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nice question

by atulmangal » Sun May 01, 2011 3:40 pm
If eleven consecutive integers are listed from least to greatest, what is the average (arithmetic mean) of the eleven integers?

(1) The average of the first nine integers is 7 (2) The average of the last nine integers is 9
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by parkman » Sun May 01, 2011 4:09 pm
I get D?

(x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8))/9 = 7
(9x +36)/9 = 7
x+4 = 7
x = 3
(11(3) + 55)/11 = (33+55)/11 = 88/11 = 8

(2) should be sufficient as well working backwards (x+10)...
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by atulmangal » Sun May 01, 2011 5:25 pm
parkman wrote:I get D?

(x+(x+1)+(x+2)+(x+3)+(x+4)+(x+5)+(x+6)+(x+7)+(x+8))/9 = 7
(9x +36)/9 = 7
x+4 = 7
x = 3
(11(3) + 55)/11 = (33+55)/11 = 88/11 = 8

(2) should be sufficient as well working backwards (x+10)...
Hi the OA is indeed D, actually i like the concept involved to solve this question, here is the concept

The only fundamental we need to remember is that for a equally spaced set , if the total number of items are odd then mean = median.

Consecutive numbers are a special case of equally spaced set.

So for 11 consecutive integers 6th number is the median so it will be the mean.

St1. the mean of first 9 integers is 7, which tells that the 5th number is 7, you know 5th
number so you know 6th number is 8.
So avg of 11 numbers is 8.
SUFFICIENT.

Similarly u can get the answer from Op B alone

Hence D is the answer
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