Solution:

The person to get off at the first floor is Paul.
The person to get off at the second floor is Christy.
The person to get off at the third floor is Mistie.
The person to get off at the fourth floor is Mia.
The person to get off at the fifth floor is David.
The person to get off at the sixth floor is Lee.

Solution:

Just One!

example
If u get Blue rock with label as "blue+red" => this is red as all labels are wrong.
In this way u can logically solve this puzzle.

Solution:

To do it, the marbles must be arranged as the vertices of an octahedron. If you are unsure, check the image below.

Solution:

30,450 miles.

Explanation:
For each 5 miles driven using this rotation, each tire will lose (2/29000)+(2/21000) of its tread.So, suppose we drive 5*N miles like this. Then, each tire will lose 2*N*((1/29000)+(1/21000)) of its tread. We wish to determine N so that the latter expression gets to be 1, representing all of each tire's tread. That happens when N=6090, which is when 5*N (the total miles driven) is 30,450.

Solution:

36 meters.

100/v1 = 80/v2
100/v2 = 80/v3

20+(20/v2)*v3=20+16 = 36

Solution:

They were playing tennis.

Tiger Phil
30 - 30
40 - 30 (Phil hit a bad shot)
Game won - 30 (Tiger hit an awesome shot)

Solution:

He should draw any chit of paper and before unfolding it, he must eat and swallow the chit. In this manner, to determine the judgment of the god, they will have to look at the other chit which will have 'Guilty' written over it. Thus everyone will believe that the chit which he ate had innocent written over it and thus they will believe that he is innocent.

Solution:

They used the planks in the manner that is shown in the figure.

Solution:

On Your lap.

Solution:

10 prisoners must sample the wine. Bonus points if you worked out a way to ensure than no more than 8 prisoners die.

Number all bottles using binary digits. Assign each prisoner to one of the binary flags. Prisoners must take a sip from each bottle where their binary flag is set.

Here is how you would find one poisoned bottle out of eight total bottles of wine.

Bottle 1 Bottle 2 Bottle 3 Bottle 4 Bottle 5 Bottle 6 Bottle 7 Bottle 8
Prisoner A X X X X
Prisoner B X X X X
Prisoner C X X X X
In the above example, if all prisoners die, bottle 8 is bad. If none die, bottle 1 is bad. If A & B dies, bottle 4 is bad.

With ten people there are 1024 unique combination so you could test up to 1024 bottles of wine.

Each of the ten prisoners will take a small sip from about 500 bottles. Each sip should take no longer than 30 seconds and should be a very small amount. Small sips not only leave more wine for guests. Small sips also avoid death by alcohol poisoning. As long as each prisoner is administered about a milliliter from each bottle, they will only consume the equivalent of about one bottle of wine each.

Each prisoner will have at least a fifty percent chance of living. There is only one binary combination where all prisoners must sip from the wine. If there are ten prisoners then there are ten more combination where all but one prisoner must sip from the wine. By avoiding these two types of combination you can ensure no more than 8 prisoners die.

One viewer felt that this solution was in flagrant contempt of restaurant etiquette. The emperor paid for this wine, so there should be no need to prove to the guests that wine is the same as the label. I am not even sure if ancient wine even came with labels affixed. However, it is true that after leaving the wine open for a day, that this medieval wine will taste more like vinegar than it ever did. C'est la vie.

Solution:

The first step that they will take will be a leap of logic. What it means is that they will deduce that every color must appear twice at least. Why? Because the master logician has assured them that the puzzle will not be impossible for anyone of them. And if a color appears only once in the circle, the person wearing it will have no clue about that color which will not be fair for him.

Then the logicians will follow the same and look for all the colors of hats in the circle. If one of them sees a color just once, he can safely assume that he is also wearing the hat of the same color as by leap of logic, no color can appear just once. Thus when the bell is rung, he will leave.

In the similar fashion, if anyone sees another color just once, he can determine that he is wearing the hat of the same color and will leave when the bell rings or will be disqualified and sent home. Unvaryingly, if a color is seen twice, they will be eliminated after the first bell. Hence, there must be at least three hats of any of the remaining color.

Assume that you are sitting in the circle and you don’t see a color once but see it twice. Then if they were the only two hats of the same color, the logicians must have left at the first bell already. But they did not. Which means that there are three hats of that color and you are wearing one. Thus you will leave after the second bell.