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Hello Bunuel,
What if : set S = {1,2,3,4} ; Mean=2.5
subset of S ={1,2} ; Mean =1.5
Then option 2 " All elements in set S are equal" does not fit in.
Please let me where am i going wrong.
Thanks,
Saba
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Re: M16-23 [#permalink]
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 24 Aug 2017, 12:25
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Bunuel wrote: saba@4010 wrote: Hello Bunuel,
What if : set S = {1,2,3,4} ; Mean=2.5
subset of S ={1,2} ; Mean =1.5
Then option 2 " All elements in set S are equal" does not fit in.
Please let me where am i going wrong.
Thanks,
Saba "The mean of set \(S\) does not exceed mean of ANY subset of set \(S\)". Your example does not work. Thanks Bunuel ,understood ! Posted from my mobile device
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Re: M16-23 [#permalink]
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25 Aug 2017, 02:47
Just need to confirm one thing.S={1,2,3} then can the subset be like A={1,1,}
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Re: M16-23 [#permalink]
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25 Aug 2017, 05:50
himanshukamra2711 wrote: Just need to confirm one thing.S={1,2,3} then can the subset be like A={1,1,} No, you don't have three 1's in S. Subsets of {1, 2, 3} are: {1, 2, 3} {1, 2} {1, 3} {2, 3} {1} {2} {3} {} - an empty set.
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Re: M16-23 [#permalink]
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29 Aug 2017, 06:04
Bunuel wrote: 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. Hi, I am confused with second option. If i didn't misunderstood the question, it state that mean of Set S should be less than mean of its subset. Lets Consider: S =(2,4,6) - mean =4 S'= (4,6) - mean = 5 Here Set S has different values. Please advise. Thanks, Arpit
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Re: M16-23 [#permalink]
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29 Aug 2017, 06:07
NeverGiveUp- Arpit wrote: Bunuel wrote: 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. Hi, I am confused with second option. If i didn't misunderstood the question, it state that mean of Set S should be less than mean of its subset. Lets Consider: S =(2,4,6) - mean =4 S'= (4,6) - mean = 5 Here Set S has different values. Please advise. Thanks, Arpit I'd advice to re-read the stem and the solution carefully and go through the discussion once more.
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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.
What are GMAT Club Tests?
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Re: M16-23 [#permalink]
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29 Aug 2017, 06:56
[quote="Bunuel"
I'd advice to re-read the stem and the solution carefully and go through the discussion once more.[/quote]
Got it. Thanks mate.
I overlooked the word 'Any '.
Posted from my mobile device
Posted from my mobile device
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Re: M16-23 [#permalink]
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19 Sep 2017, 00:48
What is S= 1,2,3,4,5
Mean=3
Median=3
Subset(4,5); Mean =4.5>3
So, Statment 3: False
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Re: M16-23 [#permalink]
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19 Sep 2017, 00:50
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Re: M16-23 [#permalink]
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22 Dec 2017, 15:18
Hello !
For case III, let's take an example where the median of set S equals the mean of set S : set S = {1,2,3}, mean=median=2. In that case, {1} is a subset of set S and 2>1. Conclusion: the mean of set S DOES exceed mean of a subset of set S
How is III still correct ?
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Re: M16-23 [#permalink]
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23 Dec 2017, 01:14
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