Solution:

Bob was in China in 2012 & he killed Anderson.

Anderson died of suffocation=>This information should lead you to statement #3 next. From here you should gather that Bob killed Anderson.
Thus leading to statement #4, Bob was in China in 2012.

Solution:

It can surely be true because it is not mentioned that it was raining.

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:

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 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:

1/2

one of two is the possibility
1. If any of the first 99 people sit in Abhishek seat, Abhishek will not get to sit in his own seat.
2. If any of the first 99 people sit in Aishwarya's seat, Abhishek will get to sit in his seat.

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:

What must be kept in mind before you start calculating through any method is that the weights can be put on both the pans.

So the weights - 1, 3, 7, 12, 43, 76, 102

The oldest man that can be weighed using these weights is a man aged 122 years

Solution:

The Winchester brothers - sam and dean

Solution:

The last marble will be yellow.

Since marbles can only be taken out in pairs and you started off with an odd number of yellows there is always going to be one yellow left over that you'll keep putting back in the box until it's left on it's own.

Solution:

192 021

If you look at it as two-digit numbers you'll see:10 11 12 13 14 15 16 17 18 19 20 21