Think of whole chickens. The second guy sold a+6 chickens after lunch for the price of a chickens before lunch. Likewise #3 sold b+10 for the price of b. Sinc

Think of whole chickens. The second guy sold a+6 chickens after lunch for the price of a chickens before lunch. Likewise #3 sold b+10 for the price of b. Since the price difference is constant, the common denominator is c+2 chickens for the price of c. The only solution (given the number of chickens) is 3 chickens for the price of 1. The chickens sold for one third the price after lunch. The rest is easy.