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Algebra

by Aman verma » Wed Feb 24, 2010 6:57 am
Q: Consider the set S = { 2,3,4.............,2n + 1} ,where n is a positive integer greater than 2007.Define X as the average of odd integers in S and Y as the average of the even integers in S . What is the value of X - Y ?

a) 0

b)1

c)(1/2)n

d) (n+1)/2n

e)2008
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by ldoolitt » Wed Feb 24, 2010 7:26 am
Aman verma wrote:Q: Consider the set S = { 2,3,4.............,2n + 1} ,where n is a positive integer greater than 2007.Define X as the average of odd integers in S and Y as the average of the even integers in S . What is the value of X - Y ?

a) 0

b)1

c)(1/2)n

d) (n+1)/2n

e)2008
So I see a quick and slow way to do this.

Slow way (but more thorough)

Let K represent the number of terms in the set. Since 2n+1 is an odd number, and the set begins with an even number, there will be an equivalent number of odds and evens in the set. And thus each subset (odds and evens) has .5K numbers in it.

Average(odd) = sum of odds / .5 * K
Average(even) = sum of evens / .5*K

Average(odd-even) = sum of odds - sum of evens / .5K

But if you start writing out the sum of odds - sum of evens you have

-2+3-4+5-6+7+...

For each negative even there is an odd that is one more than it. You will have

1+1+1+...

How many 1's will you have? .5*K, since each sum to one takes 2 numbers (an odd and an even) and thus halves the number of terms in the set. The sum will be .5*K*1 = .5*K. Putting that back into the forumla you have:

.5*K / .5*K = 1

Or you could have simply notice that -2+3-4+5-6+7+... = 1+1+1+... and that the number of terms in each subset would be half the total number of terms in the set thus giving you an average of 1.
Last edited by ldoolitt on Wed Feb 24, 2010 7:27 am, edited 1 time in total.
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by Pedros » Wed Feb 24, 2010 7:27 am
I think the difference is 1 ( B )



Because ( 2n+1 ) is an odd number , I think about them as 2 different evenly spaced sets, the midpoint of the ODD set is only one space after that of the EVEN set . Regardless of the number of elements of each set the difference between the 2 midpoints will be 1.
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by Fiver » Wed Feb 24, 2010 9:32 am
Pedros wrote:I think the difference is 1 ( B )



Because ( 2n+1 ) is an odd number , I think about them as 2 different evenly spaced sets, the midpoint of the ODD set is only one space after that of the EVEN set . Regardless of the number of elements of each set the difference between the 2 midpoints will be 1.
Well done Pedros.

Though Pedros has developed a wonderful abstract solution to this problem, I thought i could give it an algebraic approach as well.

Both the sets of even and odd nos. are in AP.

Avg. of an AP = (FT + LT)/2
Avg. of the even set (X) = (2+2n)/2 = (1+n)
Avg. of the odd set (Y) = (3+2n+1)/2 = (2+n)

Y - X = 1
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by ajith » Wed Feb 24, 2010 9:39 am
Aman verma wrote:Q: Consider the set S = { 2,3,4.............,2n + 1} ,where n is a positive integer greater than 2007.Define X as the average of odd integers in S and Y as the average of the even integers in S . What is the value of X - Y ?

a) 0

b)1

c)(1/2)n

d) (n+1)/2n

e)2008
There are 2n numbers down there

Sum of even numbers n/2(2+2n) = n(n+1)
Average of even numbers = n+1

Sum of odd numbers n/2(3+2n+1) = n(n+2)
Average of odd numbers = n+2

[spoiler]Difference = 1; B[/spoiler]
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by Pedros » Wed Feb 24, 2010 9:43 am
Fiver wrote:
Both the sets of even and odd nos. are in AP.

Avg. of an AP = (FT + LT)/2
Avg. of the even set (X) = (2+2n)/2 = (1+n)
Avg. of the odd set (Y) = (3+2n+1)/2 = (2+n)

Y - X = 1

As always FIVER......Impressive.
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by Aman verma » Thu Feb 25, 2010 2:06 am
Ans[spoiler]b)1[/spoiler].This was a sitter indeed and it was posted on purpose.There is one Expert who was constantly complaining that the questions posted are not real GMAT questions. Maybe he find this one easy and to his liking.The most realistic questions on GMAT can be found on the active GMAT question pool which ,I am afraid, nobody has access, atleast I don't have.
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