Time & Distance : Pokemon Hunter and the Rogue Brook I was working my way through some Puzzles in Discrete Maths by Rosen, when I came across the following question:

  
*
  
*A Pokemon Hunter is rowing upstream a brook
  
*As he passes under the 'bridge-of-curse', he throws a Pikachu into the brook
  
*5 minutes later, he realises that Pikachu will die and he should not have done that 
  He rows back and picks Pikachu 3km. downstream of the 'bridge-of-curse'
What is the speed of the river flow?

My Answer:

  
*
  
*Distance : 3km = 3000 m
  
*Time : 5min + 5min =  10 minutes = 600 s
  
*Speed = 3000 / 600  m/s  = 5 m/s

Doubt:
Am I correct ? Seems a bit too easy ...
 A: Let's assume the trainer rows upstream with a speed of $v_r - v_s$ where $v_s$ is the velocity of the stream and $v_r$ the velocity of the trainer in still water. Downstream his velocity will then be $v_r + v_s$. Now we know that He rows 5 mins upstream and then turns around and rows an unknown amount of time downstream to reach $-3$ km from his original position:
$$\begin{align*}
5 (v_r - v_s) - t (v_r + v_s) & = -3 \\
-(5+t) v_s = -3
\end{align*}$$
Speeds in km/min here. Now this is not so nice since we have one equation too few, but knowing that $v_r \ge v_s > 0$ will help us find a solution nonetheless:

We rewrite $v_s = \frac{3}{5+t}$ so
$$-3=5(v_r - \frac{3}{5+t}) - t(v_r + \frac{3}{5+t}) = (5-t)v_r - \frac{3(5+t)}{5+t}\\
\Rightarrow (5-t)v_r = 0$$
Wow, luckily since $v_r \ge v_s > 0$ we have $5-t = 0$ so $t=5$. This means that
$$v_s = \frac3{10} \frac{\text{km}}{\text{min}} = 5 \frac{\text m}{\text s}$$
So you are correct, but the fact that $t=5\text{ min}$ is far from obvious :)
