Last visit was: 19 Nov 2025, 06:13 It is currently 19 Nov 2025, 06:13
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
mehdiov
Joined: 22 Jun 2010
Last visit: 13 Oct 2011
Posts: 18
Own Kudos:
1,051
 [68]
Given Kudos: 1
Posts: 18
Kudos: 1,051
 [68]
7
Kudos
Add Kudos
61
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,388
Own Kudos:
778,221
 [9]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,388
Kudos: 778,221
 [9]
5
Kudos
Add Kudos
4
Bookmarks
Bookmark this Post
User avatar
BrushMyQuant
Joined: 05 Apr 2011
Last visit: 21 Oct 2025
Posts: 2,284
Own Kudos:
2,552
 [6]
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
 [6]
6
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,739
Own Kudos:
35,341
 [3]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,739
Kudos: 35,341
 [3]
2
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
mehdiov
What is the value of │x + 7│?

(1) │x + 3│= 14
(2) (x + 2)^2 = 169

Target question: What is the value of |x+7|?

Statement 1: |x+3| = 14
When solving questions involving ABSOLUTE VALUE, there are 3 steps:
1. Apply the rule that says: If |x| = k, then x = k and/or x = -k
2. Solve the resulting equations
3. Plug in the solutions to check for extraneous roots

So, x+3 = 14
OR
x+3 = -14
When we solve the two equations, we get x = 11 OR x = -17

NOTE: Although we got two different answers, we must check whether we get 2 different answers to the target question.

If x = 11, then |x + 7| = |11 + 7| = 18
If x = -17, then |x + 7| = |-17 + 7| = 10
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: (x+2)² = 169
This means EITHER (x+2) = 13 OR (x+2) = -13
When we solve the two equations, we get x = 11 OR x = -15
If x = 11, then |x + 7| = |11 + 7| = 18
If x = -15, then |x + 7| = |-15 + 7| = 8
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us that |x + 7| = 18 OR 10
Statement 2 tells us that |x + 7| = 18 OR 8
So, if BOTH statements are true, then |x + 7| must equal 18
Since we can answer the target question with certainty, the combined statements are SUFFICIENT

Answer =
RELATED VIDEO
User avatar
warriorguy
User avatar
Retired Moderator
Joined: 04 Aug 2016
Last visit: 08 Feb 2023
Posts: 378
Own Kudos:
357
 [1]
Given Kudos: 144
Location: India
Concentration: Leadership, Strategy
GPA: 4
WE:Engineering (Telecommunications)
Posts: 378
Kudos: 357
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Statement 1: |x+3| = 14

We need to consider the modulus sign and build cases

Case I : x + 3 > 0

If X + 3 > 0 then |x+3| will be positive and hence we will get x+3 = 14 => x=11

Substitute back and see if satisfies the condition; 11+3 = 14 > 0 --> Keep it

Case II: X+3 < 0

If x+3 < 0; then |x+3| will be negative => -(x+3) =14 => -x -3 =14 => -x=17 => x=-17

Substitute back and see if satisfies the condition; -17 +3 = -14 < 0 --> Keep it

Two different values => A alone is not sufficient


Statement II: \((x+2)^{2}\) = 169

\(\sqrt{(x^{2})}\) = |x| and hence we can say that |x+2|=13

Case I: x+2>0

Similar to st. 1: we can say that |x+2| will be positive and hence it will be x+2=13 => x=11

Substitute back and see if satisfies the condition; 11 +2 = 13 > 0 --> Keep it

Case II: x+2<0

Similar to st. 1: we can say that |x+2| will be negative and hence it will be -(x+2) => -x-2=13 => x=-15

Substitute back and see if satisfies the condition; -15+2 = -13 < 0 --> Keep it

Two different values => B alone is not sufficient


Combining A and B, we get a common value of 11 and hence C is sufficient to answer the question.
avatar
cruiseav
Joined: 11 Feb 2018
Last visit: 19 May 2022
Posts: 35
Own Kudos:
Given Kudos: 3
Posts: 35
Kudos: 17
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
What is the value of |x + 7|?

(1) \(|x+3|=14\) --> \(x=11\) or \(x=-17\), so \(|x+7|=18\) or \(|x+7|=10\). Not sufficient.

(2) \((x+2)^2=169\) --> \(x=11\) or \(x=-15\), so \(|x+7|=18\) or \(|x+7|=8\). Not sufficient.

(1)+(2) \(|x+7|=18\). Sufficient.

Answer: C.

The working is fairly simple here. My only doubt is we getting lX+7l=18 in each of the statement 1 and 2. So, the answer is C
But we also getting lX+7l = 10 (from statement 1 ) and lX+7l=8 (statement 2). So why we ignoring this values of lX+7l. And both of these are different values of lX+7l
Thanks
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,388
Own Kudos:
778,221
 [1]
Given Kudos: 99,977
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,388
Kudos: 778,221
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
cruiseav
Bunuel
What is the value of |x + 7|?

(1) \(|x+3|=14\) --> \(x=11\) or \(x=-17\), so \(|x+7|=18\) or \(|x+7|=10\). Not sufficient.

(2) \((x+2)^2=169\) --> \(x=11\) or \(x=-15\), so \(|x+7|=18\) or \(|x+7|=8\). Not sufficient.

(1)+(2) \(|x+7|=18\). Sufficient.

Answer: C.

The working is fairly simple here. My only doubt is we getting lX+7l=18 in each of the statement 1 and 2. So, the answer is C
But we also getting lX+7l = 10 (from statement 1 ) and lX+7l=8 (statement 2). So why we ignoring this values of lX+7l. And both of these are different values of lX+7l
Thanks

In DS when you have say a = 1 or a = 2 in (1) and a = 1 or a = 3 in (2), then when considering (1)+(2) you are taking the common value, so a = 1.

The same in the given question, both (1) and (2) give two possible values of |x+7|. When considering (1)+(2) |x+7| could only be one of those value, the one which is common for (1) and (2).
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 19 Nov 2025
Posts: 5,794
Own Kudos:
Given Kudos: 161
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,794
Kudos: 5,509
Kudos
Add Kudos
Bookmarks
Bookmark this Post
ajitsah
What is the value of │x + 7│?

(1) |x + 3│= 14
(2) (x + 2)^2 = 169

S1:
|x+3| =14
=> x = -3 +/- 14
=> x = -17 or x = 11
NOT SUFFICIENT

S2:
\((x + 2)^2 = 169\)
=> |x+2| = 13
=> x =-2 +/-13
=> x = -15 or x = 11
NOT SUFFICIENT

Combining S1 & S2
x = 11
SUFFICIENT

IMO C
User avatar
unraveled
Joined: 07 Mar 2019
Last visit: 10 Apr 2025
Posts: 2,720
Own Kudos:
Given Kudos: 763
Location: India
WE:Sales (Energy)
Posts: 2,720
Kudos: 2,258
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mehdiov
What is the value of │x + 7│?

(1) │x + 3│= 14
(2) \((x + 2)^2 = 169\)
From a strategy point of view(question specific) i am posting.

|x+7| suggests that it would vary depending on x. So statements 1 and 2 both should give a unique value in order for either of them(A or B) or both of them individually(D) or together(C or E) to be sufficient.
But before that we must realise that |x+7| calls for us to know the sign of x first(x < 0, x = 0, x > 0) and then its unique value.

Both the statements we see are not suggesting any specific sign for x or it being '0'.
Statement 1: |x+3| is more or less same as |x+7| if we are trying to find the sign of x.
Statement 2: \((x + 2)^2 = 169\) = \((|x + 2|)^2 = 169\). Again we have |x+2| which is not helpful at all.

So, we can straight eliminate A, B & D(since both statement either deal in mod or squares) and jump to get to C or E. Hence doing that i solve it, saving precious time. Solving St.1 we have x = 11 or -17 and solving St. 2: x = 11 or -15.

If we have 'a' common value of x then the answer is C, else its E.
Here the common value is 11. So Answer is C.
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,586
Own Kudos:
Posts: 38,586
Kudos: 1,079
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Moderators:
Math Expert
105388 posts
496 posts