ktrout2020 wrote:If a + b + c < 0, is c < 1 ?

(1) c < a + b - 1
(2) a + b − 1 > 0

Source: Kaplan

Given: a + b + c < 0

Target question: Is c < 1?

Statement 1: c < a + b - 1
Take: a + b + c < 0
Subtract a and subtract b from both sides to get: c < -a - b
Now add this inequality to the statement 1 inequality: c < a + b - 1
We get: 2x < -1
Divide both sides by 2 to get: c < -1/2
If c < -1/2, then c is definitely less than 1
So, the answer to the target question is YES, x IS less than 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: a + b − 1 > 0
Take: a + b − 1 > 0
Rewrite as: 0 < a + b - 1
Add c to both sides to get: c < a + b + c - 1
Add 1 to both sides to get: c + 1 < a + b + c
We already know that a + b + c < 0
So, we can add this info to our inequality to get: c + 1 < a + b + c < 0
This tells us that c + 1 < 0
Subtract 1 from both sides to get: c < -1
If c < -1, then c is definitely less than 1
So, the answer to the target question is YES, x IS less than 1
Since we can answer the target question with certainty, statement 2 is SUFFICIENT

Answer: D

Cheers,
Brent