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# If a + b + c < 0, is c < 1 ?

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If a + b + c < 0, is c < 1 ?  [#permalink]

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05 Jun 2016, 14:14
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53% (01:52) correct 47% (01:53) wrong based on 128 sessions

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If a + b + c < 0, is c < 1 ?

(1) c < a + b – 1

(2) a + b − 1 > 0

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Md. Abdur Rakib

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Re: If a + b + c < 0, is c < 1 ?  [#permalink]

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05 Jun 2016, 14:23
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If a + b + c < 0, is c < 1 ?

(1) c < a + b – 1. We can sum the inequalities with the signs in the same direction: (a + b + c) + c < 0 + a + b – 1 --> c < -1/2. Sufficient.

(2) a + b − 1 > 0. We can subtract inequalities with the signs in the opposite direction: (a + b + c) - (a + b − 1 )< 0 --> c < -1. Sufficient.

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Re: If a + b + c < 0, is c < 1 ?  [#permalink]

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22 Oct 2018, 14:21
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Top Contributor
AbdurRakib wrote:
If a + b + c < 0, is c < 1 ?

(1) c < a + b – 1

(2) a + b − 1 > 0

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

Cheers,
Brent

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Re: If a + b + c < 0, is c < 1 ?   [#permalink] 22 Oct 2018, 14:21
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