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I: Not True: If Set contains only One element, then Set S won't have any SS. II: True: When all the elements are equal, then mean of S=mean of any SS. III: Not True: Consider Consecutive No in Set S.Mean of S will always be greater than the smallest No. of the Consecutive series of Set S, and hence Mean of S becomes greater then Subset of S.

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.

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

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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! _________________

Rgds, TGC! _____________________________________________________________________ I Assisted You => KUDOS Please _____________________________________________________________________________

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.

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.

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 (e 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.

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

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23 Dec 2014, 11:26

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

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25 Dec 2014, 09:03

I did it in a more practical way, like this:

Lets say S is {5,6,7,8,9} | i used consecutive integerns to make my life a bit easier and s is {6,7,8} or {5,6,7}

Then, taking the options one by one: I: Obviously, we could have more than one elements and still have the same mean. NO II: We can see that for the first s this is wrong (same mean but different numbers). But for the second s this is true (different mean and different numbers). So, we can say that they should all be equal. III: This seems to be true too, as if we choose an s, the mean of which is 7 or less than 7, then for S, the mean equals the median.

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