Puzzle 15 | (Camel and Banana Puzzle)

A person has bananas and a camel. The person wants to transport the maximum number of bananas to a destination which is KMs away, using only the camel as a mode of transportation. The camel cannot carry more than bananas at a time and eats a banana every km it travels. What is the maximum number of bananas that can be transferred to the destination using only camel (no other mode of transportation is allowed). 

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Solution: 
Let&#;s see what we can infer from the question:

• We have a total of bananas.
• The destination is KMs
• Only 1 mode of transport.
• Camel can carry a maximum of banana at a time.
• Camel eats a banana every km it travels.

With all these points, we can say that person won&#;t be able to transfer any banana to the destination as the camel is going to eat all the banana on its way to the destination.

But the trick here is to have intermediate drop points, then, the camel can make several short trips in between. 

Also, we try to maintain the number of bananas at each point to be multiple of .

Let&#;s have 2 drop points in between the source and destination.

With bananas at the source. at a first intermediate point and at 2nd intermediate point.

Source&#;&#;&#;&#;&#;IP1&#;&#;&#;&#;&#;&#;IP2&#;&#;&#;&#;&#;&#;&#;-Destination

      x km            y km                    z km

&#;&#;&#;&#;&#;&#;&#;>  |    &#;&#;&#;&#;&#;> | &#;&#;&#;&#;&#;&#;&#;&#;>

<&#;&#;&#;&#;&#;&#;-    |    <&#;&#;&#;&#;&#;  |

&#;&#;&#;&#;&#;&#;->    |    &#;&#;&#;&#;&#;> |

<&#;&#;&#;&#;&#;&#;      |                             |

&#;&#;&#;&#;&#;&#;->     |                             |

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• To go from source to IP1 point camel has to take a total of 5 trips 3 forward and 2 backward. Since we have bananas to transport.
• The same way from IP1 to IP2 camel has to take a total of 3 trips, 2 forward and 1 backward. Since we have bananas to transport.
• At last from IP2 to a destination only 1 forward move.

Let&#;s see the total number of bananas consumed at every point.

• From the source to IP1 its 5x bananas, as the distance between the source and IP1 is x km and the camel had 5 trips.
• From IP1 to IP2 its 3y bananas, as the distance between IP1 and IP2 is y km and the camel had 3 trips.
• From IP2 to destination its z bananas.

We now try to calculate the distance between the points:

1. &#; 5x = so we get x = 200
2. -3y = so we get y = 333.33 but here the distance is also the number of bananas and it cannot be fraction so we take y =333 and at IP2 we have the number of bananas equal , so its -3y =
3. So the remaining distance to the market is -x-y =z i.e  -200-333 => z =467.

Now, there are bananas at IP2. However the camel can carry only bananas at a time, so we need to leave one banana behind.

So from IP2 to the destination point camel eats 467 bananas. The remaining bananas are -467=533.
 

So the maximum number of bananas that can be transferred is 533.

Another Approach:

If the camel doesn&#;t eat a banana while returning, which means when it doesn&#;t have a banana, then the maximum number of bananas that can be transferred will be 833.

Let us divide the journey into 3 parts. First, the camel takes bananas and travels 333 KMs and puts the remaining bananas (-333 = 667) at first intermediate place and returns back to the source. Similarly, two more times, the camel takes bananas and travel for 333 KMs and put there remaining. Now, the total numbers of bananas are at the first intermediate place. 

Again, the camel takes bananas and travels 500 KMs and puts the remaining bananas(-500=500) at second intermediate place and returns back to first intermediate place. Similarly one more times, the camel takes bananas and travels for 500 KMs and put there remaining bananas. Now, the total number of bananas is at the second intermediate place. 

At last, the camel takes bananas and travels the remaining journey to 167 KMs. Thus, the total number of bananas is 833 at the final destination.

So, the maximum number of bananas can be transferred is 833, if the camel doesn&#;t eat a banana while returning.

Refer the following link for a detailed explanation: https://puzzling.stackexchange.com/a/232 

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