Step 1: Recall the formula for equivalent weight of a redox agent.
For any oxidising or reducing agent, the equivalent weight is \[ \text{Equivalent Weight} = \frac{\text{Molecular Weight}}{n\text{-factor}} \] where the $n$-factor is the total number of electrons gained (for an oxidising agent) or lost (for a reducing agent) per formula unit.
Step 2: Find the n-factor of KMnO4 in acidic medium.
In acidic medium, the permanganate ion $MnO_4^-$ is reduced to $Mn^{2+}$: \[ MnO_4^- + 8H^+ + 5e^- \rightarrow Mn^{2+} + 4H_2O \] The oxidation state of Mn changes from $+7$ to $+2$, a change of 5. So each formula unit of $KMnO_4$ gains 5 electrons. The $n$-factor of $KMnO_4$ in acidic medium is 5.
Step 3: Calculate the equivalent weight of KMnO4.
\[ \text{Equivalent weight of } KMnO_4 = \frac{M_A}{5} \]
Step 4: Find the n-factor of K2Cr2O7 in acidic medium.
In acidic medium, the dichromate ion $Cr_2O_7^{2-}$ is reduced: \[ Cr_2O_7^{2-} + 14H^+ + 6e^- \rightarrow 2Cr^{3+} + 7H_2O \] The oxidation state of each Cr changes from $+6$ to $+3$, a gain of 3 electrons per Cr atom. Since there are 2 Cr atoms per formula unit, the total electrons gained = $2 \times 3 = 6$. The $n$-factor of $K_2Cr_2O_7$ in acidic medium is 6.
Step 5: Calculate the equivalent weight of K2Cr2O7.
\[ \text{Equivalent weight of } K_2Cr_2O_7 = \frac{M_B}{6} \]
Step 6: State the final answer.
The equivalent weights of $KMnO_4$ and $K_2Cr_2O_7$ in acidic medium are $\frac{M_A}{5}$ and $\frac{M_B}{6}$ respectively. \[ \boxed{\frac{M_A}{5},\ \frac{M_B}{6}} \]