The arithmetic mean of the list of numbers above is 4. If k

The arithmetic mean of the list of numbers above is 4.
So, (3 + k + 2 + 8 + m + 3)/6 = 4
Multiply both sides by 6 to get: 3 + k + 2 + 8 + m + 3 = 24
Simplify: 16 + k + m = 24
Subtract 16 from both sides to get: k + m = 8

If k and m are integers k ≠ m, what is the median of the list?
Let's assign some values to k and m that satisfy the above condition AND such that k + m = 8
How about k = 1 and m = 7

So, our set of values becomes {3, 1, 2, 8, 7, 3}

What is the median of the list?
Arrange numbers in ASCENDING ORDER to get: { 1, 2, 3, 3, 7, 8}
Since we have an EVEN number of values, the median will equal the AVERAGE of the 2 middlemost values
Median = (3 + 3)/2 = 6/2 = 3

Answer: C

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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)

Sol: from the given information we can say that k+m=8

Consider various values of k and m satisfying the above condition, we have

1,7
2,6
3,5

Note k and m can be any of the above numbers

Now for each combintion we see that median is average of 3rd and 4th term and in case it is 3

Ans C

Posted from my mobile device

I see that people have spent on average 2.50 mins on this problem. Here is an approach that will help you solve it within 30/45 secs. Important thing to know here is we don't need to test all the possible values. This a PS problem and the answer has to be unique. If different sets gave different answers the question would be invalid. Thus, even if we test one set of values, we can answer the question. So I just tested 1 and 7 for m and k which gives 3 as median. --> Ans: C

@



Hello Bunuel

need some help


List = {2,3,3,8,K,M}

Given that mean is more than 4

which implies 2+3+3+ 8+ K+M >4*6 --> K+M> 8

and we know K!=M

there fore

K=1 M=8 median = 3 { 1,2,3,3,8,8}

K=5 M =10 median = 4 {2,3,3,5,8,10}



I am sure I missing some thing can you please highlight my mistake

mean=4 not more that 4.

3, k, 2, 8, m, 3
The arithmetic mean of the list of numbers above (meaning above this line) is 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

Using the average formula: average = sum/number, we see that the sum of these numbers is 24. Thus we have:

3 + k + 2 + 8 + m + 3 = 24

16 + k + m = 24

k + m = 8

Since k ≠ m and assuming that k < m, then the ordered pairs of (k, m) could be (3, 5), (2, 6), (1, 7), (0, 8), etc.

Let’s investigate the possible ordered pairs further:

If (k, m) = (3, 5), then the numbers in ascending order are:

2, 3, 3, 3, 5, 8 --- with median = 3

If (k, m) = (2, 6), then the numbers in ascending order are:

2, 2, 3, 3, 6, 8 --- with median = 3

If (k, m) = (1, 7), then the numbers in ascending order are:

1, 2, 3, 3, 7, 8 --- with median = 3

If (k, m) = (0, 8), then the numbers in ascending order are:

0, 2, 3, 3, 8, 8 --- with median = 3

At this point, we can see that no matter how we “stretch” k and m (e.g., let’s say (k, m) = (-92, 100)), we would still have median = 3.

Answer: C

Why K or M cannot be negative?
Thank you.

Posted from my mobile device

MI83

Lets consider the negative ones also,
As we know, k+m = 8
then possible pair (9,-1) ; (10,-2) ; (11,-3).....

(1) In (9,-1) case ; the sequence becomes -1,2,3,3,8,9
Median will be = (3+3)/2 = 3

(2)In (10,-2) case; the sequence becomes -2,2,3,3,8,10
Median will be = (3+3)/2=3

(3)In(11,-3) case; the sequence becomes -3,2,3,3,8,11
Median will be = (3+3)/2 = 3

Median will always be "3"

Hope it helps

Neat question.

2,3,3,8,k,m

2 + 3 + 3 + 8 + k + m / 6 = 4
16 + k + m = 24

What's clear is that k < 8 and m < 8. So what could k and m possible be? We know they have to sum to 8.

1. k = 1, m = 7 (or vice versa) <--- Gives us: 1,2,3,3,7,8 so a median of 3
2. k = 2, m = 6 (or vice versa) <--- 2,2,3,3,6,8 so a median of 3
3. k = 5, m = 3 (or vice versa) <--- 2,3,3,3,5,8 ...3

Median is 3.

C IMO.

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