(x+1)(|x|-1)>0

[Needgmat]

Is x>1

1) (x+1)(|x|-1)>0

2) |x|<5

OAA

Please explain.

Many thanks in advance

[GMATGuruNY]

Is x > 1 ?

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

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

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

x > 1:
If we plug x = 2 into (x + 1)(|x| - 1) > 0, we get:
(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.

[Jay@ManhattanReview]

Is x>1

1) (x+1)(|x|-1)>0

2) |x|<5

OAA

Please explain.

Many thanks in advance

S1: (x+1)(|x|-1)>0 => EITHER both (x+1) and (|x|-1) are positive OR both (x+1) and (|x|-1) are negative.

Taking them positive:

=> (x+1) > 0 => x > -1 AND |x|-1 > 0 => |x| > 1 => x < -1 OR x > 1. Taking both ranges x > -1 AND x < -1 OR x > 1, we can conclude that x > 1.

Taking them negative:

=> (x+1) < 0 => x < -1 AND |x|-1 < 0 => |x| < 1 => -1 < x < 1. But this not a possible scenerio; x cannot the less than -1 and greater than -1 at the same time.

So, we conclude that x > 1. Sufficient!

S2: |x|<5 => -5 < x < 5. x can be negative, 0, or positive. Insufficient!

OA: A

-Jay

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[wisdomkindred]

Is x>1

1) (x+1)(|x|-1)>0

2) |x|<5

OAA

Tips: The key to unlock confusion with "and" and "or" in inequality:

- Drawing graphs and crossing off all irrelevant values.
- "and" means must; "or" means could

(1): There are 2 cases: case 1 or case 2

Case 1: x+1 > 0 AND |x|-1 >0
X+1 >0 --> x > -1 (A)
|x|-1 >0 --> |x| >1 --> x>1 OR x <-1 (B)

--> graph for A: //////////(-1)-------------(1)--------
--> graph for B: --------(-1)//////////////(1)--------

Note about the graphs:
////////////means the values crossed off
----------------- means the values are accepted

Combining the two graphs above: this is the AND case, so we must cross off all values that don't fit in any graph, represented by "//////////"
So the value of the 2 inequalities of case 1 is: x>1.

Case 2: x+1 <0 AND |x|-1 <0

X+1 < 0 --> x <-1 (C)
|x|-1 <0 --> |x| < 1 --> -1<x<1 (D)

--> graph for C: ---------(-1)/////////(1)///////
--> graph for D: /////////(-1)--------(1)////////

As we see, there is no value of x that satisfy the both graphs. That means there is no solution for the inequalities in case 2.

--> So only case 1 is appropriate to consider. That means (x+1) (|x| - 1) > 0 when x>1. Sufficient.

(2) |x| < 5 --> -5<x<5 --> graph: //////-5--------1--------5//////

Clearly, x can be bigger than 1 or less than 1. Insufficient.