let FeO = x gm and FeO+Fe₂O₃= y gm so FeO = [72/(72+160)]*y and Fe₂O₃ = [160/(160+72)]*y

2FeO + 1/2O_{2  ---------->}  Fe₂O₃

so amount of Fe₂O₃ formed by x gm FeO  = 160x/144

similarly amount of Fe₂O₃ formed by  [72/(72+160)]*y gm FeO = 160 [72/(72+160)]*y/144

so total Fe₂O₃ = 160x/144 + 160 [72/(72+160)]*y/144 +  [160/(160+72)]*y = 160x/144 + [160*72/232]y/144 + 160y/232

now according to the given condition {[160x/144 + [160*72/232]y/144 + 160y/232}/{x+y) = 105/100

now solve