Running LapsDate: 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/ |
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