# 200 ft race (with turnaround) where two people have different jump lengths but jump the same distance in equal time - who wins and by how much?

This problem is from Problem Solving Strategies - Crossing the River with Dogs and Other Mathematical Adventures by Ken Johnson and Ted Herr.

I decided to draw a picture from the segment of $90$ ft to $100$ ft of the race (the wall where they turn around). I chose to start at $90$ ft because both two and three go into 90 evenly, so neither Roo or Tigger are in the middle of a jump at that time.

Thus, at the $96$ foot mark, both Roo and Tigger are at the same spot. But, at the $98$ foot mark, Roo is about to jump to the $100$ ft mark and Tigger is mid-jump en route to the $99$ ft mark. So, when Roo jumps, he is at the $100$ ft mark and ready to turn around, while Tigger is on the $102$ ft mark and then waiting to turn back around.

I am having trouble trying to figure out how much Roo wins by then exactly. It's hard to tell from my picture, but I believe it would be $2$ feet. Is that correct, or does Roo possibly win by more? Any help would be appreciated.

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Roo travels $200$ feet to reach the finish line.

Since Tigger has to travel in $3$ foot jumps, Tigger needs to travel $102$ feet until (s)he can turn around. Then Tigger needs to travel another $102$ feet to reach the start/finish line, a total of $204$ feet. The two jumpers travel at the same rate. So Roo is $4$ feet ahead of Tigger at the end, sort of.

This is in fact not necessarily the case, because it relies on a specific model of jumping. Since we were not given a detailed description of the jump processes of either creature, we are forced to assume something. We implicitly assumed that even although the travelling is divided into jumps, the actual speeds are constant.

Suppose instead that a jump consists of standing in one place flexing the leg muscles for a while, and perhaps having a drink, while the movement part of a jump is quick. Then the answer need not be $4$ feet.

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I don't see how you got that Roo was four feet ahead. How do you see Tigger travels 204 feet? I only see him jumping 2 feet extra from the first turnaround, meaning he jumps 202 feet. –  Joe Mar 3 '12 at 2:51
He (or she) has to get back! –  André Nicolas Mar 3 '12 at 2:53
Ah, wow. Something so simple, yet I missed it; thanks. –  Joe Mar 3 '12 at 2:55
@Jay: Tigger turns round after running 102 feet, so he has to run another 102 feet to get back to the start; that’s a total of 204 feet. –  Brian M. Scott Mar 3 '12 at 2:56

You can also solve it by counting jumps. Roo reaches the halfway point in exactly $50$ jumps, so he requires $100$ jumps to finish the race. Tigger reaches the $99$ foot mark in $33$ jumps, then takes $2$ more jumps to cross the $100$ foot mark and return to the $99$ foot mark, and finally needs $33$ jumps to return from there to the starting line, for a total of $68$ jumps. At $3$ jumps per second (or whatever the basic unit of time is), Roo finishes in $\frac{100}3$ seconds. In $\frac{100}3$ seconds Tigger makes $2\cdot\frac{100}3=\frac{200}3$ jumps, which is $\frac43$ of a jump, or $4$ feet, short of the $68$ jumps that he requires. Assuming that his speed is constant, he’s still $4$ feet shy of the finish line when Roo crosses it.

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Nice alternative. –  Joe Mar 3 '12 at 3:13

Suppose that the unit of time mentioned in the question is 1 second (it doesn't really change the reasoning if it is another). Then we are told that Roo's speed is $$\frac{3\times 2 \text{ ft}}{\text{sec}} =\frac{6 \text{ ft}}{\text{sec}},$$ and that Tigger's speed is $$\frac{2\times 1 \text{ yard}}{\text{sec}} =\frac{2\times 3 \text{ ft}}{\text{sec}} =\frac{6\text{ ft}}{\text{sec}}.$$

The race is a tie.

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The fact that it says "up and back" shows that your model is not true because of how they have different jump lengths (and the turn is at 100 feet). –  Joe Mar 3 '12 at 2:48