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

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If the mean of set S does not exceed mean of any subset of [#permalink] New post 01 May 2010, 23:12
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 07 Feb 2012, 15:24, edited 1 time in total.
Edited the question and added the OA
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Re: PS Challenge! [#permalink] New post 01 May 2010, 23:46
Lets have a set S:{5,6,7} and Mean = 6 and Median = 6
Now one subset of S can be {6,7} whose mean > 6
So III is not always true!
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Re: PS Challenge! [#permalink] New post 02 May 2010, 04:00
IMO D

1 and 2nd both could be true, but in must be true we can omit which is not essential always.

2nd must be true.

Now if 2nd is true then median must be equal to mean..
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Last edited by gurpreetsingh on 02 May 2010, 07:30, edited 1 time in total.
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Re: PS Challenge! [#permalink] New post 02 May 2010, 05:53
Quote:
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


I feel all the elements should be equal.

if all the elements are equal then the mean will always be same as the element(no mater how many elements are considered for mean).

at the same time the median will be equal to mean (all elements are equal)


so I feel II&III are always true. ANS:D
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Re: PS Challenge! [#permalink] New post 02 May 2010, 05:59
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

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Re: PS Challenge! [#permalink] New post 02 May 2010, 06:25
@ykaiim

if u accept that there should be only 1 element then the rest 2 statements will by default be true.

if there is only one element a)All elements in set S are equal b)The median of set S equals the mean of set S
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Re: PS Challenge! [#permalink] New post 02 May 2010, 06:35
My take

Statement 1: if the set is 6 6 6 6 6 more than one so this is not sufficient

Statement 2: If the set is 6 or 7 than the mean of the entire set is 6.5 but a subset mean can be 7 or 6 so having 2 different numbers does not satisfy this. So sufficient

Statement 3: From statement 2 we can't have multiple different values in the set so the median most = the mean

so I chose D
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Re: PS Challenge! [#permalink] New post 02 May 2010, 08:11
Dear Ravi,

The case I have put forward meets all the criteria of the stated problem. Tell me where I m wrong. May be I miss some important part.

Answer A says none of the three qualities is necessary.

I donot deny the case if set S has just one element but it is not neccessary to meet the conditions.


RaviChandra wrote:
@ykaiim

if u accept that there should be only 1 element then the rest 2 statements will by default be true.

if there is only one element a)All elements in set S are equal b)The median of set S equals the mean of set S

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Re: PS Challenge! [#permalink] New post 02 May 2010, 09:36
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ykaiim 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 ?


Ok please provide set of number(s),which violate the given 3 choices.


Quote:
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:



the example provided by u. doesnt hold good for all the cases.

Mean of S is 4.
Mean of S1{3} is 3. S>S1...


hope this is clear. :)
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Re: PS Challenge! [#permalink] New post 02 May 2010, 11:19
The biggest problem what some of us made is that we didn't consider the points independently.
I mean to say that we can't solve the problem as per the following logic
Quote:
if all the elements are equal then the mean will always be same as the element(no mater how many elements are considered for mean)
.
We should treat each point mutually exclusive and independent.
When we'll discuss about II we don't need to interpret II on the basis of I.
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Re: PS Challenge! [#permalink] New post 04 May 2010, 13:00
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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.
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Re: PS Challenge! [#permalink] New post 05 May 2010, 01:10
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.



Why can't we consider the subset with the greatest number
Ex- S= (5,6,7) and subset s (7)
Please explain :?:
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Re: PS Challenge! [#permalink] New post 05 May 2010, 01:46
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hardnstrong wrote:

Why can't we consider the subset with the greatest number
Ex- S= (5,6,7) and subset s (7)
Please explain :?:


The questions says: the mean of set S does not exceed mean of ANY subset of set S --> Mean of S<=Mean of ANY subset of S.

I considered the example of subset with smallest number to show that set S can not have 2 (or more) different elements. In any set with 2 (or more) different elements we can pick subset that will have the smaller mean than the mean of the entire set.

In your example (S={5,6,7}, mean of S=6), there are subsets of S with smaller mean (eg s'={5} or s'={5,6}), with the mean equal to the mean of entire set (eg s'={5,7} or s'={6}) and with bigger mean (eg s'={7} or s'={6,7}).

The stem could have stated opposite thing: "the mean of ANY subset of S does not exceed mean of S" and the answer would be the same. And to demonstrate that pick the subset with biggest number of the set in a set with 2 (or more) different elements and you'll see that this subset will have the mean bigger than the mean of entire set.

Basically this part of the stem is just the complicated way of saying that S contains either A. only one element or B. more than one identical elements.

Hope it's clear.
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Re: PS Challenge! [#permalink] New post 05 May 2010, 02:14
Thanks bunuel. it clear now +1

ANY is the catch here. Sometimes a single word can take things in opposite direction, just as in this case
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Re: PS Challenge! [#permalink] New post 18 Sep 2010, 22:43
Hi Bunuel,

Why the first statement is wrong...didn't understand.
I. Set S contains only one element.
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Re: PS Challenge! [#permalink] New post 18 Sep 2010, 22:54
Because can contain the same element repeat an arbitrary number of times as well.
The set A={3,3,3,3,3} id different from B={3}

Yet in either case the condition of means is satisfied, whereas A has 5 elements and B has only 1
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Re: PS Challenge! [#permalink] New post 18 Sep 2010, 23:03
Nope. Let's go my way...let say if set S={3}....then mean/median = 3 (this does satisfy with question stem "If the mean of set S does not exceed mean of any subset of set S")....then this is true.

please correct me if I am missing anything.
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Re: PS Challenge! [#permalink] New post 19 Sep 2010, 01:03
Expert's post
appy001 wrote:
Hi Bunuel,

Why the first statement is wrong...didn't understand.
I. Set S contains only one element.

appy001 wrote:
Nope. Let's go my way...let say if set S={3}....then mean/median = 3 (this does satisfy with question stem "If the mean of set S does not exceed mean of any subset of set S")....then this is true.

please correct me if I am missing anything.


Question asks: "which of the following MUST be true about set S" (not COULD be true).

I. Set S contains only one element --> it's not necessarily true as S can contain more than one element and still satisfy the requirement in stem. For example if S={3, 3, 3, 3} then mean of S equals to 3 and it does not exceed the mean of ANY subset of S, which also will be equal to 3.

Hope it's clear.
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Re: PS Challenge! [#permalink] New post 19 Sep 2010, 02:00
oh well.. :roll:

I read "one element" as one TYPE of element --- that is only the same unique element in what ever number for the whole set which is essentially the same as II...

elements is really a reference to the data points in the set
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Re: PS Challenge! [#permalink] New post 19 Sep 2010, 07:10
gmat1011 wrote:
oh well.. :roll:

I read "one element" as one TYPE of element --- that is only the same unique element in what ever number for the whole set which is essentially the same as II...

elements is really a reference to the data points in the set


In set theory there is no requirement for uniqueness of elements, and in general elements does not refer to unique constituents

So {3,3,3} is different from {3}
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Re: PS Challenge!   [#permalink] 19 Sep 2010, 07:10
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