**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

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?

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.

So if we have stops just before every one km , the we can transport all 3000 bananas to the destinations

Very much confusing