Find all School-related info fast with the new School-Specific MBA Forum

It is currently 27 Sep 2016, 09:34
GMAT Club Tests

Close

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

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.

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

The average (arithmetic mean) of the 5 positive integers k,

  post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
Manager
Manager
avatar
Joined: 03 Oct 2008
Posts: 62
Followers: 0

Kudos [?]: 26 [0], given: 0

The average (arithmetic mean) of the 5 positive integers k, [#permalink]

Show Tags

New post 05 Oct 2008, 16:49
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?

A. 16
B. 18
C. 19
D. 20
E. 22
Manager
Manager
avatar
Joined: 30 Sep 2008
Posts: 111
Followers: 1

Kudos [?]: 18 [0], given: 0

Re: math [#permalink]

Show Tags

New post 05 Oct 2008, 21:37
k < m < r < s < t, so the median of the list = r

r is the greatest possible when k, m, s are the smallest possible or k = 0, m = 1, s = r+1

sum = 0 + 1 + r + (r+1) + 40 = 16 x 5 = 80 => r = 19

So C is the answer
Manager
Manager
avatar
Joined: 25 Aug 2008
Posts: 230
Location: India
WE 1: 3.75 IT
WE 2: 1.0 IT
Followers: 2

Kudos [?]: 50 [0], given: 5

Re: math [#permalink]

Show Tags

New post 06 Oct 2008, 02:36
k < m < r < s < t, so the median in the following list is r.

r is the greatest possible when k, m, s are the smallest possible i.e. k = 1, m = 2, s = r+1
Note: k cannot be zero because they had mentioned k, m, r, s and t in question as positive integers.

sum = 1 + 2 + r + (r+1) + 40 = 16 x 5 = 80
=> r = 18

B is the answer. :-D
_________________

Cheers,
Varun


If you like my post, give me KUDOS!!

Intern
Intern
avatar
Joined: 21 Sep 2008
Posts: 18
Followers: 0

Kudos [?]: 4 [0], given: 0

Re: math [#permalink]

Show Tags

New post 06 Oct 2008, 02:37
The anser is B

0 is neither positive nor negative.
Hence the smallest of the 2 can be 1 and 2

1+2+r+s+40 = 80
r+s = 37
The maximum value r can take is 18

Hence answer B
Manager
Manager
avatar
Joined: 30 Sep 2008
Posts: 111
Followers: 1

Kudos [?]: 18 [0], given: 0

Re: math [#permalink]

Show Tags

New post 06 Oct 2008, 03:01
yes, zero is neither negative nor positive. Thanks!!! :-D
SVP
SVP
User avatar
Joined: 05 Jul 2006
Posts: 1525
Followers: 5

Kudos [?]: 258 [0], given: 40

GMAT ToolKit User
Re: math [#permalink]

Show Tags

New post 06 Oct 2008, 03:18
albany09 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


k+m+r+s+t = 80 if t = 40 thus k+m+r+s = 40

k,m,r,s have different values not = 0 and r is in the middle of the values = median

least values of k,m= 1,2

thus s+r = 80 - 43 = 37

r<s thus , s<40

r can only be 18 so that s can be = 19 ( ie slightly bigger)
Manager
Manager
avatar
Joined: 30 Sep 2008
Posts: 111
Followers: 1

Kudos [?]: 18 [0], given: 0

Re: math [#permalink]

Show Tags

New post 06 Oct 2008, 03:25
yezz wrote:
albany09 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


k+m+r+s+t = 80 if t = 40 thus k+m+r+s = 40

k,m,r,s have different values not = 0 and r is in the middle of the values = median

least values of k,m= 1,2

thus s+r = 80 - 43 = 37

r<s thus , s<40

r can only be 18 so that s can be = 19 ( ie slightly bigger)


Let replace r with 19, can you find the other integers?
SVP
SVP
User avatar
Joined: 05 Jul 2006
Posts: 1525
Followers: 5

Kudos [?]: 258 [0], given: 40

GMAT ToolKit User
Re: math [#permalink]

Show Tags

New post 06 Oct 2008, 03:53
lylya4 wrote:
yezz wrote:
albany09 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


k+m+r+s+t = 80 if t = 40 thus k+m+r+s = 40

k,m,r,s have different values not = 0 and r is in the middle of the values = median

least values of k,m= 1,2

thus s+r = 80 - 43 = 37

r<s thus , s<40

r can only be 18 so that s can be = 19 ( ie slightly bigger)


Let replace r with 19, can you find the other integers?


I am sorry , but i do not get your question, please rephrase
Re: math   [#permalink] 06 Oct 2008, 03:53
Display posts from previous: Sort by

The average (arithmetic mean) of the 5 positive integers k,

  post reply Question banks Downloads My Bookmarks Reviews Important topics  


GMAT Club MBA Forum Home| About| Terms and Conditions| GMAT Club Rules| Contact| Sitemap

Powered by phpBB © phpBB Group and phpBB SEO

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.