I am not totally sure that I am 100% completely right (you might want to double check) but this is how I view the question you posted below.....

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

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

Can someone put a little light on this prob?


To find the Arithmetic Mean of 11 consecutive integers you must view Number 1 to start.

Number 1 tells us that the average of the "first 9" of our consecutive integers must equal 11. I first used integers 1 - 11 but found that when I averaged 1 - 9 I came up to only 6 (Arithmetic Mean). Same thing with 2 - 12 with average of 6. So I tried 3 - 13 and found that the average of 3, 4, 5, 6 , 7, 8, 9, 10, and 11 averaged 7.

I checked this against the second set to see if the average was 9. So, 5, 6, 7, 8, 9, 10, 11, 12, and 13 equal the average of 9 that we are looking for.

There is still one more step and that is to average the 3 - 13. That came out to be 8.

So your answer, as I see it, is 8 and you must use 3 - 13 to get it.

If you want the excel version of what I did then email me and I can upload it.

Hope this helps!!!!!

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

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

Can someone put a little light on this prob?

Thanks.

The key is that the integers are consecutive. So, if we can determine any one of the 11 (and know where it falls), we can answer the question.

(1) The average of the first 9 consecutive integers is 7.

We know that avg = sum of terms / # of terms.

So, 7 = sum of terms/9
sum of terms = 63.

Well, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9, we can certainly determine the last 2: sufficient.

(2) The average of the last 9 terms is 9.

Exact same reasoning as (1): sufficient.

Both (1) and (2) are sufficient: choose (D).

Stuart:
Your comment:
"...Well, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9,...."

What is this one set of 9 consecutive integers that add upto 63?
(how does one go on about it?)

Thank you

@ Deepak Shah

What is this one set of 9 consecutive integers that add upto 63?
(how does one go on about it?)

Let first number be a. since the numbers are consecutive second member of the series would be a+1, 3rd = a + 2 and so on

a + a+1 + a+2 +........+ a+8 = 63
9a + 36 = 63
9a = 27
a = 3

so the series would be

3,4,5,6,7,8,9,10,11,....

Deepak_J_Shah wrote:Stuart:
Your comment:
"...Well, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9,...."

What is this one set of 9 consecutive integers that add upto 63?
(how does one go on about it?)

Thank you

We know that the average of a set of consecutive integers is the median of the set.

In this case, avg = 63/9 = 7. Since we have an odd number of terms, 7 is the middle term. Then we just add 4 numbers to each side to get our set:

{3, 4, 5, 6, 7, 8, 9, 10, 11}

Of course, since this is DS, we didn't actually have to do all that work.

The quickest method is to know that using symmetry, the average of the total set must be the average of the two given averages.
So the answer is the average of 7 and 9 which is 8.

Likewise if you took the average of any two numbers or sets of numbers equidistant from the middle, the average of those numbers or sets of numbers would be the average. This makes intuitive sense as the average of any uniformly distributed set will be the exact center of that set.