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The average (arithmetic mean) of the 5 positive integers k, [#permalink]
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05 Oct 2008, 16:49
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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: math [#permalink]
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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



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Re: math [#permalink]
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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.
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Re: math [#permalink]
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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



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Re: math [#permalink]
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06 Oct 2008, 03:01
yes, zero is neither negative nor positive. Thanks!!!



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Re: math [#permalink]
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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)



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Re: math [#permalink]
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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?



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Re: math [#permalink]
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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










