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If the mean of set S does not exceed mean of any subset of

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If the mean of set S does not exceed mean of any subset of [#permalink] New post   Updated on: 07 Feb 2012, 16:24
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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

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D

Originally posted by angel2009 on 02 May 2010, 00:12.
Last edited by Bunuel on 07 Feb 2012, 16:24, edited 1 time in total.
Edited the question and added the OA
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   04 May 2010, 14:00
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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.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   05 May 2010, 02:46
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hardnstrong wrote:

Why can't we consider the subset with the greatest number
Ex- S= (5,6,7) and subset s (7)
Please explain :?:


The questions says: the mean of set S does not exceed mean of ANY subset of set S --> Mean of S<=Mean of ANY subset of S.

I considered the example of subset with smallest number to show that set S can not have 2 (or more) different elements. In any set with 2 (or more) different elements we can pick subset that will have the smaller mean than the mean of the entire set.

In your example (S={5,6,7}, mean of S=6), there are subsets of S with smaller mean (eg s'={5} or s'={5,6}), with the mean equal to the mean of entire set (eg s'={5,7} or s'={6}) and with bigger mean (eg s'={7} or s'={6,7}).

The stem could have stated opposite thing: "the mean of ANY subset of S does not exceed mean of S" and the answer would be the same. And to demonstrate that pick the subset with biggest number of the set in a set with 2 (or more) different elements and you'll see that this subset will have the mean bigger than the mean of entire set.

Basically this part of the stem is just the complicated way of saying that S contains either A. only one element or B. more than one identical elements.

Hope it's clear.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   01 Oct 2013, 01:59
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TGC wrote:
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!


An empty set has no mean or the median.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   17 Oct 2013, 03:05
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Yash12345 wrote:
Bunuel 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

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.

Also discussed here: ps-challenge-93565.html




hi bunuel , little confused here ..
Please explain me where am i going wrong.

I took the elements of set S={1,2,3,4)
And the subset elemets as ={2,3,4)

however this does not meet the second situation requirement. i.e. ( all elemets in set s are equal)


We are given that "the mean of set S does not exceed mean of ANY subset of set S".

Now, notice that S cannot be {1, 2, 3, 4), because it has subsets with the mean smaller than the mean of {1, 2, 3, 4):

Mean of S = 10/4 = 2.5. Mean of {1}, which is a subset of S, is 1 --> 2.5 > 1.

Does this make sense?
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   02 Nov 2014, 07:54
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vinraj wrote:
HI, Bunuel,

I dont understand how we can say that all elements in a set are equal in case of a set S={2} , which has only a single element.
Is this an established rule in Set Theory or is it a personal opinion.

If the set has just one element, then it is not comparable with anything else . So, according to English grammar & logic, it does not make sense to say that all elements are same in that Set.

Kindly tell me if this is an established Practice.. Thank you


Yes, if we have a set with just 1 element, we can say that all elements of the set are the same.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   05 May 2010, 02:10
Bunuel wrote:
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


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.



Why can't we consider the subset with the greatest number
Ex- S= (5,6,7) and subset s (7)
Please explain :?:
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   18 Sep 2010, 23:43
Hi Bunuel,

Why the first statement is wrong...didn't understand.
I. Set S contains only one element.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   18 Sep 2010, 23:54
Because can contain the same element repeat an arbitrary number of times as well.
The set A={3,3,3,3,3} id different from B={3}

Yet in either case the condition of means is satisfied, whereas A has 5 elements and B has only 1
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   19 Sep 2010, 00:03
Nope. Let's go my way...let say if set S={3}....then mean/median = 3 (this does satisfy with question stem "If the mean of set S does not exceed mean of any subset of set S")....then this is true.

please correct me if I am missing anything.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   19 Sep 2010, 02:03
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appy001 wrote:
Hi Bunuel,

Why the first statement is wrong...didn't understand.
I. Set S contains only one element.

appy001 wrote:
Nope. Let's go my way...let say if set S={3}....then mean/median = 3 (this does satisfy with question stem "If the mean of set S does not exceed mean of any subset of set S")....then this is true.

please correct me if I am missing anything.


Question asks: "which of the following MUST be true about set S" (not COULD be true).

I. Set S contains only one element --> it's not necessarily true as S can contain more than one element and still satisfy the requirement in stem. For example if S={3, 3, 3, 3} then mean of S equals to 3 and it does not exceed the mean of ANY subset of S, which also will be equal to 3.

Hope it's clear.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   19 Sep 2010, 03:00
oh well.. :roll:

I read "one element" as one TYPE of element --- that is only the same unique element in what ever number for the whole set which is essentially the same as II...

elements is really a reference to the data points in the set
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   19 Sep 2010, 08:10
gmat1011 wrote:
oh well.. :roll:

I read "one element" as one TYPE of element --- that is only the same unique element in what ever number for the whole set which is essentially the same as II...

elements is really a reference to the data points in the set


In set theory there is no requirement for uniqueness of elements, and in general elements does not refer to unique constituents

So {3,3,3} is different from {3}
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   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!
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   17 Oct 2013, 01:53
Bunuel 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

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.

Also discussed here: ps-challenge-93565.html




hi bunuel , little confused here ..
Please explain me where am i going wrong.

I took the elements of set S={1,2,3,4)
And the subset elemets as ={2,3,4)

however this does not meet the second situation requirement. i.e. ( all elemets in set s are equal)
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   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.

Please review your answer again.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   21 Oct 2014, 14:02
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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}.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   02 Nov 2014, 07:49
HI, Bunuel,

I dont understand how we can say that all elements in a set are equal in case of a set S={2} , which has only a single element.
Is this an established rule in Set Theory or is it a personal opinion.

If the set has just one element, then it is not comparable with anything else . So, according to English grammar & logic, it does not make sense to say that all elements are same in that Set.

Kindly tell me if this is an established Practice.. Thank you
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   14 Mar 2018, 16:20
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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


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


If the mean of set S does not exceed the mean of any subset of set S, then all the elements in S must be equal. That is because if there is an element that is not equal to the others, then the mean of S will exceed the mean of at least one subset of S. For example, if S = {1, 4, 4}, we see that the mean of S is 3, and the mean of a subset of S, say {1, 4}, is only 2.5.

Thus, we see that Roman numeral II is true. Furthermore, if all the elements in S are equal, then the median of S equals the mean of S, so Roman numeral III is true also.

Answer: D
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post   30 Dec 2019, 02:43
Hi Bunuel and ScottTargetTestPrep

If set S was empty, none of the conditions I, II and III must be true, yet the mean of set S would not exceed the mean of any subset of S. Hence, I chose A. I would highly appreciate if you could explain where my flaw is?

Many thanks!
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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
30 Dec 2019, 02:43
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