2018-II-10

$\begin{array}{rcl}
3a & = & 4b \\
\dfrac{a}{b} & = & \dfrac{4}{3} \\
a : b & = & 4 : 3
\end{array}$

Hence, we have

$\begin{array}{cccccccccc}
a & : & b & & & = & 4 & : & 3 & & & \\
a & & & : & c & = & 2 & & & : & 5 \\ \hline
a & : & b & & & = & 4 & : & 3 & & & \\
a & & & : & c & = & 2 \times 2 & & & : & 5 \times 2 \\ \hline
a & : & b & : & c & = & 4 & : & 3 & : & 10
\end{array}$

Let $a = 4k$, $b = 3k$ and $c = 10k$, where $k$ is a non-zero constant. Hence, we have

$\begin{array}{cl}
= & \dfrac{a + 3b}{b + 3c} \\
= & \dfrac{4k + 3(3k)}{3k + 3(10k)} \\
= & \dfrac{13k}{33k} \\
= & \dfrac{13}{33}
\end{array}$