Solution:

This is a Perception riddle and hence there is no right or wrong attached to this answer.

It can be considered as a sound
> It makes the sounds of branches snapping and the almighty thump when the trunk hits the hard ground. The sound is sound and doesn't rely on people to interpret it.
> Yes. And isn't that a rather arrogant question, as what about all the animals that presumably hear it?

It cannot be considered as a sound because
> Does it depend on what you mean by 'sound'? The airwaves would still vibrate but without an ear/listening device to receive them, there would be no sound.
> The sound is something within human experience. Outside of this, it's just air (or whatever another medium) vibrating. If someone is there to hear it, it makes a sound, if not, it causes rapid movement of particles.

Solution:

The password is U472BMT.

Read aloud the line and you will sound something like "U Four Seven 2 B M T".

Solution:

16

Explanation:
9 [1x1] squares
5 [2x2] squares
1 [4x4] square
1 [4x4] square

9 + 5 + 1 + 1 = 16 squares.

Solution:

This can be done if you draw a five pointed star and place the balls on the corners and intersection points. See the figure to understand better.

Solution:

TIF56

Explanation:
The password is formed as:
T (5th letter of Shruti)
I (5th letter of Arshika)
F(1st letter of Friday)
5(5th day of week)
6(Length of Friday)

Similarly, we will pick the 1st letter for Monday and 3rd letter for Wednesday.

Solution:

Isaac Newton picks up a piece of wood and lights it from the fire on the west end of the island.

He then quickly carries it near the east end of he island and starts a new fire. The wind will cause that fire to burn out the eastern end and he can then shelter in the burnt area.

(Funny Part : Isaac Newton survives the fire, but dies of starvation, with all the food in the forest burnt )

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:

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 best thing for you will be to shoot your first shot in air.

If you try to shoot at Jack and hits him by chance, Leo always hits on the target and you are dead.

If you choose to shoot at Leo and hits him by chance, then there is a fifty percent chance that Jack will hit you.

If you shoot in air, the next turn is of Jack and he knows that Leo is a better shooter which means that he will shoot at him. If he misses, he is definitely dead as Leo will shoot on him coz he shot at him. The next turn will be yours and you stand a 1/3rd chance of hitting him. Also if, he is able to hit Leo. The next chance is yours again and you can try the 1/3rd probability of hitting over Jack.

This is the best situation that you can face. At every other possible way, you will face a worse situation for sure.

Solution:

So that the following plan can be followed, let us number the coins from 1 to 12. For the first weighing let us put on the left pan coins 1,2,3,4 and on the right pan coins 5,6,7,8.

There are two possibilities. Either they balance, or they don't. If they balance, then the different coin is in the group 9,10,11,12. So for our second one possibility is to weigh 9,10,11 against 1,2,3

(1) They balance, in which case you know 12 is the different coin, and you just weigh it against any other to determine whether it is heavy or light.
(2) 9,10,11 is heavy. In this case, you know that the different coin is 9, 10, or 11, and that that coin is heavy. Simply weigh 9 against 10; if they balance, 11 is the heavy coin. If not, the heavier one is the heavy coin.
(3) 9,10,11 is light. Proceed as in the step above, but the coin you're looking for is the light one.

That was the easy part.

What if the first weighing 1,2,3,4 vs 5,6,7,8 does not balance? Then any one of these coins could be the different coin. Now, in order to proceed, we must keep track of which side is heavy for each of the following weighings.

Suppose that 5,6,7,8 is the heavy side. We now weigh 1,5,6 against 2,7,8. If they balance, then the different coin is either 3 or 4. Weigh 4 against 9, a known good coin. If they balance then the different coin is 3, otherwise it is 4. The direction of the tilts can tell us whwther the offending coin is heavier or lighter.

Now, if 1,5,6 vs 2,7,8 does not balance, and 2,7,8 is the heavy side, then either 7 or 8 is a different, heavy coin, or 1 is a different, light coin.

For the third weighing, weigh 7 against 8. Whichever side is heavy is the different coin. If they balance, then 1 is the different coin. Should the weighing of 1,5, 6 vs 2,7,8 show 1,5,6 to be the heavy side, then either 5 or 6 is a different heavy coin or 2 is a light different coin. Weigh 5 against 6. The heavier one is the different coin. If they balance, then 2 is a different light coin.

Solution:

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