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?**