Correct solutions were also submitted by Ray Kremer, Mahendra Nambiar, Angela Bates, Nathan Pauli, Tracy Thatcher, Courtney Pelowski.  From outside the Bradley University community, correct solutions were submitted by Aaron David Kahn, Emanuele Macri, Lorenzo Pozzoli, Massimo Brigogne, Edward Lee, Monty Gray, Burkart Venzke, Andrey Mamontov, Maxsims Ovsjanikovs, Tim Kelley, Cyril Terakopiantz, Philippe Fondanaiche, Jongmin Baek, Brian Laughlin, Denis Borris, and one other solver.

There are several possible equations to set up.  The following is from Edward Lee, a high school student.

Let c be the speed of the current, b the speed of the boat (in still water).  The time it took for the hat to float back to its starting point is 6/c.  The distance the boat traveled (upstream) in the two hour period after losing the hat is 2(b - c) miles, so the total upstream distance traveled is 6 + 2(b - c).  During the return trip the boat was moving at a rate of b+c so the time from when the hat was lost until the return of the boat to the starting point was [6 + 2(b - c)]/(b + c) + 2.  Setting this quantity equal to 6/c and solving gives c = 1.5, independent of b.

Here's another solution, this one equation-free, sent in by Tim Kelley.

Relative to the water, the hat stays still.  So, again relative to the water and hat, the rower moves just as fast upstream as downstream.  After the hat dropped, the rower rowed for four hours and the hat moved six miles, so the current is flowing at a rate of 6/4 = 1.5 miles per hour.

You are visitor number  4164 to this page.
Page last updated 3 April 2000.

ã2000 Alberto L. Delgado