Trevon made a trip to the ferry office and back. The trip there took five hours and the trip back took 4.9 hours. He averaged 1.1 km/h faster on the return trip than on the outbound trip. What was Trevon's average speed on the outbound trip?
+130511 Let R be the average rate for the outbound trip.....and we know that Rate*Time = Distance......but......the distance out and back are the same.....so we have.....
R(5) = (R + 1.1)4.9 simplify
5R = 4.9R + 5.39 subtract 4.9R from each side
0.1R = 5.39 divide both sides by 0.1
R = 53.9 km/hr = the rate for the outbound trip
