Logic Question

[DeathStalker908]

This problem has been bothering me for a long time, and I want to see if anyone else is able to find a solution. Here it goes...

You have 12 metal balls, and one of them is either heavier or lighter than the rest (you don't know which). If you have an equal-arm balance scale (the one that if you put weight on one side, that side gets pushed down, and the other side gets lifted up) that you can use only 3 times, how do you figure out which ball is lighter/heavier than the others?

If you know the answer, or are close to it, post here listing your 3 steps.

Good luck =)

[supersimons]

nice one. I am still stumped.

[DeathStalker908]

nice one. I am still stumped.

Don't worry, I was too when I first heard that question.

Is anyone else up to answer this problem?

[Skyre]

Easy - I'm a logical thinker.

Step 1) Put 6 balls on each side.
Step 2) Take the 6 balls that were on the heavier side and split them up. Place those on each side (3 and 3)
Step 3) Take the 3 balls that were on the heavier side and put 2 on the scale.

Problem solved. If one side goes down, it's the heavier one. If neither goes down, the one you didn't put on is the heavier one.

---

Got anymore?

[DeathStalker908]

Easy - I'm a logical thinker.

Step 1) Put 6 balls on each side.
Step 2) Take the 6 balls that were on the heavier side and split them up. Place those on each side (3 and 3)
Step 3) Take the 3 balls that were on the heavier side and put 2 on the scale.

Problem solved. If one side goes down, it's the heavier one. If neither goes down, the one you didn't put on is the heavier one.

---

Got anymore?

I don't think you fully understood the question, lol.
I said the ball can be EITHER LIGHTER OR HEAVIER..if you weigh 6 balls on each side, one of the sides will ALWAYS go up while the other one ALWAYS goes down. And since you DONT know if that ball is lighter or heavier than the rest, you won't know which side to choose. So, your first step is basically a waste of a turn :P. Get the question now? If you don't make another post...and dw, other people told me to repeat the question several of times.

And yes, I have a couple other logic problems which aren't nearly as brutal as this one..if you guys think this one is too hard, I can ask the other ones :P Just make a post to let me know.

PS: Thanks for trying though :P

[Piane]

umm 6 balls on each side u dont know whats heavy or light....only thing that comes to mind, is to just do it. put 6 on 1, 6 on another, and hopefully its 1 side up, 1 side down. if ur wrong, then 1 turn down, switch 1 of the balls out from side to side lol


unless im missing the whole concept of having 6 on each side, where as thats ur 1st turn and if u have to have 1 side up/down, why not take all the balls and put em on the same side....problem also solved =/


ffs gah guess my 3rd option is, dont use any of the balls. it doesnt say anywhere that the balls are already on the scale to start with @_@

[DeathStalker908]

umm 6 balls on each side u dont know whats heavy or light....only thing that comes to mind, is to just do it. put 6 on 1, 6 on another, and hopefully its 1 side up, 1 side down lol

Umm, if you weigh both sides with 6 balls at the same time, there is a 100% chance that one of the sides will be higher than the other one (which is lower). Think again :D

[Skyre]

Hmm...

I could easily do it in 4 steps ... but 3 ...

I'll keep working on it...

With the first step, I can assuredly cut the balls in half.

Step 1) Put 3 balls on each side. If the scale reads = then it's not among those 6, so remove those 6. OR if the scale reads =\= then it is among those 6, so remove the non-used 6.

Is that the correct first step? Well, you don't need to tell me, but it seems a good choice.

Random guess: You can't just hold 1 ball in one hand and another in the other hand and check that way? (Too many variables, how big, what's the weight difference.)

[pimpin002andhalf]

1-Ok here is the answer. Devide them into two groups of 6. Place group 1 on the scale(3 on each side) if they are level then put group one to the side. If not the odd ball is in group 2.

2-Take the last 6 and split them into 3 groups of two. Place each group on the scale until one group is uneven.

3-Take the group of uneven balls and split them into two groups. A & B. Take any of the other balls and weigh them with either A or B. If the scale is uneven it will not only tell you what group the odd ball is in. It will also tell you if its heavier or lighter.(Please don't make me go anymore in depth If you need me to because you don't understand any of this you don't need to be doing logic problems.)

[pimpin002andhalf]

1-Ok here is the answer. Devide them into two groups of 6. Place group 1 on the scale(3 on each side) if they are level then put group one to the side. If not the odd ball is in group 2.

2-Take the last 6 and split them into 3 groups of two. Place each group on the scale until one group is uneven.

3-Take the group of uneven balls and split them into two groups. A & B. Take any of the other balls and weigh them with either A or B. If the scale is uneven it will not only tell you what group the odd ball is in. It will also tell you if its heavier or lighter.(Please don't make me go anymore in depth If you need me to because you don't understand any of this you don't need to be doing logic problems.)

Got anymore? I eat Logic for breakfast.

[Skyre]

1-Ok here is the answer. Devide them into two groups of 6. Place group 1 on the scale(3 on each side) if they are level then put group one to the side. If not the odd ball is in group 2. Yup, easy to get to 6 balls.

2-Take the last 6 and split them into 3 groups of two. Place each group on the scale until one group is uneven. This would require you to use the scale more than the allotted 2 remaining times

3-Take the group of uneven balls and split them into two groups. A & B. Take any of the other balls and weigh them with either A or B. If the scale is uneven it will not only tell you what group the odd ball is in. It will also tell you if its heavier or lighter.(Please don't make me go anymore in depth If you need me to because you don't understand any of this you don't need to be doing logic problems.) So, this would also go above the 3 max uses of the scale.



Got anymore? I eat Logic for breakfast.

You should chew slower.

[pimpin002andhalf]

You should chew slower.

Didnt see that part. Only the part stating 3 steps.

[DeathStalker908]

1-Ok here is the answer. Devide them into two groups of 6. Place group 1 on the scale(3 on each side) if they are level then put group one to the side. If not the odd ball is in group 2.

2-Take the last 6 and split them into 3 groups of two. Place each group on the scale until one group is uneven.

3-Take the group of uneven balls and split them into two groups. A & B. Take any of the other balls and weigh them with either A or B. If the scale is uneven it will not only tell you what group the odd ball is in. It will also tell you if its heavier or lighter.(Please don't make me go anymore in depth If you need me to because you don't understand any of this you don't need to be doing logic problems.)

So close...there's just a problem in step 2...so in step one, you eliminate half of the balls, so you're left with 6. And in step two, you separate that 6 into 3 groups (of 2). Then you said "Place each group on the scale until one group is uneven.." which can take up to 3 turns and is something you CANT do because of your weighing limit. You might've made a typo or something...because this wrong :( (however, this is probably the closest anyone has ever gone to).

PS: Good try though :D

EDIT: Gah, someone already made a post correcting you while I made this one, lol.

[pimpin002andhalf]

To me it could be done with a continued use of the scale. You know not taking all the weights off.

[pimpin002andhalf]

To me it could be done with a continued use of the scale. You know not taking all the weights off.

Which would not separate the uses of the scale.

[DeathStalker908]

Got anymore? I eat Logic for breakfast.

I'm just gonna go ahead and ask other ones..this one is obviously tough as ****

You have two buckets: a 3-gallon bucket, and a 5-gallon bucket. If you have an infinite amount of water to pour into those buckets, and if you can pour water from one bucket to another or empty out a bucket, how do you get EXACTLY 4 gallons?

PS: You can't say something like "you fill in half of the bucket.." or any fraction like that...cause everyone said that too, lol. Then, when I told them they couldn't do that, they say "you just eyeball it then!"

PSS: You were probably thinking the same thing in the above statement.

PSSS: This question is very easy, rofl.

[pimpin002andhalf]

Or place all the balls on one after another until balanced.

[pimpin002andhalf]

I'm just gonna go ahead and ask other ones..this one is obviously tough as ****

You have two buckets: a 3-gallon bucket, and a 5-gallon bucket. If you have an infinite amount of water to pour into those buckets, and if you can pour water from one bucket to another or empty out a bucket, how do you get EXACTLY 4 gallons?

PS: You can't say something like "you fill in half of the bucket.." or any fraction like that...cause everyone said that too, lol. Then, when I told them they couldn't do that, they say "you just eyeball it then!"

PSS: You were probably thinking the same thing in the above statement.

PSSS: This question is very easy, rofl.

Fill up the 3 gallon. Dump it into the 5. Fill up the 3 gallon again and dump what you can(2 gallons) in the 5 gallon bucket. Leaving 1 gallon left in the 3 gallon bucket. Dump the five gallon bucket out. Dump the remainder of the 3 gallon bucket into the 5 gallon jug(which was one gallon). fill up the 3 gallon jug one more time and dump it into the 5 gallon. Wala! 4 gallons.

[DeathStalker908]

Hmm...

I could easily do it in 4 steps ... but 3 ...

I'll keep working on it...

With the first step, I can assuredly cut the balls in half.

Step 1) Put 3 balls on each side. If the scale reads = then it's not among those 6, so remove those 6. OR if the scale reads =\= then it is among those 6, so remove the non-used 6.

Is that the correct first step? Well, you don't need to tell me, but it seems a good choice.

Random guess: You can't just hold 1 ball in one hand and another in the other hand and check that way? (Too many variables, how big, what's the weight difference.)

I think that's the best first choice, to be honest. I mean, I don't know the answer (unless you guys want me to try and solve this problem again), but it sounds right :P

Random guesses aren't allowed! The weight difference is so small that you wouldn't be able to notice if any of the balls were different. However, you can see the difference if you weigh them =)

[DeathStalker908]

Fill up the 3 gallon. Dump it into the 5. Fill up the 3 gallon again and dump what you can(2 gallons) in the 5 gallon bucket. Leaving 1 gallon left in the 3 gallon bucket. Dump the five gallon bucket out. Dump the remainder of the 3 gallon bucket into the 5 gallon jug(which was one gallon). fill up the 3 gallon jug one more time and dump it into the 5 gallon. Wala! 4 gallons.

Nice, lol.

There are two ways to do this...can you figure out the other way?

[pimpin002andhalf]

Nice, lol.

There are two ways to do this...can you figure out the other way?

Just reverse the gallons. So take the five gallon fill it up. Dump the 5 gallon bucket into the 3 gallon bucket (leaving 2 gallons in the 5 gallon bucket.) Dump the 3 gallon bucket out. Dump the 5 gallon bucket into the 3 gallon bucket (putting 2 gallons in it). Fill up the 5 gallon bucket then take it and top off the 3 gallon bucket. Wala! A remainder of 4 gallons in the 5 gallon bucket.

[DeathStalker908]

Just reverse the gallons. So take the five gallon fill it up. Dump the 5 gallon bucket into the 3 gallon bucket (leaving 2 gallons in the 5 gallon bucket.) Dump the 3 gallon bucket out. Dump the 5 gallon bucket into the 3 gallon bucket (putting 2 gallons in it). Fill up the 5 gallon bucket then take it and top off the 3 gallon bucket. Wala! A remainder of 4 gallons in the 5 gallon bucket.

Correct!

I will save the other word problem for tomorrow :o

However, in the meantime, I recommend you try to complete the first one I asked :D

[pimpin002andhalf]

Number the balls 1 to 12. Then weigh 1 - 4 against balls 5 - 8.
If balls 1 - 4 and 5 - 8 are balanced then the odd ball is in group 9 - 12
So weigh ball 9 and 10 against 11 and 8 (obviously ball 8 is not the odd ball because we weighed it on the first scale usage). If ball 9 - 10 and 11 - 8 balance then 12 is the odd one. Weigh 12 against any other to find out if it is heavy or light.

-If balls 9 - 10 and balls 11 - 8 do not balance. Lets say for example balls 11 and 8 are heavier, than 9 and 10. That means then either 11 is heavy, or 9 is light, or 10 is light. Weigh 9 against 10. If they balance 11 is heavy, if they do not the lighter of 9 and 10 is the odd ball.

Same thing applies if balls 11 and 8 are lighter than 9 and 10.

-If balls 1 -4 and 5 - 8 do not balance. For example if 5 - 8 are heavier than 1 - 4. Then one of balls 1 - 4 is light, or one of balls 5 -8 is heavy. Weigh balls 1, 2, and 5 against 3, 6, and 9. If they balance then either 7 is heavy, or 8 is heavy, or 4 is light. Weigh ball 7 against 8. If they balance, 4 is the odd ball, otherwise the heavier of 7 and 8 is the odd ball.

-If balls 1, 2, 5 and balls 3, 6, 9 do not balance. For example if balls 1, 2, and 5 are lighter than 3, 6, and 9. Then either 6 is heavy, or 1 is light, or 2 is light. Weigh 1 against 2 to find out which one of the three choices is true. Otherwise, suppose 1, 2, and 5 are heavier than 3, 6, and 9. Then either ball 3 is light, or 5 is heavy. Weigh 3 against one of the othe balls. Let's say 2 to find out which of the two choices is true.

It's the same thing if 1, 2, and 5 are lighter than 3, 6, and 9.

Ive said balls more during this than I wanted to. Anything else?

[DeathStalker908]

Number the balls 1 to 12. Then weigh 1 - 4 against balls 5 - 8.
If balls 1 - 4 and 5 - 8 are balanced then the odd ball is in group 9 - 12
So weigh ball 9 and 10 against 11 and 8 (obviously ball 8 is not the odd ball because we weighed it on the first scale usage). If ball 9 - 10 and 11 - 8 balance then 12 is the odd one. Weigh 12 against any other to find out if it is heavy or light.

-If balls 9 - 10 and balls 11 - 8 do not balance. Lets say for example balls 11 and 8 are heavier, than 9 and 10. That means then either 11 is heavy, or 9 is light, or 10 is light. Weigh 9 against 10. If they balance 11 is heavy, if they do not the lighter of 9 and 10 is the odd ball.

Same thing applies if balls 11 and 8 are lighter than 9 and 10.

-If balls 1 -4 and 5 - 8 do not balance. For example if 5 - 8 are heavier than 1 - 4. Then one of balls 1 - 4 is light, or one of balls 5 -8 is heavy. Weigh balls 1, 2, and 5 against 3, 6, and 9. If they balance then either 7 is heavy, or 8 is heavy, or 4 is light. Weigh ball 7 against 8. If they balance, 4 is the odd ball, otherwise the heavier of 7 and 8 is the odd ball.

-If balls 1, 2, 5 and balls 3, 6, 9 do not balance. For example if balls 1, 2, and 5 are lighter than 3, 6, and 9. Then either 6 is heavy, or 1 is light, or 2 is light. Weigh 1 against 2 to find out which one of the three choices is true. Otherwise, suppose 1, 2, and 5 are heavier than 3, 6, and 9. Then either ball 3 is light, or 5 is heavy. Weigh 3 against one of the othe balls. Let's say 2 to find out which of the two choices is true.

It's the same thing if 1, 2, and 5 are lighter than 3, 6, and 9.

Ive said balls more during this than I wanted to. Anything else?

Omg, you're the first person to get that question right. GOOD JOB! XD

Other problem...Sally likes 225 but not 224; she likes 900 but not 800; she likes 144 but not 145. Which does she like: 1600 or 1700?

Tell me the answer and explain why...if you don't explain why, I won't say if you're right or not =)

[pimpin002andhalf]

Omg, you're the first person to get that question right. GOOD JOB! XD

Other problem...Sally likes 225 but not 224; she likes 900 but not 800; she likes 144 but not 145. Which does she like: 1600 or 1700?

Tell me the answer and explain why...if you don't explain why, I won't say if you're right or not =)

1600 She only like number with Whole numbrt square roots.

[DeathStalker908]

1600 She only like number with Whole numbrt square roots.

Correct!

Too bad I can't find anymore logic problems:(

[pimpin002andhalf]

I keep an eye out for your future posts.

[supersimons]

damn i missed out on all the fun. Awesome on someone figuring out the balls question.
but those two were cake. I just wasnt around.

[ReligionX]

Nice, lol.

There are two ways to do this...can you figure out the other way?
1.) Put the 3 gallon bucket inside of the 5 gallon bucket.
2.) Fill the 5 gallon bucket and the 3 gallon bucket in the 5 gallon bucket. The 3 gallon bucket displaces 3 gallons, putting 2 gallons in the 5 gallon bucket.
3.) Remove the 3 gallon bucket. Empty the water from the 3 gallon bucket.
4.) Pour the 2 gallons from the 5 gallon bucket into the 3 gallon bucket.
(alternate 1-4, remove 3 gallons from the full 5 gallon bucket).
5.) Put the 3 gallon bucket (with 2 gallons in it) inside of the 5 gallon bucket.
6.) Fill the 5 gallon bucket around the 3 gallon bucket. The 3 gallon bucket with 2 gallons of water will weigh the same as the allowed 2 gallons in the 5 gallon bucket, and will not float.
7.) Remove the 3 gallon bucket.
8.) Pour the 2 gallons from the 3 gallon bucket into the 5 gallon bucket with 2 gallons in it.

[ReligionX]

This is one from a game that my brother gave me. I think I figured it out a while ago.

There are 3 sheep and 3 wolves on one side of a river. You can only transport 2 animals at a time on a boat, and there has to be at least 1 animal on the boat every time. If there are a greater number of wolves than sheep on a side, they will eat the sheep. You cannot "quickly get animals on the boat before the wolves would attack sheep." How do you get them all across without losing any sheep? What is the least number of trips (from one side to another) it takes to complete this?

[Egnar]

1)Put a wolf and a sheep on the boat - On the other side leave a wolf and bring the sheep back to the beginning so it looks like this:

W S WWSS

2) Put another wolf on the boat - On the other side leave the sheep and bring the wolf back to the beginning - It should look like this now:

WS W WSS

3) Put another sheep on the boat with the wolf and bring it over to the other side - Leave the sheep and bring the wolf back again. It should look like this now:

WSS W WS

4) Put the final sheep on board and bring it over to the other side - The wolf once again stays on the boat and goes back to the begining:

WSSS W W

5) Put the final wolf on the boat and on the other side have them both get off:

WWWSSS


5 round trips, best i can go.

[DeathStalker908]

Damn, I was just about to ask a question similar to the sheep / wolf one..even though its similar, I'll just ask it anyways XD

There are two adults and two children on the left side of a river. Each child weighs 100 pounds, and each adult weighs 200 pounds. There is also a boat that can lift up to 200 pounds; how do they get across?

PS: The adults/children CAN'T swim across =)

[DeathStalker908]

Here's another really easy one.

Imagine being in a game show, where there is a host that shows you 3 really big doors. He tells you that there is 1 car, and two goats behind the doors. Then, he tells you to pick a door, and he will open the other door containing a goat. So, you choose the first door. Then, he opens up door #3, which contains a goat. Then he tells you that you can either stay with your initial choice (door 1), or choose the other door (2). What do you do?

A) Choose door 2.
B) Stay with door 1.
C) It doesn't matter..
D) Punch the game host in the face and run away from the security guards.
E) Bribe the game host before the game, hoping that he would accept it and tell you the right choice.

[supersimons]

Damn, I was just about to ask a question similar to the sheep / wolf one..even though its similar, I'll just ask it anyways XD

There are two adults and two children on the left side of a river. Each child weighs 100 pounds, and each adult weighs 200 pounds. There is also a boat that can lift up to 200 pounds; how do they get across?

PS: The adults/children CAN'T swim across =)

Send both kids. Send one kid back. Send One adult. Send the other kid back. Send Both Kids. Send one back. Send the last adult. Send the kid back. Send both kids.

[supersimons]

Here's another really easy one.

Imagine being in a game show, where there is a host that shows you 3 really big doors. He tells you that there is 1 car, and two goats behind the doors. Then, he tells you to pick a door, and he will open the other door containing a goat. So, you choose the first door. Then, he opens up door #3, which contains a goat. Then he tells you that you can either stay with your initial choice (door 1), or choose the other door (2). What do you do?

A) Choose door 2.
B) Stay with door 1.
C) It doesn't matter..
D) Punch the game host in the face and run away from the security guards.
E) Bribe the game host before the game, hoping that he would accept it and tell you the right choice.

Have you seen the movie 21. Choose Door 2.

[DeathStalker908]

Have you seen the movie 21. Choose Door 2.

No, I haven't. Now explain why someone would choose door 2 :D

[pimpin002andhalf]

Here's another really easy one.

Imagine being in a game show, where there is a host that shows you 3 really big doors. He tells you that there is 1 car, and two goats behind the doors. Then, he tells you to pick a door, and he will open the other door containing a goat. So, you choose the first door. Then, he opens up door #3, which contains a goat. Then he tells you that you can either stay with your initial choice (door 1), or choose the other door (2). What do you do?

A) Choose door 2.
B) Stay with door 1.
C) It doesn't matter..
D) Punch the game host in the face and run away from the security guards.
E) Bribe the game host before the game, hoping that he would accept it and tell you the right choice.

Option A you have a 66% chance to get it right or the 33% chance if you stay with you original guess.

[supersimons]

Four people need to cross a bridge. The bridge can only support the weight of 2 people. It is also dark and someone always needs to carry the lantern.
This is how long it takes for each person to cross the bridge. You must use the slowest time.

person a 5min
person b 10 min
person c 20 min.
person d 25 min

What is the fastest way possible to get all four people across the bridge. No you cant throw the lantern back. It has to be brough back.

[DeathStalker908]

Option A you have a 66% chance to get it right or the 33% chance if you stay with you original guess.

Mind explaining why? =)

[pimpin002andhalf]

1)Put a wolf and a sheep on the boat - On the other side leave a wolf and bring the sheep back to the beginning so it looks like this:

W S WWSS

2) Put another wolf on the boat - On the other side leave the sheep and bring the wolf back to the beginning - It should look like this now:

WS W WSS

3) Put another sheep on the boat with the wolf and bring it over to the other side - Leave the sheep and bring the wolf back again. It should look like this now:

WSS W WS

4) Put the final sheep on board and bring it over to the other side - The wolf once again stays on the boat and goes back to the begining:

WSSS W W

5) Put the final wolf on the boat and on the other side have them both get off:

WWWSSS


5 round trips, best i can go.

It's the best and only way.

[pimpin002andhalf]

Damn, I was just about to ask a question similar to the sheep / wolf one..even though its similar, I'll just ask it anyways XD

There are two adults and two children on the left side of a river. Each child weighs 100 pounds, and each adult weighs 200 pounds. There is also a boat that can lift up to 200 pounds; how do they get across?

PS: The adults/children CAN'T swim across =)



C= Child
A=Adult

2C->
1C<-
1A->
1C<-
2C->
1C<-
1A->
1C<-
2C->

Finito!

[pimpin002andhalf]

Mind explaining why? =)

Ok the original equation is 1 door= 33% So if you choose one you have a 33% chance you are right and a 66% chance that its one of the other doors. If you take one of those doors out. You still have the 66% chance that your original guess was wrong so by switching the guess to the door 2 you have then a 66% chance of getting it right instead of your original 33% right 66 % wrong.

[DeathStalker908]

Four people need to cross a bridge. The bridge can only support the weight of 2 people. It is also dark and someone always needs to carry the lantern.
This is how long it takes for each person to cross the bridge. You must use the slowest time.

person a 5min
person b 10 min
person c 20 min.
person d 25 min

What is the fastest way possible to get all four people across the bridge. No you cant throw the lantern back. It has to be brough back.

1) A + D go. A carries the lantern back. 30 minutes pass by.

2) A + C go. A carries the lantern back. 25 minutes pass by.

3) A + B go. 10 minutes pass by.

In total, it would take 1 hour and 5 minutes (just some extra info).
PS: I have NOT checked my answer.

[supersimons]

1) A + D go. A carries the lantern back. 30 minutes pass by.

2) A + C go. A carries the lantern back. 25 minutes pass by.

3) A + B go. 10 minutes pass by.

In total, it would take 1 hour and 5 minutes (just some extra info).
PS: I have NOT checked my answer.

give it a quick check.

[pimpin002andhalf]

Four people need to cross a bridge. The bridge can only support the weight of 2 people. It is also dark and someone always needs to carry the lantern.
This is how long it takes for each person to cross the bridge. You must use the slowest time.

person a 5min
person b 10 min
person c 20 min.
person d 25 min

What is the fastest way possible to get all four people across the bridge. No you cant throw the lantern back. It has to be brough back.

ab->
a<-
ac->
a<-
ad->

Total time = 65 mins or 1 Hour 5 mins. (By having the 5 Minute man cross with the latern reuduces the travel time to it minimum. )

[supersimons]

ab->
a<-
ac->
a<-
ad->

Total time = 65 mins or 1 Hour 5 mins. (By having the 5 Minute man cross with the latern reuduces the travel time to it minimum. )

not correct.
Your a smart guy you can get this one.

[DeathStalker908]

Ok the original equation is 1 door= 33% So if you choose one you have a 33% chance you are right and a 66% chance that its one of the other doors. If you take one of those doors out. You still have the 66% chance that your original guess was wrong so by switching the guess to the door 2 you have then a 66% chance of getting it right instead of your original 33% right 66 % wrong.

Let me explain it a in slightly different way..

Watch this video, it explains it all, quite well. www.youtube.com/watch?v=mhlc7peGlGg

If you don't want to, I can sum it all up right here.

For example, Door 1 has car, door 2 has a goat, and door 3 has a goat.

If you don't switch:
You choose Door 1. You have 1 right out of 1 guess. (Correct)
You choose Door 2. You have 1 right out of 2 guesses. (Fail)
You choose Door 3. You have 1 right out of 3 guesses. (Fail)
1 divided by 3 = 33%.

If you switch:
You choose Door 1, and switch. You have 0 right out of 1 guess. (Fail)
You choose Door 2, and switch. You have 1 right out of 2 guesses. (Correct)
You choose Door 3, and switch. You have 2 right out of 3 guesses. (Correct)
2 divided by 3 = 66%.

[abandit99]

1-Ok here is the answer. Devide them into two groups of 6. Place group 1 on the scale(3 on each side) if they are level then put group one to the side. If not the odd ball is in group 2.

2-Take the last 6 and split them into 3 groups of two. Place each group on the scale until one group is uneven.

3-Take the group of uneven balls and split them into two groups. A & B. Take any of the other balls and weigh them with either A or B. If the scale is uneven it will not only tell you what group the odd ball is in. It will also tell you if its heavier or lighter.(Please don't make me go anymore in depth If you need me to because you don't understand any of this you don't need to be doing logic problems.)

Thats more than 3 steps. Check where i've highlighted. Each weighing is a 'step'. You cannot just group several weighings into 1 and call them 1 step. I think its you who shouldnt be doing logic problems :)

This scale problem is one of the hardest and longest i have encountered. I had to write it all out on paper to figure it out.
This site helps sort out all the possible outcomes from each weighing.
http://www.mazes.com/scales/Scale301.html

[DeathStalker908]

Thats more than 3 steps. Check where i've highlighted. Each weighing is a 'step'. I think its you who shouldnt be doing logic problems :)

This scale problem is one of the hardest and longest i have encountered.
This site helps sort out all the possible outcomes from each weighing.
http://www.mazes.com/scales/Scale301.html

Um, he made another post saying the right answer. Look again =)

Edit: His post is on page 3, by the way.

[abandit99]

edit: my bad

[DeathStalker908]

The sentence below is true.
The sentence above is false.

Which sentence is true, and which sentence is false?

[thespartan]

1st sentence is false, 2nd is true.

[Domination]

1)If I say "Everything I tell you is a lie," am I telling you the truth or a lie?

2) A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?


lol the last one made my head hurt.

3)Which of the following statements are true?

1. At least one of these ten statements is false.
2. At least two of these ten statements are false.
3. At least three of these ten statements are false.
4. At least four of these ten statements are false.
5. At least five of these ten statements are false.
6. At least six of these ten statements are false.
7. At least seven of these ten statements are false.
8. At least eight of these ten statements are false.
9. At least nine of these ten statements are false.
10. At least ten of these ten statements are false

[alex128]

3)Which of the following statements are true?

1. At least one of these ten statements is false.
2. At least two of these ten statements are false.
3. At least three of these ten statements are false.
4. At least four of these ten statements are false.
5. At least five of these ten statements are false.
6. At least six of these ten statements are false.
7. At least seven of these ten statements are false.
8. At least eight of these ten statements are false.
9. At least nine of these ten statements are false.
10. At least ten of these ten statements are false

1-5 are true.

[alex128]

2) A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?



"You will sentence me to 6 years in prison."

I might be missing something in this one... I am either correct or halfway there.

[Egnar]

not correct.
A and B go over = 10 minutes
A goes back = 15 minutes
CD go over = 40 minutes
B goes back = 50 minutes
AB go over = 60 minutes

Or, 5 minutes faster than the previously posted answer.

[aj8560]

The sentence below is true.
The sentence above is false.

Which sentence is true, and which sentence is false?

Impossible to answer, creates a the liars paradox. Nice try.

Now for my own.

How can one clock in an east coast state and another clock in a west coast state have exactly the same time?

[ReligionX]

How can one clock in an east coast state and another clock in a west coast state have exactly the same time?
That's not really so much a logic question as it is a trivia question. I've heard this one before, and I looked it up again. The necessary information required to solve the problem isn't contained in the question.

[Egnar]

Well technically since most clocks aren't synced up to the internet or any other satellite system it's very very much possible for 24 clocks, each in a different time zone, to have the exact same time - - At least 23 of the clocks would be wrong, though. - I know that isn't the answer you're looking for, though, without the necessary information (as stated by Religion) I stand correct.

[alex128]

How can one clock in an east coast state and another clock in a west coast state have exactly the same time?

You live in New york and you move and bring your clock with you to California?

Makes sense to me!

[almightywoodygod]

Too many unknowns, what states, what coasts.

If you left out the states part and said America's east and west coasts, then the answer would be they were in panama or something of the sort.

[aj8560]

How can one clock in an east coast state and another clock in a west coast state have exactly the same time if the clocks are correctly synced and you cannot change the time on one of the clocks, the time always has to be correct for that timezone. Has to be one of the lower 48 states, no Hawaii, Alaska, or Puerto Rico. And use your common sense you know the east and west coast, and for your information the Gulf of Mexico isn't on the east coast.

[alex128]

How can one clock in an east coast state and another clock in a west coast state have exactly the same time if the clocks are correctly synced and you cannot change the time on one of the clocks, the time always has to be correct for that timezone. Has to be one of the lower 48 states, no Hawaii, Alaska, or Puerto Rico. And use your common sense you know the east and west coast, and for your information the Gulf of Mexico isn't on the east coast.


Florida... please don't say it is Florida because that would be dumb.

[ReligionX]

How can one clock in an east coast state and another clock in a west coast state have exactly the same time if the clocks are correctly synced and you cannot change the time on one of the clocks, the time always has to be correct for that timezone. Has to be one of the lower 48 states, no Hawaii, Alaska, or Puerto Rico. And use your common sense you know the east and west coast, and for your information the Gulf of Mexico isn't on the east coast.

http://wiki.answers.com/Q/How_can_one_clock_in_an_east_coast_state_and_anoth er_clock_in_a_west_coast_state_have_exactly_the_sa me_time

No. It really isn't common sense. It's trivia. Some people, like our friends in Europe, might not know where our time zones in what states are divided. I certainly wouldn't know this without looking it up, and I certainly have no idea what countries in Europe have what time zones, or if countries in Europe still use / did use daylight savings, and how it is / was observed.

[Domination]

How can one clock in an east coast state and another clock in a west coast state have exactly the same time?

Doesn't day light savings time cause this paradox to happen, day light saving time doesn't happen at the same time all over the u.s or the world for that matter. so you could have the same time on the west coast and east coast when this happens.

[almightywoodygod]

If it's zero degrees outside today and it's supposed to be twice as cold tomorrow, how cold is it going to be?

[supersimons]

well if it was 0 degrees Celsius out then tomorrow it would -8.9 degrees Celsius. and if was
0 degrees Farenheit then tomorrow it will be -31 degrees Fahrenheit.
and if you really want kelvin 0 is absolute zero and tomorrow it will still be zero.
if it matters i just used the Fahrenheit to Celsius formula to move numbers back in forth. But I think i am wrong. I typed it all in so might as well submit. If this is wrong i have another possible outcome i just dont want to do the math.

[Sexy]

Damn, I was just about to ask a question similar to the sheep / wolf one..even though its similar, I'll just ask it anyways XD

There are two adults and two children on the left side of a river. Each child weighs 100 pounds, and each adult weighs 200 pounds. There is also a boat that can lift up to 200 pounds; how do they get across?

PS: The adults/children CAN'T swim across =)

you put no limit on trips taken, so im going to go ahead and say... multiple trips.

if you specified in later posts i didn't read ahead yet.

[monaro]

This problem has been bothering me for a long time, and I want to see if anyone else is able to find a solution. Here it goes...

You have 12 metal balls, and one of them is either heavier or lighter than the rest (you don't know which). If you have an equal-arm balance scale (the one that if you put weight on one side, that side gets pushed down, and the other side gets lifted up) that you can use only 3 times, how do you figure out which ball is lighter/heavier than the others?

If you know the answer, or are close to it, post here listing your 3 steps.

Good luck =)

To be honest.. i would take 3 random steps to get it over and done with, so i could ****ing eat them =]

[Lann625]

To be honest.. i would take 3 random steps to get it over and done with, so i could ****ing eat them =]
Your going to eat big brass balls? That deserves a blow-up-ban-hammer

[wizdew]

This problem has been bothering me for a long time, and I want to see if anyone else is able to find a solution. Here it goes...

You have 12 metal balls, and one of them is either heavier or lighter than the rest (you don't know which). If you have an equal-arm balance scale (the one that if you put weight on one side, that side gets pushed down, and the other side gets lifted up) that you can use only 3 times, how do you figure out which ball is lighter/heavier than the others?

If you know the answer, or are close to it, post here listing your 3 steps.

Good luck =)

Put the balls in water. The one that it takes longest to hit the bottom(or floats) is the lightest and the one that hits the fastest is the heaviest.

[wasabi]

Put the balls in water. The one that it takes longest to hit the bottom(or floats) is the lightest and the one that hits the fastest is the heaviest.

you dont have any water.

[Kuinshii]

I'm just gonna go ahead and ask other ones..this one is obviously tough as ****

You have two buckets: a 3-gallon bucket, and a 5-gallon bucket. If you have an infinite amount of water to pour into those buckets, and if you can pour water from one bucket to another or empty out a bucket, how do you get EXACTLY 4 gallons?

Bored and going through random threads...This riddle was used in one of the Die Hard movies :D. The one with Jeremy Irons... /random

[Bignate]

This is one from a game that my brother gave me. I think I figured it out a while ago.

There are 3 sheep and 3 wolves on one side of a river. You can only transport 2 animals at a time on a boat, and there has to be at least 1 animal on the boat every time. If there are a greater number of wolves than sheep on a side, they will eat the sheep. You cannot "quickly get animals on the boat before the wolves would attack sheep." How do you get them all across without losing any sheep? What is the least number of trips (from one side to another) it takes to complete this?

6 i did it in high school, dont remember the actual route tho.


And ds, willing to bet that guy searched the answer to the balancing balls question lol

[Xaitan]

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