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The average (arithmetic mean) of the 5 positive integers [#permalink]
 28 Dec 2010, 17:54
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51% (03:15) correct
48% (01:43) wrong based on 2 sessions
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.
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Ajit
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]
28 Dec 2010, 18:15
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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. 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}=18Answer: B. Hope it's clear.
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]
28 Dec 2010, 21:19
Same approach as Bunuel's. Bunnel - Do we have more similar questions like this one for practice?
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]
29 Dec 2010, 00:58
Got 19 as the answer because forgot to consider that the integers need to be distinct and are not equal per the stem.
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Re: The average (arithmetic mean) of the 5 positive integers k, [#permalink]
29 Dec 2010, 01:30
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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
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Last edited by Bunuel on 12 Jan 2012, 07:24, edited 1 time in total.
NOTE: this question has the different average value from that of in the initial post, though the questions are basically the same. Thus merged.
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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 Given , mean is 22 total sum = 5 * mean = 5* 22 = 110 e is 40 so a+b+c+d = 70 and a<b<c<d<e Median will be the middle term so value of "c" will be median let a = 1 and b = 2 as we need to find the maximum value of median c+d = 70 -(a+b) = 70 -3 = 67 so c < 33.5 and d >33.5 (on dividing 67 by 2) . As c and d are integers c = 33 and d =34 Hence median is 33
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Thanks for the explanation kp 
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Arithmeti mean is 22
so sum of numbers = 5* 22 = 110
e is 40 so a+b+c+d = 70 and a<b<c<d<e
Median will be the middle term so value of "c" will be median
since a+b+c+d = 70, sum is fix. to get C max a and b need to be min
let us chose a = 1 and b = 2 as we need to find the maximum value of median
c+d = 70 -(a+b) = 70 -3 = 67
As c and d are integers in increasing order so
c = 33 and d =34
Hence median is 33
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sandeeepsharma wrote: Arithmeti mean is 22
so sum of numbers = 5* 22 = 110
e is 40 so a+b+c+d = 70 and a<b<c<d<e
Median will be the middle term so value of "c" will be median
since a+b+c+d = 70, sum is fix. to get C max a and b need to be min
let us chose a = 1 and b = 2 as we need to find the maximum value of median
c+d = 70 -(a+b) = 70 -3 = 67
As c and d are integers in increasing order so
c = 33 and d =34
Hence median is 33 Thank you Sandeep. Cheers, Rajesh
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Thanks rajesh it is my pleasure that i am able to help you
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Rajesh, can you please list the answer choices instead of the including the answer with the first post.
thanks for posting these questions.
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Nice question. Thanks. 33 is the answer. Thanks for the explanations as well guys.
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To find the greatest median, you need have smallest values for the numbers below the median. That is a=1, b=2. Now, subtract from the sum of numbers = (22 x 5) - (1 + 2 + 40) = 68. So, we need to fit in c and d within a sum of 68 such that c is the largest it can be. We get c = 33 and d = 35.
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]
28 Jun 2012, 11:01
Why cant we take the smallest number as zero
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Re: The average (arithmetic mean) of the 5 positive integers [#permalink]
28 Jun 2012, 11:03
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Re: The average (arithmetic mean) of the 5 positive integers
[#permalink]
28 Jun 2012, 11:03
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