Statement 1 : M-3Z>0
Case 1
Let M = 5 and Z = -6
M-3Z -> 5-3(-6) = 23 >0
M+Z -> 5-6 = -1 <0
Case 2
Let M = 5 and Z = -3
M-3Z -> 5-3(-3) = 14 >0
M+Z -> 5-3 = 2 >0
Since both > and < are possible, statement 1 is insufficient
Statement 2 : 4Z-M>0
Case 1
Let M = -5 and Z = 2
4Z-M -> 4x2 -(-5) = 13 >0
M+Z -> -5+2 = -3 <0
Case 2
Let M = -5 and Z = 6
4Z-M -> 6x2 -(-5) = 17 >0
M+Z -> -5+6 = 1>0
Since both > and < are possible, statement 2 is insufficient
Taking Statement 1&2 together
If we were to plug numbers here, we would have to come up with numbers that could fulfill both the equations 1 & 2. In all such cases we would find that M & Z are positive and therefore M+Z will be >0
Examples
M = 5, Z = 1.5
M = 10, Z = 3
M = 15, Z = 4
M = 15, Z = 4.5
Therefore Statement 1 & 2 together are sufficient.
A better way to solve this would be simplifying equation and deriving the properties
From 1 -> M>3Z
From 2 -> 4Z>M
Therefore 4Z>3Z; now this can only be possible when z is positive
In addition when z is positive M has to be positive (using statement 1 M>3Z)
Therefor 1&2 together are sufficient
C
