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:

I: Not True: If Set contains only One element, then Set S won't have any SS. II: True: When all the elements are equal, then mean of S=mean of any SS. III: Not True: Consider Consecutive No in Set S.Mean of S will always be greater than the smallest No. of the Consecutive series of Set S, and hence Mean of S becomes greater then Subset of S.

If the mean of set S does not exceed mean of any subset of set S , which of the following must be true about set S?

I. Set S contains only one element II. All elements in set S are equal III. The median of set S equals the mean of set S

A. none of the three qualities is necessary B. II only C. III only D. II and III only E. I, II, and III

"The mean of set S does not exceed mean of any subset of set S" --> set S can be: A. \(S=\{x\}\) - S contains only one element (eg {7}); B. \(S=\{x, x, ...\}\) - S contains more than one element and all elements are equal (eg{7,7,7,7}).

Why is that? Because if set S contains two (or more) different elements, then we can always consider the subset with smallest number and the mean of this subset (mean of subset=smallest number) will be less than mean of entire set (mean of full set>smallest number).

Example: S={3, 5} --> mean of S=4. Pick subset with smallest number s'={3} --> mean of s'=3 --> 3<4.

Now let's consider the statements:

I. Set S contains only one element - not always true, we can have scenario B too (\(S=\{x, x, ...\}\));

II. All elements in set S are equal - true for both A and B scenarios, hence always true;

III. The median of set S equals the mean of set S - - true for both A and B scenarios, hence always true.

Re: If the mean of set S does not exceed mean of any subset of [#permalink]

Show Tags

30 Sep 2013, 11:17

Hi Bunuel,

I am still unable to get the solution.

When we can have an empty set {0} as a subset of each set, in that case we would have a average as 0 and thus consider the below example.

You have a set : {1,1,1}

One possible subset : {0}

Average of the set : 1

Average of subset:0

So still it exceeds the average of subset .

Can you advise on that?

Rgds, TGC!
_________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

If the mean of set S does not exceed mean of any subset of set S , which of the following must be true about set S?

I. Set S contains only one element II. All elements in set S are equal III. The median of set S equals the mean of set S

A. none of the three qualities is necessary B. II only C. III only D. II and III only E. I, II, and III

"The mean of set S does not exceed mean of any subset of set S" --> set S can be: A. \(S=\{x\}\) - S contains only one element (eg {7}); B. \(S=\{x, x, ...\}\) - S contains more than one element and all elements are equal (eg{7,7,7,7}).

Why is that? Because if set S contains two (or more) different elements, then we can always consider the subset with smallest number and the mean of this subset (mean of subset=smallest number) will be less than mean of entire set (mean of full set>smallest number).

Example: S={3, 5} --> mean of S=4. Pick subset with smallest number s'={3} --> mean of s'=3 --> 3<4.

Now let's consider the statements:

I. Set S contains only one element - not always true, we can have scenario B too (\(S=\{x, x, ...\}\));

II. All elements in set S are equal - true for both A and B scenarios, hence always true;

III. The median of set S equals the mean of set S - - true for both A and B scenarios, hence always true.

If the mean of set S does not exceed mean of any subset of set S , which of the following must be true about set S?

I. Set S contains only one element II. All elements in set S are equal III. The median of set S equals the mean of set S

A. none of the three qualities is necessary B. II only C. III only D. II and III only E. I, II, and III

"The mean of set S does not exceed mean of any subset of set S" --> set S can be: A. \(S=\{x\}\) - S contains only one element (eg {7}); B. \(S=\{x, x, ...\}\) - S contains more than one element and all elements are equal (eg{7,7,7,7}).

Why is that? Because if set S contains two (or more) different elements, then we can always consider the subset with smallest number and the mean of this subset (mean of subset=smallest number) will be less than mean of entire set (mean of full set>smallest number).

Example: S={3, 5} --> mean of S=4. Pick subset with smallest number s'={3} --> mean of s'=3 --> 3<4.

Now let's consider the statements:

I. Set S contains only one element - not always true, we can have scenario B too (\(S=\{x, x, ...\}\));

II. All elements in set S are equal - true for both A and B scenarios, hence always true;

III. The median of set S equals the mean of set S - - true for both A and B scenarios, hence always true.

If the mean of set S does not exceed mean of any subset of set S , which of the following must be true about set S?

I. Set S contains only one element II. All elements in set S are equal III. The median of set S equals the mean of set S

A. none of the three qualities is necessary B. II only C. III only D. II and III only E. I, II, and III

"The mean of set S does not exceed mean of any subset of set S" --> set S can be: A. \(S=\{x\}\) - S contains only one element (e B. \(S=\{x, x, ...\}\) - S contains more than one element and all elements are equal (eg{7,7,7,7}).

Why is that? Because if set S contains two (or more) different elements, then we can always consider the subset with smallest number and the mean of this subset (mean of subset=smallest number) will be less than mean of entire set (mean of full set>smallest number).

Example: S={3, 5} --> mean of S=4. Pick subset with smallest number s'={3} --> mean of s'=3 --> 3<4.

Now let's consider the statements:

I. Set S contains only one element - not always true, we can have scenario B too (\(S=\{x, x, ...\}\));

II. All elements in set S are equal - true for both A and B scenarios, hence always true;

III. The median of set S equals the mean of set S - - true for both A and B scenarios, hence always true.

Re: If the mean of set S does not exceed mean of any subset of [#permalink]

Show Tags

23 Dec 2014, 11:26

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: If the mean of set S does not exceed mean of any subset of [#permalink]

Show Tags

25 Dec 2014, 09:03

I did it in a more practical way, like this:

Lets say S is {5,6,7,8,9} | i used consecutive integerns to make my life a bit easier and s is {6,7,8} or {5,6,7}

Then, taking the options one by one: I: Obviously, we could have more than one elements and still have the same mean. NO II: We can see that for the first s this is wrong (same mean but different numbers). But for the second s this is true (different mean and different numbers). So, we can say that they should all be equal. III: This seems to be true too, as if we choose an s, the mean of which is 7 or less than 7, then for S, the mean equals the median.

Perhaps it makes absolutely no sense and could be random, but it led me to the correct answer in about half a minute...

After days of waiting, sharing the tension with other applicants in forums, coming up with different theories about invites patterns, and, overall, refreshing my inbox every five minutes to...

I was totally freaking out. Apparently, most of the HBS invites were already sent and I didn’t get one. However, there are still some to come out on...

In early 2012, when I was working as a biomedical researcher at the National Institutes of Health , I decided that I wanted to get an MBA and make the...