$4 \cdot 8$ Given $6<a<10, \frac{a}{2} \leqslant b \leqslant 2 a, c=a+b$, then
(A) $9 \leqslant c \leqslant 30$.
(B) $15 \leqslant c<30$.
(C) $9<c \leqslant 18$.
(D) $9 \leqslant c<30$.
(E) $9<c<30$.
(Chongqing, Sichuan Province, Junior High School Mathematics Competition, 1984)
[Solution] Given $6<a<10$, we know $3<\frac{a}{2}<5$, and $12<2 a<20$.
$\therefore \quad 3<\frac{a}{2} \leqslant b \leqslant 2 a<20$,
$\therefore \quad 9<a+b<30$, that is $9<c<30$.
Therefore, the answer is $(E)$.