rochak22 wrote:
. The sum of the integers in list S is the same as the sum of the integers in list T. Does S contain more integers
than T?
(1) The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
(2) The median of the integers in S is greater than the median of the integers in T.
Okay the Question asks if the integers in set S is more than the integers in set T ??
Statement1 :: The average (arithmetic mean) of the integers in S is less than the average of the integers in T.
Now, use smart number plugin ..... lets say if S = 2, 2, 2 and T = 3, 3, then the sums for both is 6, & the average of S is less than T and S has more integers than T.
similarly, if S = -3, -3 and T = -2, -2, -2, then the sums for both is -6, the average of S is less than T and S has fewer integers than T. Therefore, Insufficient
Statement 2 :: If T = 2, 2, 2 and S = 3, 3, then the sums for both is 6 & the median of S is greater than median of T and S has fewer integers than T.
& If T = -3, -3 and S = -2, -2, -2, then the sums for both -6 & the median of S is greater than median of T and S has more integers than T. Therefore, Insufficient.
1+2 .......... Lets say If S = -7, 9, 10 and T = 6, 6, then the sums for both is 12 & the average of S is less than the average pf T & the median of S is greater than the median of T and S has more integers than T.
if S = -6, -6 and T = -10, -9, 7, then the sums for both is -12 & the average of S is less than the average of T & the median of S is greater than the median of T and S has fewer integers than T.
Therefore, Insufficient.
Hence, E .................