Monday, June 30, 2008

Catch That Bus!

The 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?


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

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

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

    11 am. :)

  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!

  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.

  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.

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

  8. This comment has been removed by the author.

  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.)

    11:00 am


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