Solution:

I am letter 'E'

Solution:

40 balls

Solution:

== Mathematical Answer ==
Mathematical Answer = number of seconds in a minute * number of minute in an hour * number hour in a day * number of day in year
60 * 60 * 24 * 365 = 31536000
60 * 60 * 24 * 365.25 = 31557600 (if you want account for leap year)


== We are looking for trick answer (which is 24) ==
Every month contains only 2 second like
A) 2nd January ("second" January)
B) 22nd January (twenty"second" January)

Therefore 2 * 12 months = 24.

Solution:

The figure tells that B is located to the North-East of E.

Solution:

In the corners of the square, tighten two ropes. Tighten another rope between them by putting rings on each rope. That rope should be tightened on the sheep's neck on the other end.

Solution:

The two numbers are 4 and 13 whose product is 52 and sum is 17.

Solution:

I + IX = X

The equation read from the other side becomes X = XI + I which is false.

Solution:

Flyball should pay Lavy $11 instead of accepting his counter offer as in that case the club will end up paying almost double the amount.

Solution:

Everyone will done one day or other.

Abraham Lincoln era is too old and we expect most of them(if not all) to be dead by now.

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:

Fifteen minutes can easily be measure using these two hour glasses.

Step 1: He will start both the hourglass.

Step 2: The moment the eleven minute hourglass is empty, he will invert it.

Step 3: When the thirteen minutes hourglass is empty, he will invert the eleven minute hourglass.
In step 3, we will have counted thirteen minutes. Since we inverted the eleven minute hourglass in step 2, it started from fresh and was inverted just for two minutes (13-11=2). In this manner when it is reversed when the thirteen minute hourglass is finished, it will have two minutes of sand left. This time when the sand finishes, he will have measured fifteen minutes. (13+2=15)