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What is the average (arithmetic mean) of eleven consecutive

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Senior Manager
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What is the average (arithmetic mean) of eleven consecutive [#permalink] New post 04 Apr 2007, 23:36
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 100% (00:16) wrong based on 1 sessions
What is the average (arithmetic mean) of eleven consecutive integers?

1. The average of the first 9 integers is 7.

2. The average of the last 9 intergers is 9.
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 [#permalink] New post 04 Apr 2007, 23:54
I think the answer should be D.

What is the OA?
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 [#permalink] New post 05 Apr 2007, 00:22
Hi,
Think that D is not correct.
The consecutive integers may be in increasing or decreasing order which will give different results.
Think it is E
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 [#permalink] New post 05 Apr 2007, 00:24
Agree D.

integers may be x, x+1, ... x+10 then
A => 9x+36=63 => x=3 => suff
B => 9x+54=81 => x=3 => suff
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 [#permalink] New post 05 Apr 2007, 17:37
Juaz,

How did you get A => 9x+36=63 and 9x+54=81
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 [#permalink] New post 05 Apr 2007, 20:44
nitinneha wrote:
Juaz,

How did you get A => 9x+36=63 and 9x+54=81


A is first nine integars
x, x+1, ... , x+8 = 9x + (1+2+...+8) = 9x + 36

And in my opinion, we should not worry about calculating the answer. We should concentrate on finding if we can calculate it or not.

Two equations and two variables will give answer. I will leave the question the moment i have the equations.
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 [#permalink] New post 06 Apr 2007, 01:20
nitinneha wrote:
Juaz,

How did you get A => 9x+36=63 and 9x+54=81


x + (x+1) + (x+2) + ...(x+8) = 9x+ (1+2+3+...+8) = 9x + (8)(9)/2 (sum of consecutive intergers)= 9x+36

hope this clears it up for u
  [#permalink] 06 Apr 2007, 01:20
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