Question

Three pipes can fill the tank in 18 hours. One of the pipes can fill it in 18 hrs and the other pipe can empty in 9 hours. At what rate does the third pipe work?

Answer 1

I think 9 hrs is the answer

Answer 2

Answer: 9 hours

Step-by-step explanation:

Since, one pipe can fill the tank alone in 18 hours,

⇒ Work done by one pipe in 1 hour = $\frac{1}{18}$

Also, the other pipe can empty the tank by own in 9 hours,

⇒ Work done by other pipe in 1 hour = $-\frac{1}{9}$

( since, we take filling tank is a work that's why we take negative sign for the outlet pipe )

Let the time taken by third pipe to fill or empty the tank = x hours,

⇒ Work done by third pipe in one hour = $\frac{1}{x}$

Hence, the total work in one our = $\frac{1}{18}-\frac{1}{9}+\frac{1}{x}$

According to the question,

When all pipe work simultaneously, then total time taken = 18 hours,

⇒ Total work done by all pipe in one hour = $\frac{1}{18}$

⇒ $\frac{1}{18}-\frac{1}{9}+\frac{1}{x}=\frac{1}{18}$

⇒  $\frac{1}{x}=\frac{1}{18}-\frac{1}{18}+\frac{1}{9}$

⇒ $\frac{1}{x}=\frac{1-1+2}{18}$

⇒ $\frac{1}{x}=\frac{2}{18}=\frac{1}{9}$

⇒ Work done  by third pipe in one hour = 1/9

⇒ When it work alone the time taken by it = 9 hours.