## Bananas

Posted by Chris on March 23, 2011 – 7:44 pm

A camel has to transport as many as possible of an available 3000 bananas. For every complete mile it walks, it must consume 1 banana (regardless of its load). It can only carry up to 1000 bananas at a time. What is the maximum number of bananas it can transport to a place a 1000 miles away?

Assume the camel eats its banana continuously. e.g. if it walks 0.102 miles, it eats 0.102 bananas.

SOLUTION: See post 15 (answer: 533) or an expensive three camel solution, post 68 (answer 833).

March 23rd, 2011 at 8:02 pm

1000 bananas

March 23rd, 2011 at 8:10 pm

Nope. You should show your reasoning.

March 23rd, 2011 at 8:49 pm

that camel wont be able to transport any banana. B’coz its consumption is 1 Banana per mile it walks. So first time when 1000 Bananas carried by camel…. all of them will b finished while reaching destination. and after that camel wont have any banana left to travel any single mile back to source…. plz tell me i m right or wrong…. i m waiting for ur replies

March 23rd, 2011 at 8:51 pm

I’ll fall for it – none. For it to cover the 1000 miles it will need 1000 bananas, it’s maximum load – and have no bananas to get back to try again.

March 23rd, 2011 at 8:53 pm

It’s much more than none.

March 23rd, 2011 at 8:58 pm

1000 – start with 1000 take 998 1 mile and go back – consuming 2, repeat, now have 2994 with 999 miles to go. Re-do multiple times until you are down to the 1000 or so at the end.

March 23rd, 2011 at 9:12 pm

Hi Eketahuna. You’re guessing.

March 23rd, 2011 at 11:04 pm

533.25 – used similar method as Eketahuna except after going 1 mile 3 times you only need to use 5 bananas, i.e. you have 2995 left with 999 miles to go. This continues until you reach 800 miles to go and you have 2000 bananas left – now you only need 2 trips per mile and use up 3 bananas each time. Then at 467 miles you have 1001 bananas left – you can go 1/4 mile further and still come back with 1000 bananas and 466.75 miles left to go. At this point you subtract 467.75 miles from 1000 bananas to get the remaining bananas = 533.25.

March 24th, 2011 at 12:20 am

I may be guessing, too.

Ist trip: take 1000 to the one third mark (i.e. 333.3 miles) and leave one third of the bananas there (333.3 bananas) and return.

2nd trip take 1000 to the halfway mark, leave 333.3 bananas there and return, collecting the original stash of 333.3 bananas at the one third maek to return home.

3rd trip: take 1000 to the halfway mark, collect the 333.3 there and take them to the destination.

Therefore the camel can transport 333.3 bananas to the destination, having consumed 266.7 in doing so.

March 24th, 2011 at 2:34 am

. . . I mean the camel consumed 2666.7 bananas.

March 24th, 2011 at 3:16 am

Hi.

Could it bo so simple:

Pick-up point is point A, delivery point is B. Distance between A and B is 1000miles. The camel need 0,1 banana per mile, ergo 0,1×1000=100 bananas to get from A to B

Trip one:

Load up 1000 bananas at point A. When at point B remaining cargo is 900 bananas. However, the camel will need 100 of these in order to get back to A. ):unload 800 bananas.

Assuming that the camel does not need to return to A after last delivery, this gives a total of 800+800+900=2500 bananas delivered to B

March 24th, 2011 at 4:22 am

It seems smart to me to create a situation where you have travelled x miles such that you have 1000 bananas at point x. Then you can just walk to the end.

This is the case when you walk 400 miles carrying 1000, leaving 200 at that point, then returning. Now we repeat, leaving a total of 400 at that point. Now we take the final 1000 en we are at 400 miles with 1000 bananas.

Now we just walk to the endpoint, leaving us with 400 bananas.

(Still didnt prove it is optimal)

March 24th, 2011 at 4:28 am

I guess it is already pretty clear it is optimal. Since you cant improve on just walking to the end from mile 400 to mile 1000. And you cant improve on the number of bananas you have at mile 400, since you clearly arent walking any ‘extra’ miles.

I am interested to see a clear proof though.

March 24th, 2011 at 4:32 am

ah I see what I did wrong. not optimal at all lol

March 24th, 2011 at 6:41 am

It would be a waste if the camel returned for the next batch if it ended up at the base with any bananas (on the camel). It is also pretty obvious that, at the outset, it shouldn’t leave any of the original 3000 bananas.

Assuming that the camel takes the 3000 bananas a dsistance m, then it would have to make three outward and two return trips. So altogether it could transport 3000-5m bananas a distance m from the start point. NB it doesn’t make any difference if that is done in a series of small journeys or in one go.

My (better than an) assumption is that the best strategy is to end up with 2000 bananas (notionally) at the second base. So that suggests that 3000-5m = 2000 => m = 200. That’s nice; we don’t have any fractional bananas to complicate the situation. So, the the camel can deposit (1000-400) + (1000-400) + (1000-200) = 600 + 600 + 800 = 2000 bananas at the first base.

Now move the 2000 bananas a further distance n. This time we want to end up with 1000 bananas (notionally) at the third base. So 2000-3n = 1000 => n = 333.333 miles (NB that’s two outward and one return trip). The camel can only leave 333 bananas (assuming a part banana is not acceptable) after the first leg. When the camel returns to the second base, it’ll have 1000 – (333 + 666.667) = 0.333 bananas. It has to discard that, and pick up the remaining 1000 bananas. When it gets back to the third base, it’ll have a total of 333 + (1000 – 333.333) = 999.667 bananas. It can then continue with all of those.

The remaining distance is 1000 – (200 + 333.333) = 466.667 miles. So it gets to the 1000 mile destination with 999.667 – 466.667 = 533 bananas.

Let’s see if we can rescue any of that wasted 0.333 banana. If the camel walked 333.5 miles, then again, it could only deposit 333 bananas, but would end up at the second base with 1000 – (333+2*333.5) = 0 bananas plus the 1000 at the second base). It could then take those 1000 bananas to the 333.5 mile base, ending up there with a total of 333 + (1000-333.5) = 999.5. So it could then take them the remaining 1000 – (200 + 333.5) = 466.5 miles, ending up with 999.5 – 466.5 = 533 bananas, yet again.

If anyone thinks that they’ve got a cunning plan to get more than that across, I’d be delighted to hear it.

Addendum: 533.333 is achievable. See post 17.

March 24th, 2011 at 7:24 am

1000. The camel starts out by carrying the 1000 bananas and walks all the way to the end while continuously eating 1 banana per mile. When the camel reaches the end it is carrying the maximum 1000 bananas in his belly.

Seriously, we have to determine how far the camel can go and still maintain the optimal number of bananas, 2000, for the 2nd leg. Since we can assume 5 bananas per mile will be consumed trying to transport 3000 bananas in groups of 1000 to the farthest distance, we just need to divide (3000-2000)/5 = 200 miles covered in 1st leg.

Now using similar logic to get the remaining 2000 bananas to their furthest point and still maintain optimal bananas, 1000, for 3rd leg we divide (2000-1000) / 3 = 333.33 where 3 is the number of bananas burned per mile.

1000 bananas left and 533.33 miles covered. Eat 466.67 of the 1000 and leaving 533.33 bananas left at the end of the journey.

March 24th, 2011 at 7:27 am

Hi Josh. You did pretty good. But you should have shown have been more careful. In the second stage, you have arbitraily added 0.25, rather than 0.333 miles.

Hi Wiz. You got it wrong last time too. But you did manage to get 500 bananas across that time

Hi Kenneth. You jumbled up the details of the camel’s consumption. Otherwise you’d have been right.

Hi Eketahuna. The devil is in the details.

—

I was a twit. We can leave 333.333 bananas at the third base (after the first trip) – the camel can eat that later. We’d then end up with 1000 bananas and 466.667 miles to go. So we’d get 533.333 bananas across (and only one part chewed). Although I can’t think of a proof, I’m pretty sure that that is the best that can be done.

March 24th, 2011 at 7:40 am

its is 3 loads of 1000 coz if it can carry 1000 bananas 1000 miles away it will go 6000 miles there 1000 back 1000 then like that three times so its 6000 so im agineus

chris if u even try to say im rong i worked it out im a mathmatical person so i work it out in my head

thnxxxx im right *:* *=* *=*

March 24th, 2011 at 7:44 am

Hi John24. Yeah. You did good

March 24th, 2011 at 8:00 am

Hi gemma. LOL, I can’t even tell what your answer is, so I can’t say that you’re wrong.

March 24th, 2011 at 8:41 am

I don’t like the part with the 1/3 of a banana. Once the camel started eating it, the fruit is spoiled and has no practical value. Therefore, the answer is 533. How to find it? Just follow the general formula: Let the road be N miles and we have rN bananas. Then the optimal (integer!) number of bananas that can be transported is:

\sum_{i from 1 to r-1}[N/(2*i+1)]=[N/3]+[N/5]+[N/7}+...+[N/(2*r-1)],

where [] is the integer part of the number (the highest integer that does not exceed the number).

For the particular problem we have N=1000 and r=3, i.e., [1000/3]+[1000/5]=333+200=533.

March 24th, 2011 at 9:49 am

Hi slavy. You mustapha fussy camel.

March 24th, 2011 at 1:11 pm

1000, the banana’s energy is still in his cells. and in hiss stomach

March 24th, 2011 at 2:54 pm

Ah yes, I must admit the actual answer I came up with was a guess. It was the end of a long day, beer O’Clock and the door was about to be closed and locked in 30 seconds so I typed a quick answer – I’ll try and do some proper calculations now…

March 24th, 2011 at 3:06 pm

Ah, it’s been done already – I’ll go back to my happy place

March 24th, 2011 at 8:21 pm

i’m going to go with gemma on this one…he is agineus

March 24th, 2011 at 9:26 pm

Hi Knightmare. Are you implying that I’m not agineus?

March 25th, 2011 at 10:47 pm

If it were to take the trip, it wouldnt be able to transport any the bananas. it would take the entire load as fuel just to reach the destination not to mention the return trip.

March 26th, 2011 at 8:27 am

Hi Dan. Read the above and learn how to get 533 (and a bit) bananas across.

March 26th, 2011 at 8:48 pm

it will be 1000 bananas it will transport and rest of the bananas it will be consumed in during its transportation…….

March 28th, 2011 at 4:24 am

Literally, it would be no bananas delivered. But if this is a trick question, I have no say.

March 28th, 2011 at 10:54 am

You would have 0 left

March 28th, 2011 at 12:13 pm

A few of you peeps could benefit by reading the solution, instead of saying 0. It’s 533.

March 29th, 2011 at 9:31 am

i no 100 bananas add them three times = 300 times 300 =600 so if u take 6 0.1 off of 600 u will get 533 easy well i think. *:* *:*

March 29th, 2011 at 11:12 am

Hi gemma. I’ve no idea what any of that meant.

March 29th, 2011 at 11:05 pm

hey Chris…don’t feel bad about gemma’s answer going over your head (and everyone else’s). the fact is that his level of thinking is just beyond our understanding.

March 30th, 2011 at 4:47 am

She’s definitely agineus.

March 30th, 2011 at 7:21 pm

1 billion

March 30th, 2011 at 10:34 pm

500 bananas as he will travel 500 miles taking 1000 bananas and at 500th mile he wud have consumed 500 bananas so he will drop it there and move back and will bring 1000 more bananas so at 500th mile he will also use 500 bananas so he will go back so he will make 3 trips from start to mid and than at mid point there will be 1500 bananas so he will take 1000 to the end point and in the way he will eat 500 bananas so at the end he will be ablle to take 500 bananas

March 31st, 2011 at 5:49 am

Hi Rafa. If it takes 1000 bananas 500 miles, it will have used 500, leaving 500. If it drops them, then it has 0 to get back with – but it needs 500.

March 31st, 2011 at 1:20 pm

hi there

the answer is 533.

the reason:miles=m,bananas=b

0 m======200 m====533 m=====1000 m

3000 b==> 2000b

==-1000b==2000b==>1000b

=============-1000==1000b====>533 b

========================-467b

this is my logical reason so if it has complications let me know.

March 31st, 2011 at 1:29 pm

Hi keihan. You hven’t explained why 200 and 533 miles are the optimium drop off points.

April 1st, 2011 at 12:23 pm

The answer is none. It will eat all the bananas so none will be delivered.

April 2nd, 2011 at 7:47 am

The camel cannot transport any bananas, because if it can carry 1000, but has to go 1000 miles, and eats one every mile, the camel will wind up eating every banana.

April 3rd, 2011 at 11:05 am

The camel cann transport up to 1000 bananas max.

April 3rd, 2011 at 11:07 am

^the question asks how many can the camel transport so the answer is 1000. Reasoning to above comment..^

April 3rd, 2011 at 11:13 am

Hi SPowell. You have misread the question. It does not ask how many bananas can the camel transport. It asks have many can it transport to a place 1000 miles away.

April 3rd, 2011 at 10:07 pm

0 bananas

April 4th, 2011 at 11:00 pm

1000 bananas it wil transfer

April 4th, 2011 at 11:10 pm

It is zero if the camel can only hold 1000 and eats 1 every mile and is traveling 1000 miles then there is nothing left. Also the reasoning for the various number of trips puts you in the hole. If you have to travel back to pick up more bananas then the camel has had to travel to one point then back to the first then back the “drop off” the camel has had to eat more than it can take. Also, I do not understand how you can get 5m ( 5 bananas per mile??) that puts you deep in the hole and once again you have the camel carring 3000 bananas it can only hold 1000 fail it is 0.

April 4th, 2011 at 11:15 pm

reason is first load of 1000 b’s and walks upto 666.6 mile eats 666.6 bananas and left the bananas load thier then

journey of first load is 666.6 no of bananas left = 333.3

second load of 1000 b’s and walks upto 666.6 mile eats 666.6 bananas and left the bananas load thier then

journey of second load is 666.6 no of bananas left = 333.3

third load 1000 b’s and walks upto 666.6 mile eats 666.6 bananas and left the bananas load thier then

journey of third load is 666.6 no of bananas left = 333.3

total no of bananas there at a distance of 666.6 m = 333.3+333.3+333.3 =999.9 bananas are there at present at a distance of 333.3 miles from destination

and again these total 999.9 bananas r carried from 666.6 to 1000 and the difference between the miles is 333.3 and the camel eats 333.3 bananas now the total no of bananas transfered to the destination is 999.9-333.3 = 666.6

i.e nearly equal to 667 this is the maximum number no of bananas it can tranfer……………..

answer is 667

April 5th, 2011 at 3:57 am

Hi Channy. Starting with the 3000 bananas with 1000 miles to go, let the camel take bananas to a point m miles away from the start. To get there, it will eat m bananas, but it has to come back to get the next lot, so the camel needs m more bananas to do that. Same for the next batch of bananas. But for the third batch it doesn’t return to the base (as there are no bananas there any more). That’s three trips out, and two trips back, giving five trips altogether, so the camel will have eaten 5m bananas. Altogether that means it will have taken 3000-5m bananas. Now realise that the best strategy requires the camel should end up with 2000 bananas at a distance m from the base, so m = (3000-2000)/5 = 200 miles.

Hi harish. If the camel walked 666.6 miles and then dropped 333.3 bananas there, it cannot go back as it needs to eat 666.6 bananas to do that, but you’ve got 0 bananas left.

April 5th, 2011 at 5:07 am

it cant transfer any bananas.Because aflike you said,if it has to carry all banana’s to m miles distance it has to eat 5m bananas.In such case it will left with 0 bananas after travelling 3000/5 = 600 miles.So no chances of transferring bananas to 1000miles.

April 5th, 2011 at 5:41 am

Srinu, read the solution, post 15. It can transport 533 bananas over the 1000 miles.

For the first part, m = 200, so the camel then ends up with 2000 bananas and 800 miles to go. The next stage it does a further 333.3 miles, ending up with 1000 bananas and 466.7 miles to go, which it then does with 533.3 bananas at the end.

April 5th, 2011 at 8:51 pm

no naners for da ppl bc he eats a 0.1 of a naner every 0.1 miles so after 1000 miles camle has eaten 1000 naners. its quite obvious

April 5th, 2011 at 8:54 pm

#39 wow, thats actualy rly smart :I ive never thought a camale could be that smart

April 5th, 2011 at 8:58 pm

… but post 39 is very wrong.

April 6th, 2011 at 1:04 am

The Answer is 500 bananas !

Solution:

1) 1st trip – The camel takes 1000 bananas (hereafter called as ‘B’) to 250 miles (Point A) consuming the same number of ‘B’ drops 500 there and return, so totally consuming 500 ‘B’ for a return journey.

2) Repeat the same as above, so now we have 1000 ‘B’ (500+500) at 250 miles and 1000 ‘B’ at starting point.

3) 3rd trip the camel takes 1000 ‘B’ to point 250 miles consuming 250 ‘B’. Now at 250 miles we have 1750 ‘B’.

4) again from 250 mile camel takes 1000 ‘B’ to 500 mile (Point B), leaves 500 ‘B’ there and return to 250 mile spot where 750 ‘B’ are balance. In this return journey it consumed 500 ‘B’.

5) it picks up balance 750 ‘B’ and takes it to Point B at 500 miles consuming 250 ‘B’. Now it has total of 1000 ‘B’ left at 500 miles at point B.

6) final trip – it takes all 1000 ‘B’ from Point B to finish point consuming 500 ‘B’ and delivers 500 bananas.

April 6th, 2011 at 4:45 am

Hi Pallav. Close. It can get 533 across. See post 15.

—

Unbelievable. This problem has just got into the top 5.

April 6th, 2011 at 6:05 pm

the camel can carry 500 bananas across how?

the camel will walk 500 miles it will have 500 bananas left it will drop the 500 the rest of the bananas and walk back. by the time it gets back it will have no more so it takes 100 more and then walks back to the 500 he dropped by then he will have 500 bananas left he takes the 500 and then he’ll have 1000 and he’ll walk to the destanition and give them the 500 bananas that are left. i’m smart! and would u believe i’m only 10 and i didn’t read any of the other posts

April 6th, 2011 at 7:32 pm

Hi abbi. See post 15. Your answer doesn’t work. If the camel took 1000 bananas 500 miles, then dropped off the 500 that were left, then it cannot go back to the pick-up point as it needs 500 bananas to do that.

April 7th, 2011 at 6:49 pm

This is a dumb awnser but I’m 12 so don’t judge. 1001 because he eats one banana at first to get him the 1 mile and in the end he ends up with 1 bannana because he eats the rest.

April 8th, 2011 at 7:18 pm

Answer is camel transports 4 bananas.

Camel carries 1000 bananas,goes 499 miles,leaves 2 there and comes back to start pointTakes another 1000 and does the same.Now there are total 4 bananas at mile 499.Does the same a third time.On reaching 499 miles, it picks up the 4 bananas there and adds to 501 remaining from its last 1000 pickup.So camel has to go last 501 miles with 505 bananas.Consumes 501 bananas and reaches destinaton with 4 bananas.

April 8th, 2011 at 9:00 pm

Aaaargh. It 533, can’t anybody read? (post 15 say).

April 10th, 2011 at 4:26 pm

easy one mile from the destination there will be a rest (refilling the load, stated any sircumstance) so the answer is 999

April 11th, 2011 at 2:15 pm

1000…?

April 11th, 2011 at 3:37 pm

… I suppose it won’t help if I say see post 15 to see why it’s 533 bananas

April 13th, 2011 at 10:48 pm

I believe its says how many can a camel transport.One to transport 2 more to feed.So three camels 1000 each mile 333 one camel is gone leaving two camels 2000 bananas and 666 miles to go.The next leg 500 miles leaving 1000 bananas one camel and 166 miles to go.Leaving you with 833 bananas at the end discarding the one partial.One camel can transport 833 bananas.

April 13th, 2011 at 11:09 pm

Hi Joel. I like that answer , but I think of a couple of camels and their owner(s) won’t.

April 14th, 2011 at 1:18 am

thnx knightmare for agreeing with me i think the same as u to..

April 14th, 2011 at 8:00 pm

ZERO because it carrys 1000 and travels 1000km and eats 1 per km

April 14th, 2011 at 8:32 pm

Hi Cody. It’s 533 for the reasons given in post 15. Or 833 if we allow post 68.

But thanks for helping to keep this in the top 5 popular posts.

April 16th, 2011 at 11:39 am

I am sure when camel walk 1000 miles the zero banan can reach there the reasons is 1000 banana can load and 1000 miles walk. he eat one banana on every miles.

April 16th, 2011 at 1:52 pm

Hi mohsin. Did you even read the question?

April 22nd, 2011 at 9:48 am

the total is ZERO!!!

April 25th, 2011 at 7:51 am

999.9

as it will be at the place by the 1000th mile

May 2nd, 2011 at 6:53 pm

I’ve gotten quite sick of bananas after reading all 76 of these comments.

August 10th, 2011 at 1:23 pm

get a more efficient camel. This one eats too much of its cargo

August 12th, 2011 at 6:19 am

Trick is to loose cargos one by one, so that it reduces iteration of travelling to and fro the same distance.

Transferring 3 cargos to any point will take (3+2) trips.

By exhausting 1 cargo(1000 bananas), we will go as far as (1000/5) = 200 miles with 2 cargos in kitty.

Transferring 2 cargos to any point will take (2+1) trips.

By exhausting 1 cargo, we will go as far as (1000/3) = 333.33 miles with 1 cargo in hand.

So at 1000 mile mark we will be left with 533.33 bananas.

Hmmm, so 3 Cargos can make us travel 1533.33 miles.

Just a thought, What should be the minimum number of bananas in each cargo so that we travel 1000 miles exact?