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:

12 steps:

Steps 11 Liter jug 6 Liter Jug
1. 11 -
2. 5 6
3. 5 0
4. 0 5
5. 11 5
6. 10 6
7. 10 0
8. 4 6
9. 4 0
10. 0 4
11. 11 4
12. 9 6 ==> i found 9 liter in 11liter jug(as required)

Solution:

Simple strategy: Elect one person to be the guesser, the other two pass. The guesser chooses randomly 'green' or 'blue'. This gives them a 50% chance of winning.


Better strategy: If you see two blue or two green hats, then write down the opposite color, otherwise write down 'pass'.

It works like this ('-' means 'pass'):

Hats: GGG, Guess: BBB, Result: Lose
Hats: GGB, Guess: --B, Result: Win
Hats: GBG, Guess: -B-, Result: Win
Hats: GBB, Guess: G--, Result: Win
Hats: BGG, Guess: B--, Result: Win
Hats: BGB, Guess: -G-, Result: Win
Hats: BBG, Guess: --G, Result: Win
Hats: BBB, Guess: GGG, Result: Lose

Result: 75% chance of winning!

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:

From left to right
#1 Mr. Reed Richard – Blue Marbles
#2 Ms. Sue Storm – Red Marbles
#3 Mr. Johnny Storm – White Marbles
#4 Mr. Ben Grimm – Green Marbles.

Let us label the houses from left to right as #1, #2, #3, #4. Now let us split the hints in different category.
a) Mr. Reed Richard’s house is to the left of the green marbles one
b) The third one is made of white marbles
c) Ms. Sue Storm owns a red marbles house
d) Mr. Johnny Storm does not live on either ends
e) Mr. Johnny Storm lives right of the blue marbles house
f) Mr. Ben Grimm lives in the fourth house
g) The first house is not made up of red marbles

By seeking all the (g) and (b), we are sure that the #1 is not made up of red marbles and nor is the #3. Now with (f) we know that Mr. Ben Grimm lives in #4 made up of green marbles and keeping in mind (c) #2 must be made of Red Marbles and Ms. Sue Storm lives there. Now moving to the hint (d) Mr. Johnny Storm must be living in #3 and adding up with hint (b), it must be made from white marbles. Now by hint (a) #4 must be made up of green marbles (or Mr. Reed Richard’s house can’t be on its left). Also by (f) Mr. Ben Grimm lives there. All of it leaves Mr. Reed Richards living in #1 with blue marbles construction.