2018-II-10

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

Hence, we have

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

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

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