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

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.)