If the average of 6 positive integers is 55 and the difference between the largest and the smallest of these 6 numbers is 24, what is the maximum value possible for the largest of these 6 integers?

Correct option is B

Given:

Number of positive integers (n) = 6

Average of these 6 integers = 55

Difference between the largest and the smallest integer = 24

Solution:

Let the numbers in non-decreasing order be:

S ≤ a ≤ b ≤ c ≤ d ≤ L

Given L− S = 24 and sum = 330.

To maximize L, minimize S ,a ,b ,c ,d.
The minimal case is when S = a = b = c = d, so:

5S + L = 330

But L = S + 24, so:

5S + S + 24 = 330

6S = 306

S = 51

L = 75