Let the speed of the car starting from point A be .v km/h and speed of the car starting from point B be y km/hr.

Case I. When they travel in the same direction, let they meet at point P (Fig. 3.12).
So, AP = distance covered by first car from point A to P
= speed x time
= x × 5 = 5
and    BP = distance covered by second car from point B to P
= speed × time
= y × 5 = 5y
Therefore, AB = AP - BP
⇒ 100 = 5x - 5y
⇒ 5a - 5y = 100
⇒ x - y = 20
Case II. (Fig. 3.13)
When they travel in opposite direction, let they meet at point Q.
So, AQ = Distance covered by first car from point A to Q
= Speed × Tinte
= x × 1 = x
and    BQ = Distance covered by second car from point B to Q
= Speed × Time
= y × 1 = y
Therefore, AB = AQ + BQ
⇒    100 = x + y
⇒ x + y = 100
Thus, we have following eqn.
x - y = 20
x + y = 100
⇒ x - y - 20 = 0
x + y - 100 = 0

Hence, speed of the car that starts from point A = 60 km/hr
Speed of the car that starts from point B = 40 km/hr.