Puzzle:
The owner of a banana plantation has a camel. He wants to transport his 3000 bananas to the market, which is located after the desert. The distance between his banana plantation and the market is about 1000 kilometer. So he decided to take his camel to carry the bananas. The camel can carry at the maximum of 1000 bananas at a time, and it eats one banana for every kilometer it travels.
What is the most bananas you can bring over to your destination?
Solution:
First of all, the brute-force approach does not work. If the Camel starts by picking up the 1000 bananas and try to reach point B, then he will eat up all the 1000 bananas on the way and there will be no bananas left for him to return to point A.
So we have to take an approach that the Camel drops the bananas in between and then returns to point A to pick up bananas again.
Since there are 3000 bananas and the Camel can only carry 1000 bananas, he will have to make 3 trips to carry them all to any point in between.
<---p1---><--------p2-----><-----p3---->
A---------------------------------------->B
When bananas are reduced to 2000 then the Camel can shift them to another point in 2 trips and when the number of bananas left are <= 1000, then he should not return and only move forward.
In the first part, P1, to shift the bananas by 1Km, the Camel will have to
- Move forward with 1000 bananas – Will eat up 1 banana in the way forward
- Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
- Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
- Leave 998 banana after 1 km and return with 1 banana – will eat up 1 banana in the way back
- Will carry the last 1000 bananas from point a and move forward – will eat up 1 banana
Note: After point 5 the Camel does not need to return to point A again.
So to shift 3000 bananas by 1km, the Camel will eat up 5 bananas.
After moving to 200 km the Camel would have eaten up 1000 bananas and is now left with 2000 bananas.
Now in the Part P2, the Camel needs to do the following to shift the Bananas by 1km.
- Move forward with 1000 bananas – Will eat up 1 banana in the way forward
- Leave 998 banana after 1 km and return with 1 banana – will eat up this 1 banana in the way back
- Pick up the next 1000 bananas and move forward – Will eat up 1 banana in the way forward
Note: After point 3 the Camel does not need to return to the starting point of P2.
So to shift 2000 bananas by 1km, the Camel will eat up 3 bananas.
After moving to 333 km the camel would have eaten up 1000 bananas and is now left with the last 1000 bananas.
The Camel will actually be able to cover 333.33 km, I have ignored the decimal part because it will not make a difference in this example.
Hence the length of part P2 is 333 Km.
Now, for the last part, P3, the Camel only has to move forward. He has already covered 533 (200+333) out of 1000 km in Parts P1 & P2. Now he has to cover only 467 km and he has 1000 bananas.
He will eat up 467 bananas on the way forward, and at point B the Camel will be left with only 533 Bananas.
This puzzle has multiple interfaces, like some checkpoints where u need to pay the half of the load…. Really interesting puzzles… it’s difficult to answere, if you do not have looked it earlier……
What about its return journey of 1000 Kms. Since the camel its owner is required to come back its banana plantation site.
Sell the camel also at the market
What about the owner ?
nice idea…. i too guess the same
Its confusing
1st ( make trip to 200 km ) – Remaining bananas at A point is 2000.
2nd (make trip to next 333.33 km )- Remaining bananas at B point is 1000 .
Now travel next (1000-200-333.33= 467.67 km ) – Finally (1000-467.67= 533.33) bananas reached at destination
Sorry for language , didn’t want to type long .
Thanks. I summarize with 3 simple stages:
1. With 3000 bananas, Camel requires 5 bananas for each kilometer. This happens till the point Camel has 2000 bananas [ for A(0)->B(1)->A(1)->B(1)->A(1)->B(1) = 5 bananas ]
2. Camel will have 2000 bananas at 200th Kilometer. From now on, it requires 3 bananas for each kilometer. This happens till the point Camel has 1001 bananas [ for A(0)->B(1)->A(1)->B(1) = 3 bananas ]
3. Camel will have 1001 bananas at 533rd kilometer. From now on it requires 1 banana for each kilometer. [ Now here you may wish to send back your Camel back from 534th->533rd km just to eat 1 banana if you don't want to waste it, else continue with 1000 bananas. Nevertheless it doesn't make any difference whether you send or not ]
At 534th kilometer, Camel will have 999 bananas. This will end up having 533 bananas at 1000th kilometer (of course after Camel consuming 1 banana here )
533+ some fractional part
this question is based on the fact of maximum efficiency…..
carring capicity is 1000 so you have always to think how u can do it.
so,
initially i have 3000 i want it to remain 2000 when i will be at next stoppage..
so,
3000-5x=2000,it gives us x=200km,
again i want …i will have 1000 banana @ my next stoppage
so,
2000-3x=1000;it gives x=333.3333;
now i have 1000 banana and i have covered 200+333.33=533.3333km
so at the end i will left with
533.333 banana or 533 banan.
how u decided ’5x’ and ’3x’. why not some other numbers, instead of 5 and 3? Just want to understand. Still, not clear on sloution
Total d = 3000
No of banana = 3000
If camel
Total d = 3000
No of banana = 3000
If camel
Damn camel, eat up all my bananas!
can anyone provide explanation for neglecting decimals logically?
You cannot have a fraction of a banana because it is assumed that the camel eats 1 whole banana every kilometer.
great explanation
The above answer is wrong
if we have stops at points 333,555,703,837 and 1000 we will have 837 banana remains.
please explain how you get 837 bananas from this
So if we have stops just before every one km , the we can transport all 3000 bananas to the destinations
awesome bro great explanation according to you, u will always sleep before u feel hungry awesome :’)
Very much confusing
Why you left the owner hungry let him also eat the banana at every kilometer and make the puzzle more confusing.
What are the mathematical equation used for this and what are necessary variables taken? Please answer me on my mail id [email protected]
amazing ques
532 is right as we cant seperate or divide banana in decimal
So they don’t want to come back home
ANS: 500 Bananas
Travel 1st 500 kms in 3 trips with 1000 bananas each time.
So, totally camel will carry 3000 bananas and will be transporting 1500 bananas when it reaches half way.
Next 500 kms, it will carry 1000 bananas and will be transporting 500 bananas to the market.
It will return back and carry 500 bananas and will not be transporting any bananas to the market.
So, only 500 bananas can be transported to the market
Reach the market and sell that camel first
667 km will camel caries banana so.. Consumed is 667 remaining 333 bananas, in 3 trip means (333*3) =999 bananas… So then further 333km.. (999-333) =666 Banana.. .. But may @end point shop keeper won’t allow camel to eat banana means 667 bananas
667 km will camel caries banana so.. Consumed is 667 remaining 333 bananas, in 3 trip means (333*3) =999 bananas… So then further 333km.. (999-333) =666 Banana.. .. But may @end point shop keeper won’t allow camel to eat banana means 667 bananas
i like the way of explanation
At the starting point, instead of moving 1 km at a time, move 200 km at a time
1. Move forward with 1000 bananas – Will eat up 200 banana in the way forward
2. Leave 600 banana after 200 km and return with 200 banana – will eat up 200 banana in the way back
3. Pick up the next 1000 bananas and move forward – Will eat up 200 banana in the way forward
4. Leave 600 banana after 200 km and return with 200 banana – will eat up 200 banana in the way back
5. Will carry the last 1000 bananas from point A and move forward – will eat up 200 banana
After step 5 the Camel doesn’t need to return to point A again. So to shift 3000 bananas by 200 km, the Camel will eat up 1000 bananas.
Now in the Part P2, the Camel needs to do the following to shift the Bananas by 300 km.
1. Move forward with 1000 bananas – Will eat up 300 banana in the way forward
2. Leave 400 banana after 300 km and return with 300 banana – will eat up this 300 banana in the way back
3. Pick up the next 1000 bananas and move forward – Will eat up 300 banana in the way forward
After step 3 the Camel does not need to return to the starting point of P2. So to shift 2000 bananas by 300 km, the Camel will eat up 900 bananas and is now left with the last 1100 bananas.
Distance covered so far = 200 + 300 = 500 km
To cover remaining distance of 500 km, camel will eat 500 bananas
So, we are left with 1100 – 500 = 600 bananas
And how will finally owner and camel come back home after selling/not selling the leftover bananas in the market? ?
for 200km 200*5 bananas 1000, 5 bananas for 1km
for 334km 334*3 bananas 1002, 3 bananas for 1km
rest 1000-(200+334) 466km 998-466 =
532 is final answer not 533
Let the camel carry 1000 bananas to the market in round one and leave him to die on the way back. That stupid camel is a liability. At least the owner can peacefully live the rest if his life, without having to worry about that animal
It is really a tricky but a good one