SOLUTION: A trucker drove 140 miles to make a delivery and returned home on the same route. Because of foggy conditions, his average speed on the return trip was 15 mph less than his average

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Question 1167191: A trucker drove 140 miles to make a delivery and returned home on the same route. Because of foggy conditions, his average speed on the return trip was 15 mph less than his average speed going. If the return trip took 3 hours longer, how fast did he drive in each direction?
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


Let the speed going be x and the speed going be x-15.

The time going is then 140/x and the time returning is 140/(x-15).

The time returning is 3 hours more than the time going:

140%2F%28x-15%29+=+140%2Fx%2B3

140%2F%28x-15%29+=+%28140%2B3x%29%2Fx

140x+=+%28140%2B3x%29%28x-15%29
140x+=+3x%5E2%2B95x-2100
3x%5E2-45x-2100+=+0
x%5E2-15x-700+=+0

%28x-35%29%28x%2B20%29+=+0

x+=+35 or x+=+-20

Clearly the negative solution makes no sense in the problem....

ANSWERS:
speed going = x = 35mph
speed returning = x-15 = 20mph

CHECK:
time going = 140/35 = 4 hours
time returning = 140/20 = 7 hours = 3 hours more than time going

That's a formal algebraic solution, which involves a lot of work -- especially factoring the quadratic into two linear factors.

If a formal algebraic solution is not required, a little mental arithmetic will find the answer much faster and with much less effort. Simply look for combinations of hours and miles per hour that make 140 miles.

Two easy combinations that satisfy all the conditions of the problem are
(1) 4*35 = 140 --> 4 hours at 35mph
(2) 7*20 = 140 --> 7 hours at 20mph