Two point charges +4q and -3q are placed at points A and B...

The electric potential at the origin $O$ is the sum of the potentials due to the individual point charges.

Using the formula $V = \dfrac{KQ}{r}$, the total potential at $O$ is:
$V_O = V_A + V_B$

Substituting the given values with distances converted to meters ($3\text{ cm} = 3 \times 10^{-2}\text{ m}$ and $4\text{ cm} = 4 \times 10^{-2}\text{ m}$):
$V_O = \dfrac{K(4q)}{3 \times 10^{-2}} + \dfrac{K(-3q)}{4 \times 10^{-2}}$

Factoring out the common terms:
$V_O = Kq \times 10^2 \left( \dfrac{4}{3} - \dfrac{3}{4} \right)$

$V_O = Kq \times 10^2 \left( \dfrac{16 - 9}{12} \right)$

$V_O = \left( \dfrac{7}{12} \right) qK \times 10^2\text{ V}$

Thus, the numerical value of the electric potential at the origin, when expressed in multiples of $10^2\text{ V}$, is \left( \dfrac{7}{12} \right) qK$.<br><br>Answer: \left(\dfrac{7}{12}\right)qK$