Question 1210285: A farmer has both pigs and chickens on his farm.
There are 78 feet and 27 heads. How many pigs and chickens are there?
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)   (Show Source):
You can put this solution on YOUR website!
First a typical standard algebraic solution....
p = # of pigs
c = # of chickens
Each pig has four feet and each chicken has two; each pig and each chicken has one head.
The total number of heads is 27:
p+c = 27 [1]
The total number of feet is 78:
4p+2c = 78 [2]
Solve the pair of equations [1] and [2] by your favorite method. When the two equations are in this form, elimination is my preference.
Multiply equation [1] by 2 and compare to equation [2] to eliminate c and solve for p.
2p+2c = 54
4p+2c = 78
2p = 78-54 = 24
p = 24/2 = 12
The number of pigs is 12, so the number of chickens is 27-12 = 15.
ANSWER: 12 pigs, 15 chickens
And now an informal solution, done mentally and quickly.
If all 27 animals were chickens, the number of feet would be 27*2 = 54.
The actual number of feet is 78, which is 24 more than that.
Each pig has 2 more feet than each chicken, so the number of pigs is 24/2 = 12.
ANSWER: 12 pigs and 17-12 = 15 chickens
For studying math, understanding the formal algebraic solution is important.
But solving problems like this informally using logical reasoning is excellent brain exercise.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website!
The 27 animal heads can't be all be chicken heads, for if they were,
they'd have only
27x2=54
feet under them. So where did the other
78-54=24
feet come from?
From the 24/2=12 pigs' right and left hind feet.
So there were 12 pigs and 27-12=15 chickens.
Or, you can solve it this way:
The 27 animal heads can't all be pig heads for if they were, they'd have
27x4=108
feet under them. So how did the 108 feet get reduced by 108-78=30?
By the 30/2=15 chicken heads' failure to have a pair of hind feet
under them.
So there were 15 chickens and 27-15=12 pigs.
Edwin
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