Answer and Explanation:
The rate of change of the volume of the brine is given by:
[tex]\Delta V = (\text{rate in}) - (\text{rate out})[/tex]
[tex]\Delta V = (4 \text{ L/min}) - (3\text{ L/min})[/tex]
[tex]\Delta V = +1\text{ L/m}[/tex]
Since [tex]\Delta V[/tex] is the same thing as [tex]V'(t)[/tex], which is the derivative of [tex]V(t)[/tex], we can integrate [tex]\Delta V[/tex] to get [tex]V(t)[/tex] (the total volume in the tank), with the constant of integration being the initial volume:
[tex]\displaystyle \int \Delta V\ dt = \int (1\text{ L/min}) \ dt[/tex]
[tex]V(t) = \left(1{\rm\frac{L}{m}}\right) t + C[/tex]
[tex]V(t) = t + 100[/tex]
We can model the volume of brine by taking out the initial volume:
[tex]B(t) = t[/tex]
The mass of brine there will be is:
[tex]\boxed{B(1) \cdot \dfrac{0.5\text{ kg}}{L} = 1 \cdot 0.5 = 0.5\text{ kg}}[/tex]
We can solve for when the concentration of the salt is 0.1 kg/L by relating our equations and the definition of concentration (mass per volume):
[tex]M = \dfrac{B(t) \cdot 0.5\text{ kg/L}}{V(t)}[/tex]
[tex]0.1 = \dfrac{t}{t + 100}[/tex]
[tex]0.1(t + 100) = t[/tex]
[tex]0.1t + 10 = t[/tex]
[tex]0.9t = 10[/tex]
[tex]\boxed{t \approx 11.11\text{ min}}[/tex]
