120(x+y)= 2
(3- 120/x)x=3y
X = 50，y = 10。
2。 According to the meaning of the question, let the speed of a be x km/h and the speed of b be y km/h.
( 120- 1y)÷(x+y)=2
(3- 120/x)x=( 1+3)y
X = 840/ 17，Y = 360/5 1。
3。 Suppose A starts from B, and after X hours, the distance between A and B is 20 kilometers, because it has been found that A's speed is 50 kilometers per hour and B's speed is 10 kilometers per hour.
10×( 120÷50+x)-20 = 50x
The solution is x =110.
4。 Let the speed of C be x km/h, 50 minutes equals 5/6 hours, and 70 minutes equals 7/6 hours.
( 120÷ 10-5/6-7/6)x = 120
The solution is x = 15.
The speed of c is 15 km/h, assuming that c and a meet for y hours, catch up with B.
[5/6+( 120-50×5/6)÷(50+ 15)+y]× 10 = 15y
Y = 53/ 13。
5。 According to the meaning of the question, let the speed of a be x km/h and the speed of b be y km/h.
1.5(x+y)=300-30
300-(3- 1)x=3y- 120
X = 120，Y = 60。
(300- 150)÷( 120+60)=5/6
A: After 5/6 hours of departure, the distance between the two cars is 150km.