Solution:

The time at first observation:
1:12

The time at second observation:
3:24

Thus, two hours and twelve minutes have passed between the two observations.

Solution:

Slumdog Millionare = Slum + Dog + Million + air

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:

It can be done in 8 moves as shown in the animated image below.

Solution:

There are two things to keep in mind:
Firstly there is at least one red hat. (There can be two or three as well).
Secondly the competition is fair for everyone.

Thus if there is only one red hat, that person will see two green hats on other heads and will be able to deduce his own color as red. However the other students will see one red and one green hat and can never be sure. In such manner, the competition will prove to be partial for one student.

Suppose if there are two red hats. Then the students who are wearing red hats will see one red and one green hat on others. Now they must have deduced that there can’t be just one red hat. Thus they will know that they are also wearing a red hat. But the one who is wearing a green hat will see two red hats and can never be sure of his own color. In this case as well, the competition will not be fair.

Thus the only possible and fair means is if all of them are wearing a red hat. The one who is able to deduce the situation first, will raise his hand and will tell the correct answer.

Solution:

The teacher will use a simple trick to form a perfect impression. He will ask all his students to raise their hands on each of the question that is asked. But the only catch will be that those who knows the answer correctly will raise their right hand and others will raise their left hand.

In this way, the principal will see all the hands being raised for each question even though all won't be knowing the correct answer. The teacher will ask only those who know the answer and they will always be correct. So the principal will be impressed to full extent.

Solution:

Move and then pour all juice from second glass to fifth glass.

Solution:

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

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:

1/2

one of two is the possibility
1. If any of the first 99 people sit in Abhishek seat, Abhishek will not get to sit in his own seat.
2. If any of the first 99 people sit in Aishwarya's seat, Abhishek will get to sit in his seat.

Solution:

If you are thinking to hold one chocolate from each box in hand and then balancing weight in bare hands, you are thinking all wrong.

Let us begin by labelling boxes as 1, 2, 3 and so on till 10.

Now pick one chocolate from box 1, two chocolates from box 2, three from box 3 and so on. In total, you will have 55 chocolates now. (1 + 2 + 3 + ..... + 10)

The ideal weight of the chocolates should be 55 * 20 = 1100. However, somewhere in there are the defected chocolate/s.

You can judge that clearly by noting down the result of 1100 - total weight of chocolates. If the weight is less than 1 gram, the defected box is box 1, if the weight is less than 2 grams, the defected box is box 2 and so on.