can you please help me with the following algebra problem? i don't know how to start.
A man raised chickens. He sold all but 8 of them. The first buyer bought 1/3 of all the chickens + 2/3 of a single chicken. The second buyer bought 1/3 of the remaining chickens + 1/3 of a single chicken. The third buyer bought 1/3 of the remainder + 2/3 of a chicken. How many chickens did the seller own before selling any? All chickens were alive and whole when the buyers received them.
thank you very much, rick
name = rick
I like this problem. To get started is to realize that the one time when you know exaclty how many chichens there are is after all the sales are made. That should make you think about working backwards. That is start with the third buyer.
The third buyer bought 1/3 of the chickens offered to him plus 2/3 of a chicken, and left 8 chickens. Thus 1/3 of the chickens offered to him would have left 8 + 2/3 = 26/3 chickens. That is 2/3 of the chickens offered to him was 26/3 chickens. Since 2/3 of the chickens offered to him is 26/3 chickens, he was offered 13 chickens. Therefore after the second buyer there were 13 chickens remaining.
Now deal with the second buyer. The second buyer bought 1/3 of the chickens offered to him plus 1/3 of a chicken, and left 13 chickens.
When you have worked back through the second and first buyers to see how many chickens the man started with, make sure you check your answer. Work forward through the three buyers to see that ther are 8 chickens remaining.Cheers,