Sean and Colleen are raking leaves in their yard. Working fogether, they can clear the yard of leaves in 24 minutes. Work. 20 minutes longer to clear the yard than it would take Colleen working alone. When \( c \) is the number of minutes it would take Colleen to finish the job when working alone, the situation is modeled by th \( \frac{1}{c}+\frac{1}{c+20}=\frac{1}{24} \). How long would it take Colleen alone to clear the yard of leaves? A. 12 minutes B. 14 minutes C. 28 minutes D. 40 minutes

To determine how long it would take Colleen to clear the yard of leaves alone, let's analyze the given information and solve the equation step by step. **Given:** - Colleen's time alone: \( c \) minutes. - Sean takes 20 minutes longer than Colleen: \( c + 20 \) minutes. - Working together, they clear the yard in 24 minutes. **Modeling the Rates:** - **Colleen's rate:** \( \frac{1}{c} \) (yards per minute). - **Sean's rate:** \( \frac{1}{c + 20} \) (yards per minute). - **Combined rate:** \( \frac{1}{24} \) (yards per minute). The relationship between their rates is: \[ \frac{1}{c} + \frac{1}{c + 20} = \frac{1}{24} \] **Solving the Equation:** 1. **Multiply both sides by \( 24c(c + 20) \)** to eliminate the denominators: \[ 24(c + 20) + 24c = c(c + 20) \] 2. **Expand and simplify:** \[ 24c + 480 + 24c = c^2 + 20c \] \[ 48c + 480 = c^2 + 20c \] 3. **Rearrange all terms to one side to form a quadratic equation:** \[ c^2 - 28c - 480 = 0 \] 4. **Solve the quadratic equation using the quadratic formula:** \[ c = \frac{28 \pm \sqrt{28^2 + 4 \times 480}}{2} \] \[ c = \frac{28 \pm \sqrt{784 + 1920}}{2} \] \[ c = \frac{28 \pm \sqrt{2704}}{2} \] \[ c = \frac{28 \pm 52}{2} \] We discard the negative solution because time cannot be negative: \[ c = \frac{28 + 52}{2} = 40 \text{ minutes} \] **Answer:** **D. 40 minutes**