Solution:

Since the beer bottle is closed, No glass will Ever fill.

The video-version of the puzzle is added below.

Solution:

The stranger said that he thought the room was his. If so, he must have had the keys to the room and thus he must not have knocked.

Solution:

There are six unique answers

Ans1:
7 x 4 = 28
36 / 4 = 9
1 + 4 = 5

Ans2:
9 x 4 = 36
28 / 4 = 7
1 + 4 = 5

Ans3:
3 x 8 = 24
56 / 8 = 7
1 + 8 = 9

Ans4:
7 x 8 = 56
24 / 8 = 3
1 + 8 = 9

Ans5:
4 x 8 = 32
56 / 8 = 7
1 + 8 = 9

Ans6:
7 x 8 = 56
32 / 8 = 4
1 + 8 = 9

Solution:

10

Explanation:
Equation 1 + 9 + 11 = 1 can be derived from
One (o) + nine (n) + eleven (e) = one => 1

Similarly for equation,
12 + 11 + 9
Twelve (t) + eleven (e) + nine (n) => ten (10)

Solution:

A cigar stuck in the wall

Solution:

The code word is TIE.

Let us assume that the word is MOD. Now if you are told one character of this word, you would not be able to identify the number of vowels in the word. For example if you are told M, then the word can be MOD or AIM but they both have different number of vowels. If you are told about O, then there are two words MOD and OAT which again have different number of vowels. If you are told about D, there are two words again MOD and DUE which have different numbers of vowels again.

In such manner, all the words that have M, O or D in them can be excluded. Then only word that remains is TIE.
If you are told about T, there are two words OAT and TIE which both have two vowels.
If you are told about I, there are two words AIM and TIE which both have two vowels.
If you are told about E, there are two words DUE and TIE which both have two vowels.

Thus TIE is the code word.

Solution:

All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.

Solution:

23rd day

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20 + 21 + 22 + 23 = 276

Thus our anniversary falls on the 23rd day of the month.

You can apply other formulas to shorten the process but that is the simplest way to do it.

Solution:

25 Fruits

Remember we told you that Birbal is a witty trader. So his sole motive is to get rid of the sacks as fast as he can.

For the first sack:
He must be able to fill fruits from one sack to other two sacks. Assume that he is able to do that after M check points. Now to find M,

(Space in first sack) M + (Space in second sack) M = (Remaining fruits in Third Sack) 30 – M
M = 10

Thus after 10 checkpoints, Birbal will be left with only 2 sacks containing 30 fruits each.

Now he must get rid of the second sack.
For that, he must fill the fruits from second sack to the first sack. Assume that he manages to do that after N checkpoints.

(Space in First Sack) N = (Remaining fruits in second sack) 30 – N
N = 15

Thus after he has crossed 25 checkpoints, he will be left be one sack with 30 fruits in it. He has to pass five more checkpoints where he will have to give five fruits and he will be left with twenty five fruits once he has crossed all thirty check points.

Solution:

Solution can be verified from the picture below.

Solution:

It has been clearly stated that none of the two students gave same number of correct answers which leads us to two different possibilities of correct answers:
Either (1 + 2 +3 + 4 + 5 = 15) or (0 + 1 + 2 + 3 + 4 = 10) stands true.

After analyzing the question, the maximum number of correct answers possible with the answers Gerry, Billy, Clark, Peeta and Jonathan gave:
Question 1 = 2 (b or c)
Question 2 = 2 (b or c)
Question 3= 4 (True)
Question 4 = 4 (True)
Question 5 = 3 (True)

Let us first assume with the maximum number of correct answers possible according to our data which are 15 (2+2+4+4+3). According to it, Billy must have given all correct answers as he was the only one who answered True for questions 3, 4 and 5. But in that case, Gerry and Clark must have given exactly three correct answers. Furthermore, Peeta and Jonathan must have given two correct answers. Therefore none of them got all the five correct answers which is wrong according to what we have analyzed.

Thus we move on to our next assumption according to which, the total number of correct answers are 10 (0+1+2+3+4). Now we must acknowledge that Questions 3 as well as Question 4 cannot be False as it will mean that the number of correct answers would not be 10. So the students giving wrong answers cannot be Gerry, Billy and Peeta.

Assume that Jonathan got all wrong, it will suggest that Gerry, Clark and Peeta each would have at least two correct answers. Then, Billy would have to be the person with only one correct answer and in that situation, the correct answers for questions 1 and 2 would be a and a respectively. Since that is not possible as then the total number of correct answers would be 1 (a) + 1 (a) + 1 (False) + 4 (True) + 2 (Flase) = 9.

Thus it must be Clark who has given all wrong answers. The correct answers are then b, a, True, False and False. Also, the scores are Clark (0), Billy (1), Gerry (2), Jonathan (3) and Peeta (4).