math
posted by Suzy .
A battalion 20 miles long advances 20 miles.
During this time, a messenger on a horse travels from the rear of the battalion to the
front and immediately turns around, ending up precisely at the rear of the battalion
upon the completion of the 20mile journey. How far has the messenger traveled?

An interesting problem!
We have to realize that the speed of the battalion is irrelevant. It is the ratio of the speed of the messenger, with respect to the battalion that matters.
Assume the battalion's speed is 1 mph, thus we look for the speed of the messenger y mph.
Time for messenger to go to the front,
t1= 20 miles / (y1)
Time for messenger to go back to the rere,
t2 = 20 miles / (y+1)
So the messenger has advanced (t1t2)y miles over the mission, or
y(t1t2)=20
20y/(y1)20y/(y+1) = 20
y cannot be 1 (or he'll never catch up), so we can multiply both sides by (y1)(y+1) to get:
20y(y+1)20y(y1) = 20(y1)(y+1)
Simplifying
y²2y1=0
Solve the quadratic to get:
y=1+√2
Check:
y(20/(y+1)20/(y1))=20 OK.
Solve for total distance travelled
= y(t1+t2)
(approx. 48.3 miles)