~The BrAiN tEaSeR Thread~

[Carcharodon]

Double Post.

 

[wiegeabo]

He lives in a houseboat on the sea.

Yes he does.

 

[Carcharodon]

Ah badass total guess.

 

[wiegeabo]

Riddle #68:

A horse is tied to a rope 15 feet long. A bail of hay is 25 feet away from him. Yet the horse is able to eat from the bail of hay. How is this possible?

 

[Carcharodon]

The other end of the rope isn't attached to anything.

 

[unstoppable]

Riddle #68:

A horse is tied to a rope 15 feet long. A bail of hay is 25 feet away from him. Yet the horse is able to eat from the bail of hay. How is this possible?

the rope isn't tied down

 

[unstoppable]

The other end of the rope isn't attached to anything.

crap to late

 

[Danalys]

the horse can move in a circle with a 30 foot diameter. so the hay was inside that circle at the oposite side from the horse.

 

[Carcharodon]

That works, too.

 

[wiegeabo]

The other end of the rope isn't attached to anything.

Yep.

 

[xwolverine2]

the horse can move in a circle with a 30 foot diameter. so the hay was inside that circle at the oposite side from the horse.

had to be the smart one

 

[xwolverine2]

Riddle #66:

There are 8 marbles that weigh 1 ounce each, and 1 marble that weighs 1.1 ounces. The marbles are all uniform in size, appearance, and shape. You have a balance that contains 2 trays. You are only able to use the scale 2 times. How do you determine which marble is the heaviest using only the scale and marbles in 2 weighings.

i cant even read that

 

[November Rain]

edit: already solved

 

[November Rain]

Take 6 of the marbles and split them in half, then weigh them. If the 6 marbles are equal weight, then you know one of the 2 left over is the heavier one. From there you just weigh between those two. But, if the 6 marbles aren't equal take the side that is heavier and only weigh 2 of the 3. If those two are equal, then you know the heavier one is the one left over. If not then the heavier one on the scale is the one that weighs 1.1 oz

although your logic still works to get the correct answer, there are actually 9 balls instead of 8

8 all weaigh one ounce and one weights 1.1

but because of your second stage analysis, it still pretty much works.

 

[Ahura Mazda]

here is one which is a bit more difficult (If I am allowed to post it) and a amendment to the above:

There are 12 stones, one is of a different weight of the others. You have the same scale and you can now use it 3 times.

Good luck.

Ps. Note I said different which means you do not know if it is lighter or heavier.

 

[Majik1387]

Whats with all these weight riddles?

 

[TNC9852002]

So, Ahura, what exactly are we supposed to figure out?

-TNC

 

[Enriquespy]

3 groups of 4 you weigth 2...

if equal, then take the other one and dvide it in 2 of 2.

and then in 2 of 1...


if diferent then... you canot know wich one has the diferent one, so i get lost here, because you would then have to compare with the other one lft

 

[wiegeabo]

The stone answer is a ***** to explain. It's like one big logic table. But here goes:

Break the stones into 3 groups of 4 and mark them AAAA, BBBB, and CCCC (just to make the explaination easier to). Weigh AAAA and BBBB. If they balance, the odd stone is in the third group and those 8 stones are of standard weight.

So take 3 of the unweighed stones and weigh it against known stones; CCC and BBB. If they balance, the remaining C is the oddball. Weigh it against a B. If it rises, it is lighter, if it falls, it is heavier.
If CCC falls, one of the C's is heavy. Weigh 2 C's. If they balance, the third is heavy. If one falls, it is heavy.
If CCC rises, one of the C's is light. Weigh 2 C's. If they balance, the third is light. If one rises, it is light.


What if AAAA and BBBB don't balance? If AAAA falls, then either an A is heavy, or a B is light, and the 4 C's are of standard weight. So create 3 new groups; AAAC, BBBA, and CCCB. (The goal is to get 3 suspect stones grouped with a standard stone.)
Weigh BBBA and CCCB. If they balance, one of the A's in AAAC is heavy. Weigh two of the A's. If they balance, the third is heavy. Otherwise the heavy A will fall.
If BBBA falls, then the A is heavy, or the B in CCCB is light. Weigh A and C. If they balance, the B is light. If A falls, it is heavy. (C can't fall.)
If CCCB falls, then a B in BBBA is light. Weigh B and B. If they balance, the other B is light. Otherwise, the light B will rise.


Similarly to the last section, if BBBB falls, then either a B is heavy, or an A is light, and the 4 C's are of standard weight. So create 3 new groups; AAAB, BBBC, CCCA.
Weigh AAAB and CCCA. If they balance, one of the B's in BBBC is heavy. Weight two of the B's. If they balance, the third is heavy. Otherwise the heavy B will fall.
If AAAB falls, then the B is heavy, or the A in CCCA is light. Weigh B and C. If they balance, the A is light. If B falls, it is heavy. (C can't fall.)
If CCCA falls, then an A in AAAB is light. Weight A and A. If they balance, the other A is light. Otherwise, the light A will rise.

 

[Ahura Mazda]

The stone answer is a ***** to explain. It's like one big logic table. But here goes:

Break the stones into 3 groups of 4 and mark them AAAA, BBBB, and CCCC (just to make the explaination easier to). Weigh AAAA and BBBB. If they balance, the odd stone is in the third group and those 8 stones are of standard weight.

So take 3 of the unweighed stones and weigh it against known stones; CCC and BBB. If they balance, the remaining C is the oddball. Weigh it against a B. If it rises, it is lighter, if it falls, it is heavier.
If CCC falls, one of the C's is heavy. Weigh 2 C's. If they balance, the third is heavy. If one falls, it is heavy.
If CCC rises, one of the C's is light. Weigh 2 C's. If they balance, the third is light. If one rises, it is light.


What if AAAA and BBBB don't balance? If AAAA falls, then either an A is heavy, or a B is light, and the 4 C's are of standard weight. So create 3 new groups; AAAC, BBBA, and CCCB. (The goal is to get 3 suspect stones grouped with a standard stone.)
Weigh BBBA and CCCB. If they balance, one of the A's in AAAC is heavy. Weigh two of the A's. If they balance, the third is heavy. Otherwise the heavy A will fall.
If BBBA falls, then the A is heavy, or the B in CCCB is light. Weigh A and C. If they balance, the B is light. If A falls, it is heavy. (C can't fall.)
If CCCB falls, then a B in BBBA is light. Weigh B and B. If they balance, the other B is light. Otherwise, the light B will rise.


Similarly to the last section, if BBBB falls, then either a B is heavy, or an A is light, and the 4 C's are of standard weight. So create 3 new groups; AAAB, BBBC, CCCA.
Weigh AAAB and CCCA. If they balance, one of the B's in BBBC is heavy. Weight two of the B's. If they balance, the third is heavy. Otherwise the heavy B will fall.
If AAAB falls, then the B is heavy, or the A in CCCA is light. Weigh B and C. If they balance, the A is light. If B falls, it is heavy. (C can't fall.)
If CCCA falls, then an A in AAAB is light. Weight A and A. If they balance, the other A is light. Otherwise, the light A will rise.

That is correct and yes it was very difficult. I actually got asked in an interview at Mackenzie to be a strategy consultant. I was given an hour to figure it out. The interviewer was called away which is why he gave me that question. I also got a few strategy case studies to do.


Ended up going into investment banking though.

 

[R_3]

RIDDLE 69
A man wanted to enter an exclusive club but did not know the password that was required. He waited by the door and listened. A club member knocked on the door and the doorman said, "twelve." The member replied, "six " and was let in. A second member came to the door and the doorman said, "six." The member replied, "three" and was let in. The man thought he had heard enough and walked up to the door. The doorman said ,"ten" and the man replied, "five." But he was not let in. What should have he said?

 

[wiegeabo]

RIDDLE 69
A man wanted to enter an exclusive club but did not know the password that was required. He waited by the door and listened. A club member knocked on the door and the doorman said, "twelve." The member replied, "six " and was let in. A second member came to the door and the doorman said, "six." The member replied, "three" and was let in. The man thought he had heard enough and walked up to the door. The doorman said ,"ten" and the man replied, "five." But he was not let in. What should have he said?

He should have said three because that's the number of letters in the word ten.

 

[R_3]

wow. ok. nice job.

 

[TNC9852002]

That one was so easy, though..

-TNC

 

[Ahura Mazda]

Here is another riddle.

You have 2 peices of rope which will each burn in exactly 30 minutes. However, there is no correlation betwen the two ropes and they do not burn uniformly. Therefore 1/4 of the rope could burn in 28 minutes and the rest in 2 minutes.

So how do you tell using the ropes and a lighter when 45 minutes have passed?