One pipe takes 6 hours working alone.
The other pipe takes 12 hours working alone.
This is a rate of work problem. The formula utilized is
[tex]w=r\times t[/tex]
where
- [tex]w[/tex] is the work done
- [tex]r[/tex] is the rate of work done
- [tex]t[/tex] is the time taken to do the work
We are looking for how long it will take each pipe to completely fill the pool
For pipe 1 working alone to fill 1 pool, the work done is 1;
[tex]w=r\times t\\\\1 = r_1 \times t_1\\\\t_1=\frac{1}{r_1}[/tex]
For pipe 2 working alone to fill 1 pool, the work done is 1;
[tex]w=r\times t\\\\1 = r_2 \times t_2\\\\t_2=\frac{1}{r_2}[/tex]
From the question, it would take 9 hours if each pipe took turns to fill half the pool. That is;
[tex]\frac{t_1}{2}+\frac{t_2}{2}=9\\\\t_1+t_2=18[/tex]
However, if both pipes worked together, it would take 4 hours for each pipe. That is;
[tex]w_1+w_2=w\\\\r_1t_1+r_2t_2=w\\\\4r_1+4r_2=1\\\\r_1+r_2=\frac{1}{4}[/tex]
Remember that [tex]r_1=\frac{1}{t_1}[/tex] and [tex]r_2=\frac{1}{t_2}[/tex]. So,
[tex]\frac{1}{t_1}+\frac{1}{t_2}=\frac{1}{4}\\\\\frac{t_1+t_2}{t_1t_2}=\frac{1}{4}[/tex]
Also recall that [tex]t_1+t_2=18[/tex]. So,
[tex]\frac{18}{t_1t_2}=\frac{1}{4}\\\\t_1t_2=72[/tex]
The only factors of 72 that satisfy the conditions
[tex]t_1+t_2=18\\\\t_1t_2=72[/tex]
are 6 and 12.
Therefore, pipe 1 will take 6 hours, and pipe 2 will take 12 hours to fill the pool if working alone.
The link below has a similar problem on rate of work
https://brainly.com/question/21485928
