# Camel Puzzle

## How to cross the desert?

Today, you are facing a unique problem. You have to cross a large desert covering a total distance of 1,000 km. You have a camel and 3,000 bananas. The camel can carry a maximum of 1,000 bananas at any time. For every kilometer that the camel needs to travel, it eats one banana before it can start moving. What is the maximum number of uneaten bananas that the camel can transport to the other end of the desert?

On the face of it, it seems that you can’t transport even a single banana uneaten to the other end of the desert. However, if that would have been the case, I wouldn’t have posted it on www.QuickerMaths.com and wasted yours and my time. That means you can solve this but you have to use a little **lateral thinking. **So** **go ahead and use your grey cells. Please provide detailed explanations with your answer.

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400

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 5331/3…

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===>

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===>

A

===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.

It finally carries 533 bananas to the other end

oops…itz534 1001-467

d most ridiculous answer s r givn by Ankitxtreme n saloni…….Lol

@ Saptarshi chaudhuri : whats ridiculous in it …..833 bananas can be carried …dont u agree ?

i got it to be 536….now lets see how i th8 it up…first 200 km the camel takes 5 bananas each km for transporting 3 1000 bunches(camel goes back 2 tyms n comes forth 3 times equaling to 5 times transport to transport all d three bunches)…NOW 2000 left….lets see wen 999 of dem finishes….dis tym 3 bananas for each km(camel goes back 1 tym n advances 2 tyms for 1 km transportr of 2 bunches..sequence -advance :go back :advance)………..so 333 km is covered at d expense of 999 bananas more…so 1001 bananas left and (1000-(200+333)) Km ie 467 km left…..hence 1001bananas-467bananas=536 bananas left at END…..::::::::::::::))))))))))))))))))))))))))

400 i think.

the camel takes the max of 1000 bananas and starts going. after 400km there will be 600 bananas left. camel drops 200 out of them and goes back with the remaining 400 (he needs to eat on the way back too right?). when he gets to the beginning point, he will be totally empty so he will repeat the same trip 2 more times. result: camel is on the 400th km with 600 bananas on his back + 400 bananas waiting there (200 left from first trip + 200 left from second trip + 600 on his back from third, cause he didnt need to go back).

so now he loads all the 1000 bananas and proceeds the remaining 600km.

so he arrives with 400 bananas.

I think 832 is right answer…..it travels one km and consume 3 banana for transporting whole banana….it take 333 km for consuming 999 banana…now there is nearly 2000 banana…for transporting it it takes 2 banana for 1 km …thus for consuming 1000 banana it require 500 km…now it travels 833 km and 2000 banana….remaining 167 km can be travelled at cost of 167 banana…banana left is 833

3 bananas will be left.

i think only 2 bananas will be left

before starting he will eat one banana… from total 3000 banans

then he will carry another 1000 banana ( no. of max banans he can carry).

after one kilometer ie. at 2nd kilometer he will eat his second banana..but the first from the carried 1000 banana.

998th banana he will eat at 999th kilometer,,,,

so only 2 banas will be left ….

He will eat 1000 bananas before starting the journey. It is enough to cover next 1000 km. As per his capacity he will carry 1000 banans to other end. He has to forgo 1000 bananas at the begining for another hungry camel……….so simple…at last..

the optimum method is as under

for the first km min 5 bananas are required this process continues up to 200 km

now he is left with 2000 bananas

for next km min 3 bananas are required this process continues up to next 334 km it means total distance covered is 200+334=534km and total bananas consumed are 200*5+334*3=2002.finally he is left with 998 bananas and 466km to be covered so max no. of bananas he is left with is 998-466=532

three banana will be at the other end as 999 banana will be consumed/trip

He will transport 2000 Bananas.

He will carry 1000 Bananas. He will drag another 2000 Bananas.

He will eat 1000 Bananas in transit.

So he ends up transporting 2000 bananas….

the owner will eat the remaining bananas since he/she wil be travelling with the camel and wil obviously feel hungry:)

my comment was with reference to vishal khardehar and chandana’s conversation

nyc one.. liked it..

Wher r u from?

Hey buddies I got Answer!!

firstly camel travel to 100km he eats 200 bananas, for 1st & 2nd around for 3rd. then remaining distance is 900km(1000-100=900 km) and remaining banana are 2500(3000-200-200-100=2500).

he Repeat this rounds until remaining 1000 bananas therefore remaining distance will be 300. Thus (1000-300=700) starting point to ending point maximum uneaten Bananas are 700 left.

hey if we do it the way u said the answer will be 500 not 700.. after it becomes 2500 at the end of next 100km there will be 2000 left. To transfer these 2000 we need only 3rounds not 5. so we will be left with 1700. next 1400. that next 1100.till here we covered 500km.we take 1000 and travel anoder 100km.we will be left with 900. It doesnt make sence in going back wasting 200bananas so as to get back 100. so instead we continue to go forward. Till this point we covered 600km.So for another 400km 400bananas are used up. so we will be left with 500 bananas

hey CHANDU I said he repeat that rounds until 1000 bananas LEFT & other wise the puzzle is Maximum bananas uneaten to another end of desert Do this again briefly…….

Hi you are correct I do this puzzle BRIEFLY I accepted my Mistake you are correct answer is 500 maximum uneaten banana:) & we waste 200 banana in Travel

Hey i am really impressed with your website

Thanks Prajakta!

Lets make the nos camels LCM of 2,3,9 i.e 18

Now eldest 1/2=> 9

middle 1/3=> 6

youngest 1/9=> 2

add the no of camels(9+6+2)=17:)

833….

condition don’t make the camel move by more than 500km at a stretch..

first cover 333km three trips…2001 bananas left

then take two trips and cover 500km….1000 bananas and 167km left

so and 1000-167=833

to cover 333km (and come back for the next portion) the camel will need 666 banana’s so at the end of the 333 u will only be able to leave 334 banana’s behind (at a given trip).

the math doesnt work thus.

1. To carry 3000 bananas in lots of 1000 each, the camel needs to make 3 trips. In addition it has to go back 2 times.

In the above 5 trips of p km each, it consumes 5p bananas

The left over bananas = 3000 – 5p.

Let this be not more than 2000.

So p can be 200 or more.

2. To carry 3000 – 5p bananas in lots of 1000 each, the camel needs to make 2 trips. In addition it has to go back 1 time.

In the above 3 trips of q km each, it consumes 3q bananas

The left over bananas = 3000 – 5p – 3q

Let this be not more than 1000.

If p = 200, q can be 334.

3. To carry 3000 – 5p – 3q bananas in lots of 1000 each, the camel needs to make 1 trip of r km.

In this trip, it consumes r bananas.

4. p + q+ r = 1000 km

If p = 200, q = 334 then r = 466 km.

Bananas consumed = 5p + 3q + r

The left over bananas = 3000 – 5p – 3q – r = 532

Gr8 to see you active on QuickerMaths.com

2970 bananas can be carried comfortably……..use ur grey cells a bit more …

hint : there is no restriction on the no of trips camel does…..it can be as many as u want …

sorry, i posted 2970 bananas assuming 10 km to cover up…

it is 833 bananas and no more than this…..same way ankitextreme did it..

i have another camel problem.

once sheik died leaving his will of 17 camels which has to b distributed accordingly to his 3 children.

eldest will get- 1/2 {half}

middle will get – 1/3 {third}

youngest get- 1/9 {ninth}

Now distribute.

lets make the nos camels LCM of 2,3,9 i.e 18

Now eldest 1/2=> 9

middle 1/3=> 6

youngest 1/9=> 2

add the no of camels(9+6+2)=17:)

i think its ne banana as shahnawaz said ,, camel starts by eating one banana ,,,

else even more can be transported if traveled to and froth by deposting fruits at intervals ,,,

Divide the journey into three legs:

Leg 1:

ending at point A, which is 200 km from the starting point.

Leg 2:

ending at point B, which is 333 km further away from point A.

Leg 3:

of 467 km ending at the other end of the desert.

Leg 1:

Start with transporting 1000 apples to Point A. The horse will

consume 200 apples on the way to point A. Deposit 600 apples at point

A and return with the remaining 200 apples to the origin. After the 3rd

trip you will end up with 600+600+800 = 2000 apples at point A.

Leg 2:

Now start transporting 1000 apples from point A to point B.

Horse will consume 333 apples on the way, deposit 334 apples at point

B and return to A for another trip. After two t rips you will end up with

334+667 = 1001 apples at point B.

Leg 3:

Load the horse with 1000 apples; give it 1 apple at the start of

this leg. The horse will consume 467 apples on the way to the other end

of the desert. You will end up with 1001-467 = 534 apples at the other

end of the desert

hey we need to replace the horse and apples with camel and bananas

only one banana will get transported to other end of the desert. Let the camel eat one banana at first and then start the journey .this way only 999 bananas will be required further to be eaten