Ferry Boat Problem

Two ferry boats serve the same route on a river, but travel at different speeds. They depart from opposite ends of the river at the same time, meeting at a point 720 yards from the nearest shore.

When each boat reaches the other side, it takes 10 minutes to unload and load passengers, then begins the return trip. This time, the boats meet at a point 400 yards from the other shore.

How wide is the river?

The ferry boat problem is created by well-known puzzle author Sam Loyd.


At the first meeting point, the total distance traveled by both boats is the width of the river. When the boats meet again, they’ve each traveled the full width of the river individually, plus another width of the river combined – so they’ve traveled 3 widths of the river combined at that point, thus each traveling 3 times as far as when they first met.

So at this second meeting point, one boat has traveled 3 x 720 yards, leaving it 400 yards away from shore. So the width of the river is the distance traveled by that boat at the second meeting point minus the extra distance it is from the shore: 3 x 720 – 400 = 1,760.

Alternatively, you can set up a proper system of 4 equations that gives you the same result. You may notice that the 10 min loading time is a complete red herring, it does not factor into the calculation at all.


  1. Boat A Boat B
    first trip 720 width of river (wr -720)
    second trip (wr-720) +400 720 + (wr-400)

    3wr = 720 + (wr-720) + (wr-720) + 400 +720 + (wr-400)

    How do you get 3×720 -400?

    • Boat A first trip 720 Boat B first trip=width of river (wr -720) Boat A second trip= (wr-720) +400 Boat B second trip=720 + (wr-400) So 3wr = 720 + (wr-720) + (wr-720) + 400 +720 + (wr-400) How do you get 3×720 -400?

      • Unfortunately the equation you set up doesn’t tell us anything – if you simplify it, you get 0 = 0.

        To get 3×720 – 400:
        1. When the boats first meet, they have traveled WR combined (that’s how they met) and Boat A traveled 720.
        2. When the boats meet again, they have traveled 3WR combined. The key is, this means each boat traveled 3 times as far as when they met the first time. So Boat A has traveled 3×720 total.
        3. Also, just by following Boat A, we know it has traveled WR + 400 by the 2nd time the boats meet. So:
        WR + 400 = 3×720
        WR = 3×720 – 400

  2. This solution assumes that the slower boat made it to the other shore and turned around. It is possible, however, that it was too slow to do this.

    • What do you mean? The boats travel at constant speed. When they meet the 1st time, they have traveled some initial distance individually, but they have traveled 1WR combined. When they meet the 2nd time, they have traveled 3WR combined, so if their speed is constant, they must have also traveled 3 times the initial distance, individually.

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