Question

A boat travel 312 miles each way downstream and back. The trip downstream took 13 hours. the trip back took 26 hours. What is the speed of the boat in still water. what is the speed of the current.

Answer 1

Remember that $Speed= \frac{distance}{time}$. Since the distance of the complete trip is 312 miles, the distance of each way is $\frac{312}{2} =156$ miles. 
Let  $S _{b}$ be the speed of the boat in still water and $S_{c}$ the speed of the current.

Now, for the downstream trip the boat is traveling with the current, so:
$S_{b} +S_{c} = \frac{156}{13}$
$S_{b} +S_{c} =12$ this will be our equation (1)

For the trip back the boat is traveling against the current, so:
$S _{b} -S_{c} = \frac{156}{26}$
$S_{b} -S_{c} =6$ this will be our equation (2)
 
Next, lets add equation (1) and equation (2) to get rid of $S _{c}$:
$\left \{ {{S_{b}+S_{c} =12} \atop {+(S_{b}-S_{c} =6})} \right.$
$2S_{b} =18$
$S _{b} = \frac{18}{2}$
$S _{b} =9$

Finally, now that we know the speed of the boat in still water, lets replace that value in our equation (1) to find the speed of the current:
$9+S_{c} =12$
$S_{c} =12-9$
$S_{c} =3$

We can conclude that the speed of the boath in still water is 9$\frac{mi}{h}$, and the current of the stream is 3$\frac{mi}{h}$.