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What is the average (arithmetic mean ) of eleven consecutive

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What is the average (arithmetic mean ) of eleven consecutive [#permalink] New post 23 Apr 2010, 18:05
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What is the average (arithmetic mean ) of eleven consecutive integers?

(1) The avg of first nine integers is 7
(2) The avg of the last nine integers is 9
[Reveal] Spoiler: OA
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink] New post 23 Apr 2010, 18:11
Some how i got E

1. (63+ X10 + X11) / 11 = ?? ..sitill missing two numbers? so insuff??
2. (X1+X2 +81)/ 11 = ?? Still missing two numbers so Insuff???

and combining 1 and 2 still missing some info ? .. Sorry friends, i need help here.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink] New post 24 Apr 2010, 06:10
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zz0vlb wrote:
What is the average ( arithmetic mean ) of eleven consecutive integers?

1.The avg of first nine integers is 7
2. The avg of the last nine integers is 9
[Reveal] Spoiler:
D



friends, please explain this to me.


Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case mean=median=x_6.

(1) x_1+x_2+...+x_9=63 --> there can be only one set of 9 consecutive integers to total 63. Sufficient.

If you want to calculate: (x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63 --> x_6=8.

OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> x_5*9=63 --> x_5=7 --> x_6=7+1=8

(2) x_3+x_4+...+x_{11}=81 --> there can be only one set of 9 consecutive integers to total 81. Sufficient.

If you want to calculate: (x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81 --> x_6=8.

OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> x_7*9=81 --> x_7=9 --> x_6=9-1=8


Answer: D.
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Re: Average ( arithmetic mean ) of eleven consecutive integers? [#permalink] New post 24 Apr 2010, 07:11
Thanks for clarifying this Bunuel. +1Kudos to you.
Re: Average ( arithmetic mean ) of eleven consecutive integers?   [#permalink] 24 Apr 2010, 07:11
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