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:
VeritasKarishma generis AjiteshArunHow 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.