Solution:

Let us analyze the statements given by Peeta, Eugene and Harry.

If Peeta is right by saying that there are at least four Rembrandt, then Harry will also be right (he said that there is at least one).

If Harry is right by saying that there is at least one Rembrandt, then Peeta and Eugene will also be right.

In that case, only Eugene's statement can be true that Leonardo has less than four Rembrandt and that too in the case that he has no painting of Rembrandt at all (otherwise, Harry's statement will be true as well).

Thus Leonardo has no painting of Rembrandt at all.

Solution:

Now there can be two solutions to the given situation because there can be two situations:
Suppose if B and C have same colors, A will know that his color is the other one and he will be able to speak up the first.
Now if B and C do not have same colors, A will stay silent. This will tell B that his and C's colors are different. Thus he will speak up first.

So either A or B will speak up first depending on the situation.

Solution:

99 matches.

The number of matches will always sum up to one less than the number of players in a knock out tournament. You may calculate it in any manner. Thus 99 matches were played.

Solution:

The Lady sitting in the chair has the detector attached to the back of her blouse.

Solution:

Let us assume that all of them are standing on the X shore and they have to travel to the Y shore.

X = mother, first daughter, second daughter, father, first son, second son, maid and dog
Y = None

Maid and dog go first and maid returns back
X = mother, first daughter, second daughter, father, first son, second son and maid
Y = dog

Maid and first son go and maid returns with dog
X = mother, first daughter, second daughter, father, second son, maid and dog
Y = first son

Father and second son go and father returns back
X = mother, first daughter, second daughter, father, maid and dog
Y = first son and second son

Mother and father go and mother comes back
X = mother, first daughter, second daughter, maid and dog
Y = father, first son and second son

Maid and Dog go and Father returns back
X = mother, first daughter, second daughter and father
Y = first son, second son, maid and dog

Father and Mother go and mother comes back
X = mother, first daughter and second daughter
Y = father, first son, second son, maid and dog

Mother and first daughter go and maid comes back with dog
X = second daughter, maid and dog
Y = mother, first daughter, father, first son and second son

Maid and second daughter go and maid comes back
X = Maid and dog
Y = mother, first daughter, second daughter, father, first son and second son

Maid and dog go
X= none
Y = mother, first daughter, second daughter, father, first son, second son, maid and dog

Thus everyone has reached the other end.

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:

Four

As the window was closed it is obvious that the six candles that were not extinguished were lit till they melted away entirely. Thus only four candles that were extinguished remain at the end.

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.

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:

17 mins

The initial solution most people will think of is to use the fastest person as an usher to guide everyone across. How long would that take? 10 + 1 + 7 + 1 + 2 = 21 mins. Is that it? No. That would make this question too simple even as a warm up question.

Let’s brainstorm a little further. To reduce the amount of time, we should find a way for 10 and 7 to go together. If they cross together, then we need one of them to come back to get the others. That would not be ideal. How do we get around that? Maybe we can have 1 waiting on the other side to bring the torch back. Ahaa, we are getting closer. The fastest way to get 1 across and be back is to use 2 to usher 1 across. So let’s put all this together.

1 and 2 go cross
2 comes back
7 and 10 go across
1 comes back
1 and 2 go across (done)

Total time = 2 + 2 + 10 + 1 + 2 = 17 mins

Solution:

Think deeply about the door number 56, people will visit it for every divisor it has. So 56 has 1 & 56, 2 & 28, 4 & 14, 7 & 8. So on pass 1, the 1st person will open the door; pass 2, 2nd one will close it; pass 4, open; pass 7, close; pass 8, open; pass 14, close; pass 28, open; pass 56, close.
Thus we can say that the door will just end up back in its original state for each pair of divisor. But what about the cases in which the pair of divisor has analogous number for example door number 16? 16 has the divisors 1 & 16, 2 & 8, 4&4. But 4 is recurrent because 16 is a perfect square, so you will only visit door number 16, on pass 1, 2, 4, 8 and 16… leaving it open at the end. So only perfect square doors will remain open at the end.