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 →

MGMAT Inequality problrm

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Source: — Data Sufficiency |
User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members
Thanked: 215 times

by kvcpk » Wed Jul 07, 2010 5:23 am
selango wrote:Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1

OA later
Let x>-3
then
x+3 = 4x-3
3x=6
x=2...YES
let x<-3
-x-3=4x-3
ILLEGAL Expression

Hence x=2 SUFF

|x + 1| = 2x - 1

Let x>-1
x+1=2x-1
x=2...YES
Let x<-1
-x-1=2x-1
ILLEGAL EXPRESSION
SUFF

Pick D
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members
Thanked: 173 times
GMAT Score:710

by gmatmachoman » Wed Jul 07, 2010 5:30 am
kvcpk wrote:
selango wrote:Is x > 0?

(1) |x + 3| = 4x - 3

(2) |x + 1| = 2x - 1

OA later
Let x>-3
then
x+3 = 4x-3
3x=6
x=2...YES
let x<-3
-x-3=4x-3
ILLEGAL Expression

Hence x=2 SUFF

|x + 1| = 2x - 1

Let x>-1
x+1=2x-1
x=2...YES
Let x<-1
-x-1=2x-1
ILLEGAL EXPRESSION
SUFF

Pick D
Bhai..dont u think the question is itself dubious?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1893
Joined: Sun May 30, 2010 11:48 pmFollowed by: 7 members
Thanked: 215 times

by kvcpk » Wed Jul 07, 2010 5:36 am
gmatmachoman wrote: Bhai..dont u think the question is itself dubious?
Yeah.. I am seeing this type of question for the first time.. But if the source is MGMAT, then it might be correct. :roll:
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 265
Joined: Mon Dec 28, 2009 9:45 pmFollowed by: 2 members
Thanked: 26 times
GMAT Score:760

by mj78ind » Wed Jul 07, 2010 6:26 am
I think it is a pretty good question, I mean there is no ambiguity etc ....... hence fair game for me

Plus I solved it right :)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 1460
Joined: Tue Dec 29, 2009 1:28 amFollowed by: 7 members
Thanked: 135 times

by selango » Wed Jul 07, 2010 6:52 am
This question is from MGMAT cat test.

I solved it as below,

From stmt1,

|x + 3| = 4x - 3

There are 2 ways to solve this,

x+3=4x-3

x=2

-(x+3)=4x-3 or -x-3=4x-3

x=0

Either x is 2 or 0.But only x=2 satisfies the Inequality.

So x=2.Sufficient.

From stmt2,

|x + 1| = 2x - 1

x+1=2x-1,x=2

-(x+1)=2x-2,x=0

But only x=2 satisfies the Inequality.

So x=2.Sufficient.

Hence D

First I solved the equation I thought to select option E as there are 2 values for x.

But I checked the Inequality with x values,only x=2 satisfy the equation.First time I am experiencing this strange thing.
--Anand--
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 748
Joined: Sun Jan 31, 2010 7:54 amFollowed by: 3 members
Thanked: 46 times

by outreach » Wed Jul 07, 2010 10:48 am
1
sQ BOTH SIDES
X2+6X+9=16X2-24X+9
15X2-30X=0
15X(X-2)=0
X=2,0


2
sQ BOTH SIDES
X2+2X+1=4X2+-4X+1
3X2-6X=0
X=2,0


NOT SURE HOW CAN OA BE d SINCE X HAS 2 VALUES. oNE SATISFIES THE COND WHEREAS OTHER DOES NOT
-------------------------------------
--------------------------------------
General blog
https://amarnaik.wordpress.com
MBA blog
https://amarrnaik.blocked/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 2326
Joined: Mon Jul 28, 2008 3:54 amFollowed by: 2 members
Thanked: 173 times
GMAT Score:710

by gmatmachoman » Wed Jul 07, 2010 11:06 am
selango wrote:This question is from MGMAT cat test.

I solved it as below,

From stmt1,

|x + 3| = 4x - 3

There are 2 ways to solve this,

x+3=4x-3

x=2

-(x+3)=4x-3 or -x-3=4x-3

x=0

Either x is 2 or 0.But only x=2 satisfies the Inequality.

So x=2.Sufficient.

From stmt2,

|x + 1| = 2x - 1

x+1=2x-1,x=2

-(x+1)=2x-2,x=0

But only x=2 satisfies the Inequality.

So x=2.Sufficient.

Hence D

First I solved the equation I thought to select option E as there are 2 values for x.

But I checked the Inequality with x values,only x=2 satisfy the equation.First time I am experiencing this strange thing.
Ok try putting x= 0 in st 1..u would see a illlogical condition..
IMO question seems wierd...need to ask stacey
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
GMAT Instructor
Posts: 1302
Joined: Mon Oct 19, 2009 2:13 pmFollowed by: 164 members
Location: Toronto
Thanked: 539 times
GMAT Score:800

by Testluv » Wed Jul 07, 2010 6:20 pm
Hi,

I've discussed this question before but it was taking too long to find my old post. We can solve more quickly by reasoning through the concept of absolute value rather than relying on pure algebra:

Is x > 0?

(1) |x + 3| = 4x - 3

We see the absolute value bars on the left. We know that absolute value is always positive or zero. We also know that, by definition, the left hand side of an equation must equal the right hand side. Therefore:

4x - 3 >= 0

x >= 3/4

If x is greater than or equal to 3/4, then it is definitely greater than 0.
SUFFICIENT.

(2) |x + 1| = 2x - 1

Similarly,

2x - 1>=0
x>= 1/2
SUFFICIENT.

Choose D.
Kaplan Teacher in Toronto
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 214
Joined: Mon Mar 29, 2010 1:46 pm
Location: Houston, TX
Thanked: 37 times
GMAT Score:700

by sk818020 » Wed Jul 07, 2010 6:26 pm
Testluv wrote:Hi,

I've discussed this question before but it was taking too long to find my old post. We can solve more quickly by reasoning through the concept of absolute value rather than relying on pure algebra:

Is x > 0?

(1) |x + 3| = 4x - 3

We see the absolute value bars on the left. We know that absolute value is always positive or zero. We also know that, by definition, the left hand side of an equation must equal the right hand side. Therefore:

4x - 3 >= 0

x >= 3/4

If x is greater than or equal to 3/4, then it is definitely greater than 0.
SUFFICIENT.

(2) |x + 1| = 2x - 1

Similarly,

2x - 1>=0
x>= 1/2
SUFFICIENT.

Choose D.
Great tip! Thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1