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In the list above, k, m, and n are three distinct positive i [#permalink]

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14 Jan 2013, 06:25

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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!!!!!!

Re: In the list above, k, m, and n are three distinct positive i [#permalink]

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14 Jan 2013, 13:15

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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.

Re: In the list above, k, m, and n are three distinct positive i [#permalink]

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28 Mar 2014, 07:13

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]

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26 May 2015, 04:09

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Re: In the list above, k, m, and n are three distinct positive i [#permalink]

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26 May 2015, 20:19

Expert's post

daviesj wrote:

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!!!!!!

I would use the method of elimination of options here since the question says "cannot be the value". All I have to do is prove that the rest of the options can be values of k, m and n.

Since median is 6.5, the first options I will try are 6 and 7 - two numbers will be eliminated if I can find a case where this works.

3, 4, 6, 7, 20 - the avg of these numbers is 8. If the last number is also 8, the mean will remain 8 as desired. So in fact, we eliminated 3 options here. k, m and n can be 6, 7 and 8.

Let's try 5 now, not 9 because 9 is more complicated. 9 gives us two number less than 6.5 and 2 more than 6.5. So there will be many different options. If instead 5 is in the list, we now have 3 numbers less than 6.5, so the other 3 numbers must be greater than 6.5 and the average of one of those numbers with 5 must be 6.5. So the fourth number should be 8 on the list to give the median 6.5. These 5 numbers (3, 4, 5, 8, 20) give an average of 8. The sixth number must be 8 to keep the average 8 but numbers must be distinct. So this is not possible. Hence none of k, m and n can be 5.

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