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

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The average (arithmetic mean) of the 5 positive integers k, [#permalink] New post 04 Mar 2006, 11:34
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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
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Re: PS: Greatest possible median [#permalink] New post 04 Mar 2006, 12:25
vivek123 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

total sum = 80
t = 40
rest (k+m+r+s) = 40

given that, median = r.
r is gretest when (k+m) is least and r = s but r cannot be equal to s because s>r.

so lets suppose k+m = 1+2 = 3, then r+s = 37.

the gretest r = 18
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Re: PS: Greatest possible median [#permalink] New post 04 Mar 2006, 12:32
vivek123 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


B.

k+m+r+s+t/5 = 16
since t=40 then k+m+r+s = 40

Median of k,m,r,s, and t is r since they are increasing order.

so r+s<40
R cannot be >20 because s which is > r has to be > 20.
If r = 19, s will be 20 and then m can be 1 and k will be 0. But 0 is not a positive integer.
So r = 18, which makes s=19 and m=2 and k = 1.
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 [#permalink] New post 04 Mar 2006, 19:07
I started with answer choices ;)

t = 40.

k < m < r < s < t

For 'r' to be greatest possible, s-r = 1
if r = 16 then s = 17 => r+s = 33 (still there is room for k & m)
if r = 18 then s = 19 => r+s = 37 (still there is room for k & m, boundary case & hence the answer)
if r = 19 then s = 20 => r+s = 40 (no room for k & m)

Therefore answer is B
  [#permalink] 04 Mar 2006, 19:07
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