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Marching Messenger (Posted on 2002-07-05) Difficulty: 5 of 5
(This is from CTK exchange)

A column of soldiers is 25 miles long and they march 25 miles a day. One morning a messenger started at the rear of the column with a message for the guy up front. The messenger began to march and gave the message to the guy up front and then returned to his position by the end of the day. Assume that the messenger marched at the same rate of speed the whole time. How many miles did the messenger march?

See The Solution Submitted by Dulanjana    
Rating: 4.1667 (12 votes)

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Solution re: Forced March | Comment 7 of 13 |
(In reply to Forced March by TomM)

Actually this answer isn't completely correct as the rate of the messenger is actually (25 + 2d)t because his rate has to include the time he is marching back to his position in line or he will only get to the point where he meets the officer. If we make 2 linear equations representing the distance each marcher has marched from the officers initial position at t=0
we get

officer: D = 25t
messenger: D = (25 + 2d)t - 25

-25 represents the fact that the messenger starts 25 miles behind the officer

Now we know that the Messenger will catch the officer at
0 = (25 + 2d)t - 25 - 25t

solving for d we get:
d = 25/(2t)

substituting in for d into rate of the messenger we get and using Distance = Rate * Time

Distance = (25 + 2(25/(2t))t
Distance = 25t + 25
t = 1 at end of day so
Distance = 50

The messenger actually marches 50 miles.
  Posted by Barry Knapp on 2003-12-08 10:49:42

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