$4 \cdot 31$ Real numbers $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ satisfy the system of equations
$$
\left\{\begin{array}{l}
x_{1}+x_{2}+x_{3}=a_{1}, \\
x_{2}+x_{3}+x_{4}=a_{2}, \\
x_{3}+x_{4}+x_{5}=a_{3}, \\
x_{4}+x_{5}+x_{1}=a_{4}, \\
x_{5}+x_{1}+x_{2}=a_{5} .
\end{array}\right.
$$

where $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are real constants, and $a_{1}>a_{2}>a_{3}>a_{4}>a_{5}$. Then the order of $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ is
(A) $x_{1}>x_{2}>x_{3}>x_{4}>x_{5}$.
(B) $x_{4}>x_{2}>x_{1}>x_{3}>x_{5}$.
(C) $x_{3}>x_{1}>x_{4}>x_{2}>x_{5}$.
(D) $x_{5}>x_{3}>x_{1}>x_{4}>x_{2}$.
(China Junior High School Mathematics League, 1993)
[Solution] From (1)-(2) we have $x_{1}-x_{4}=a_{1}-a_{2}>0$, which means $x_{1}>x_{4}$.
From (2) - (3) we have $x_{2}-x_{5}=a_{2}-a_{3}>0$, which means $x_{2}>x_{5}$.
From (3) - (4) we have $x_{3}-x_{1}=a_{3}-a_{4}>0$, which means $x_{3}>x_{1}$.
From (4) - (5) we have $x_{4}-x_{2}=a_{4}-a_{5}>0$, which means $x_{4}>x_{2}$.
In summary, $\quad x_{3}>x_{1}>x_{4}>x_{2}>x_{5}$.
Therefore, the answer is $(C)$.