algebra
posted by amie .
alberto drove to the lake and back. the trip there took two hours and the trip back took three hours. he averaged 20 km/h faster on the trip there than on the return trip. What was alberto's average speed on the outbound trip?

speed on first trip  x km/h
speed on return trip  x20
let the distance be d
time for first part = d/x
time for 2nd part = d/(x20)
d/x = 2 > d = 2x
d/(x20) = 3 > d = 3x  60
3x60 = 2x
x = 60
So the distance each way was 120 km
average speed = 240/5 = 48 km/h
Notice that the two speeds were 60 km/h and 40 km/h, and to get the average speed we CANNOT just average the two speeds. 
since distance = speed * time, if the outbound speed was x, then
2x = 3(x20)
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