bek_gmat wrote:Help needed with the following question:

1. If two sets of numbers, P and Q, have the same number of elements, 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.

IMO B

St1: Sets can involve any consecutive number. (For even set : 2,4,6,8,10 or 100,102,104,106,108) it is not possible to say whose mean is bigger?

St2: Because 2 of the sets have same number of elements and because the numbers are consecutive, we can conclude that middle number (median) of othe set, also will be the mean of the same set. So we can determine the mean. IMO STATEMENT 2 is SUFF.
For ex:
1
3
5
7
9
Median : 5
Mean : 5

Careful!

St2: Because 2 of the sets have same number of elements and because the numbers are consecutive, we can conclude that middle number (median) of othe set, also will be the mean of the same set. So we can determine the mean. IMO STATEMENT 2 is SUFF.

This is one of the greatest tricks that the GMAT has at its disposal - the power of suggestion. If you consider what you know ONLY about statement 2, we don't know at all that the terms are consecutive. The only place the word "consecutive" appears is in statement 1. Without that term, we could easily use:

Situation 1:

P: 0, 1, 2, 3, 4
Q: 1, 2, 3, 4, 5 (the median is higher as stipulated by statement 2)

And here the mean of Q is higher than that of P

Situation 2:

P: 0, 1, 2, 3, 4
Q: -1,000,000, 2, 3, 4, 5 (again, the median is higher as stipulated by statement 2, and we have the same number of elements, as required by the question stem)
Here Q clearly has a much lower mean.

The most common way to miss a DS question is to think that you have more information than you really do. This is often because you:

-Assume something about values that isn't necessarily true (that they're integers, or positive, or within a certain band - i.e. not approaching infinity on either end)
-Incorporate part of one statement into the other

Be careful here! One helpful strategy is to always challenge yourself to get "the other" answer. Here, I think it's pretty easy to get the mean of Q higher than that of P just by using quick, consecutive integers. But if your goal, next, is to get the other answer, you challenge yourself to use all available options - you want to push the limits of the given information. And if you're looking for a situation like I did above - a negative million to really skew the results - you'll probably soon realize that the numbers don't have to be consecutive. You *want* to use unique numbers so you'll double-check the given information to try to use something crazy.

bek_gmat wrote:Help needed with the following question:

1. If two sets of numbers, P and Q, have the same number of elements, 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.

the answer should be C.
from statement 1 :
eg: P may be 3,5,7 and Set Q may be 0,2,4. Thus avg of P is greater than Q
if P may be 1,3,5 and set Q may be 4,6,8. Thus avg. of P is less than Q.

From statement: Nothing help.

From C: From both statement- from first statement and from the question we know that both the set P and Q are in A.P series as well as they contain the same no. of integer. So Median is = Average.
from second statement- Median of q is > than p Thus if Median is greater than mean is also greater(as they contain the same no of elements)