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
Gmat prep# DS AM
Source: Beat The GMAT — Data Sufficiency |
Here is the solution:aditi_bc wrote:pl solve
AM = (x+2+7+11+16)/5 = (x+36)/5
given AM(x,2,7,11,16) = Median
so lets think the possibilities of x from the (x+36)/5:
x (x+36)/5
-- ----------
4 8 --8 is not there in the set
9 9 --9 can be included in place of x which can also be the median
14 10 --10 is not there in the set and so on..
19 11 --19 can be included in the place of x as 11 can be median
Statement 1 --> 7<x<11
Only 9 is fits in x(as per above possibilities). Hence this statement is sufficient.
Statement 2 --> x is the median of 5 numbers.
From the above possibilities only 9 and 19 can be included in the set. But as x is also a median..19 cannot be median. Because if 19 is included in the set, then 2,7,11,16,19 --> 19 is AM and 11 is median which fails. If you consider 9, then 2,7,9,11,16 --> 9 is both AM and median
hence this statement is sufficient.
So OA is D
Given
The average of x,2,7,11,16 is equal to the median.
1) 7<x<11 from this we know x is the median
Total count of numbers given is odd then after arranging them in ascending order the middle value is the median
2,7,x,11,16
2+7+x+11+16/5 = x
We can solve for x
SUFFICIENT
2)x is the median so we know the order is again
2,7,x,11,16
2+7+x+11+16/5 = x
We can solve for x
SUFFICIENT
Hence D)
The average of x,2,7,11,16 is equal to the median.
1) 7<x<11 from this we know x is the median
Total count of numbers given is odd then after arranging them in ascending order the middle value is the median
2,7,x,11,16
2+7+x+11+16/5 = x
We can solve for x
SUFFICIENT
2)x is the median so we know the order is again
2,7,x,11,16
2+7+x+11+16/5 = x
We can solve for x
SUFFICIENT
Hence D)
