$\mathrm{x} m \mathrm{mol}$ of ${\mathrm{KIO}}_{3 }$reacts completely with$\mathrm{y} \mathrm{m} \mathrm{mol}$ of KI to give ${\mathrm{I}}_2$ quantitively. If $\mathrm{z} \mathrm{m} \mathrm{mol}$of hypo are  required for complete titration against this ${\mathrm{I}}_2$, then, which statement is incorrect?

The reaction is,

${\mathrm{KIO}}_3+5\mathrm{KI}⟶3{\mathrm{I}}_2$

$\mathrm{x} \mathrm{y} \frac{Z}{2}=3x=\frac{3y}{5}$

n factor for  ${\mathrm{KIO}}_3\to 5$

n factor for $\mathrm{KI}\to 1$

n factor for $12\to 5/3$

Here, $5\mathrm{x}=\mathrm{y}$

Also, 

${\mathrm{I}}_2+2{\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3⟶{\mathrm{Na}}_2{\mathrm{S}}_4{\mathrm{O}}_6+2\mathrm{NaI}$

$Z\frac{Z}{2}$

With these values, the relation given in option D becomes incorrect.