Question

a hose with a larger diameter working alone can fill a swimming pool in 9 hours. a hose with a smaller diameter working alone can fill a swimming pool in 18 hours. working together, how long would it take the two hoses to fill the swimming pool? the rate of the hose with the large diameter is 1/9. complete what is the unknown in the problem? the rate of smaller hose the time it takes for the smaller hose working alone to fill the pool the time it takes for the hoses working together to fill the pool retry

Explanation:

Step1: Determine the rate of the smaller hose

The smaller - hose fills the pool in 18 hours. So its rate is $\frac{1}{18}$ of the pool per hour.

Step2: Find the combined rate of the two hoses

The rate of the large - hose is $\frac{1}{9}$ and the rate of the small - hose is $\frac{1}{18}$. The combined rate $r$ is $\frac{1}{9}+\frac{1}{18}=\frac{2 + 1}{18}=\frac{3}{18}=\frac{1}{6}$ of the pool per hour.

Step3: Calculate the time taken when working together

Let $t$ be the time taken when working together. Since rate $\times$ time = work done (and work done is 1 pool), and $r=\frac{1}{6}$, we have $\frac{1}{6}t = 1$. Solving for $t$, we get $t = 6$ hours.

Answer:

6 hours