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Re: 8, 5, x, 6 The median of the list of positive integers [#permalink]

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27 Mar 2013, 03:40

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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"
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Re: The median of the list of positive integers above is 5.5 [#permalink]

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27 Mar 2013, 04:26

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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.
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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

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 can not be the average because in thisc ase 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 can not 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).

Re: 8, 5, x, 6 The median of the list of positive integers [#permalink]

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03 Aug 2013, 05:43

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Zarrolou wrote:

n - 19 must be an integer, option (C)6.25 is out.

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
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Re: The median of the list of positive integers above is 5.5 [#permalink]

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22 Nov 2014, 15:22

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Re: The median of the list of positive integers above is 5.5 [#permalink]

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20 Aug 2015, 22:46

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 median of the list of positive integers above is 5.5 [#permalink]

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26 Dec 2015, 11:17

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

gmatclubot

The median of the list of positive integers above is 5.5
[#permalink]
26 Dec 2015, 11:17

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