another mind boggling puzzle...just give it a try!
May 28, 2007 5:36am CST
hello friends! here is another great puzzle from my side. do try it and keep posting! there is a desert which is 1000 kilometers long and a camel which can carry a maximum of 1000 bananas. there are 3000 bananas on one side, to be taken to the other side of the desert. however, the condition is that the camel eats one banana for each kilometer that it travels. what is the maximum number of bananas that the camel can transport to the other side of the desert? please post your reasoning as well along with the solution or the answer.
8 Jun 07
Trip 1, one third of the way and back, eats 666 bananas leaves 333 deposited Trip 2, one third of the way it has eaten 333 bananas but replenishes these from cache of trip 1, walks a further sixth of the way leaves a cache of 333 and walks back eating the remaining bananas it carries There is no a cache of 333 bananas at the half way point. By the time it reaches the half way point it has consumed 500 of the 1000 but picks up 333 so it has 833, it walks the remaining 500 km eating 500 leaving 333 bananas which it delivers. + 1/3 of a banana as I have rounded down throughout. So the answer is 333 and a 1/3 The owner of the bananas wishes he had chosen a more efficient from of transport! all the best urban
15 Jun 07
For me, camel needs to travel with 4 pit stop (every 250kms) For the first 250kms=camel carry 1000 bananas less the 250 bananas (eaten by camel, since evry km 1 banana will be eat by the camel) with 750 bananas left, then less another 250 for wayback to get the additional 1000 bananas. Left is 500 bananas, in total 1500 bananas will be left for the first 250 kms the camel travels. for the next 250kms (500 kms total travel) same procedure. first 1000 bananas left 500 bananas (two-way). With 500 bananas left in the first pit stop. To get these, camel will consume 250 bananas with 250 bananas left plus the 500 bananas. Total of 750 bananas in the second pit stop. Another 250 kms (total travel is 750 kms) Camel will now travel only once with 250 bananas eaten. Left if 500 bananas. For the last travel, additonal 250 kms (total travel is 1000kms) camel will now only eat additional 250 bananas with 250 bananas left. So the maximum number of bananas the camel will transport to the other side of the dessert is 250. This is my conclusion for other possibilities: If the camel use one pit stop (1000 kms to carry the 1000 bananas) camel will eat all the bananas and cannot go back to the other side because the camel already ate all the bananas. If the camel use two pit stop (regardless of kms for every pit stop) same will happen using the one pit stop. Try these and same result will happen. If the camel use 3 pit stop (lets say 300,300,and 400 kms). Camel will carry only 1200 bananas after the first pit stop. To go the next pit stop. Camel will transport first the first 1000 and left only 400 and the camel can no longer carry the 200 bananas left in the first pit stop because the camel needs 300 bananas to be eat. Same will happen if other combination of kms will be used.
29 May 07
The answer would be an optimum number based on a mathematical model. Optimum solution can be achieved by taking arbitrary points x,y.. along the way collecting the bananas on those points and load the camal to it's fullest capacity i.e. 1000 bananas per trip. Model will be as follows: Starting with 3000 bananas with full loading of camal there will be 5 trips to reach a point say "x" along the way to get 2000 bananas left, i.e: 3000 - 5x(bananas consumed by camel)=2000; x = 200 kilometers Applying same method, next stop at distance "y" from "x" would be with 1000 bananas left to load the camel, to it's fullest capacity, to be optimum: so there will be 3 trips now: 2000 - 3y (bananas consumed by camel) = 1000; y = 333.33 kilometers Now camel has traveled 533.33 kilometers consuming 2000 bananas and there are 467.67 kilometers left with 1000 bananas in hand, Now again fulling loading the camel and consuming 467.67 bananas along the way we will remain with at least 533.33 bananas which is the optimum solution for this problem. So the answer is 533.33 bananas:-)
28 May 07
As the camel can take a maximum of 1000 and can eat 1 every 1 km, i think that the camel will have to eat one before he goes, he can then carry the 1000, he will only be able to leave one at the end of the journey though. I think you have completely lost me.
• Sri Lanka
28 May 07
The camel has 1000km to go 3000 bananas to take with a maximum load of 1000bananas and mileage 1banana per kilo meter. The only way to achieve this is to keep the bananas somewhere in the middle and return. The camel has to do 2 return trips and 1 one way trip to take the 3000 bananas. So first it will move 200 kilometers consuming 1000 bananas. Now it is left with 2000 bananas, and 800 kilo meters. To transport this it will do one return trip and one one-way trip. This needs 3 bananas per kilo meter. So it will go until 1000 bananas are left. It moves 333 kilo meters with another 1000 bananas. Now it is left with 1000 bananas, and 467 kilo meters. So it can carry all 1000 and go on. By the time the trip is over it will eat another 467 bananas. It has now carried 533 bananas to the other side.
29 May 07
hey that was a great try buddy! now i m not going to post whether it is the right or wrong answer, else it is going to spoil the fun of this puzzle. but then you need to see why you have assumed the 200+800 division. what if the camel goes 300+700 or 500+500? how can you be sure that in those cases the answer is not going to be more bananas transported? how can you be sure that your division is the optimal one? also, why just 2 stages, the camel can as well go in 5 phases of 200km each! or 3 phases. there are many such cases possible. so how do you know that your case is the best one?
• Sri Lanka
29 May 07
I think the answer is correct, but you have a standard way of arriving at it and wants to have it. But it is difficult to explain it unles we are chatting with a web cam. If the camel does anything less than 200 kilo meters in the first journey, then it cannot limit the number of bananas to 2000 for the next journey. Even if there are 2001 bannanas left it will have to do 3 trips again. So the answer of 533 is the optimum.