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Is x > 1 ? GMAT Prep Pack 1 QDS08380

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Source: — Data Sufficiency |
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by GMATGuruNY » Sat Jul 06, 2013 2:52 am
jainpiyushjain wrote:I am having difficulty understanding the official explanation, can you please help.

Is x > 1 ?

(1) (x + 1)(|x| - 1) > 0
(2) |x| < 5

OA A

Thank you
Statement 1: (x + 1)(|x| - 1) > 0
The CRITICAL POINTS are -1 and 1.
These are the only values where (x + 1)(|x| - 1) = 0.
To determine the ranges where (x + 1)(|x| - 1) > 0, test one value to the left and right of each critical point.

x < -1:
Plug x = -2 into (x + 1)(|x| - 1) > 0
(-2 + 1)(|-2| - 1) > 0
-1 * 1 > 0.
- 1 > 0.
Doesn't work.
x < -1 is NOT a valid range.

-1 < x < 1:
Plug x = 0 into (x + 1)(|x| - 1) > 0
(0 + 1)(|0| - 1) > 0
1 * -1 > 0.
- 1 > 0.
Doesn't work.
-1<x<1 is NOT a valid range.

x > 1:
Plug x = 2 into (x + 1)(|x| - 1) > 0:
(2 + 1)(|2| - 1) > 0
3 * 1 > 0.
3 > 0.
This works.
x > 1 is a valid range.

Since x>1 is the only valid range, SUFFICIENT.

Statement 2: |x| < 5
If x=4, then x>1.
If x=-4, then x<1.
INSUFFICIENT.

The correct answer is A.
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by Matt@VeritasPrep » Sat Jul 13, 2013 10:09 am
Let me add a mathematical approach too:

If (x+1) * (|x| - 1) > 0, then either both (x+1) and (|x| - 1) are positive or both (x+1) and (|x| - 1) are negative.

Let's examine the first case: both are positive.
Since (x+1) > 0, we know x > -1.
Since |x| - 1 > 0, we know |x| > 1.
When |x| > 1, either x > 1 or x < -1. But we know x > -1, so x < -1 is impossible.
Hence, x > 1.

Now the second case: both are negative.
Since (x+1) < 0, we know x < -1.
Since |x| - 1 < 0, we know |x| < 1.
When |x| < 1, we know -1 < x < 1.
But we can't have both x < -1 and -1 < x < 1.
Hence the second case is impossible: we can't have both (x+1) and (|x| - 1) negative.

Hence both (x+1) and (|x| - 1) are positive, and x must be greater than 1.
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