Set S consists of the integers

[aaron1981]

Set S consists of the integers {1, 2, 3, 4 ... „(2n +‚ 1)…}, where n is a positive integer. If X is the average of the odd integers in set S and Y is the average of the even integers in set S, what is the value of „(X - Y)…?

(A) 0
(B) 1/2
(C) 1
(D) 3/2
(E) 2

OA A

[Jay@ManhattanReview]

Set S consists of the integers {1, 2, 3, 4 ... „(2n +‚ 1)…}, where n is a positive integer. If X is the average of the odd integers in set S and Y is the average of the even integers in set S, what is the value of „(X - Y)…?

(A) 0
(B) 1/2
(C) 1
(D) 3/2
(E) 2

OA A

Since there is no restriction on n, let us try to find „(X - Y)… using a few values of n.

Case 1: n = 2, an even number

Thus, Set S = {1, 2, 3, 4, 5}

=> X = (1 + 3 + 5)/3 = 3;

=> Y = (2 + 4)/2 = 3

=> (X - Y) = 3 - 3 = 0.

Case 2: n = 3, an odd number

Thus, Set S = {1, 2, 3, 4, 5, 6, 7}

=> X = (1 + 3 + 5 + 7)/4 = 4;

=> Y = (2 + 4 + 6)/3 = 4

=> (X - Y) = 4 - 4 = 0.

The correct answer: A

Hope this helps!

Relevant book: Manhattan Review GMAT Number Properties Guide

-Jay
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[GMATGuruNY]

Set S consists of the integers {1, 2, 3, 4 ... „(2n +‚ 1)…}, where n is a positive integer. If X is the average of the odd integers in set S and Y is the average of the even integers in set S, what is the value of „(X - Y)…?

(A) 0
(B) 1/2
(C) 1
(D) 3/2
(E) 2

For any evenly spaced set, AVERAGE = MEDIAN.
Thus:
X = the average of the odd integers in S = the MEDIAN of the odd integers in S.
Y = the average of the even integers in S = the MEDIAN of the even integers in S.

Let n=2, with the result that 2n+1 = (2*2) + 1 = 5.
Resulting set:
{1, 2, 3, 4, 5}.

X = the median of odd integers {1, 3, 5} = 3.
Y = the median even integers {2, 4} = 3.
Thus, X-Y = 3-3 = 0.

The correct answer is A.

[Jay@ManhattanReview]

Set S consists of the integers {1, 2, 3, 4 ... „(2n +‚ 1)…}, where n is a positive integer. If X is the average of the odd integers in set S and Y is the average of the even integers in set S, what is the value of „(X - Y)…?

(A) 0
(B) 1/2
(C) 1
(D) 3/2
(E) 2

OA A

Another take on this question...

We have an evenly spaced set S = {1, 2, 3, 4 ... „(2n +‚ 1)…}.

Thus, X = Average of {1, 3, 5, ..., (2n - 1), „(2n +‚ 1)…}; since (2n + 1) is an odd number for any positive integer

And Y = Average of {2, 4, 6, ..., (2n - 2), „2n…}; since 2n is an even number for any positive integer

Since the sets {1, 3, 5, ..., (2n - 1), „(2n +‚ 1)…} and {2, 4, 5, ..., (2n - 2), „2n…} are evenly spaced, their average would be the average of their first and the last term.

Thus, X = [1 + (2n+1)]/2 = n + 1

and Y = [2 + 2n]/2 = n + 1

=> X - Y = (n+1) - (n+1) = 0.

The correct answer: A

Hope this helps!

-Jay

Download free ebook: Manhattan Review GMAT Quantitative Question Bank Guide
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