Hiking the Franconia Ridge

Let t=time of sunrise, v=constant speed of Alex and u=constant speed of Bob .

1 . distance that Alex covered from t to 11:00 am = distance Bob covered from 11:00 am to 8:00 pm.

2 . distance that Alex covered from 11:00 am to 3:00 pm = distance Bob covered from t to 11:00 am.

d=vt

3 . from 1 : v(11-t) = u (9) so v/u = 9/(11-t)

4 . from 2 : v(4) = u(11-t) so v/u = (11-t)/4

5 . comparing 3 and 4 : 9/(11-t) = (11-t)/4
36 = (11-t)(11-t)
6=11-t
t= 5 or t=05:00 ot 0+5+0+0=5

Let the distances of each section (before and after meeting) be $a$ and $b$ . Let the speeds be $u$ and $v$ . Let $t$ be the time interval between sunrise and 11 pm.

Then $\begin{cases} a = ut & b = 4u & \text{Alex} \\ b = vt & a = 9v & \text{Bob}\end{cases}$ Equate the equations for $a$ and $b$ : $\begin{cases} ut = 9v & (a) \\ vt = 4u & (b) \end{cases}$ Multiply the two equations: $uvt^2 = 36uv$ and we see that $t = 6$ . The sunrise was at $\boxed{\text{05:00}}$ am.

Let $a=$ the distance traveled by one of them before they meet. Let $b=$ the distance traveled by the other before they meet. Then:

$\frac { a }{ 11-x } =\frac { b }{ 4 }$

and

$\frac { a }{ 9 } =\frac { b }{ 11-x }$

We divide the first expression by the other, to get:

${ (11-x) }^{ 2 }=36$

$\therefore \boxed { x=5 }$