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 →

Mod Problem

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 458
Joined: Sun Aug 31, 2008 10:44 amFollowed by: 1 member
Thanked: 3 times
by gmat009 » Tue Oct 07, 2008 4:06 pm
Is |x| + |x -1| = 1
(1) x >= 0
(2) x <= 1
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is
sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient.

OA is C, IMO E
Plz. explain
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
User avatar
Master | Next Rank: 500 Posts
Posts: 246
Joined: Mon May 19, 2008 7:34 am
Location: Texaco Gas Station
Thanked: 7 times

by cubicle_bound_misfit » Tue Oct 07, 2008 5:54 pm
i try to solve this as great 'guru' Ian Stewart taught me long time back .

try to draw a numbe line.

now we have to prove if

distance of x from 0 + distance of x from 1 is equal to 1

try to draw all the different scenarios where this is possible.

if x <0 you will find in the number line however near x is to the 0 (left side of the zero) it will always be greater than 1 try with -.99 , -1/2 etc.

So, this equation to be true x should be >0

now think of two ranges in the number line

0<= x<=1 and x>1 for any value of x > 1 the mod itself will be greater than 1 so only possible range is [0,1] which we get by C.

hope this will help.
please let me know if this approach is wrong.
Cubicle Bound Misfit
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 458
Joined: Sun Aug 31, 2008 10:44 amFollowed by: 1 member
Thanked: 3 times

by gmat009 » Tue Oct 07, 2008 6:21 pm
cubicle_bound_misfit wrote:i try to solve this as great 'guru' Ian Stewart taught me long time back .

try to draw a numbe line.

now we have to prove if

distance of x from 0 + distance of x from 1 is equal to 1

try to draw all the different scenarios where this is possible.

if x <0 you will find in the number line however near x is to the 0 (left side of the zero) it will always be greater than 1 try with -.99 , -1/2 etc.

So, this equation to be true x should be >0

now think of two ranges in the number line

0<= x<=1 and x>1 for any value of x > 1 the mod itself will be greater than 1 so only possible range is [0,1] which we get by C.

hope this will help.
please let me know if this approach is wrong.
I have not understood this solution, so I am not sure this approach is right or wrong but there must be more easy way to solve this problem
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 105
Joined: Tue Jul 15, 2008 7:03 pm
Thanked: 5 times

by pre-gmat » Tue Oct 07, 2008 7:07 pm
I think there's an easy solution to this.

a and b all by itself is not sufficient.

Combining both X>=0 and X<=1 we get

0<=x<=1

so x is basically anywhere from 0 to 1.

Try any value between 0 and 1 (try 0 and 1 as well) you will get the equation

|x| + |x-1| = 1
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Wed May 14, 2008 11:09 pm

by lilweezy » Tue Oct 07, 2008 8:15 pm
you have to look at each statement on it's own for data sufficiency. i think the answer is A.

|x| + |x-1| = 1

(1) x>=0
(2) x<=1

for option (1), any value you put for x that satisfies the equation works in the formula. for option (2), you get different answers for x=1 or x=-1. therefore statement (2) is not sufficient, but statement (1) is.

Answer is A.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 45
Joined: Sun Mar 23, 2008 7:33 amFollowed by: 1 member
GMAT Score:750

by schakiiieee » Wed Oct 08, 2008 3:47 pm
lilweezy wrote:you have to look at each statement on it's own for data sufficiency. i think the answer is A.

|x| + |x-1| = 1

(1) x>=0
(2) x<=1

for option (1), any value you put for x that satisfies the equation works in the formula. for option (2), you get different answers for x=1 or x=-1. therefore statement (2) is not sufficient, but statement (1) is.

Answer is A.
No it does not. Take x=5. --> 5 + 4 = 9 which is not 1.
Any value that is between 0 and 1:
e.g. x=1/4
1/4 + |1/4-1| = 1/4 + |-3/4| = 1.
Works.
(as it is continuous between 0 and 1 --> ANS C)
For any x > 1 it fails!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Junior | Next Rank: 30 Posts
Posts: 25
Joined: Thu Oct 02, 2008 2:11 pm
Thanked: 3 times

by namitrajiv » Wed Oct 08, 2008 9:28 pm
should be C

|X| = X when x>0

|X-1| = 1-X when x<1

combine x+ 1-x = 1 , thats what we need

first statement says x>=0 therefore we can conclude that |x| = x
but we are not sure about |x-1|
second statemnet says x<=1 thus we knw |x-1| =1-x
thus combining the two we can have the ans...
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 67
Joined: Tue Sep 16, 2008 9:29 pm
Location: new york
Thanked: 2 times

by Gmatss » Thu Oct 09, 2008 8:41 am
IMO C

Combining both statment

0<=x<=1

Any positive fraction between 0 and 1 inclusive satifies the condition
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Wed May 14, 2008 11:09 pm

by lilweezy » Thu Oct 09, 2008 2:08 pm
schakiiieee wrote:
lilweezy wrote:you have to look at each statement on it's own for data sufficiency. i think the answer is A.

|x| + |x-1| = 1

(1) x>=0
(2) x<=1

for option (1), any value you put for x that satisfies the equation works in the formula. for option (2), you get different answers for x=1 or x=-1. therefore statement (2) is not sufficient, but statement (1) is.

Answer is A.
No it does not. Take x=5. --> 5 + 4 = 9 which is not 1.
Any value that is between 0 and 1:
e.g. x=1/4
1/4 + |1/4-1| = 1/4 + |-3/4| = 1.
Works.
(as it is continuous between 0 and 1 --> ANS C)
For any x > 1 it fails!
you're right, i was totally off on this one.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 2623
Joined: Mon Jun 02, 2008 3:17 amFollowed by: 355 members
Location: Montreal
Thanked: 1090 times
GMAT Score:780

by Ian Stewart » Thu Oct 09, 2008 2:47 pm
gmat009 wrote:
I have not understood this solution, so I am not sure this approach is right or wrong but there must be more easy way to solve this problem
The approach is very easy, and it is the most useful approach to know for the more difficult absolute value questions on the GMAT. It relies on the following:

|x| = the distance between x and zero on the number line
|x - y| = the distance between x and y on the number line

So, when the question asks if |x| + |x-1| = 1, it's just asking "Is the distance from x to zero plus the distance from x to 1 equal to 1?" And if you draw a number line, it's easy to see that this will be true if (and only if) x is between zero and one. So C.

I do think it takes a bit of time to get used to thinking this way about absolute value, but this question literally takes ten seconds if you can approach it as a distance question, not as an algebra question.
For online GMAT math tutoring, or to buy my higher-level Quant books and problem sets, contact me at ianstewartgmat at gmail.com

ianstewartgmat.com
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 1578
Joined: Sun Dec 28, 2008 1:49 amFollowed by: 9 members
Thanked: 82 times
GMAT Score:720

by maihuna » Wed Apr 15, 2009 7:58 am
my algebra approach will be:

Using 1: x + |x-1| =1
if x<1 x + (-(x-1)) = 1 or 1=1 true
if x>=1 x+x-1 =1 or x=1 but we dont know this

Using 2: If x<=1 we know we will get part 1 as above and so the answer C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Master | Next Rank: 500 Posts
Posts: 246
Joined: Mon May 19, 2008 7:34 am
Location: Texaco Gas Station
Thanked: 7 times

by cubicle_bound_misfit » Wed Apr 15, 2009 10:01 am
nother blast from the past
Cubicle Bound Misfit
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 38
Joined: Tue Jan 27, 2009 10:24 am

by sanjay_dce » Wed Apr 15, 2009 10:46 am
Ian Stewart wrote:
gmat009 wrote:
I have not understood this solution, so I am not sure this approach is right or wrong but there must be more easy way to solve this problem
The approach is very easy, and it is the most useful approach to know for the more difficult absolute value questions on the GMAT. It relies on the following:

|x| = the distance between x and zero on the number line
|x - y| = the distance between x and y on the number line

So, when the question asks if |x| + |x-1| = 1, it's just asking "Is the distance from x to zero plus the distance from x to 1 equal to 1?" And if you draw a number line, it's easy to see that this will be true if (and only if) x is between zero and one. So C.

I do think it takes a bit of time to get used to thinking this way about absolute value, but this question literally takes ten seconds if you can approach it as a distance question, not as an algebra question.
Thank you Ian you Rock!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Legendary Member
Posts: 621
Joined: Wed Apr 09, 2008 7:13 pmFollowed by: 4 members
Thanked: 33 times

by vittalgmat » Wed Apr 15, 2009 10:59 am
Thanks Ian ,
as u said it takes a little time to get used to looking at absolute as a distance and imagine in terms of number line.
But the idea is paying off.
Many thanks for that.

Needless to say my ans is a C
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1