［Solution］From the given conditions, we have
$$
x-a \geqslant 0, \quad x-15 \leqslant 0, \quad-15 \leqslant x-a-15 \leqslant-a<0 \text {. }
$$

Then $y=(x-a)+(15-x)+(15+a-x)=30-x$.
Therefore, when $x=15$, $y$ attains its minimum value of 15.
Hence, the correct choice is $(C)$.
13・10 Let $y=|x-a|+|x-15|+|x-a-15|(0<a<15)$. If $a \leqslant x \leqslant 15$, then the minimum value of $y$ is
(A) 30.
(B) 25.
(C) 15.
(D) 7.5.
(Anhui Province, China Junior High School Mathematics Competition, 1990)