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

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

The average (arithmetic) mean of the 5 positive integers [#permalink]
06 May 2008, 11:07

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The average (arithmetic) mean of the 5 positive integers k,m,r,s, and t is 16, and k<m<r<s<t. If t is 40, what is the greatest possible value of the median of the 5 integers?

Re: arithmetic mean [#permalink]
06 May 2008, 11:15

hmm let me try this

the sum of the 5 integers = 16*5 = 80

we know that t is 40 .. so k+m+r+s = 40

we are asked to maximize the median of the 5 numbers which means that we are really looking for 'r', in that case lets try minimizing k&m. Since they have to be integers let use k=1, m=2 .. that means that r+s = 37 ... since we want to maximize r ... and s>r ... the possible set of numbers could be 18,19.

Therefore r = 18.

OA? _________________

INSEAD Sept 2010 Interview Invite Nov 5, 2009Admit & Matriculating Wharton Sept 2010 Interview Invite Oct 30, 2009Waitlisted & Ding Harvard Sept 2010 Ding without Interview Ivey May 2010 Interview Invite Nov 23, 2009Admit + $$

Re: arithmetic mean [#permalink]
06 May 2008, 11:51

puma wrote:

The average (arithmetic) mean of the 5 positive integers k,m,r,s, and t is 16, and k<m<r<s<t. If t is 40, what is the greatest possible value of the median of the 5 integers?

a) 16 b) 18 c) 19 d) 20 e) 22

We want to make k,m = 1,2. We can't go any small b/c they must be integers and must be positive.

now we want to me s=s+1 b/c that way s is as great as it can be.

so now its just (1+2+40+r+r+1)/5=16 --> 80. 2r+44=80 --> 2r=36

B.

I originally made the same mistake and forgot that the integers cannot be the same.

Re: arithmetic mean [#permalink]
07 May 2008, 06:15

let r = 19 (plug it in ) => s = 20 (allw for max) => min man could be (40+20+19+2+1)/5 => 82/5 which is > 16. so rule out c,d,e now, consider r=18 => s => 19 => min mean (40+19+18+2+1)=> 80/5 => 16 ! => ans is b.

Re: arithmetic mean [#permalink]
08 May 2008, 11:17

puma wrote:

The average (arithmetic) mean of the 5 positive integers k,m,r,s, and t is 16, and k<m<r<s<t. If t is 40, what is the greatest possible value of the median of the 5 integers?

a) 16 b) 18 c) 19 d) 20 e) 22

Good question to make you think.... well, at least me...

Anyway, you have five values ___, ___, ___, ___, 40

Since the mean of the 5 numbers is 16, 16 x 5 = 80. We know the last value is 40, so the average of the remaining 4 numbers to find is 40 as well.

Let's assume the following:

1, 2, ___, ___, 40

so now we have 43 total, out of a possible 80. 80 - 43 = 37. 37 / 2 = 18.5, or since we need two increasing numbers, 18 and 19. Filling in the remaining numbers in our patterns yields:

1, 2, 18, 19, 40. Thus the median number must be 18.

B.

gmatclubot

Re: arithmetic mean
[#permalink]
08 May 2008, 11:17