It is currently 19 Oct 2017, 08:10

# STARTING SOON:

Live Chat with Cornell Adcoms in Main Chat Room  |  R1 Interview Invites: MIT Sloan Chat  |  UCLA Anderson Chat  |  Duke Fuqua Chat (EA Decisions)

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# M16-23

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

16 Sep 2014, 00:59
Expert's post
7
This post was
BOOKMARKED
00:00

Difficulty:

75% (hard)

Question Stats:

37% (01:10) correct 63% (00:55) wrong based on 115 sessions

### HideShow timer Statistics

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
[Reveal] Spoiler: OA

_________________

Kudos [?]: 128862 [0], given: 12183

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

16 Sep 2014, 00:59
Expert's post
6
This post was
BOOKMARKED
Official Solution:

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 $$S$$" implies that set $$S$$ could be:

A. $$S=\{x\}$$ - $$S$$ contains only one element (e.g. {7 });

B. $$S=\{x, x, ...\}$$ - $$S$$ contains more than one element and all elements are equal (e.g. {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$$\gt$$smallest number).

Example: if $$S=\{3, 5\}$$, then mean of $$S=4$$. Pick subset with the smallest number: $$s'=\{3\}$$, mean of $$s'=3$$. As we see $$3 \lt 4$$.

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.

_________________

Kudos [?]: 128862 [0], given: 12183

Current Student
Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 958

Kudos [?]: 1845 [0], given: 229

Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

### Show Tags

26 Dec 2014, 21:02
statement 2 seems to assume that we have multiple elements in the set but what if we have a condition where by we have just one element in the set . i feel that “single element” SET cannot be classified for the statement : “all elements in the set are equal”—————> the word “all + plural” linguistically implies more than “one”
_________________

Piyush K
-----------------------
Our greatest weakness lies in giving up. The most certain way to succeed is to try just one more time. ― Thomas A. Edison
Don't forget to press--> Kudos
My Articles: 1. WOULD: when to use? | 2. All GMATPrep RCs (New)
Tip: Before exam a week earlier don't forget to exhaust all gmatprep problems specially for "sentence correction".

Kudos [?]: 1845 [0], given: 229

Intern
Joined: 18 Feb 2015
Posts: 9

Kudos [?]: 5 [0], given: 43

Location: India
Concentration: Technology, General Management

### Show Tags

17 Aug 2015, 22:41
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I do not understand as to why have not considered an empty set, which is also a subset of any set ?

Kudos [?]: 5 [0], given: 43

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

18 Aug 2015, 01:35
fetchnitin wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I do not understand as to why have not considered an empty set, which is also a subset of any set ?

An empty set has no mean or the median.
_________________

Kudos [?]: 128862 [0], given: 12183

Intern
Joined: 17 Mar 2014
Posts: 5

Kudos [?]: 2 [0], given: 6

### Show Tags

15 Sep 2015, 20:44
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

Kudos [?]: 2 [0], given: 6

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [1], given: 12183

### Show Tags

16 Sep 2015, 03:45
1
KUDOS
Expert's post
prodigypringle wrote:
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Thus, the median of {1,1,1,1} is (1 + 1)/2 = 1.
_________________

Kudos [?]: 128862 [1], given: 12183

Manager
Joined: 27 Feb 2015
Posts: 59

Kudos [?]: 9 [0], given: 56

Concentration: General Management, Economics
GMAT 1: 630 Q42 V34
WE: Engineering (Transportation)

### Show Tags

07 Oct 2016, 08:18
Bunuel wrote:
prodigypringle wrote:
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Thus, the median of {1,1,1,1} is (1 + 1)/2 = 2.

(1+1)/2 = 1 right?
so statement 3 should not be true.

Kudos [?]: 9 [0], given: 56

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

08 Oct 2016, 03:10
deepak268 wrote:
Bunuel wrote:
prodigypringle wrote:
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Thus, the median of {1,1,1,1} is (1 + 1)/2 = 2.

(1+1)/2 = 1 right?
so statement 3 should not be true.

2 there was clearly a typo. III must be true. Do you have an example when it's not?
_________________

Kudos [?]: 128862 [0], given: 12183

Intern
Joined: 21 May 2016
Posts: 8

Kudos [?]: [0], given: 3

### Show Tags

27 Dec 2016, 19:27
Please elaborate on choice 1.If a set contains 1 element,then its subset would have 1 element any time,then its mean is equal to its subset mean always.

Kudos [?]: [0], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

28 Dec 2016, 06:42
GouthamNandu wrote:
Please elaborate on choice 1.If a set contains 1 element,then its subset would have 1 element any time,then its mean is equal to its subset mean always.

The question asks which of the options MUST be true not could be true. While I could be true it's not necessary to be true. Set S can have more than 1 identical elements: {x, x, ...}
_________________

Kudos [?]: 128862 [0], given: 12183

Intern
Joined: 26 Feb 2017
Posts: 17

Kudos [?]: 6 [0], given: 27

### Show Tags

03 Mar 2017, 15:14

Statement 1 - Could have this 3 possibilities below .
1- {x,x,x...x}
2- {x}
3- { }

So, The statement 1 is no Right because if { } there is no mean?

Kudos [?]: 6 [0], given: 27

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

04 Mar 2017, 01:58
vitorpteixeira wrote:

Statement 1 - Could have this 3 possibilities below .
1- {x,x,x...x}
2- {x}
3- { }

So, The statement 1 is no Right because if { } there is no mean?

No. The question asks which of the options MUST be true not could be true. While I could be true it's not necessary to be true. Set S can have more than 1 identical elements: {x, x, ...}
_________________

Kudos [?]: 128862 [0], given: 12183

VP
Joined: 26 Mar 2013
Posts: 1260

Kudos [?]: 285 [0], given: 163

### Show Tags

05 Mar 2017, 01:24
Bunuel wrote:
Official Solution:

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 $$S$$" implies that set $$S$$ could be:

A. $$S=\{x\}$$ - $$S$$ contains only one element (e.g. {7 });

B. $$S=\{x, x, ...\}$$ - $$S$$ contains more than one element and all elements are equal (e.g. {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$$\gt$$smallest number).

Example: if $$S=\{3, 5\}$$, then mean of $$S=4$$. Pick subset with the smallest number: $$s'=\{3\}$$, mean of $$s'=3$$. As we see $$3 \lt 4$$.

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.

Dear Bunuel,

Why do not you count evenly spaced set in your consideration?

if there is a set $$S=\{1, 3, 5\}$$ and subset $$S=\{1, 3\}$$

Mean of S is 3 & Mean of subset is 2. So statement II does not need to have equal elements.

Where did I go wrong?

Kudos [?]: 285 [0], given: 163

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

05 Mar 2017, 03:49
Mo2men wrote:
Bunuel wrote:
Official Solution:

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 $$S$$" implies that set $$S$$ could be:

A. $$S=\{x\}$$ - $$S$$ contains only one element (e.g. {7 });

B. $$S=\{x, x, ...\}$$ - $$S$$ contains more than one element and all elements are equal (e.g. {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$$\gt$$smallest number).

Example: if $$S=\{3, 5\}$$, then mean of $$S=4$$. Pick subset with the smallest number: $$s'=\{3\}$$, mean of $$s'=3$$. As we see $$3 \lt 4$$.

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.

Dear Bunuel,

Why do not you count evenly spaced set in your consideration?

if there is a set $$S=\{1, 3, 5\}$$ and subset $$S=\{1, 3\}$$

Mean of S is 3 & Mean of subset is 2. So statement II does not need to have equal elements.

Where did I go wrong?

You did not read carefully. The first sentences of the solution makes important word in caps and bold!

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$" implies that set $$S$$ could be...

For your example, if $$S=\{1, 3, 5\}$$, consider a subset {1}.
_________________

Kudos [?]: 128862 [0], given: 12183

VP
Joined: 26 Mar 2013
Posts: 1260

Kudos [?]: 285 [0], given: 163

### Show Tags

05 Mar 2017, 06:53
Bunuel wrote:

You did not read carefully. The first sentences of the solution makes important word in caps and bold!

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$" implies that set $$S$$ could be...

For your example, if $$S=\{1, 3, 5\}$$, consider a subset {1}.

Thanks a lot Bunuel.

I have a question about terminology meaning not related to this question.

When I say 'Value X does not exceed value Y', How should methamaticllay represnt it?

is it X <= Y or only x=Y?

Thanks

Kudos [?]: 285 [0], given: 163

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

05 Mar 2017, 06:56
Mo2men wrote:
Bunuel wrote:

You did not read carefully. The first sentences of the solution makes important word in caps and bold!

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$" implies that set $$S$$ could be...

For your example, if $$S=\{1, 3, 5\}$$, consider a subset {1}.

Thanks a lot Bunuel.

I have a question about terminology meaning not related to this question.

When I say 'Value X does not exceed value Y', How should methamaticllay represnt it?

is it X <= Y or only x=Y?

Thanks

It's the same as x is less than or equal to y, so it's $$x \leq y$$.
_________________

Kudos [?]: 128862 [0], given: 12183

Intern
Joined: 21 Dec 2012
Posts: 13

Kudos [?]: 6 [0], given: 12

Location: India
Concentration: Finance, Finance
GMAT 1: 480 Q27 V22
GPA: 3.99
WE: Consulting (Consulting)

### Show Tags

22 Mar 2017, 23:55
Here is the justification:

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 SS?

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

Option 1: If set S contains only one element then it will not satisfy "If the mean of set S does not exceed mean of any subset of set S". Hence I. is not an answer.
Option 2: All elements in set S are equal. - Eg: {5,5,5,5} is S, subset of S = {5,5,5}. Hence this satisfies the condition "mean of S does not exceed mean of subset of S". II is an answer.
Option 3:The median of set S equals the mean of set S. Taking the same eg. as mentioned above,Mean of S = 5 , the median of S = 5. This same also satisfied the condition that if there is a set of this sort exists then - it satisfies the condition "mean of S does not exceed mean of subset of S".
Hence the option is D.

Kudos [?]: 6 [0], given: 12

Intern
Joined: 03 May 2014
Posts: 24

Kudos [?]: 10 [0], given: 43

### Show Tags

14 Aug 2017, 17:00
1. Set S contains only one element.
Does it mean count of element or value of element?
My understanding is that set S has only one element by count not by value... therefore {X,X...} is not possible. Please help understand.

Kudos [?]: 10 [0], given: 43

Math Expert
Joined: 02 Sep 2009
Posts: 41893

Kudos [?]: 128862 [0], given: 12183

### Show Tags

14 Aug 2017, 22:08
ManSab wrote:
1. Set S contains only one element.
Does it mean count of element or value of element?
My understanding is that set S has only one element by count not by value... therefore {X,X...} is not possible. Please help understand.

Option 1, Set S contains only one element means that there is only one elements in S: {X}. This option is not necessarily true. Please re-read the solution.
_________________

Kudos [?]: 128862 [0], given: 12183

Re: M16-23   [#permalink] 14 Aug 2017, 22:08

Go to page    1   2    Next  [ 30 posts ]

Display posts from previous: Sort by

# M16-23

Moderators: Bunuel, Vyshak

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.