The median of the list of positive integers above is 5.5

8, 5, x, 6

The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?
A. 3
B. 5.5
C. 6.25
D. 7
E. 7.5

Notice that we are told that x is a positive integer.

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order.

Since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}. Now, since \(x\leq{5}\) then the maximum sum of the numbers is 5+5+6+8=24, so the maximum average is 24/4=6 (options C, D, and E are out). 3 cannot be the average because in this case the sum of the numbers must be 4*3=12, which is less than the sum of just two numbers 6 and 8 (remember here that since x is a positive integer it cannot decrease the sum to 12). So we are left only with option B (for x=3 the average is (3+5+6+8)/4=22/4=5.5).

Answer: B.
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Given "The median of the list of positive integers above is 5.5"

So arranging the numbers from lowest to highest, avg of 2 middle terms needs to be 5.5

so the sequence will be x 5 6 8

Lets say x = 4, which gives us mean = sum/4 = 5.5

B is correct

This is amazing soln
I solved by plugging in..

(8+5+x+6)/4 . .
plug nos from options 1 by 1..

when ans 5.5 is plugged in,you get x=3. .
so you have 3,5,6,8 as the order which confirms with the given data
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8, 5, x, 6

The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?

If the median is 5.5, then x cannot be >8 ( 5,6,8,9 median 7>5,5). It cannot be > 6 (5,6,7,8 median 6.5>5.5).
So we have found that \(x<=5\)

\(\frac{x+5+6+8}{4}=mean\)
\(x=mean*4-19\)
x is an integer so 4*mean - 19 must be an integer, option (C)6.25 is out.
(D)7 (E)7.5 these options return values of x > 9, so they cannot be right.
Option (A)3 returns a negative integer, but x must be positive.
Option (B)5.5 is correct, and the value of x is 3, which is possible given the initial condition of "The median of the list of positive integers above is 5.5"

The median will be the average of the two elements in the middle as the list contains even number of elements.
So,the sum of the two numbers in the middle will be 5.5 * 2 = 11.

With the above data,we can write the list as follows :

x 5 6 8 where 1 < x < 6.
Avg = (x+5+6+8)/4 = (19+x)/4.
So, 5< Avg <6

Hence, B will be the answer.

** Alternatively, we can work out this problem taking options one by one.

The reason c is out is because if the average was 6.25, x = 6 and the media would become 6

(6.25 x 4) = 25

x = 25 - 19

8, 5, x, 6

The median of the list of positive integers above is 5.5. Which of the following could be the average (arithmetic mean) of the list?
A. 3
B. 5.5
C. 6.25
D. 7
E. 7.5

Notice that we are told that x is a positive integer.

Experts your view please??

In my opinion this q is flawed unless it should explicitly mentioned "list of different positive no"

list could be 5,6,6,8 and in that case we have mean 6.25

the question directly says that the median is 5.5. this can be only if 5 and 6 are the middle numbers. thus, x must be <=5.
now, let's take x=5 = the sum of the numbers is 24, and the average is 6. since the maximum value of x can be 5, it must be that the maximum value of the mean can't be greater than 6. thus, we can eliminate C, D, and E.

now, we have the sum = 24.
we have in the answer choices 3 and 5.5
since these numbers should be the mean, it means that 12 or 22 is the sum of the integers.
Now, we have 5,6, and 8. the sum of these numbers is way above 12. We are told that we have only positive numbers. thus, 12 can't be the sum and 3 can't be the average. A is out, and B must be the possible average. to test, let's say that x=3. now, we have 3+5+6+8=22. 22/4 = 5.5.

all good.

looks like I got to the right answer the way bunuel did

Excellent Question this one.
Here is my solution -->

Notice that the number of terms here are even => The median of a set with even number of terms is the average of two middle terms when arranged in either increasing or decreasing order.

Median =5.5

Hence the sum of two middle terms must be 11

For that allowed values of positive integer x => 1,2,3,4,5

Hence the mean => 19+x/4 = Value
x=4*Value -19

This value is nothing but the option.

Lets put in each value and see which on of them gives us x as {1,2,3,4,5)
A)Value=3 => x=12-19=-7 => Rejected
B)Value=>5.5=> x=22-19=3=> Allowed
C)Value=6.25=>x=25-19=6=> Rejected

D)Value=7=>x=28-19=9=> Rejected
E)Rejected


Hence B


Method 2=>
Back solving

Let Mean = 3
Hence x=-7 => Rejected
Mean = 5.5 => x=3 => Median of set = 5.5 => Accepted.

Hence B
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Median = 5.5 = Average of two middle numbers in case the number of terms in teh set is even.

i.e. 5 and 6 are the two middle terms of the set

i.e. x must be smaller than 5 for it to be the second term when terms are arranged in the ascending order

so the sum of the terms in the set = x+5+6+8 = x+19

average = (x+19)/4 = (x/4)+4.75 and since x is smaller than or equal to 5

average ≤ (5/4)+4.75

i.e. average ≤ (1.25)+4.75

average ≤ 6

since x is positive hence average > 4.75

ANswer: Option B

The median is above 5.5, shouldn't we consider options like x,5,6,8 & 5,6,8,x & 5,6,x,8 ?

What Am I missing?

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. So, if the set in ascending order is {5, 6, x, 8} (meaning if \(6\leq x \leq 8\)), then the median would be the average of 6 and x, and it cannot be 5.5, it could be at least 6, for x = 6. Similarly if if the set in ascending order is {5, 6, 8, x} (meaning if \(x \geq 8\)), then the median would be the average of 6 and 8, so 7, not 5.5.

Thus, since given that the median is 5.5 (the average of 5 and 6) then integer x must be less than or equal to 5, in order 5 and 6 to be two middle numbers. So our list in ascending order is: {x, 5, 6, 8}.

Check complete solution here: https://gmatclub.com/forum/the-median-o ... l#p1064955

Hope it helps.
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Solution:

Since the median is 5.5, which is (5 + 6)/2, we see that x must be the smallest number in the list, i.e., x ≤ 5. However, we are also given that x is a positive integer, so x ≥ 1. We can create the following inequality for the average:

(1 + 5 + 6 + 8)/4 ≤ (x + 5 + 6 + 8)/4 ≤ (5 + 5 + 6 + 8)/4

5 ≤ (x + 5 + 6 + 8)/4 ≤ 6

Since only 5.5 is between 5 and 6, we see that choice B is the correct answer.

Answer: B
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median = 5.5, so middle numbers MUST be 5 and 6
x 5 6 8; x<=5

24/4=6
23/4=5.75
22/4=5.5 ---> answer
21/4=5.25
20/4=5

The arithematic meani should be greater than 3 since 19/4 =4.25 therefore A is out
and it should be less than 6.25 since the median is 5.5
and plugging in 5.5 we get x=3 which satisfies

Therefore IMO B

let's go by the options

A if 3 is the mean of 4 numbers then total is 12. This is not possible as x is >0 and rest of the numbers total more than 12. Out
B if 5.5 is the mean then total is 22 which means x is 22-(8+5+6) =22-19 which is 3. the set now looks like {3,5, 6, 8} and median of this set is 5.5
C if 6,25 then total is 25 which means x is 25-19 =6. the set looks like ( 5,6,6,8) but the median of this set is 6. therefore out

D and E are > 6.25 and therefore the median in these cases would result in >=6 out

B is our answer

All the solutions have considered median to be equal to 5.5, where as the question clearly says that the median is above 5.5. can any one please clarify

I think you misunderstood this part



The word "above" there refers to the list of numbers (8, 5, x, 6), which is above the stem. So, we are told that the median of the list is 5.5.

Hope it's clear.
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given median is 5.5 i.e. set is x,5,6,8
5+6 /2 = 5.5
if x = 5 then max avg = 6
so answer has to be <=6
avg cannot be 3 as sum then would be 12
left with option B 5 ; i.e. sum would be 22 and x = 3
option B is correct