The excursion boat on the river takes 2 1/2 hours to make the trip to a point 12 miles up stream and to return. if the rate at which the boat travels in still water is 5 times the rate of the river current, what is the rate of the current?

Let V, be the rate in still water and let C = rate river current
If the boat is going :
upstream, its rate is V-C and if going 
downstream, its rate is V+C,
But V = 5C, then
Upstream Rate: 5C - C = 4 C
Downstream rate: 5C+C = 6C

Time = distance/Rate,  (or time = distance/speed) , then:
Upstream time 12/4C and
Downstream time: 12/.6C

Upstream time +downstream time:= 2h30 ' then:

12/4C + 12/.6C = 2.5 hours
3/C + 2/C = 5/2 (2.5 h = 5/2)
Reduce to same denominator :
 5C = 10 and Rate of the current = 2 mi/h