Monday, June 30, 2008

Catch That Bus!

Express BusThe local bus leaves Ashwood at 9:21 am and arrives in Baytree at 12:06 pm on the same day. The express bus leaves Ashwood at 10:00 am, traveling the same route, and arrives in Baytree at 11:40 am. At what time does the express bus pass the local bus if each is traveling at a constant speed?

9 comments:

  1. 11:39am by my calculations
    meeting at exactly 3/5 of the way to Baytree

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  2. Your offset is correct, and your proportion is correct, but you made one minor error.

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  3. Oops! heh, 99 minutes from the wrong bus.

    11 am. :)

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  4. Bummer. I wish I hadn't read those Posts. I was so close as I had 39/65 and 26/65 written down (but no reduced) and I knew they would get me to the proper offset but I hadn't 100% settled on how to apply them. Now I know where my logic should have lead me but now I can never be sure. So it goes. Maybe next time.

    Blaine: Do you have a distro list set up to inform others when you put up a new puzzler (other than Cartalk and WESUN ones, I catch them via Broadcast). If you do, I'd love to be on it. I promise not to post any premature spoilers again!

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  5. I don't have an e-mail distribution when I post, but you can subscribe to the RSS feed. Unfortunately you'll get the NPR posts too, but just ignore them if you like.

    I don't prohibit spoilers in my puzzles, but it seems like people are good about trying to wait a few days before posting an outright answer. And you can always just not read the comments.

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  6. When will you give the solution to the Catch That Bus puzzle? I think I have the correct answer but do not want to post it is respect for others still solving it.

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  7. A correct answer has been posted by HappySteve...

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  8. This comment has been removed by the author.

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  9. Here's how I solved it. The slow bus takes 165 minutes. The fast bus takes 100 minutes. The slow bus has a 39 minute headstart.

    Take T to be the number of minutes since the *fast* bus left. After T minutes it will be T/100 of the way to Baytree. Accounting for the headstart, the slow bus will be (T+39)/165 of the way to Baytree. Now you can find the point where they have covered the same fraction of the distance by equating these:

    T/100 = (T+39)/165

    Cross multiply and you have:
    165T = 100(T+39)
    165T = 100T + 3900
    165T - 100T = 3900
    65T = 3900
    T = 3900/65
    T = 60

    In other words, 60 minutes after the fast bus leaves, it will have covered 60/100 (3/5) the distance and caught up with the slow bus.

    (Note: you can also solve this for the number of minutes since the slow bus left in which case you get T = 99 as the offset from the time the *slow* bus started.)

    Answer:
    11:00 am

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