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

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Manager
Joined: 03 Oct 2008
Posts: 61
The average (arithmetic mean) of the 5 positive integers k, [#permalink]

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05 Oct 2008, 15:49
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
Joined: 30 Sep 2008
Posts: 111

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05 Oct 2008, 20:37
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

Manager
Joined: 24 Aug 2008
Posts: 213
Location: India
WE 1: 3.75 IT
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06 Oct 2008, 01:36
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

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Intern
Joined: 21 Sep 2008
Posts: 17

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06 Oct 2008, 01:37
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

Manager
Joined: 30 Sep 2008
Posts: 111

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06 Oct 2008, 02:01
yes, zero is neither negative nor positive. Thanks!!!
Retired Moderator
Joined: 05 Jul 2006
Posts: 1747

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06 Oct 2008, 02:18
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
Joined: 30 Sep 2008
Posts: 111

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06 Oct 2008, 02:25
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?
Retired Moderator
Joined: 05 Jul 2006
Posts: 1747

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06 Oct 2008, 02:53
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, 02:53
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# The average (arithmetic mean) of the 5 positive integers k,

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