Riddle 10 points for first correct answer?!
Question: Riddle 10 points for first correct answer!?
There are 8 people standing on one side of a cliff - 3 Adults and 5 Children!. They all want to cross to the other side of the cliff using the connecting bridge!.
The problem is that the children are too scared to cross the bridge alone so they must be accompanied by an adult!. The Bridge itself can only carry the weight of 1 adult and 1 child at any one time!.
What is the minimum amount of "cross trips" required to completely transfer the 8 people!?
thanksWww@Enter-QA@Com
The problem is that the children are too scared to cross the bridge alone so they must be accompanied by an adult!. The Bridge itself can only carry the weight of 1 adult and 1 child at any one time!.
What is the minimum amount of "cross trips" required to completely transfer the 8 people!?
thanksWww@Enter-QA@Com
Answers:
5Www@Enter-QA@Com
For those ppl saying 4 trips READ THE QUESTION
the children are too scared to cross the bridge alone so they must be accompanied by an adult!.
I have worked out how to do it in 5 trips but as YourVoice is looking for people who can enter their competition I'm not telling you how it is done!.Www@Enter-QA@Com
the children are too scared to cross the bridge alone so they must be accompanied by an adult!.
I have worked out how to do it in 5 trips but as YourVoice is looking for people who can enter their competition I'm not telling you how it is done!.Www@Enter-QA@Com
7 I think!.
One adlut bring a child and so on and then all that adults cross!. So!. 1&1, 1&1, 1&1,1&1,1&1, 1,1!.
ummmmm!. ow my head!.Www@Enter-QA@Com
One adlut bring a child and so on and then all that adults cross!. So!. 1&1, 1&1, 1&1,1&1,1&1, 1,1!.
ummmmm!. ow my head!.Www@Enter-QA@Com
after further review I'm thinking 12
one child and one adult = 2 x3 = 6
then the adult going back to get child =1
child and adult going back = 2
then adult going back for last child =1
child and adult going back = 2Www@Enter-QA@Com
one child and one adult = 2 x3 = 6
then the adult going back to get child =1
child and adult going back = 2
then adult going back for last child =1
child and adult going back = 2Www@Enter-QA@Com
it's 10
1&1 x 3 = 6 adult comes back after the 3rd for the next kid thats 7, they both cross, thats 8, another adult comes back thats 9, and then they both cross again, making it 10Www@Enter-QA@Com
1&1 x 3 = 6 adult comes back after the 3rd for the next kid thats 7, they both cross, thats 8, another adult comes back thats 9, and then they both cross again, making it 10Www@Enter-QA@Com
7 > 3A take 3C first trip!. 2A go back (thats two plus the first 3=5)!. Then the 2A take last 2C (Add two more to 5) = 7Www@Enter-QA@Com
4 - each adult goes with one child, and 2 children go together without an adult!.Www@Enter-QA@Com
I believe that the minimum amount of cross trips is 4!.Www@Enter-QA@Com
Seven!.
Poor parent that decides to go a few times, unless of course they take turns!.Www@Enter-QA@Com
Poor parent that decides to go a few times, unless of course they take turns!.Www@Enter-QA@Com
4Www@Enter-QA@Com
7Www@Enter-QA@Com
4Www@Enter-QA@Com
4Www@Enter-QA@Com
7Www@Enter-QA@Com
10,0000000000000Www@Enter-QA@Com
5Www@Enter-QA@Com
4Www@Enter-QA@Com
7!?Www@Enter-QA@Com
Well that depends!.!. Are they fat!?Www@Enter-QA@Com
It's 4!.Www@Enter-QA@Com
!? i think it is 7!.
E xWww@Enter-QA@Com
E xWww@Enter-QA@Com
is it 7Www@Enter-QA@Com
it is 6Www@Enter-QA@Com
i'd say 11 times cross trips!.!. that's my count!.!.Www@Enter-QA@Com