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In the list above, k, m, and n are three distinct positive i

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Manager
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Kudos [?]: 60 [0], given: 35

In the list above, k, m, and n are three distinct positive i [#permalink] New post 14 Jan 2013, 05:25
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D
E

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  45% (medium)

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60% (03:52) correct 40% (02:55) wrong based on 30 sessions
3, k, 20, m, 4, n
In the list above, k, m, and n are three distinct positive integers and the average (arithmetic mean) of the six numbers in the list is 8. If the median of the list is 6.5, which of the following CANNOT be the value of k, m, or n ?

A.9
B.8
C.7
D.6
E.5

I didn't get the answer.My answer came out to be A.9..Help!!!!!!
[Reveal] Spoiler: OA

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Kudos [?]: 13 [1] , given: 4

Re: In the list above, k, m, and n are three distinct positive i [#permalink] New post 14 Jan 2013, 12:15
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sum of all numbers = 8*6=48

and so k+m+n =21----------- (1)

in this list it can be safely assumed that 20 is the biggest number, because even if we assume 21 to be one of the nos, it would not satisfy the sum =48 condition.

median is (sum of middle two terms)/2, if no of terms is even.

So 3rd term + 4th term = 13 ---(2)

So we have to arrange the six number in increasing order. both 3 and 4 cannot be 3rd and 4th terms so that means, there is at max one term which is less than 3. So then there are two cases:

1st case. let us assume that k is less than 3: the order is k,3,4,m,n,20

4+m=13 from (2)
m=9

so from--(1) k+n=12
which means out of the five options 9 is already in.
But since k is less than 3, the values of n would be either 11 or 12 (not in options)

2nd Case: if k is greater than 4

3,4,k,m,n,20

in this case k+m=13 and hence n = 8
now k is greater than 4, if k =5 then m=8 which cant be because already n=8 and as per question all k, m and n are distinct integers.

hence 5 cant be the choice.

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Kudos [?]: 60 [0], given: 35

Re: In the list above, k, m, and n are three distinct positive i [#permalink] New post 14 Jan 2013, 17:48
DISTINCT...I missed this word...so i eliminated choice E,as I got the values
3-4-5-8-8-20 ,which had median as 6.5...thnks for the explaination...+1

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Re: In the list above, k, m, and n are three distinct positive i [#permalink] New post 28 Mar 2014, 06:13
Option E.
Sum of 6 nos.=8*6=48
So k+m+n=48-27=21
Also,median in case of even no. of terms=average of mid two terms=>6.5*2=13=third+fourth terms.
Actually,when we take any of k,m,n to be 5,we get the other two values to be equal which is not possible since
the question specifically asks for DISTINCT integers.
Re: In the list above, k, m, and n are three distinct positive i   [#permalink] 28 Mar 2014, 06:13
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In the list above, k, m, and n are three distinct positive i

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