Let the speed of the car leaving A to be ‘x’ km/h and the speed of the car leaving B to be ‘y’ km/h.
Total distance=80 km
Also 1 hour 20 mins= 4/3 hour
When an automobile goes in the same direction as a specified condition,
Distance travelled by y= 8y km
Distance travelled by x= 8x= 80+8y
Now, Equation formed is 8x=80+8y
$\Rightarrow$ 8x-8y=80
$\Rightarrow$ x-y=10 —-(1)
Similarly, For opposite direction,
Distance travelled by x= (4/3)x
Distance travelled by y= (4/3) y
Also total distance covered= 4/3 $\times$ x+4/3y
$\Rightarrow$ 80= 4x+4y/3
$\Rightarrow$ 240= 4x+4y
$\Rightarrow$ 60=x+y —-(2)
Solving Equation (1) and (2)
$\Rightarrow$ 60=x+y —-(2)
$\Rightarrow$ x-y=10 —-(1)
$\Rightarrow$ x= 10+y —-(3)
Substituting the value of (3) in (2) we get,
$\Rightarrow$ 60= 10+y+y
$\Rightarrow$ 60-10=2y
$\Rightarrow$ 50=2y
$\Rightarrow$ 25=y
Also we know that, x=10+y=> x= 10+25=> x=35
Thus, Speed of car starting from A is 35km/h
Speed of car starting from B is 25km/h