If M is a sequence of consecutive integers which contains

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.

Rule:
For any set that is SYMMETRICAL ABOUT THE MEDIAN, average = median.



Statement 1:
Here, set M is symmetrical about a median of 15.5 -- the value halfway between 10 and 21.
Thus:
average = median = 15.5
SUFFICIENT.

Statement 2:
Since the median is unknown, the average cannot be determined.
INSUFFICIENT.



For another DS problem that can be solved with the blue concept above, check my post here:
https://gmatclub.com/forum/if-x-is-a-nu ... 94343.html

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.

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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

Or am I missing something. Please help explain.
Thank you all in advance

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.

VeritasKarishma generis AjiteshArun

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.

The average for a set of consecutive integers is equal to the median ( since the common difference of the any adjacent terms is equal)
(1) In M, the number of terms that are less than 10 is equal to the number of terms greater than 21.

Here the average will always be equal to the median = 15.5 SUFFICIENT

(2) There are 20 terms in M.

Here we can not determine the median of the average. Hence, INSUFFICIENT

Answer Option A