Finding out the value of rate constant on titration
From a mole balance, we find $C = n_x(t) + n_y(t) + n_z(t) = n_x(t) + 2n_y(t)$, where the second equality follows from the stoichiometry of the reaction. In particular, our boundary conditions are $C = n_x(0) = 2n_y(\infty).$ 

We assume that the reaction is first-order, so $n_x(t) = n_x(0)\exp(-kt) = C\exp(-kt).$ The two measurements give us the set of equations 
\begin{align*} 
80D &amp;= n_x(5) + 5n_y(5) &amp;&amp;= C\exp(-5k) + \frac{5}{2}C(1-\exp(-5k)) \\
100D &amp;= 5n_y(\infty) &amp;&amp;= \frac{5}{2}C,
\end{align*}
where $D$ is an undetermined proportionality constant for the titration, and the coefficient of 5 in $n_y$ accounts for the fact that 2 equivalents of $A$ react with $n_y$ and 3 equivalents with $n_z$. Dividing one equation by the other produces the final result $$\frac{4}{5} = \frac{2}{5}\exp(-5k)+1-\exp(-5k),$$ which can be solved for $k$ and hence $t_{1/2}$.