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If M is a sequence of consecutive integers which contains

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If M is a sequence of consecutive integers which contains [#permalink] New post 20 May 2012, 21:49
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If M is a sequence of consecutive integers which contains more than 11 terms, what is the average of M?

(1) In M, the number of terms that are less than 10 is equal to the number of terms greater than 21.
(2) There are 20 terms in M.
[Reveal] Spoiler: OA
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 21 May 2012, 00:06
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Smita04 wrote:
If M is a sequence of consecutive integers which contains more than 11 terms, what is the average of M?

(1) In M, the number of terms that are less than 10 is equal to the number of terms greater than 21.
(2) There are 20 terms in M.


kinda tricky question... got me confused for a while....

answer is (A)

from statement (1), the series is as follows:
(x no.of terms), 10, 11, 12, 13,....., 21, (x no.of terms)

no matter how many terms are there before 10 or after 21, the average always has to be the median (for consecutive integers, AVG=MEDIAN). in this case the median is the average of 6th term and 7th term of 10 to 21 series.... which is 15+16/2 = 15.5
SUFFICIENT...

from statement (2), we only know there are 20 terms.... that information is not enough to the median/mean....
INSUFFICIENT...

therefore (A)
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 21 May 2012, 00:07
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If M is a sequence of consecutive integers which contains more than 11 terms, what is the average of M?

Since M is a sequence of consecutive integers then M is an evenly spaced set, so its average equals to its median.

(1) In M, the number of terms that are less than 10 is equal to the number of terms greater than 21 --> consider the following set of consecutive integers {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}. Now, if we place equal number of consecutive integers before 10 and after 21 then in any case the median would be the average of two middle numbers 15 and 16 so average=median=(15+16)/2=15.5. Sufficient.

(2) There are 20 terms in M. M can be any set of 20 consecutive integers. Not sufficient.

Answer: A.

Hope it's clear.
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 23 May 2012, 10:55
hey very nice question! I picked C and erred!
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 02 Jun 2012, 19:47
Nice question, reveals an important aspect of mean/avg
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 03 Jun 2012, 02:25
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Smita04 wrote:
If M is a sequence of consecutive integers which contains more than 11 terms, what is the average of M?

(1) In M, the number of terms that are less than 10 is equal to the number of terms greater than 21.
(2) There are 20 terms in M.


A nice tricky question. Testing concepts and logic. :)
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 03 Jun 2012, 05:37
Hi,

For an Arithmetic progression (sequence of consecutive integers in this case) the mean as well as median is equal to the middle term of the sequence.

For even number of terms:
1,2,3,4
Mean = (2+3)/2 = 1.5

For odd number of terms:
1,2,3,4,5
Mean = 3

Thus, only a symmetrical distribution is required around the middle term.

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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 03 Jun 2013, 01:42
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 04 Jun 2013, 01:51
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I am not sure that more could be added to the solution by Bunuel, but i will try to provide my understanding and solution.

st 1) this statement tells us that this middle number between 10 and 21 should be median and the average since numbers are consequetive and the total quantity of numbers is >=12 . There are 10 numbers between 21 and 10, so the average of the numbers will be 15.5, nomatter howmany numbers are in the set. Sufficient.

St. 2) these 20 numbers can start at any point in a number line so it is not sufficient to find the final answer.

Answer is A.
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Re: If M is a sequence of consecutive integers which contains [#permalink] New post 19 Nov 2013, 06:25
I would also like to add to the different solutions.

1) For a set with consecutive integers we have: (First + Last) / 2 = average = median

Just test some cases in line with the given constraints in the stem --> (22+9)/2 = 31/2, (23+8) / 2 = 31/2,
From here we can clearly see a pattern that this will repeat no matter how many numbers there are.

Visually one could think that the 11 #s from 10 to 21 are fixed, and we are only making equal terms with equal length form the mean increase, which has no effect on the mean itself.


As a side note: By adding more #s to the "fixed set", the std. increases..


Statement 1 sufficient.
Re: If M is a sequence of consecutive integers which contains   [#permalink] 19 Nov 2013, 06:25
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