praetorian123 wrote:

Is the median of set A bigger than the mean of set B(same number

of digits)?

1) set A is consecutive even numbers and Bis consecutive odd numbers

2) the sum of A is bigger than that of B

please explain ?

I think answer is C.

Statement I - Insufficient.

case (1)Imagine A is a set of all negative even consecutive numbers. So, median would be negative. B is a set of all positive consecutive odd numbers. So, mean would be positive. So, median (A) < mean (B)

case (2) Imagine A is a set of all positive even consecutive numbers. And B is a set of all negative odd consecutive numbers. Then, median (A) > mean (B)

Statement II - Insufficient

From this statement, all we can derive is that mean (A) > mean (B) because sum of numbers in set A is greater than that of numbers in B. However, median (A) can be greater than mean (A) i.e. median (A) > mean (B).

Or way less than mean (A) so that median (A) < mean (B)

Now consider both the statements together. In a set of even numbers, mean is always equal to median. And since mean (A) > mean (B) from statement II, median (A) > mean (B)