Matematicas pow 6 imp 1

Juan Pablo Santa Maria
27\1\11 P: 7

POW #6 The Hay baler problem

Problem Statement: A hay baler has 5 bales these bales have to be weighed but they are not weighed individually but indifferent combinations. There is bale: 1,2,3,4,5 so the point is to find the weigh of each bale and it has to make sense.

1-2 |
1-3 |
1-4 |
1-5 |
2-3 |
2-4 |
2-5 |
3-4 |
3-5 |
4-5 |Process: At the beginning of the POW I was very confused because I didn’t really understand the problem because It didn’t give me much info. First I knew I had to find the weights of bales#1,2,3,4,5 so I started to do the combinations:

80 | 1+2 |
82 | 1+3 |
83 | 1+4 |
84 | 1+5 |
85 | 2+3 |
86 | 2+4 |
87 | 2+5 |
88 | 3+4 |
90 | 3+5 |
91 | 4+5 |
Then what I did was toadd all the weights 80,82,83,84,85,86,87,88,90,91 which gave me a result of 856kg then I divided this by 20 to get the average of 42,8kg, which is the average weight per bale. They weights have arange of 80-91 so from this I can say that the weight of each bale will be between 38 and 48 kilograms. So with the help of my dad I started to find each combination I started with 38kg and I made thistable. This table was wrong because the order of the combinations was wrong which made my results not accurate when I combined the kg of each hay bales.

1: | 38kg |
2: | 42kg |
3: | 44kg |
4:| 44/45kg |
5: | 46 |



Then I did the exact same thing with 39 but still the results didn’t match. So I started to make easy conclusions, which are that bales 1-2 are the lightest, and 4-5are the heaviest so you can say that combination 1-2 is the first number 80 and the combination 4-5 will be the last number 91 so then ill have to find their values. So after thinking a lot I camewith an easy conclusion, which made me think that the problem is very easy. The solution was to put each combination in order from least to greatest the only difficult part is to find the value of hay...