Can anyone please help me with this question.

FINALLY I SOLVED IT:
m+n=50 so, n=50-m
Extra Weight per New Student= n-40
so, Total Extra Weight= m(n-40)= m(10-m).
THINK OF THIS "TOTAL EXTRA WEIGHT" LIKE A BUCKET OF WATER. TO FIND OUT HOW MUCH THE AVERAGE WEIGHT OF THE CLASS INCREASES, YOU HAVE TO POUR THE EXTRA WEIGHT EQUALLY OVER EVERYONE NOW IN THE CLASS.

Increase in Avg. = total extra weight/ total number of students. = m(10-m) / 40+m.

To maximize avg., we need INCREASE to its highest.
Imagine you have 2 numbers that always add upto 10.(whose total is 10).

1*9=9
2*8=16
3*7=21
4*6=24
5*5=25
6*4=24.....

Here we've (m & 10-m)
PRODUCT is always maximized when the numbers are as close to equal as possible.

FINAL STEP:
m= 5
INCREASE= m(10-m) / 40+m
=25/45 =5/9
5/9 = 0.55555555...

NEW AVERAGE = OLD + INCREASE = 40 + 0.55555

FINAL ANSWER = 0.55555/ 40.56kgs

You are given: m+n=50
To maximize the final average,
For a fixed sum, the product m×n is maximum when both are equal. So, m=25,n=25
Step 1: Original total weight = 40×40=1600
Step 2: Added total weight = 25×25=625
Step 3: New average

Total students = 40+25=65

Average= (1600+625) / 65 = 2225 / 65
≈34.23

Initial avg= 40
Initial students=40
Total initial weight= 1600 kgs
m new students added, avg weight is n. Total new weight added = m*n.
Avg after addition of m students is (1600+m*n) Kgs / (40+m) i.e. total weight/ total students.
For this fraction to be maximum, denominator should be minimum. i.e. m=1. => n= 49
average= (1600+ 1*49)/ 40+1 = 1649/41 = 40.21 kgs.