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Running Laps

Date: 07/03/2002 at 12:29:55
From: Jiahe
Subject: Math Olympiad Contest Problem

Hello,

My dad bought me a Math Olympiad book which he says would help my 
problem solving skills.  Right now, I am stuck on one of the 
problems.  

Two dogs run around a circular track 300 feet long.  One dog runs at 
a steady rate of 15 feet per second.  The other dog runs at a steady 
rate of 12 feet per second.  Suppose they start at the same time and 
point.  What is the least number of seconds that will elapse before 
they are again together at the starting point?

The answer in the book says that it will be 100 seconds.  The part 
that I don't understand is why it is 100 seconds because the problem 
only mentioned that the dogs start at the same point and time but 
never mentions how many laps they run.  

Thank you so much for your time and consideration for reading this!

-Jiahe


Date: 07/03/2002 at 14:09:46
From: Doctor Ian
Subject: Re: Math Olympiad Contest Problem

Hi Jiahe,

A dog that runs at 15 feet per second will go once around the 
track in 300/15 = 20 seconds.  Does that make sense?  

So we can make a table of the times when he'll be back at the 
starting point:

        Dog 1
     -------------
     lap   seconds
     ---   -------
      1      20
      2      40
      3      60
      4      80

Now, if the other dog runs at 12 feet per second, how many 
seconds will it take to run around the track?  (I'll leave that 
for you to figure out.)  

We can use that to fill in the information for that dog in our 
table:

        Dog 1          Dog 2
     -------------  -------------
     lap   seconds  lap   seconds
     ---   -------  ---   -------
      1      20      1        ?
      2      40      2      2*?
      3      60      3      3*?
      4      80      4      4*?

If you keep adding rows to the table, you'll find some number of 
seconds for which they're BOTH back at the starting line.  Go 
ahead and try that, and compare what you get with the answer from 
the book. 

For another way to look at this kind of problem, see

  Laps and Least Common Multiples
  http://mathforum.org/library/drmath/view/60770.html 

Does this help? 

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 


Date: 07/04/2002 at 12:19:20
From: Jiahe
Subject: Thank you (Math Olympiad Contest Problem)

Thank you so much!!  The problem is much clearer to me now.


Date: 07/04/2002 at 12:52:51
From: Doctor Anthony
Subject: Re: Math Olympiad Contest Problem

An easy way to answer this is to impose a speed of -12 feet per 
second on both dogs. That means that one dog is stationary and the 
other is moving at 3 ft/sec.  The moving dog has to cover 300 feet to 
get back to the start and time = Distance/speed = 300/3 = 100 secs.

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Basic Algebra
High School Puzzles

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