RisingForceX wrote:
If x + y + z > 0, is z > 1?
(1) z > x + y +1
(2) x + y + 1 < 0
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 > 0Take 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 > 0ADD them to get: 2z > 1
Divide both sides by 2 to get: z > 1/2
So, z COULD equal 2, in which case
z > 1Or z COULD equal 3/4, in which case
z < 1Since 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 > 0ADD them to get: z - 1 > 0
Add 1 to both sides to get
z > 1Perfect!!!
Since we can answer the
target question with certainty, statement 2 is SUFFICIENT
Answer: B
Cheers,
Brent
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