Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 24 May 2017, 12:57

### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# What is the value of |x + 7|? (1) |x + 3| = 14 (2) (x + 2)^2

Author Message
Manager
Joined: 23 Nov 2008
Posts: 77
Followers: 1

Kudos [?]: 51 [0], given: 0

What is the value of |x + 7|? (1) |x + 3| = 14 (2) (x + 2)^2 [#permalink]

### Show Tags

04 Dec 2008, 14:33
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 1 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the value of |x + 7|?

(1) |x + 3| = 14
(2) (x + 2)^2 = 169
SVP
Joined: 29 Aug 2007
Posts: 2476
Followers: 70

Kudos [?]: 774 [0], given: 19

### Show Tags

04 Dec 2008, 15:28
twilight wrote:
What is the value of |x + 7|?

(1) |x + 3| = 14
(2) (x + 2)^2 = 169

(1) |x + 3| = 14

if x is +ve: x + 3 = 14. x = 11
if x is -ve: x + 3 = - 14. x = -17

(2) (x + 2)^2 = 169
if x is +ve: (x + 2)^2 = 169
x + 2 = -13 or 13
x = either -15 or 11.

C: 1&2: x = 11.
_________________

Gmat: http://gmatclub.com/forum/everything-you-need-to-prepare-for-the-gmat-revised-77983.html

GT

Intern
Joined: 01 Dec 2008
Posts: 3
Followers: 0

Kudos [?]: 1 [0], given: 0

### Show Tags

05 Dec 2008, 12:10
whenever u have to slove abs and quadratics, you'll always have to expect two answers. there is a very similar problem up today on [url]gmatgremath.blogspot.com[/url]

jsut remember to set up two diff eqs whenyou have the abs value question
Manager
Joined: 27 May 2008
Posts: 201
Followers: 1

Kudos [?]: 42 [0], given: 0

### Show Tags

05 Dec 2008, 21:59
the Q is |x + 7| right,

why cant the answer be D

1) |x + 3| = 14 so |x + 7| = 18
2) (x + 2)^2 = 169 can be written as |x + 2| = 13 so |x + 7| = 18

Senior Manager
Joined: 30 Nov 2008
Posts: 489
Schools: Fuqua
Followers: 10

Kudos [?]: 290 [0], given: 15

### Show Tags

05 Dec 2008, 22:23
Answer is C. Here is my explanation.

Clue 1: |x+3| = 14 ==> x+3 = 14 or -(x+3) = 14. By solving them, x can be 11 or -17. So clue is INSUFFICIENT.
Clue 2: (x+2)^2 = 169. Taking the square root, x+2 = 13 or x+2 = -13. By Solving them, x can be 11 or -15. Clue alone is INSIFFICIENT.

Now considering clues 1 and 2, the common value of x is 11. We need to both the clues to solve the answer.
Manager
Joined: 30 Sep 2008
Posts: 111
Followers: 1

Kudos [?]: 20 [0], given: 0

### Show Tags

06 Dec 2008, 03:03
selvae wrote:
the Q is |x + 7| right,

why cant the answer be D

1) |x + 3| = 14 so |x + 7| = 18
2) (x + 2)^2 = 169 can be written as |x + 2| = 13 so |x + 7| = 18

|x + 7| = |x + 3 + 4|

|x+3| + 4 = 14 + 4 = 18

As |x + 3 + 4| <> |x+3| + 4 so |x + 7| <>18

Note that |x + y| <> |x| + |y| (you can prove by squaring both sides)
Re: DS: abs   [#permalink] 06 Dec 2008, 03:03
Display posts from previous: Sort by