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What is the average (arithmetic mean) of eleven consecutive

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Senior Manager
Joined: 02 Sep 2006
Posts: 254
What is the average (arithmetic mean) of eleven consecutive [#permalink]

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05 Apr 2007, 00:36
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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.
Senior Manager
Joined: 20 Feb 2007
Posts: 256

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05 Apr 2007, 00:54
I think the answer should be D.

What is the OA?
Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

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05 Apr 2007, 01:22
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
Director
Joined: 18 Jul 2006
Posts: 524

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05 Apr 2007, 01:24
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
Manager
Joined: 15 Nov 2006
Posts: 218
Location: Ohio

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05 Apr 2007, 18:37
Juaz,

How did you get A => 9x+36=63 and 9x+54=81
Intern
Joined: 04 May 2004
Posts: 47
Location: India

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05 Apr 2007, 21:44
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.
Manager
Joined: 29 Nov 2006
Posts: 56

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06 Apr 2007, 02:20
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
06 Apr 2007, 02:20
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