Word Problem| Algebra| Problem 1

Two pipes were installed in a water tank. One pipe serves as an inlet pipe while another, a multi-purpose pipe, serves as either inlet pipe or a drain pipe. When both pipes are used to fill up the tank, they can fill 7/9 of the tank in 4 hours. In filling the tank, the operator sometimes pushes the wrong button and the multi-purpose pipe functions as a drain pipe. In such cases, only 1/6 of the tank can be filled up in 6 hours. How long will it take the inlet pipe to fill the tank alone?

SOLUTION:

Let x be the number of hours the inlet pipe can fill the tank alone, and y be the number of hours the multi-purpose pipe can either fill or drain the tank alone.

The equations are 

\left(\frac{1}{x}+\frac{1}{y}\right)\cdot 4=\frac{7}{9}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:Equation\:1

\left(\frac{1}{x}-\frac{1}{y}\right)\cdot 6=\frac{1}{6}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:Equation\:2

From equation 1

\frac{1}{x}+\frac{1}{y}=\frac{7}{36}

\frac{1}{x}=\frac{7}{36}-\frac{1}{y}

Substitute 1/x to the second equation.

\frac{1}{x}-\frac{1}{y}=\frac{1}{36}

\frac{7}{36}-\frac{1}{y}-\frac{1}{y}=\frac{1}{36}

\frac{2}{y}=\frac{7}{36}-\frac{1}{36}

\frac{2}{y}=\frac{6}{36}

6y=72

y=12

Solve for x

\frac{1}{x}=\frac{7}{36}-\frac{1}{12}

\frac{1}{x}=\frac{4}{36}

4x=36

x=9

 

Therefore, the inlet pipe can fill the tank alone in 9 hours.

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