# SOLUTION: Please help me solve this problem: A boat goes downstream a distance of 5 miles in 15 minutes, but it takes 20 minutes for the return trip. Find the right of the boat in still wate

Algebra ->  Algebra  -> Matrices-and-determiminant -> SOLUTION: Please help me solve this problem: A boat goes downstream a distance of 5 miles in 15 minutes, but it takes 20 minutes for the return trip. Find the right of the boat in still wate      Log On

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 Click here to see ALL problems on Matrices-and-determiminant Question 156805: Please help me solve this problem: A boat goes downstream a distance of 5 miles in 15 minutes, but it takes 20 minutes for the return trip. Find the right of the boat in still water and the rate of the current.Answer by nerdybill(6962)   (Show Source): You can put this solution on YOUR website! Please help me solve this problem: A boat goes downstream a distance of 5 miles in 15 minutes, but it takes 20 minutes for the return trip. Find the right of the boat in still water and the rate of the current. . First, convert all times given from minutes to hours. 20 mins * 1hr/60mins = 20/60 = 1/3 = .333 hours 15 mins * 1hr/60mins = 15/60 = 1/4 = .25 hours . Let b = rate of the boat and c = rate of the current . we have two unknowns, so we'll need two equations. We will need to apply the "distance formula": d = rt where d is distance r is rate t is time . from: "A boat goes downstream a distance of 5 miles in 15 minutes" we get equation 1: (b+c)(1/4) = 5 . from: "but it takes 20 minutes for the return trip" we get equation 2: (b-c)(1/3) = 5 . Solve equation 1 for c: (b+c)(1/4) = 5 b+c = 20 c = 20-b . Plug the result above into equation 2 and solve for b: (b-c).333 = 5 (b-(20-b))(1/3) = 5 (b-20+b) = 15 2b-20 = 15 2b = 35 b = 17.5 mph (rate of boat) . Plug the result above back into equation 1 and solve for c: (b+c)(1/4) = 5 (17.5+c)(1/4) = 5 17.5+c = 20 c = 2.5 mph (rate of current)