Using Different Problem Solving Methods (Pigs and Chicken Problem)

Often times experienced mathematicians tend to get comfortable with certain problem-solving strategies. For example, in a problem one might use a system of equations to solve a problem rather than employing a simpler more easy way to solve it. Though using system of equations are great, knowing how to solve problems using different approaches is important, not just for oneself, but for their students.
Take for example the following problem: A farmer has both pigs and chickens on his farm. There are 78 feet and 27 heads. How many pigs and how many chickens are there?
Solution 1: (Using Algebra System of Equations)
4p+2c=78 (pigs have 4 feet and chickens have 2 feet with 78 feet in total)
p+c=27 (27 heads mean that the number of chicken and pigs total 27)
Then by algebra p=27-c. Therefore by substitution, 4(27-c)+2c=78. 108-2c=78. 2c=30. c=15. Since, c=15, p+c=p+15=27. p=12. Therefore, the farmer has 15 chickens and 12 pigs.
In this first solution it may be unclear for students as to why we set up the algebra as such. This, in my experience, is the preferred method of problem-solving of many experienced mathematicians. However, often it is the least understood among math students. Another and even more eloquent way to solve this problem is by simple reasoning.
Solution 2:
We know that there are 27 total animals. So there are 27 total spots. Now since we are talking about pigs and chickens, each spot must have at least 2 feet each. So by giving each spot two feet we have given 27(2)=54 total feet so far. Now we have 78-54=24 feet left. Then we give out the remaining feet in pairs to each of the 27 spots until we run out. We can only give out 24/2=12 pairs of feet. Therefore only 12 spots have 4 feet which mean only 12 are pigs and the remaining 27-12=15 are chickens since they only have 2 feet. This can easily be seen through a picture displaying each step of the process.
This is the best thing about mathematics. There is only one answer (usually) but several creative ways to arrive at it. To limit yourself and students to only one way is horrible and only cheats yourself and your students. In the latter solution, there is deeper more engaged reasoning going on. Understanding the situation and employing simple logic without loss of understanding. Also, with the ability to do math using various methods, it creates a way to discuss with students the connections between both solution methods.
Now, there are other solution strategies that work great with this problem and similar formatted problems. If you have some others feel free to comment below.


Andrew R.

Patient and Knowledgeable Math Tutor

50+ hours
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