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.

Solution:

You can put one precious stone in any one of the bag and all the rest in the other bag. If you do it, you will get just a shade under ¾ probability of picking up a precious stone.

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:

1. Start both the 7 minute hour glass & 11 minute hour glass.
2. Wait till the 7 minute hour glass times out. Time is 7 minute!
3. Restart the 7 minute hour glass. At this time 11 minute hour glass will have 4 minutes left to time out.
4. As soon as 11 minute glass times out invert the 7 minute hour glass. Total time now is 11 minutes.
5. After inverting 7 minute hour glass, it will now have 4 minutes left for time out.
6. After these 4 minutes times out, the total time is 15 minutes.

Solution:

Only twenty-three people need be in the room, a surprisingly small number. The probability that there will not be two matching birthdays is then, ignoring leap years, 365x364x363x...x343/365 over 23 which is approximately 0.493. this is less than half, and therefore the probability that a pair occurs is greater than 50-50. With as few as fourteen people in the room the chances are better than 50-50 that a pair will have birthdays on the same day or on consecutive days.