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Question 1027166: 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 20-mile journey. How far has the messenger traveled?
Found 3 solutions by josmiceli, ikleyn, MathTherapy:
Answer by josmiceli(19441)   (Show Source):
You can put this solution on YOUR website! I think the answer is 60 miles.
Anyone at the rear will advance 20 miles. the
messenger rides 20 more miles to the front.
Then the messenger rides 20 more miles to
where the rear stops ( that is 20 miles behind front )
 miles
Hope I got it!
Answer by ikleyn(52814)  (Show Source):
You can put this solution on YOUR website! .
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 20-mile journey.
How far has the messenger traveled?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Let "b" be the batalion' speed (in mph), and "m" be the messenger speed (relative the ground, of course).
Let "t" be the time the messenger spent for the journey from the rear of the batalion to its front.
During this time, the front moved forward to the distance b*t, so the messenger covered the distance L + bt,
where L is the batalion length, which is 20 miles according the condition.
Thus the equation for this part of the journey is
L + b*t = m*t. (1)
Now let us consider the second part of the journey: the messenger is moving back from the instant front position
of the batalion to its rear, while the rear is moving forward towards the messenger.
Let "s" be the time when they "met" each other and the messenger get the current (updated) rear position.
The equation for this part of the journey is
m*s + b*s = L. (2)
We are almost at the finish line. Now re-write the equations (1) and (2) in the equivalent form:
m*t - b*t = L, (1')
m*s + b*s = L. (2')
Add equations (1') and (2') (both sides). You will get
m*(t+s) = 2L. (3)
t+s is the total time of the messenger journey forward and back.
Hence, m*(t+s) is the total distance traveled by the messenger.
The equation (3) says that this distance is 2L = 2*20 = 40 miles.
Answer. Messenger traveled 40 miles.
Notice. Interesting that the answer does not depend on how far the batalion had advanced.
In other words, the answer does not depend on the speed of the batalion.
As well as does not depend on the speed of the messenger (assuming it is higher then the speed of the batalion).
Answer by MathTherapy(10555)  (Show Source):
You can put this solution on YOUR website! 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 20-mile journey. How far has the messenger traveled?
Let the number of miles the battalion traveled, when the messenger got to its front, be M
Then number of miles the battalion traveled, when the messenger traveled from the front to the rear = 20 - M
The messenger traveled the length of the battalion, or 20 miles, plus the distance the battalion had traveled when he/she got to the front
He/She then traveled the length of the battalion (20 miles), once again, plus the distance the battalion had traveled when he/she got to the rear
So, the messenger traveled a total of: 20 + M + 20 + 20 - M, or 
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