Posted - 02/15/2012 : 17:16:51 I am having trouble setting this problem up. So far I have x=walking speed, 15x=train and x/2=oxcart. Where do I go from here if this is correct. No distance given or times. Once upon a time, a traveler was heading home from a large city. She went halfway by train, at a speed 15 times as fast as she could walk. She then rode the rest of the way in the back of an oxcart. Because of the rough road, she could have walked twice as fast as the oxcart. Would she have saved time if she had traveled the entire way on foot? Thanks

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the_hill1962

Posted - 02/16/2012 : 13:38:17 Yes, you are correct in what you are thinking so far. I might suggest to use "r" instead of "x" since "r" is what we usually use for speed (rate). It would be tough to know how to continue if you you didn't know how to work with the d=rt equation. You didn't mention it in your post so I am wondering if you are familiar with it. You should know that distance = rate x time. This equation is equivalent to r=d/t (that is rate = distance time and also t=d/r (that is time = distance rate. Hopefully you have seen these also and know how they are arrived at. If not, please ask us for clarification. Now, you problem can be solved using a couple different approches. The easiest, I think, is to use t=d/r since the question is asking about how much "time". Of course, if she was to walk the whole distance, t=d/r. So the problem consists of trying to figure out if the time on the train plus the t on the oxcart is more than d/r Remember, each part is half the distance. So, the time on the train is t=0.5d/15r and the time on the oxcart is t=0.5d/0.5r Add up the two times and see if it is more or less than just d/r. It is interesting to note (and sort of a hint here) that the time on the oxcart (0.5d/0.5r) simplifies to d/r itself! You may not find that interesting but it makes sense when you realize that since the oxcart is moving half as fast as walking BUT she is just going half the distance, it comes out that the time would just be the same to do that as traveling twice as fast as the oxcart twice that distance it traveled. With that hint, the answer makes sense. Let us know if you can't get the answer.