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Mathematics Challenge – Highlighting the Standards for Mathematical Practice
Problem: How much does each bale weigh?
Amanda Lambertus, Arkansas State University
Rachel Fegtly, Arkansas State University
The problem for this issue’s mathematical challenge is from Driscoll’s book Fostering algebraic thinking: A guide for teachers grades 6-10 (1989). It is recommended that you attempt to solve the problem before considering the approach below.
The Problem
You have five bales of hay. For some reason, instead of being weighed individually, they were weighed in all possible combinations of two: bales 1 and 2, bales 1 and 3, bales 1 and 4, bales 1 and 5, bales 2 and 3, bales 2 and 4, and so on. The weights of each of these combinations were written down and arranged 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? (Driscoll, 1999)
In order to think about the Mathematical Practices, pre-service teachers at Arkansas State University are involved in solving non-routine problems in a problem-solving journal. Students keep records of their thinking and attempts at completing the problem. These problems give them insight to their own mathematical thinking. As a result, each problem gives students the opportunity to be deeply involved in the Standards of Mathematical Practice. The Hay Bale Problem highlights, making sense of problems and persevere in solving them, constructing viable arguments, using appropriate tools, attending to precision, and looking for and making use of structure. Outlined below is a summary of how a student might approach this problem. It combines several students thinking presented as one attempt to solve the problem.
Making Sense of the Problem and Persevering in Solving
Initially, students needed to carefully examine the problem and make sense of what it is asking. First, the bales are weighed in pairs and the combinations of weights were recorded and arranged in numeric order. The students need to find a method for describing the bales in a non-confusing manner. Throughout the following dialogue, the reader can see that the student is using and stating assumptions and results from those assumptions. She is constructing an argument that is logical and easy to follow.
Student Solution: Let the bales be x, y, z, p, and q. The bales are named in numeric order, so that x is the lightest bale, y is the second lightest, z is the third lightest, p the fourth lightest, and q the heaviest bale. Therefore the following possible combinations exist, {x+y, x+z, x+p, x+q, y+z, y+p, y+q, z+p, z+q, p+q} – 10 combinations.
By naming and ordering the bales, the student is able to associate pairs of bales with possible weights.
Since x and y are the two lightest bales their sum should be equal to 80 kgs i.e. x+y = 80. Also, if p and q are the two heaviest then their sum is equal to 91 kg. i.e. p+q = 91. This means that the following sums may be true.
x+y = 80
x+z = 82
x+p = 83
x+q = 84
y+z = 85
y+p = 86
y+q = 87
z+p = 88
z+q = 90
p+q = 91
It is difficult to tell which combinations will sum to which values, but this seems logical.
In an additional approach, the student examined the weights of the bales and possible number combinations. The student is looking for and making use of structure in the number system and properties of addition.
80 – 2 even or two odd weights needed
82 – 2 even or two odd
83 – 1 even and one odd
84 – 2 even or 2 odd
85 – 1 even and one odd
86 – 2 even or 2 odd
87 – 2 even or 2 odd
88 – 2 even or 2 odd
90 – 2 even or 2 odd
91 – 1 even and one odd
This means that at least one bale has an odd weight value and at least one bale has an even weight value AND at least 4 odd weight values or 4 even weights values. Therefore, either 4 even weight values and 1 odd OR 4 odd weight values and 1 even.
If I combine the two ideas, the following is true…
x+y = 80, 2 even or two odd weights needed
x+z = 82, 2 even or two odd
x+p = 83, 1 even and one odd
x+q = 84, 2 even or two odd
y+z = 85, 1 even and one odd
y+p = 86, 2 even or two odd
y+q = 87, 1 even and one odd
z+p = 88, 2 even or two odd
z+q = 90, 2 even or two odd
p+q = 91, 1 even and one odd
By examining a subset of these equations, the following conclusions can be drawn.
x+y = 80, 2 even or two odd weights needed
x+z = 82, 2 even or two odd
x+p = 83, 1 even and one odd
x+q = 84, 2 even or two odd
Let’s look at the first two equations…
x+z = 82
x+y = 80 (Subtract the second equation from the first)
→ z-y = 2 i.e. z = y + 2 or that bale z weighs two more kilograms than bale y.
x+p = 83
x+z = 82 (again subtract the second equation from the first)
→ p-z = 1 i.e. p = z + 1 or that bale p weighs one more kilogram than bale z.
x+q = 84
x+p = 83 (Subtract the second equation from the first)
→ q-p = 1 i.e. q = p + 1 or that bale q weights one more kilogram than bale p.
We can conclude that three of the weights are close together and unique. The only bale that we don’t know about is bale x.
At this point, the student knew enough information, that she could choose a weight for bale x and see if her assumptions held.
I think the lowest number must be 37, 38, or 39. I don’t think it can be 40.
Let x = 38, then y = 42, z = 44, p = 45, and q = 46. Now we can check the equations.
x+y = 80 38 + 42 = 80
x+z = 82 38 + 44 = 82
x+p = 83 38 + 45 = 83
x+q = 84 38 + 46 = 84
y+z = 85 42 + 44 = 86**
y+p = 86 42 + 45 = 87**
y+q = 87 42 + 46 = 88**
z+p = 88 44 + 45 = 89**
z+q = 90 44 + 46 = 90
p+q = 91 45 + 46 = 91
**My values do not hold for these equations based on our previous assumptions. So I am going to conclude that x does not equal 38.
Next, I tried 37 kgs for the lightest bale.
Let x = 37, then y = 43, z = 45, p = 46, and q = 47. Now we can check the equations.
x+y = 80 37 + 43 = 80
x+z = 82 37 + 45 = 82
x+p = 83 37 + 46 = 83
x+q = 84 37 + 47 = 84
y+z = 85 43 + 45 = 88**
y+p = 86 43 + 46 = 89**
y+q = 87 43 + 47 = 90**
z+p = 88 45 + 46 = 91**
z+q = 90 45 + 47 = 92**
p+q = 91 46 + 47 = 93**
**My values do not hold for these equations based on our previous assumptions. So I am going to conclude that x does not equal 37.
I tried 39 kilograms for the lightest bale.
Let x = 39, then y = 41, z = 43, p = 44, and q = 45. Now we can check the equations.
x+y = 80 39 + 41 = 80
x+z = 82 39 + 43 = 82
x+p = 83 39 + 44 = 83
x+q = 84 39 + 45 = 84
y+z = 85 41 + 43 = 84**
y+p = 86 41 + 44 = 85**
y+q = 87 41 + 45 = 86**
z+p = 88 43 + 44 = 87**
z+q = 90 43 + 45 = 88**
p+q = 91 44 + 45 = 89**
**My values do not hold for these equations based on our previous assumptions. So I am going to conclude that x does not equal 39.
Something is still not correct. It might be in my assumption that q = p + 1
I still believe that x is 37, 38, or 39, but there seems to be a problem with the higher bale weights. I know that p + q = 91, and if x = 39 then p = 44. This would imply that q = 47.
Next, I tried 39 kilograms for the lightest bale and 47 for the heaviest.
Let x = 39, then y = 41, z = 43, p = 44, and q = 47. Now we can check the equations.
x+y = 80 39 + 41 = 80
x+z = 82 39 + 43 = 82
x+p = 83 39 + 44 = 83
x+q = 84 39 + 47 = 86##
y+z = 85 41 + 43 = 84##
y+p = 86 41 + 44 = 85##
y+q = 87 41 + 47 = 88##
z+p = 88 43 + 44 = 87##
z+q = 90 43 + 47 = 90
p+q = 91 44 + 47 = 91
##My values do not hold for these equations based on our previous assumptions. Therefore, I am going to conclude that x does not equal 39.
Something is still not correct.
While, my assumptions for the middle 5 equations, (##) were not true, I did find all my weight values and correlated them with unique pairs of bales.
Therefore, their weights are 39kg, 41kg, 43kg, 44kg, and 47kg.
Through the student’s solution, we can observe her thinking and the different methods she used to reason through the problem.
When using problems similar to this one, it is important that the teacher consider the entire process, not just the solution. We are looking for evidence of the Standards for Mathematical Practice in the student work. It is also important to consider how the students are connecting their ideas. As a follow-up activity, teachers may try to have their students find the mistake in the student’s reasoning above, or to highlight the different mathematical practices that they see highlighted in the task. §
Reference:
Driscoll, M. (1989). Fostering algebraic thinking: A guide for teachers grades 6-10. Heinemann, Portsmouth, NH, pp. 80-81.
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