Question 706091
Explaining this seems hard, but I am able, with thought, to SEE how to handle the first part.


First, we do not have specific facts on the time or distance, but we know that way-out and way-back are equal distances.  Average speed will be total distance divided by sum of time for the whole trip, out and back.


Trip Part__________speed_____________time_____________distance
way out____________10________________d/10_____________d
way back___________20________________d/20_____________d


total distance = 2d
total time = d/10 + d/20


Average Speed Entire Trip = {{{2d/((d/10)+(d/20))}}}
Do the Arithmetic, algebraic steps.