Suppose that at the beginning, the speed of A is A, the speed of B is B, and the first speed of A is X.

Accord to that known available equation:

Derive from ②

5(a+x)=6a

x=0.2a

Substitute 3 to get

5(a+50)=6( 1.2a+5)

The solution is a= 100.

Substitute (1)

2=240/b- 1

b=80

So the speed of B car before departure is 80 km/h.