Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Jun 1006:00 AM PDT
-06:15 PM PDT
Register for the GMAT Club Virtual MBA Spotlight Fair – the world’s premier event for serious MBA candidates. This is your chance to hear directly from Admissions Directors at nearly every Top 30 MBA program..
03 Oct 2004, 22:29
Kudos
Bookmarks
can any one explain the answer
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
04 Oct 2004, 01:07
Kudos
Bookmarks
For stem 1:
We can think of Two scenarios:
First scenario: S=(-3,-1,1,3,5). Here the mean = 5/5 = 1. Median =1
Second scenarion: S =(-3,-1,1,3,5,7). Here the mean = 12/6 = 2. Median =(1+3)/2 = 2. Same results. We can conclude that the avg = median. A firm Yes.
Thus, sufficient.
For stem 2:
Its given that the range is 2(n-1). Lets consider a 4 member set. We will then have the range as 2(4-1) =6. The set can be S= (1,0,0,7) which has Range = 6. The avg is not equal to Median here. The set can also be (0,3,3,6) again the range being 6. However, here, the avg = median. We cannot conclude that avg = median for sure with this stem. Thus Insufficient.
The answer is A.
We can think of Two scenarios:
First scenario: S=(-3,-1,1,3,5). Here the mean = 5/5 = 1. Median =1
Second scenarion: S =(-3,-1,1,3,5,7). Here the mean = 12/6 = 2. Median =(1+3)/2 = 2. Same results. We can conclude that the avg = median. A firm Yes.
Thus, sufficient.
For stem 2:
Its given that the range is 2(n-1). Lets consider a 4 member set. We will then have the range as 2(4-1) =6. The set can be S= (1,0,0,7) which has Range = 6. The avg is not equal to Median here. The set can also be (0,3,3,6) again the range being 6. However, here, the avg = median. We cannot conclude that avg = median for sure with this stem. Thus Insufficient.
The answer is A.
04 Oct 2004, 07:00
Kudos
Bookmarks
GREAT EXPLANATION!
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
