Acomputer training institute has 625 students that are paying a course fee of $400. their research shows that for every $20 reduction in the fee, they will attract another 50 students. what fee should the school charge to maximize their revenue?
$275
$380
$320
$325

1.
If no changes are made, the school has a revenue of :

625*400$/student=250,000$

2.
Assume that the school decides to reduce n*20$.

This means that there will be an increase of 50n students.

Thus there are 625 + 50n students, each paying 400-20n dollars.

The revenue is: 

(625 + 50n)*(400-20n)=12.5(50+n)*20(20-n)=250(n+50)(20-n)

3.

check the options that we have, 

a fee of $380 means that n=1, thus 

250(n+50)(20-n)=250(1+50)(20-1)=242,250   ($)

a fee of $320 means that n=4, thus

 250(n+50)(20-n)=250(4+50)(20-4)=216,000    ($)

the other options cannot be considered since neither 400-275, nor 400-325 are multiples of 20.

Conclusion, neither of the possible choices should be applied, since they will reduce the revenue.

It is 6% because if you subtract 69-58=6 so 6% is the answer to your question.