Solution:

1)The first bit is easy, as the first letter of each plate spells WHODUNNIT IX(A challenge to our Detective.)
2)The second bit is a little trickier, but I gave you the solution.If you read the last three letters in each plate from the bottom up and right to left you get ADVANCE EACH LETTER BY THE NUMBER SHOWN, so advance W by 5 to get B, H by 13 to get U and so on until you spell BUTLER DID IT.

Solution:

Tuesday

We know that Christanio Ronaldo tells the truth on only a single day of the week. If the statement on day 1 is untrue, this means that he tells the truth on Monday or Tuesday. If the statement on day 3 is untrue, this means that he tells the truth on Wednesday or Friday. Since Christanio Ronaldo tells the truth on only one day, these statements cannot both be untrue. So, exactly one of these statements must be true, and the statement on day 2 must be untrue.
Assume that the statement on day 1 is true. Then the statement on day 3 must be untrue, from which follows that Christanio Ronaldo tells the truth on Wednesday or Friday. So, day 1 is a Wednesday or a Friday. Therefore, day 2 is a Thursday or a Saturday. However, this would imply that the statement on day 2 is true, which is impossible. From this we can conclude that the statement on day 1 must be untrue.

This means that Christanio Ronaldo told the truth on day 3 and that this day is a Monday or a Tuesday. So day 2 is a Sunday or a Monday. Because the statement on day 2 must be untrue, we can conclude that day 2 is a Monday.

So day 3 is a Tuesday. Therefore, the day on which Christanio Ronaldo tells the truth is Tuesday.

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:

Step 5 is invalid, because we are dividing by (x-y), and since x=y, we are thus dividing by 0. This is an invalid mathematical operation (division by 0), and so by not following basic mathematical rules

Solution:

f the last elf is taking some time to answer the question then it shall mean that the elf's before him are all wearing distinct coloured hats. However sufficient time shall be given to the last elf to give the answer.
If he views the similar coloured hats in front of him , he shall quickly tell the answer

Solution:

1/3

Explanation
Assuming T1 is the Toy for brother1, T2 is the toy for brother2 and T3 is the toy for brother3.

Following are the possible cases for toys distribution:

Boy1 Boy2 Boy3
T1 T2 T3
T1 T3 T2
T2 T1 T3
T2 T3 T1 .... (A)
T3 T1 T2 .... (B)
T3 T2 T1

In both steps (A) & (B), no one gets the correct toy.
Therefore probability that none brother can get the own toy is 2/6 = 1/3

Solution:

It can be solve as shown below.

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:

I am Spice

Explanation:
Spice is hot if you remove first two alphabets then it will become "ice", which is really cold.

Solution:

You will ask, I have a number in my mind that i.e. either 2 or 3. Is the number that I have in my mind smaller or equal to the number that you are thinking of?

If Alan replies with no, he is thinking of the number 1.
If Alan replies with yes, he is thinking of the number 3.
If Alan replies with don't know, he is thinking of the number 2.

Solution:

The answer is very simple. All she had to do is take the fifteen cards from the top and reverse them. This would make another pile out of that and there will be two piles - one of 15 cards and one of 37 cards. Also both of them will have the same number of inverted cards.

Just think about it and if the mathematical explanation will help you understand better, here it is.
Assume that there were p inverted cards initially in the top 15 cards. Then the remaining 37 cards will hold 15-p inverted cards.
Now when she reverses the 15 cards on the top, the number of inverted cards will become 15-p and thus the number of inverted cards in both of the piles will become same.