(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?
Let distance be in miles, time in days, and velocity in miles/day. Let A be the messenger's starting point at the end of the column, B the head of the column when he starts and the end of the colomn when he finishes, and C the point where he reaches the head of the column. Let t be the time for him to reach point C and v his velocity. Thus,
vt = 25 + 25*t for forward trip ( AC = AB + BC )
v*(1t) = 25t for backward trip ( BC = BC )
Eliminating t from these equations gives
v^2  50*v  25^2 = 0
Only the positive root is meaningful
v = 25*(1+sqrt(2)) miles/day
Since his trip is for one day,
trip distance = 25*(1+sqrt(2)) miles ~= 60.36 miles

Posted by Bractals
on 20040824 11:48:09 