Need Algebra help, 10 POINTS IF YOU CAN GET A RIGHT ANSWER!!!?

Question

ITS A WORD PROBLEM AND BEGINS LIKE THIS:

A camel rests by a pile of 3,000 bananas at the edge of a 1,000 mile desert. He plans to travel across the desert, transporting as many bananas as possible. He can carry up to 1,000 bananas at any given time but he must eat one banana every mile.

The most amount of bananas he can come up with at the end of the 1,000 miles is 533, how do you get this answer and what equation or method would you use?

Going back and fourth every miles he must eat a banana even when not carrying anything

He can stop at a halfway point like 250 or anything miles because he can't go the full 1,000

Answer 1

Using the given information, no matter what is tried, (I even tried "dropping bananas" off at various point to attemp to make it to the other side with any amount of bananas. The net result, however, is either you run out of bananas at the other end of the 1000 miles, or you just don't have enough bananas to go back. What I'm guessing is, you either are allowed to disobey reality (Eating -bananas during your journey) or the consumption rate is actually lower than 1 banana/mile. I don't know, but I think there might be an error in the given information somewhere, sorry I couldn't help. :(

Answer 2

Not possible.

The camal will have to travel about 5000 miles.

1000 miles to the other end.

1000 miles back.

1000 miles to the other end.

1000 miles back.

1000 miles to the other end.

Answer 3

you questions hard. I'm guessing Either 1,000 or 1,004  

Answer 4

I heard this puzzle on a podcast called the mathfactor. (yes I am a geek :) )

The camel has to set up caches because it can't do it in one trip. We don't want to leave any bananas behind. The first cache should have 2000 bananas and it will take 5 trips to do this (there and back, there and back, there) 1/5 of 1000 bananas is 200. we can go 200 miles with 200 bananas. The first cache is at 200 miles. We are now 800 miles away and have 2000 bananas. We need a 2nd cache of 1000 bananas which will take 3 trips/ 1/3 of 1000 is 333 1/3 so the second cache should be 333 miles from the 1st one. We have now traveled a total of 533 miles.  So we load up the last 1000 bananas travel 467 miles and arrive with 533 bananas.

I hope that makes sense but you can also listen to the podcast

Answer 5

The bananas must be 1000 at 533 miles or near it.

1) First stop at point 200

Carry 1000 bananas to point 200 miles

Net carriage = 800 ( 1000 - 200)

Leave 600 bananas and take back 200 bananas to point 0.

2) Carry 1000 bananas to point 200 miles.

Net carriage = 800 ( 1000 -200)

Leave 600 bananas and take 200 bananas to point 0.

Total bananas = 600 + 600 = 1200

3) Carry 1000 bananas to pint 200 miles.

Net carriage = 800 ( 1000 - 200)

Total bananas at point 200 miles = 1200 + 800 = 2000

3) Bananas should only be 1000 at point 533 miles or near it.

Distance between 533 miles and 200 miles is 333 miles.

1000 bananas should be consumed before 533 miles.

Since we need 3 trips to transport the excess 1000 bananas, 333.33 bananas must be eaten during the trip.

It means that the distance between 200 and the point to stop is

200 + 333.33 = 533.33 miles.

4) Carry 1000 bananas to point 533.33 miles

Net carriage = 666.67 ( 1000 - 333.33)

Leave 333.33 bananas and take 333.33 bananas to point 200 miles.

5) Carry 1000 bananas to pint 533.33 miles

Net carriage = 666.67 ( 1000 - 333.33)

Total bananas at point 533.33  = 1000 ( 333.33 + 666.67)

6) Distance to point 1000 miles = 1000 - 533.33 = 466.67

Number of bananas left after reaching point 1000 miles =

1000 - 466.67 =  533.33 bananas

Answer 6

I am getting 500, not 533.. okay let me explain how.

It is a desert right? so no one is there to steal bananas from the Camel. And as you said it can stop at a distance of 250 mile each, so the Camel can stop at 3 places inbetween. Also I assume that the Camel is greedy :)) doesn't mind walking in desert as long as it gets 1 banana /mile to eat.

It starts with 1000 bananas and reaches 1st mile stone i.e. 250 miles, stops, unloads 500 Bananas and comes back to rest 2000 bananas. Now again it starts with 1000 bananas and then comes back after unloading 500 bananas at 250 mile distance. Finally it takes left 1000 bananas and reaches at 250 mile and it will have 1,750 bananas at 250mile inside desert.

Now it takes 1000 bananas and reaches to 2nd mile stone (500 miles), unloads 500 bananas and come back to take rest 750 bananas. When it reaches 500 miles this time, it will have 1,000 bananas. Now it has 1000 bananas and it takes all and reaches 3rd mile stone at 750 mile, rests a bit and finally crosses rest 250 miles. During the last leg of journey it will consume 500 bananas and will be left with 500.....

This is not a truly algebra problem, rather it is a problem to test logical and lateral thinking ability :))

Hope this come to your help

cheers!