Solution:

The ball will take infinite bounces before it is stopped. As per the question it will keep on bouncing half way up every time it hits the ground. But gravity will play its part at certain point of time which will make it stop.

Solution:

Since the door is not visible in this picture, the bus is moving in the left direction.

Solution:

It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.

Solution:

From top to bottom, the flavors are :

nutty-chocolate
peppermint
orange
Jamaica
walnut

Solution:

You must put labels on the tablets as A and B before using. In that case, if you pour tablets together, you will get 3A, 2A 1B, 1A 2B or 3B. If they are from the same bottles you can take one from another bottle and save the remaining two for another day. If you get two from same and one from other, you can draw one from another bottle and you will have two pairs of which you can eat one and save the other.

Solution:

Six.

There are only five colors, so six socks guarantee that two will be the same color.

Solution:

The probability of two numbers being relatively prime is 60%. Thus if you keep playing the bet, you will hypothetically make 20 cents on every play.

Thus, you should accept the bet.

Solution:

Put one black rock in one sack and put all the rest in the other one. In such manner you will get about 3/4 probability of surviving.

Solution:

7

Explanation:
8th is bald man head, check the video below.

Solution:

Colors:
Y B R G
G R B Y
B Y G R
R G Y B


Signs:
x + - /
- / x +
/ - + x
+ x / -

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.