ciprocal of the number is \( \frac{1}{3} \). Find the number. 5. A town's old street sweeper can clean the streets in 60 h . The old sweeper together with a new sweeper can clean the streets in 15 h . How long would it take the new sweeper to do the job alone?

Let's solve both problems step by step. --- ### **Problem 1:** **Find the number whose reciprocal is \( \frac{1}{3} \).** **Solution:** Let the number be \( x \). The reciprocal of \( x \) is \( \frac{1}{x} \). According to the problem: \[ \frac{1}{x} = \frac{1}{3} \] To find \( x \), take the reciprocal of both sides: \[ x = 3 \] **Answer:** The number is **3**. --- ### **Problem 2:** **A town's old street sweeper can clean the streets in 60 hours. The old sweeper together with a new sweeper can clean the streets in 15 hours. How long would it take the new sweeper to do the job alone?** **Solution:** Let's denote: - **Old Sweeper's Rate (\( R_{\text{old}} \)):** \[ R_{\text{old}} = \frac{1}{60} \text{ (jobs per hour)} \] - **New Sweeper's Rate (\( R_{\text{new}} \)):** \[ R_{\text{new}} = \frac{1}{n} \text{ (jobs per hour)}, \text{ where } n \text{ is the time in hours for the new sweeper alone.} \] - **Combined Rate:** \[ R_{\text{combined}} = \frac{1}{15} \text{ (jobs per hour)} \] According to the problem: \[ R_{\text{old}} + R_{\text{new}} = R_{\text{combined}} \] \[ \frac{1}{60} + \frac{1}{n} = \frac{1}{15} \] Subtract \( \frac{1}{60} \) from both sides to solve for \( \frac{1}{n} \): \[ \frac{1}{n} = \frac{1}{15} - \frac{1}{60} \] \[ \frac{1}{n} = \frac{4 - 1}{60} = \frac{3}{60} = \frac{1}{20} \] Now, take the reciprocal to find \( n \): \[ n = 20 \text{ hours} \] **Answer:** The new sweeper alone can clean the streets in **20 hours**. ---