Puzzle | Camel and Banana Puzzle

A person has 3000 bananas and a camel. The person wants to transport the maximum number of bananas to a destination that is 1000 km away, using only the camel as a mode of transportation. The camel cannot carry more than 1000 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 a camel (no other mode of transportation is allowed)?

Things to note:

• Bananas available: 3000
• Distance to destination: 1000 KM
• Camel's capacity: 1000 bananas at a time
• Camel eats: 1 banana per kilometer

The goal is to find the maximum number of bananas that can be transferred to the destination.

Key Insights:

• The camel will need to make multiple trips because it can only carry 1000 bananas at a time.
• The camel consumes bananas during each trip, so you need to minimize waste by strategically using intermediate drop points where bananas can be stored.
• You need to calculate how many trips the camel will make and how many bananas it will consume at each stage.

Check if you were right - full answer with solution below.

Approach:

Step 1: From Source to Intermediate Point 1 (IP1)

• Initially, we have 3000 bananas.
• The camel needs to carry 3000 bananas but can only take 1000 at a time. It will need to make multiple trips (both forward and backward) to transport bananas to the first intermediate point.
• For every trip forward, the camel eats 1 banana per kilometer.
• The camel needs to make 5 trips between the source and IP1:
• 3 trips forward (to carry the bananas) and 2 trips backward (to pick up more bananas).
• For every kilometer covered, the camel consumes:
• 5 bananas per kilometer
• So, for the first intermediate point at distance 𝑥 from the source:
• The number of bananas left at IP1: 3000−5x

To maximize bananas at the intermediate point, let’s set the number of bananas left at IP1 to 2000 (as the camel cannot make more than 5 trips with the available bananas).

This gives us: 3000−5x = 2000 ⟹ x = 200 KM

Step 2: From Intermediate Point 1 (IP1) to Intermediate Point 2 (IP2)

• Now, 2000 bananas are at IP1, and the camel needs to transport them further.
• Here, the camel will make 3 trips (2 forward trips and 1 backward trip) between IP1 and IP2.
• For every kilometer covered between IP1 and IP2, the camel consumes:
• 3 bananas per kilometer
• So, for the second intermediate point at distance y from IP1:
• The number of bananas left at IP2: 2000−3y
• To maximize bananas at IP2, we set this value to 1000.

This gives us:

2000 − 3y = 1000 ⟹ y = 333.33 ≈ 333 KM

Step 3: From Intermediate Point 2 (IP2) to Destination

• Now, the camel has 1001 bananas left at IP2, but it can only carry 1000 at a time.
• So, it will leave 1 banana behind and proceed with 1000 bananas.
• The camel will travel the remaining distance z, which is: z = 1000 − (200 + 333) = 467 KM
• During this trip, the camel will consume 467 bananas, leaving:

1000 − 467 = 533 bananas

Final Answer:

The maximum number of bananas that can be transferred to the destination is 533.

This solution shows that through the use of two drop points and managing the number of trips carefully, the camel can transport a maximum of 533 bananas over the 1000 KM distance.