rakeshd347 wrote:If x+y+z > 0, is z > 1 ?

(1) z > x + y + 1
(2) x + y + 1 < 0

IMPORTANT: If two inequalities have inequality signs facing the same direction, we can ADD those inequalities.
For example, if A < B and C < D, then we can add them to get A+C < B+D

Target question: Is z > 1 ?

Given: x + y + z > 0

Statement 1: z > x + y + 1
We already have x + y + z > 0
Since the two inequality signs are facing the same direction, we can ADD the inequalities to get:
x + y + 2z > x + y + 1
Subtract x and y from both sides: 2z > 1
Divide both sides by 2: z > 1/2
This does not provide enough information to determine whether or not z > 1
So, statement 1 is NOT SUFFICIENT

Aside: we can show that statement 1 is NOT SUFFICIENT with the these two conflicting cases that both satisfy statement 1:
Case a: x = -0.1, y = -0.1, z = 0.9, in which case z < 1
Case b: x = 1, y = 1, z = 10, in which case z > 1
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: x + y + 1 < 0
Rewrite this as 0 > x + y + 1
We already have x + y + z > 0
Since the two inequality signs are facing the same direction, we can ADD the inequalities to get:
x + y + z > x + y + 1
Subtract x and y from both sides: z > 1
PERFECT!
It must be the true that z > 1
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer = B

Cheers,
Brent