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M12-14

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M12-14  [#permalink]

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New post 16 Sep 2014, 00:46
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38% (00:48) correct 62% (00:44) wrong based on 104 sessions

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If the mean of set \(S\) is 20, what is the median of set \(S\)?


(1) In set \(S\) there are as many numbers larger than 20 as there are numbers smaller than 20.

(2) All numbers in set \(S\) are even integers.

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Re M12-14  [#permalink]

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New post 16 Sep 2014, 00:46
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Re: M12-14  [#permalink]

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New post 16 Mar 2015, 18:31
hello Bunuel,

This explanation seems a bit confusing to me .
The question says that the mean is "20".
Then it should include "20" in set S,
But the examples you took do not include 20.
can you please explain each statement more clearly?
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New post 17 Mar 2015, 05:35
minhaz3333 wrote:
hello Bunuel,

This explanation seems a bit confusing to me .
The question says that the mean is "20".
Then it should include "20" in set S,
But the examples you took do not include 20.
can you please explain each statement more clearly?


The mean of {1, 2} is 1.5, which is NOT in the set. So, the mean of a set is not necessarily a member of the set.
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Re: M12-14  [#permalink]

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New post 17 Mar 2015, 19:23
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(1) In set S there are as many numbers larger than 20 as there are numbers smaller than 20

=> Dont know the number of numbers smaller and larger 20 => Cannot know the middle point of the number set => Insufficient

(2) All numbers in set S are even integers => Just know there are two Median for this number set => Insufficient

(1)+(2): Also insufficient

=> Answer: E
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Re: M12-14  [#permalink]

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New post 17 Aug 2016, 07:10
Initially got confused by statement
(1) In set SS there are as many numbers larger than 20 as there are numbers smaller than 20.

Was under the impression that number 20 has to in the set..
for example : {-30, -25, 20, 25, 30}, here there are two numbers are below 20 and two numbers above 20.. Mean = 20, Median = 20
from Bunuel's example set : {14, 18, 24, 24}, here again two numbers are below 20 and two above 20 .. Mean = 20, Median = 21

Clearly Statement (1) is Not Sufficient !!!


Bunuel wrote:
Official Solution:


Statements (1) and (2) combined are insufficient. Consider:

\(\{14, 18, 24, 24\}\). The median is 21;

\(\{18, 18, 22, 22\}\). The median is 20.


Answer: E
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Re: M12-14  [#permalink]

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New post 28 Dec 2016, 21:26
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Can we apply a simple logic here, that to find the Median we shud have no of terms and also an assurance that the set is in AP ??
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Re M12-14  [#permalink]

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New post 24 Apr 2018, 16:15
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.
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Re: M12-14  [#permalink]

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New post 24 Apr 2018, 21:03
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Re: M12-14  [#permalink]

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New post 19 Aug 2018, 03:08
Great question . I fell for the trap that 20 will be a part of the set and hence marked statement 1 as sufficient. Now I realised my mistake.Thanks
Re: M12-14 &nbs [#permalink] 19 Aug 2018, 03:08
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