tank problem Differential equation A tank is partially filled with 100 gallons of coffee in which 10 lbs of sugar is dissolved. Coffee containing 1/3 lb of sugar per gallon is pumped into the tank at rate 3 gal/min. The yummy well-mixed solution is then pumped out at a slower rate of 1 gal/min. 
A- What is the rate at which the tank is increasing before the tank is full? 
B- Set up a differential equation for finding the number of pounds A(t) of sugar in the tank at anytime? 
for part A i'm not sure but for part B i think that it should be something like that 
dA(t)/t= (1/3)(3)-(A(t)/100)(3) but I'm not sure how to finish this 
 A: Let $A(t)$ be the number of pounds of coffee in the tank at time $t$. We find an expression for $A'(t)$.
There is a standard pattern for setting up the appropriate differential equatiom. We look separately at the rate sugar is (i) entering the tank and (ii) leaving the tank.
Entering: Liquid is entering at $3$ gallons per minute, and each gallon has $\frac{1}{3}$ pound of sugar. Thus sugar is entering at the rate $\frac{1}{3}\cdot 3$, that is, $1$.
Leaving:  If we set $t=0$ at the beginning, then the amount of liquid in the tank at time $t\ge 0$ is $100+2t$. The concentration of sugar at time $t$ is $\frac{A(t)}{100+2t}$. Since liquid is leaving at rate $1$, sugar is leaving at rate $\frac{A(t)}{100+2t}$. 
A suitable differential equation for $A(t)$ is therefore
$$A'(t)=1-\frac{A(t)}{100+2t}.$$
One needs to write down an appropriate initial condition.
The differential equation now can be solved in any of the usual ways, for example by first considering the related homogeneous equation.
A: So you have $dA/dt = (\text{A in}) - (\text{A out})$. For $\text{A in}$, you want to find the concentration of the input (lbs/gal) and multiply it by the volume input (gal/min) to get the A input (lbs/min). You did this correctly. 
For $\text{A out}$ you want to do the same. Find the concentration of the output (lbs/gal) and multiply it by the volume output (gal/min) to get the A output (lbs/min). Your volume output is 1 gal/min. Your output concentration is A(t)/V(t), where V(t) is the current volume of the tank. In a simpler problem, this volume would be constant, but here it isn't, because the input and output flow rates are different. What is it here?
