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PondScm007 Registered User
Joined: 13 Oct 2002 Location: central nj Posts: 963
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Posted: Wed Mar 05, 2003 10:31 am Post subject: evil bridge problem |
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NJ has high-school standardized testing this week, and in the time after whilst being stuck sitting in class for liek a half hour waiting for everyone throughout the school to finish (gah!) somone told me this problem... allegedly it IS possible to do, but i cant figure it out...anyone have any better luck?
there are four people- 1, 2, 5, and 10. they have to all get across a bridge.
the bridge can only hold 2 people at a time. someone walking on the bridge must be
carrying a flashlight. the people's names correspond to how long it takes
them to cross (1 takes 1 minute to get across, 5 takes 5 minutes to get across etc)
they must all cross in 17 minutes. if two people cross the bridge, they move at
the slower person's rate (so if 1 and 5 went across, it would take 5 minutes)
how do they do it?
???????????????????????????? _________________ my pic host-age died!
BR, yo |
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Thunder Registered User
Joined: 08 May 2002
Posts: 77
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Posted: Wed Mar 05, 2003 10:45 am Post subject: |
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-10 and 1 (1 has flash light) cross together and 10's speed.
-1 turns back to start point and 5 begins to cross toward other side where 10 now is (both of course moving at 5's speed).
-1 turns back around with 2 in tow and bothe cross together at 2's speed.
10 + 5 + 2=17 minutes and all across.
_________________ I put on my robe and Wizard's hat!!! HARRRRR |
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The Nanite Server Admin
Joined: 26 Apr 2002 Location: lost Posts: 6606
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Posted: Wed Mar 05, 2003 11:06 am Post subject: |
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Why can't they all have flashlights?
1 and 10 cross at the same time = 10 minutes
2 goes alone = 2 minutes
5 goes alone = 5 minutes
total time = 17 minutes. _________________
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Usually Dead Registered User
Joined: 14 Jan 2003 Location: Salt Lake City, Utah Posts: 680
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Posted: Wed Mar 05, 2003 11:07 am Post subject: |
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There was another problem like this I heard about in grade school. It went something like this:
A farmer wants to take his dog, goose, and a head of lettuce across the river to sell at the marketplace. There's a small boat to get across river, but there's only enough room for the farmer and one of his items at a time.
The farmer can't leave the goose alone with the head of cabbage, becuase the goose will eat it. He can't leave the dog alone with the goose, becuase the dog will eat that poor bird. So how does he get himself and all his things across the river intact?
The answer was this:
The farmer takes the goose -----------> over the river and drops it off on the far side. It's okay to leave the dog alone with the cabbage since dogs are carnivores.
Then the farmer comes back <------------- and gets the dog. He takes the dog over ----------> to the far side of the river and drops the dog off, then gets the goose and comes back again <----------- to the side of the river he started on. Now the farmer, the goose, and the cabbage are on the side of the river he started on. The dog is alone on the far side.
The farmer takes the cabbage --------> over to the far side of the river and sets it next to the dog, leaving the goose alone. Then he comes back
<--------- and gets the goose, then goes to the far side one last time --------->. So now he finally has the dog, goose and cabbage all on the marketplace side of the river. Yay!
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sharkbyte Ville Supporter
Joined: 16 Feb 2002 Location: Massachusetts Guild: ={jFf}-USV= & Ville $upporter Posts: 2887
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Posted: Wed Mar 05, 2003 2:34 pm Post subject: |
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Thunder wrote: | -10 and 1 (1 has flash light) cross together and 10's speed.
-1 turns back to start point and 5 begins to cross toward other side where 10 now is (both of course moving at 5's speed).
-1 turns back around with 2 in tow and bothe cross together at 2's speed.
10 + 5 + 2=17 minutes and all across.
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You actually end up w/ 19 minutes. Each time #1 turns back, you need to add another minute for them to cross. _________________
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sharkbyte Ville Supporter
Joined: 16 Feb 2002 Location: Massachusetts Guild: ={jFf}-USV= & Ville $upporter Posts: 2887
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Posted: Wed Mar 05, 2003 2:38 pm Post subject: Re: evil bridge problem |
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PondScm007 wrote: |
there are four people- 1, 2, 5, and 10. they have to all get across a bridge.
the bridge can only hold 2 people at a time. someone walking on the bridge must be
carrying a flashlight. the people's names correspond to how long it takes
them to cross (1 takes 1 minute to get across, 5 takes 5 minutes to get across etc)
they must all cross in 17 minutes. if two people cross the bridge, they move at
the slower person's rate (so if 1 and 5 went across, it would take 5 minutes)
how do they do it?
???????????????????????????? |
The answer is simple, or else there is information missing....
First of all, there is nothing that says there is only 1 flashlight. So we can assume that they each have flashlights.
1 & 2 cross together. Total time > 2 minutes
5 crosses. Total time > 5 minutes
10 crosses. Total time > 10 minutes
2+5+10 = 17 minutes
Can anyone see what the shortest amount of time it would take, for all to cross?
_________________
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The Nanite Server Admin
Joined: 26 Apr 2002 Location: lost Posts: 6606
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Posted: Wed Mar 05, 2003 2:41 pm Post subject: Re: evil bridge problem |
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sharkbyte wrote: | PondScm007 wrote: |
there are four people- 1, 2, 5, and 10. they have to all get across a bridge.
the bridge can only hold 2 people at a time. someone walking on the bridge must be
carrying a flashlight. the people's names correspond to how long it takes
them to cross (1 takes 1 minute to get across, 5 takes 5 minutes to get across etc)
they must all cross in 17 minutes. if two people cross the bridge, they move at
the slower person's rate (so if 1 and 5 went across, it would take 5 minutes)
how do they do it?
???????????????????????????? |
The answer is simple, or else there is information missing....
First of all, there is nothing that says there is only 1 flashlight. So we can assume that they each have flashlights.
1 & 2 cross together. Total time > 2 minutes
5 crosses. Total time > 5 minutes
10 crosses. Total time > 10 minutes
2+5+10 = 17 minutes
Can anyone see what the shortest amount of time it would take, for all to cross?
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sure
1 & 2 = 2 minutes
5 & 10 = 10 minutes
total time = 12 minutes _________________
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sharkbyte Ville Supporter
Joined: 16 Feb 2002 Location: Massachusetts Guild: ={jFf}-USV= & Ville $upporter Posts: 2887
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Posted: Wed Mar 05, 2003 2:45 pm Post subject: Re: evil bridge problem |
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Bushido wrote: | sharkbyte wrote: |
Can anyone see what the shortest amount of time it would take, for all to cross?
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sure
1 & 2 = 2 minutes
5 & 10 = 10 minutes
total time = 12 minutes |
Hooray. You get a cookie. _________________
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Fred Astaire Registered User
Joined: 03 Jan 2003 Location: -=New York=- Guild: =B3-USV= Posts: 846
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Posted: Wed Mar 05, 2003 2:55 pm Post subject: |
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ok now lets assume they have one flashlight and have to share it.
1+2 cross= 2 mins
1 comes back= 3 mins
5+10 cross= 13 mins
this time 2 comes back= 15 mins
1+2 cross= 17 mins
_________________ for Blue |
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Thunder Registered User
Joined: 08 May 2002
Posts: 77
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Posted: Wed Mar 05, 2003 2:59 pm Post subject: |
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I believe 17 minutes. For this to be a problem at all I think you have to go with the "one flashlight".....
*****
Also, when one begins his trip back, 5 is going at the same time in the other direction:
5 --------->
1 <---------
So you only count 5 min., they are crossing at the same time....just in opposite directions. The whole idea is to get you to think of a clever way to solve the puzzle.
*****
Everyone having a flashlight doesn't make since....if evryone had one then why even mention it at all. _________________ I put on my robe and Wizard's hat!!! HARRRRR |
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Thunder Registered User
Joined: 08 May 2002
Posts: 77
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Posted: Wed Mar 05, 2003 3:05 pm Post subject: |
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Also before you write back and tell me that 5 will not be able to see because 1 is going the other direction with the flashlight....don't.
If you read the prob. it says: "someone walking on the bridge must be
carrying a flashlight". That is all it says about the flashlight....and it makes for a good clue as to the solution.
And that's about all i got 2 say abot that.
Your friend,
-Forest _________________ I put on my robe and Wizard's hat!!! HARRRRR |
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CyC0Dad Registered User
Joined: 19 Jul 2001 Location: Phoenix, AZ Posts: 1382
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Posted: Wed Mar 05, 2003 5:14 pm Post subject: |
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Fr3d Astaire wrote: | ok now lets assume they have one flashlight and have to share it.
1+2 cross= 2 mins
1 comes back= 3 mins
5+10 cross= 13 mins
this time 2 comes back= 15 mins
1+2 cross= 17 mins
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That is the corect answer. Well actually there are 2 variations of the answer. The trick is both 5 and 10 must cross the bridge together only 1 time, so someone faster must cross first in order to bring the flashlight back across. Congrats to Fr3d.....
Other variation:
1+2 ---> 2 mins
2 <------ 4 mins
5+10 --> 14 mins
1 <------ 15 mins
1+2 ---> 17 mins |
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PondScm007 Registered User
Joined: 13 Oct 2002 Location: central nj Posts: 963
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Posted: Wed Mar 05, 2003 5:17 pm Post subject: |
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neg you can only be walking if you have the flashlight, or the person next to you has the light.. and i found it out a bit later, i gave in and begged for the answer...gj fred and dad _________________ my pic host-age died!
BR, yo |
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Replica Server Admin
Joined: 01 Jul 2002 Location: i'm a loner dottie, a rebel! Posts: 6144
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Posted: Mon Mar 10, 2003 12:51 pm Post subject: |
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this reminds me ( not near as hard as) of the ap calc test last year... man some fo those problesm sucked. got a lousy 3 on it, lol. but not that this school takes them anyways _________________ |
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