum
Create Topic
User avatar
rahulms
User avatar
Joined: 17 Jan 2010
Last visit: 10 May 2010
Posts: 16
Own Kudos:
173
 [5]
Given Kudos: 8
Posts: 16
Kudos: 173
 [5]
Is |x| + |x - 1| = 1? 22 Jan 2010, 07:42
3
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

35% (medium)

Question Stats:

69% (01:27) correct 31% (01:29) wrong based on 325 sessions

History

Date
Time
Result
Not Attempted Yet
Is \(|x| + |x - 1| = 1?\)

(1) \(x \geq 0\)
(2) \(x \leq 1\)
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,802
Own Kudos:
810,902
 [6]
Given Kudos: 105,868
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,802
Kudos: 810,902
 [6]
Is |x| + |x - 1| = 1? 22 Jan 2010, 09:18
1
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
rahulms
Is \(|x| + |x - 1| = 1?\)

1) \(x \geq 0\)
2) \(x \leq 1\)
Show more

Let's work on the stem. In which ranges equation \(|x| + |x - 1| = 1\) holds true?

Two check points \(0\) and \(1\), for \(|x|\) and \(|x - 1|\) respectively. Thus three ranges to check:

A. \(x<0\) --> \(|x| + |x - 1| = 1\) --> \(-x-(x-1)=1\) --> \(x=0\), not OK as \(0\) is not in the range we are looking;

B. \(0\leq{x}\leq1\) --> \(|x| + |x - 1| = 1\) --> \(x-(x-1)=1\) --> \(1=1\), which is true. This means that for all x-es in the range \(0\leq{x}\leq1\) given equation holds true;

C. \(x>1\) --> \(|x| + |x - 1| = 1\) --> \(x+(x-1)=1\) --> \(x=1\), not OK as \(1\) is not in the range we are looking.

So we got that equation \(|x| + |x - 1| = 1\) holds true only in the range \(0\leq{x}\leq1\).

(1) \(x \geq 0\), not sufficient
(2) \(x \leq 1\), not sufficient.

(1)+(2) Together two statements give us exactly the range in which the given equation holds true. Sufficient.

Answer: C.
General Discussion
User avatar
EMPOWERgmatRichC
User avatar
Major Poster
Joined: 19 Dec 2014
Last visit: 31 Dec 2023
Posts: 21,777
Own Kudos:
13,047
 [3]
Given Kudos: 450
Status:GMAT Assassin/Co-Founder
Affiliations: EMPOWERgmat
Location: United States (CA)
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Expert
Expert reply
GMAT 1: 800 Q51 V49
GRE 1: Q170 V170
Posts: 21,777
Kudos: 13,047
 [3]
Is |x| + |x - 1| = 1? 27 Jan 2015, 20:27
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
Hi All,

This question can be solved by TESTing VALUES.

We're asked if the |X| + |X-1| = 1. This is a YES/NO question.

Fact 1: X >= 0

IF....
X = 1, then |1| + |1-1| = 1 and the answer to the question is YES.

IF...
X = 2, then |2| + |2-1| = 3 and the answer to the question is NO.
Fact 1 is INSUFFICIENT

Fact 2: X <= 1

IF....
X = 1, then the answer to the question is YES (we already did the work above).

IF....
X = -2, then |-2| + |-2 - 1| = 5 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, we have a range of values for X:
0 <= X <= 1

IF....
X = 1, then the answer to the question is YES.

IF...
X = 0, then |0| + |0-1| = 1 and the answer to the question is YES.

Note, that we are NOT told that X has to be an integer, so we have to consider NON-integer possibilities....

IF....
X = 1/3, then |1/3| + |1/3 -1| = 1 and the answer to the question is YES.
Combined, SUFFICIENT.

Final Answer:
Show Spoiler
C

GMAT assassins aren't born, they're made,
Rich
avatar
amlan009
avatar
Joined: 07 Jul 2010
Last visit: 26 Jul 2020
Posts: 5
Own Kudos:
9
Given Kudos: 4
Posts: 5
Kudos: 9
Is |x| + |x - 1| = 1? 29 May 2015, 01:21
Kudos
Add Kudos
Bookmarks
Bookmark this Post
rahulms
Is \(|x| + |x - 1| = 1?\)

(1) \(x \geq 0\)
(2) \(x \leq 1\)
Show more


Firstly mode is the distance of number from zero on the number line
So |-2|=2 and |2|=2 ...it's wonderful to think of mode and square as villains because they hide the sign -2^2= 4 and so is 2^2 ....that way cube is such a sweet guy , it really doesn't play with emotions much

The best way to solve this problem in quick time is by putting in values
Immediately looking at option 1 , we can figure out that testing a certain test of values 1,2,3 will lead to yes/no answer
Bye bye , A and D

Likewise , if we take negative values in statement 2 , we can quickly arrive that statement 2 also shall give yes/no answers
sorry , B , you are not my friend here !

Hmmmmm...if the value of the variable x lies between 0 and 1 though, and we have a 1-x , can you actually visualise what is happening
Whatever lack we have of 1 is substituted by 1-x and the fact that we are moduling(lol, dunno whether this term exists ) it , it means we shall always have 1 !

Sweet little boy C is our friend here !
User avatar
Temurkhon
User avatar
Joined: 23 Jan 2013
Last visit: 06 Apr 2019
Posts: 408
Own Kudos:
325
 [1]
Given Kudos: 43
Schools: Cambridge'16
Schools: Cambridge'16
Posts: 408
Kudos: 325
 [1]
Is |x| + |x - 1| = 1? 04 Jun 2015, 03:58
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Is |x| + |x-1| = 1?

means the distance from 0 to x + distance from x to 1=1? The only possibility for that is when 0<=x<=1

Number line: --------------0-----x-------1------>

St1. x>=0, INSUFF

St.2. x<=1, INSUFF

St1+St2, SUFF

C
User avatar
mejia401
User avatar
Joined: 15 Sep 2011
Last visit: 26 Nov 2018
Posts: 251
Own Kudos:
1,438
Given Kudos: 46
Location: United States
WE:Corporate Finance (Manufacturing)
Posts: 251
Kudos: 1,438
Is |x| + |x - 1| = 1? 27 Jul 2015, 10:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
rahulms
Is \(|x| + |x - 1| = 1?\)

1) \(x \geq 0\)
2) \(x \leq 1\)

Let's work on the stem. In which ranges equation \(|x| + |x - 1| = 1\) holds true?

Two check points \(0\) and \(1\), for \(|x|\) and \(|x - 1|\) respectively. Thus three ranges to check:

A. \(x<0\) --> \(|x| + |x - 1| = 1\) --> \(-x-(x-1)=1\) --> \(x=0\), not OK as \(0\) is not in the range we are looking;

B. \(0\leq{x}\leq1\) --> \(|x| + |x - 1| = 1\) --> \(x-(x-1)=1\) --> \(1=1\), which is true. This means that for all x-es in the range \(0\leq{x}\leq1\) given equation holds true;

C. \(x>1\) --> \(|x| + |x - 1| = 1\) --> \(x+(x-1)=1\) --> \(x=1\), not OK as \(1\) is not in the range we are looking.

So we got that equation \(|x| + |x - 1| = 1\) holds true only in the range \(0\leq{x}\leq1\).

(1) \(x \geq 0\), not sufficient
(2) \(x \leq 1\), not sufficient.

(1)+(2) Together two statements give us exactly the range in which the given equation holds true. Sufficient.

Answer: C.
Show more

Hello, question goes out to everyone.

If the mod is opened up, how do I know which condition must be greater than or equal to, or just greater than? Had you switched the operators (\(x\leq{0}\) for \(x<0\) mistakely, of course) then the answer could have been different. Walker's piece on mods doesn't touch on how operators are decided, and I haven't come across an explanation in the forum yet. So, anybody, what's the logic behind the operators?

Kudos to whom can shed the light first. Thanks,
User avatar
ENGRTOMBA2018
User avatar
Joined: 20 Mar 2014
Last visit: 01 Dec 2021
Posts: 2,319
Own Kudos:
3,890
Given Kudos: 816
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE:Engineering (Aerospace and Defense)
Products:
My Reviews
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
Posts: 2,319
Kudos: 3,890
Is |x| + |x - 1| = 1? 27 Jul 2015, 10:59
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mejia401
Bunuel
rahulms
Is \(|x| + |x - 1| = 1?\)

1) \(x \geq 0\)
2) \(x \leq 1\)

Let's work on the stem. In which ranges equation \(|x| + |x - 1| = 1\) holds true?

Two check points \(0\) and \(1\), for \(|x|\) and \(|x - 1|\) respectively. Thus three ranges to check:

A. \(x<0\) --> \(|x| + |x - 1| = 1\) --> \(-x-(x-1)=1\) --> \(x=0\), not OK as \(0\) is not in the range we are looking;

B. \(0\leq{x}\leq1\) --> \(|x| + |x - 1| = 1\) --> \(x-(x-1)=1\) --> \(1=1\), which is true. This means that for all x-es in the range \(0\leq{x}\leq1\) given equation holds true;

C. \(x>1\) --> \(|x| + |x - 1| = 1\) --> \(x+(x-1)=1\) --> \(x=1\), not OK as \(1\) is not in the range we are looking.

So we got that equation \(|x| + |x - 1| = 1\) holds true only in the range \(0\leq{x}\leq1\).

(1) \(x \geq 0\), not sufficient
(2) \(x \leq 1\), not sufficient.

(1)+(2) Together two statements give us exactly the range in which the given equation holds true. Sufficient.

Answer: C.

Hello, question goes out to everyone.

If the mod is opened up, how do I know which condition must be greater than or equal to, or just greater than? Had you switched the operators (mistakely, of course) then the answer could have been different. Walker's piece on mods doesn't touch on how operators are decided, and I haven't come across an explanation in the forum yet. So, anybody, what's the logic behind the operators?

Kudos to whom can shed the light first. Thanks,
Show more

"Kudos to whom can shed the light first".. this is not a competition!!

The mod or || sign represents distance of a number from 0. What this means is that |x| = 4 shows that x is at a distance of 4 units from 0 (from + direction towards and from - direction towards 0).

Generally, |x-a| = b shows that the distance of 'x' from 'a' is b units.

Now, when you open the modulus, the sign would change depending on whether x is \(\geq\) or < 0 .

|x| = x for x \(\geq\) 0

|x| = -x for x<0.

For modulus or absolute value questions, you need to take the equality with the '>' sign. That is the convention and follows the definition of an absolute value. Additionally, you only need to account for equality once in your question.

Also, think of modulus or absolute value this way:

|x| = x for x=0,1,2,3,4,5... or true for all NON-NEGATIVE numbers. Remember the 'non-negative' part. This includes 0 as well.

But |x| = -x for x<0, x=-1,-2,-0.25 ...

This is the reason why we put equality with the '>' sign to account for all non-negative numbers. The 'nature' of 'x' does not change if it is 0 and above but it does change (multiply x by -1) if x is <0.

Per the question given , |x| + |x-1| =1 should be looked at x<0 , 0<=x<1 and x>1 . The reason why we have chosen 0 and 1 ---> compare |x| and |x-1| to |x-a|

For x<0 , |x| = -x and |x-1| = -(x-1)
For 0<= x<1 , |x| = x and |x-1| = -(x-1)
For x>=1 , |x| = x and |x-1| = (x-1)
User avatar
MSIMBA
User avatar
Joined: 04 Jun 2017
Last visit: 14 Aug 2020
Posts: 73
Own Kudos:
40
Given Kudos: 180
Location: India
Concentration: Strategy, Operations
GMAT 1: 500 Q39 V20
GPA: 3.82
GMAT 1: 500 Q39 V20
Posts: 73
Kudos: 40
Is |x| + |x - 1| = 1? 14 Jun 2019, 20:20
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Is |x| + |x-1| = 1?

A. X => 0 , put X =0 in |x| + |x-1|, Comes out be -1
X > 0 Say 1 in |x| + |x-1| , comes out to be 1


Not sufficient

B. X = < 1 ,put 0 and 1,, Answer would be -1 and 1

Not sufficient

From A and B We can say it equals to 1

C is answer
Moderators:
Bunuel
Math Expert
109802 posts
kingbucky
498 posts
Gmat860sanskar
212 posts