What is the average of eleven consecutive integers? 1) The : DS Archive
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# What is the average of eleven consecutive integers? 1) The

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What is the average of eleven consecutive integers? 1) The [#permalink]

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13 Apr 2007, 21:11
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What is the average of eleven consecutive integers?

1) The average of the first nine integers is 7.
2) The average of the last nine integers is 9

OA: D

I just setup the equation X/9 = 7, to get X = 63...but then after that I just started plugging in numbers that would add up to 63. Took a bit of time and just wondering if there is another method.
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13 Apr 2007, 21:48

Explanation

lets say first number in the sequence is X then the sum of first 9consequitive number will be 9X+ 8 but its avg is 7 hence 9X + 8 = 63 so we can get the value of X and hence our answer. 1 is sufficient

Similarly 2 is also sufficient... hence answer is D

regards,

Amardeep
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13 Apr 2007, 23:34
1. Average of first nine integers is 7
[ x+(x+1)+(x+2)+...+(x+8) ] / 9 = 7

this implies, (9x + 36)/9 = 7 => 9x+36=63 =>9x = 27 =>x=3

then the sequence is : 3,4,5,6,7,8,9,10,11,13,15 => 1. sufficient

2. Average of last nine integers is 9

In either case, you can get the full sequence, therefore, D.
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14 Apr 2007, 03:27
Use this formula rather than adding up the consecutive integers:

lets consider the numbers are x, x+1, x+2.....

average of first 7 CONSECUTIVE integers = (First term+last term)/2 = (x +x+6)/2 = 7
solve for x
now the average of 11 integers can be found out. SUFF

Statement2:
The same way, the average of the last 9 CONSECUTIVE integers = [(x+2)+(x+10)]/2 =9
solve for x , average of 11 integers can be found out. SUFF

Hence it's 'D'
14 Apr 2007, 03:27
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