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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]

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20 May 2012, 22:49
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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.
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21 May 2012, 01:07
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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.

Hope it's clear.
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Re: If M is a sequence of consecutive integers which contains  [#permalink]

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21 May 2012, 01:06
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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....

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]

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03 Jun 2012, 06:37
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.

Regards,
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03 Jun 2013, 02:42
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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Re: If M is a sequence of consecutive integers which contains  [#permalink]

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04 Jun 2013, 02:51
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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.

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Re: If M is a sequence of consecutive integers which contains  [#permalink]

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19 Nov 2013, 07:25
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.
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Re: If M is a sequence of consecutive integers which contains  [#permalink]

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12 Sep 2017, 17:07
Since the set contains consecutive integers ( positive or negative but no of terms are more than 11), the mean of the numbers is the median.

Stmt 1:
Let's say there are X terms that are less than 10 then there are same x terms that are greater than 21. So between 10 and 21 there are 12 numbers. So median would always be in between these numbers

Say then corresponding pairs would be ( 10,21), (11,20),(12,19), (13,18), (14,17), (15,16)

If we take average of any corresponding pairs that will be the average of the seq or numbers

STMT 2: There are 20 Terms in M. Clearly Inssuf because we can form the 20 terms with only positive or only negative or half of them positive and another half negative.
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If M is a sequence of consecutive integers which contains  [#permalink]

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22 May 2018, 07:46
Hi,

Can somebody please clarify my understanding?

1) Does "a sequence of consecutive integer" have to be in 1,2,3,4,5 sequence? or it could be skipped like 1,3,5,7, ....?
A questions below, I've done earlier said it could be skip sequence.
Sorry, I wasn't allow to post link. So here's the question.

Set X consists of seven consecutive integers, and Set Y consists of nine consecutive integers. Is the median of the numbers in set X equal to the median of the numbers in set Y?

(1) The sum of the numbers in set X is equal to the sum of the numbers in set Y.
(2) The median of the numbers in set Y is 0.

-------------------------------------------------------------------------------------------------------------------

This leads to my 2nd question.
2) if it can be skipped, then 1) wouldn't be sufficient.
For example: 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 >> Mean is 15
But for -6,-2,2,6,10,14,18,22,26,30,34 >> then the number is 14

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If M is a sequence of consecutive integers which contains  [#permalink]

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22 May 2018, 11:02
omeee wrote:
Hi,

Can somebody please clarify my understanding?

1) Does "a sequence of consecutive integer" have to be in 1,2,3,4,5 sequence? or it could be skipped like 1,3,5,7, ....?
A questions below, I've done earlier said it could be skip sequence.
Sorry, I wasn't allow to post link. So here's the question.

Set X consists of seven consecutive integers, and Set Y consists of nine consecutive integers. Is the median of the numbers in set X equal to the median of the numbers in set Y?

(1) The sum of the numbers in set X is equal to the sum of the numbers in set Y.
(2) The median of the numbers in set Y is 0.

-------------------------------------------------------------------------------------------------------------------

This leads to my 2nd question.
2) if it can be skipped, then 1) wouldn't be sufficient.
For example: 1,3,5,7,9,11,13,15,17,19,21,23,25,27,29 >> Mean is 15
But for -6,-2,2,6,10,14,18,22,26,30,34 >> then the number is 14

Hello

A sequence of consecutive integers means that only - consecutive only... so NO integers can be skipped in between.
So 1,2,3,4,5.. OR -5, -4, -3, -2, -1, 0, 1, 2.. OR 65, 66, 67, 68,.. are ALL sequences of consecutive integers.
But 1, 3, 5, 7.. is NOT a sequence of consecutive integers.
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Re: If M is a sequence of consecutive integers which contains  [#permalink]

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26 May 2018, 01:30
@manvermagmat
Got it. Thanks!
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Re: If M is a sequence of consecutive integers which contains  [#permalink]

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26 Jul 2019, 21:33

How come we conclude that the set is of integers {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}??
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If M is a sequence of consecutive integers which contains  [#permalink]

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27 Jul 2019, 00:11
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 [that are] greater than 21.
(2) There are 20 terms in M.

warrior1991 wrote:

How come we conclude that the set is of integers {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}??

warrior1991 , we don't have to use only these numbers, but we do have to use at least these numbers. The restriction says that some number of terms below 10 must be equal to some number of terms above 21.

10 is the lower benchmark. 21 is the higher benchmark. If there is some number of terms that are [that exist] below 10 and above 21, then 10 must be in the set and so must 21.

The integers are consecutive. We could decide that we want two terms below 10 and thus two terms above 21:

{8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23}

We know that the average is $$\frac{first + last}{2}$$. That average is the same as if we had not included 8, 9, 22, and 23.
Bunuel decided that there would be zero terms below 10 and zero terms above 21. He could do so because there are 12 consecutive integers from 10 to 21.

The sequence has more than 11 terms. There are zero terms below 10 and zero terms above 21. The conditions are satisfied, and we know the average. Sufficient.

Hope that helps.
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