In acute angled $∆$PQR if 3 sinP + 4 cosQ = 6 and 4 sinQ + 3 cosP = 1,then the angle R equal to 

3 sinP + 4 cosQ = 6 ..............(i)
4 sinQ + 3 cosP = 1 .............(ii)
Squaring and adding (i) and (ii)
We get $\sin (\mathrm{P}+\mathrm{Q})=\frac{1}{2}\Rightarrow \mathrm{P}+\mathrm{Q}=\frac{\mathrm{\pi }}{6} \mathrm{or} \frac{5\mathrm{\pi }}{6}$
$\Rightarrow 3\sin \mathrm{P}+4\cos \mathrm{Q}<\frac{11}{2}$, which is not possible;
So, $\mathrm{R}=\frac{\mathrm{\pi }}{6}$