GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 25 Mar 2019, 07:08 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here. ### Request Expert Reply # M16-23

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

2
3 00:00

Difficulty:   75% (hard)

Question Stats: 40% (01:04) correct 60% (01:00) wrong based on 146 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

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

1
4
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.

_________________
Director  Status: Everyone is a leader. Just stop listening to others.
Joined: 22 Mar 2013
Posts: 757
Location: India
GPA: 3.51
WE: Information Technology (Computer Software)

### Show Tags

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".
Intern  Joined: 18 Feb 2015
Posts: 7
Location: India
Concentration: Technology, General Management

### Show Tags

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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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.
_________________
Intern  Joined: 17 Mar 2014
Posts: 5

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

1
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.
_________________
Intern  Joined: 27 Feb 2015
Posts: 44
Concentration: General Management, Economics
GMAT 1: 630 Q42 V34 WE: Engineering (Transportation)

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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?
_________________
Intern  B
Joined: 21 May 2016
Posts: 8

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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, ...}
_________________
Intern  S
Joined: 26 Feb 2017
Posts: 30
Location: Brazil
GMAT 1: 610 Q45 V28 GPA: 3.11

### Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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, ...}
_________________
SVP  D
Joined: 26 Mar 2013
Posts: 2101

### Show Tags

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?
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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}.
_________________
SVP  D
Joined: 26 Mar 2013
Posts: 2101

### Show Tags

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
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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 $$x \leq y$$.
_________________
Intern  Joined: 21 Dec 2012
Posts: 11
Location: India
Concentration: Finance, Finance
GMAT 1: 480 Q27 V22 GPA: 3.99
WE: Consulting (Consulting)

### Show Tags

The answer is D.
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.
Intern  B
Joined: 03 May 2014
Posts: 26

### Show Tags

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.
Math Expert V
Joined: 02 Sep 2009
Posts: 53831

### Show Tags

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

Go to page    1   2    Next  [ 38 posts ]

Display posts from previous: Sort by

# M16-23

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

Moderators: chetan2u, Bunuel 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®.  