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 Marching Messenger (Posted on 2002-07-05)
(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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 Forced March | Comment 1 of 13
Consider three times during the march:

At T = 0, the messenger is at point A (0 miles from point A) the lead officer is at point B (25 miles from point A)

At T = t (when they meet), the lead officer is at point C (d miles from point B); the messenger is also at point C (d + 25 miles from point A.
The officer's speed was d/t. The messenger's speed was (d + 25)/t

At T = 1 days march, the officer is at point D (25 miles from point B); the messenger is at point B (d miles from point C) The officer's speed was 25/(1 day's march)

The officer marched at a steady pace, so d/t = 25 or d = 25t

The total distance covered by the messenger was 25 + 2d; the time was 1 days march; the speed was (d + 25)/t

So we have two equations:
d = 25t
25 + 2d = 1 * (d + 25)/t

Substituting for d, we get
25 + 50t = (25 + 25t)/t

25t + 50 t² = 25 + 25t
50 t² = 25
2t² = 1
t² = ½
t = 1 /√2

so d = 25t = 25/√2
and 2d = 2(25/√2) = 25(√2)

and the messenger travelled 25 + 2d = 25 (1 + √2), or approx 60.355 miles

 Posted by TomM on 2002-07-04 16:56:30

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