1. It takes 10÷60 minutes for another car to take four people to the railway station, and four people from the faulty car have already left 10÷60×5 kilometers.

The remaining (10- 10÷60×5) kilometers, another car will meet these four people again.

We need (10-10 ÷ 60× 5) ÷ (60+5) more points, and these four people will walk again (10-/kloc-0 ÷ 60× 5) ÷.

Then go to the railway station by bus.

That is,10 ÷ 60× 5+(10-10 ÷ 60× 5) ÷ (60+5 )× 5 = 20/13 is about/

When the arrival time is 20 ÷13 ÷ 5+(10-20 ÷13) ÷ 60, the component minutes are converted at ×60: about 27 minutes.

2. The scheme is decomposed at a distance of 10km from the railway station, and this distance is set to AB. The location where four people get off the faulty car is A, the location of the railway station is B, and the route of the other car is: When the four people on the car are sent to the railway station (this point is set to C), stop and let them walk to the railway station. At point C, they turned around to pick up four people from the faulty car (the boarding point of these four people is D), and both groups arrived at the railway station at the same time.

Analysis shows that the walking distance of these two kinds of people is equal. That is, AD=CB. The other lane is ADCDCB.

Let the walking distance be Xkm, and the equation can be listed according to the fact that the time for another car to walk ADCDCB is equal to the time for four people to arrive at the railway station with the faulty car.

X÷5+( 10-X)÷60 =[ 10+( 10-2X)]÷60

X 60 on both sides of the equation simultaneously gives12x+(10-x) =10+(10-2x).

1 1X+ 10=20-2X

13X= 10

X= 10÷ 13

The time is x ÷ 5+(10-x) ÷ 60 =10 ÷13 ÷ 5+(10-10 ÷/)

Note: These 8 people can arrive at the railway station before checking in.