To solve this problem, let's assume that Farmer Charlie initially had "x" number of chickens on his farm.
According to the given information, if he sells 75 chickens, he can feed the remaining chickens for 20 days longer. This means that the amount of chicken food he has can last for (x - 75) chickens for (20 + 15) days.
Similarly, if he buys 100 extra chickens, he would run out of chicken food 15 days earlier. This means that the amount of chicken food he has can last for (x + 100) chickens for (20 - 15) days.
Now, let's set up the equations based on the above information:
Equation 1: (x - 75) * (20 + 15) = x * 20
Equation 2: (x + 100) * (20 - 15) = x * 20
Let's solve these equations to find the value of x, which represents the initial number of chickens.
Expanding the equations:
Equation 1: (x - 75) * 35 = 20x
Equation 2: (x + 100) * 5 = 20x
Simplifying the equations:
Equation 1: 35x - 2625 = 20x
Equation 2: 5x + 500 = 20x
Solving for x:
Equation 1: 35x - 20x = 2625
15x = 2625
x = 175
Equation 2: 20x - 5x = 500
15x = 500
x = 33.33
Since the number of chickens cannot be fractional, we can conclude that Farmer Charlie initially had 175 chickens on his farm.
Therefore, the correct answer is option A) 350.