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 →

What is the median (GMAT PREP 1)

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Master | Next Rank: 500 Posts
Posts: 435
Joined: Wed Nov 16, 2011 7:27 amFollowed by: 16 members
Thanked: 48 times
by alex.gellatly » Sun Sep 09, 2012 12:15 am
If 0<x<1, what is the median of the values x, x^-1, x^2, squareroot(x), and x^3?

x
x^-1
x^2
squareroot(x)
x^3

I figured it out after the test, but got it wrong originally. Just throwing it out for conversation..
A useful website I found that has every quant OG video explanation:

https://www.beatthegmat.com/useful-websi ... tml#475231
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
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 » Sun Sep 09, 2012 1:25 am
easiest way is to plug in.
Let x = 1/4, then
x = 1/4
x^-1 = 4
x^-2 = 16
√x = 1/2
x^3 = 1/64

So, 1/64,1/4,1/2,4,16 are in ascending order. i.e x^3 < x < √x < 1/x < 1/(x^2). So the median is [spoiler]√x[/spoiler].

IMO 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: 101
Joined: Sun Jun 03, 2012 10:10 pmFollowed by: 1 member
Thanked: 10 times

by everything's eventual » Sun Sep 09, 2012 6:21 pm
As x is a fraction, the the more times you multiply the number by itself the smaller it gets.

Therefore the smallest number will be x^3 and then the next smallest is x ^ 2. The largest number will be x ^ -1 as this will be a real number greater than 1.

So we have the following till now in ascending order : x^3 , x^2 , _ , _ , x^-1

squareroot (x) > x ( as x is a fraction, the squareroot will be greater than the number itself)

So we finally have the following :

x^3 , x^2 , x , sqrrt (x) , x ^-1

So the median is x

Above is the logical way of doing it. We can ofcourse use neelgandham's method of plugging-in values as it is a much quicker option. The only mistake in my humble opinion was that neelgandham has considered x ^ -2 whereas the problem states x ^2.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1