Puzzle:
On Bagshot Island, there is an airport. The airport is the home base of an unlimited number of identical airplanes. Each airplane has a fuel capacity to allow it to fly exactly 1/2 way around the world, along a great circle. The planes have the ability to refuel in flight without loss of speed or spillage of fuel. Though the fuel is unlimited, the island is the only source of fuel.
What is the fewest number of aircraft necessary to get one plane all the way around the world assuming that all of the aircraft must return safely to the airport? How did you get to your answer?
Notes:
(1) Ignore extra fuel consumption as a result of acceleration, evaporation of fuel, bleeding-heart-liberal fiscal policies, etc.
(2) All the planes have to make it back safely, so you can’t give all your fuel away to another plane.
(3) Assume that refueling is an extremely fast process.
Puzzle 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 refueling process is applied analogously as for the first phase of the flight.
Simple answer:
A, B, and C starts. After burning 1/4th fuel each C fills A and B and returns. Now A and B tanks are full and C reaches island safely. After burning another 1/4th fuel each B fills A and left with half tank and returns home. Now A is tank full and is 1/4th the way. Now A can go 3/4th way before it is empty. As and when C reaches island it starts flying in reverse direction and reaches 1/4th way with 1/2 fuel burnt and meets A with empty tank. C fills 1/4th fuel in A and both starts flying towards the island. While B is coming to meet them at 1/8th distance away from island with 3/4th fuel left. Fills 1/4th fuel in each plane and all planes reach home safely with plane A covered the whole earth.
Used 3 planes and 5 fuel tanks.
I feel, 2 flights are enough.
A, B starts and at 1/4th of the distance, B fuels 1/4th of fuel to A and goes back.
With this A can travel 3/4th of total distance.
Now B starts in opposite direction (Earth is round). It just need to travel 1/4th to catch flight A.
Fuels it for 1/4th and both reach home.
Explanation with example:
Distance = 15+15 (kms)
A,B,C -> 15 (lts) each
{1} At 1/6 point, [ 1/6(30)=5 ] B,C have 10 each and B gives [1/3(15)=5 ] 5 to C, and can return with 5. (B=5;C=15;A=10)
{2} At 1/4 point, [ 1/4(30)=7.5 ] C=12.5 (20-7.5); A=7.5, C gives 7.5 to A. (C=5;A=15)
{3} C returns 1/4 to 1/6 (7.5-5=2.5), B is now at base and returns to help C, At 1/12 point, C=0, now B helps refuel, and both come base.
At this point A is at 2/4 point (half of circle), and B,C start traversing opposite side to help A.
At 5/6 point, B helps C (opposite of {1}) and returns.
At 3/4 point, [ 3/4(30)=22.5 ] A=0, and C helps A (opposite of {2}) and returns along A.
At 11/12 point, B helps C (opposite of {3})
Now all 3 returns to base.
This explanation is best for understanding…thanks
No doubt, only 3 plane is needed for completing full circle .
This can be solved with two planes only,
Explanation : considering the clock as measurement criteria.
A,B (considering both have 6ltrs capacity of fuel), both starts from 12 and reach 3,
B refuels A and goes back to 12 and refuels up.
Now Fuel : Position = A (6ltrs : 3) , B (6ltrs : 12)
Now A travels 6 points and reaches 9 , while B travels 3 points in anticlockwise direction and waited for A there with 3 ltrs of fuel.
Now Fuel : Position = A (0ltrs : 9) , B (3ltrs, 9)
B transfers 3 ltrs of fuel to 3 , So fuel left
A (3 : 9), B (3,9)
Both comfortably goes back to starting point 12 with 3 ltrs of petrol.
BINGO !
This is wrong please don’t consider this answer
Thanks!!!
Will post my version later