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 →

Modulus+inequality:GMATPREP

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 217
Joined: Sun Jan 10, 2010 1:39 pmFollowed by: 1 member
Thanked: 7 times
by zaarathelab » Mon Dec 19, 2011 1:26 am
If y is an integer and y = |x| + x, is y=0?
(1) x < 0
(2) y < 1

I am getting A as the answer whereas the OA is D

Statement 1

x<0

If x less than 0 the |x| + x = 0 (simple logic). This can be proved algebraically as well.

Hence y = 0. sufficient.

Statement 2

y<1

algebraically

case 1
|x| + x <1
x<1-x
x<1/2 - - - -1
case 2

x >- (1 -x) - - -- not possible. Discard
Hence x<1/2

Now x can take 1/3,1/4 value...in that case y<1 , but not 0, but if x=0 then y = 0.

Hence statement 2 is insufficient

Where am I going wrong?
Success = Max(Hardwork) + Min(Luck)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Data Sufficiency |
Legendary Member
Posts: 966
Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members
Thanked: 230 times

by shankar.ashwin » Mon Dec 19, 2011 1:38 am
Statement 2:

|x| + x can never be -ve for any value of 'x'

It can take positive values or 'zero'

given 'y' is an integer and < 1. Hence y = 0
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Community Manager
Posts: 1060
Joined: Fri May 13, 2011 6:46 amFollowed by: 52 members
Location: Utrecht, The Netherlands
Thanked: 318 times

by neelgandham » Mon Dec 19, 2011 2:13 am
If y is an integer and y = |x| + x, is y=0? The function y =f(x) = |x| + x can also be defined by
y = 2x, if x > 0 and
y = 0 if x < 0
(1) x < 0
If x < 0, According to the function, y = 0. hence sufficient!
(2) y < 1
From the function definition, We know that 0<y<1.y is an integer between 0 and 1(0 included).So, y = 0 is the only solution. hence sufficient!

Answer D
Anil Gandham
Welcome to BEATtheGMAT | Photography | Getting Started | BTG Community rules | MBA Watch
Check out GMAT Prep Now's online course at https://www.gmatprepnow.com/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Master | Next Rank: 500 Posts
Posts: 217
Joined: Sun Jan 10, 2010 1:39 pmFollowed by: 1 member
Thanked: 7 times

by zaarathelab » Mon Dec 19, 2011 2:19 am
shankar.ashwin wrote:Statement 2:

|x| + x can never be -ve for any value of 'x'

It can take positive values or 'zero'

given 'y' is an integer and < 1. Hence y = 0
i guess i overlooked the integer bit

Thanks for pointing it out
Success = Max(Hardwork) + Min(Luck)
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 934
Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members
Location: AAMCHI MUMBAI LOCAL
Thanked: 63 times

by [email protected] » Fri Feb 17, 2012 11:40 pm
If y is an integer and y = |x| + x, is y=0?
(1) x < 0
(2) y < 1

This is a yes or no question and statement 1 everybody got as sufficient.

There is a strong contradiction for statement 2 whether it is sufficient or not.

According to me, statement 2 is sufficient but because of a different reason. According to me y can never be equal to zero from statement 2 and that is what I got...

statement 2 says y < 1.

hence can we say that |x| + x < 1, from the equation that we are given...

Now try putting all the values of x.

if x = o then the inequality does not stand. x < 0 is also not possible as it does not stand in the inequality...

According to the prephrase that we received from the question, that y = 0 only if x < 0 or x = 0...

Both the conditions do not meet. statement 2 gives us an inequality and we have to assume that it is true and from that see if our condition is met.

It does not meet and therefore the answer is a confirm no...

actually from y < 1 it is proved that y is not negative not zero but between 0 and 1 only


Hence D....



If you see any mistake then please point out my mistake....
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 588
Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members
Location: New Delhi, India
Thanked: 130 times
GMAT Score:720

by rijul007 » Sat Feb 18, 2012 4:50 am
[email protected] wrote:If y is an integer and y = |x| + x, is y=0?
(1) x < 0
(2) y < 1

This is a yes or no question and statement 1 everybody got as sufficient.

There is a strong contradiction for statement 2 whether it is sufficient or not.

According to me, statement 2 is sufficient but because of a different reason. According to me y can never be equal to zero from statement 2 and that is what I got...

statement 2 says y < 1.

hence can we say that |x| + x < 1, from the equation that we are given...

Now try putting all the values of x.

if x = o then the inequality does not stand. x < 0 is also not possible as it does not stand in the inequality...

According to the prephrase that we received from the question, that y = 0 only if x < 0 or x = 0...

Both the conditions do not meet. statement 2 gives us an inequality and we have to assume that it is true and from that see if our condition is met.

It does not meet and therefore the answer is a confirm no...

actually from y < 1 it is proved that y is not negative not zero but between 0 and 1 only


Hence D....



If you see any mistake then please point out my mistake....
Acc to the stem, we infer the following-

If x>0,
y = 2x

If x<= 0,
y = 0

St(1)
x<0
y = 0

Suff


St(2)-
y < 1

For x<=0,
y =0

Lets check other possible values of y
For x= 1/2, y = 1
for x= 1, y=2
for x=3/2 , y = 3

If we make a set of all possible values of y
{0, 1, 2, 3, 4, 5, 6, 7.....}

Now acc to st(2)
The only value of y less than 1 is 0

Hence, suff

Option D


Amit -
How does x <= 0 not stand in the inequality, x+|x|<1?

For x = 0
0<1

For x= -1
-1+1<0
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 934
Joined: Tue Nov 09, 2010 5:16 amFollowed by: 14 members
Location: AAMCHI MUMBAI LOCAL
Thanked: 63 times

by [email protected] » Sat Feb 18, 2012 5:46 am
Yes I got my mistake brother Rijul... Thank You very much...

I am convinced that y = 0 satisfies the equation...
IT IS TIME TO BEAT THE GMAT

LEARNING, APPLICATION AND TIMING IS THE FACT OF GMAT AND LIFE AS WELL... KEEP PLAYING!!!

Whenever you feel that my post really helped you to learn something new, please press on the 'THANK' button.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1