Question

A tank initially holds $100gal$ of a brine solution containing $20Ib$ of salt. At $t=0$, fresh water is poured into the tank at the rate of $5gal/min$, while the well-stirred mixture leaves the tank at the same rate. Find the amount of salt in the tank at any time $t$.

• A. $20e^{\frac{-t}{40}}$
• B. $20e^{\frac{t}{40}}$
• C. $40e^{\frac{-t}{20}}$
• D. $20e^{\frac{-t}{20}}$

Answer 1

The correct option is D $20e^{\frac{-t}{20}}$

Let $Q$ be the amount of sait in the tank and $\frac{dQ}{dt}$ be the rate of change of amount in the tank

We get the equation as $\frac{dQ}{dt}+\frac{1}{20}Q=0$

It is linear equation and the solution of it is $Q=C{e}^{-\frac{t}{20}}$

At $t=0$ , the value of $Q$ is $20$

$\Rightarrow C=20$

Therefore the equation is $Q=20e^{-\frac{t}{20}}$

Therefore the correct option is $D$