Lets take some simple cases first
c = 1: in this case person who has hat on his head can see that no one else has hat on their head, so he will understand that he is the one with hat.

c = 2: in this case person with hat will see one other person with hat, rest all will see two hats. Now had there been only 1 hat, this case would have been solved on very first day, but in this case no one will go on first night, so the guy who see one hat will understand that there must be one hat on his head, so both of them will go underwater on second night.

c = 3: in this case person with hat will see two hats and rest all will see two hats, now had there been 1 or 2 hats some guys would have gone on first or second night, thus on third day guys who see 2 hats will understand that they have hats on their head and they will all go underwater on third night.

Similarly we can see that it will take c days to remove all hats.

Given that the number of red and blue marbles are the same, we would tend to think that the odds are 50-50. we would try different combinations, such as 25 of each colored marble in a jar or putting all red marbles in one jar and all the blue in the other. we will still end up with a chance of 50%.

So lets think of a better way to distribute the marbles. What if we put a single red marble in one jar and the rest of the marbles in the other jar? This way, we are guaranteed at least a 50% chance of getting a red marble (since one marble picked at random, doesn’t leave any room for choice).  Now that you have 49 red marbles left in the other jar, you have a nearly even chance of picking a red marble (49 out of 99).

So let’s calculate the total probability.

P( red marble ) = P( Jar 1 ) * P( red marble in Jar 1 ) + P( Jar 2 ) * P( red marble in Jar 2 )
P( red marble ) = 0.5 * 1 + 0.5 * 49/99
P( red marble ) = 0.7474

Thus, we end up with ~75% chance of picking a red marble.

We can follow these steps.
Steps to measure 15 minutes

1. Light Rope #1 from both ends.
2. At the same time light Rope #2 from one end.
3. When Rope #1 burns fully, it will be 30 minutes and Rope #2 will take 30 more minutes to burn, light the other end of Rope #2 at this time, thus it will take 15 minutes to burn.
4. Start cooking your egg at the same point.
5. Stop cooking when Rope #2 burns fully.

The fewest number of cuts will be 2.

If we cut it into 3 part where the parts are 1, 2 and 4 segments long.
Day 1 – Give 1 segment bar
Day 2 – Give 2 segment bar ask for 1 segment bar back
Day 3 – Give 1 segment bar
Day 4 – Give 4 segment bar ask for 1 + 2 back
Day 5 – Give 1 segment bar
Day 6 – Give 2 segment bar ask for 1 segment bar back
Day 7 – Give 1 segment bar

The solution will be N.

Let’s start with the two person say A, B.
If A knows B then A can not be a celebrity.
If A does not know B then B can not be a celebrity.

So we strike out one name from our list, so on each question we can reject one name.

When there are only two people left in the list, then we will ask first person, “Do you know the second person ?”. If he says yes then second person is the celebrity and if he says no then the first person is the celebrity.

Thus we need to ask a maximum of N questions to correctly figure out celebrity name.

The answer will be 2 and 2.

If 1st person belongs to Truth tribe, he wud tell the truth that I belong to truth tribe,
And in 2nd scenario if 1st person belongs to Lie tribe then he’ll lie and would say I belong to Truth tribe.
So in both scenarios he wud say the same thing,
that means the second was person was lying,
which implies that the third person was telling the Truth..
So, the third person belongs to Truth tribe!

so the answer is Truth Tribe.

The probability of getting your seat is directly proportional to Steve remembers his seat no.
As we know that probability of Steve remembers and forgets seat no. is 1/2 i.e 50%-50%.

so the same with probability of getting your own seat.

so the answer is 1/2 i.e 50%.

If he (the smartest philosopher) had nothing on his head, then he realized that the second smartest philosopher would have quickly worked out that the third smartest was laughing only at the second smartest philosopher, and thus the second smartest philosopher would have stopped laughing.

The solution will be LCM (Lowest Common Multiple) of 10,9,8,7,6,5,4,3,2,1 -1 (for the monkey’s coconut). LCM would give the least number which is divisible by all of these number and subtracting one would give us the number of coconuts which were initially there.

So the answer is 2519