I don't know where to start but I will watch and learn if someone else wants to answer.
I just found this. Looks like it should help, I have not watched it yet but it looks relevant.
The solution is being added quicker than it is being drained away. So the amount in the container is constantly increasing.
At a given point in time (t minutes after start) the amount in the tank will be 5000+50t-30t = 5000+20t
At the start 5000L of brine which has 100Kg sale
Entering 0.3Kg/L at 50L/min
Leaving The concentation in the tank at any given moment at 30L/min
Let A(t) be the amount of salt in the tank immediately after any given time t (minutes).
I will call this A for short but remember that it is a function of time.
\(\frac{dA}{dT}=\text{rate in - rate out}\\ \frac{dA}{dT}=\text{concentation in* rate of flow - concentation out *rate of flow}\\ \frac{dA}{dT}= \left(\frac{0.3Kg}{1L}*\frac{50L}{1minute}\right)- \left(\frac{A Kg}{(5000+20t)L}*\frac{30L}{1minute}\right)\\ \frac{dA}{dT}= \left(\frac{15Kg}{1minute}\right)- \left(\frac{30A Kg}{(5000+20t)minute}\right)\\ \frac{dA}{dT}= 15- \frac{30A}{5000+20t}\;\;\quad\frac{Kg}{min} \)
I am not sure if it is finished. I probably should go further ? what do you think ?
I don't have time right now.
By the way, the video question was simpler than this one but that is the only resource that I used to help me.
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ok I have looked more and I do not know how to go further. Maybe the question does not require me too.
Does anyone else know how to take if further?