It takes Party A one day to complete this project, and it costs X yuan every day;

B it takes b days to complete this project, and it costs y yuan every day.

4/A+ 15/B= 1

4X+ 15Y=420

8/A+(B- 10)/B= 1

8X+(B- 10)Y=440

By solving the above equations, we can get: A= 16, B=20, X=30, Y=20.

1. It takes Party A and Party B 16, 20 days to complete this project.

2. Party A and Party B each need 480 yuan to complete this project, and 400 yuan.

3. Set M days for Party A and N days for Party B..

M/ 16+N/20= 1

m+N & lt； = 18

Cost =30M+20N=400+5M

It is equivalent to finding the minimum value of m (if you have studied linear programming, you should use its method to do this problem, if not, you can simply analyze it)

Because N=20-5M/4, so

m+N & lt； = 18 is equivalent to 20m/4.

So A works at least 8 days.

4. Let's choose one of the three.

M/ 16+N/20= 1

30M+20N & lt； =430

Solution: m

So A works for six days at most.