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

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

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New post 22 Oct 2014, 13:17
awal_786@hotmail.com wrote:
I found it my self, from other forum. key word in this question is "ANY" sub set.
Sorry for bothering to much


That's why it's important to read the question and solutions carefully.

In my solution here: if-the-mean-of-set-s-does-not-exceed-mean-of-any-subset-of-93565.html#p720756 Crucial word ANY is in bold. Here: if-the-mean-of-set-s-does-not-exceed-mean-of-any-subset-of-93565.html#p720904 it's bold and in capital letters. And in my reply to your doubt here: if-the-mean-of-set-s-does-not-exceed-mean-of-any-subset-of-93565.html#p720904 it's in italic and underlined.
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Re: If the mean of set S does not exceed mean of any subset of  [#permalink]

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New post 02 Nov 2014, 06:49
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]

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New post 02 Nov 2014, 06:54
1
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]

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New post 16 Sep 2015, 23:36
Suggest some algebraic notation:

sum of S is x
number of elements in S is n
number of subtracted value from x is y (for subset)
number of subtracted elements from n is z (for subset)

have inequality:
x/n=<(x-y)/(n-z)
xn-xz=<nz-ny

so, ny=<xz

Test options:
"Set S contains only one element": in this case ny>xz (always equal to 0), Out
"All elements in set S are equal": ny=<xz, Remains
"The median of set S equals the mean of set S": second option is feasible here, Remains

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

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New post 14 Mar 2018, 15:20
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 &nbs [#permalink] 14 Mar 2018, 15:20

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