Answer: (C) 3
Given $2\tan A = 3 \Rightarrow \tan A = \dfrac{3}{2}$. Dividing numerator and denominator by $\cos A$:
$$\frac{4\sin A + 3\cos A}{4\sin A - 3\cos A} = \frac{4\tan A + 3}{4\tan A - 3} = \frac{4\times\frac{3}{2}+3}{4\times\frac{3}{2}-3} = \frac{6+3}{6-3} = \frac{9}{3} = 3$$
The key trick is to divide both numerator and denominator by $\cos A$, converting everything to $\tan A$. Since $2\tan A = 3$, substitute $\tan A = 3/2$ directly. This avoids finding individual sin and cos values. Examiners expect this manipulation to be shown clearly in one or two steps.