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The answer above is entirely wrong, as it fails to take into account the bananas Cleo ate on her return trips. If Cleo takes 1000 bananas 750 miles and then drops off the remaining 250, she'll starve to death on the way home.

I don't know how to solve this algebraicly, but I solved it with a computer program, using an algorithm developed by my son, who is now in pre-calc. The algorithm goes like this:

Decide on a movement increment (how many miles Cleo will go before dropping off bananas). For example, 100 Miles. So.

Cleo picks up as many bananas as she can carry (1000) at the plantation, goes 100 Miles, and drops off the remaining 900.

She grabs 100 bananas to fuel her return, goes back, and picks up another 1000.

She goes 100 miles again, fills to 1000 again by picking up another 100 bananas, and continues on another 100 Miles. She then drops all her bananas.

She continues this way, going back to the earliest pile of bananas that still has enough for 2 moves in it, picking up as many as she can carry, and carrying it to the first empty multiple of 100 miles before dropping it off. She fuels her trips by filling up her pack every 100 Miles along the way.

When she reaches the point where the second pile of bananas back has less than enough to go 200 miles (her trip would be useless), she grabs all the bananas that are where she is and heads for market.

A movement of 100 miles gets only 500 bananas to market, but you can get 666.6666666... if you pick the right movement length. It took a computer for me, though.

supposibly the answer is 530 so my teacher says but who knows we all could be wrong !

I don't agree with that answer, because you forgot that Cleo must eat one banana for every mile that she walks, including the walk back. Therefore, Cleo would be able to get 333 and one third bananas to the market.

This answer does not fit the requirements :0( Cleo needs to eat 1 banana for every mile travelled (that means to and FROM any location.) If she travels 750 miles with 1000 bananas, she's stuck! She's eaten 750 bananas, but can't leave any. In fact, she doesn't have enough to get back to the plantation, either.

Try starting with 1000 bananas, and going to the 250 mile mark. She can leave 500 there, because she'll need 250 to get back to the plantation. Repeat with the remaining 2000 bananas in two more trips.......

Actually, your solution is incorrect. Cleo needs to eat to get home. So, if she takes 100 (I think you meant 1000) bananas (one thousand), she can only go 500 miles and return home empty handed (500 bananas eaten each way). What you need to do is improve her per-mile rate of consumption. Your first way point is 200 miles. You bring your first batch of 1000, and drop off 600 (bringing 200 for the trip back home with you). Repeat for the second thousand. On the third trip, you get to the way point with 800 bananas. This gives you 2000 bananas at the 200 mile marker. Your next way point would be 333 miles even further (mile marker 533). This will get you 1001 bananas to this mile market. You then go straight to market (another 467 miles) and deliver 533 bananas. IF you count portions of a mile, you can get that 1 banana (of the 1001) to market as well by increasing your second way point's distance by 1/3 of a mile. The best answer is 533 to 534 bananas. ;)

Debemos de recordar que Cleo tiene que comerse un guineo por cada milla. si cleo sale con mil bananas y camia 400 millas, deja 200,porque para regresar nesecita comerse 400 bananas mas,en el sgundo viaje camina 400 millas mas y deja 200,porque para regresar nesecita comerse 400 mas , recoge las ultimas 1000 y camina 400 millas ycomo no tiene que regresar le quedan 600 y sumandole el resto que habia dejado en sus dos viajes anteriores le suman 1000 , camina las ultimas 600 y deja en el mercado 400 bananas {kino}

That answer is not correct, because cleo eats a banana EACH AND EVERY mile she walks, which means when she goes home, she will eat bananas too, say she walks with 1000 bananas to the 500 mile mark, she has eaten 500 bananas, but going back 500 miles, she will eat another 500 bananas, so she has eaten 1000 bananas.

You are wrong the Cleo needs to eat on the way back too so try again.

I do not believe that the awnsser is 750 bananas because the problem clearly states that Cleo eats a banana every mile she walks. So in the beginning how could she walk 750 miles and leave 250 bananas if she still needs to walk 750 miles to get back. That would mean that she had 1,750 bananas and she can carry only 1,000 bananas total, so that wouldn't work. I believe that the awnser is 500 bananas because she can walk 250 miles and leave 500 bananas and walk back. Then she woul pick up another 1,000 bananas and walk 250 miles then pick up 250 of the 500 bananas that are already there and then continue walking another 250 miles. [That is 500 miles total.] and then she can leave 500 bananas. Then she walks 250 miles towards the PLANTATION and picks up the remaining 250 bananas and walks the rest of the way to the plantation. Then she picks up her last 1,000 bananas and walks 500 miles. She picks up the 500 bananas there. [Putting her at 1,000 again] and walks the 500 miles to the market leaving her with 500 bananas. Hope that helped.

You can not get 750 as she eats a banana every mile walked forward or backward

Algebra

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