Misplacing ones excrement

In the somewhat more colourful language of this site: The Internet is losing its shit over this second grade math problem.

What is the problem?

“There are 49 dogs signed up to compete in the dog show. There are 36 more small dogs than large dogs signed up to compete. How many small dogs are signed up to compete?”

The answers that are discussed on the site are

1. 36, with an argument not worthy of the name: 49-36=13, so 13 large dogs, so the answer is 36
2. 42.5, because 49-36=13, 13/2=6.5 and 36+6.5=42.5
3. 36 once more, but with the added information that 36-13=23, that’s 23 more small dogs

Answers 1 and 3 suffer from something that I observe with many first-year students: I ask something, in a Calculus class it is quite often a primitive function, and many suggestions come flying in, with the expectation that I will deem one of them correct. My reaction then is: you could have checked your answer yourself, in the case of the primitive by differentiation. That is why I asked that question in the first place: to show that they can verify many of their solutions by just substituting back into whatever it was they were solving.

You can verify whether 36 is correct: 36 small dogs is 23 more than 13 large dogs and 23 is definitely not equal to 36. So, no, 36 is not the correct answer.

The second answer — “That’s how I did it in my head.” — is correct, albeit unfortunate for one dog. It is also how I would (try to) explain it to a second-grader.
There are 36 more small dogs than large dogs, so we can set 36 small dogs aside and then look at the remaining dogs. If we want to keep the difference equal to 36 then we must be able to take away two dogs at a time (small and large) from those remaining until we have a group of small and a group of large dogs left. Most likely the second-grader will tell you that that is not going to happen with an odd number of dogs. The conclusion of course is that this is an ill-posed problem.

You can solve this problem — or rather, show it’s unsolvability — in other ways.
The sum of the two numbers in question is odd, that means that they have different parity, i.e., one is odd the and other is even, but then their difference is odd as well and it can’t be 36.
Or you can introduce letters, S and L, and express the demands as equations: S+L=49 and S-L=36. Solving these will lead to the solution done in the head: 2S=85, hence S=42.5 and L=6.5.

I hope that the teacher will admit that there was an error in the problem and that they will turn this into a valuable lesson: you can often check whether your answer is correct or makes sense.

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