SOLUTION: boat will go 15 mph in still water,he can go 12 miles down stream in the same time as it takes him to go 9 miles upstream,,then what is the speed of the current

Algebra.Com
Question 280230: boat will go 15 mph in still water,he can go 12 miles down stream in the same time as it takes him to go 9 miles upstream,,then what is the speed of the current
Answer by stanbon(75887)   (Show Source): You can put this solution on YOUR website!
boat will go 15 mph in still water,he can go 12 miles down stream in the same time as it takes him to go 9 miles upstream,,then what is the speed of the current
----------------------------------
Downstream DATA:
distance = 12 miles ; rate = 15+c mph ; time = 12/(15+c) hrs
--------------------------------------
Upstream DATA:
distance = 9 miles ; rate = 15-c mph ; time = 9/(15-c) hrs
--------------------------------------
Equation:
time = time
12/(15+c) = 9/(15-c)
12*15-12c = 9*15+9c
21c = 3*15
current = 2.14 mph
========================
Cheers,
Stan H.

RELATED QUESTIONS

Upstream, downstream, Katherine's boat will go 15 miles per hour in still water. If she... (answered by stanbon)
Junior's boat will go 15 mph in still water. If he can go 12 miles downstream in the... (answered by solver91311)
Junior's boat will go 15 mph in still water. If he can go 12 miles downstream in the same (answered by ankor@dixie-net.com)
Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream (answered by Nate,mathdude2)
Juniors boat will go 15 miles per hour in still water. If he can go 12 miles downstream... (answered by Grinnell)
Junior's boat will go 15 miles per hour in still water. if he can go 12 miles downstream... (answered by solver91311)
Junior's boat will go 15 miles per hour in still water. If he can go 12 miles downstream... (answered by nyc_function)
junior's boat will go 15 miles per hour in still water. if he can go 12 miles downstream... (answered by mananth)
Juniors boat will go 15 miles per hour in still water. If he can go 12 miles downstream... (answered by lwsshak3)