Question : A banana plantation is located next to a desert. The plantation owner has 3000 bananas that he wants to transport to the market by camel, across a 1000 kilometre stretch of desert. The owner has only one camel, which carries a maximum of 1000 bananas at any moment in time, and eats one banana every kilometre it travels. What is the largest number of bananas that can be delivered at the market?

The Solution: 533 1/3 bananas. Explanation: Since there are 3000 bananas and the camel can carry at most 1000 bananas, at least five trips are needed to carry away all bananas from the plantation P (three trips away from the plantation and two return trips): P (plantation) ===forth===> <===back==== ===forth===> <===back==== ===forth===>

A Point A in the abouve picture cannot be the market. This is because the camel can never travel more than 500 kilometres into the desert if it should return to the plantation (the camel eats a banana every kilometre it travels!). So point A lies somewhere in the desert between the plantation and the market. From point A to the next point, less than five trips must be used to transport the bananas to that next point. We arrive at the following global solution to the problem (P denotes the plantation, M denotes the market): P (plantation) ===forth===> <===back==== ===forth===> <===back==== ===forth===>

A ===forth===> <===back==== ===forth===> B ===forth===> M (market) Note that section PA must be in the solution (as explained above), but section AB or section BM might have a length of 0. Let us now look at the costs of each part of the route. One kilometre on section PA costs 5 bananas. One kilometre on section AB costs 3 bananas. One kilometre on section BM costs 1 banana. To save bananas, we should make sure that the length of PA is less than the length of AB and that the length of AB is less than the length of BM. Since PA is greater than 0, we conclude that AB is greater than 0 and that BM is greater than 0. The camel can carry away at most 2000 bananas from point A. This means the distance between P and A must be chosen such that exactly 2000 bananas arrive in point A. When PA would be chosen smaller, more than 2000 bananas would arrive in A, but the surplus can't be transported further. When PA would be chosen larger, we are losing more bananas to the camel than necessary. Now we can calculate the length of PA: 3000-5*PA=2000, so PA=200 kilometres. Note that this distance is less than 500 kilometres, so the camel can travel back from A to P. The situation in point B is similar to that in point A. The camel can't transport more than 1000 bananas from point B to the market M. Therefore, the distance between A and B must be chosen such that exactly 1000 bananas arrive in point B. Now we can calculate the length of AB: 2000-3*AB=1000, so AB=333 1/3. Note that this distance is less than 500 kilometres, so the camel can travel back from B to A. It follows that BM=1000-200-333 1/3=466 2/3 kilometres. As a result, the camel arrives at the market with 1000-466 2/3=533 1/3 bananas. The full scenario looks as follows: first, the camel takes 1000 bananas to point A. There it drops 600 bananas and returns with 200 bananas. Then the camel takes again 1000 bananas to point A. Again, it drops 600 bananas and returns with 200 bananas. After this, the camel takes the last 1000 bananas from the plantation to point A. From point A, it leaves with 1000 bananas to point B. In point B, it drops 333 1/3 bananas and returns with 333 1/3 bananas. Then it takes the second load of 1000 bananas from point A to point B. Finally, it carries the 1000 bananas from point B to the market, where it arrives with 533 1/3 bananas.

Answer is 533 and 1/3 banana

If Camel just picks up a load of 1,000 bananas and heads out across the desert, she will eat them all up by the time she gets to the other side. She will also leave 2,000 bananas, unused, to rot back at the oasis. The trick is to use those 2,000 bananas as fuel to get the remaining 1,000 bananas as far across the desert as possible, before Camel makes her final dash for the market.

Camel needs to eat five bananas per mile so long as she is trying to ferry more than 2,000 bananas. Later, when she's hauling between 1,000 and 2,000 bananas, she needs three bananas per mile. And after that, she only eats one banana per mile.

To understand why, let's start at the beginning.

Camel is standing there in the oasis with 3,000 bananas. She picks up the first 1,000. Say she carries them just one mile into the sand, eating one banana. She could drop 999 bananas there, but then she couldn't walk back. So, being a camel with foresight, she drops 998 bananas and keeps one to eat on the return trip.

Now she can pick up the second 1,000 bananas and do the same thing, dropping 998 at the one-mile marker and shambling back to the oasis.

With the third load, there's no return trip: all her bananas have been moved one mile.

How many did she eat up? Five: two on the first round trip, two on the second, and one on the last trip, which is one-way.

She could keep this up, one mile at a time, for 200 miles, by which time she would have used up 1,000 bananas. Or she could just take the first load 200 miles, drop 600 bananas, go back, pick up the next 1,000, etc. Either way, she will find her self at the 200-mile marker with 2,000 bananas.

(Note that there are no monkeys or hungry humans out there in the sand dunes, and no other camels, either. Camel feels her bananas will be safe when she drops a load in the desert and goes back for more.)

Once she has the 2,000 bananas out in the desert, Camel the Mathematical Camel reasons that she now needs three bananas per mile to push her stash farther: 1 round trip for the first load of bananas and 1 one-way trip for the second load. Either with her calculator or with mental math, she determines that she will use up the second 1,000 bananas moving the supply forward 333 1/3 miles. She can either proceed in one-mile increments, or go the whole 333 1/3 miles at once, or anything in between. In the end, Camel finds herself with 1,000 bananas 533 1/3 miles (200 + 333 1/3) into her journey.

It's hot, but Camel takes a deep breath, picks up the 1,000 bananas, and slogs on. This time she can just keep going with no return trips, because she hasn't left any bananas in the desert - just in her stomach.

433 2/3 miles farther on, and lighter by 433 2/3 bananas (she's a nibbler), Camel pads out of the desert and into the market, where a mob of camel-lovers and mathematicians is waiting to pay her handsomely for the 533 1/3 bananas (1,000 - 433 2/3) she has left. She even sells that last 1/3 of a banana to a souvenir hunter from the Annenberg Channel.

In short

533 1/3 bananas well 533 anyway

First leg-out, back, out, back, out. 5 one way trips bananas consumed 1000, bananas moved 2000 1000/5=200 units. Status 2000 bananas at unit 200

Second leg out, back, out. 3 one way trips Bananas consumed 1000, bananas moved 1000 1000/3=333 1/3 units Status 1000 bananas at unit 200+333 1/3=533 1/3

Third les one trip 466 2/3 units (1000-533 1/3) Bananas consumed 466 2/3 Bananas delivered on far side 533 1/3.

The number of Bananas that were transported was "0". The camel starts with 1000 bananas and it eats 1 banana for each km. So by the moment it reaches the market the number of bananas wil be "0".

the camel can never end up carrying any bananas as it wud eat one banana for one km, and the dist being 1000 km, there wont be any banana left, even though there are 3000 bananas , since it cannot carry more than 1000 at a time , theres no possibility of any bananas reaching the destination

There will be 0 banana.bcause camel can carry only 1000 banana at a time and there are 3000 banana so to reach the market camel has to travel 3 times and camel eats one banana per kilometer thus no banana will left.

There will be no bananas left to be transported to the market....coz the camel eats 1 banana for every 1km, thus all 1000 bananas are over, by the time the the market is reached...

0 banana will be transported bcoz camel can carry 1000 bananas maximum and he eats one banana at one km so there are two cases 1) suppose camel carries less than 1ooo bananas to mkt he ate one banana at each one km so wat will be reached to mkt.............. nothing since he has to cover 1000 km distance even his own need did not fulfill bcoz he needed minimum 1000 bananans for himself only but he had less than 1000 bananas 2) now if he carries 1000 bananas to mkt now wat will happened his own need will be fulfilled bat he has nothing left to carry to mkt 3) he cannot carry more than 1000 bananas if it is possible then some bananas wounld reach to mkt

3 banana will be transported because in first trip camel carry maximum 1000 banana and he eat 1 banana per 1km when he reach 999km after that he has eaten 999 banana and rest 1 banana dilever to distance so that in three trip he eat =999*3 rest banana=1000*3-2997=3

as the camel could carry only 1000 banannas and eat 1 bananna in every 1 km of 1000 km .the bananna that will reach in the market are all eaten by the camel.1000 bananna will be reaching along with the camel.but their will be no more bananna to be sold in the market.3000 banannas will be raching in the market

Consider, the camel travels 250kms from plantation with 1000 bananas on its each trip. So, 1.500 bananas will be kept at 250km and 500 will be eaten(to and fro) in its first trip. 2.second trip is same as above. 3.As 1000 left, it will eat 250 and 750 bananas will be brought at 250km. Total = 1750 bananas

Reaching next 250km(i.e 500th km) it takes 1.1000 bananas on 1st trip, 500 banana will be placed on 500th km. 2.2nd trip, eating 250 bananas, 500 will be brought. Total = 1000 bananas

Then, with 1000 bananas, he can travel to the market with 500 bananas where 500 are eaten by the camel by its way.

First of all travel 500 km (3 times) first take 1000 bananas and put in half way and return again take 1000 banana drop in half way return and again take 1000 banana and drop it half way, i.e., at 500 km distance. so now you have 1500 banana left at half distance.

Now take 1000 bananas and travel 250 km, drop it there and return and take 500 remaining bananas and travel 250 km and drop it. so now you have 1000 bananas left and a distance of 250 km is remaining. so take 1000 bananas and go. So at last you left with 750 bananas.

the max no of banana will be 500 first the camel will take 1000 bananas to 250km upto three times so the total no of banana he eat in three times going and one time returning will be 1250... now it will move another 250 km by taking 1000 bananas and will eat 250 in going and 250 in returning and drop another 500 after 500 km and will take the remaining 750 to that 500 km and will eat 250 out of them now the total bananas left after 500 km will be 1000 and camel will take it to another 500 km and will eat 500 thust the total no of banana left will be 500

800 bananas. Travel the distance in 4 steps. 1. Travel 150km 300 times with 1000 bananas each time.finally we have 2550 bananas. 2. Travel next 250km 3 times.first 2 times trael with 1000 bananas each time and third time travel with remaining 550 bananas.here finally we have 1800 bananas. 3. Travel next 400 km 2 times.1st time with 1000 bananas and 2nd time with remaining 800 bananas. hereat the distance of total 800km (150+250+400=800)from the banana plantation we are left with 1000 bananas. 4. Now travel the remaining 200km with 1000 bananas and at the destination we have 800 bananas.

400 Bananas. 0==>200==>400==>800==>1000 1. The remaining bananas after transporting 200 km its 2000 2. And at 400km transport it will be 1400 3. And at 600km it will be 800 4. So only 800 bananas left to transport to 400kms. so 800-400=400 bananas.

initially the camel starts with 1000 bananas and it will travel for 250km and deposits 500 bananas there let us say that point as A .now it with the remaining 250 bananas it will travel back to the starting point and again starts with 1000 bananas and deposits another 500 bananas at point A .Now again it will go back and starts with last 1000 bananas and will travel for 500km and deposits 250 bananas there and with the remaining 250 bananas it will return to point A where there is a stock of 1000 bananas .Now starting from A it will travel to point B where the camel is left with 750 bananas and there is a stock of 250 bananas which are previously deposited so it will again start at point B with a load of 1000 bananas and reach the destination which is 500km away with 500bananas

First move: 998+998+999(For fst KM=1000+1000+995) So on ..., after 1000/5 KM total bananas are 1000+1000 from here next KM=999+998=1000+997 So on, ... after 200+1000/3 KM total bananas left are 1000+1

Here after next KM i.e 534th KM total bananas left are 1000

So, left KM are 1000-534=466

From remaining 1000 bananas 466 are eaten by Cammel remaining bananas are===534

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