A gorilla harvests 3,000 bananas and needs to carry them 1,000 miles to
the supermarket. He can only carry 1,000 at a time. Since he is a
gorilla he eats 1 banana every mile he goes in any direction. He can
(and will have to) leave bananas anywhere along the way. Once all his
bananas have reached the end he DOES NOT need any to eat to get back.
Remember he eats 1 banana every mile he goes even if he is going back to
pick up more bananas. What is the maximum number of bananas he can get to
the market ?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions.

- anonymous

- katieb

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your **free** account and access **expert** answers to this

and **thousands** of other questions

- vishal_kothari

When he has more than 2000 bananas, it costs him 5 bananas / mile to
transport them, since he makes 3 trips forward and 2 trips back.
When he has more than 1000 bananas, it costs him 3 bananas / mile, for 2
trips forward and 1 trip back.
And when he has 1000 or fewer, it's 1 banana / mile.
I think the way to do it is: see how far he can get 2000 bananas, then 1000
bananas, and then make one final trip.
200 miles at the rate of 5 bananas / mile = 1000 bananas cost.
So if the first stopping point is at 200 miles,
he would carry 1000, eat 200, drop 600, and eat 200.
Then he would do that again.
On the third trip, he doesn't eat the second 200, so at the 200 mile
marker he has 600 + 600 + 800 = 2000 bananas.
If the second leg is 333 miles, it works out as:
carry 1000, eat 333, drop 334, eat 333.
carry 1000, eat 333, drop 667.
Now he has 1001 bananas, and it's just as well to throw one away. To
make it come out exact, we'd need to use fractional miles and fractional
bananas.
Now we have 1000 bananas at the 533 mile marker.
For the last leg, he just carries the last 1000 bananas,
eating 467 along the way, and arriving with 533 bananas.
If the initial legs are shorter, it works out the same.
If the initial legs are longer, he does more miles at higher banana /
mile rates, ending up with fewer bananas.....
old one...but good one....

- experimentX

is he eating when he's coming back to take banana back ..?

- anonymous

superb, @vishal

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- perl

vishal stole the answer online

- perl

http://www.gottfriedville.net/mathprob/misc-bananas.html

- perl

take back your medals, we have a fraud here

- perl

@vishal_kothari
you are a fraud.

- vishal_kothari

i have solved it earlier...

- perl

lol

- vishal_kothari

it is too long to solve...

- perl

the whole point is to solve. why dont you say you got it online?

- vishal_kothari

i have a file in my pc ..where i have solved all this questions..so i just got it from there...

- perl

stop lying

- perl

i have exposed you

- vishal_kothari

the link that u have given consists all questions that sritama is asking..

Looking for something else?

Not the answer you are looking for? Search for more explanations.