Elementary POW, February 3-7,1997


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Dr. John Galt is studying six frogs he found in the swamp near Starnesville. These frogs are identical to normal frogs except that the distances they jump are determined by a pattern that is unique to each frog. For example, Frog A jumps 2 feet every time, but Frog B jumps 2 feet on odd numbered jumps and 3 feet on even numbered jumps.

Dr. Galt writes this in his notes as Frog A (2) and Frog B (2,3). The other frogs are in Dr. Galt's notes as follows: Frog C (1,5,2), Frog D (2,3,2,4), Frog E (4,3,1,3,2) and Frog F (1,1,1,1,1,4).

The cage in which the frogs are kept is outside Dr. Galt's lab. One night the cage door pops open. Two hours later, the open door is discovered. If each frog leaves the cage when the door pops open, jumps once every minute and starts by jumping the first jump in its pattern as its first jump out of the cage, which frog will be the greatest distance from the cage when their escape is discovered?

This week's mentors are Sandy Norman's graduate classes at the University of Texas at San Antonio.

Correct Solutions with good explanations were submitted by:

Highlighted Solutions

Bronwyn, Bethany, Robert, Seirra, Briana, Caitlin,
Kate, Valerie, Jane and Francesca
4th Grade - Ms. Wheeler
mvislh@warwick.net

We think that Frog D will be the farthest distance from the cage in 2 hrs. This is how we solved it.


Frog A (2)           60*2 = 120 feet in 1 hr
                      2 hours 120*2 = 240feet

Frog B (2,3)         (30*2) + (30*3) = 150 feet in 1 hr
                      2 hours = 150*2 = 300 feet

Frog C (1,5,2)       (20*1)+(20*5)+(20*2)= 160 feet in 1 hr
                      2 hours = 160*2 = 320 feet

Frog D (2,3,2,4,)    (15*2)+(15*3)+(15*2)+(15*4) = 165 feet in 1 hr
                      2 hours = 165*2 = 330 feet

Frog E (4,3,1,3,2)   (12*4)+(12*3)+(12*1)+(12*3)+(12*2) = 156 feet in 1 hr
                      2 hours = 156*2 = 312 feet

Frog F (1,1,1,1,1,4) (10*1)+(10*1)+(10*1)+(10*1)+(10*1)+(10*4) = 90ft/hr
                      2 hours = 90*2 = 180 feet
Grace McCants
5th grade at Emerson Elementary School
Madison, Wisconsin
(Mrs. Dunlap's class)

I think frog D got the greatest distance from the cage. It traveled 330 feet from the cage. Here's how I got it:

  1. The answer for frog A is 240 feet. I got that by multiplying 120 (minutes in 2 hours) by 2 (feet per jump) and got 240.

  2. For frog B, I got 300 feet. I divided 120 by 2 (because there were two numbers in the pattern for frog B) and got 60. That is the number of times he jumped 2 feet, and 3 feet. Then I multiplied 60x2 and 60x3 and added the two answers together and got 300.

  3. For frog C, I got 320. First I divided 120 by 3 (number in the pattern) and got 40. Then I multiplied 40x1, 40x5, and 40x2. Adding up these results, I got 320.

    I used the same method for the remaining frogs, dividing 120 each time by the number of jumps in the pattern, and then adding up the feet for each number in the pattern. Here are the results for all six frogs:

      A=240
      B=300
      C=320
      D=330
      E=312
      F=180

    So frog D got the furthest.

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