Solution:

73

Explanation:
It's a very famous puzzle known as "Josephus Problem Puzzle". It can be solved in various ways but the trick to solve this problem is illustrated below.

Binay number of 100 is: 1100100
complement pf 100 is 0011011, whose decimal equivalent is 27.
The desired answer will be 100 - 27 = 73

Solution:

It has been clearly mentioned that all the boxes are labeled incorrectly. If he opens the Box3, then he will get either Dark Chocolates or Milk Chocolates as it is labeled incorrectly. Let us suppose he finds Dark Chocolates in there. Now since all are labeled incorrectly, Box B A must contain Milk Chocolates and Box B must contain Milk Chocolates and Dark Chocolates.

Solution:

40 balls

Solution:

1

Explanation:
4 + 9 = 1
The trick to solve is to assume the first number as the time in AM and the second number is the number of hours you have to add to it. You'll find the time in PM when you do that.

For example
7 AM + 7 hours = 14:00 or 2 PM

4 AM + 9 hours = 13:00 or 1 PM.

Solution:

Karate Kid (car + at + tea + kid)

Solution:

Since Jasmine and Thibault ate equal amount, they should pay the same amount. But since that's not possible, they can decide with 'rock, paper, scissor' or an equivalent measure to determine who will pay the extra dollar. Now, let's assume tha Jasmine pays $1 extra after losing

Jasmine will pay $5, Thibault will pay $4 and Noah will pay $2.

If the pizza's cost was $11, we can simply keep the same offset and divide. Jasmine is paying 6% more, Thibault is paying 15% more and Noah is paying 27% more.
Keeping the offset same both Jasmine and Thibault will pay $5 each and Noah will pay $1.

Solution:

The ball will take infinite bounces before it is stopped. As per the question it will keep on bouncing half way up every time it hits the ground. But gravity will play its part at certain point of time which will make it stop.

Solution:

All Digits [0-9] follow the following pattern: There is one occurrence of all digits in every 10 numbers Example. Digit-9 occurred one time between 10 and 19 in 9, between 20 and 29 in 29
However, for tens(91,92,93...99) and hundreds (900,901.... 999), the occurrence is much more. Since we have included 1000, Therefor 1 is the most occurred digit.
Note: Despite three '0s' in 1000, 0 is not most occurred digit as they don't have tens and hundreds like other digits.

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.

Solution:

4 6 5 3 (from magician perspective)
3 5 6 4 (viewer perspective)

Detail:

From user perspective:
3 5 6 4 => 5 6 4 3 => 6 5 4 3

For magician perspective
3 4 5 6 (ascending order as)

Solution:

Both will rotate at Same speed.

Explanation:
Gears of the same size will always rotate at constant speeds despite any smaller or larger gears in between them.