Solution to:
Charlie's Chickens
Let n be the number of chickens and let d be the number of days in which Charlie will run out of chickenfood with this number of chickens.
The total amount of chickenfood always equals the number of chickens multiplied by the number of days the food lasts.
Therefore, the total amount of chickenfood equals n × d.
If Charlie would sell 75 of his chickens, he could feed the remaining chickens twenty days longer, so the total amount of chickenfood would then equal
( n  75 ) × ( d + 20 ).
But the total amount of chickenfood does not change, so
( n  75 ) × ( d + 20 ) = n × d
which can be rewritten to equation 1:
d = ^{4}/_{15} × n  20.
If Charlie would buy 100 extra chickens, he would run out of chickenfood fifteen days earlier, so the total amount of chickenfood would then equal
( n + 100 ) × ( d  15 ).
But the total amount of chickenfood does not change, so
( n + 100 ) × ( d  15 ) = n × d
which can be rewritten to equation 2:
d = ^{3}/_{20} × n + 15.
Combining equations 1 and 2 gives:
^{4}/_{15} × n  20 = ^{3}/_{20} × n + 15.
Solving this equations gives n = 300, so farmer Charlie has 300 chickens.
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