Are unit conversions unnecessary when constructing this differential equation? 
A 5-gallon bucket is full of pure water. Suppose we begin dumping salt
  into the bucket at a rate of 1/4 pounds per minute. Also, we open the
  spigot so that $1/2$ gallons per minute leaves the bucket, and we add
  pure water to keep the bucket full. If the saltwater solution is
  always well mixed, what is the amount of salt in the bucket after 
(a) one minute? (b) 10 minutes?
Solution:
The rate of change in the amount of salt in the bucket is $
 \frac{dS}{dT}$.  The amount that comes in per unit time minus what
  goes out per unit time so, $ \frac{dS}{dT}$ = $ \frac{1}{4}$
  -$\frac{S}{10}$.

I have no why we're using $ \frac{1}{4}$.  $ \frac{1}{4}$ comes from that fact that we dump salt at a rate of .25 pounds per minute. However, we're talking about gallons and pounds here which are not even the same unit. I really think we should do some unit conversion to make to .25 pounds per minute into gallons per minute. What I'm saying here is if I were a physicist, I can't subtract apples from oranges. I can't subtract weight from coulombs. I can't subtract mass from weight. Pounds per minute and gallons per minute cannot be compared unless we modify one to be in the same unit as the other and consider density differences and use the fact that density = $ \frac{m}{v}$ 
Why is it that the solution claims we can subtract different units there?
I have no idea how $ S/10$  was derived. I think .5 gallon is 10 percent of the total amount of water plus salt that the container holds. So Salt amount denoted by "S" divided by 10 gives us the amount of salt we have in the container? Are we assuming that the container always has 5-gallons of pure water? What's the maximum capacity of this container? 
I'm really confused on how we know that 10 percent of the total amount of salt in the bucket is leaving each minute. 
The problem claims that .5 gallons per minute leaves the bucket. Does the problem intend to say that .5 gallons of water and salt leave per minute or .5 gallons of salt leave per minute or .5 gallons of water leave per minute? The saltwater solution is well mixed, so I assume that .5 gallons of a overall uniformly distributed salt water mix leave. 
 A: First of all, $S(t)$ is the amount of salt in the bucket at time $t$ and this has units of pounds instead of gallons. Consequently, the rate of change of the amount of salt is $\frac{dS}{dt}$ which has units of pounds per minute. By definition,
$$ \frac{dS}{dt} = \textrm{rate of salt entering the tank} - \textrm{rate of salt leaving the tank}. $$
The rate of salt entering the tank was given as $1/4$ pounds per minute. The rate of salt leaving the tank is a little complicated, but it is given by
$$ \textrm{rate of water exiting the tank}\times\textrm{concentration of salt in the water}. $$
Let us check the units:
$$ \left(\frac{\textrm{gallons}}{\textrm{minute}}\right)\left(\frac{\textrm{pounds}}{\textrm{gallon}}\right) = \frac{\textrm{pounds}}{\textrm{minute}}. $$
Now, the flow rate of water leaving the tank is $1/2$ gallons per minute, and the concentration of salt in the water at time $t$ is
$$ \frac{S(t)}{\textrm{volume of water at time $t$}} = \frac{S(t)}{5}, $$
since the volume of water is unchanged at any time because we add the same amount of pure water to keep the tank full. Summarising everything, we obtain
$$ \frac{dS}{dt} = \frac{1}{4} - \left(\frac{1}{2}\right)\left(\frac{S(t)}{5}\right) = \frac{1}{4} - \frac{S(t)}{10}. $$
