Question Analysis:
The question provides information about the transfer of 5 people from E to R and another independent set of 5 people from R to E. We have to find the maximum possible average achieved by class E at the end of the operation.

Given Information:
Transfer of 5 people from E to R.
Transfer of another independent set of 5 people from R to E.
The set transferred from R to E contains none from the set of students that came to R from E.

Solution:
Let us assume that there are n students in class E before the transfer. Then, the average of class E before the transfer can be calculated as:

Average of class E = Sum of Marks of all students in E / Total number of students in E

After transferring 5 students from E to R, the new number of students in class E will be (n-5). Similarly, after transferring 5 students from R to E, the new number of students in class E will be (n-5+5) = n.

Let us assume that the sum of marks of all students in class E before the transfer is S. Then, the sum of marks of remaining students in class E after the transfer will be (S - 5x), where x is the average marks of the 5 students transferred from E to R.

Similarly, let us assume that the sum of marks of all students in class R before the transfer is T. Then, the sum of marks of remaining students in class R after the transfer will be (T + 5y), where y is the average marks of the 5 students transferred from R to E.

Now, the new average of class E after the transfer can be calculated as:

New average of class E = Sum of Marks of remaining students in E / Total number of students in E

= (S - 5x) / n

We have to find the maximum possible value of this expression. To maximize this expression, we have to minimize the value of x (the average marks of the 5 students transferred from E to R) and maximize the value of y (the average marks of the 5 students transferred from R to E).

Since the set transferred from R to E contains none from the set of students that came to R from E, we can assume that the average marks of the 5 students transferred from R to E is greater than or equal to the average marks of the 5 students transferred from E to R.

Therefore, to maximize the new average of class E, we have to assume that the average marks of the 5 students transferred from E to R is minimum, i.e., 1. In this case, the sum of marks of remaining students in E after the transfer will be (S - 5), and the sum of marks of remaining students in R after the transfer will be (T + 5y).

To maximize the new average of class E, we have to assume that all the 5 students transferred from R to E have scored 100 marks (maximum possible marks). In this case, the sum of marks of remaining students in R after the transfer will be (T + 500).

Therefore, the new average of class E after the transfer will be:

New average of class E = (S - 5) / n

= (S / n) - 5/n

Since we have to maximize this expression, we have to assume that the original average of