Cross that bridge
Three people (A, B and C) need to cross a bridge. A can cross the bridge in 10 minutes, B can cross in 5 minutes, and C can cross in 2 minutes. They have a bicycle and any person can cross the bridge in 1 minute using it. All three men and the bicycle start together on one side of the bridge. Only one person can use the bicycle at a time. The bicycle can be mounted and dismounted anywhere. All three men can be crossing together. What's the shortest time that all three men can cross the bridge in?
Labels: logic, mathschallenge
posted by Chris at
6:21 AM
11 Comments:
I've posted (on Sep 27, 5:07PM) a very nice explanation of the "more prisoner problems".
3.5 minutes.
A on bike and C walks. wait for C to reach other side. (2 mins)
C returns to center of bridge on bike. (30 Seconds)
B rides bike from there and C walks. (1 mins)
(in the 2.5 minutes it takes C to walk across and ride back half way, B will be half way across, so they meet exactly in the center)
Hi EB. They can do better than that. Greetz.
yep.. i have worked out all the numbers yet, but it's going to be pretty close to 3 mins (i think).
They all start together. A will ride somewhere about 76% to 79% of the way and walk the rest. When C reaches the bike, he'll ride back to B. B will then ride the rest and C will walk the rest. They will all arrive pretty close to the same time. I just haven't worked out the exact spot that A gets off the bike. 80% is too long and 75% is too short.
You're getting closer. As it's you, I'll give a clue. They can do it in better than 3 minutes.
General clue. Always use algebraic methods rather than guessed actual values if possible.
simple answer, if c rides the bike and everyone hops on the back it will only take them 2 minutes
Hi Anonymous. Ooops, I had accidentally deleted that exclusion when I posted. It would only have then taken 1 minute though, not 2.
All. No piggy-back rides etc.
I've just posted an excellent soluton to the "More Prisoner Problems" blog. The post is dated SEPT 28, 5:42 PM. I think that you'll like it.
In bridges per minute A's speed is 1/10, B's is 1/5, C's is 1/2, and the bicycle's is 1. It is fairly obvious that all three have to cross the bridge in the same time.
Again it is fairly obvious that A (the slowest) should take the bicycle first. He must dismount at some point y, and then walk the rest of the way. C will eventually get to the bike at y. He then rides it back to some point x, dismounts and walks the rest of the way. Eventually B gets to the bike at x. He then rides it the rest of the way.
Let t be the time taken for any/all to cross the bridge. Then:
(Using time = distance / speed).
For A: t = 1*y + 10*(1-y) = 10 - 9y
For B: t = 5*x + 1*(1-x) = 1 + 4x
For C: t = 2*y + (y-x) + 2*(1-x) = 2 + 3y - 3x
Solving for x and y => x = 12/25, y=59/75.
So t = 73/25 = 2.92 minutes.
A doesn't have to take the bicycle first, he could take it last and they still could all make it in 2.92 minutes.
Mo, well spotted (again). I hadn't tried that way, but you're right.
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