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

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Re: PS Challenge! [#permalink] New post 24 Sep 2010, 00:40
One specific question about this type of problems:

If we are given a set of integers, say S=(1,2,3,1,7,19,-86,8....), the integers in the set DO NOT HAVE to be in ascending order, right?
I understand, that the order does not have any bearing on mean and average, but still.... I'm asking because if integers are randomly distributed and the guestion asks me to find, say, second member of the set, I would not know what to do :(

Last edited by Financier on 24 Sep 2010, 00:50, edited 1 time in total.
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Re: PS Challenge! [#permalink] New post 24 Sep 2010, 00:46
Financier wrote:
One specific question about this type of problems:

If we are given a set of integers, say S=(1,2,3,1,7,19,-86,8....), the integers in the set DOES NOT HAVE to be in ascending order, right?
I understand, that the order does not have any bearing on mean and average, but still.... I'm asking because if integers are randomly distributed and the guestion asks me to find, say, second member of the set, I would not know what to do :(


The numbers in a set can be written in any order. If a question were to ask you about a specific number in the set, more information would have to be provided.
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Re: PS Challenge! [#permalink] New post 24 Sep 2010, 06:20
Financier wrote:
One specific question about this type of problems:

If we are given a set of integers, say S=(1,2,3,1,7,19,-86,8....), the integers in the set DO NOT HAVE to be in ascending order, right?
I understand, that the order does not have any bearing on mean and average, but still.... I'm asking because if integers are randomly distributed and the guestion asks me to find, say, second member of the set, I would not know what to do :(


Sets are unordered collections.

"second member" is an ambiguous concept
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Re: PS Challenge! [#permalink] New post 29 Sep 2010, 09:26
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


* none of the three qualities is necessary
* II only
* III only
* II and III only
* I, II, and III


The mean of S must be less than or equal to the mean of EVERY subset of S. For any set S, you could take as a subset the single smallest element in S. The mean of any set of numbers must be greater than its smallest element - unless there's only one element, OR if all the elements are the same. These are the only ways the condition can be true. Since any set S with multiple, equal elements will satisfy the condition, I is not necessarily true, but II is, and III follows, because the mean and median of any set of identical elements will both be equal to that element. (D).
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Re: PS Challenge! [#permalink] New post 25 May 2012, 01:37
ykaiim wrote:
IMO A.

Say S = {3, 3 ,6} and S1 = {6}.
Mean of S = 4 while Mean of S1 = 6
Median of S = 3 while Median of S1 = 6. Let's check the three statements:

I. Set S contains only one element [We have three elements here. Incorrect]

II. All elements in set S are equal [No, not neccessary. Incorrect]

III. The median of set S equals the mean of set S [Not neccessary. We have Median of S = 3 and Mean of S = 4. Incorrect]


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


* none of the three qualities is necessary
* II only
* III only
* II and III only
* I, II, and III


@ykaiim
The key word here is ANY.

Your set S1= {3,3,6} your Subset S1={6}

so here we can see mean of s=4 and mean of s1= 6

here mean of s does not exceed the mean of s1

but we can also take the subset as s1= {3}

in this case mean of S = 4 exceeds the mean of s1= 3,

so condition ii must be true for the statement to hold , and if a set contains elements which are all equal , then mean = median hence iii must also be true.

so this set S1= {3,3,6} fails condition ii as subset {3} will violate the given terms.

courtesy: Bunuel :)
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post 25 May 2012, 01:44
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

* none of the three qualities is necessary
* II only
* III only
* II and III only
* I, II, and II

lets take a set S= {1,2,3,4,5}
mean= 3 and mean of its its subset ( 4,5) will be 4.5 which would be greater that 3.
even for set of three values ( 1,2,3) mean would be 2 and its subset (3) mean would be 3 which is greater than 2
therefore for mean of any subset not to be greater than set mean ..
all values should be equal.
and if all values are equal than its median and mean would also be equal..
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post 25 May 2012, 07:00
I answered E. missed reading that the question says must be true and not could be!
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Re: PS Challenge! [#permalink] New post 11 Feb 2013, 09:29
Bunuel wrote:
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


"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.

Answer: D.



Hi,

i couldn't able to understand option 1. since there is only one elt in the set then the subset may contain that same element or zero huh?? Please explain option 1
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Re: PS Challenge! [#permalink] New post 11 Feb 2013, 09:36
Expert's post
FTG wrote:
Bunuel wrote:
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


"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.

Answer: D.



Hi,

i couldn't able to understand option 1. since there is only one elt in the set then the subset may contain that same element or zero huh?? Please explain option 1


Not sure I understand your question.

Anyway, option I says: set S contains only one element. This statement is not always true, for example consider set {1, 1}.
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Re: PS Challenge! [#permalink] New post 11 Feb 2013, 10:05
I am sorry, was that not considered as two same element ?
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Re: PS Challenge! [#permalink] New post 12 Feb 2013, 03:18
FTG wrote:
I am sorry, was that not considered as two same element ?


bunuel, I have the same doubt..
The example you quoted is a 2 element set.. agreed that the element is same though..

question is dubious ..
Answer should be A..


All elements in set S are equal - true for both A and B scenarios

This can be true as situation A : 1 element set but this statement says All elements in set S are equal .. When there is only 1 element, question of all elements doesn't arise. .
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post 11 Nov 2013, 05:43
Hi,
I didn't think 3 statement must be true.
Let s : {5,35,65} Its true mean and median are same but the it doesn't satisfy that Set S mean doesn't exceed any subset of S mean.

Please explain.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink] New post 11 Nov 2013, 05:46
Expert's post
vikrantgulia wrote:
Hi,
I didn't think 3 statement must be true.
Let s : {5,35,65} Its true mean and median are same but the it doesn't satisfy that Set S mean doesn't exceed any subset of S mean.

Please explain.


Can you please explain what you mean? Thank you.
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Re: If the mean of set S does not exceed mean of any subset of   [#permalink] 11 Nov 2013, 05:46
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