1180.41 – Sherman and Cyrus


Sherman is chasing his pet frog Cyrus. Sherman takes 2 steps while Cyrus takes 3 jumps, but Sherman's steps are twice as long as the frog's jumps. The frog had taken 10 jumps before Sherman noticed and started after him. How many more jumps will Cyrus make before Sherman catches him?


Solution

Discussion: This problem illustrates the value of thinking visually as well as numerically.

Solving the Problem: One might well try to work out an equation for this problem, since both Sherman and Cyrus travel the same total distance. And, of course, it can be done. But this problem illustrates the value of making a sketch. Then the sketch can be labeled somehow to keep track of the action. This is best done by acting it out, using the drawn model. It turns out that Cyrus makes an additional 30 jumps.

Heuristics: There are several, working together: #6: make a model, which in this case is a sketch (#5: draw a picture), and #8: act the problem out. Or use #10: write an equation. Two solutions are illustrated in the figure below.

Using the Problem with Students:

a. as a class activity, with homework. I'll read the problem through with the class and ask, "How can all of this be organized? How can we figure out what happens?" Students might come up with the sketch idea, and/or the equations idea, and/or something else too. Once those leads are out there, whether the students or I provide them, I turn the class loose to work on it in class, probably in pairs, or overnight as homework.

The next day in class we'll go through whatever solution methods the students have come up with. I make sure to see that both approaches (as in the Solving section above) are displayed. It might be interesting to take a poll and see which students prefer which method.

Modifying or Extending the Problem: What if the numbers are different? Or, further, what if Sherman takes s steps while Cyrus takes c jumps? And what if Sherman's steps are n times a long as Cyrus's jumps? And what if Cyrus takes j initial jumps? Is there a general formula lurking here?

Similar Problem: 1100.51: Agatha the cow.

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