mgmt_gmat wrote:Lists S and T consist of the same number of positive integers. Is the median of the
integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are
consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

Take n (S) = n (T). Is Me (S) > AM (T)?

(1) When S and T have an odd number of elements under these strict conditions, Me (S) is never the same as AM (T), it could be either less or more than each other. Insufficient

(2) While S and T consist of the same number of positive integers, if the sum of the integers in S is greater than the sum of the integers in T, AM of S will always be greater than the AM of T, but nothing concrete can be made about "Is Me (S) > AM (T)?"

If S = {1, 2, 11} and T = {1, 3, 8}, Me (S) < AM (T), whereas, when S = {5, 9, 11} and T = {1, 3, 8}, Me (S) > AM (T). Insufficient

Taken together

If the sum of the integers in S (that contains a number of positive consecutive even integers) is greater than the sum of the integers in T (that contains the same number of positive consecutive odd integers as S does), the median of S will always be greater than the mean of T. Sufficient

[spoiler]C[/spoiler]

Agree with sanju09,

IMO D

what';s OA?

The Median and Arithmetic Mean will be same in r/o Consequtive even or odd numbers.
The comparative total of the odd and even nos. helps us decide the Mean of each group of nos.
Therefore the Median or Mean of group S will be > Mean/Median of T.

mgmt_gmat wrote:Lists S and T consist of the same number of positive integers. Is the median of the
integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are
consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

St. 1: Integers in S are consecutive even integers, and the integers in T are
consecutive odd integers.
but we don't know the exact sequence of the integers from where it is starting ..not sufficient..
St 2: sum of integers in S = sum of integers in T so we can say their avg are equal but we can't compare median of S with avg of T

as we don't know the sequence of numbers or rather the no. might be random

Combined: since S are consecutive even integers and integers in T are consecutive odd integers. so their common difference is same. Hence their median and average are same..
now , sum of the integers in S is greater than the sum of the integers in T.
hence median of the integers in S greater than the average (arithmetic mean) of the integers in T

Ans C

mgmt_gmat wrote:Lists S and T consist of the same number of positive integers. Is the median of the
integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are
consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

This question is wonderfully explained by Ian Stewart here :

https://www.beatthegmat.com/ds-median-an ... 12554.html

mgmt_gmat wrote:Lists S and T consist of the same number of positive integers. Is the median of the
integers in S greater than the average (arithmetic mean) of the integers in T?
(1) The integers in S are consecutive even integers, and the integers in T are
consecutive odd integers.
(2) The sum of the integers in S is greater than the sum of the integers in T.

Q: is Median(S)>Mean(T)? Given no. of terms in S=no. of terms in T say N

(1) Clearly insufficient. As we can have set S{12,14,16} and set T{31,33,35} OR S{42,44,46} and T{1,3,5}.
two possibiliteis

(2) Sum(S)>Sum(T). Also insufficient. As we can have set S{1,1,10} and set T{3,3,3} OR S{20,20,20} and T{1,1,1}.
two possibilities

(1)+(2) From (1) question became Mean(S)>Mean(T) true? --> As there are equal # of term in sets and mean(average)=(Sum of terms)/(no. of terms), then we have: is Sum(S)/N>Sum(T)/N true? --> Is Sum(S)>Sum(T) true? This is exactly what is said in statement (2) that Sum(S)>Sum(T). Hence sufficient.

C