25 distinct positive integers are arranged in ascending order. The average of first 12 is 30 while the average of last 16 is 60. What is the maximum possible average of 10th, 11th, and 12 th element in the series?
Solution
The 10th, 11th, and 12th elements in the series are the first three elements of the last 16. And the average of 16 numbers is 60. To maximize the first three of these 16, we should keep the numbers as close to 60 as possible. So, the numbers will be 52, 53, 54, 55, ..., 59,61, ...,65, 66, 67, 68.
The 10th, 11th, and 12th elements in the series are the last three elements of the first 12. Let us check if 52, 53, and 54 will satisfy the average or not.
The sum of the first 12 will be 30*12 = 360.
The sum of the first nine will be 360 - (52+53+54) = 201.
The sum of nine distinct positives can be 201.
So, 52, 53, and 54 will be the maximum possible values for the 10th, 11th, and 12th elements in the series. Their average will be 53.
