Is the number x positive?

[himu]

Is the number x positive?

(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.

[GMATGuruNY]

Is the number x positive?

(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.

|x| = the distance between 0 and x.
|x-y| = the distance between 0 and x-y.
|x+y| = the distance between 0 and x+y.

Statement 1: On the number line, 0 is closer to x - 1 than to x.
|x-1| < |x|.
Since there is absolute value notation on each side, we can square the inequality.
(x-1)² < x²
x² - 2x + 1 < x²
-2x < -1
x > 1/2.
Thus, x must be positive.
SUFFICIENT.

Statement 2: On the number line, 0 is closer to x than to x + 1.
|x| < |x+1|.
Since there is absolute value notation on each side, we can square the inequality.
x² < (x+1)²
x² < x² + 2x + 1
-2x < 1
x > -1/2.
Thus, x could be negative or positive.
INSUFFICIENT.

The correct answer is A.

[Brent@GMATPrepNow]

Is the number x positive?

(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.

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

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.


Target question: Is x positive?

Statement 1: On the number line, 0 is closer to x – 1 than to x.
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: A

Cheers,
Brent