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 probability of two numbers being relatively prime is 60%. Thus if you keep playing the bet, you will hypothetically make 20 cents on every play.

Thus, you should accept the bet.

Solution:

The bridge won't break.

When the car has travelled half of the bridge, that is 15km, the car must have used more than 300 grams of fuel. Therefore, the impact of bird's added weight will be zero literally.

Solution:

They used the planks in the manner that is shown in the figure.

Solution:

The password is U472BMT.

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

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:

Yes, you should switch the envelope. In the beginning when you picked up the envelope, you had a 1/3 probability of finding an offer letter in the envelope. There was 2/3 chance that the letter was there in the two envelopes on the table.

If you keep your selected envelope, you still have a 1/3 chance of finding an offer letter in that. However, since the interviewer has removed one empty envelope from the table, if you switch, you have a probability of 2/3 that the offer letter is inside that.

Solution:

The figure tells that B is located to the North-East of E.

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.

Solution:

The stranger said that he thought the room was his. If so, he must have had the keys to the room and thus he must not have knocked.