Farmer Charlie has a chicken farm. On a certain day, Charlie calculates in how many days he will run out of chicken-food. He notices that if he would sell 75 of his chickens, he could feed the remaining chickens twenty days longer with the chicken-food he has, and that if he would buy 100 extra chickens, he would run out of chicken-food fifteen days earlier.

First Question: How many chickens does farmer Charlie have?

Second Question: One chicken lays two eggs in three days. How many eggs do three chickens lay in seven days?

First Answer: We call the amount of chicken-food that one chicken eats per day a "portion". Let f be the number of "portions" of chicken-food that farmer Charlie has and let c be the number of chickens.

Normally, farmer Charlie would run out of chicken-food in f / c days.

If he would sell 75 of his chickens, he could feed the remaining chickens twenty days longer, so

f / (c-75) = f / c + 20

which can be rewritten to equation 1:

20c2 - 1500c - 75f = 0.

If he would buy 100 extra chickens, he would run out of chicken-food fifteen days earlier, so

f / (c+100) = f / c - 15

which can be rewritten to equation 2:

f = 3/20×c2 + 15c.

Combining equations 1 and 2 gives:

20c2 - 1500c - 75×( 3/20×c2 + 15c ) = 0.

Solving this equations gives c=300, so farmer Charlie has 300 chickens.

Second Answer: 1 chicken lays 2 eggs in 3 days, so 3 chickens lay 3×2=6 eggs in 3 days. In 7 days, these 3 chickens lay 7/3 times as much eggs as in 3 days. Conclusion: 3 chickens lay 7/3×6=14 eggs in 7 days.

## Saturday, July 26, 2008

### Charlie's Chickens

Posted by Covert Bay at 1:46 AM

Labels: Riddles/Puzzles

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