# 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.