Algebra Problem. No Algebra allowed.

There are 50 animals. Some are parakeets and some are dogs. The 50 animals have collectively 140 legs. There are no gimped animals with less than (or more than) the usual number of legs. How many are dogs and how many are parakeets?

Explain in detail how you solved it. This problem was given to elementary school kids who had not yet taken any algebra, so no algebra allowed.

50 animals, 100 legs minimum, extra 40 legs must be dog legs, which means 20 dogs, 30 parakeets.

Yeah, there was math involved, but that's how I solved it.

Originally posted by geepamoogle
50 animals, 100 legs minimum, extra 40 legs must be dog legs, which means 20 dogs, 30 parakeets.

Yeah, there was math involved, but that's how I solved it.

Oddly enough, math on that level is invisible algebra. You're just not directly referring to x's and y's, but you're using many of the same techniques.

Originally posted by geepamoogle
Oddly enough, math on that level is invisible algebra. You're just not directly referring to x's and y's, but you're using many of the same techniques.

Originally posted by XanthosNZ
All math is visible or invisible algebra.

Originally posted by geepamoogle
50 animals, 100 legs minimum, extra 40 legs must be dog legs, which means 20 dogs, 30 parakeets.

Yeah, there was math involved, but that's how I solved it.

Originally posted by XanthosNZ
All math is visible or invisible algebra.

Originally posted by sonhouse
First assume how many legs if all dog, 50*4=200. 200-140=60
60/2=30 parakeets, 50-30=20 Dogs (20*4=80), 80 legs +60 legs=140

140 legs divided by 4 legs equals 35 fourlegged animals. Since total animals =50, then you are 15 animals short.

Therefore take away 15 four legged animals from 35 to get 20.

Therefore 20 four legged animals means 30 two legged animals for a total of 50 animals.

Originally posted by sonhouse
So it's easy to do either starting with minimum or maximum legs.