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**ThePerfectHacker** Let $\displaystyle A(t)$ be the amount in kilograms of salt in water at time $\displaystyle t$. Thus, we know that $\displaystyle A(0)=20$.

Thus,

$\displaystyle \frac{dA}{dt} = \mbox{ rate in }-\mbox{ rate out} $

The rate is is $\displaystyle (.03)(25) = .75 \mbox{kg}/\mbox{min}$.

The rate out is $\displaystyle \mbox{concentration} \times 25 \mbox{L}/\mbox{min}$. The concentration is given by $\displaystyle \mbox{amount}/\mbox{volume}$ the amount is $\displaystyle A(t)$ and volume is $\displaystyle 5000$. Thus we have $\displaystyle A(t)/200$.

The proper differencial equation is,

$\displaystyle \frac{dA}{dt} = .75 - \frac{A(t)}{200}$.