## 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 chicken-food with this number of chickens.
The total amount of chicken-food always equals the number of chickens multiplied by the number of days the food lasts.
Therefore, the total amount of chicken-food 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 chicken-food would then equal

(n- 75 ) × (d+ 20 ).

But the total amount of chicken-food 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 chicken-food fifteen days earlier, so the total amount of chicken-food would then equal

(n+ 100 ) × (d- 15 ).

But the total amount of chicken-food 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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