Chapter 12.6, Problem 15P

To determine

To find: time required for pipe C to empty the tank.

Answer to Problem 15P

The time required was approximate $9$ hours to empty the tank.

Explanation of Solution

Given information:

Pipe A can fill the tank in $4h$ and pipe B can fill the tank $9h$ less than time taken by pipe C to empty the tank.

When all pipe are opened it take $2h$ to fill the tank.

Formula:

The quadratic equation $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Consider that the time taken by pipe C to empty the tank was $xh$ Where $h$ is the hours.

Since as mentioned above that pipe B can fill the tank $9h$ less than time taken by pipe C to empty the tank.

Therefore time taken by pipe B is $\left(x-9\right)$ .

Also when all pipe are opened it take $2h$ to fill the tank.

Now for time taken for pipe C will be as mentioned below.

  $A+B+C=2$ .

Therefore $\frac{1}{4}+\frac{1}{x-9}-\frac{1}{x}=2$ .

Recall that we will take L.C.M of the denominator.

  $\frac{1}{4}+\frac{1}{x-9}-\frac{1}{x}=2\Rightarrow \frac{x\left(x-9\right)+4x-4x+36}{4x\left(x-9\right)}=2\Rightarrow x^2-9x+36=8x^2-72x\Rightarrow 7x^2-63x-36=0.$

Recall that the quadratic equation $\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ where $a=7,b=-63,c=-36$ .

Therefore $\begin{array}{l}x=\frac{-\left(-63\right)\pm \sqrt{{\left(-63\right)}^2+4\times 7\times 36}}{2\times 7}\Rightarrow x=\frac{-\left(-63\right)\pm \sqrt{3969+1008}}{14}\Rightarrow x=\frac{63\pm \sqrt{4977}}{14}\Rightarrow x=\frac{63\pm 70.54}{14}\\ \Rightarrow x=\frac{63+70.54}{14},\frac{63-70.54}{14}\Rightarrow x=\frac{130.54}{14},\frac{-7.54}{14}\Rightarrow x=9.32,-\mathrm{0.54.}\end{array}$

Therefore time required was $9.32$ hours to empty the tank.

Therefore time required was approximate $9$ hours to empty the tank.

Here we reject $x=-0.54$ as the time cannot be in negative.