If x + y + z > 0, is z > 1?

[BTGModeratorVI]

If x + y + z > 0, is z > 1?

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

Answer: B
Source: Official Guide

[deloitte247]

Target question: is Z >1?
Statement 1: 2 > x+y+1
x+y+z > 0 --- eqn (1)
z > x+y+1 ------ eqn (2)
Add eqn (1) + eqn (2) because they have the same sign
(x+y+z) + z > x+y+1 + 0
x+y+2z > x+y+1
Subtract (x+y) from both sides
x + y - (x+y) + 2z > x+y - (x+y) +1
2z > 1
$$z>\frac{1}{2}$$
z can be any number between 1/2 and infinity.
So, the target question cannot be answered with certainty, hence, statement 1 is NOT SUFFICIENT.

Statement 2: x+y+1<0
x+y+z>0 --- eqn (1)
x+y+1<0 --- eqn (2)
Subtract eqn (1) from eqn (2) because they have
(x+y+z) - (x+y+1) > 0 - 0
z - 1>0
z>1
Therefore, the target question can be answered with certainty here; hence, statement 2 is SUFFICIENT. Thus, the correct answer is option B.

[Brent@GMATPrepNow]

If x + y + z > 0, is z > 1?

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

Answer: B
Source: Official Guide

Target question: Is z > 1

Given: x + y + z > 0

Statement 1: z > x + y +1
Let's create a similar inequality to x + y + z > 0
Take z > x + y +1 and subtract x and y from both sides to get: z - x - y > 1
We now have two inequalities with the inequality signs facing the same direction.
z - x - y > 1
x + y + z > 0
ADD them to get: 2z > 1
Divide both sides by 2 to get: z > 1/2
So, z COULD equal 2, in which case z > 1
Or z COULD equal 3/4, 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
Let's use the same strategy.
This time, let's multiply both sides by -1 to get: -x - y - 1 > 0
We now have two inequalities with the inequality signs facing the same direction.
-x - y - 1 > 0
x + y + z > 0
ADD them to get: z - 1 > 0
Add 1 to both sides to get z > 1
Perfect!!!
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: B

Cheers,
Brent