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?

Answer 1

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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