Faucets

A large tank is partially filled with a solution. The tank has a faucet that allows the solution to enter the tank at a rate of 16 3/4 liters per minute. The tank also has a drain that allows the solution to leave the tank at a rate of 19 4/5 liters per minute. (a) What expression represents the change in the volume of solution in the tank in 1 minute? (b) Simplify and give your answer as a simplified mixed number. (c) What is the change in volume of the solution after 10 minutes? Show necessary work. Give your answer as a simplified mixed number.

Correct answer:

V₁ =  -61/20 l
c =  -61/2 l

Step-by-step explanation:

$$\begin{gathered} t_1=1\text{min} \\ {Q}_1=16\frac{3}{4}=16+\frac{3}{4}=\frac{16\cdot 4+3}{4}=\frac{64+3}{4}=\frac{67}{4}=16.75\text{l/min} \\ {Q}_2=19\frac{4}{5}=19+\frac{4}{5}=\frac{19\cdot 5+4}{5}=\frac{95+4}{5}=\frac{99}{5}=19.8\text{l/min} \\ Q=Q_1-Q_2=\frac{67}{4}-\frac{99}{5}=\frac{335}{20}-\frac{396}{20}=\frac{335-396}{20}=\frac{-61}{20}=-\frac{61}{20}=-3.05\text{l/min} \\ {V}_1=t_1\cdot Q=1\cdot \frac{-61}{20}=\frac{1\cdot (-61)}{20}=\frac{-61}{20}=-\frac{61}{20}\text{l}=-3\frac{1}{20}\text{l}=-3.05\text{l} \end{gathered}$$

$$\begin{gathered} t_2=10\text{min} \\ c=t_2\cdot Q=10\cdot \frac{-61}{20}=\frac{10\cdot (-61)}{20}=\frac{-610}{20}=-\frac{61}{2}\text{l}=-30.5\text{l} \end{gathered}$$

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