Salty water question Here's the question, from my $10 mathbook that everyone is growing to love and hate:

A saltwater solution initially contains 5lb of salt in 10 gal of fluid.  If water flows in at 0.5 gal/min and the mixture flows out at the same rate, how much salt is present after 20 mins?

I get that this is a differential equation problem, and that the quantity we are observing is the concentration of the salt in the water.
The rate of change of concentration ($\frac{dSolute}{dt}$) is proportional to:  the current concentration, and the rate at which the fluid is (being replaced) by fresh water.
And so I of course have $y =y_0 e^{kt}$.
How do I think about this problem?  
There is a solution given, which is
$ \frac{dS}{dt} = -\frac{1}{2}(\frac{S}{10})$
At $t=20$, $S=5e^{-1}$
I can guess where the constants came from ($-\frac{1}{2}$, 10) but I don't see what's going on here, especially with the $S=5e^{-1}$ quantity.
Working backwards, it says
$$\frac{\mbox{change in concentration}}{\mbox{unit of time}} = \mbox{(rate at which we are losing solution)}\left(\frac{\mbox{current salt concentration}}{\mbox{volume of water}}\right).$$
Why is that the formula?
 A: The differential equation is $\frac{dS}{dt} = -\frac{S}{20}$, which has solution $S=Ce^{\frac{-t}{20}}$.  S(0) is given as 5, so $S(t)=5e^{\frac{-t}{20}}$  At t=20, this is $5e^{-1}$
Added:  I just took the equation from your post.  In a small amount of time, $\delta t$ you have $\frac{1}{2}\delta t$ pure water added and the same quantity of salty water removed.  This represents $\frac{1}{20}\delta t$ of the volume of the salty water, so you should remove that much salt.  The change in salt is $-\frac{1}{20}S\delta t$.  This gives $\frac{dS}{dt} = -\frac{S}{20}$
A: Ross's answer was good, but here's a bit more clarity.
The question is "how much salt is lost after 20 minutes (of above described loss scenario)?"
In $y = y_0 e^{kt}$, we have to find $k$.
So how much salt is lost?
Well, 


*

*You lose $\frac{1}{2}$ gallons of water per minute

*The concentration of salt per gallon is $\frac{S}{10}$, because there is S salt in the 10 gallons of water (at any instant of time)

*Thus the net rate of change of salt in the vat is
$ -\frac{1}{2} $ [gallons/min] $ \frac{S}{10} $ [salt/gallon]
$ = -\frac{S}{20} $ [salt/min]

*Thus the differential equation is 
$ \frac{dS}{dt} = -\frac{S}{20} $
Matching with
$ \frac{dy}{dt} = ky $
for a differential equation, this means k = $-\frac{1}{20}$.
Then
$y = 5 e^{-\frac{1}{20}t}$
$t = 20$ so
$y = 5 e^{-1}$
