11. Market transactions. A farmer bought 100 heads of cattle at a total cost of 1000 dollars on the market. One cow cost 50 dollars, one sheep - 10 dollars, and one rabbit - 50 cents. How much money did the farmer spend on buying cows, sheep, and rabbits separately?

This problem can be solved through more or less meticulous trial and error, but there are methods that allow for a quick solution.
11. A farmer bought 19 cows for 950 dollars, 1 sheep for 10 dollars, and 80 rabbits for 40 dollars, which totals 100 heads with a combined value of 1000 dollars. The problem can be easily solved arithmetically using the method of averages: the average cost per head of livestock is the same as the cost of one sheep. Algebraically, the problem can be solved as follows. Since \(x+y+z=100\), then \(\frac{1}{2} x+\frac{1}{2} y+\frac{1}{2} z=50\)

\[
\begin{aligned}
50 x+10 y+\frac{1}{2} z & =1000 \\
-\quad \frac{1}{2} x+\frac{1}{2} y+\frac{1}{2} z & =50 \\
\hline 49 \frac{1}{2} x+9 \frac{1}{2} y & =950
\end{aligned}
\]

(prices are given in dollars), or \(99 x+19 y=1900\). Thus, the problem reduces to solving the indeterminate equation. The only* set of \(x\) and \(y\) that satisfies this equation is \(x=19\), \(y=1\). To make the total number of heads equal to 100, \(z\) must be 80.