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Aeroplane Hardest Quiz

Aeroplane Hardest Quiz - 21 july

The puzzle question is : On Bagshot Island, there is an airport. The airport is the homebase 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?
(a) Each airplane must depart and return to the same airport, and that is the only airport they can land and refuel on ground.
(b) Each airplane must have enough fuel to return to airport.
(c) The time and fuel consumption of refueling can be ignored. (so we can also assume that one airplane can refuel more than one airplanes in air at the same time.)
(d) The amount of fuel airplanes carrying can be zero as long as the other airplane is refueling these airplanes. What is the fewest number of airplanes and number of tanks of fuel needed to accomplish this work? (we only need airplane to go around the world)

For Solution: Click Here


  1. This Is a Real Hard Quiz
    My ans is 7 which surely is wrong

    Nice title hardest quiz

    1. I got 7 planes as well. I dont see how you can do it with less given no plane can hold more than 4/8 fuel at any given point (if 8/8 is total distance) and all planes must return to the airport.

    2. Well, After looking at this closer.... If you reuse planes and can fill up more than one plane at a time, you can do this with 3 planes.

  2. Replies
    1. 3 will refuel it 50 % when it covers 25% of the distance, it will cover 75% of the total distance then, as world in round in shape, the third one will refuel it from the other side (50%). so each and every one will return safely.

    2. Your theory is correct, However... divide trip into 8 parts... plane one needs 8/8 of fuel to make trip, but can carry only 4/8ths at a time, since a plane must have gas to return, the farthest a plane can travel and still give fuel is 1/8th of the trip, but that only enables a plane to give 2/8ths in donation, as it needs one to get there, and one to get back... so, two planes gets one plane 6/8ths of the way around the globe. so, if we come from the other side with two planes, the first going 1/8th, and then donating 2/8ths to the other before turning and going home... then the 2nd plane from this side can meet the original plane at the 6/8ths mark, with enough fuel to get both planes home... thus 4 planes is the fewest planes possible.

    3. oh, except that the original plane can only accept one 8th in donation at the first fueling rendezvous, so the original plane can only make it 5/8ths... then, coming the other way, a plane travels 1/8th, before giving each of two accompanying planes 1/8th worth of fuel each, and returning home...those two in turn travel 1/8th further, putting them at 2/8ths around... one of them can donate then, 1/8th to the other and then return home. This would accomplish getting a single plane to meet the original, and return, but have no extra fuel to donate to the original... we still need 3/8ths for him, so, I will guess 8 planes, as each is able to receive 1/8th at... but now I'm seeing a better way... not all planes must leave at the same time...

    4. 3 planes, two planes starts the journey and fly until they have used close to 1/3 of their fuel capacity . This point is more than 1/4 of the journey but still one plane can refill the other one and safely return to starting point. The refilled plane will be able to fly 1/2 total distance, just passing the 3/4 point at which the third plane can then give it 1/3 of its fuel and both planes can land safely

  3. Replies
    1. 2 Planes.

      Why do I need fewer planes? Because of a simple loophole. The riddle states the the plane's fuel consumption is based upon distance traveled, not time spent in the air, allowing a plane to sit still, not consuming fuel.

      Using this, imagine two planes setting out from the airport, plane A and B. Both planes use fuel from plane A, keeping B at max. They go until plane A barely has enough fuel to make it home. Then plane B sits while plane A returns home, refuels, and comes back to continue the journey.
      At 25% around the world, plane A goes home, and plane B flies as far as it can (75%). Plane A then comes around from the other side of the world and they journey back slowly like they did before.
      Technically speaking, there is a problem as distance approaches 25%, as it would take infinitely many trips from plane A to fuel plane B an increasingly small amount. To avoid this technicality, I am assuming the airstrip is not a single point in space.

    2. When plane B is at 75% it has zero fuel. Since plane A must use half its fuel to get to 75%, when it gets there it won't be able to give any fuel to B and still get back. So how do you get A back without a plane C to help A get to 75% with enough fuel for A to get back too?

    3. Sorry, how do you get *B back?

  4. consider the fact that the earth is round so that the refuelling plane can fly from either way to refuel the plane..

  5. 4 planes. that is the correct answer.

    3 planes start and travel half their capacity.
    1st plane is filled completely from 2nd and half the fuel from 3rd is poured in 2nd. they travel back but they still require 25% fuel of total capacity to come back. hence the 4th plane is required which provides them 25% each and still has enough left to return back to airport.
    same case follows when the 1st planes makes the 3/4th trip round at other side.

    hence proved...:)

    1. That just blew my mind :o

    2. One of those planes that returned home could be the one to provide the second refueling.They said any plane can refuel at the island, so 3, not 4

    3. At 25% distance, refuel plane #1 can give 25% of its fuel to traveler plane. Traveler plane then has enough fuel to travel to 75% distance. Refuel plane #1 returns to thee island with 25% fuel to spare, and refills its tanks instantly. Refuel plane #1 then travels to meet traveler plane at 75% using 25% of it's fuel in the process, and gives 25% more fuel to get traveler plane home. So 2, not 3.

  6. 3. Planes go 1/8th the distance and 1 tops up the other 2 before returning to base arriving with no fuel. The other 2 make another 1/8th of the way before one of the pair tops up the other and returns to base again arriving with no fuel. The remaining plane then makes a half circuit of the world and is met by 1 of the other two at the 3/4 mark where the two planes share the remaining fuel. These two are met at the 7/8ths mark by the third whch has just enough fuel on board to get all 3 home.

    1. exactly what i was thinking :p

  7. the answer is two... first plane sets off and at 1/4 distance the plane is refuelled by the 2nd plane that then returns to base, the first plane can then travel to 3/4 of the world, the 2nd plane refuels and meets the plane at 3/4 in the opposite direction, refuelling the plane and both returning to the base

    1. When plane is refueled at 1/4 distance the plane
      Which gives fuel can't return to the island
      since 1/2 the fuel is required to travel 1/4 distance
      and remaining fuel is given to other net
      fuel is 0 at 1/4 distance of one plane so it can't
      return to island.

    2. 3 at most, they all take off at the same time, each of the fueling planes gives the actual one a quarter tank of fuel, it then flies around the world

  8. as fuel wasnt dictated as being finite in terms of carrying on board and wasnt specified as being taken out of the refuelling planes actual fuel, obviously time constraints this answer would be very unlikely if not impossible but in a theoretical world it is possible, slip streams etc

  9. My best answer is 6 tanks of fuel, from 4 planes, but setting off at the beginning at from different times. From the previous answers I don't see any proofs that show me how it this was done with only 6 tanks of fuel. Time will be measured by how far 1 tank of fuel enables a plane to fly. So 1 unit of time (1T) is how long it takes a plane to fly 1/2 the circumference of the earth
    Step - 1(time elapsed = 1/4T). 3 planes fly east (east or west, I just chose east, will be abbreviated to E or W) 1/8 the circumference of the earth (will be abbreviated to 1/8C). This takes 25% of their fuel, leaving 3 planes at 1/8CE with 75% fuel.
    Step - 2(no time elapsed). 1 of the 3 planes refills the other 2 planes. This leaves 1 plane with 25% fuel and 2 planes with 100% fuel at 1/8CE.
    Step - 3(time elapsed = 1/4T). The plane with 25% fuel returns home (a 1/8C journey) and runs out of fuel. The 2 other planes travel on 1/8C and arrive at 1/4CE with 75% fuel remaining in each.
    Step - 4(no time elapsed). 1 of the planes refuels the other. Now we have 1 plane with 50% fuel and 1 plane with 100% fuel at 1/4CE
    Step - 5(time elapsed = 1/2T). The plane with 50% fuel returns home (1/4C takes 50% fuel). The plane with 100% fuel flies on another 1/4C. This puts him at 1/2C with 50% fuel left.
    Step - 6(time elapsed = 1/4T). The 2 planes that have returned home refuel and take off with another plane, the fourth plane. These 3 planes travel 1/8CW, leaving them with 75% fuel. The plane at 1/2C travels on another 1/8C leaving him with 25% fuel and putting him at 5/8CE or what we may now call 3/8CW.
    Step - 7(no time elapsed). 1 of the planes at 1/8CW refuels the other 2 leaving 1 plane with 25% fuel and 2 planes with 100% fuel.
    Step - 8 (time elapsed = 1/4T). The plane with 25% fuel at 1/8C returns home and is finished. The other 2 planes at 1/8CW fly on another 1/8C to 1/4CW, leaving them with 75% fuel each. The plane at 3/8CW now flies 1/8C to join them at 1/4CW, leaving him with 0% fuel.
    Step - 9 (no time elapsed). Each of the 2 planes with 75% fuel give 25% to the third plane with 0% fuel. There are now 3 planes with 50% fuel at 1/4CW.
    Step - 10(time elapsed = 1/2T). Now the 3 planes at 1/4CW return home with 1 having traversed the entire circumference of the earth.

    Time elapsed = 2T Tanks of Fuel used = 6 Planes used = 4
    (2T = the duration of a single flight around the earth)
    I believe this is the fewest tanks used and fewest planes used possible.

  10. This comment has been removed by a blog administrator.

  11. 1.

    You can always carry extra fuel tank.

  12. This comment has been removed by the author.

  13. 3
    How about this.. B and C travel first 1/8th distance with A. then they pass on the fuel to A, keeping sufficient enough for B and C to travel back home. So A has its fuel share + 1/4 th from B+ 1/4 th from C. Therefore it travels half of the distance on its own fuel. then half of remaining half(1/4) using B's fuel and then remaining 1/4 using C's fuel.
    Answer: 3
    I find no flaw. Simple is better :)

    1. That means that at 1/8th distance, A has enough fuel to travel halfway around the world(from B and C) AND has some extra fuel(it's own). The limit on fuel a plane can is only enough to take it halfway across the world, so I think that solution is wrong...

  14. A,B,C travel 1/8th the distance consuming 25% fo their fuel,at this point C shares 50% of its fuel with A & B, thus A & B have 100% of the fuel and C comes back with the remaining 25% fuel.
    Now A & B travel futher 1/8th of the distance, at ths point B share 25% of it's fuel with A, thus A have 100% fule and B comes back with remaining 50%.
    Now A continues its journey and when A reaches midway, it will have 50% fo the fuel. at this point B & C will start their journey in opposite direction. when A has covered 5/8th of journey, B & C would have covered 1/4th journey from opposite side, thus C shares 25% of its fuel with B and goes back. B continues its journey towards A. when A and B meet at 3/4th of the journey, A will not have any fuel and B will have 75% fuel. At C leaves again with and B shares 25% fuel with A and B goes back to airport with remaining 50% fuel.
    When A and C meet at 7/8th of the journey C shares 25% of its fule with A.
    This way 3 planes and 5.5 tanks of fuel is needed.

    1. BINGO !! the best answer .. 3 planes and 5.5 tanks of fuel, this is the most efficient solution

    2. Seconded! Anuj has the best answer I've seen

    3. You are correct! This is the solution I came up with too! Kudos!

  15. 3

    two A,B planes start from east to west...

    after 1/3 distance B refuels A with 1/3 fuel
    B has 1/3rd fuel left and can return 1/3 distance

    Hence A now has full capacity fuel and can travel 1/2 globe
    Total travel = 1/3 + 1/2 = 5/6 distance
    remaining = 1/6 of globe

    Another aircraft C takes off from airport from west to east (opposite of first)

    It refuels A and 1/6 distance can be covered easily

  16. The answer is 2 airplanes with 2.5 tanks of gas.

    Plane A will circle the globe.
    Plane B will provide fuel.
    Refuel time = 0 (from puzzle).
    Both planes travel at same speed (because they are identical)

    Plane A and B take off heading east, at 1/4 plane B refills plane A. This allows plane A to travel from 1/4 to 3/4. Plane B returns to the airport, re-fuels and travels west to meet plane A at 3/4.

    At 3/4 Plane B re-fuels plane A and they both return to the airport.

    We can use two airplanes because while plane A is traveling from 1/4 to 3/4 (east to west half way around the globe) plane B can travel from 1/4 to 0 to 3/4 (West to east half way around the globe).

    We only need 2.5 tanks of gas because plane B travels from 0-> 1/4 ->0 and from 0-> 3/4 -> 0 for a total of 1 tank and transfers 1/4 of a tank twice once at 1/4 and once at 3/4.

    1. after refilling Plane A at 1/4 distance Plane B has no fuel left to return back.

  17. only 2 planes are needed. each with 1 extra tank of gas, and they refuel each other at 1/4 trip intervals.

    example: plane1 refuels half of plane2's tank when they are 1/4 of the way around the world, thus expending half of plane1's extra tank. plane2 refuels all of plane1's tank when they are 1/2 way, thus expending plane2's extra tank. at the 3/4 mark plane1 once again refuels plane2, and when they arrive at Bagshot Island both planes will have just run out of fuel.

  18. 2 planes lets say x and y ..first x and y start with full tanks right after takeoff y starts refuellng x until 1/4th the dist is covered and the fuel lost is refilled in the course for onwards x continues to fly until it reaches 3/4th of the distance in the meantime y returns to the base and flies the other way round until it meets x at 3/4 mark and starts returning with it refuelling it for the rest of the course

  19. 4 planes Suppose it takes 360 l and distance is 360 kms. Fuel rate is 1km/l. Total fuel for each plane is 180 l. A,B goes to 60km (1/3 of 180 km), transfers 60 l from B to A. B comes back. A reaches (180+60)240 km.

    Other way round B,C flies 60 km ie 300th km. C transfers fuel to C and returns. B meets A at 240th km spending 60 l. B now has 120 l. B transfers 60 l to A. Both A, B reaches 300th km.

    C, D reaches 300th km. C has 120 l, D has 120 l, A has 0 l, B has 0 l. C, D transfers 60 l each to A, B. All returns to 360th km.

  20. A small modification can be made to the solution provided by Anuj Jain.
    After A has completed its half journey, B will start from the island in opposite direction. It will meet A at 3/4th of the journey. At this point B will transfer 25% of his fuel to A and now B will also have 25% fuel. They will continue their journey and at the same time C will take off from the island. At 7/8th of the journey both A and B haven't have any fuel. C will transfer 25% fuel to A and 25% fuel to B and left with 25% fuel. Using this all three can safely return to the island.
    Hence the solution takes 3 planes and 5 tanks.

  21. I got 6:
    - Plane 1 (**who will go around the world**, travelling EAST) flies 1/6th of the way (consumes 1/3 of his fuel).
    - Plane 2 travels 1/6th of the way (with Plane 1), transfers 1/3 of his fuel to Plane 1 and returns to base (0% fuel remaining).
    - Plane 1 continues flying east as far as he can (4/6ths or 2/3 around the world) (0% fuel remaining).
    - Plane 3 and 4 travel WEST. Plane 4 transfers 1/3 of his fuel to plane 3, then Plane 4 returns home (0% fuel remaining).
    - Plane 3 continue west and meets up with Plane 1 (2/3 fuel remaining), gives 1/3 to Plane 1 and they both fly west 1/6th of the way around the world. They are both out of fuel and at the 5/6th mark aroudn the world.
    - Planes 5 and 6 (same as Planes 2 and 4) fly west to meet them and transfer 1/3 of their fuel to Planes 1 and 3 respectively. All 4 planes are at the 5/6th mark with 1/3 tank of fuel.
    - Planes 1,3,5 and 6 fly east back to base (all 0% fuel) and Plane 1 has made it around the world.

    Step 0 1/6 2/6 1/2 4/6 5/6 0
    0 12 3456
    1 12 3456
    2 2 1
    3 2 1 34 56
    4 2 13 456
    5 2 13 456
    6 2 1356 4
    7 2 13456


    1. Should read East.. oops

      Plane 3 continue west and meets up with Plane 1 (2/3 fuel remaining), gives 1/3 to Plane 1 and they both fly >>>WEST<<< 1/6th of the way around the world. They are both out of fuel and at the 5/6th mark aroudn the world.


    2. I think yours is right. Same principle (using 1/3) as the answer I just submitted, but I used 14 planes while you used only 6. The difference I think is that I went only east to west, and you took advantage of going in both directions.

  22. damn lost the diagram formatting..



    1. scale up the top is slightly off, stupid font.. anyway, at step 0, planes 1&2 are in base (0, heading east) and planes 3,4,5&6 are MEANT to be in base also (0 at the end, heading west).

  23. 3 Planes, 5 Tanks of fuel
    Planes A, B, and C leave the island. At 1/8th of the way around the world all planes have 3/4 tanks. Plane C fills Plane A and Plane B. At this point A and B are full, and Plane C is at 1/4. Plane C returns home. Plane A and B travel to 1/4th of the way around, and at this point both planes are at 3/4 full. Plane B tops off A, leaving Plane A full and Plane B at 1/2. At this point, Plane B returns home. Plane A then travels halfway around to 3/4th of the way around the world. Here he is met by Plane C which refueled at the island, and is now at 1/2 tank (Plane A is empty). Plane C shares fuel with Plane A, leaving both planes with 1/4 tank. They fly to 7/8th around the world and meet with Plane B which refueled at the island and is now at 3/4 a tank. Plane B shares 1/4 tank with both Plane A and C, leaving all with 1/4 tank, which is enough to fly home.

    So, 3 planes: A, B, and C
    And, 5 tanks: All 3 started with full tanks, and B and C had to each refuel once.

  24. two planes piggy back so the piggy back plane pumps its fuel to the one flying. there you go all around the world.

  25. *I presume that each plane has ONE fuel tank that it can fly or refuel from.
    One plane can make it around the world half way, so we just need to send planes to go fetch that one exactly half way round.
    I calculate 26 planes are needed to fetch the 27th.
    Basically when one eastward plane embarks, 8 leave westward (1st sortie), one third of the way to the rendezvous point, half these planes give half their remaining fuel to the other half. The first half now has one third fuel remaining (enough to get home) the other half now has full tanks again and keep going. Those 4 do the same after another third of the way and the last 2 again after another third. That leaves one plane with a full tank reaching the rendezvous point and immediately refueling the plane coming from ahead. A 2nd sortie of 12 planes leaves a while later to fetch the turn around's from the 1st sortie and a 3rd sortie of 6 planes a while later still to fetch the final 2 coming home.

  26. Answer is 3 Planes and 5 tanks of feul...

  27. YOUre all WAY overthinking this one

    Plane 1&2&3 leave together Heading East

    @ the 25% mark (half tank of fuel left) 2&3 give half of their remaining fuel (1/4 tank each) to plane 1 topping it off and leaving them each 1/4 tank since they will only make it half way back one more plane, plane 4, leaves and meets them halfway home giving them 1/4 tank each to make the safe return for all 3

    plane 1 now has a full tank and can make it to 75%

    Now, planes 5&6 leave Heading West to meet plane 1 at the 75 % and do the same thing, giving plane 1 a half tank to make the last 1/4 of the journey, and plane 7 to meet 5&6 half way back, giving them all 1/4 tank and enough to safely return,

    NOW that is 7 planes and 7 tanks of fuel, it can be cut to 4 if planes 2,3,4 refuel at the base and make the second leg of the trip.

    ALSO, since the riddle says the airport is the only place on ground that the planes can refuel, it leaves the option for an aircraft carrier ON WATER for the plane to refuel reducing the number of planes to 1,

    Correct answer is technically 4 planes and 7 tanks of fuel if you follow the rules.

    1. Ans is 3 planes.
      3 planes A,B,C start together.After reaching 1/4 of the distance c will refuel both A & B(each 1/4th of fuel) and c returns.
      now A & B after going another 1/4 distance, B will refuel A(1/4th fuel) and B now A has 1/2 of fuel xtra,that means it can go for an other half distance. and the same two planes B&C will refuel A from the other the ans is 2 aircrafts needed for a plane to go around.totally 3planes.
      consider a semicircle and make it 4parts.u can easily understand

  28. 14 planes is the answer

    Think of there being two way stations located one-third and two-thirds of the way around the semi-great-circle to Bagshot Island's "antipode". (I.e the stations are 1/6 of the great circle apart, or 1/3 of a tank of gas.)

    At any time 2 planes can start from the island and travel to the first way station, using 1/3 their fuel. One plane can then fill the tank of the other plane using the second third of its fuel and return to the island while exhausting the final third -- leaving one plane at the first way station with a full tank.

    Repeat this procedure a second time (using 4 planes in all) and you end up with 2 planes with full tanks at the first station.

    Using the same technique, these planes can both fly to the second way station, transfer a third of a tank of fuel from one to the other, filling one tank while leaving in the other the third of a tank needed to return to the first way station -- where this plane will meet plane number 5, which has just flown from the island and has a spare third of a tank of fuel to transfer and a final third for itself so that both these planes can get back to the island.

    So all of the above results in a total of 5 planes having been flown, one of which is at station two with a full tank and the other 4 have returned safely to the island.

    Now repeat all of the above with a second set of 5 planes. This leaves two planes with full tanks of gas at station two (2/3 of half way around).

    Now fly both of these planes to the "antipode", requiring a third of a tank of gas from each, and transfer 1/3 of a tank from one to the other so that one plane is full and the other has a third of a tank left.

    The plane at the antipode that is full now has enough gas to complete its journey around the great circle.

    All that remains is to safely bring home the other plane, which was last seen at the antipode with 1/3 tank of gas, which is enough to enable it to return to the second way station.

    To get back to the first way station, it will need to meet another plane at the second station that has left from the first station with a full tank, so after a transfer of 1/3 tank both planes can make it back to the first station. To get that other plane to the first station would have required two planes to fly from the island and transfer 1/3 of a tank of gas (see paragraph 3 above).

    So we have now used 12 planes in total and still have two empty planes at the first way station needing to get home to the island. Each of these requires a single "rescue" plane each flying from the island to share a third of a tank of fuel. Making 14 in all

    1. Oops, Anonymous October 17, 2012 at 5:39 AM did it better. In 6.

  29. The answer is 3 , when planes reaches 25% a plane comes to refill it and enables it to reach 75% and another plane flies in opposite direction to give it fuel at 75% (earth is round) 25% more fuel to enable it reach its destination

  30. One plane. Why couldn't it refuel itself?

  31. This comment has been removed by the author.

  32. 6 planes are needed, sounds stupid, but it is true. They all have 4/8 fuel. Start with 3 planes going 1 direction.

    distance 1/8: nr 3 (has 3/8) refuels nr 1 and 2 (need 1/8) and flies back (need 1/8).
    2/8: nr 2 (has 3/8) refuels nr 1 (need 1/8) and flies back (need 2/8).
    nr 1 with full fuel goes to 6/8 distance.

    The problem is getting back. Nr 4, who he meets at 6/8, and who has already refuled at 7/8 with nr 5, would have only 3/8 fuel. While nr 1 needs 2/8 to get back and nr 5 is already going back. So nr 1 needs to meet with nr6 at 7/8 distance, to get another 1/8 fuel. So though it sounds weird, you need 6 planes.

  33. I read the given answer. I has a big error. Firstly, a plane can travel half the world. Let us call a full tank say 6/12 (=1/2). You say B passes 1/3 of its fuel to C after 1/3 max distance. So in other words. It passes 2/12 fuel to C after 2/12 distance. So C has a full thank again. After 1/4 distance world you say C fully refuels A. So after 3/12 distance. At this point A has 3/12 tank and C has 5/12 tank. In order to fill A,it needs 3/12, leaving C it with 2/12 and unable to return home.

    Why did it go wrong. Because you idiots should have made B fly less distance. At 1,5/12 distance. It could have given 1,5/12 to A and 1,5/12 to C, leaving it 1,5/12 to fly back. So B traveled less and gave more fuel (by also filling A), making it possible.

    In short. You fail, please correct it.

  34. 3
    It is certain that no plane has the tank capacity to carry fuel beyond what is needed to make a halfway trip
    A+B travel 1/4th of the circle. A gives B its remaining 1/4th so that B can cover 3/4th in total of the whole circle while A waits there.
    So the scenario is A and B are waiting at the opposite ends of the circle [A at 1/4th and B at 3/4th]. Now 2 other planes (C and D) are required to bring back A from 1/4 th position. Now the same set of C and D can bring back B from 3/4th position. So B makes the full circle trip while 3 planes (A,C and D) assist it.

  35. The answer is 3. You must watch the timing and where each plane A, B, and C is at each time. Let’s say it will take 12 hours to fly around the world and a full tank holds 6 units of fuel. In this example plane A will make the full trip.
    0:00 o’clock all planes leave the island together eastward at 0°longitude.
    A fuel = 6 units; B fuel = 6 units; C fuel = 6 units; Total 18 units of fuel

    1:00 o’clock all planes continue together eastward at 30°E longitude.
    A fuel = 5 units; B fuel = 5 units; C fuel = 5 units; Total 15 units of fuel

    2:00 o’clock all planes are together eastward at 60°E longitude.
    A fuel = 4 units; B fuel = 4 units; C fuel = 4 units; Total 12 units of fuel
    Plane B gives 2 units of fuel to plane A and plane B doubles back heading westward.
    A fuel = 6 units; B fuel = 2 units; C fuel = 4 units; Total 12 units of fuel

    3:00 o’clock plane A and plane C are together eastward at 90°E; plane B westward at 30°E longitude
    A fuel = 5 units; B fuel = 1 unit; C fuel = 3 units; Total 9 units of fuel
    Plane C give plane A 1 unit of fuel and plane C doubles back heading westward.
    A fuel = 6 units; B fuel = 1 unit; C fuel = 2 units; Total 9 units of fuel

    4:00 o’clock plane A eastward at 120°E; plane B lands at 0°; plane C westward at 60°E longitude.
    A fuel = 5 units; B fuel = 0 units; C fuel = 1 unit; Total 6 units of fuel
    Plane B refuels and heads eastward.
    A fuel = 5 units; B fuel = 6 units; C fuel = 1 unit; Total 12 units of fuel

    5:00 o’clock plane A eastward at 150°E; plane B eastward at 30°E; plane C westward at 30°E longitude.
    A fuel = 4 units; B fuel = 5 units; C fuel = 0 units; Total 9 units of fuel
    Plane B doubles back heading westward and gives plane C 1 unit of fuel.
    A fuel = 4 units; B fuel = 4 units; C fuel = 1 unit; Total 9 units of fuel

    6:00 o’clock plane A eastward at 180° longitude; plane B and plane C land together at 0° longitude.
    A fuel = 3 units; B fuel = 3 units; C fuel = 0 units; Total 6 units of fuel
    Plane B and Plane C refuel and both planes head westward together.
    A fuel = 3 units; B fuel = 6 units; C fuel = 6 units; Total 15 units of fuel

    7:00 o’clock plane A eastward at 150°W; plane B and plane C together westward at 30°W longitude.
    A fuel = 2 units; B fuel = 5 units; C fuel = 5 units; Total 12 units of fuel

    8:00 o’clock plane A eastward at 120°W; plane B and plane C together westward at 60°W longitude.
    A fuel = 1 units; B fuel = 4 units; C fuel = 4 units; Total 9 units of fuel
    Plane B gives 2 units of fuel to Plane C and Plane B doubles back eastward.
    A fuel = 1 unit; B fuel = 2 units; C fuel = 6 units; Total 9 units of fuel

    9:00 o’clock plane A eastward at 90°W; plane B eastward at 30°W; plane C westward at 90°W longitude.
    A fuel = 0 units; B fuel = 1 unit; C fuel = 5 units; Total 6 units of fuel
    Plane C doubles back eastward and gives plane A 3 units of fuel.
    A fuel = 3 units; B fuel = 1 unit; C fuel = 2 units; Total 6 units of fuel

    10:00 o’clock plane A and plane C eastward together at 60°W; plane B lands at 0° longitude.
    A fuel = 2 units; B fuel = 0 units; C fuel = 1 unit; Total 3 units of fuel
    Plane B refuels and heads westward.
    A fuel = 2 units; B fuel = 6 units; C fuel = 1 unit; Total 9 units of fuel

    11:00 o’clock plane A eastward at 30°W; plane B westward at 30°W; plane C eastward at 30°W longitude
    A fuel = 1 unit; B fuel = 5 units; C fuel = 0 units; Total 6 units of fuel
    Plane B doubles back eastward and gives 1 unit of fuel to plane C.
    A fuel = 1 unit; B fuel = 4 units; C fuel = 1 unit; Total 6 units of fuel

    12:00 o’clock all three planes land together at 0° longitude.
    A fuel = 0 units; B fuel = 3 units; C fuel = 0 units; Total 3 units of fuel

    Fuel used
    0:00 18 units
    4:00 6 units
    6:00 9 units
    10:00 6 units
    12:00 -3 units (the fuel remaining in plane B)
    A total of 36 units of fuel were used. At 6 units per tank this equals 6 tanks.
    Or you can think of it this way. The three planes were flying all of the time for 12 hours consuming 1 unit of fuel per hour per plane. 3 x 12 = 36 units or 6 tanks.

  36. The problem said "In a long circle" which means we have the liberty to move any of the planes used for fueling purposes to move in opposite direction to meet the plane at the other end.. in that case 3 will become the answer..

  37. Since time is not of the essence...put the plane on a boat...

  38. 3

    Plane A and B fly clockwise.
    At 1/4th distance, Plane B gives 50% fuel to Plane A, and returns to the base (using remaining 1/4th)

    Plane A can travel to 3/4th distance now.
    Plane B refuels and goes in anti-clockwise direction and meets Plane A at 1/4th distance, fuels it. and both return happily to base

    1. At 1/4th distance pane B will only have 50% of its fuel. If it gives this to plane A then Plane B will be out of fuel and not get back home.

    2. At 1/4 distance B cannot give 2/4 fuel to A because A already has 3/4 fuel and would be full...

  39. Because the earth rotates, it can be done with 2 planes.

    Assume Bagshot Island is on the equator, pick a route that goes over the poles and fly the plane at a speed fast enough for it to use all its fuel in 12 hrs.

    Thus it goes,

    Plane 1 leaves, flies north, over north pole and then due south. Since we apparently can determine the rate, let's say this journey takes either 12 hrs... guess where the Bagshot Island is now. Plane 2 flies up and refuels Plane 1 when it is directly over the island, and lands. Since refueling time and consumption during refueling is said to be negligible, Plane 2 leaves Plane 1 a full tank to go over the south pole and back, again taking exactly 12 hrs... landing back on Bagshot Island. QED, Plane 1 circuits the globe, 2 planes total, 2 tanks of gas.

    The only problem with this approach that I can see is that the distance traveled in the air would indeed take the plane half way around the world allowing it to make it back over the island, but the route as measured on the ground would be more than half the circumference of the globe. Its not clear which interpretation the author intended.

    BTW, an argument could made for a 1 plane journey... If the island is on the equator and the plan flew due west using it's fuel at a rate that would make it last >12 hrs, the island would appear underneath the plane again before it runs out of fuel. The plane would have passed over the entire equator (its route thus appearing as a great circle) but it would have only traveled (in the air) < or = half the circumference of the globe (when viewed from above, the plane does a half circle). This solution is less than satisfactory though as, by extension, you could argue for "a lift off, enter orbit, and then wait 24 hrs" expending zero fuel, waiting for the island to appear below again clearly not what the author had in mind.

    If we are to disregard the rotation of the earth, then 3 planes/5 tanks is the min

  40. %100 : 7 PLANE

    1. Following the prerequisites of the puzzle, 2 planes and 4 tanks of fuel are required.
      Plane A flies around the world. Plane B flies alongside plane A and transfers ½ its fuel to plane A at the ¼ mark, then returns to base to refuel. Plane A now has enough fuel to get to the ¾ point. Plane B now flies in the opposite direction to meet plane A at the ¾ distance mark and again transfers ½ its load of fuel to Plane A and then flies alongside plane A back to base. Plane A uses a total of 2 tanks of fuel. Plane B uses a total of 2 tanks of fuel.

  41. 2 planes.
    First, one plane is used to refill the plane when they have covered 1/3rd of the earth. Then the plane returns safely.
    The main plane will now travel another 1/2 of the world as its full.
    Another plane will be sent from other direction to meet this empty plane and fill it.

    1. After 1/3 of the earth the plane only has 1/3 of fuel left... Because it only can cover half the earth in the first place. It can't pass fuel it does not have.

  42. Lets assume the total distance as 2000 miles and as mentioned each plane can fly half of the journey i.e 1000 miles.
    Assuming the plane fuel consumption is 1000miles/ 1000 liter
    Now see below
    Plane flying journey total Remaining fuel Remaining fuel given to other plane
    1 1 999 998->pass it to third plane
    2 1 999 998-> pass it to third plane
    3 1 2996

    Total journey is 2000 so the third plane will be able to complete the whole journey successfully. Thus the answer should be 3

  43. My answer is, 1 plane.

    If distance is the only factor in fuel consumption, then altitude could play a very important role.
    What if, at the half way point, the planes altitude is high enough to glide the rest of the trip.
    The airplane being used was never described or defined. If we imagine this airplane has the capability to glide, and that the weather permits optimal travel conditions, then you only need 1

  44. 1 plane is required, earth is also rotating about is axis, let us move the plane is opposite direction to what earth rotate and in the same speed of earth rotation. so when plane completes its half rotation, earth will also rotate half. And due to relative velocity, the plane will be at its original place, i.e. at the island, so no concept of fuel tank blah blah blah...

  45. Total 6 Planes and 6Tanks required 3 from one side to send one plane upto 3/4th of total distance and 3 planes from other side to pull this empty plane to the end point.

    Like: 3 planes go upto 1/8th distance and each have 3/4th fuel fill two planes full and remains one with 1/4th fuel and it go back.....2 full planes go upto 1/4th distance and have fuel 3/4th each and fill one planes full and remains one with 1/2 fuel and it go one plane can go upto 3/4th distance and tank is the path is circular so from other side if we send 3 planes then as above 2 planes at 1/4 distance have fuel 3/4th each and from other side with empty tank are at same point. we fill all three plane with 1/2 tank and will go to the point.

    1. we can reuse planes so total planes required 3+1=4 and 6tanks fuel

    2. This comment has been removed by the author.

    3. again good one 3 planes and 5tanks fuel

      3 will push one plane upto 1/4th distance and full tank....again same 1 plane will go 1/4th in opposite side and meet the plane now both plane can have 1/4tank fuel last one plane come upto 1/8th with 3/4tank fuel meet these 2 planes and can have all planes 1/4th tank fuels so can reach back to the point. :)

  46. again good one 3 planes and 5tanks fuel

    3 will push one plane upto 1/4th distance and full tank....again same 1 plane will go 1/4th in opposite side and meet the plane now both plane can have 1/4tank fuel last one plane come upto 1/8th with 3/4tank fuel meet these 2 planes and can have all planes 1/4th tank fuels so can reach back to the point. :)

  47. This comment has been removed by the author.