Going against the current, a boat takes 6 hours to make a 120 -mile trip. When the boat travels with the current on the return trip. It takes 5 hours. If \( x= \) the rate of the boat in still water and \( y= \) the rate of the current, which of the following expressions represents the rate of the boat going with the current? \( x-y \) \( x+y \) \( x y \)

To solve the problem, we need to analyze the situation step by step. 1. **Identify the known conditions:** - The boat takes 6 hours to travel 120 miles against the current. - The boat takes 5 hours to travel 120 miles with the current. - Let \( x \) be the rate of the boat in still water (in miles per hour). - Let \( y \) be the rate of the current (in miles per hour). 2. **Determine the rates:** - When the boat is going against the current, its effective speed is \( x - y \). - When the boat is going with the current, its effective speed is \( x + y \). 3. **Set up the equations based on the time and distance:** - For the trip against the current: \[ \text{Distance} = \text{Speed} \times \text{Time} \] \[ 120 = (x - y) \times 6 \] This simplifies to: \[ x - y = \frac{120}{6} = 20 \quad \text{(Equation 1)} \] - For the trip with the current: \[ 120 = (x + y) \times 5 \] This simplifies to: \[ x + y = \frac{120}{5} = 24 \quad \text{(Equation 2)} \] 4. **Identify the expression for the rate of the boat going with the current:** - The rate of the boat going with the current is represented by the expression \( x + y \). Thus, the correct expression that represents the rate of the boat going with the current is: \[ \boxed{x + y} \]