$1 \cdot 15$ If $\frac{b}{a}=2, \frac{c}{b}=3$, then $\frac{a+b}{b+c}$ equals
(A) $\frac{1}{3}$.
(B) $\frac{3}{8}$.
(C) $\frac{3}{5}$.
(D) $\frac{2}{3}$.
(E) $\frac{3}{4}$.
(39th American High School Mathematics Examination, 1988)
[Solution] If $\frac{b}{a}=2$, then $\frac{a}{b}=\frac{1}{2}$, $\therefore \quad \frac{a+b}{b+c}=\frac{\frac{a}{b}+1}{1+\frac{c}{b}}=\frac{\frac{1}{2}+1}{1+3}=\frac{3}{8}$. Therefore, the answer is $(B)$.