harsh.champ wrote:Consider the set S = 2, 3, 4, ...., 2n + l, where n is a positive integer larger than 2007. Define X as the average of the odd integers in S and Y as the average of the even integers in S. What is the value of X - Y?
(1)0
(2)1
(3)n/2
(4)n+1/2n
(5)2008
We know that we're starting with an even number and ending with an odd number (since 2n is even, 2n+1 will be odd). Therefore, we have an equal number of odds and evens in the set.
Every even term in the set is one less than it's "partner" odd term (i.e. 2 is one less than 3; 4 is one less than 5; 6 is one less than 7; ...). In other words, we could write the subsets as follows:
{2, 4, 6, 8, 10, ..., 2n}
{2+1, 4+1, 6+1, 8+1, 10+1, ..., 2n +1}
As we can see, since each subset has n terms, the sum of the odd set will be n more than the sum of the even set.
Average = sum of terms / # of terms, so:
Difference in averages = difference in sums/number of terms = n/n = 1
* * *
Now, while that approach would get you a whopping 10/10 on your grade 10 math test (maybe even 11/10, since it's so beautiful), it's certainly not the quickest way to generate the right answer. On the GMAT we couldn't care less about 10/10, we just want 1/1, which we get for clicking the right bubble on the screen no matter how we get there. So, let's talk about some other, quicker, approaches.
First, we could reason it out. The odd set is just the even set shifted one spot to the right, so it's average should also be shifted one spot to the right. Done!
Second, this is a perfect question for picking numbers. Picking numbers might seem impossible because of the "n is larger than 2007" restriction, but if we understand the concepts we quickly recognize that the restriction is completely irrelevant to generating the answer and we can ignore it.
So, let's try a couple of sets:
{2,3}
Average of odd terms is 3, average of even terms is 2, n=1; X-Y = 1.
(2) works, (4) works... eliminate (1), (3) and (5).
{2, 3, 4, 5}
Average of odd terms is 4, average of even terms is 3, n=2; X-Y = 1.
(2) works, (4) doesn't... eliminate (4).
Only (2) is left, done!
