Step 1: Set up coordinates for P and Q.
Let $P = (x_1, y_1)$ lie on $3x + 4y - 4 = 0$ and $Q = (x_2, y_2)$ lie on $5x - y - 4 = 0$. Since the midpoint of $PQ$ is $(1, 5)$, we have $x_1 + x_2 = 2$ and $y_1 + y_2 = 10$.
Step 2: Write the line conditions.
$P$ on $3x + 4y = 4$ gives $3x_1 + 4y_1 = 4$. $Q$ on $5x - y = 4$ gives $5x_2 - y_2 = 4$.
Step 3: Substitute midpoint relations.
Using $x_2 = 2 - x_1$ and $y_2 = 10 - y_1$ in the second equation: \[5(2-x_1)-(10-y_1)=4 \implies y_1-5x_1=4 \implies y_1=4+5x_1.\]
Step 4: Solve for $x_1$ and $y_1$.
Substituting into $3x_1+4y_1=4$: $3x_1+4(4+5x_1)=4 \implies 23x_1=-12 \implies x_1=-12/23$ and $y_1=32/23$. So $P=(-12/23,\, 32/23)$.
Step 5: Find coordinates of Q.
$x_2=2+12/23=58/23$ and $y_2=10-32/23=198/23$. So $Q=(58/23,\, 198/23)$.
Step 6: Compute the slope of PQ.
\[\text{slope} = \frac{198/23-32/23}{58/23-(-12/23)} = \frac{166}{70} = \frac{83}{35}.\] \[ \boxed{\dfrac{83}{35}} \]