let n-5, n-4, n-3, n-2, n-1, n, n+1, n+2, n+3, n+4, n+5 be the 11 consecutive numbers. The average of these numbers is n. Then the question can be rephrased to 'What is the value of n?'

1) Average of the 1st nine integers is 7

Average of the first 9 integers n-5, n-4, n-3, n-2, n-1, n, n+1, n+2, n+3 is n-1 = 7. So, the value of n is 8. Statement I alone is sufficient to answer the question.

2) Average of last 9 integers is 9

Average of the last 9 integers n-3, n-2, n-1, n, n+1, n+2, n+3, n+4, n+5 is n+1 = 9. So, the value of n is 8. Statement II alone is sufficient to answer the question.

Option D is the right choice!

DFritschy wrote:What is the average (arithmetic mean) of eleven consecutive integers?

1) Average of the 1st nine integers is 7

2) Average of last 9 integers is 9

For a very fast approach, its worthwhile to remember some interesting things about consecutive integers. One such thing is that if you know the average of any of the integers you can find the average of all of the integers. Since this is DS you don't even need to solve , just remember that and your in. Choose D