Solution:

120 times

Solution:

Tilt the Mug until the coke barely touches the lip of the Mug. If the bottom of the Mug is visible then it is less than half full. If the Mug bottom is still completly coverd by the coke, then it is more than half full.

Solution:

Forty.

The number of decks is irrelevant; the answer is the same if one or one-hundred decks are used.

Any card drawn will be a A,2,3,4,5,6,7,8,9,10,J,Q, or K, so there are 13 possibilities each time a card is drawn.

The fastest way to draw a four of a kind is if the first four cards all have the same 'value.' The slowest way, which provides the solution, is to first draw 13 three of a kinds, and then one more card.

Since 13 x 3 + 1 = 40, if 40 cards are drawn it is guaranteed that those forty cards contain at least one four of a kind.

Solution:

As the message contains one truth, the third says that the cash is in the second bag, if it is to be true, then the first bag message will also become true. So cash cannot be in second and third bags. cash is in the first bag.

Solution:

For one Friend Stewie will give Chocolate in a bag.

Solution:

This is simply the numbers 2^0,2^1,2^2 ... that is 1,2,4,8,16... So for making 1000 kg we need up to 1, 2, 4, 8, 16, 32, 64, 128, and 512

Solution:

Light both the ends of the first string and one end of the second string. 30 minutes will have passed when the first string is fully burned, which means 30 minutes have burned off the second string. Light the end of the second string and when it is fully burned, 45 minutes will have passed.

Solution:

The spirit lies behind Door 4.


If the spirit is behind Door 1, then both 2 and 3 are true.

If the spirit is behind Door 2, then both 1 and 4 are true.

If the spirit is behind Door 3, then 1, 3, and 4 are all true.

If the spirit is behind Door 4, then the statements on all the doors are false, except for that on Door 2. This matches the rules, and therefore, the resort hotel spirit lurks behind Door 4.

Solution:

I have written 'your exact weight' on the paper.

Solution:

As per the puzzle given above The fewest number of aircraft is 3! Imagine 3 aircraft (A, B and C). A is going to fly round the world. All three aircraft start at the same time in the same direction. After 1/6 of the circumference, B passes 1/3 of its fuel to C and returns home, where it is refueled and starts immediately again to follow A and C.

C continues to fly alongside A until they are 1/4 of the distance around the world. At this point C completely fills the tank of A which is now able to fly to a point 3/4 of the way around the world. C has now only 1/3 of its full fuel capacity left, not enough to get back to the home base. But the first 'auxiliary' aircraft reaches it in time in order to refuel it, and both 'auxiliary' aircraft are the able to return safely to the home base.

Now in the same manner as before both B and C fully refueled fly towards A. Again B refuels C and returns home to be refueled. C reaches A at the point where it has flown 3/4 around the world. All 3 aircraft can safely return to the home base, if the refueled process is applied analogously as for the first phase of the flight.