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average of 1st eleven integers - didnt understand the quest

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by vishal.pathak » Sat Nov 19, 2011 1:45 am
What is the average of 1st eleven integers?
1) The average of 1st nine integers is 7.
2) The average of last nine integers is 9.
OA D


I didnt get this question. 1st 11 integers are 1 to 11 and the average is 6. If the question is about any 11 consecutive integers then using the formula of summation of an AP on 1st statement gives us the 1st term as -29/9 which is not an integer.

Is this question incorrect
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by vishal chugh » Sat Nov 19, 2011 1:55 am
3,4,5,6,7,8,9,10,11(here the average is 7); when the average becomes 9 , it means you move to the two nos. to right of 3,4......cause there 2 nos. are making ur average less... cause of eliminating these two numbers( 3,4) and putting (12,13) , we are adding total of 12+13-3-4=18; which is divided among 9 integers,,, so our average increases by 2. so average of 11 integers is
3,4,5,6,7,8,9,10,11,12,13, is 8

he need not to say whether these are consecutive nos. or not.. cause when average increases by 2, and u r changing 2 nos.. that necessarily means that these are consecutive nos.
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by vishal.pathak » Sat Nov 19, 2011 2:04 am
vishal chugh wrote:3,4,5,6,7,8,9,10,11(here the average is 7); when the average becomes 9 , it means you move to the two nos. to right of 3,4......cause there 2 nos. are making ur average less... cause of eliminating these two numbers( 3,4) and putting (12,13) , we are adding total of 12+13-3-4=18; which is divided among 9 integers,,, so our average increases by 2. so average of 11 integers is
3,4,5,6,7,8,9,10,11,12,13, is 8

he need not to say whether these are consecutive nos. or not.. cause when average increases by 2, and u r changing 2 nos.. that necessarily means that these are consecutive nos.
got it, thanks
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