92xj Posted October 12, 2010 Report Share Posted October 12, 2010 The 5 members of the Family family and their 5 dogs (1 for each family member) were hiking when they encountered a river to cross. They rented a boat that could hold 3 living things, people or dogs. Unfortunately, the dogs were tempermental. Each was comfortable only with it;s owner present and could not be near another person, not even momentarily, without its owner present. The dogs could be with other dogs however, without their owners. The crossing would have been impossible except that Lisa's dog could drive the boat. How was the crossing arranged? How many trips did it take? Quote Link to comment Share on other sites More sharing options...
Dawg Posted October 12, 2010 Report Share Posted October 12, 2010 Is this mutiple choice? Quote Link to comment Share on other sites More sharing options...
gzilla45 Posted October 12, 2010 Report Share Posted October 12, 2010 I come up with 5 round trips so long as Lisa's dog does all the driving. Lisa's dog takes person 1 with dog 1 and so on until just Lisa is on the other side alone. Her dog then goes to pick her up last. The dogs can be by other dogs but not people. That was a fun way to kill a lunch break. Quote Link to comment Share on other sites More sharing options...
buckee Posted October 12, 2010 Report Share Posted October 12, 2010 (edited) It took 4 crossings 1- Lisa's dog took 2 other dogs across (3 across) 2- one family member took their dog across (2 across) 3- another family member took their dog across (2 across) 4- then Lisa took the other 2 family members across (3 across) If I'm wrong, just shoot me ..LOL:bang: Edited October 12, 2010 by buckee Quote Link to comment Share on other sites More sharing options...
gzilla45 Posted October 12, 2010 Report Share Posted October 12, 2010 It took 4 crossings 1- Lisa's dog took 2 other dogs across (3 across) 2- one family member took their dog across (2 across) 3- another family member took their dog across (2 across) 4- then Lisa took the other 2 family members across (3 across) If I'm wrong, just shoot me ..LOL:bang: This would work so long as there is a rope that they can bring the boat back across the river with. Quote Link to comment Share on other sites More sharing options...
SourthenILdeerhunter Posted October 12, 2010 Report Share Posted October 12, 2010 :gun2: It took 4 crossings 1- Lisa's dog took 2 other dogs across (3 across) 2- one family member took their dog across (2 across) 3- another family member took their dog across (2 across) 4- then Lisa took the other 2 family members across (3 across) If I'm wrong, just shoot me ..LOL:bang: Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 I come up with 5 round trips so long as Lisa's dog does all the driving. Lisa's dog takes person 1 with dog 1 and so on until just Lisa is on the other side alone. Her dog then goes to pick her up last. The dogs can be by other dogs but not people. That was a fun way to kill a lunch break. Lisa's dog can not be with another dog without Lisa with it. Your idea does not work. Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 It took 4 crossings 1- Lisa's dog took 2 other dogs across (3 across) 2- one family member took their dog across (2 across) 3- another family member took their dog across (2 across) 4- then Lisa took the other 2 family members across (3 across) If I'm wrong, just shoot me ..LOL:bang: It would be a long shot from here. Quote Link to comment Share on other sites More sharing options...
gzilla45 Posted October 12, 2010 Report Share Posted October 12, 2010 Lisa's dog can not be with another dog without Lisa with it. Your idea does not work. Crap. Back to the drawing board. :hammer1: Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 Crap. Back to the drawing board. :hammer1: haha, i spent the last hour figuring it out. draw it out. use A,B,C,D,E for people and a,b,c,d,e for dogs. Also, work backwards...... Quote Link to comment Share on other sites More sharing options...
gzilla45 Posted October 12, 2010 Report Share Posted October 12, 2010 I was using 1d and 1p which is the same thought path but just a little harder to read. My new answer is 9 total trips. I had to steal some of Buckee's logic but I think this works now. If not, I'm going to need a new pen and pad of paper to draw this out on. 1. Lisa's dog takes 2 dogs. 2. lisa's dog comes back. 3. Lisa's dog takes the other 2 dogs. 4. Lisa's dog comes back. 5. Lisa and her dog cross. 6. Lisa comes back. 7. Lisa takes 2 people. 8. Lisa comes back. 9. Lisa takes the last 2 people. Quote Link to comment Share on other sites More sharing options...
cinch314 Posted October 12, 2010 Report Share Posted October 12, 2010 1. AaB 2. B comes back 3. BbC 4. C comes back 5. CcD 6. D comes back 7. Dde 8. Aa comes back 9. AaE Ok so I dont have the dog driving but he is with Lisa the whole time. Lisa's dog can not be with another dog without Lisa with it. Your idea does not work. Quote Link to comment Share on other sites More sharing options...
Ohiobucks Posted October 12, 2010 Report Share Posted October 12, 2010 Are any of the siblings identical twins? Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 I was using 1d and 1p which is the same thought path but just a little harder to read. My new answer is 9 total trips. I had to steal some of Buckee's logic but I think this works now. If not, I'm going to need a new pen and pad of paper to draw this out on. 1. Lisa's dog takes 2 dogs. 2. lisa's dog comes back. 3. Lisa's dog takes the other 2 dogs. 4. Lisa's dog comes back. 5. Lisa and her dog cross. 6. Lisa comes back. 7. Lisa takes 2 people. 8. Lisa comes back. 9. Lisa takes the last 2 people. Step 7. two people can not be with 5 dogs. New sheet of paper for you. Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 1. AaB 2. B comes back 3. BbC 4. C comes back 5. CcD 6. D comes back 7. Dde 8. Aa comes back 9. AaE Ok so I dont have the dog driving but he is with Lisa the whole time. Step 3. Little c can not be left alone from C while being near other people. Start over. Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 Are any of the siblings identical twins? nope, there is no hidden things like the path you are thinking down. It is like the riddle of having a 3 gallon bucket and a 5 gallon bucket but you have to end with exactly 4 gallons of water. Quote Link to comment Share on other sites More sharing options...
gzilla45 Posted October 12, 2010 Report Share Posted October 12, 2010 When I first started this riddle I thought it was going to be fun. Now it has consumed most of my Tuesday afternoon and caused mass frustration!! Quote Link to comment Share on other sites More sharing options...
doubledrop Posted October 12, 2010 Report Share Posted October 12, 2010 OK WHO FARTED????? Just kidding i needed a laugh because this is was over my head and its giving me a head ache. Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 When I first started this riddle I thought it was going to be fun. Now it has consumed most of my Tuesday afternoon and caused mass frustration!! This is good for the brain. Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 OK WHO FARTED????? Just kidding i needed a laugh because this is was over my head and its giving me a head ache. You really dont want to know where this question came from. Quote Link to comment Share on other sites More sharing options...
gzilla45 Posted October 12, 2010 Report Share Posted October 12, 2010 You really dont want to know where this question came from. I'm guessing a 5th grade math book. :bang: Quote Link to comment Share on other sites More sharing options...
doubledrop Posted October 12, 2010 Report Share Posted October 12, 2010 probably hunter109 Quote Link to comment Share on other sites More sharing options...
92xj Posted October 12, 2010 Author Report Share Posted October 12, 2010 I'm guessing a 5th grade math book. :bang: College finite math. But dont let that turn you away. Its a fairly easy problem that 4th graders have figured out. probably hunter109 See above, I doubt it. Quote Link to comment Share on other sites More sharing options...
doubledrop Posted October 12, 2010 Report Share Posted October 12, 2010 thats why I cant get it im still in 3rd grade. Quote Link to comment Share on other sites More sharing options...
cinch314 Posted October 12, 2010 Report Share Posted October 12, 2010 Ok let me try again 1. abc 2. a goes back 3. ad 4. a goes back 5. BCD 6. Dd go back 7. Aa 8. Cc go back 9. CDE 10. a goes back 11. acd 12. a goes back 13. ae Quote Link to comment Share on other sites More sharing options...
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