Last visit was: 19 Nov 2025, 13:48 It is currently 19 Nov 2025, 13:48
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
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
5,511
 [55]
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,511
 [55]
2
Kudos
Add Kudos
53
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [18]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [18]
7
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
User avatar
BrushMyQuant
Joined: 05 Apr 2011
Last visit: 21 Oct 2025
Posts: 2,284
Own Kudos:
2,552
 [7]
Given Kudos: 100
Status:Tutor - BrushMyQuant
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Expert
Expert reply
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
Posts: 2,284
Kudos: 2,552
 [7]
2
Kudos
Add Kudos
5
Bookmarks
Bookmark this Post
General Discussion
User avatar
prabsahi
Joined: 09 Jun 2014
Last visit: 24 Mar 2025
Posts: 216
Own Kudos:
289
 [1]
Given Kudos: 205
Location: India
Concentration: General Management, Operations
Products:
Posts: 216
Kudos: 289
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
VeritasKarishma
Kinshook
How many integer values of x satisfy the equation |x-3|+|x-20|<17?

A. 2
B. 4
C. 0
D. 1
E. 3

|x - 3| + |x - 20| signifies sum of 'distance from 3' and 'distance from 20'. Note that distance between 3 and 20 is 17. So whichever point we take on the number line, the sum of the distance will be at least 17. So the sum will never be less than 17.

Answer (C)

Check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... s-part-ii/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/1 ... -part-iii/


I generally go for the convetional approach wherein I start taking the modulus out after finding critical points and its really time
consuming.

I tried reading this a few times.However,I am not be able to understand 100% .

Please help me with my understanding.


|x - 3| + |x - 20| signifies sum of 'distance from 3' and 'distance from 20'.

>> I interpret x is some point which is at a distance 3 and 20 from number number line based on the blog link you sent.

Next you wrote ,
Note that distance between 3 and 20 is 17...

>>Does this mean x lies somewhere between 3 and 20 inclusive.

So whichever point we take on the number line, the sum of the distance will be at least 17
>> say we take x=5 then |x-3|=2 and |x-20| = 15 so sum is 17 ??
is this what we are trying to interpret ?

I am sorry.I just want to be crystal clear with this new concept.

I will re-read your blogs multiple times so to understand and grasp the concept better.

Please help .


Thanks
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [3]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [3]
1
Kudos
Add Kudos
2
Bookmarks
Bookmark this Post
prabsahi
VeritasKarishma
Kinshook
How many integer values of x satisfy the equation |x-3|+|x-20|<17?

A. 2
B. 4
C. 0
D. 1
E. 3

|x - 3| + |x - 20| signifies sum of 'distance from 3' and 'distance from 20'. Note that distance between 3 and 20 is 17. So whichever point we take on the number line, the sum of the distance will be at least 17. So the sum will never be less than 17.

Answer (C)

Check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... s-part-ii/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/1 ... -part-iii/


I generally go for the convetional approach wherein I start taking the modulus out after finding critical points and its really time
consuming.

I tried reading this a few times.However,I am not be able to understand 100% .

Please help me with my understanding.


|x - 3| + |x - 20| signifies sum of 'distance from 3' and 'distance from 20'.

>> I interpret x is some point which is at a distance 3 and 20 from number number line based on the blog link you sent.

Next you wrote ,
Note that distance between 3 and 20 is 17...

>>Does this mean x lies somewhere between 3 and 20 inclusive.

So whichever point we take on the number line, the sum of the distance will be at least 17
>> say we take x=5 then |x-3|=2 and |x-20| = 15 so sum is 17 ??
is this what we are trying to interpret ?

I am sorry.I just want to be crystal clear with this new concept.

I will re-read your blogs multiple times so to understand and grasp the concept better.

Please help .


Thanks

IF you are new to this concept, start from here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... edore-did/
Then go on to the two posts mentioned above.
User avatar
prabsahi
Joined: 09 Jun 2014
Last visit: 24 Mar 2025
Posts: 216
Own Kudos:
289
 [2]
Given Kudos: 205
Location: India
Concentration: General Management, Operations
Products:
Posts: 216
Kudos: 289
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
VeritasKarishma
prabsahi
VeritasKarishma
How many integer values of x satisfy the equation |x-3|+|x-20|<17?

A. 2
B. 4
C. 0
D. 1
E. 3

|x - 3| + |x - 20| signifies sum of 'distance from 3' and 'distance from 20'. Note that distance between 3 and 20 is 17. So whichever point we take on the number line, the sum of the distance will be at least 17. So the sum will never be less than 17.

Answer (C)

Check:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... s-part-ii/
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2016/1 ... -part-iii/


I generally go for the convetional approach wherein I start taking the modulus out after finding critical points and its really time
consuming.

I tried reading this a few times.However,I am not be able to understand 100% .

Please help me with my understanding.


|x - 3| + |x - 20| signifies sum of 'distance from 3' and 'distance from 20'.

>> I interpret x is some point which is at a distance 3 and 20 from number number line based on the blog link you sent.

Next you wrote ,
Note that distance between 3 and 20 is 17...

>>Does this mean x lies somewhere between 3 and 20 inclusive.

So whichever point we take on the number line, the sum of the distance will be at least 17
>> say we take x=5 then |x-3|=2 and |x-20| = 15 so sum is 17 ??
is this what we are trying to interpret ?

I am sorry.I just want to be crystal clear with this new concept.

I will re-read your blogs multiple times so to understand and grasp the concept better.

Please help .


Thanks

IF you are new to this concept, start from here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2011/0 ... edore-did/
Then go on to the two posts mentioned above.[/quote]


Thanks Karishma,though I read the blog but I am not new to this concept.

The way you have applied here without solving the equations is something that I need to get better at.

Thanks a lot though,the other two blogs that you have written (In fact all your blogs) are a gem.
avatar
shyamalim
Joined: 27 May 2017
Last visit: 27 Jul 2019
Posts: 1
Posts: 1
Kudos: 0
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I understood the Veritas logic , but can someone please solve this using the conventional(critical points ) method too . I am stuck at a point .

Thanks.:)
avatar
Rakesh2019
Joined: 05 Sep 2018
Last visit: 11 Jun 2020
Posts: 8
Own Kudos:
17
 [4]
Given Kudos: 33
GMAT 1: 610 Q44 V30
GMAT 1: 610 Q44 V30
Posts: 8
Kudos: 17
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Ok. Here is the conventional method ( though lengthy one).

Firstly plot the zero points. In this case we have 2 points i.e. 3 & 20
With these 2 points we have 3 regions a) x<3 b)3<=x<20 c) x>=20

a) x<3
The equation becomes 3-x+20-x < 17
=> 2x>6
=> x>3 However this does not satisfies initial condition (x<3) so No solution.

b) 3<=x<20
The equation becomes x-3+20-x<17
=> the x cancels out; we do not have any solution.

c) x>=20
The equation becomes x-3+x-20< 17
=> 2x<40
=> x<20 Again the solution does not satisfies initial condition (x>= 20). Hence no solution

Answer is C.
User avatar
Ankur17000
Joined: 29 Nov 2023
Last visit: 20 Dec 2024
Posts: 16
Own Kudos:
Given Kudos: 39
Posts: 16
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have a question here, the general definition is -a when a<0. But here we are using -a when a<=0. The answer comes out right, but is this correct? And if it is, is this some exception to the rule?
BrushMyQuant
We need to find How many integer values of x satisfy the equation |x-3|+|x-20|<17?

Let's solve the problem using two methods:

Method 1: Algebra

As we have two absolute values so we will have three cases and the number line will be divided into three points by equating following equations to 0

x - 3 =0 and x - 20 = 0

Number line is divided into three parts x ≤ 3, 3 ≤ x ≤ 20, x ≥ 20
-Case 1: x 3

Now when x 3 then both x - 3 and x - 20 will be non-positive
=> | x-3| = -(x-3) and |x-20| = -(x-20)
=> -(x-3) - (x-20) < 17
=> -2x + 23 < 17
=> 2x > 23 - 17
=> x > 6/2
=> x > 3

But the range was x 3
=> NO SOLUTION in this case
-Case 2: 3 ≤ x ≤ 20

Now when 3 ≤ x ≤ 20 then x - 3 will be non-negative and x - 20 will be non-positive
=> | x-3| = (x-3) and |x-20| = -(x-20)
=> (x-3) - (x-20) < 17
=> 17 < 17





=> NO SOLUTION in this case
Case 3: x ≥ 20

Now when x ≥ 20 then both x - 3 and x - 20 will be non-negative
=> | x-3| = (x-3) and |x-20| = (x-20)
=> (x-3) + (x-20) < 17
=> 2x - 23 < 17
=> 2x < 40
=> x < 40/2
=> x < 20

But the range was x ≥ 20
=> NO SOLUTION in this case

Method 2: Graph

|x-3| = Distance between x and 3
|x-20| = Distance between x and 20

=> |x-3| + |x-20| = Distance between x and 3 + Distance between x and 20

Now, there can be 5 cases

Case 1: x is smaller than 3

Distance between x and 20 will be greater than the distance between 3 and 20, so will be more than 17
=> Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17

Attachment:
case 1.jpg


Case 2: x = 3

Distance between x and 20 = 17 and distance between x and 3 = 0
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
case 2.jpg

Case 3: x is between 3 and 20

Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
case-3.jpg

Case 4: x = 20

Distance between x and 20 = 0 and distance between x and 3 = 17
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
case 4.jpg

Case 5: x is greater than 20

Distance between x and 3 will be greater than the distance between 3 and 20, so will be more than 17
=> Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17

Attachment:
case-5.jpg

So, Answer will be C
Hope it helps!

Watch the following video to MASTER Inequality + Absolute value Problems

User avatar
BrushMyQuant
Joined: 05 Apr 2011
Last visit: 21 Oct 2025
Posts: 2,284
Own Kudos:
Given Kudos: 100
Status:Tutor - BrushMyQuant
Location: India
Concentration: Finance, Marketing
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
GPA: 3
WE:Information Technology (Computer Software)
Expert
Expert reply
Schools: XLRI (A)
GMAT 1: 700 Q51 V31
Posts: 2,284
Kudos: 2,552
Kudos
Add Kudos
Bookmarks
Bookmark this Post
To open absolute value where |a| = -a
we take a ≤ 0 as a can be 0 also.
Hope it helps.
Ankur17000
I have a question here, the general definition is -a when a<0. But here we are using -a when a<=0. The answer comes out right, but is this correct? And if it is, is this some exception to the rule?
BrushMyQuant
We need to find How many integer values of x satisfy the equation |x-3|+|x-20|<17?

Let's solve the problem using two methods:

Method 1: Algebra

As we have two absolute values so we will have three cases and the number line will be divided into three points by equating following equations to 0

x - 3 =0 and x - 20 = 0

Number line is divided into three parts x ≤ 3, 3 ≤ x ≤ 20, x ≥ 20
-Case 1: x 3

Now when x 3 then both x - 3 and x - 20 will be non-positive
=> | x-3| = -(x-3) and |x-20| = -(x-20)
=> -(x-3) - (x-20) < 17
=> -2x + 23 < 17
=> 2x > 23 - 17
=> x > 6/2
=> x > 3

But the range was x 3
=> NO SOLUTION in this case
-Case 2: 3 ≤ x ≤ 20

Now when 3 ≤ x ≤ 20 then x - 3 will be non-negative and x - 20 will be non-positive
=> | x-3| = (x-3) and |x-20| = -(x-20)
=> (x-3) - (x-20) < 17
=> 17 < 17





=> NO SOLUTION in this case
Case 3: x ≥ 20

Now when x ≥ 20 then both x - 3 and x - 20 will be non-negative
=> | x-3| = (x-3) and |x-20| = (x-20)
=> (x-3) + (x-20) < 17
=> 2x - 23 < 17
=> 2x < 40
=> x < 40/2
=> x < 20

But the range was x ≥ 20
=> NO SOLUTION in this case

Method 2: Graph

|x-3| = Distance between x and 3
|x-20| = Distance between x and 20

=> |x-3| + |x-20| = Distance between x and 3 + Distance between x and 20

Now, there can be 5 cases

Case 1: x is smaller than 3

Distance between x and 20 will be greater than the distance between 3 and 20, so will be more than 17
=> Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17

Attachment:
case 1.jpg


Case 2: x = 3

Distance between x and 20 = 17 and distance between x and 3 = 0
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
case 2.jpg

Case 3: x is between 3 and 20

Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
case-3.jpg

Case 4: x = 20

Distance between x and 20 = 0 and distance between x and 3 = 17
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
case 4.jpg

Case 5: x is greater than 20

Distance between x and 3 will be greater than the distance between 3 and 20, so will be more than 17
=> Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17

Attachment:
case-5.jpg

So, Answer will be C
Hope it helps!

Watch the following video to MASTER Inequality + Absolute value Problems

User avatar
Ankur17000
Joined: 29 Nov 2023
Last visit: 20 Dec 2024
Posts: 16
Own Kudos:
Given Kudos: 39
Posts: 16
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post

This is the general definition of what happens when an absolute value is opened. |a| =! -a when a = 0. It is non-negative when a = 0. But here you made a = 0 as negative I need to know why that is correct.

I am actually saying your intervals should be X<3, X>=3 to X<20, X>=20

BrushMyQuant
To open absolute value where |a| = -a
we take a ≤ 0 as a can be 0 also.
Hope it helps.
Ankur17000
I have a question here, the general definition is -a when a<0. But here we are using -a when a<=0. The answer comes out right, but is this correct? And if it is, is this some exception to the rule?
BrushMyQuant
We need to find How many integer values of x satisfy the equation |x-3|+|x-20|<17?

Let's solve the problem using two methods:

Method 1: Algebra

As we have two absolute values so we will have three cases and the number line will be divided into three points by equating following equations to 0

x - 3 =0 and x - 20 = 0

Number line is divided into three parts x ≤ 3, 3 ≤ x ≤ 20, x ≥ 20
-Case 1: x 3

Now when x 3 then both x - 3 and x - 20 will be non-positive
=> | x-3| = -(x-3) and |x-20| = -(x-20)
=> -(x-3) - (x-20) < 17
=> -2x + 23 < 17
=> 2x > 23 - 17
=> x > 6/2
=> x > 3

But the range was x 3
=> NO SOLUTION in this case
-Case 2: 3 ≤ x ≤ 20

Now when 3 ≤ x ≤ 20 then x - 3 will be non-negative and x - 20 will be non-positive
=> | x-3| = (x-3) and |x-20| = -(x-20)
=> (x-3) - (x-20) < 17
=> 17 < 17





=> NO SOLUTION in this case
Case 3: x ≥ 20

Now when x ≥ 20 then both x - 3 and x - 20 will be non-negative
=> | x-3| = (x-3) and |x-20| = (x-20)
=> (x-3) + (x-20) < 17
=> 2x - 23 < 17
=> 2x < 40
=> x < 40/2
=> x < 20

But the range was x ≥ 20
=> NO SOLUTION in this case

Method 2: Graph

|x-3| = Distance between x and 3
|x-20| = Distance between x and 20

=> |x-3| + |x-20| = Distance between x and 3 + Distance between x and 20


Now, there can be 5 cases

Case 1: x is smaller than 3

Distance between x and 20 will be greater than the distance between 3 and 20, so will be more than 17
=> Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17

Attachment:
The attachment case 1.jpg is no longer available


Case 2: x = 3

Distance between x and 20 = 17 and distance between x and 3 = 0
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
The attachment case 2.jpg is no longer available

Case 3: x is between 3 and 20

Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
The attachment case-3.jpg is no longer available

Case 4: x = 20

Distance between x and 20 = 0 and distance between x and 3 = 17
=> Distance between x and 3 + Distance between x and 20 = 17 => it is NOT < 17

Attachment:
The attachment case 4.jpg is no longer available

Case 5: x is greater than 20

Distance between x and 3 will be greater than the distance between 3 and 20, so will be more than 17
=> Distance between x and 3 + Distance between x and 20 > 17 => it is NOT < 17

Attachment:
The attachment case-5.jpg is no longer available

So, Answer will be C
Hope it helps!

Watch the following video to MASTER Inequality + Absolute value Problems

Attachment:
GMAT-Club-Forum-hep3bhuf.png
GMAT-Club-Forum-hep3bhuf.png [ 6.77 KiB | Viewed 773 times ]
Moderators:
Math Expert
105390 posts
Tuck School Moderator
805 posts