SOLUTION: A boat travels 30 miles against a current in 5 hours. The return trip with the current takes 3 hours. If the boat travels at the same rate on both parts of the trip, what is the ra

Algebra ->  Equations -> SOLUTION: A boat travels 30 miles against a current in 5 hours. The return trip with the current takes 3 hours. If the boat travels at the same rate on both parts of the trip, what is the ra      Log On


   



Question 389070: A boat travels 30 miles against a current in 5 hours. The return trip with the current takes 3 hours. If the boat travels at the same rate on both parts of the trip, what is the rate of the water current, in miles per hour?
Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
A boat travels 30 miles against a current in 5 hours. The return trip with the current takes 3 hours. If the boat travels at the same rate on both parts of the trip, what is the rate of the water current, in miles per hour?

distance = rate * time
30 miles = RA * 5 hours
30/5 = 6 mph = RA, rate against current
30 miles = RW * 3 hours
30/3 = 10 mph = RW, rate with current
BR = boat rate, CR = current rate
RA = BR - CR, RW = BR + CR
6 = BR - CR, 10 = BR + CR
replace 10 - CR for BR in 1st
6 = 10 - CR - CR = 10 - 2CR
-4 = -2CR
4 = 2CR
2 mph = CR = rate of the water current