Energy per atom in a parallel radioactivity equation
You may state that, in $60$ s, one A atom has liberated $20$ MeV by the first reaction. In the same time, three A atoms have liberated $3·40$ MeV by the $2$nd reaction. The total energy liberated in $60$ seconds, is $120 + 20 = 140 $ MeV.- As a consequence, the average energy liberated by one A atom is $140$ Mev$  /4 = 35$ MeV