Hay Baleer Problem
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 …show more content…
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 a solution. The weights of the bales were 39, 41, 43, 44, and 47. Here are the calculations:
80 = 39 + 41 (1 & 2)
82 = 39 + 43 (1 & 3)
83 = 39 + 44 (1 & 4)
84 = 41 + 43 (2 & 3)
85 = 41 + 44 (2 & 4)
86 = 39 + 47 (1 & 5)
87 = 43 + 44 (3 & 4)
88 = 41 + 47 (2 & 5)
90 = 43 + 47 (3 & 5)
91 = 44 + 47 (4 & 5)
We could not find any other set of weights and we believe that this is the only set of possible weights.
Extensions
Some extensions to this problem might be: v What if the weights were different? v Find the combined weights if there was a sixth bale that weighed 48 kgs.
Evaluation
When I first read this problem, my initial thoughts were "Oh my god, this is really hard", "This is too confusing", and "This is going to take so long to do". I think this is a good P.O.W. because it teaches you how to solve things, but I personally think in reality, the hay baler could learn to not be lazy and just reweigh the bales. From this problem, I learned to problem solve and keep experiment through trial and error until you get your answer. I think the problem could have given us the weight of one bales so we could use an algebraic expression to figure out the weights of the other bales. I did not enjoy working on it because it kind of bothered me that the hay baler was too lazy to reweigh the bales. It was not too hard or too easy, it was appropriate for what we've been learning. In terms of math, I learned about different mixes of numbers and problem solving. I expect to use the problem solving in future problems.
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 …show more content…
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 a solution. The weights of the bales were 39, 41, 43, 44, and 47. Here are the calculations:
80 = 39 + 41 (1 & 2)
82 = 39 + 43 (1 & 3)
83 = 39 + 44 (1 & 4)
84 = 41 + 43 (2 & 3)
85 = 41 + 44 (2 & 4)
86 = 39 + 47 (1 & 5)
87 = 43 + 44 (3 & 4)
88 = 41 + 47 (2 & 5)
90 = 43 + 47 (3 & 5)
91 = 44 + 47 (4 & 5)
We could not find any other set of weights and we believe that this is the only set of possible weights.
Extensions
Some extensions to this problem might be: v What if the weights were different? v Find the combined weights if there was a sixth bale that weighed 48 kgs.
Evaluation
When I first read this problem, my initial thoughts were "Oh my god, this is really hard", "This is too confusing", and "This is going to take so long to do". I think this is a good P.O.W. because it teaches you how to solve things, but I personally think in reality, the hay baler could learn to not be lazy and just reweigh the bales. From this problem, I learned to problem solve and keep experiment through trial and error until you get your answer. I think the problem could have given us the weight of one bales so we could use an algebraic expression to figure out the weights of the other bales. I did not enjoy working on it because it kind of bothered me that the hay baler was too lazy to reweigh the bales. It was not too hard or too easy, it was appropriate for what we've been learning. In terms of math, I learned about different mixes of numbers and problem solving. I expect to use the problem solving in future problems.