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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.
Re: GMat Club Tests: M16 Q23 [#permalink]
15 Jan 2012, 07:40
Expert's post
1
This post was BOOKMARKED
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]
30 Sep 2013, 10:17
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 _____________________________________________________________________________
Re: GMat Club Tests: M16 Q23 [#permalink]
17 Oct 2013, 00:53
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.
Re: GMat Club Tests: M16 Q23 [#permalink]
17 Oct 2013, 02:05
Expert's post
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.
Re: GMat Club Tests: M16 Q23 [#permalink]
17 Oct 2013, 02:22
Bunuel wrote:
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 (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]
23 Dec 2014, 10:26
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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
25 Dec 2014, 08:03
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.
Perhaps it makes absolutely no sense and could be random, but it led me to the correct answer in about half a minute...
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