Solution:

3rd

Explanation:
In the 3rd image, all the three arrows curved in the same direction while in other images they curved in different directions.

Solution:

This puzzle has no answer if you are considering this figure in a 2d domain. However if you follow it on a three-dimensional plane, you can do it like shown in the figure given.

Solution:

Steps:
1) Light fire on first bar from both the ends and second bar from one end.

2) After 30 seconds(when first bar gets completely burned out) , burn the second bar from second end as well.

3) When second run completely gets burned , you know its 45 seconds.

Solution:

On Your lap.

Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

The numbers that have been written over the calendar are:
12, 4, 8 and 3

A few numbers have been circled under particular months. The curious thing is that the numbers that have been written are the same number of corresponding months under which the dates have been marked.

What we can judge by the overall calendar is that the dates that have been marked may refer to the letters at that position in the month's name. In addition, the written numbers can be the position of those letters. Let's simplify below:

12th month is December
The number marked under December = 5
The fifth letter of December = M

4th month is April
The number marked under April = 4
The fourth letter of April = I

8th month is August
The number marked under August is 6
The sixth letter of August = T

3rd month is March
The numbers marked under March are 4 and 5
The fourth and fifth letter of March = C and H

When you combine the letters, you get the name as MITCH.
Therefore, Mitch is the killer.

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:

He askes the god to beat him half to death.

Solution:

17 min

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 minutes. Is that it? No. That would make this question too simple even as a warm up question.

Let's brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let's put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 minutes

Solution:

White should move the rook from
A1 to D1 and hence complete the castling move.

Solution:

There are six unique answers

Ans1:
7 x 4 = 28
36 / 4 = 9
1 + 4 = 5

Ans2:
9 x 4 = 36
28 / 4 = 7
1 + 4 = 5

Ans3:
3 x 8 = 24
56 / 8 = 7
1 + 8 = 9

Ans4:
7 x 8 = 56
24 / 8 = 3
1 + 8 = 9

Ans5:
4 x 8 = 32
56 / 8 = 7
1 + 8 = 9

Ans6:
7 x 8 = 56
32 / 8 = 4
1 + 8 = 9