Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

27 Mar 2013, 02:40

2

This post received KUDOS

1

This post was BOOKMARKED

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"
_________________

It is beyond a doubt that all our knowledge that begins with experience.

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

Show Tags

27 Mar 2013, 03:26

1

This post received KUDOS

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.
_________________

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]

Show Tags

03 Aug 2013, 04:43

1

This post was BOOKMARKED

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
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

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

Show Tags

22 Nov 2014, 14:22

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

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

Show Tags

20 Aug 2015, 21:46

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]

Show Tags

26 Dec 2015, 10:17

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

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

Show Tags

15 Dec 2016, 15:26

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.

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Post-MBA I became very intrigued by how senior leaders navigated their career progression. It was also at this time that I realized I learned nothing about this during my...