The setup sounds simple but I fumbled the direction signs at first.
First, set up equations for the rower's speed using the day 1 information: let r be the still-water speed. Upstream speed is r-2, downstream is r+2. The total distance is 4(r-2) + 5(r+2) = 23, solve for r. Then compute the return distance (23 miles) and the effective speed from camp to S (which is upstream, so r-2), find the time, and subtract from 18:00 to get departure time.
Pro tip: After solving, verify the answer by checking that the computed departure time makes sense (e.g., not negative) and that the speeds are positive. Also, mention that in a real-world scenario, factors like fatigue or changing currents could affect the result, but for this problem we assume constant conditions.
Let r be the rower's still-water speed in mph. Upstream speed = r - 2, downstream speed = r + 2. Use day 1: distance upstream = 4(r-2), distance downstream = 5(r+2), total = 23.
Combine terms: 4r - 8 + 5r + 10 = 23 => 9r + 2 = 23 => 9r = 21 => r = 21/9 = 7/3 mph ≈ 2.33 mph.
Return is from camp to S, which is upstream (against current). Effective speed = r - 2 = 7/3 - 2 = 1/3 mph. Distance = 23 miles. Time = distance / speed = 23 / (1/3) = 69 hours.
Arrival at 18:00 on day 2. Subtract 69 hours: 69 hours = 2 days and 21 hours. 18:00 minus 21 hours = 21:00 previous day, then minus 2 days = 21:00 two days before arrival. So departure was at 21:00 on the day before day 2? Clarify: If arrival is day 2 at 18:00, subtract 69 hours: 18:00 - 69h = 18:00 - (48h + 21h) = 18:00 - 21h = 21:00 on day 0 (two days prior). So departure was at 21:00 two days before arrival.
Verify: 69 hours is about 2.875 days, so departure is indeed 2 days and 21 hours before 18:00, which is 21:00 two days earlier. Ensure no arithmetic errors.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.