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Gomez
candid chimera

 Posted: Thu Aug 05, 2010 8:37 pm    Post subject: 1 Was on the train home the other day and I found this puzzle in the paper. Now, maybe I was having a slow day but I just couldn't work it out. Maybe you guys will have more luck. A cyclist is riding from town A to town B. On the first day he covers one fifth of the total distance. On the second day he covers one third of the remaining distance. On the third day he covers one quarter of the remaining distance. On the fourth day he covers half of what's left. He now has 25 miles left to ride. How far apart are A and B? For the sake of my self-esteem, I politely request you refrain from submitting any answers within the next four minutes.
Internet Stranger
Paragon of Mafia Hunters

 Posted: Thu Aug 05, 2010 9:03 pm    Post subject: 2 The closest I could get is 120, but that put me at 24 to go, not 25._________________"I didn't know she was your sister I swear!" www.InternetStranger.com
Quailman
His Postmajesty

 Posted: Thu Aug 05, 2010 9:18 pm    Post subject: 3 I get 125, if I backed into it correctly. Each day he covered 25, 33 1/3, 16 2/3 and 25, leaving 25.
Zag
Unintentionally offensive old coot

 Posted: Thu Aug 05, 2010 9:26 pm    Post subject: 4 I suspect that you've misquoted it, that it is supposed to be this chestnut: 1st day 1/5 distance. 2nd day 1/4 remaining 3rd day 1/3 remaining 4th day 1/2 remaining 25 miles left. In THIS version of the puzzle, all the legs were the same distance, 25 miles, so the total was 125 miles. ------------ For your version of the puzzle: At the start of the 4th day, he had 50 miles to go (because going half left 25 miles) At the start of the 3rd day, he had x miles to go, where x = x/4 + 50 = 200/3 miles At the start of the 2nd day, he had y miles to go, where y = y/3 + 200/3 = 100 miles At the start of the 1st day, he had z miles to go, where z = z/5 + 100 = 125 miles. Interesting that it does work out to be the same. I guess I shouldn't have been surprised.
groza528
No Place Like Home

 Posted: Thu Aug 05, 2010 9:37 pm    Post subject: 5 Indeed it will always work out the same regardless of the order. The algebra works out so from the original: T = 25 * (2/1) * (3/2) * (4/3) * (5/4) = 125 In Gomez's version: T = 25 * (2/1) * (4/3) * (3/2) * (5/4) = 125 Organized in that fashion it should be clear that regardless of the order the fractions come in, they will always yield the same result (though with different lengths for the legs, as you pointed out)
L'lanmal
Daedalian Member

 Posted: Mon Aug 09, 2010 2:12 am    Post subject: 6 Maybe he was a commuter.
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