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# M12-14

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Math Expert
Joined: 02 Sep 2009
Posts: 55635

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16 Sep 2014, 00:46
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Difficulty:

85% (hard)

Question Stats:

40% (00:50) correct 60% (00:51) wrong based on 121 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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16 Sep 2014, 00:46
1
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.

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Joined: 11 Nov 2014
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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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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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Joined: 11 Sep 2013
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17 Mar 2015, 19:23
1
1
(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

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Joined: 21 Jul 2015
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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.

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Joined: 04 Mar 2016
Posts: 43
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28 Dec 2016, 21:26
1
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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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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24 Apr 2018, 21:03
Ritz1605 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate.

Check more solutions here: https://gmatclub.com/forum/if-the-mean- ... 34780.html

Hope it helps.
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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
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Joined: 22 Oct 2018
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19 Dec 2018, 19:00
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. This explanation doesn't take you through why each part is wrong. It is insufficient.
Re M12-14   [#permalink] 19 Dec 2018, 19:00
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# M12-14

Moderators: chetan2u, Bunuel