The time taken for a committee meeting is partly constant and partly varies as the square of the number of members present.If there are 12 members present the meeting lasts only 56minutes but with 20 it takes exactly two hours.How long will it last if there are 16 members there
Let's assume that the constant time taken for the committee meeting is represented by "c" and the time taken that varies as the square of the number of members present is represented by "k * n^2", where "k" is a constant and "n" is the number of members present.
From the given information:
When there are 12 members present, the meeting lasts 56 minutes:
c + k * 12^2 = 56 ----(1)
When there are 20 members present, the meeting lasts 2 hours (120 minutes):
c + k * 20^2 = 120 ----(2)
We need to solve these two simultaneous equations to find the values of "c" and "k".
From equation (1):
c + 144k = 56
From equation (2):
c + 400k = 120
Now, we subtract equation (1) from equation (2):
(400k - 144k) = (120 - 56)
256k = 64
k = 64/256
k = 1/4
Substituting this value of k in equation (1):
c + 144k = 56
c + 144 * (1/4) = 56
c + 36 = 56
c = 56 - 36
c = 20
Therefore, the equation representing the time taken is:
Time = 20 + (1/4) * n^2
If there are 16 members present:
Time = 20 + (1/4) * 16^2
Time = 20 + (1/4) * 256
Time = 20 + 64
Time = 84
Hence, if there are 16 members present, the committee meeting will last 84 minutes.