By definition y vol H_2O_2 means that sample solution of H_2O_2 from each ml of which yml O_2 at STP is produced on thermal decomposition.
So y ml H_2O_2 solution will produce y^2 ml O_2 at STP.
Now the balanced equation of thermal decomposition reaction of H_2O_2 is as follows
2H_2O_2" "" "->" "" "O_2+2H_2O
=2" "mol" "" "=22400mL" ""at STP"
This equation reveals that 22400mL O_2 at STP is produced from 2mol H_2O_2
So y^2 mL O_2 will be produced from (2y^2)/22400 mole H_2O_2 and this amount of H_2O_2 is present in ymL solution.
Again x mL y(M) KMnO_4 solution will contain (xyxx10^-3) mole KMnO_4
Now the balanced redox reaction is
5H_2O_2+2KMnO_4+3H_2SO_4->K_2SO_4+8H_2O+4O_2 comparing the mole ratio of reactants calculated in the reacting solutions with that of balanced redox reaction we can write
(xyxx10^-3)/((2y^2)/22400)=2/5
=>x/y=0.4/11.2~~0.036
