According to the positional values of the alphabets, 1) ACE9 → A = 1; C = 3; E = 5 → A + C + E = 1 + 3 + 5 = 9 2) GIK27 → G = 7; I = 9; K = 11 → G + I + K = 7 + 9 + 11 = 27 3) QSU50 → Q = 17; S = 19; U = 21 → Q + S + U = 17 + 19 + 21 = 57 but not 50. 4) VXY71 → V = 22; X = 24; Y = 25 → V + X + Y = 22 + 24 + 25 = 71 Hence, the odd one among the given options is QSU50.