The median of the list of positive integers above is 5.5

Question

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

Bunuel · Accepted Answer

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.

Sachin9 · Answer

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 :lol:

piyatiwari · Answer

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

Zarrolou · Answer

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"

anbuKp · Answer

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?

Bunuel · Answer

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.

ScottTargetTestPrep · Answer

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

ashKing12 · Answer

0\leq{x}\leq{5}

adding 19 both side

0+19\leq {x+19}\leq{5+19}

dividing by 4 both sides

\frac{19}{4}\leq{\frac{(x+19)}{4}}\leq{\frac{24}{4}}

4.7\leq{mean}\leq{6}

option B

Kashu18 · Answer

I was stuck at the same point, but as you'll see the question very smartly gives the median more than 5.5 which could be 5.51, 5.500000001 and so on. Hence, 5.5 is the safest scenario to be considered to rule out options.

anbuKp

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?

alamin8 · Answer

I want to know where it is written "x" is less than or equal to 5, i.e., x<= 5! Can you enlighten me?

Bunuel · Answer

It is not written directly, it is inferred from the median.

For the median to be 5.5, the two middle numbers must be 5 and 6.

So  x  cannot be greater than 5. If  x  were greater than 5, then 5 and 6 would no longer be the two middle numbers.

Therefore,  x \leq 5 .

The median of the list of positive integers above is 5.5

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The median of the list of positive integers above is 5.5

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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