20ml of 0.1 Msolution of compound Na2CO3 is tirated against 0.05M HCl , x ml of HCl is used when phenophthalein is a used as an indicator and y ml of HCl is used when methyl orange is the indicator in two separate titration . hence ( y- x) is

Titration of Na2CO3 using phenolphthalein as indicator involves the reaction: Na2CO3 + HCl = NaHCO3 + NaCl

so 1 Mole Na2CO3 consumes 1 Mole HCl 

or 20 ml 0.1M Na2CO3 soln consumes 40 ml 0.05 M HCl soln. so x = 40.

Titration of Na2CO3 using methyl orange as indicator involves the reactions: Na2CO3 + 2HCl = H2CO3 + 2NaCl

and also   NaHCO3 + HCl = H2CO3 + NaCl

so 1 Mole Na2CO3 consumes 2 Moles of HCl 

and 1 Mole NaHCO3 consumes 1 Mole of HCl 

or 20 ml 0.1M Na2CO3 soln consumes 40x2 ml 0.05 M HCl soln. and 20 ml 0.1M NaHCO3 soln consumes 40 ml 0.05 M HCl soln. (as the compound contains both Na2CO3 and NaHCO3 in equimolar proportion)

Hence y = 40x2+40 = 120

so y - x = 120 - 40 = 80.

Regards