Solution:

By keeping 8 boxes in a box, the number of boxes that are left empty increases by (8 - 1) i.e. 7.

Now let us assume that x is the number of times the 8 boxes have been put in a box, we will come to know that
11 + 7x = 102
Or x = 13.

Thus in total
11 + 13 * 8 = 115 boxes have been used.

Solution:

The human in front will count the number of green hats and if the count is an odd number, he will say green and if the number is even, he will say pink.

The next human behind him definitely saw either an odd or an even number of hats. If the number is still odd, then he has a pink hat. That human says Pink and survives. Thereafter, there will be an odd number of green hats forward and the third person can decide whether he is wearing a green hat or a pink hat.

If the second person sees an even number of green hats, then he must have a green hat as well and will be freed. In this fashion, the humans will just have to keep the track if the number of green hats in front of them are odd or even in number and then they can spontaneously conclude with the color of their hats.

Since the first human will be guessing, he will have a 50/50 chance of surviving but the other 9 people will definitely be saved.

Solution:

Since there are 100 logicians, the lights are turned on and off hundred times and after the 100th time, all the logicians leave the room.

If all of them sees 99 black foreheads, the lights get turned off. When the light is turned on again, everyone is able to see 99 black foreheads again. This happens 100 times and then every logician leaves the room.

Let us make it more simplified for you. Assume that only one person has a black forehead. The people who invited them said that at least one person has a black forehead and as they turn off the lights, that logician leaves.

Imagine the scenario with two logicians having black forehead. One of the logician sitting there must be thinking "I have a black forehead or I don't have a black forehead. If I am not the one with black forehead, the other logician with black forehead will deduce that he is having a black forehead and then he will leave."

"If I don't have a black forehead, the other logician will stay and thus I must be having a black forehead as well. Thus we must leave when the lights are turned off for the second time"

When you repeat the logic to 100 times, you will get the answer.

Solution:

If you have come up with the fact that the bugs can avoid the collision only when they decide to move in the same direction be it clockwise or anti clockwise, you have already solved half of it. In any other way, the bugs will definitely run into each other.

Now the bugs can move either clockwise or anti-clockwise. The probability of choosing a specific direction is then 1/2.

We will utilize the simple equation of probability to find out our answer

P(No Collision) = P (The bugs move in clockwise direction) + P(The bugs move in anti-clockwise direction)
= 0.5 * 0.5 * 0.5 + 0.5 * 0.5 * 0.5
= 0.25

Thus the probability that the bugs won't collide into each other is 0.25 or 1/4.

Solution:

There are many ways in which you can think a solution for the problem. Your approach should be to minimize the efforts. What if you took out one candy from the box labeled Peach and Grape?

What you have to consider is the fact that every box is labelled incorrectly. Thus if you pick a candy from the box labelled with Peach and Grape, it is evident that the box will not have a mixture of both (as it is labelled wrong). Thus you will get to know which candies are present in that box.

Suppose the candy you take out turns out to be peach flavored, then the box labeled as Grapes is the one which contains a mixture of both peach and grapes. This is because it is labelled incorrectly and since we have already figured out the box with Peach candies, we can be confident of the fact. And then the jar labelled as peach will be the one containing grape flavored candies.

If the candy you take out from the mixed labeled box is Grape flavored, you can use the same logic to find out the rest.

Consider picking a candy from the other two boxes instead of what is told in the solution and you will find out that you will have to take out a candy from each of the boxes before being sure.