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

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Joined: 20 Aug 2010
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If the mean of set S does not exceed mean of any subset of  [#permalink]

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15 Jan 2012, 08:28
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Question Stats:

41% (01:15) correct 59% (01:39) wrong based on 460 sessions

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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

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.

Why the answer is Not B??
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Joined: 02 Sep 2009
Posts: 49496
Re: GMat Club Tests: M16 Q23  [#permalink]

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15 Jan 2012, 08:40
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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.

Also discussed here: ps-challenge-93565.html
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Re: If the mean of set S does not exceed mean of any subset of  [#permalink]

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03 Jun 2013, 03:08
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

*New project from GMAT Club!!! Check HERE
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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 .

Rgds,
TGC!
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Joined: 02 Sep 2009
Posts: 49496
Re: If the mean of set S does not exceed mean of any subset of  [#permalink]

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01 Oct 2013, 01:59
1
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 .

Rgds,
TGC!

An empty set has no mean or the median.
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Re: GMat Club Tests: M16 Q23  [#permalink]

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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.

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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Joined: 02 Sep 2009
Posts: 49496
Re: GMat Club Tests: M16 Q23  [#permalink]

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17 Oct 2013, 03:05
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.

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: GMat Club Tests: M16 Q23  [#permalink]

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17 Oct 2013, 03:22
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.

So statements II and III are always true.

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?

Yes bunuel my doubt is solved . Thus it is compulsory that all the elements of set s are equal.

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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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.

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

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22 May 2018, 08:21
Hi,

I'm considering an extreme scenario.

What if S = {1,1,2,3,7} Mean = 7
subset of S = {1,1,2} or {1,2,3} or {3,7}. All have lesser mean than 7

If above is true, II would be wrong then???

Re: If the mean of set S does not exceed mean of any subset of &nbs [#permalink] 22 May 2018, 08:21
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# If the mean of set S does not exceed mean of any subset of

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