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Re M1623 [#permalink]
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16 Sep 2014, 00:59
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. Answer: D
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Re: M1623 [#permalink]
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26 Dec 2014, 21:02
Could you please help me with statement 2 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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Re M1623 [#permalink]
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17 Aug 2015, 22:41
I think this is a highquality 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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Re: M1623 [#permalink]
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18 Aug 2015, 01:35



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Re: M1623 [#permalink]
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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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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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Re: M1623 [#permalink]
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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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Re: M1623 [#permalink]
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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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Re: M1623 [#permalink]
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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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Re: M1623 [#permalink]
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28 Dec 2016, 06:42



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Re: M1623 [#permalink]
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03 Mar 2017, 15:14
Please Help me find out if I understood.
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



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Re: M1623 [#permalink]
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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.
Answer: D 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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Re: M1623 [#permalink]
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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.
Answer: D 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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Re: M1623 [#permalink]
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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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Re: M1623 [#permalink]
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22 Mar 2017, 23:55
The answer is D. 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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Re: M1623 [#permalink]
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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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