ruplun wrote:If 2 sets of numbers P and Q have same # of elemnts, is the mean of set Q lower than the mean of set P?

(1) Set Q consists of consecutive even integers and set P of consecutive odd integers.
(2) The median of set Q is higher than the mean of set P.

There's an important rule that says, "If the numbers in a set are evenly spaced, then the mean and median of that set are equal"

So, for example, the mean and median of this set {1, 4, 7, 10, 13, 16} are equal since the numbers are evenly/equally spaced.

Since consecutive odd numbers are evenly spaced (each number is 2 greater than the number before it) the mean and median will be equal in a set of consecutive odd integers.

The same applies to a set of consecutive even integers.

Statement 1:
We can show that this is INSUFFICIENT by using counter examples.
Given the conditions in statement 1, we could have these two cases:

case a) Q:{4,6} P:{1,3} --> Mean of set Q is greater than mean of set P
case b) Q:{4,6} P:{11,13} --> Mean of set P is greater than mean of set Q

Statement 2:
We can show that this is INSUFFICIENT by using counter examples.
Given the conditions in statement 2, we could have these two cases:

case a) Q:{1,5,7} P:{1,4,7} --> Mean of set Q is greater than mean of set P
case b) Q:{1,5,7} P:{1,4,100} --> Mean of set P is greater than mean of set Q

Statements 1 AND 2:
From statement 1, we know that the mean and median in set Q are equal (since the numbers are evenly spaced).
Similarly, we know that the mean and median in set P are equal (since the numbers are evenly spaced).

Statement 2 tells us that set Q has the greater median. Since the mean and median in set Q are equal (and the same applies to set P), we can conclude that the mean of set Q MUST BE greater than the mean of set P.

So, statements 1 and 2 combined are SUFFICIENT and the answer is C