Solution:

A keyhole

Explanation:
It goes right through the door but is stuck to the door i.e never comes in or goes out.

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:

It can be solved as below.

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.

Solution:

Just 2

Day One: You make your first cut at the 1/7th mark and give that to the worker.
Day Two: You cut 2/7ths and pay that to the worker and receive the original 1/7th in change.
Day three: You give the worker the 1/7th you received as change on the previous day.
Day four: You give the worker 4/7ths and he returns his 1/7th cut and his 2/7th cut as change.
Day Five: You give the worker back the 1/7th cut of gold.
Day Six: You give the worker the 2/7th cut and receive the 1/7th cut back in change.
Day Seven: You pay the worker his final 1/7th.

Solution:

Slumdog Millionare = Slum + Dog + Million + air

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

111111111

Explanation:
It's a maths magical square root series as :
Square root of number 121 is 11
Square root of number 12321 is 111
Square root of number 1234321 is 1111
Square root of number 123454321 is 11111
Square root of number 12345654321 is 111111
Square root of number 1234567654321 is 1111111
Square root of number 123456787654321 is 11111111
Square root of number 12345678987654321 is 111111111 (answer)

Solution:

There is no mention if you can use sign or not and thus the only possible way to get this is as follows:

(15 - 9) + (13 - 7) + (7 - 1) + (9 - 1) + (13 - 9)

If you solve inside the brackets, you will get the following equation
6 + 6 + 6 + 8 + 4
Adding all these numbers, you will get 30 which is the required sum.

Solution:

The competition was held on 11 November 2013. The logic is simple. In America, the dates are mentioned as Month/Day/Year and in Britain, dates are written as Day/Month/Year. But on 11 November 2013, the date and month will be the same. And thus both will write in the same manner i.e. 11/11/2013. If it would have been any other date, the evaluators would have been able to identify it.

Solution:

See the figure for the answer. Wasn't it a bit tricky?