B
y>0, what is the value of x?
stmnt 1: |x-3|>y
or |x-3|>0 (y>0), also absolute value is always positive or 0.
x can be 3 or any other number. it gives no unique value. not sufficient
stment 2: |x-3|<-y
or |x-3|<0, absolute value can never be negative. the least it can be is 0, or x=3 suffient, hence B
BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course
Redeem
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
x value
Source: Beat The GMAT — Data Sufficiency |
(i) Insufficient, x and y can take up many different values
(ii) Since the absolute value of a number is 0 or positve, it can never be negative.
So |x-3|> -y and can never be less than y.
So the only possibility is the |x-3|=-y (or 0), which is possible when x=3
Ans = B
Whats the OA
(ii) Since the absolute value of a number is 0 or positve, it can never be negative.
So |x-3|> -y and can never be less than y.
So the only possibility is the |x-3|=-y (or 0), which is possible when x=3
Ans = B
Whats the OA
Wanna know what I'm upto? Follow me on twitter: https://twitter.com/harikrish
it is obvious stmt 1 is insufficient.
I used Ian's distance method.. it is so cool!!! I seem to have removed an "absolute" mindblock!!
Stmt 2 is tricky and at the outset looks like a wrong stmt.
but |x - 3| is less than or equal to negative value of a VARIABLE, not a constant. So it is ok.
As ppl have explained above, y can only take 0.
we can as well equation to 0
ie | x -3 | = 0. Ie the distance between x and 3 is 0 on the numberline.
This can happen only if x = 3.
B
Million thanks to Ian for pushing the distance idea for absolutes, in multiple posts.
HT helps
I used Ian's distance method.. it is so cool!!! I seem to have removed an "absolute" mindblock!!
Stmt 2 is tricky and at the outset looks like a wrong stmt.
but |x - 3| is less than or equal to negative value of a VARIABLE, not a constant. So it is ok.
As ppl have explained above, y can only take 0.
we can as well equation to 0
ie | x -3 | = 0. Ie the distance between x and 3 is 0 on the numberline.
This can happen only if x = 3.
B
Million thanks to Ian for pushing the distance idea for absolutes, in multiple posts.
HT helps
