When two pipes fill a pool together, they can finish in 4 hours. If one of the pipes fills half the pool then the other takes over and finishes filling the pool, it will take them 9 hours. How long will it take each pipe to fill the pool if it were working alone?

One pipe takes __ hours alone.
The other pipe takes __ hours alone.

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

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

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