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Set S consists of the integers {1, 2, 3, 4 . . . (2n + 1)}, where n is

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Set S consists of the integers {1, 2, 3, 4 . . . (2n + 1)}, where n is  [#permalink]

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New post 25 Sep 2018, 04:53
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Question Stats:

77% (01:33) correct 23% (01:51) wrong based on 31 sessions

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Set S consists of the integers {1, 2, 3, 4 . . . (2n + 1)}, where n is  [#permalink]

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New post 25 Sep 2018, 16:31
Bunuel wrote:
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

Set S ends with an odd integer, a fact we know because the last term is \(2n + 1\).

\(2n + 1\) must be an odd integer. Plug in any value to check.
\(2n\) is always Even. \(1\) is Odd.
[Even + Odd] = ODD

Let set S be {1, 2, 3, 4, 5, 6, 7}

X is the average of the FOUR odd integers:
1, 3, 5, 7
In an evenly spaced set with an even number of terms the average of the set is the average of the two middle terms:*
X = \(\frac{3+5}{2} =\frac{8}{2}=4\)
X \(=4\)

Y is the average of the THREE even integers
2, 4, 6
In an evenly spaced set with an odd number of terms, the average is the middle term:
Y \(=4\)

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

Answer

*OR average is
X = \(\frac{1+3+5+7}{4}=\frac{16}{4}=4\)

Y = \(\frac{2+4+6}{3}=\frac{12}{3}=4\)

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Re: Set S consists of the integers {1, 2, 3, 4 . . . (2n + 1)}, where n is  [#permalink]

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New post 27 Sep 2018, 17:33
Bunuel wrote:
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


In set S, there are 2n + 1 integers: n + 1 of them are odd and n of them are even. Recall that the sum of the first k positive odd integers is k^2, so the sum of the first n + 1 positive odd integers is (n + 1)^2. Likewise, the sum of first k positive even integers is k(k + 1), so the sum of the first n positive even integers is n(n + 1). Therefore, we have:

X = (1 + 3 + … + (2n + 1))/(n + 1) = (n + 1)^2/(n + 1) = n + 1

and

X = (2 + 4 + … + 2n))/n = n(n + 1)/n = n + 1

Therefore, the difference is X - Y = (n + 1) - (n + 1) = 0.

Alternate Solution:

The odd numbers in set S are 1, 3, 5, … , 2n + 1. Since these numbers form an evenly spaced sequence of integers, the average of these numbers is equal to the average of the first and last terms, which is ((2n + 1) + 1) / 2 = (2n + 2) / 2 = n + 1.

Similarly, the even numbers in set S, which are 2, 4, 6, … , 2n, also form an evenly spaced sequence of integers. Thus, the average of the even numbers in set S is (2n + 2) / 2 = n + 1.

We see that the difference is (n + 1) - (n + 1) = 0.

Second Alternate Solution:

We know that the number (2n + 1) must be odd. Therefore, the set {1, 2, 3, 4, …, 2n + 1} will begin and end with an odd integer. So let’s assume that set S = {1,2,3,4,5}. The average of the odd numbers is (1 + 3 + 5)/3 = 3. The average of the even numbers is (2 + 4)/2 = 3.

Since the two averages are equal, their difference will be 3 - 3 = 0. This difference of 0 will hold for any chosen positive integer value of n.

Answer: A
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Re: Set S consists of the integers {1, 2, 3, 4 . . . (2n + 1)}, where n is   [#permalink] 27 Sep 2018, 17:33
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