To solve this problem, we use the concept of combined work rates.

Let:

• $R_L =$ the rate of the larger hose, $\frac{1}{60}$ pools per minute.
• $R_S =$ the rate of the smaller hose (to be determined).
• Together, their combined rate is $\frac{1}{35}$ pools per minute.

Step 1: Write the combined work rate equation

The total rate is the sum of the individual rates: $R_L + R_S = \frac{1}{35}.$

Step 2: Substitute $R_L$

Since the larger hose’s rate is $R_L = \frac{1}{60}$, we substitute: $\frac{1}{60} + R_S = \frac{1}{35}.$

Step 3: Solve for $R_S$

Subtract $\frac{1}{60}$ from both sides: $R_S = \frac{1}{35} - \frac{1}{60}.$

Find a common denominator for $\frac{1}{35}$ and $\frac{1}{60}$. The least common denominator is 420: $\frac{1}{35} = \frac{12}{420}, \quad \frac{1}{60} = \frac{7}{420}.$

Subtract: $R_S = \frac{12}{420} - \frac{7}{420} = \frac{5}{420}.$

Simplify: $R_S = \frac{1}{84}.$

Step 4: Calculate time for the smaller hose

The rate $R_S = \frac{1}{84}$ means the smaller hose can fill the pool in 84 minutes on its own.

Final Answer:

The smaller hose will take 84 minutes to fill the pool by itself.

Would you like a step-by-step breakdown or have further questions? Here are some related questions for practice:

1. How do you calculate the rate of two workers completing a task together?
2. What if the larger hose takes 80 minutes instead of 60 minutes?
3. Can this problem be solved using fractions alone, without converting rates?
4. How does finding a common denominator help when subtracting fractions?
5. How would you solve this problem if the smaller hose’s rate was given instead?

Tip: When working with rates, think of each rate as a fraction of the job completed per unit of time. This makes combining them straightforward.