Neutralization of alkali solution by a mix of strong acids
I think you over-complicated things a bit by focusing on the formation of sodium salts, when in reality all you need to take into account is the net neutralization reaction:

$$\ce{H3O+ + OH- -&gt; 2 H2O}$$

From the equivalence point condition of any neutralization reaction one can find volume of the base $V_\mathrm{b}$ expressed via the volume of acid(s) $V_\mathrm{a}$:

$$V_\mathrm{b} = \frac{C(\ce{H3O+})}{C(\ce{OH-})} \cdot V_\mathrm{a} \tag{1}$$

Since there are two moderately diluted strong acids and a base, one can conclude that the dissociation of each acid and a base is independent and complete:

\begin{align}
\ce{HCl + H2O &amp;&lt;=&gt;&gt; H3O+ + Cl-} \\
\ce{H2SO4 + 2H2O &amp;&lt;=&gt;&gt; 2H3O+ + SO4^2-} \\
&amp; \\
\ce{NaOH &amp;&lt;=&gt;&gt; Na+ + OH-}
\end{align}

Note that concentration of protons is doubled in case of sulfuric acid. Total concentration of protons in the solution is 

$$C(\ce{H3O+}) = \frac{n(\ce{H3O+})}{V(\ce{H3O+})} = \frac{n(\ce{HCl}) + 2 \cdot n(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \tag{2}$$

Taking into account that $n = CV$ and that two identical volumes of acid were mixed ($V(\ce{HCl}) = V(\ce{H2SO4}) = V(\text{acid})$) (2) can be transferred into:

$$
\require{cancel}
\begin{align}
C(\ce{H3O+}) &amp;= \frac{C(\ce{HCl}) \cdot V(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}) \cdot V(\ce{H2SO4})}{V(\ce{HCl}) + V(\ce{H2SO4})} \\
&amp;= \frac{\cancel{V(\text{acid})} \cdot (C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4}))}{2 \cdot \cancel{V(\text{acid})}} \\
&amp;= \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2} \tag{3}
\end{align}
$$

Now let's plug (3) in (1) and finally find the volume of alkali:

$$V_\mathrm{b} = \frac{C(\ce{HCl}) + 2 \cdot C(\ce{H2SO4})}{2 \cdot C(\ce{OH-})}  \cdot V_\mathrm{a} = \frac{\pu{0.1 M} + 2 \cdot \pu{0.2 M}}{2 \cdot \pu{0.1 M}} \cdot \pu{20 mL} = \pu{50 mL}$$