Ans should be A.

1. x=2, x-1 =1 Yes
x=-1, x-1 =-2 Yes

2. x=-3,x+1 =-2 No
x=-1/3, x+1=2/3 Yes

Hence A is sufficient.
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|x| denotes the distance between 0 and x on the number line.

F.S 1 states that |0-(x-1)|<|0-x| --> |x-1|<|x| --> Square on both sides, and subtract -->\(1*(2x-1)>0 --> x>\frac{1}{2}\). Sufficient.

F.S 2 states that |x|<|x+1| --> Just as above, this gives --> 1*(2x+1)>0 --> \(x>\frac{-1}{2}\). Insufficient.

A.
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All that is equal and not-Deep Dive In-equality

Hit and Trial for Integral Solutions

Bumping up for any other explanation. This is how I approached it -

(1) On the number line, 0 is closer to x – 1 than to x.
(2) On the number line, 0 is closer to x than to x + 1

(1) assume x= -2; => x-1 = -3. (Given answer choice is not true/satisfied, so x can't be negative)
Consider x= 22; => x-1 =21 (Given answer choice is true/satisfied)

So, it is true that x is positive.

(2) same assumption, x=-32 => x+1 = -31(Given answer choice is not true/satisfied, so x can't be negative)
Consider x=1 => x+1 = 2 (Given answer choice is true/satisfied)

So, it is true that x is positive.

I am not sure what you meant by -1/2 (not inclusive) and 0 (inclusive) and I'd appreciate if you can shed some light on it. But yeah, I messed up not considering fractions. So ...

For option 2 - if x were -1/2 then x+1 would be 1/2. Given statement of 0 closer to x than x+1 is busted with two answers (0 is as much closer to -1/2 as it is to 1/2). Hence option 2 has multiple answers.

It goes to show again how much I suck at picking magic numbers (or numbers at all). I have to keep reminding myself to test for fractions too. Ugh. Is there any other general approach to this problem besides picking numbers?

1. A is positive. otherwise, if x is negative, then negative minus negative makes an even bigger negative, and the new number would be even farther from 0.
2. it can be x=1 and x+1 =2, and 0 is closer to x, or it can be:
x=-0.1 and x+1=0.9. x is closer to x than to x+1. since we have 2 outcomes, B is not sufficient.

A.

Another approach is the sketch the cases on a number line.

Target question: Is x positive?

Statement 1: On the number line, 0 is closer to x – 1 than to x.
First, recognize that x-1 will always be to the left of x.
Second, recognize that there are 3 possible ways to place x-1 and x with relation to zero.

If zero is closer to x-1 than to x, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.
If case #1 is true, we can see that x must be positive
If case #3 is true, we can see that x must be positive
Since both possible cases yield the same answer to the target question, we can answer the target question with certainty.
So, statement 1 is SUFFICIENT

Statement 2: On the number line, 0 is closer to x than to x + 1.
Recognize that x+1 will always be to the right of x.
Also recognize that there are 3 possible ways to place x and x+1 with relation to zero.

If zero is closer to x than to x+1, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios.
If case #1 is true, we can see that x is negative
If case #3 is true, we can see that x is positive
Since the two possible cases yield different answers to the target question, we cannot answer the target question with certainty.
So, statement 2 is NOT SUFFICIENT

Answer =
Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com