Panaac wrote:Can someone please solve this problem for me?

3 K 2 8 M 3

The arithmetic mean of the list of numbers above is 4. If K and M are integers, and K ≠ M, what is the median of the list?

a) 2
b) 2.5
c) 3
d) 3.5
e) 4

The answer is C.

aritmetic mean=(3+k+2+8+m+3)/6
4=(3+k+2+8+m+3)/6
24=16+k+m
8=k+m
k +m=8 so we can say k,m<=8 k,m=(1.7) (3,5) (2,6) we know K ≠ M
so arranging this number in ascending order 3,5
2,3,3,3,5,6 median 3+3/2=3
now let us take 1,7
1,2,3,3,6,7 so median again 3+3/2=3
now let us take 2,6
then 2,2,3,3,6,6 again median 3+3/2=3 answer c

Thank you so much, amissing6!

I just have 1 question. When it says K and M are integers, does it means K and M can be negative or 0?

Panaac wrote:Thank you so much, amissing6!

I just have 1 question. When it says K and M are integers, does it means K and M can be negative or 0?

k and m integer means it can be 0 or -negative but in this case it wont be there

Panaac wrote:Thank you so much, amissing6!

I just have 1 question. When it says K and M are integers, does it means K and M can be negative or 0?

well it can mean. however result won't change.

Panaac wrote:Can someone please solve this problem for me?

3 K 2 8 M 3

The arithmetic mean of the list of numbers above is 4. If K and M are integers, and K ≠ M, what is the median of the list?

a) 2
b) 2.5
c) 3
d) 3.5
e) 4

The answer is C.

median is the middle number in odd set, or the average of two middle numbers in even set, then on some level answer has to be 3, which is C. K and M would have to be elements that are both below three or both above three to change the value of the median. but since both K+M=8, all combinations of K and M that can add up to 8 don't disturb the median i.e. 1+7, 2+6, 3+5, 5+3, 6+2, 7+1, etc.

For a lot of average problems, it is helpful to think of the "balanced scale".

We know the average of all the numbers is 4. In the set, we see the number "2". This number takes us 2 to the left of the average of 4. Each "3" takes us 1 to the left. The "8" takes us 4 to the right. To sum up:

LEFT-----------RIGHT
-2----------------+4
-1
-1

We have -4 on the left, and +4 on the right. Thus, the numbers on the left balance out the ones on the right. Thus, K and M cannot collectively deviate from the set's average of 4 (not without changing the average). Thus, the average of K and M must be 4. Thus, their sum must be 8. Because we are told that K does not equal M, one possible solution set is 5 and 3 (two unequal numbers that add to 8).

Let's observe the median if K and M are 5 and 3:

{2, 3, 3, 3, 5, 8}

We have an even number of numbers in the set (ie, 6). Thus, the median is the average of the middle 2. Clearly, here the median is 3.

Because there can be only one correct answer, and because we don't have "it cannot be determined" as one of our answer choices, the answer must be "3", and there is no need to consider other solution sets (6 and 2, 7 and 1, 8 and 0, etc.).

Choose C.

took me 4 mins to work this out!!

Panaac wrote:Can someone please solve this problem for me?

3 K 2 8 M 3

The arithmetic mean of the list of numbers above is 4. If K and M are integers, and K ≠ M, what is the median of the list?

a) 2
b) 2.5
c) 3
d) 3.5
e) 4

We can create the following equation:

(3 + 2 + 8 + 3 + K + M)/6 = 4

16 + K + M = 24

K + M = 8

Since K cannot equal M, there is no way for K and M to be 4. Also, since both K and M are integers, one of the numbers must be less than or equal to 3 and the other greater than or equal to 5.

If one of the numbers is 3 and the other is 5, then in ascending order, the list would be:

2, 3, 3, 3, 5, 8

We see that the median is 3.

If one of the numbers is less than 3 and the other is greater than 5 (say K < 3 and M > 5), we see that when we list the numbers in ascending order, K will be before the two 3's and M will be after the two 3's. Thus, the two 3's must be in the 3rd and 4th positions, making the the median to be 3 again.

Answer: C