The Haybaler Problem

heather moir

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Oct 14, 2011
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You have five bales of hay. For some reason, instead of being weighed individually, they wereweighed in all possible combinations of two: bales 1 and 2, bales 1 and 3, bales 1 and 4, bales 1and 5, bales 2 and 3, bales 2 and 4, and so on. The weights of each of these combinations werewritten down and arranged in numerical order, without keeping track of which weight matchedwhich pair of bales.​
The weights in kilograms were 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91. Howmuch does each bale weigh? Is there more than one possible solution? If so, find out what all
of the solutions are and explain how you know.

Can someone please help me start this problem!?

What I know:

I have five bales of hay. When combined together, 10 combinations are possible.
 
My guess

If I had to guess, I would say divide each weight by 2 and get a decimal. Then find out which two numbers go with that decimal. like say you got, 2.5, the numbers would be 2 and 3.
 
My first thought is to assign a symbol to represent the weight of each bale, going from lightest to heaviest.

We know that 80 must be the average weight of the two lightest bales.

(For example, if symbols A and B represent the two lightest weights, then we can write A + B = 160.)

We know that 91 must be the average weight of the two heaviest bales.

(For example, if symbols D and E represent the two heaviest weights, then we can write D + E = 182.)

For an initial case, we could assume that no two bales weigh the same, and go from there to see whether there is a solution.

For subsequent cases, we could assume that some bales weigh the same.

Do these thoughts help you to get started?
 
Hello, heather moir!

You have five bales of hay. For some reason, instead of being weighed individually, they were weighed
in all possible combinations of two: 1 & 2, 1 & 3, 1 & 4, 1 & 5, 2 & 3, 2 & 4, and so on.
The weights of each of these combinations were written down in numerical order,
without keeping track of which weight matched which pair of bales.
The weights in kilograms were 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91.

How much does each bale weigh?
Is there more than one solution?
If so, find all the solutions and explain how you know.
\(\displaystyle \text{Let }A,B,C,D,E\text{ be the individual weights}\) . \(\displaystyle \text{ of the 5 bales, with: }\,A < B < C < D < E\)

\(\displaystyle \text{I will list the pairs of bales and the weighings.}\)
\(\displaystyle \text{Note that they are }not\text{ listed next to their corresponding weights.}\)

. . \(\displaystyle \begin{array}{cc}\text{Bales} & \text{Weight} \\ \hline A+B & 80 \\ A+C & 82 \\ A+D & 83 \\ A+E & 84 \\ B+C & 85 \\ B+D & 86 \\ B+E & 87 \\ C+D & 88 \\ C+E & 90 \\ D+E & 91 \end{array}\)

\(\displaystyle \text{Adding, we get: }\:4A + 4B + C + 4D + 4E \:=\:856\) . . \(\displaystyle \rightarrow\quad A+B+C+D+E \:=\:214 \;\;{\bf(a)}\)


\(\displaystyle \text{It can be shown that two smallest weighings (80, 82)}\) .\(\displaystyle \text{belong to }A\!+\!B \text{ and }A\!+\!C.\)
. . \(\displaystyle \text{and that two largest weighings (90, 91) belong to }C\!+\!E\text{ and }D\!+\!E.\)

\(\displaystyle \begin{array}{ccccccc}\text{We have:} & A+B &=& 80 & [1] \\ & A+C &=& 82 & [2] \end{array}\)

\(\displaystyle \text{Subtract [2] - [1]: }\:C - B \,=\,2 \quad\Rightarrow\quad C \,=\,B+2\;\;{\bf(b)}\)


\(\displaystyle \begin{array}{cccccc}\text{We have:} & C + E &=& 90 & [3] \\ & D +E &=& 91 & [4] \end{array}\)

\(\displaystyle \text{Subtract [4] - [3]: }\:D - C \,=\,1 \quad\Rightarrow\quad D \:=\:C+1\:=\:B+3\;\;{\bf(c)}\)


\(\displaystyle \text{Substitute }{\bf(b)}\text{ and }{\bf(c)}\text{ into }{\bf(a)}:\;A + B + (B+2) + (B+3) + E \:=\:214\)

\(\displaystyle \text{We have: }\:A + 3B + E \:=\:209 \quad\Rightarrow\) . . \(\displaystyle B \:=\:\dfrac{209-(A+E)}{3} \:=\:69 + \dfrac{2-(A+E)}{3}\;\;{\bf(d)}\)

\(\displaystyle \text{Since }B\text{ is an integer, }\,2-(A+E)\text{ must be a multiple of 3.}\)
. . \(\displaystyle \text{Hence: }\,2-(A+E) \:=\:3k \quad\Rightarrow\quad A+E \:=\:-3k+2\)
. . \(\displaystyle \text{That is, }A+E\text{ is two more than a multiple of 3.}\)

\(\displaystyle \text{The only choices are 83 and 86.}\)
\(\displaystyle \text{Since }A+E\text{ is at least the 4th weight on the list, then: }\,A+E \,=\,86\)

\(\displaystyle \text{Substitute into }{\bf(d)}:\;B \:=\:69 - \frac{2-86}{3} \quad\Rightarrow\quad \boxed{B \:=\:41}\)

\(\displaystyle \text{Substitute into }{\bf[1]}:\:A + 41 \,=\,80 \quad\Rightarrow\quad \boxed{A \:=\:39}\;\;\boxed{E \:=\:47}\)

\(\displaystyle \text{Substitute into }{\bf(c)}:\:D \:=\:41 + 3 \quad\Rightarrow\quad \boxed{D \:=\:44}\)

\(\displaystyle \text{Substitute into }{\bf(b)}:\:C \:=\:41 + 2 \quad\Rightarrow\quad \boxed{C \:=\:43}\)
 
Soroban selectively-omitted the following from his quote of the original post. I can certainly understand why.

Can someone please help me start this problem!?
 
Soroban selectively-omitted the following from his quote of the original post. I can certainly understand why.

Yes, heather, it is important to note that
mmm4444bot is the last person in this thread to have given you \(\displaystyle help.\)

After that post, you did not receive \(\displaystyle help.\)
 
Just a few other thoughts with the grandfather of this oldie.

Suppose five bales of hay are weighed two at a time in all possible ways. The weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. How much does each bale weigh?


Allow me to lead you through the logic and see if you can get to the right answer yourself.
1--With 5 bales, a-b-c-d-e, being weighed 2 at a time in all possible combinations, leads to each bale being weighed ? times.
2--The sum of all the weighings is ____?
3--What then is the weight of the 5 individual bales?
4--Which bale can you determine the weight of right away, knowing the sum of the 5 bales and the sums of the lightest two, a + b, and the heaviest two, d + e?
5--Having this one bale weight, how can you proceed to derive the others?


The easy solution--
One of the simpler solutions is as follows:
If you add the list of ten different weighings, you get a total of 1156.
Since each bale was weighed four times, 4A + 4B + 4C + 4D + 4E = 1156.
Dividing the 1156 total by four yields A+B+C+D+E=289.
Since A<B<C<D<E, A + B = 110 and D + E = 121.
Substituting into A+B+C+D+E=289 yields C=58.
The second lightest combined weight, A+C must be 112. Thus A=54. From here he rest is pretty self explanatory.


The longer solution--
Solution: Clearly each bale weighs a different amount or there would be less than 10 different weighings. Assign the weight of each bale to A through E such that A < B < C < D < E . Clearly, A + B = 110 and D + E = 121. Since A + B = 110 and A < B, the greatest A can be is 54 and the least B can be is 56. Similarly, since D +
E = 121 and D < E, the greatest D can be is 60 and the least E can be is 61. As C > B and < D, A + C = 112 and it follows that C = B + 2. Similarly, since D > C, A + D = 113 and it follows that D = C + 1. A boundry of our problem is therefore A = 54, B = 56, D = 60 and E = 61. Since D - B = 4 at this boundry, it follows that B could be
greater than 56 and/or D could be less than 60 , as long as we meet the two conditions derived earlier, that C = B + 2 and D = C + 1. If the boundry condition were the answer, then C = 58, D = 60 and E = 61. But C + E = 119, which is not one of the given totals and no sum gives us 120, which is a requirement. Therefore B is greater than 56 and/or D is less than 60. If B = 57, then C = 59, D = 60 and E = 61. But C + D = 119, which is not one of the given totals. Therefore B is not equal to 57. If B = 58, then C = 60, D = 61 and E = 60. But D must be less than E and/or both C + D and D + E add up to 121, which violates our given weighings. Thus, B is not equal to 58 nor can it be greater than 58. If D = 59, then C = 58, B = 56, A = 54 and E = 62. This solution satisfies all the given weighings, and is therefore a valid solution. What if D = 58 however? If D = 58, then E = 63, C = 57, B = 55 and A must equal 55, which violates our condition that A < B. Thus A = 54, B = 56, C = 58, D = 59 and E = 62. Not as elegantly straight forward but just another viewpoint.






 
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