The key here is to realize that in consecutive integers the mean = median.

Statement 1:

We know the mean = 10 = median. We also know that the range is 14. Therefore there are 7 greater than the median and 7 below it. The greatest is therefore (10+7) = 17 and the least is (10 - 7) = 3.

Statement 2:

The same logic to solve 1 can be used to solve 2. Again we know the mean = 10 = median. If the greatest integer is 17 (which is 7 from the median), then the lest must be 3 (which is 7 from the median)

So both statements 1 and 2 are sufficient. Hence, D

Not able to get the solution. Could you please elaborate ?

svd.kumar wrote:1) If he average (arithmetic mean) of n consecutive odd integers is 10, what is the least of integers?

1) The range of n integers is 14.
2) The greatest of n integers is 17.

OA is D

Please explain how to solve this.

Let b = the biggest integer and s = the smallest.

Range = b-s.

In any set of evenly spaced integers:
average = median = (b+s)/2.

Since the average here = 10:
(b+s)/2 = 10
b+s = 20.

Statement 1: range = 14.
Thus, b-s = 14.
Subtracting this equation from b+s=20, we get:
(b+s) - (b-s) = 20-14
2s=6
s=3.
SUFFICIENT.

Statement 2: b=17.
Since b+s = 20, s=3.
SUFFICIENT.

The correct answer is D.