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Re: PS Challenge! [#permalink]
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24 Sep 2010, 01:40
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
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Re: PS Challenge! [#permalink]
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24 Sep 2010, 01:46
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]
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24 Sep 2010, 07:20
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]
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29 Sep 2010, 10:26
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]
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25 May 2012, 02:37
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]
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25 May 2012, 02:44
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]
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25 May 2012, 08:00
I answered E. missed reading that the question says must be true and not could be!



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Re: PS Challenge! [#permalink]
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11 Feb 2013, 10:29
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]
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11 Feb 2013, 10:36
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]
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11 Feb 2013, 11:05
I am sorry, was that not considered as two same element ?



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Re: PS Challenge! [#permalink]
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12 Feb 2013, 04:18
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 scenariosThis 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]
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11 Nov 2013, 06:43
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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11 Nov 2013, 06:46



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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
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21 Oct 2014, 13:49
I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set
Same Example: Set S: 5,6,7
Sub set A: 5 Sub set B: 7
for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.
Please review your answer again.



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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
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21 Oct 2014, 14:02
awal_786@hotmail.com wrote: I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set
Same Example: Set S: 5,6,7
Sub set A: 5 Sub set B: 7
for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.
Please review your answer again. S cannot be {5, 6, 7}. The stem says that the mean of set S does not exceed mean of any subset of set S. The mean of S is 6. One of the subsets of S is {5}, with mean of 5. So, the mean of S, which is 6, is greater than the mean of its subset {5}.
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Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
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22 Oct 2014, 03:51
Bunuel wrote: awal_786@hotmail.com wrote: I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set
Same Example: Set S: 5,6,7
Sub set A: 5 Sub set B: 7
for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.
Please review your answer again. S cannot be {5, 6, 7}. The stem says that the mean of set S does not exceed mean of any subset of set S. The mean of S is 6. One of the subsets of S is {5}, with mean of 5. So, the mean of S, which is 6, is greater than the mean of its subset {5}. Agree, mean of set S doesn't exceed mean of Sub set S, but Mean of Sub set of S can exceed mean of set S, That's what I mean, your question doesn't have 2 way condition, then why you are taking it as two one condition?



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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
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22 Oct 2014, 03:58
awal_786@hotmail.com wrote: Bunuel wrote: awal_786@hotmail.com wrote: I read all explanation but I still think question should ask " what could be true" instead of what must be true because we can have few sets where mean of set S doesn't exceed the mean of Sub set of S without Same number in whole set
Same Example: Set S: 5,6,7
Sub set A: 5 Sub set B: 7
for Sub set A set S doesn't exceed mean of its sub set and it is not dependent all similar integers in set. ( condition of question is met ) If we take both your choices correct II, III then we should not be able to find alternate sets consisting Dissimilar numbers and sub set has lower mean than set itself.
Please review your answer again. S cannot be {5, 6, 7}. The stem says that the mean of set S does not exceed mean of any subset of set S. The mean of S is 6. One of the subsets of S is {5}, with mean of 5. So, the mean of S, which is 6, is greater than the mean of its subset {5}. Agree, mean of set S doesn't exceed mean of Sub set S, but Mean of Sub set of S can exceed mean of set S, That's what I mean, your question doesn't have 2 way condition, then why you are taking it as two one condition? Sorry, but that does not make any sense. The mean of set S does not exceed mean of any subset of set S means that none of the subsets of S has the mean less than the mean of S. Stem gives some condition on set S. Your example ({5, 6, 7}) does not satisfy it, so S cannot be {5, 6, 7}.
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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
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22 Oct 2014, 14:05
Agree, mean of set S doesn't exceed mean of Sub set S, but Mean of Sub set of S can exceed mean of set S, That's what I mean, your question doesn't have 2 way condition, then why you are taking it as two one condition?[/quote]
Sorry, but that does not make any sense. The mean of set S does not exceed mean of any subset of set S means that none of the subsets of S has the mean less than the mean of S.
Stem gives some condition on set S. Your example ({5, 6, 7}) does not satisfy it, so S cannot be {5, 6, 7}.[/quote]
Basically you are sticking to your answer, and not trying to understand. try solving this question without keeping your previous answer in mind. Set S {5,6,7} Sub set of set s {7} here mean of set S doesn't exceed mean of its sub set and your condition is obtained without taking all similar numbers in set S



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Re: If the mean of set S does not exceed mean of any subset of [#permalink]
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22 Oct 2014, 14:10
I found it my self, from other forum. key word in this question is "ANY" sub set. Sorry for bothering to much



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If the mean of set S does not exceed mean of any subset of [#permalink]
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22 Oct 2014, 14:12
awal_786@hotmail.com wrote: Basically you are sticking to your answer, and not trying to understand. try solving this question without keeping your previous answer in mind. Set S {5,6,7} Sub set of set s {7} here mean of set S doesn't exceed mean of its sub set and your condition is obtained without taking all similar numbers in set S Seems that you don't understand neither the question nor the solution. Last attempt: the stem says that the mean of set S does not exceed mean of ANY subset of set S. This means that S CANNOT be {5, 6, 7} because its mean is more than the mean of one of the subsets of S. Sorry, but I cannot explain any better than this.
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