Chapter 7, Problem 79RE

Solution

To calculate: The distance would take each pipe running alone to fill the pool.

The obtained time for the pipe A , B , and C is $15 \text{ hr}, 5.45 \text{hr, and }12\text{ hr}$ .

Given Information:

Three pipes, A, B, and C, are connected to a swimming pool. When all three pipes are running, the pool can be filled in 3 hr. When only A and B are running, the pool can be filled in 4 hr. When only B and C are running, the pool can be filled in 3.75 hr.

Calculation:

Consider the given information,

Suppose that x be  the  rate  of  pipe   A , in filled pool per hour, y is representing the rate of pipe  B , in filled pool per hour, and z is representing be  the rate  of pipe  C ,  in filled pool per hour.

Set up the equations. When all three pipes are running, the pool can be filled in 3 hr:

  $\begin{array}{c}\left(x+y+z\right)3=1\\ x+y+z=\frac{1}{3}\end{array}$ …… (1)

When only A and B are running, the pool can be filled in 4 hr:

  $\begin{array}{c}\left(x+y\right)4=1\\ x+y=\frac{1}{4}\end{array}$ …… (2)

When only B and C are running, the pool can be filled in 3.75 hr:

  $\begin{array}{c}\left(y+z\right)\left(3.75\right)=1\\ y+z=\frac{1}{3.75}\end{array}$ …… (3)

Substitute equation 2 value in equation.

  $\begin{array}{c}\frac{1}{4}+z=\frac{1}{3}\\ z=\frac{1}{3}-\frac{1}{4}\\ =\frac{1}{12}\end{array}$

Now, find the value of y by using equation 3.

  $\begin{array}{c}y+\frac{1}{12}=\frac{1}{3.75}\\ y=\frac{1}{3.75}-\frac{1}{12}\\ =\frac{4}{15}-\frac{1}{12}\\ =\frac{11}{60}\end{array}$

Substitute value of y in equation 2.

  $\begin{array}{c}x+\frac{11}{60}=\frac{1}{4}\\ x=\frac{1}{4}-\frac{11}{60}\\ =\frac{4}{60}\\ =\frac{1}{15}\end{array}$

To find the time, taking the reciprocals of the rates of the pipes, the times they can fill the pool alone are:

  $\begin{array}{l}\text{Pipe} A: 15 \text{ hr}, \\ \text{Pipe} B: \frac{60}{11}\approx 5.45 \text{hr} \\ \text{Pipe} C: 12\text{ hr}\end{array}$

Therefore, the obtained time for the pipe A , B , and C is $15 \text{ hr}, 5.45 \text{hr, and }12\text{ hr}$ .