# 11 - Distance Traveled By A Messenger From Rear To Front Then Back To Rear Of A Marching Battalion

**Problem**

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?

**Answer Key**

**Solution**

*x*= speed of the messenger

*y*= speed of the battalion

Recall that distance / speed = time

$\dfrac{20}{y}$ = time for the battalion to advance 20 miles

$\dfrac{20}{x - y}$ = time for messenger to go from rear to front

$\dfrac{20}{x + y}$ = time for messenger to go from front to rear

$\dfrac{20}{y} = \dfrac{20}{x - y} + \dfrac{20}{x + y}$

$(x - y)(x + y) = y(x + y) + y(x - y)$

$x^2 - y^2 = (xy + y^2) + (xy - y^2)$

$x^2 - 2xy - y^2 = 0$

$\dfrac{x^2}{y^2} - 2\dfrac{x}{y} - 1 = 0$

Let *z* = *x*/*y*

$z^2 - 2z - 1 = 0$

$z = \dfrac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)}$

$z = \dfrac{2 \pm \sqrt{8}}{2} = \dfrac{2 \pm 2\sqrt{2}}{2}$

$z = 1 + \sqrt{2} ~ \text{ and } ~ 1 - \sqrt{2}$

Use *z* = 1 + sqrt(2)

$\dfrac{x}{y} = 1 + \sqrt{2}$

Time of Messenger = Time of Battalion

$\dfrac{d}{x} = \dfrac{20}{y}$

$d = 20\left( \dfrac{x}{y} \right)$

$d = 20\left( 1 + \sqrt{2} \right)$

$d = 48.28 ~ \text{miles}$ ← *answer*