Question

A large tank contains 200 liters of a salt solution which has a concentration of 0.35 kilograms per liter. Pure water is added to the solution at a rate of 40 liters per minute. At the same time, the solution drains from the tank at 40 liters per minute

Answer 1

The question is incomplete, but from the given information it's clear that this is a standard mixing problem. Let $A(t)$ be the amount of salt (in kilograms) in the tank at time $t$.

Then the rate of change of the amount of salt in the tank is described by the differential equation

$\dfrac{\mathrm dA(t)}{\mathrm dt}=\left(0\dfrac{\text{kg}}{\text{L}}\right)\left(40\dfrac{\text{L}}{\text{min}}\right)-\left(\dfrac{A(t)}{200}\dfrac{\text{kg}}{\text{L}}\right)\left(40\dfrac{\text{L}}{\text{min}}\right)$
$\drac{\mathrm dA}{\mathrm dt}=-\dfrac A5$

Separating the variables, we get

$\dfrac{\mathrm dA}A=-\dfrac{\mathrm dt}5$

and integrating both sides yields

$\ln|A|=-\dfrac t5+C$
$\implies A=Ce^{-t/5}$

Initially, the tank contains $\left(0.35\dfrac{\text{kg}}{\text{L}}\right)\left(200\text{ L}\right)=70\text{ kg}$ of salt, i.e. $A(0)=70$. This means you have

$70=Ce^{-0/5}\implies C=70$

and so the amount of salt in the tank as a function of time is

$A(t)=70e^{-t/5}$