Jesse is late for work and speeds to get to the office, which is 60 miles away from his house. On the way back, he drives at a more relaxed pace. Jesse drove 30 mph faster on the way to his office than on his way back. What was his overall average driving speed for the day?
speed on first way --- x
speed on return ---- x-30
time for first part = 60/x
time for other part = 60/(x-30)
total time = 60/x + 60/(x-30)
= (60(x-30) + 60x)/(x(x-30)
= (60x - 1800 + 60x)/(x-30)
=120(x - 15)/(x(x-30)
avg speed = 120/(120(x - 15)/(x(x-30) )
= 120x(x-30)/(120(x-15)
= x(x-30)/(x-15)
There is no constant answer, it will depend on his initial speed.
e.g if his first speed = 60 , then his return speed = 30
avg speed = 60(30)/(45) = 40
if his first speed = 80 , then his return speed = 50 and
avg speed = 80(50)/(65) = appr 61.5
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