Solution:

13 x 3 + 1 = 40

The number of decks is absolutely not relevant here which means whether we mix 5 or 500 cards still results would be same.
Any card drawn will be a Ace,2,3,4,5,6,7,8,9,10,Jack,Queen or King, so there are 13 possibilities each time a card is drawn.
If you are lucky just 5 cards of the same kind can be obtained in 4 steps

The unluckiest(worst) way is our solution as we need to guarantee a four of a kind.
Draw 3 of each kind =>now we have 39 cards. Next card will guarantee 4 of a kind.

Solution:

1 2 3 4 5 6 and 7 8 6

Since summation of numbers 1-9(1 2 3 4 5 6 7 8 9) is 45 , so its impossible to divide into two equal parts.
Since we cant flip over 6 to make it 9 but we can flip 9 to make it 6 and hence the solution

Solution:

This statement is true but only in a specific case i.e. if both of the siblings continue to live till the day the younger is as old as the elder one was during the time of their birth.

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:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.

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:

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

Solution:

9 - 8 + 76 - 5 + 4 + 3 + 21 = 100


Other ways of making number 100 (withiout using number 32)
98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
98 - 7 - 6 - 5 - 4 + 3 + 21 = 100
9 - 8 + 76 - 5 + 4 + 3 + 21 = 100
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1 = 100
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1 = 100
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1 = 100
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1 = 100
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1 = 100
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1 = 100
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1 = 100

Solution:

The attached picture shows you how you can do it. If you were able to deduce any other way, feel free to comment.

Solution:

36 meters.

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

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

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.