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December 20, 2014

December 20, 2014

Posted by **m-cook** on Monday, August 17, 2009 at 4:20am.

- 6th grade math -
**MathMate**, Monday, August 17, 2009 at 8:46amThere are two ways to do it, the farmer's way and the mathematicians's. I prefer the farmer's way.

First, the farmer's way:

We have 18 animals and 52 legs.

If we have 9 of each, that makes an average of 3 legs each animal, we get 3*18=54 legs. We only have 52. So change one pig (9-1=8) for a chicken (9+1=10) to get 52 legs (two less). That makes 10 chickens and 8 pigs. No calculators, no pain either.

Now the mathematicians way:

Let p be the number of pigs, then the number of chickens is (18-p).

The total number of legs is thus

4*p + 2(18-p) = 52

2p = 52-36

p = 16/2

=8

(18-p) = 18-8 = 10

Answer: there are 8 pigs and 10 chickens.

- 6th grade math -
**john**, Monday, August 17, 2009 at 9:57pmmake a funny sentense for each math word

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