Solution:

The correct answer to this twisted problem is four times.

You can try solving the problem with the help of two coins and if you are unable to understand, you may even disassemble some clockwork to find out.

Now to explain it, it is evident that the small one will turn 3 times against the big one. The reason is because it has 8 teeth which is 1/3 of the big one (24). But now we want the big wheel to stop and the small one to turn around the big one. In this process, the small one will make an additional turn on its way around the big one. This makes a total of four turns.

Solution:

The 100th statement for sure is incorrect because it says that at least 100 of the statements are incorrect.

Suppose if that is correct, then 100 statements cannot be true.
This suggests that the 100th statement is incorrect and that the first statement is true.

Similarly 99 statements cannot be true because if they were true, then two statements would become correct i.e. the 1st and the 99th.
But the 99th statement says that at least 99 are incorrect.
This suggests that the 99th statement is incorrect and that 2ndone is true.

If we keep analyzing is the same way till the end, we will find out that only the first fifty statements are true and all the remaining ones are incorrect.

Solution:

20
Just count the nines in the numbers: 6, 16, 26, 36, 46, 56, 76, 86, 96, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69.
Note: 66 contains TWO nines

Solution:

A) Yes

Explanation"
There are two scenarios.

Scenario-1: Millie is married
Millie is looking at Stones(not married)

Scenario-2: Millie is not married
Terry(married) is looking at Millie

Solution:

Four

three placed flat on the table in a triangle(touching each other) and put the fourth one on top of them in the middle.

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:

Never

Explanation:
In the question, the distance of x miles is given at a particular moment. The middle car is running at that time at 50 MPH. The distance will keep increasing with time as three are running at different speeds. It will increase by x miles every hour till a certain point of time and then it will start decreasing till the fastest car meets the slowest again.

As per question, all the cars stop then. Thus, the distance will never be x miles again.

Solution:

A cigar stuck in the wall

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:

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:

At 01:05