Limits and Derivatives This is a homework question where I don't quite understand what I am being asked to do: 
A tank contains $5000$ L of pure water. Brine that contains $30$ g of salt per litre of water is pumped into the tank at a rate of $25$ L/min. Show that the concentration of salt after t minutes (in grams per litre) is: 
$$ C(t)= 30t/200+t $$
I am not sure what I am being asked to do? 
 A: $30$ grams of salt per litre and $25$ litres per minute mean that salt is being pumped into the tank at a rate of $30\cdot25=750$ grams per minute. After $t$ minutes this will add $750t$ grams of salt to the tank. Since the water started out pure, the total amount of salt in the tank after $t$ minutes must be $750t$ grams.
The concentration in grams per litre is the total number of grams of salt in the tank divided by the number of litres of solution. How many liters of solution will be in the tank after $t$ minutes? Express that in terms of $t$, and you have everything you need.
(By the way, the answer that you list is incorrect; it contains a typo.)
A: Your expression seems to be incorrect. The correct expression seems to be $$C(t) = \frac{30t}{20 + t}$$.
Now, let us say 1 minute has passed after the pumps were started. So, you added 25L of water to the tank. That 25L of water contains 30g per litre. Thereby, in total it contains $30 \times 25 = 750$g of salt. The total amount of water in the tank after 1 minute is 525L. Therefore, the concentration of salt (amount of salt per litre) in the tank after 1 minute is given by $\frac{750}{525} = \frac{30}{21}$. This matches the above expression.
Now, you should be able to generalize this.
A: salt after $t$ minutes = $30 \cdot 25 \cdot t$
water after $t$ minutes = $500 + 25 t$
compute concentration $30 \cdot 25 \cdot t \over 500 + 25 t$
