[Billthedrill]

They should reword the thing, because people who deal in half-dogs get reported to the RSPCA.

x + (x +36) = 49
2x = 13
x=6.5
small dogs = x+36, so 42.5

That's obviously not what they had in mind, but if you can't define a problem correctly nobody can solve it correctly.

[Morgana]

42.5

Sooooooooooooooooo how does 1/2 a dog compete in a dog show???

[beejaa]

That’s exactly how I worked it out. The x refers to the big dogs and the x + 36 refers to the small dogs. The problem is that there are 6.5 big dogs and 42.5 small dogs, and since there are two half dogs, they both must be really small...

[CurlyDave]

...small dogs = x+36, so 42.5...

I would postulate that there were Z medium dogs, where Z is an odd number between 1 and 13. No dogs are harmed if this is done.

While this removes the need for a chainsaw, it is way beyond the age level of the assignment.

[scrabblehack]

Hmmm....it must be that
l = [0,6]
s = [36,42] then
m = [1,13].
That is, there are somewhere between 0 and 6 large dogs,
36 and 42 small dogs, and 1 and 13 medium dogs.

For that matter,
http://www.trainpetdog.com/dog-breed-size-chart.html
has giant and very large dogs as well. So it could be between 1 and 13 medium/very large/giant dogs.

[Celerity]

Well, you’re supposed to phrase the answer as “There are 6 large dogs and a dog that isn’t small or large, it’s medium.”

I had a similar question when I was in school. It was small, medium and large (Fries or something, I don’t recall) the .5 was an exercise. The problem listed smalls and larges and that .5 carry over was “something that isn’t small or large” like extra large or medium.