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# The mean of set S does not exceed mean of any subset of set

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

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23 May 2008, 02:27
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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

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Last edited by Bunuel on 07 Feb 2012, 16:23, edited 1 time in total.
Edited the question and added the OA
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23 May 2008, 05:58
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sondenso wrote:
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 only necessary condition seems to be II.
If all nos are equal.... them mean of any subset is the same.
I is obviously ruled out ( e.g. S = {1,1,1} )

It is not necessary in itself, rather, is automatically implied by II.
All sets that satsify II satisfy III.

pls help...
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23 May 2008, 06:57
The questions asks for"must be" hence II & III is the answer as II has to be true and also II would obviously imply III. Case I need not be the case.

Ans: II & III
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23 Nov 2011, 19:51
What is wrong with the below example?

Set S be {0,1} and subset of set S be {1}. Here Mean of S is 0.5 and Mean of subset of S is 1, which satisfies the question "The mean of set S does not exceed mean of any subset of set S". In this case option 2 fails.

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07 Feb 2012, 15:57
What is the median of a set {2,2,2,2} .....a set with all elements with the same value?

Median of a set that has a single element will be the same as the element that is if set = {2} then median = 2 as well.

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07 Feb 2012, 16:20
Expert's post
teal wrote:
What is the median of a set {2,2,2,2} .....a set with all elements with the same value?

Median of a set that has a single element will be the same as the element that is if set = {2} then median = 2 as well.

Generally:
The median of a set with odd # of terms is just a middle term (when ordered in ascending/descending order).

The median of a set with even # of terms is the average of two middle terms (when ordered in ascending/descending order).

So, the median of {2,2,2,2} is (2+2)/2=2 and median of {2} is 2. Basically the median of a set with all equal numbers is this number itself (in this case it does not matter whether a set has an odd or even # of elements).

Hope it's clear.

As for the question:
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.

This question is also discussed here: ps-challenge-93565.html

Hope it helps.
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Re: M16-23   [#permalink] 07 Feb 2012, 16:20
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