in general, there are 3P3 ways possible to arrange a,b,c
st(1) a*c=b implies product of any two numbers, Not Sufficient
a=1/2,b=3,c=1.5 -> 1/2*3=1.5 <> a,c,b
a=2,b=3,c=6 -> 2*3=6 <> a,b,c
st(2) a+c=b implies addition of any two numbers, Not Sufficient
a=-3,b=1,c=2 -> a,b,c
a=-3,b=-5,c=-2 -> b,a,c
Combined st(1&2): b=ac, a+c=ac, ac-a=c, a(c-1)=c
it's evident that c=1 turns this expression into 0, hence c cannot be 1
a=!b implies a<b or a>b
different cases
0<c<1 --> c=1/2, a=-1, b=-1/2 AND a,b,c (Here also possible to test c<1 including c<0 values which are set aside for sol.timeliness)
c>1 --> c=3, a=1.5, b=4.5 AND a,c,b
enough and stop here, as two medians are possible b or c
Not Sufficient
e
kakz wrote:Is b the median of the three numbers a,b,c?
(1)a*c=b
(2)a+c=b
