The term in x in the first expansion is \[\binom31(4x)^2\frac{10}{x}=480x,\qquad p=480.\] In the second expansion, choosing j reciprocal terms gives power \(x^{5-3j}\). For \(x^{-1}\), \(5-3j=-1\), so \(j=2\), giving \[\binom52(2x)^3\left(\frac{k}{x^2}\right)^2=\frac{80k^2}{x},\qquad q=80k^2.\] Hence \[480=6(80k^2),\qquad\boxed{k=\pm1}.\]