The Camel and Banana 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?

As many as 'these' numbers of bananas can be saved!

The Camel and Banana Puzzle : Solution

What was the puzzle? 

 Let A be the starting point and B be the destination in this transportation. If the camel is taken with 1000 bananas at start, to reach the point B which is 1000 km away from A, it needs 1000 bananas. So there will be no bananas left to return back to point A.

That's why we need to break down the journey into 3 parts.

Part 1 :

For every 1 km the camel needs to -

1. Move ahead 1 km with 1000 bananas but eat 1 banana in a way.

2. Leave 998 bananas at the point and take 1 banana to return back to previous point.

3. Pick up another 1000 bananas and move forward while eating 1 banana.

4. Drop 998 bananas at the same point. Return back to previous point by consuming 1 banana.

5. Pick left over 1000 bananas and move 1km forward while consuming 1 banana to same point where 998 + 998 bananas are dropped. Now, the camel doesn't need to  return back to previous point. So, 998 + 998 + 999 are carried to the point.

That is for every 1km, the camel needs 5 bananas.

After 200 km from point A, the camel eats of 200x5 = 1000 bananas and at this point the part 1 ends.

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PART 2 :

1. Move ahead 1 km with 1000 bananas but eat 1 banana in a way.

2. Leave 998 bananas at the point and take 1 banana to return back to previous point.

3. Pick up another 1000 bananas and move forward to the point where 998 bananas left while eating 1 banana.

Now, the camel needs only 3 bananas per km.

So for next 333 km, the camel eats up 333x3 = 999 bananas.

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PART 3 :

So far, the camel has travelled 200 + 333 = 533 km from point A and needs to cover 1000 - 533 = 467 km more to reach at B.

Number of bananas left are 3000 - 1000 - 999 = 1001.

Now, instead of wasting another 3 bananas for next 1 km here, better drop 1 banana at the point P2 and move ahead to B with 1000 bananas. This time the camel doesn't need to go back at previous points & can move ahead straightaway.

For the remaining distance of 467 km, the camel eats up another 467 bananas and in the end 1000 - 467 = 533 bananas will be left.

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A Check Post At Each Mile

A poor villager grows mango in his land and sells them in the town. The town is 1000 miles away from the village. He has rented a truck for transporting the mangoes to the town. The truck can carry 1000 mangoes at one time and this season, he was able to yield 3000 mangoes.

There is a problem. At each mile till the town, there is a check post at which he must give one mango each while traveling towards the town. However, if he is traveling from the town towards his village, he won’t have to give anything.

Transportation Truck

Tell a way in which the villager can take highest possible number of mangoes to the town.

Know here that highest possible number!

Source 

Smart Saving At Check Posts

How much each check post charging?

Obviously, he can't make 3 trips from town to village straightaway as in that case he wouldn't have anything left (3 x 1000 mangoes paid).

So he need to divide the journey into parts. While breaking journey into parts he has to make sure that after each part he will need less trips to complete the next part.

Now if somehow he pays 1000 mangoes in first part of the journey then for next part he has to make only 2 trips to carry 2000 mangoes.

Part 1 : Hence, he should first make 3 trips till 333 miles. In this part, he would pay 3 x 333 = 999 mangoes leaving 3000 - 999 = 2001 mangoes in stock.

Part : 1

Part 2 : He should leave 1 mango here & take 2000 mangoes further. For next part, he need to make at least 2 trips for 2000 mangoes. In order to save number of trips in next part some how he need to make mangoes in stock less than 1000. For that he should make 2 trips 500 mile further. So he will pay 2 x 500 = 1000 mangoes but having 2000 - 1000 = 1000 mangoes in stock. Still he has to travel 1000 - 500 - 337 = 167 miles.

Part : 2

Part 3 : For next 167 miles, he need to make only 1 trip of 1000 mangoes where he will pay 167 mangoes leaving 1000 - 167 = 833 mangoes. 

Part : 3

This is how he can save 833 mangoes in entire journey. 

Logic Problem: The Trainee Technician

A 120 wire cable has been laid firmly underground between two telephone exchanges located 10km apart.Unfortunately after the cable was laid it was discovered to be the wrong type, the problem is the individual wires are not labeled. There is no visual way of knowing which wire is which and thus connections at either end is not immediately possible.

You are a trainee technician and your boss has asked you to identify and label the wires at both ends without ripping it all up. You have no transport and only a battery and light bulb to test continuity. You do have tape and pen for labeling the wires.

What is the shortest distance in kilometers you will need to walk to correctly identify and label each wire?

Know here the efficient way! 

Source 

To Be A Skilled Technician

What was the task to test the skill? 

The shortest distance is 20 km! Surprised? Read further.

Let's name the two exchanges as a 1 & 2 respectively. Now at end 1, let's make a groups of wires having 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15 number of wires. Now somebody might ask why not 15 groups having 8 wires in each. After reading the entire process here, we'll get the answer of it.So we 15 groups have total 1 + 2 + 3....+ 15 = 120 wires. Let's name these groups as A, B, C, D ...... O. That means group A has 1, B has 2, C has 3 wires & so on.

Now join together all the wires of the particular group. For example, 2 wires of group B should be joined together, 7 wires of G tied together & so on. The sole wire of A is left as it is.

We will take the battery & bulb to other end traveling 10 km. We will say a wire is paired with the other if the bulb gets illuminated if battery & bulb connected in between.

Now let's take any wire at the other end & find the number of wires that are pairing with that particular wire under test. We will group such wires & label with those exactly how we labeled at end 1.

For example, if we find 2 wires pairing with particular wire then that wire & 2 paired wires together to be grouped in 3 wires & labeled as C.The sole wire not getting paired with any will be labeled as A. And group with wire pairing with 7 other wires together should be labeled as H.

In this way, we will have the exact group structure that we have at end 1. By now, we have identified & labeled correctly wires in groups of 1,2,3,....15 wires at both ends.

Now, we are going to label each wire of group by it's group & count number. For example, the only wire in group A labeled as A1, 2 wires in B are labeled as B1,B2, wires in G are labeled as G1,G2,G3,G4,G5,G6,G7 and so on.

Now, at end 2 itself, what we are going to do is connecting first wire of each group to A1. Second wire of each group to be connected to B2, third of each to be connected to C3 and so on. (refer the diagram above, where labels of wires that are to be connected together are written in same color).