Scheme 1: suppose the speed of bus 18 is x m/min, the walking speed of Xiaocong is y m/min, and the distance between two adjacent cars driving in the same direction is s m. ..

If a 18 bus passes every 6 minutes, then 6x-6y = S 1.

Every 3 minutes, a 18 bus comes from the top, and then 3x+3y = S2.

X=s/4 and y=s/ 12 can be obtained from ① and ②.

So s/x=s/(s/4) =4.

That is, the departure interval of bus terminal 18 is 4 minutes.

Solution 2: Similarly, s=4x can be obtained from ① and ②.

So s/x=4.

A: The departure interval of bus terminal 18 is 4 minutes.

Option 3: departure interval t, speed x and bus speed y.

yt=6(y-x)

yt=3(y+x)

6(y-x)=3(y+x)

3y=9x

y=3x

Substitute yt=3(y+x):

3xt=3(3x+x)

3xt= 12x

t=4

The departure interval is 4 minutes.