Corey Camel has 3,000 bananas in her small banana grove in the desert. The marketplace where Corey sells her bananas is 1,000 miles away. Corey has to walk to the market to sell her bananas. For every mile that Corey walks, she must eat 1 banana. Corey can only carry 1,000 bananas at one time. For POW 13, I have to figure out how many bananas Corey can get to the market.

Process

To solve POW 13, I first worked on a mini-POW to help me figure out POW 13. I used the same process in POW 13 as I did in a mini-POW. The process is as follows:

1. Corey picks up 1,000 of the bananas, travels 200 miles, and drop 600 bananas at the 200 mile point. There are 200 bananas left for the trip back.

2. Pick up 1,000 more bananas, travel 200 miles, pick up 200 of the bananas that Corey dropped earlier. (Corey now has 1,000 bananas again.)

3. Travel an additional 333 1/3 miles. (533 1/3 mile point) Corey now has 666 2/3 bananas. Drop 333 1/3 bananas there and travel back 333 1/3 miles to the 200 mile point.

4. Pick up the 200 bananas that were dropped there earlier and head back to the banana grove.

5. Pick up another 1,000 bananas. Travel back to the 200 mile point. Drop 800 bananas and pick up the remaining 200 bananas that were dropped there earlier.

6. With the 1,000 bananas, travel 333 1/3 miles to the 533 1/3 mile point. You now have 666 2/3 bananas. Pick up the bananas you dropped at this point earlier and now you have 1,000 bananas again.

7. Travel the remaining 466 2/3 mile trip to the market. 1,000 - 466 2/3 = 533 1/3 bananas.

Solution:

a.Corey Camel will have 533 1/3 bananas when she gets to the market. b.Yes, I think that this solution is the best possible one, because if you try it with Corey leaving less bananas at the 200 mile mark or if you try it with Corey traveling out farther than the 200 mile mark for the first step then Corey always ends up falling short of bananas on the way to the market. 533 1/3...

...POW13: CoreyCamel
By Alex Cohen
Problem Statement: CoreyCamel has 3,000 bananas to take to the market. He can only carry 1,000 at a time. The market is 1,000 miles away. Every mile Corey eats a banana. My job is to find out; how many bananas he can get to the market place?
Process: To do this POW I started with Mini Camel to find a good strategy. The strategy we found was this: take as many bananas as you can to a certain distance and drop the rest, minus enough to get back. For example go 100 miles, drop 800, then come back 100 miles, you have used all of your bananas. Next I applied this method to CoreyCamel, and since our teacher said the most number of bananas he can bring back was 533 1/3 I realized that after the 1st 1,000 were gone, I would use the other 2,000 by dividing them into 1/3s since the only way to get 33 1/3 is to divide something by 3, like 1,000 divided by 3 is 333 1/3. 533 333 is 200, so I know that the 1st trip has to evolve 200, since I need 200 to be left over when I'm done to get me the rest of the way. I drew out a time line of events to show Corey's movements as he progresses.
Solution: Corey starts by picking up 1,000 bananas, and walking them 200 miles. Next he drops 600 and goes back 200. After that he picks up another 1,000 and goes 200 miles. He then picks up...

...Jason Vo
Ms. King
3/11/15
Period 1
Due Date: 3/13/15
Corey’s CamelsPOW
Problem Statement
The problem basically is about a camel named Corey that has to carry 3000 bananas to a place 1000 miles away. But, Corey can only carry 1000 bananas and Corey has to eat one banana for every mile she walks/runs/skips/hops, etc. In addition to that, there is a refrigerator at every mile. So the question being asked is how many bananas can Corey get to the market with all these requirements?
Strategy
My strategy for this problem is to make some sort of a chart and/or explain the POW in steps. I will explain the problem in a clear and understandable way to make sure the person reading it will understand it and how the problem is very complicated/easy.
Problem Process
Once I read the problem, I knew you had to think outside the box, but I wasn’t sure how. So I started off like this:
Number of Bananas Corey is carrying
Number of Bananas dropped off and at which mile
Number of bananas left/needed for the returning trip
1000
200th: 600 bananas
200 bananas
1000
200th: 600 bananas
200 bananas
1000
200th: 800 bananas
0 bananas
1000
600th: 200 bananas
400 bananas
1000
600th: 200 bananas
0 bananas
Solution
Evaluation
I think I would rate the...

...POW #13CoreyCamel
Problem Statement: Corey the Camel has a small banana grove in the desert, her harvest this year was 3,000 bananas. The market where Corey sells her bananas is 1,000 miles away. Corey has to walk to the market to sell her bananas, for each mile Corey walks, she eats one banana. Corey can only carry 1,000 bananas at a time. In thisPOW the goal is to find the number of bananas that Corey can get to the market.
Process: To find the answer to this POW I did the mini POW, as suggested. I used the same process for POW13, as I did for the mini POW. The process is as follows:
1. Corey starts the trip with 1,000 bananas.
2. She travels 200 miles, she’s left with 800 bananas. She stashes 600 bananas at 200 mile point, keeping 200 the trip back.
3. Corey picks up another 1,000 bananas.
4. She travels 200 miles, she has 800 left. She then picks up 200 from the bananas stashed. She now carries 1000 bananas and has 400 more stashed.
5. She travels an additional 333 1/3 miles, she’s left with 666 2/3 bananas, she stashes 333 1/3 there (533 1/3 mile point), she has 333 1/3 bananas left.
6. She then travels back 333 1/3 miles to 200 mile point. She has no bananas left, so picks up 200 stashed...

...CoreyCamel
Problem Statement CoreyCamel owns a banana grove with 3000 bananas on it. She can only carry 1000 bananas at a time. The market where she can sell her bananas is 1000 miles away. Sounds easy right? Wait there’s a catch. For every mile she walks she has to eat one banana. See now it got hard. So with all of this said how many bananas will he be able to take to the market and sell?
Process Doing the miniPOW helped a lot with this one. Because knowing that CoreyCamel can drop bananas on the way to the market is an important part, because of he couldn’t the answer would be an obvious zero. Since you can drop them I decided to drop 800 bananas at the 100 mile point. That way she can eat her 100 bananas that she needs to on her way there and still have another 100 to get back to her banana grove. When she gets there she’ll have 900 new bananas. The only problem is that she can only carry 1,000 bananas at a time. So she takes 1,000 bananas the rest of the way. This would leave her with 100 bananas. Then I realized that wasn’t the largest amount that she could sell. So I started to work backwards from 500 miles. I did this because 500 is half of 1,000. The only problem with 500 is that she would use 500 bananas to get there and wouldn’t be able to drop any bananas because she would have to use the last 500 bananas to get back. So I started to work with 400...

...Problem Statement
Corey the camel has 3,000 bananas. He has to deliver the bananas 1,000 miles to the market. For every mile Corey travels, he eats one banana. Corey can only carry 1,000 bananas at one time. For this problem I have to find out how many bananas Corey delivers to the market.
Process
For my process I used the maximum number of bananas Corey could carry per trip, which were 1,000 bananas in three trips. The first Trip I stopped at 250 miles and stashed 500 bananas. Corey ate 500 bananas on that trip because he walked to 250 miles and back to zero which means Corey ate 500 bananas, leaving him at zero bananas when he reached start. In the second trip Corey traveled with 1,000 bananas. He stopped at 250 miles, and picked up 250 bananas, which leaves him at 1,000. He continued to travel 250 more miles making him reach 500 miles; at 500 miles Corey stashed 250 bananas. On this trip Corey lost 750 bananas leaving him at 250. On trip three, Corey picked up 250 bananas at 250 miles, leaving him at 1,000 bananas because the banana he ate per mile was regained when Corey picked up the 250 bananas. Corey continued to 500 miles where he picked up 250 bananas, 250 is the number of miles that Corey traveled, leaving him at 1,000 bananas. Corey had...

...POW13
Problem Statement:
The problem of the week states how many bananas can corey the camel get to the market if he has to eat one banana every mile and it s 1000 miles to the market and he has 3000 bananas and he is able to hold only 1000 bananas at a time.
Process:
I knew that corey had to eat one banana every mile and he had to go 1000 miles but could only carry 1000 bananas at a time and there was 3000 miles so i knew he would have to drop off bananas at certain places to be able to go back and get more bananas. I was writing different amount of miles corey could and how many bananas he had to drop off and pick up and how he could go back and get more bananas from the place where he started. This is the process i did to get my answer.
1. start trip with 1000 bananas
2. travel 200 miles, you're left with 800 - stash 600 at 200 mile point, keep 200 for 200 mile trip back.
3. pick up another 1000
4. travel 200 miles, you have 800 left, pick up 200 from stashed, you now carry 1000 and have 400 more stashed.
5. travel an additional 333 ⅓ miles, you're left with 666 ⅔ , stash 333 ⅓ there (533 ⅓ mile point), you have 333 ⅓ left
6. travel back 333 ⅓ miles to 200 mile point, you have no bananas left, pick up 200 stashed (leaving 200 still at 200 mile point), go back the other 200 miles.
7. pick up another 1000
8. travel to 200 mile point, leaving 800 bananas, pick up...

...POW #13
Problem: This POW’s problem was to have seven spaces, three of which were shaded, 1 empty space, then three spaces that were plain. The object was to switch the colors to the opposite side, but while doing this you could only move your marker to an adjacent space, or jump over ONE marker to another open space.
Process: I used pennies in place of the plain markers, and nickels in place of the shaded markers. I started by randomly moving blocks around and then when I eventually succeeded in getting the colors on the opposite sides, I immediately went back and tried to do it again. I then tried to do it with more and more coins. My diagram is on the other page.
Solution: A. The best illustration of my solution is my diagram on the following page. The illustration only shows the moves for when I used 3 of each coin, but when I used 2 I got 8, when I used 3(like in the diagram) I got 15, when I used 4 I got 24, and when I used 5 I got 35.
B. I am confident in this solution because I did each multiple times and got the same answer each time, my answer didn’t change once.
Evaluation: I thought this problem was fairly easy. It didn’t take too long to move the coins to opposite sides and I found I entertaining and fun. I though that it was just about right for me....

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