Solution:

AMAN

Explanation:
The numbers actually corresponds to the months of the year.

4 - April
5 - May
8 - August
11 - November

Now when you combine the initials of these months, you will get the killer's name i.e. AMAN

Solution:

White should move the rook from
A1 to D1 and hence complete the castling move.

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).

Solution:

I will be thrown at lions. Now if Akbar threw him at lions, Birbal’s statement will stand true and he will have to bury him alive. But if he bury him, the statement will emerge as false. Thus he had no choice left

Solution:

13 numbers

Explanation:
Since we are looking to find odd reminder which implies that the remainder will be either 1 or 3
Therefore number must follow below two conditions:
A) We can list number 5n + 1 or 5n + 3. (where n = 0,1,2,3.....)
B) The number must end in 1 or 3.

List of such numbers less than 100 are:
1, 3, 11, 13, 21, 23, 31, 33, 41, 43, 51, 53, 61, 63, 71, 73, 81, 83, 91, 93

Let's remove the nonprime ones which are 1, 21, 33, 51, 63, 81 and 91

So, I will be left with 13 numbers which fit the bill.
3, 11, 13, 23, 31, 41, 43, 53, 61, 71, 73, 83 and 93

Solution:

It was only a trick

I borrowed my parents (Pound 50 + Pound 50 = Pound 100).
I bought the book for Pound 97 and i had Pound 3 change.
So, the money that i borrowed and bought the book are Pound 100 (borrowing money)
and the change Pound 3 also the borrowing money the book is Pound 97
Total: Pound 97 + Pound 3(the change) = Pound 100
The money that i used to buy the book all i borrowed.
Pound 97 is the borrowing money, change: Pound 3 also the borrowing money. Pound 97 + Pound 3 = Pound 100

Solution:

888 + 88 + 8 + 8 + 8 = 1000

Solution:

Read the question carefully again. New Year do falls after Christmas but that happens if two different years. The question is put up against the year 2050 and thus there will be 51 weeks and 2 days in between them as New Year will appear on 1 January 2050 and Christmas will happen on 25 December 2050.

Solution:

one fish, one cat and one dog.

Explanation:
The question might appear a bit difficult in the starting but if you analyze the statements, you will realize that it is just a tricky one.

All except two were fishes and all except two were cats. With these statements we can assume that two of the animals were not fishes and two were not cats. Now one of those animals that are not fishes can be a cat and one of those two animals that are not cats can be a fish. Just carry out the same analysis for the statement that all except two animals were not dogs and you will come across the result i.e.:
In that competition, there was one fish, one cat and one dog.

Solution:

7 Races.

Solution:

C7 => B8
Exchange pawn for a knight

Check-Mate

GameOver