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!

Stuart Kovinsky wrote:

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!

__________
Thanks for sharing the shortcut approach though that also takes some time.Isn't it?
Another question in my mind:Suppose ,if we get a small set ,will it be better to take the 1st approach suggested by you and be 100%sure or is it beneficial to apply the shortcut technique all the time??
I think for smaller sets,the time taken would be equivalent for both the sets.
Seeking advice!!

harsh.champ wrote: __________
Thanks for sharing the shortcut approach though that also takes some time.Isn't it?
Another question in my mind:Suppose ,if we get a small set ,will it be better to take the 1st approach suggested by you and be 100%sure or is it beneficial to apply the shortcut technique all the time??
I think for smaller sets,the time taken would be equivalent for both the sets.
Seeking advice!!

The first shortcut approach is about as quick as it's humanly possible to answer the question; if you reasoned it out that way immediately, you'd be done in 15-20 seconds.

Picking numbers may or may not be time consuming based on the question. Like any other skill, if you're not used to using it, it will take you longer. Accordingly, you should practice picking numbers whenever you get the chance to determine:

1) On which questions is it useful to pick numbers?

2) When is picking numbers faster for me than algebra?

and

3) Which numbers should I pick for which types of questions?

As for which approach to take for which question, that's a very personal decision. You know what your strengths and weaknesses are and roughly how long it takes you to solve things algebraically. A good general guideline is that if the algebra jumps out at you and you can see a relatively quick way to solve, then math is probably quicker; on the flip side, if the algebra isn't clear or if you know from experience that it's going to take too long, picking numbers is the way to go.

Sum of 1st n odd numbers = n^2 ( 1+ 3+ 5+ ... + (2n-1) = n^2)

Similarly, Sum of the 1st n even numbers = n(n+1)


Here 1 is missing in the odd numbers and (2n+1) is added

So sum of the odd numbers in the set = n^2 -1 + 2n +1 = n^2 +2n

Also, in the set, odd numbers = even numbers = n

So, X-Y = ((n^2 +2n) - (n(n+1)))/n
= (n^2 + 2n -n^2 -n)/n
= n/n =1