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 500,000 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]

Show Tags

28 Dec 2010, 17:54

8

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

55% (hard)

Question Stats:

60% (02:39) correct
40% (01:37) wrong based on 413 sessions

HideShow timer Statistics

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?

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

Please could someone explain why we cant take the smallest value as 0.

Note that k, m, r, s, and t are positive integers, thus neither of them can be zero.

Given: \(0<k<m<r<s<t=40\) and \(average=\frac{k+m+r+s+40}{5}=16\). Question: \(median_{max}=?\) As median of 5 (odd) numbers is the middle number (when arranged in ascending or descending order) then the question basically asks to find the value of \(r_{max}\).

\(average=\frac{k+m+r+s+40}{5}=16\) --> \(k+m+r+s+40=16*5=80\) --> \(k+m+r+s=40\). Now, we want to maximize \(r\):

General rule for such kind of problems: to maximize one quantity, minimize the others; to minimize one quantity, maximize the others.

So to maximize \(r\) we should minimize \(k\) and \(m\), as \(k\) and \(m\) must be distinct positive integers then the least values for them are 1 and 2 respectively --> \(1+2+r+s=40\) --> \(r+s=37\) --> \(r_{max}=18\) and \(s=19\) (as \(r\) and \(s\) also must be distinct positive integers and \(r<s\)). So, \(r_{max}=18\)

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

04 Apr 2014, 02:43

ajit257 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

Please could someone explain why we cant take the smallest value as 0.

Average of 5 integers = 16

So total = 80

Given that t = 40; so k+m+r+s = 80-40 = 40

Make two groups so that k+s = 20 & m+r = 20

We require to find the highest possible value of r which is the median, so take values of k & m the least possible & value of s maximum possible

1 < 2 < r < 20

20-2 = 18

So 18 is the maximum possible value to r which satisfies all the conditions given _________________

Kindly press "+1 Kudos" to appreciate

Last edited by PareshGmat on 04 Apr 2014, 02:56, edited 2 times in total.

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

04 Apr 2014, 02:47

sairajesh063 wrote:

Please help me out in this question.

The arithmetic mean of 5 positive integers a,b,c,d and e is 22 , and a<b<c<d<e.if e is 40,what is the greatest possible value of the median of the 5 integers?

Ans is : 33

a+b+c+d+e = 110

e = 40

a+b+c+d = 70

1<2<c<34

70/2 - 1 = 34 = Max possible value of d

34-1 = 33 = Maximum possible value of c = Answer

Details of calculation as provided in the post above _________________

Kindly press "+1 Kudos" to appreciate

Last edited by PareshGmat on 04 Apr 2014, 02:57, edited 1 time in total.

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

Please could someone explain why we cant take the smallest value as 0.

Average of 5 integers = 16

So total = 80

Given that t = 40; so k+m+r+s = 80-40 = 40

Make two groups so that k+s = 20 & m+r = 20

We require to find the highest possible value of r which is the median, so take value of k the least possible & value of s maximum possible

1 < m < r < 20

20-1 = 19

So 19 is the maximum possible value to r which satisfies all the conditions given

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

04 Apr 2014, 02:58

Bunuel wrote:

PareshGmat wrote:

ajit257 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

Please could someone explain why we cant take the smallest value as 0.

Average of 5 integers = 16

So total = 80

Given that t = 40; so k+m+r+s = 80-40 = 40

Make two groups so that k+s = 20 & m+r = 20

We require to find the highest possible value of r which is the median, so take value of k the least possible & value of s maximum possible

1 < m < r < 20

20-1 = 19

So 19 is the maximum possible value to r which satisfies all the conditions given

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

18 Jun 2014, 20:51

Since last digit t is 40 and arithmatic mean of all 5 digits is 16 k+m+r+s = 16*5 - 40 = 40

Questions is asking maximum possible value of median which is nothing but maximum possible value of r. median is middle number if there are odd number of numbers. here 5 are there, so middle one is median.

possible numbers, since they are in ascending positive integers, giving least value to k start with k=1, then m = 2(should be greater than k) sum of r+s = 40-3 = 37

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

A. 16 B. 18 c. 19 d. 20 e. 22

Using the logical approach:

We need to find the median which is the third value when the numbers are in increasing order. Since k<m<r<s<t, the median would be r.

The average of the positive integers is 16 which means that in effect, all numbers are equal to 16. If the largest number is 40, it is 24 more than 16. We need r to be maximum so k and m should be as small as possible to get the average of 16. Since all the numbers are positive integers, k and m cannot be less than 1 and 2 respectively. 1 is 15 less than 16 and 2 is 14 less than 16 which means k and m combined are 29 less than the average. 40 is already 24 more than 16 and hence we only have 29 - 24 = 5 extra to distribute between r and s. Since s must be greater than r, r can be 16+2 = 18 and s can be 16+3 = 19.

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

18 Jun 2014, 21:16

Given that k<m<r<s<t and t =40, we are required to maximize the median which is r. R can be maximized if k, m and s is least. As all are +ive integers the least k and m are 1 & 2 respectively. 's' is greater than r so minimum value of s should 1 more than r, therefore s = r + 1. Sum of all no. = mean * 5 = 16*5 = 80. It means k + m + r + S + t = 80, we can write 1 + 2 + r + r + 1 + 40 = 80 , it gives 2r=36, therefore r=18.

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

18 Jun 2014, 21:20

1

This post received KUDOS

jyothesh wrote:

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

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

19 Jul 2015, 06:17

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________

Re: The average (arithmetic mean) of the 5 positive integers [#permalink]

Show Tags

19 Jul 2015, 07:20

Same as Bunuel's..

\(\frac{(k+m+r+s+40)}{5} = 16\)

\((k+m+r+s) = 40\)

The only way by which the median can be maximised is by minimising the values lesser than the median and keeping the values greater than the median in such a way that their differences are minimum.

\(1+2+r+s+40 = 80\)

\(r+s = 80-40-2-1 = 37\)

Maximum value of \(r\) can be 18 (such that \(s = 19\)). So Ans (B). _________________

I used to think the brain was the most important organ. Then I thought, look what’s telling me that.

gmatclubot

Re: The average (arithmetic mean) of the 5 positive integers
[#permalink]
19 Jul 2015, 07:20

This is the kickoff for my 2016-2017 application season. After a summer of introspect and debate I have decided to relaunch my b-school application journey. Why would anyone want...

Check out this awesome article about Anderson on Poets Quants, http://poetsandquants.com/2015/01/02/uclas-anderson-school-morphs-into-a-friendly-tech-hub/ . Anderson is a great place! Sorry for the lack of updates recently. I...

“Oh! Looks like your passport expires soon” – these were the first words at the airport in London I remember last Friday. Shocked that I might not be...