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:

One specific question about this type of problems:

If we are given a set of integers, say S=(1,2,3,1,7,19,-86,8....), the integers in the set DO NOT HAVE to be in ascending order, right? I understand, that the order does not have any bearing on mean and average, but still.... I'm asking because if integers are randomly distributed and the guestion asks me to find, say, second member of the set, I would not know what to do

Last edited by Financier on 24 Sep 2010, 01:50, edited 1 time in total.

One specific question about this type of problems:

If we are given a set of integers, say S=(1,2,3,1,7,19,-86,8....), the integers in the set DOES NOT HAVE to be in ascending order, right? I understand, that the order does not have any bearing on mean and average, but still.... I'm asking because if integers are randomly distributed and the guestion asks me to find, say, second member of the set, I would not know what to do

The numbers in a set can be written in any order. If a question were to ask you about a specific number in the set, more information would have to be provided.

One specific question about this type of problems:

If we are given a set of integers, say S=(1,2,3,1,7,19,-86,8....), the integers in the set DO NOT HAVE to be in ascending order, right? I understand, that the order does not have any bearing on mean and average, but still.... I'm asking because if integers are randomly distributed and the guestion asks me to find, say, second member of the set, I would not know what to do

Sets are unordered collections.

"second member" is an ambiguous concept _________________

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

* none of the three qualities is necessary * II only * III only * II and III only * I, II, and III

The mean of S must be less than or equal to the mean of EVERY subset of S. For any set S, you could take as a subset the single smallest element in S. The mean of any set of numbers must be greater than its smallest element - unless there's only one element, OR if all the elements are the same. These are the only ways the condition can be true. Since any set S with multiple, equal elements will satisfy the condition, I is not necessarily true, but II is, and III follows, because the mean and median of any set of identical elements will both be equal to that element. (D).

Say S = {3, 3 ,6} and S1 = {6}. Mean of S = 4 while Mean of S1 = 6 Median of S = 3 while Median of S1 = 6. Let's check the three statements:

I. Set S contains only one element [We have three elements here. Incorrect]

II. All elements in set S are equal [No, not neccessary. Incorrect]

III. The median of set S equals the mean of set S [Not neccessary. We have Median of S = 3 and Mean of S = 4. Incorrect]

angel2009 wrote:

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

* none of the three qualities is necessary * II only * III only * II and III only * I, II, and III

@ykaiim The key word here is ANY.

Your set S1= {3,3,6} your Subset S1={6}

so here we can see mean of s=4 and mean of s1= 6

here mean of s does not exceed the mean of s1

but we can also take the subset as s1= {3}

in this case mean of S = 4 exceeds the mean of s1= 3,

so condition ii must be true for the statement to hold , and if a set contains elements which are all equal , then mean = median hence iii must also be true.

so this set S1= {3,3,6} fails condition ii as subset {3} will violate the given terms.

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

Show Tags

25 May 2012, 02:44

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

* none of the three qualities is necessary * II only * III only * II and III only * I, II, and II

lets take a set S= {1,2,3,4,5} mean= 3 and mean of its its subset ( 4,5) will be 4.5 which would be greater that 3. even for set of three values ( 1,2,3) mean would be 2 and its subset (3) mean would be 3 which is greater than 2 therefore for mean of any subset not to be greater than set mean .. all values should be equal. and if all values are equal than its median and mean would also be equal..

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.

So statements II and III are always true.

Answer: D.

Hi,

i couldn't able to understand option 1. since there is only one elt in the set then the subset may contain that same element or zero huh?? Please explain option 1

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.

So statements II and III are always true.

Answer: D.

Hi,

i couldn't able to understand option 1. since there is only one elt in the set then the subset may contain that same element or zero huh?? Please explain option 1

Not sure I understand your question.

Anyway, option I says: set S contains only one element. This statement is not always true, for example consider set {1, 1}. _________________

I am sorry, was that not considered as two same element ?

bunuel, I have the same doubt.. The example you quoted is a 2 element set.. agreed that the element is same though..

question is dubious .. Answer should be A..

All elements in set S are equal - true for both A and B scenarios

This can be true as situation A : 1 element set but this statement says All elements in set S are equal .. When there is only 1 element, question of all elements doesn't arise. . _________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

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

Show Tags

11 Nov 2013, 06:43

Hi, I didn't think 3 statement must be true. Let s : {5,35,65} Its true mean and median are same but the it doesn't satisfy that Set S mean doesn't exceed any subset of S mean.

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

Show Tags

11 Nov 2013, 06:46

Expert's post

vikrantgulia wrote:

Hi, I didn't think 3 statement must be true. Let s : {5,35,65} Its true mean and median are same but the it doesn't satisfy that Set S mean doesn't exceed any subset of S mean.

Please explain.

Can you please explain what you mean? Thank you. _________________

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

Show Tags

21 Oct 2014, 13:49

I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set

Same Example: Set S: 5,6,7

Sub set A: 5 Sub set B: 7

for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.

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

Show Tags

21 Oct 2014, 14:02

Expert's post

awal_786@hotmail.com wrote:

I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set

Same Example: Set S: 5,6,7

Sub set A: 5 Sub set B: 7

for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.

Please review your answer again.

S cannot be {5, 6, 7}. The stem says that the mean of set S does not exceed mean of any subset of set S.

The mean of S is 6. One of the subsets of S is {5}, with mean of 5. So, the mean of S, which is 6, is greater than the mean of its subset {5}. _________________

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

Show Tags

22 Oct 2014, 03:51

Bunuel wrote:

awal_786@hotmail.com wrote:

I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set

Same Example: Set S: 5,6,7

Sub set A: 5 Sub set B: 7

for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.

Please review your answer again.

S cannot be {5, 6, 7}. The stem says that the mean of set S does not exceed mean of any subset of set S.

The mean of S is 6. One of the subsets of S is {5}, with mean of 5. So, the mean of S, which is 6, is greater than the mean of its subset {5}.

Agree, mean of set S doesn't exceed mean of Sub set S, but Mean of Sub set of S can exceed mean of set S, That's what I mean, your question doesn't have 2 way condition, then why you are taking it as two one condition?

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

Show Tags

22 Oct 2014, 03:58

Expert's post

awal_786@hotmail.com wrote:

Bunuel wrote:

awal_786@hotmail.com wrote:

I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set

Same Example: Set S: 5,6,7

Sub set A: 5 Sub set B: 7

for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.

Please review your answer again.

S cannot be {5, 6, 7}. The stem says that the mean of set S does not exceed mean of any subset of set S.

The mean of S is 6. One of the subsets of S is {5}, with mean of 5. So, the mean of S, which is 6, is greater than the mean of its subset {5}.

Agree, mean of set S doesn't exceed mean of Sub set S, but Mean of Sub set of S can exceed mean of set S, That's what I mean, your question doesn't have 2 way condition, then why you are taking it as two one condition?

Sorry, but that does not make any sense. The mean of set S does not exceed mean of any subset of set S means that none of the subsets of S has the mean less than the mean of S.

Stem gives some condition on set S. Your example ({5, 6, 7}) does not satisfy it, so S cannot be {5, 6, 7}. _________________

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

Show Tags

22 Oct 2014, 14:05

Agree, mean of set S doesn't exceed mean of Sub set S, but Mean of Sub set of S can exceed mean of set S, That's what I mean, your question doesn't have 2 way condition, then why you are taking it as two one condition?[/quote]

Sorry, but that does not make any sense. The mean of set S does not exceed mean of any subset of set S means that none of the subsets of S has the mean less than the mean of S.

Stem gives some condition on set S. Your example ({5, 6, 7}) does not satisfy it, so S cannot be {5, 6, 7}.[/quote]

Basically you are sticking to your answer, and not trying to understand. try solving this question without keeping your previous answer in mind. Set S {5,6,7} Sub set of set s {7} here mean of set S doesn't exceed mean of its sub set and your condition is obtained without taking all similar numbers in set S

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

Show Tags

22 Oct 2014, 14:12

Expert's post

awal_786@hotmail.com wrote:

Basically you are sticking to your answer, and not trying to understand. try solving this question without keeping your previous answer in mind. Set S {5,6,7} Sub set of set s {7} here mean of set S doesn't exceed mean of its sub set and your condition is obtained without taking all similar numbers in set S

Seems that you don't understand neither the question nor the solution.

Last attempt: the stem says that the mean of set S does not exceed mean of ANY subset of set S. This means that S CANNOT be {5, 6, 7} because its mean is more than the mean of one of the subsets of S. Sorry, but I cannot explain any better than this. _________________

MBA Admission Calculator Officially Launched After 2 years of effort and over 1,000 hours of work, I have finally launched my MBA Admission Calculator . The calculator uses the...

Final decisions are in: Berkeley: Denied with interview Tepper: Waitlisted with interview Rotman: Admitted with scholarship (withdrawn) Random French School: Admitted to MSc in Management with scholarship (...

The London Business School Admits Weekend officially kicked off on Saturday morning with registrations and breakfast. We received a carry bag, name tags, schedules and an MBA2018 tee at...