# Algebra Problem. No Algebra allowed.

AThousandYoung
Posers and Puzzles 27 Feb '07 08:10
1. AThousandYoung
All My Soldiers...
27 Feb '07 08:103 edits
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.
2. 27 Feb '07 08:30
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.
3. AThousandYoung
All My Soldiers...
27 Feb '07 08:36
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.
Yep. That's the solution basically.
4. 27 Feb '07 09:07
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.
5. XanthosNZ
Cancerous Bus Crash
27 Feb '07 09:51
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.
All math is visible or invisible algebra.
6. 27 Feb '07 11:28
Originally posted by XanthosNZ
All math is visible or invisible algebra.
Except for the visibly invisible algebra. Or the invisibly visible algebra.
7. sonhouse
Fast and Curious
27 Feb '07 14:56
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.
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
8. 27 Feb '07 15:00
Originally posted by XanthosNZ
All math is visible or invisible algebra.
All algebra is evincible or invincible math.
9. sonhouse
Fast and Curious
27 Feb '07 20:29
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
So it's easy to do either starting with minimum or maximum legs.
10. uzless
The So Fist
27 Feb '07 21:23
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.
11. AThousandYoung
All My Soldiers...
07 Mar '07 23:38
Originally posted by sonhouse
So it's easy to do either starting with minimum or maximum legs.
You can in fact do it with any starting guess, though assuming one or the other maximum is easier.