Question 1: Speed A and B mean 4A+4B=7A+70, (4-3)*(A+B)= 10+80, so A=50 and B=40. So the total distance is 4a+4b = 360, [360-(360/50) * 40]/40 =1.8.

Question 2: the original a machine, b hours. (A+2)*7/8*B=AB, (A-2)*(B+2/3)=AB, so A= 14, B=4.

Total time of a machine: A*B=56

Question 3: The time of speed A, B and C. A is A, then the time of B is a+ 10, and the time of C is C.

(A-20)A/[( 1- 1/3)A]=(A+ 10-20)B/[( 1- 1/4)B]=(C-20)