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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]

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

What is the OA?
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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
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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
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05 Apr 2007, 18:37
Juaz,

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