2020-I-04

Ans: $\dfrac{4}{5}$
$\begin{array}{rcl}
\dfrac{a}{b} & = & \dfrac{6}{7} \\
a : b & = & 6 : 7
\end{array}$

Also,

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

Hence, we have

$\begin{array}{lllllll}
a & : & b & & & = & 6 & : & 7 & & \\
a & : & & & c & = & 4 & : & & & 3 \\ \hline
a & : & b & & & = & 6 \times 2 & : & 7 \times 2 & & \\
a & : & & & c & = & 4 \times 3 & : & & & 3 \times 3\\ \hline
a & : & b & & & = & 12 & : & 14 & & \\
a & : & & & c & = & 12 & : & & & 9 \\ \hline
a & : & b & : & c & = & 12 & : & 14 & : & 9
\end{array}$

Let $a= 12k$, $b = 14k$ and $c = 9k$, where $k \neq 0$. Therefore, we have

$\begin{array}{cl}
& \dfrac{b + 2c}{a + 2b} \\
= & \dfrac{14k + 2(9k)}{12k + 2(14k)} \\
= & \dfrac{32k}{40k} \\
= & \dfrac{4}{5}
\end{array}$