Ans: C
$\begin{array}{rcl}
\dfrac{2\alpha+3\beta}{3\alpha+2\beta} & = & \dfrac{7}{10} \\
10(2\alpha +3\beta) & = & 7(3\alpha +2\beta) \\
20\alpha +30\beta & = & 21 \alpha +14\beta \\
\alpha & = & 16\beta
\end{array}$
Therefore, we have
$\begin{array}{cl}
& \dfrac{2\alpha+\beta}{\alpha+2\beta} \\
= & \dfrac{2(16\beta) +\beta}{(16\beta) +2\beta} \\
= & \dfrac{33\beta}{18\beta} \\
= & \dfrac{11}{6}
\end{array}$
