The question is really asking if x, y and z are evenly spaced.
statement 1: The mean of the set {x, y, z, 4} is greater than the mean of the set {x, y, z}.
(x+y+z+4)/4 > (x+y+z)/3
3x + 3y + 3z + 12 > 4x + 4y + 4z
x + y + z < 12
if x=1, y=2, z=3, then x, y and z are evenly spaced. YES.
if x=1, y=2, z=4, then x, y and z are not evenly spaced. NO.
statement 2: The median of the set {x, y, z, 4} is less than the median of the set {x, y, z}.
let's say the median of the set {x, y, z, 4} = M.
the median of set {x, y, z} must be y, since x<y<z.
we don't know where 4 is positioned relative to x, y and z. we could have either one of these 3 cases:
case 1: x <= 4 <= z
in this case, M = (4+y)/2, and from statement 2 we have:
(4+y)/2 < y
4+y < 2y
y > 4
we still have insufficient info to answer the question stem. insufficient.
case 2: 4 < x
in this case, M = (x+y)/2, and from statement 2 we have:
(x+y)/2 < y
x + y < 2y
y > x
we know already know this from the question stem. insufficient.
case 3: 4 > z
in this case, M = (y+z)/2, and from statement 2 we have:
(y+z)/2 < y
z + y < 2y
z < y
this case is not possible since we know that z must be greater than y.
hence, statement 2 on its own is insufficient.
both statements together:
x + y + z < 12
4 < y (from case 1)
since y>4, the the minimum possible value for y is 5. if y = 5, then z must be at least 6. so now we have y + z = 11, which means x must 0 or less (in order to satisfy statement 1). clearly, x, y and z are not evenly spaced. sufficient.
choose C.
I hope I didnt miss anything.
-BM-
