hatever
the
total distance to be traveled, the train in this puzzle will have used up
all of the time allowed for completing the trip at 60
miles per hour during the first half at 30 miles per
hour. So the solution is  well, there is .
 Most people guess 90 mph, inasmuch
as the average of 30 mph and 90 mph is given by (30 + 90) / 2 = 60.
A train that does 30 mph for
the first half of a trip and 90 mph for the second
half will average only 45 mph for the whole trip.
 Here are other popular guesses: 120
(48 mph), 180 (51 mph), 240 (53 mph).
Try modifying the problem so that the
train is required to average only 59 miles per hour for
the whole trip after traveling at 30 mph during the
first half.
Surprise: The train would have
to go 1,770 mph!
Supersonic trains are still the stuff of science fiction.
Time in the Numerator
The sophisticated person puts time in
the numerator (see Bucketing
Bandwidth). Examples abound. Here is
a message received in 2003 from Ketan Bhaidasna
that makes the point...
Ben and Carl
were in a 100meter race. When Ben
crossed the finish line, Carl was only at the
90meter mark.
Ben suggested they run
another race. This time Ben would start ten
meters behind the starting line. Will
Carl win or lose  or will the second race
end in a tie?
Ketan's solution...
Carl will lose
again. In the second race, Ben started
ten meters back. By the time Carl
reaches the 90meter mark, Ben will have
caught up to him. Therefore, the final
ten meters will belong to the faster of the
two. Since Ben is faster than Carl, he
will win the final 10 meters and the race.

The analytic approach...
Let the respective velocities
be V_{B} and V_{C }
with the first race ending at time T.
We have that T = 100 / V_{B} =
90 / V_{C}.
Thus, V_{B} =
100/T and V_{C} = 90/T.
Ben finishes the second race
at 110 / V_{B} = 110/100 / T = 1.1T
Carl finishes the second race
at 100 / V_{C} = 100/90/T = 1.111T,
slower by about 1%.
So then, how much of a
headstart should Ben give Carl?
This question will be
pertinent in the World's
Fastest
Dragster.

