7・75 Let $f(x)$ be a function defined on the set of real numbers $R$, and satisfies the following relations:
$$
f(10+x)=f(10-x), f(20-x)=-f(20+x) .
$$

Then $f(x)$ is
(A) an even function, and also a periodic function.
(B) an even function, but not a periodic function.
(C) an odd function, and also a periodic function.
(D) an odd function, but not a periodic function.
(China High School Mathematics League, 1992)
[Solution 1] From the problem, we have $f[10+(10-x)]=f[10-(10-x)]$, which means
$$
f(x)=f(20-x).
$$

Also,
$$
f(20-x)=-f(20+x)=f(x),
$$

thus
$$
\begin{aligned}
f(40+x) & =f[20+(20+x)] \\
& =-f(20+x)=f(x),
\end{aligned}
$$

so $f(x)$ is a periodic function with a period of 40.
Furthermore, $f(-x)=f(20+x)=-f(x)$, indicating that $f(x)$ is an odd function.
Therefore, the answer is $(C)$.
[Solution 2] The graph of the function given in the problem is shown on the right.
Clearly, $f(x)$ is an odd periodic function.
Therefore, the answer is $(C)$.