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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