BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

If ||x+3| - 12| < 13, what is the range of x? Options

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Moderator
Posts: 7187
Joined: Thu Sep 07, 2017 4:43 pmFollowed by: 23 members
by BTGmoderatorDC » Tue Aug 13, 2019 4:50 pm

Timer

00:00

Answers

A

B

C

D

E

Stats

Difficulty—

If ||x+3| - 12| < 13, what is the range of x?

Options

a) (-4, 22)
b) (-28, -2)
c) (-28, 22)
d) (-28, 2) U (2, 22)
e) (-28, -2) U (4, 22)

OA C

Source: e-GMAT
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
Legendary Member
Posts: 2214
Joined: Fri Mar 02, 2018 2:22 pmFollowed by: 5 members

by deloitte247 » Sat Aug 17, 2019 8:50 am
If || x + 3 | - 12 | < 13 what is the range of x ?
Let | x + 3 | = y
| y - 12 | < 13
Range of x for which | y - 12 | < 13 is -13 < y - 12 < 13
-13 + 12 < y - 12 + 12 < 13 + 12
-1 < y < 25 where y = | x + 3 |
-1 < | x + 3 | < 25 even if x = 0, the value of x + 3 will always be greater than zero
$$Therefore,\ 0\le\left|x+3\right|<25$$
Then the range of x will be ; -25 < x + 3 < 25
-25 - 3 < x + 3 - 3 < 25 - 3
-28 < x < 22
Hence, range of x = ( -28, 22 )
Answer = option C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2548
Joined: Sun Oct 29, 2017 2:04 pmFollowed by: 6 members

by swerve » Sat Aug 17, 2019 2:41 pm
\(|x+3|\) will always be positive. Therefore \(|x+3|<25\) in order for \(||x+3|- 12|< 13\) to be true!

\(|x+3|<25 \Rightarrow -28 < x < 22\)

Therefore, __C__
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1