Last visit was: 18 Nov 2025, 14:11 It is currently 18 Nov 2025, 14:11
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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
LamboWalker
User avatar
Joined: 06 Jun 2021
Last visit: 01 Jul 2025
Posts: 251
Own Kudos:
852
 [37]
Given Kudos: 304
GMAT Focus 1: 675 Q86 V81 DI83
GMAT Focus 2: 735 Q90 V85 DI84
GMAT 1: 690 Q48 V35
GMAT Focus 2: 735 Q90 V85 DI84
GMAT 1: 690 Q48 V35
Posts: 251
Kudos: 852
 [37]
On the number line, which of the following specifies the set of all 04 Mar 2024, 02:39
Kudos
Add Kudos
37
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:

75% (01:50) correct 25% (02:03) wrong based on 479 sessions

History

Date
Time
Result
Not Attempted Yet
­On the number line, which of the following specifies the set of all numbers x such that |x + 1| + |x - 2| < 5?

A. -3 < x < 2
B. -2 < x < 2
C. -2 < x < 3
D. -2 < x < -1
E. -1 < x < 3­
ShowHide Answer
Official Answer
C
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,062
 [8]
Given Kudos: 99,964
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: 105,355
Kudos: 778,062
 [8]
On the number line, which of the following specifies the set of all 04 Mar 2024, 02:57
5
Kudos
Add Kudos
3
Bookmarks
Bookmark this Post
LamboWalker
­On the number line, which of the following specifies the set of all numbers x such that |x + 1| + |x - 2| < 5?
A. -3 < x < 2
B. -2 < x < 2
C. -2 < x < 3
D. -2 < x < -1
E. -1 < x < 3­
Show more
­
The critical points (aka key points or transition points) for \(|x + 1| + |x - 2| < 5\) are -1 and 2 (those are the values of x for which the expressions in the absolute values become 0). This gives us the following three ranges to consider:
    \(--------(-1)--------(2)--------\)
    \(x < -1\)
    \(-1 ≤ x < 2\)
    \(2 ≤ x\)
When x moves from one range to another, one of the absolute values we have, |x + 1| or |x - 2|, changes sign when the absolute value is dropped.
1. When \(x < -1\), then:
    \(x + 1 < 0 \), and thus \(|x + 1| = -(x + 1)\) (recall that \(|a| = -a\), when \(a ≤ 0\)).
    \(x - 2 < 0 \), and thus \(|x - 2| = -(x - 2)\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \(-(x + 1) - (x - 2) < 5\). This gives \(x > -2\). Since we consider
\(x < -1\) range, then combining \(x > -2\) with \(x < -1\) gives a set for which the inequality holds for the considered range: \(-2 < x < -1\).
2. When \(-1 ≤ x < 2\), we are moving to the second range, then:
    \(x + 1 > 0 \), and thus \(|x + 1| = x + 1\).
    \(x - 2 < 0 \), and thus \(|x - 2| = -(x - 2)\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \((x + 1) - (x - 2) < 5\). This gives \(3 < 5\), which is true. So, when \(-1 ≤ x < 2\) our inequqlity always holds.
3. When \(2 ≤ x\), we are moving to the thrid range, then:
    \(x + 1 > 0 \), and thus \(|x + 1| = x + 1\).
    \(x - 2 > 0 \), and thus \(|x - 2| = x - 2\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \((x + 1) + (x - 2) < 5\). This gives \(x < 3\). Since we consider
\(2 ≤ x\) range, then combining \(x < 3\) with \(2 ≤ x\) gives a set for which the inequality holds for the considered range: \(2 ≤ x < 3\).
Therefore, \(|x + 1| + |x - 2| < 5\) holds for the following three ranges:
    \(-2 < x < -1\)
    \(-1 ≤ x < 2\)
    \(2 ≤ x < 3\)
Observe that these three ranges comprise one continuous range \(-2 < x < 3\), so that's our final answer.
Answer: C.­
General Discussion
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,062
Given Kudos: 99,964
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: 105,355
Kudos: 778,062
On the number line, which of the following specifies the set of all 04 Mar 2024, 02:58
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Check vey similar question, also from GMAT Prep Focus, here: https://gmatclub.com/forum/on-the-numbe ... 16667.html
User avatar
LamboWalker
User avatar
Joined: 06 Jun 2021
Last visit: 01 Jul 2025
Posts: 251
Own Kudos:
852
Given Kudos: 304
GMAT Focus 1: 675 Q86 V81 DI83
GMAT Focus 2: 735 Q90 V85 DI84
GMAT 1: 690 Q48 V35
GMAT Focus 2: 735 Q90 V85 DI84
GMAT 1: 690 Q48 V35
Posts: 251
Kudos: 852
On the number line, which of the following specifies the set of all 04 Mar 2024, 23:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hi Bunuel,

Thanks for your solution. I used a different approach, let me know if this is also correct.

Case 1: x is positive, so sign & inequality doesn't change.

\(|x + 1| + |x - 2| < 5\\
x + 1 + x - 2 <5\\
2x -1<5\\
2x<6\\
x<3                 ............... (1)\\
\)­­

Case 2: x is negative, so sign & inequality changes.­

\(\\
|x + 1| + |x - 2| < 5\\
-(x+1)+(-(x-2))>5\\
-x-1-x+2>5\\
-2x+1>5\\
-2x>4\\
x>-2                 ............... (2)\\
\)­­­­

From (1) and (2), -2<x<3

Answer: C
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,062
 [2]
Given Kudos: 99,964
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: 105,355
Kudos: 778,062
 [2]
On the number line, which of the following specifies the set of all 05 Mar 2024, 00:29
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
LamboWalker
Hi Bunuel,

Thanks for your solution. I used a different approach, let me know if this is also correct.

Case 1: x is positive, so sign & inequality doesn't change.

\(|x + 1| + |x - 2| < 5\\
x + 1 + x - 2 <5\\
2x -1<5\\
2x<6\\
x<3                 ............... (1)\\
\)­­

Case 2: x is negative, so sign & inequality changes.­

\(\\
|x + 1| + |x - 2| < 5\\
-(x+1)+(-(x-2))>5\\
-x-1-x+2>5\\
-2x+1>5\\
-2x>4\\
x>-2                 ............... (2)\\
\)­­­­

From (1) and (2), -2<x<3

Answer: C
Show more

­This is totally wrong.

When \(some \ expression\leq{0}\), then \(|some \ expression|={-(some \ expression)}\). For example: \(|-5|=5=-(-5)\);

When \(some \ expression\geq{0}\), then \(|some \ expression|={some \ expression}\). For example: \(|5|=5\).

So, when considering how would |x + 1| and |x - 2| behave in different ranges, we should consider the expression in the modulus, x + 1 and x - 2, not x itself.


10. Absolute Value



For more check Ultimate GMAT Quantitative Megathread



Hope it helps.­­­­­­­­­­­­­­­­
User avatar
xsanj
User avatar
Joined: 05 Jan 2024
Last visit: 25 Jun 2024
Posts: 2
Own Kudos:
1
 [1]
Given Kudos: 31
Location: India
Posts: 2
Kudos: 1
 [1]
On the number line, which of the following specifies the set of all 02 Apr 2024, 02:08
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Would this approach work (why or why not please, as I'm getting the correct answer. Can't seem to understand the longer explanation above):

|x + 1| + |x - 2| < 5

1.
x + 1 + x - 2 < 5
2x - 1 < 5
2x < 6
x < 3

2.
-x-1 + -x +2 < 5
-2x + 1 < 5
x > -2

So answer is: -2 < x < 3­
User avatar
Catman
User avatar
Joined: 03 Aug 2017
Last visit: 12 Feb 2025
Posts: 320
Own Kudos:
328
 [3]
Given Kudos: 219
Products:
My Reviews
Posts: 320
Kudos: 328
 [3]
On the number line, which of the following specifies the set of all 19 Jul 2024, 03:50
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
xsanj
Would this approach work (why or why not please, as I'm getting the correct answer. Can't seem to understand the longer explanation above):

|x + 1| + |x - 2| < 5

1.
x + 1 + x - 2 < 5
2x - 1 < 5
2x < 6
x < 3

2.
-x-1 + -x +2 < 5
-2x + 1 < 5
x > -2

So answer is: -2 < x < 3­
Show more
Your approach is working because f(x) less than 5 between x=-1 and x=2. There is no range break. And x >-2 is the subset of x<-1.

­There are three test cases at two values of x=-1 and x=2

1. when x>2
f(x)= x+1+x-2 < 5
x < 3 
"Range 1" = 2<x<3 

2. when -1<x<2
f(x)= x+1-x+2 < 5
f(x)=3 < 5 
"Range 2"=  -1<x<2

3. when x<-1
f(x)= -x-1-x+2 <5
f(x)=-2x+1 < 5
"Range 3= "x>-2
Total Range = Range1+ Range2+ Range 3

Range of x will be -2 <x< 3 ­
User avatar
ravjaz
User avatar
Joined: 18 Mar 2025
Last visit: 16 Nov 2025
Posts: 15
Own Kudos:
1
Given Kudos: 46
Posts: 15
Kudos: 1
On the number line, which of the following specifies the set of all 01 Apr 2025, 06:38
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I'm sorry but I can't understand how you get to
x < -1
-1 ≤ x < 2
2 ≤ x

Why is it <-1 and not less than or equal to and why is it more than or equal to 2? Where did we get this?
Bunuel
LamboWalker
­On the number line, which of the following specifies the set of all numbers x such that |x + 1| + |x - 2| < 5?
A. -3 < x < 2
B. -2 < x < 2
C. -2 < x < 3
D. -2 < x < -1
E. -1 < x < 3­
­
The critical points (aka key points or transition points) for \(|x + 1| + |x - 2| < 5\) are -1 and 2 (those are the values of x for which the expressions in the absolute values become 0). This gives us the following three ranges to consider:

\(--------(-1)--------(2)--------\)
\(x < -1\)
\(-1 ≤ x < 2\)
\(2 ≤ x\)
When x moves from one range to another, one of the absolute values we have, |x + 1| or |x - 2|, changes sign when the absolute value is dropped.
1. When \(x < -1\), then:

\(x + 1 < 0 \), and thus \(|x + 1| = -(x + 1)\) (recall that \(|a| = -a\), when \(a ≤ 0\)).
\(x - 2 < 0 \), and thus \(|x - 2| = -(x - 2)\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \(-(x + 1) - (x - 2) < 5\). This gives \(x > -2\). Since we consider
\(x < -1\) range, then combining \(x > -2\) with \(x < -1\) gives a set for which the inequality holds for the considered range: \(-2 < x < -1\).
2. When \(-1 ≤ x < 2\), we are moving to the second range, then:

\(x + 1 > 0 \), and thus \(|x + 1| = x + 1\).
\(x - 2 < 0 \), and thus \(|x - 2| = -(x - 2)\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \((x + 1) - (x - 2) < 5\). This gives \(3 < 5\), which is true. So, when \(-1 ≤ x < 2\) our inequqlity always holds.
3. When \(2 ≤ x\), we are moving to the thrid range, then:

\(x + 1 > 0 \), and thus \(|x + 1| = x + 1\).
\(x - 2 > 0 \), and thus \(|x - 2| = x - 2\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \((x + 1) + (x - 2) < 5\). This gives \(x < 3\). Since we consider
\(2 ≤ x\) range, then combining \(x < 3\) with \(2 ≤ x\) gives a set for which the inequality holds for the considered range: \(2 ≤ x < 3\).
Therefore, \(|x + 1| + |x - 2| < 5\) holds for the following three ranges:

\(-2 < x < -1\)
\(-1 ≤ x < 2\)
\(2 ≤ x < 3\)
Observe that these three ranges comprise one continuous range \(-2 < x < 3\), so that's our final answer.
Answer: C.­
Show more
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 18 Nov 2025
Posts: 105,355
Own Kudos:
778,062
Given Kudos: 99,964
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: 105,355
Kudos: 778,062
On the number line, which of the following specifies the set of all 01 Apr 2025, 07:12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ravjaz
I'm sorry but I can't understand how you get to
x < -1
-1 ≤ x < 2
2 ≤ x

Why is it <-1 and not less than or equal to and why is it more than or equal to 2? Where did we get this?
Bunuel
LamboWalker
­On the number line, which of the following specifies the set of all numbers x such that |x + 1| + |x - 2| < 5?
A. -3 < x < 2
B. -2 < x < 2
C. -2 < x < 3
D. -2 < x < -1
E. -1 < x < 3­
­
The critical points (aka key points or transition points) for \(|x + 1| + |x - 2| < 5\) are -1 and 2 (those are the values of x for which the expressions in the absolute values become 0). This gives us the following three ranges to consider:

\(--------(-1)--------(2)--------\)
\(x < -1\)
\(-1 ≤ x < 2\)
\(2 ≤ x\)
When x moves from one range to another, one of the absolute values we have, |x + 1| or |x - 2|, changes sign when the absolute value is dropped.
1. When \(x < -1\), then:

\(x + 1 < 0 \), and thus \(|x + 1| = -(x + 1)\) (recall that \(|a| = -a\), when \(a ≤ 0\)).
\(x - 2 < 0 \), and thus \(|x - 2| = -(x - 2)\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \(-(x + 1) - (x - 2) < 5\). This gives \(x > -2\). Since we consider
\(x < -1\) range, then combining \(x > -2\) with \(x < -1\) gives a set for which the inequality holds for the considered range: \(-2 < x < -1\).
2. When \(-1 ≤ x < 2\), we are moving to the second range, then:

\(x + 1 > 0 \), and thus \(|x + 1| = x + 1\).
\(x - 2 < 0 \), and thus \(|x - 2| = -(x - 2)\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \((x + 1) - (x - 2) < 5\). This gives \(3 < 5\), which is true. So, when \(-1 ≤ x < 2\) our inequqlity always holds.
3. When \(2 ≤ x\), we are moving to the thrid range, then:

\(x + 1 > 0 \), and thus \(|x + 1| = x + 1\).
\(x - 2 > 0 \), and thus \(|x - 2| = x - 2\).
Hence, in this range \(|x + 1| + |x - 2| < 5\) becomes \((x + 1) + (x - 2) < 5\). This gives \(x < 3\). Since we consider
\(2 ≤ x\) range, then combining \(x < 3\) with \(2 ≤ x\) gives a set for which the inequality holds for the considered range: \(2 ≤ x < 3\).
Therefore, \(|x + 1| + |x - 2| < 5\) holds for the following three ranges:

\(-2 < x < -1\)
\(-1 ≤ x < 2\)
\(2 ≤ x < 3\)
Observe that these three ranges comprise one continuous range \(-2 < x < 3\), so that's our final answer.
Answer: C.­
Show more

When determining the intervals around the critical points, it's important to include each critical point in one of the intervals. It doesn't matter which interval includes the equal sign for a critical point, as long as all critical points are accounted for. The key is to ensure every critical point is included without any gaps or overlaps.

Thus, as long as you include the equal sign at each critical point, it doesn't matter which interval it’s assigned to
Moderators:
Bunuel
Math Expert
105355 posts
Krunaal
Tuck School Moderator
805 posts