# Problem Of The Week #8

*This*

**essay Problem Of The Week #8**is available for you on Essays24.com! Search Term Papers, College Essay Examples and Free Essays on Essays24.com - full papers database.Autor: anton • December 24, 2010 • 809 Words (4 Pages) • 1,533 Views

**Page 1 of 4**

Problem Statement

This weeks P.O.W. was named, "The Hay Baler Problem", it was about a hay baler who had just finished weighing five bales of hay. Unfortunately, he weighed the bales was in combinations of two. He weighed bale 1 with 2, 1 with 3, 1 with 4, and every other two-baled combination. In the end, he came up with a set of 10 different weights. However, he did not remember which order he weighed the bales and was now supposed to determine the weight of each individual bale, not the bale combinations. Using the final weights (in kilograms) he came up with from all the bale combinations: 80, 82, 83, 84, 85, 86, 87, 88, 90 and 91, it is your job to figure out the final weight of each of the five individual bales of hay. Also, you may want to find out if there is possibly more than one set of weights for this problem, and how you know this.

Process

When I first had to work on this problem, I was confused and asked Mr. Schenck for help (He was my old math teacher) because I did not really get it. He didn't understand it either so I gave him the problem so he could think about it over night. I decided to try it again later on over night also. The next day I went to see Mr. Schenck and we first started by trying to solve for the lowest weight 80. We decided we would just try and start adding up two numbers and see if the weights would all work out. We figured that since 80 divided by 2 was 40, and two bales would not weigh the same, we would start with one up from the halfway point 40, and go with 41. This is what we got:

80= 41+ 39

This worked out and made sense. Figuring that bale 1 was 41 kgs, we moved on to the next weight.

82= 41+41

This is where we came to a problem, seeing as two bales wouldn't weigh the same, we could not have 41 as the weight of bale 1.

Seeing as we came to that problem, we decided to reverse it around and this is what we got:

80 = 39+41 (1&2)

82 = 39+43 (1&3)

83 = 39+44 (1&4)

84 = 39+ 45 (1&5)

85 = 41+44 (2&4)

86 = 41+45 (2&5)

87 = 44+43 (3&4)

88 = 45+43 (3&5)

90 = ? + ?

Process

90 is where we came to a problem. Even if we added my two highest weights together, we still only got a combined weight of 89. We decided since most of the answers were right we would just try changing a few numbers. Since my highest numbers would not add up high enough, we decided we would try to increase the top number. We changed the 45 to 46 and it still did not work out, so we increased it once again to 47 where we got my solution.

Solution

After going through the process of experimentation, we finally reached

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