Let the number of students in the sections P, Q, R and S be p, q, r and s, respectively.
Thus, the average weight of all students together in the four sections

\(\frac{Total weight of all the students combined}{Total # of students}\)


\(\frac{45*p+50*q+55*r+65*s}{p+q+r+s}=55\)

Solve

10p+5q=10s

2p+q=2s


Since we need to maximize r , we need to find the minimum possible values of p, q and s so that the above equation holds true.

Since the RHS is 2s, it is even. Also, in the LHS, 2p is even. Thus, q must be even.

Since the smallest even number that we can consider for q is 2, as we have at least one student in each section. Thus, we have q = 2.

Thus, the equation gets modified to:


2p + 2 = 2s

=> p + 1 = s

Thus, we use the minimum possible values: p = 1, s = 2.

Thus, we have p = 1, q = 2 and s = 2.

Since there are a total of 40 students in all sections combined, the maximum value of students in section \(R = r = 40 - (p+q+s) = 40 − 5 = 35\)

The correct answer is Option D