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.
Answer (E)
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