$A$ slab of material of dielectric constant $K$ has the same area as the plates of a parallel-plate capacitor but has a thickness $(3/4)d$,where $d$ is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates?

(N/A) Let $E_{0} = V_{0} / d$ be the electric field between the plates when there is no dielectric and the potential difference is $V_{0}$.
If the dielectric is now inserted,the electric field in the dielectric will be $E = E_{0} / K$.
The potential difference $V$ will then be:
$V = E_{0} \left(\frac{1}{4} d\right) + \frac{E_{0}}{K} \left(\frac{3}{4} d\right)$
$V = E_{0} d \left(\frac{1}{4} + \frac{3}{4K}\right) = V_{0} \left(\frac{K+3}{4K}\right)$
The potential difference decreases by the factor $(K+3) / 4K$ while the free charge $Q_{0}$ on the plates remains unchanged.
The capacitance $C$ thus increases as:
$C = \frac{Q_{0}}{V} = \frac{Q_{0}}{V_{0} \left(\frac{K+3}{4K}\right)} = \left(\frac{4K}{K+3}\right) C_{0}$