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: Find the rate of the smaller hose

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

Step2: Find the combined rate

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

Step3: Find the time taken when working together

Let $t$ be the time taken when they work together. We know that rate $\times$ time = work. Since the work is 1 (filling 1 pool) and the rate $r=\frac{1}{6}$, and $r\times t = 1$. So $t=\frac{1}{r}$. Substituting $r = \frac{1}{6}$, we get $t = 6$ hours.

Answer:

6 hours