If the overall average weight has to increase after the new people are added, the average weight of the new entrants has to be higher than 40
So n > 40
Consequently, m has to be < 10 (as n + m = 50).
Working with the "differences’’ approach, we know that the total additional weight added by "m" students would be (n - 40) each, above the already exiting average of 40.
m(n - 40) is the total extra additional weight added, which is shared amongst 40 + m students.
So,
has to be maximum for the overall m + 40 average to be maximum
At this point, use the trial and error approach to arrive at the answer
The maximum average occurs when m = 5 and n = 45 And the average is
=40+ | extra weight×extra number of people |
| total number of people |
=40+=40.56