Originally Posted by

**HallsofIvy** Have you not at least **tried** this yourself? You have posted several problems where you just state the problem. That violates the rules of this forum.

Start by setting it up as a "differential equation". Let X(t) be the amount of salt, in pounds, in the tank at time t in minutes. The rate of change of X, dX/dt, is the rate at which salt is coming into the tank minus the rate at which salt is going out of the tank. The rate at which salt is going into the tank is 2 pounds per gallon times 2 gallons per minute= 4 pounds per minute.

The rate at which salt is going out of the tank is a little more complicated. It is the number of pounds per gallon times 4 gallons per minute.

One complication is that, because water is coming in at 2 gallons per minute but going out of the tank at 4 gallons per minute so the amount of water in the tank is reducing at 2- 4= -2 gallons per minute. The tank originally contains 200 gallons so after t minutes there are 200- 2t gallons of water in the tank. With X(t) pounds of salt in the tank there are $\displaystyle \frac{X(t)}{200- 2t}$ "pounds per gallon". The salt is going out of the tank at the rate of $\displaystyle -4\frac{X(t)}{200- 2t}$.

Putting those together, $\displaystyle \frac{dX}{dt}= 4- \frac{X}{200- 2t}$. Solve that differential equation with the initial condition X(0)= 20.

Once you know X(t)

a) What is X(25)?

b) Solve X(t)= 100 for t.

c) What is X(100)?