Solution:

146900

Explanation:
Whenever you face such type of questions, it is wise to begin from the last thing. Here in this question the last thing will be the 9th temple. He climbed down 100 steps and thus you know, he had Rs. 100 before beginning climbing down. Thus, he must have offered Rs. 100 to the god in that temple too (he offered half of the total amount). Also, he must have dropped Rs. 100 while climbing the steps of the ninth temple. This means that he had Rs. 300 before he begand climbing the steps of the ninth temple.

Now, we will calculate in the similar manner for each of the temples backwards.
Before the devotee climbed the eight temple: (300+100)*2 + 100 = 900
Before the devotee climbed the seventh temple: (900+100)*2 + 100 = 2100
Before the devotee climbed the Sixth temple: (2100+100)*2 + 100 = 4300
Before the devotee climbed the fifth temple: (4300+100)*2 + 100 = 8900
Before the devotee climbed the fourth temple: (8900+100)*2 + 100 = 18100
Before the devotee climbed the third temple: (18100+100)*2 + 100 = 36,500
Before the devotee climbed the second temple: (36500+100)*2 + 100 = 73300
Before the devotee climbed the first temple: (73300+100)*2 + 100 = 146900

Therefore, the devotee had Rs. 146900 with him initially.

Solution:

Yes it is possible in Roman Numerals. In Roman Numerals,
9 = IX
10 = X
11 = XI

Thus by removing one from 9, you are getting 10 and by removing one from 11, you are getting 10 again.

Solution:

There are three roses and four bees.

Solution:

1000

Solution:

No

Explanation:
7 is the only prime followed by a cube.
Assume that n3-1 is a prime for some n.
N3-1 will be equal to (n−1)(n2+n+1)
Now, n-1 divides n3-1.
If n-1>1, then we will have a contradiction to n3-1 being a prime.

Solution:

* First, the fireman and the athlete will go spending two minutes.
* The fireman will come back taking one minute.
* The drunk guy and the old lady will go spending five minutes.
* The athlete will come back taking two minutes.
* The athlete and fireman will go spending two minutes.

Two + One + Five + Two + Two = Twelve.

Thus all of them will be out of the house before it collapses.

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:

43

Let us break the problem into different parts.
We know that after six, all the numbers that are divisible by three can be ordered. This is because we can expand them as the sum off sixes and nines.
After twenty six, the numbers that are divisible by three if subtracted by twenty can also be ordered.
After forty six, all the numbers will fit in one of the already discussed category. Thus all the numbers can be ordered.
So forty three is the last number that does not come under any of the already discussed category.
44 = 20 + 6 * 4, 45 = 6 * 6 + 9

Solution:

I am DAVID.

At the core I have 5
If we talk in roman numeric system, 5 = V

Similarly, 500 stands in the beginning and end
500 = D

I also include the first letter i.e. A and the first number i.e. I (in roman numeric system)

All of them combine to form DAVID.

Solution:

54

Explanation:
7^2 + 2^2 + 1^1
6^2 + 3^2 + 3^2
2^2 + 5^2 + 5^2