Average of Set - Please help

[rahulvsd]

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.

[spoiler]OA: A. How is 1 Sufficient? [/spoiler]

[aneesh.kg]

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.

[spoiler]OA: A. How is 1 Sufficient? [/spoiler]

Very good problem.

Before going ahead solving this, let us discuss a little bit about the mean of an Arithmetic Progression.

Lets take a simple AP: 4, 6, 8 whose Mean = 6.
If I add a 2 to this AP and a 10 to this AP, the Mean of (2, 4, 6, 8, 10) is the same.
If I now add a (0, -2) to this and then a (12, 14) the mean of (-2, 0, .... 12, 14) is still the same.

A more visual method to digest this is to plot these numbers on the number line. If you add the same number of terms before and after the first and the last term respectively, you will notice that the mean is still mid-way.

Moral of the Story: If we add/subtract the same number of terms before and after an AP with the same constant difference, the Mean does not change.

Lets look at this from another angle.
If you have an AP: a1, a2,... an, then the mean = (a1 + an)/2 = [(a1 - d) + (an + d)]/ 2 = (a2 + an-1) / 2 = (ak + an - k)/2

Does Statement(1) make sense now?

A set of consecutive integers is an AP with constant difference = 1.
If the number of terms before 10 are same as the number of terms after 21, the mean or the average will remain (10 + 21)/2.

Statement(1) is SUFFICIENT. You don't need the exact number of terms.
Statement(2) is not.

[spoiler](A)[/spoiler] is the answer.

[neelgandham]

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.

The average of a set of consecutive numbers is the mid term(If the number of terms is Odd) or the average of the midterms.
Let us have a look at the sequence

n,n+1,n+3..................(t numbers),10,11,12,13,14,15,16,17,18,19,20,21, n+t+12, n+t+13...(t numbers).

There are a total of t(first t numbers) + 12(10 to 21) + t (last t numbers) = 2(t+6). Since, the number of terms in the sequence is even, the average of the set of the midterms i.e (15+16)/2 = 15.5

So, statement 1 is sufficient to answer the question

(2) There are 20 terms in M.

The average of the terms varies with the magnitude of the terms. i.e Though they have the same number of terms(20), The average of 1,2,3,4,5,6,7,8,9,10,11,12,13,14,154,16,17,18,19,20 is not equal to the average of the sequence 2,3,4,5,6,7,8,9,10,11,12,13,14,154,16,17,18,19,20,21.
So, statement 2 is insufficient to answer the question[/quote]

[Anurag@Gurome]

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.

[spoiler]OA: A. How is 1 Sufficient? [/spoiler]

Since M is a sequence of consecutive integers then M is an evenly spaced set, so its average is equal 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.
If the set of consecutive integers is {10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21}.
Now, if we place an 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.

The correct answer is A.