I went with A.
Based on statement 1, I let x=1988, y=1989 and z=1990. (x+y+z)/3=a and x=2a.
Substituted 2a for x (2a+y+z)=3a
Solved for a, a=y+z
Since, x=2a, x=2y+2z, given sales are positive, x, which is 1988’s sales must be larger than 1989 or 1990 for that matter. Sufficient.
Based on statement 1, I let x=1988, y=1989 and z=1990. (x+y+z)/3=a and x=2a.
Substituted 2a for x (2a+y+z)=3a
Solved for a, a=y+z
Since, x=2a, x=2y+2z, given sales are positive, x, which is 1988’s sales must be larger than 1989 or 1990 for that matter. Sufficient.









