Sum of the oxidation states of  “Cr” in  $K_{3}Cr{O}_8$ and  $Cr{O}_5$ is A Than $(A-5)$ is

In  $K_{3}Cr{O}_{8}Cr\text{\hspace{0.17em}\hspace{0.17em}}oxdationState=\pm 5$
In    $Cr{O}_{5}Cr\text{\hspace{0.17em}\hspace{0.17em}}oxdationState=+6$
                                $6+5=11-5=6$