Given, inequality is
$$
\begin{aligned}
& \frac{6 x}{4 x-1} < \frac{1}{2} \Rightarrow \frac{12 x}{4 x-1} < 1 \\
\Rightarrow \quad & \frac{12 x}{4 x-1}-1 < 0 \Rightarrow \frac{8 x+1}{4 x-1} < 0 \quad \text{...(i)}
\end{aligned}
$$
For above inequality Eq. (i), we have two cases either
Case II $8 x+1>0,4 x-1 < 0$
Case II $8 x+1 < 0,4 x-1>0$
From Case $(\mathrm{I})$, we get $x>\frac{-1}{8}, x < \frac{1}{4}$
$\therefore x \in\left(-\frac{1}{8}, \frac{1}{4}\right) \quad \text{...(ii)}$
From Case $(\mathrm{II}), x < \frac{-1}{8}, x>\frac{1}{4}$, both of which can not be true.
$\therefore E_{\text {G. }}$. (ii) represent the solution set of given inequality.