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M16-23

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16 Sep 2014, 00:59
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75% (hard)

Question Stats:

41% (01:06) correct 59% (00:59) wrong based on 138 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

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16 Sep 2014, 00:59
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Official Solution:

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 $$S$$" implies that set $$S$$ could be:

A. $$S=\{x\}$$ - $$S$$ contains only one element (e.g. {7 });

B. $$S=\{x, x, ...\}$$ - $$S$$ contains more than one element and all elements are equal (e.g. {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$$\gt$$smallest number).

Example: if $$S=\{3, 5\}$$, then mean of $$S=4$$. Pick subset with the smallest number: $$s'=\{3\}$$, mean of $$s'=3$$. As we see $$3 \lt 4$$.

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.

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26 Dec 2014, 21:02
statement 2 seems to assume that we have multiple elements in the set but what if we have a condition where by we have just one element in the set . i feel that “single element” SET cannot be classified for the statement : “all elements in the set are equal”—————> the word “all + plural” linguistically implies more than “one”
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17 Aug 2015, 22:41
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I do not understand as to why have not considered an empty set, which is also a subset of any set ?
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18 Aug 2015, 01:35
fetchnitin wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. I do not understand as to why have not considered an empty set, which is also a subset of any set ?

An empty set has no mean or the median.
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15 Sep 2015, 20:44
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1
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16 Sep 2015, 03:45
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prodigypringle wrote:
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Thus, the median of {1,1,1,1} is (1 + 1)/2 = 1.
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07 Oct 2016, 08:18
Bunuel wrote:
prodigypringle wrote:
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Thus, the median of {1,1,1,1} is (1 + 1)/2 = 2.

(1+1)/2 = 1 right?
so statement 3 should not be true.
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08 Oct 2016, 03:10
deepak268 wrote:
Bunuel wrote:
prodigypringle wrote:
What if the Subset had even number of elements vs odd number of elements

S = {1,1,1,1} mean = 1 median = 0.5
S = {1,1,1} mean = 1 median = 1

The median of a set with even number of terms is the average of two middle terms when arranged in ascending/descending order. Thus, the median of {1,1,1,1} is (1 + 1)/2 = 2.

(1+1)/2 = 1 right?
so statement 3 should not be true.

2 there was clearly a typo. III must be true. Do you have an example when it's not?
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27 Dec 2016, 19:27
Please elaborate on choice 1.If a set contains 1 element,then its subset would have 1 element any time,then its mean is equal to its subset mean always.
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28 Dec 2016, 06:42
GouthamNandu wrote:
Please elaborate on choice 1.If a set contains 1 element,then its subset would have 1 element any time,then its mean is equal to its subset mean always.

The question asks which of the options MUST be true not could be true. While I could be true it's not necessary to be true. Set S can have more than 1 identical elements: {x, x, ...}
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03 Mar 2017, 15:14

Statement 1 - Could have this 3 possibilities below .
1- {x,x,x...x}
2- {x}
3- { }

So, The statement 1 is no Right because if { } there is no mean?
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04 Mar 2017, 01:58
vitorpteixeira wrote:

Statement 1 - Could have this 3 possibilities below .
1- {x,x,x...x}
2- {x}
3- { }

So, The statement 1 is no Right because if { } there is no mean?

No. The question asks which of the options MUST be true not could be true. While I could be true it's not necessary to be true. Set S can have more than 1 identical elements: {x, x, ...}
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05 Mar 2017, 01:24
Bunuel wrote:
Official Solution:

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 $$S$$" implies that set $$S$$ could be:

A. $$S=\{x\}$$ - $$S$$ contains only one element (e.g. {7 });

B. $$S=\{x, x, ...\}$$ - $$S$$ contains more than one element and all elements are equal (e.g. {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$$\gt$$smallest number).

Example: if $$S=\{3, 5\}$$, then mean of $$S=4$$. Pick subset with the smallest number: $$s'=\{3\}$$, mean of $$s'=3$$. As we see $$3 \lt 4$$.

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.

Dear Bunuel,

Why do not you count evenly spaced set in your consideration?

if there is a set $$S=\{1, 3, 5\}$$ and subset $$S=\{1, 3\}$$

Mean of S is 3 & Mean of subset is 2. So statement II does not need to have equal elements.

Where did I go wrong?
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05 Mar 2017, 03:49
Mo2men wrote:
Bunuel wrote:
Official Solution:

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 $$S$$" implies that set $$S$$ could be:

A. $$S=\{x\}$$ - $$S$$ contains only one element (e.g. {7 });

B. $$S=\{x, x, ...\}$$ - $$S$$ contains more than one element and all elements are equal (e.g. {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$$\gt$$smallest number).

Example: if $$S=\{3, 5\}$$, then mean of $$S=4$$. Pick subset with the smallest number: $$s'=\{3\}$$, mean of $$s'=3$$. As we see $$3 \lt 4$$.

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.

Dear Bunuel,

Why do not you count evenly spaced set in your consideration?

if there is a set $$S=\{1, 3, 5\}$$ and subset $$S=\{1, 3\}$$

Mean of S is 3 & Mean of subset is 2. So statement II does not need to have equal elements.

Where did I go wrong?

You did not read carefully. The first sentences of the solution makes important word in caps and bold!

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$" implies that set $$S$$ could be...

For your example, if $$S=\{1, 3, 5\}$$, consider a subset {1}.
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05 Mar 2017, 06:53
Bunuel wrote:

You did not read carefully. The first sentences of the solution makes important word in caps and bold!

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$" implies that set $$S$$ could be...

For your example, if $$S=\{1, 3, 5\}$$, consider a subset {1}.

Thanks a lot Bunuel.

I have a question about terminology meaning not related to this question.

When I say 'Value X does not exceed value Y', How should methamaticllay represnt it?

is it X <= Y or only x=Y?

Thanks
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05 Mar 2017, 06:56
Mo2men wrote:
Bunuel wrote:

You did not read carefully. The first sentences of the solution makes important word in caps and bold!

"The mean of set $$S$$ does not exceed mean of ANY subset of set $$S$$" implies that set $$S$$ could be...

For your example, if $$S=\{1, 3, 5\}$$, consider a subset {1}.

Thanks a lot Bunuel.

I have a question about terminology meaning not related to this question.

When I say 'Value X does not exceed value Y', How should methamaticllay represnt it?

is it X <= Y or only x=Y?

Thanks

It's the same as x is less than or equal to y, so it's $$x \leq y$$.
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22 Mar 2017, 23:55
Here is the justification:

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 SS?

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

Option 1: If set S contains only one element then it will not satisfy "If the mean of set S does not exceed mean of any subset of set S". Hence I. is not an answer.
Option 2: All elements in set S are equal. - Eg: {5,5,5,5} is S, subset of S = {5,5,5}. Hence this satisfies the condition "mean of S does not exceed mean of subset of S". II is an answer.
Option 3:The median of set S equals the mean of set S. Taking the same eg. as mentioned above,Mean of S = 5 , the median of S = 5. This same also satisfied the condition that if there is a set of this sort exists then - it satisfies the condition "mean of S does not exceed mean of subset of S".
Hence the option is D.
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14 Aug 2017, 17:00
1. Set S contains only one element.
Does it mean count of element or value of element?
My understanding is that set S has only one element by count not by value... therefore {X,X...} is not possible. Please help understand.
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14 Aug 2017, 22:08
ManSab wrote:
1. Set S contains only one element.
Does it mean count of element or value of element?
My understanding is that set S has only one element by count not by value... therefore {X,X...} is not possible. Please help understand.

Option 1, Set S contains only one element means that there is only one elements in S: {X}. This option is not necessarily true. Please re-read the solution.
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