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Date: 09/02/2000 at 19:19:37 From: Lucas Marshall Subject: Pre-Calculus Word Problem, but it seems like Algebra My problem is a word problem. You've answered a problem kind of like this already, but mine has different information and different nuances. I have an answer, but I think my logic is too simple. Here's the problem. A column of soldiers 25 miles long marches 25 miles a day. One morning, just as the day's march began, a messenger started at the rear of the column with a message for the man at the front of the column. During the day he marched forward, delivered the message to the first man in the column and returned to his position just as the day's march ended. How far did the messenger march? My answer is 50 miles, but it seems too easy. I find it odd that there was no speed given for either the column or the messenger. How would you solve this? Thanks a lot.
Date: 09/02/2000 at 23:14:35 From: Doctor Peterson Subject: Re: Pre-Calculus Word Problem, but it seems like Algebra Hi, Lucas. Often a simplistic line of reasoning can solve a problem that is hard to solve by more advanced reasoning. If you'd told me your reasoning, you might have convinced me, though I have a feeling you are in fact wrong. Let's see how I would do it. I like to picture this sort of thing in a graph of position vs. time. The front and rear of the column will be represented by lines with the same slope; the messenger must be going faster, starting at the rear line, meeting the front line, and then returning (I presume at the same speed) to the rear line: 50+ * front | +* | */ \ | * / \ messenger | * / \ * 25+ / + rear | / * | / * | / * |/ * 0+--------------+ 0 1 The first thing I see is that the messenger wasn't going twice as fast as the column (which is going 25 miles per day - its speed IS given, though in rather vague terms with regard to time); if he were, he would have gone 50 miles in a day, and would have met the front just as they stopped, with no time to get back. I also see that he would have gone 50 miles if he had met them halfway along their march, going 37.5 miles forward and 12.5 miles back. But then he clearly wouldn't have gone the same speed both ways. So your answer is wrong. I'll just suggest where I would go from here. Suppose he meets them at time x (measuring time as a fraction of a "day," however long their march takes). Where will he meet the front of the line? What is his speed going forward? What is it going back? Set those expressions equal and solve for x. This is a nice problem, with an interesting solution. If you need more help, write back and show me how far you got. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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