1. Given real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
2 a+13 b+3 c=90,3 a+9 b+c=72 . \\
\text { Then } \frac{3 b+c}{a+2 b}=(\quad) .
\end{array}
$$
(A) 2
(B) 1
(C) 0
(D) -1
$$
\begin{array}{l}
2(a+2 b)+3(3 b+c)=90, \\
3(a+2 b)+(3 b+c)=72. \\
\text { Then }\left\{\begin{array}{l}
a+2 b=18, \\
3 b+c=18
\end{array} \Rightarrow \frac{3 b+c}{a+2 b}=1\right. \text {. } \\
\end{array}
$$