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. How many chickens does farmer Charlie have?

Let C = number of chickens, D = days food lasts. Total food F = C × D × (food per chicken per day). After selling 75: (C-75)(D+20) = CD. After buying 100: (C+100)(D-15) = CD. Expanding: CD + 20C - 75D - 1500 = CD, so 20C - 75D = 1500. And: CD - 15C + 100D - 1500 = CD, so -15C + 100D = 1500. Solving this system: from first, C = (1500 + 75D)/20. Substituting: -15(1500+75D)/20 + 100D = 1500. -11250 - 56.25D + 100D = 1500. 43.75D = 12750. D = 12750/43.75 = 291.43... that's not right. Let me solve differently: Multiply first eq by 3: 60C - 225D = 4500. Multiply second by 4: -60C + 400D = 6000. Adding: 175D = 10500, D = 60. Then 20C - 75(60) = 1500, so 20C - 4500 = 1500, 20C = 6000, C = 300.