This question was reported to be poorly attempted by majority of the candidates.  They were reported not to have applied the principle of mean and standard deviation correctly.  Candidates were expected to show that if 6 is the mean of 2, 5, 6, 8, x and y then  = 6.  This implied that                      + y = 15 ... (eqn.1).  Also, since   was the standard deviation, then   =   i.e.  = 5.  Simplifying gave                (= 6)2 + (y – 6)2 = 9.....(eqn 2).  From (eqn.1), y = 15 - . Therefore eqn. 2 became                                  ( – 6)2 + (15 –  – 6)2 = ( -6)2 + (9 – )2 = 9.  Expanding this equation and bringing like terms together gave ² – 15 + 54 = 0.  Solving this quadratic equation gave  = 6 or 9.  Substituting these values into eqn. 1 gave y = 9 or 6.  This implied that the two unknown numbers were 9 and 6 i.e when  = 6,  = 9 or when  = 9, y = 6.