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
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Is x > 1 ? GMAT Prep Pack 1 QDS08380
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jainpiyushjain
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Statement 1: (x + 1)(|x| - 1) > 0jainpiyushjain 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
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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As a tutor, I don't simply teach you how I would approach problems.
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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.
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.
