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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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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

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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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.

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.

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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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.

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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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.

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

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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