If M is a sequence of consecutive integers which contains

Question

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.

Bunuel · Accepted Answer

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.

nrmlvrm · Answer

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)

ziko · Answer

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.

omeee · Answer

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

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

amanvermagmat · Answer

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.

warrior1991 · Answer

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}??

generis · Answer

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     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.

GMATGuruNY · Answer

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

Smita04

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.

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.

Show Spoiler

A

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

rxb266 · Answer

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

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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

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