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Cow, Pig, and Sheep Problem

Link: Cow, Pig, and Sheep Problem

When I first read through this problem I thought it was so easy. I completed it so quickly and wondered even why we had been assigned this. Then I realized that I didn’t read the whole thing and that I needed to have 100 animals. So then I went about solving the ACTUAL problem. I started off the way I start off any problem: write down what I know. Doing this led to two equations and three variables, which I knew I couldn’t solve. There had to be something else that I could extract from the problem. Combing through it again, this time much more frustrated, I decided to try and take the constraint on having at least one of each animal and turn that into something that I could use. All that resulted in, though, was my changing my original equations so that at least one of each animal would be accounted for. The only way I could think to solve this was to start substituting guesses in for how many sheep there were, thus reducing my two equations to two variables and therefore making them solvable. I started off by guessing there were 51 sheep. I knew that I needed an odd number of sheep because I had already accounted for one and I needed to bring the cost to $100 even. This gave me a weird answer that didn’t work within the constraints of the situation so I moved on and tried 85 sheep. This didn’t work either. I had the feeling that there was something I was missing, so I took a break and went back to the problem a few hours later. That break was exactly what I needed. After looking at the algebra that I had used to solve my previous attempts I saw that I could generalize this situation to help me make better guesses for the number of sheep. My previous solutions kept giving me negative answers, so I needed to rework my guesses so that I could produce positive answers. The cost of my pigs, for example, had to be less than the remaining total after I purchased all my sheep (and one of each animal). So that meant that 3(97-s)<86.5-0.5s. If I solved this then I could find a restriction on s, thus narrowing down the possibilities for me to begin my guessing. This led me to the fact that I had to have more than 81.8 sheep, so I began the remainder of my tests there. After some time I found my solution. It was annoying to solve this through an educated guess and check method, and I wish I could have remembered a more efficient way to do so, but sometimes math is all about the long, tedious, number crunching!

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About Nicole Bencze

I am an awesome newish math teacher who calls Delta School District her home. I'm crazy in love with math and there's no place in this world I'd rather be than in a classroom showing my kidlets how amazing it can be! Outside of teaching I'm quite the coffee addict and I live inside the fandom universe falling in love again and again with Lord of the Rings, Harry Potter, Mortal Instruments, Twilight, Hunger Games, Percy Jackson, Game of Thrones, Sherlock, Doctor Who, and a ton more. My cat is my best friend and I'm way too invested in TV shows!

2 responses »

  1. Hi Nicole. I gave my kids a similar problem called “Farmer’s Dilemma.” It had horses, cows, and sheep, priced at $10, $1, and $0.50 respectively.

    We used Excel to do this problem 🙂

    Reply

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