You drive to the mailbox down the road at 35 mph slip a lett

You drive to the mailbox down the road at 35 mph, slip a letter in, turn around, and drive back at 25 mph. What is your average speed? (Note that it is not 30 mph, and you do not need to know the distance.) I tried this but don\'t know where I am going wrong.
T (going)= D/35
and for the return trip
T(returning) = D/25
The total time for the trip is T(going) +T(returning) = D/35 + D/25 and you can find the average speed by dividing the total distance 2*D, by the total time:
V(average) = 2D/(D/35+D/25)
V(average)= 1/(1/35 + 1/25)

V(average)= (1*60/1)/(1/60*60/1)
V(average)= 60 mph I know this answer is wrong. It doesn\'t make sense that it should be the combined speeds.

Solution

You drive to the mailbox down the road at 35 mph, slip a letter in, turn around, and drive back at 25 mph. What is your average speed? (Note that it is not 30 mph, and you do not need to know the distance.) I tried this but don\'t know where I am going wrong.
T (going)= D/35
and for the return trip
T(returning) = D/25
The total time for the trip is T(going) +T(returning) = D/35 + D/25 and you can find the average speed by dividing the total distance 2*D, by the total time:
V(average) = 2D/(D/35+D/25)
V(average)= 1/(1/35 + 1/25) **** You changed the NUM to 1, should be 2 *****
Avg = 2/(1/35 + 1/25)
Avg = 2/(5/175 + 7/175) = 2/(12/175)
Avg = 175/6
Avg = ~ 29.1667 mph
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This is the basis of the Michelson-Morley experiment that disproved the existence of \"the ether\" (not exactly the same, but similar).
V(average)= (1*60/1)/(1/60*60/1)
V(average)= 60 mph I know this answer is wrong. It doesn\'t make sense that it should be the combined speeds.