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by ariz » Tue Jan 10, 2012 9:01 am
What is the value of X?

1. X^4 = |X|

2. X^2 > X

OA: C

I was confused about the 2 statements because I thought they contradict each other

Source: another GMAT site

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by chieftang » Tue Jan 10, 2012 9:38 am
ariz wrote:What is the value of X?

1. X^4 = |X|

2. X^2 > X
1. X is {-1, 0, 1}

INSUFFICIENT

2. X < 0 or X > 1

INSUFFICIENT

1,2.

The intersection of the sets {-1, 0, 1} and {... 4, 3, 2, -1, -2, -3, ...} is {-1}

SUFFICIENT

Answer C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
Last edited by chieftang on Tue Jan 10, 2012 10:55 am, edited 1 time in total.

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by GMATGuruNY » Tue Jan 10, 2012 10:30 am
ariz wrote:What is the value of X?

1. X^4 = |X|

2. X^2 > X

OA: C

I was confused about the 2 statements because I thought they contradict each other

Source: another GMAT site
Statement 1: x� = |x|.
Case 1: x� = x.
x� - x = 0.
x(x³ - 1) = 0.
x = 0 or x=1.

Case 2: x� = -x.
x� + x = 0.
x(x³ + 1) = 0.
x = 0 or x= -1.
Since it's possible that x=-1, x=0, or x=1, INSUFFICIENT.

Statement 2: x² > x.
x² - x > 0.
x(x-1) > 0.

The critical points are x=0 and x=1.
These are the only values where x² = x.
When x is any other value, x² > x or x² < x.
Thus, there are 3 ranges to consider: x<0, 0<x<1, and x>1.

To determine the range of x, test one value to the left and right of each critical point.
When x = -1, x² > x. Thus, x<0 is part of the range.
When x = 1/2, x² < x. Thus, 0<x<1 is not part of the range.
When x = 2, x² > x. Thus, x>1 is part of the range.
Thus, two ranges satisfy statement 2:
x<0 or x>1.
INSUFFICIENT.

Statements 1 and 2 combined:
Of the three values that satisfy statement 1, only x=-1 is included in the ranges that satisfy statement 2 (x<0 or x>1).
Thus, x=-1.
SUFFICIENT.

The correct answer is C.
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by chieftang » Tue Jan 10, 2012 10:51 am
GMATGuruNY wrote:
ariz wrote:What is the value of X?

1. X^4 = |X|

2. X^2 > X

OA: C

I was confused about the 2 statements because I thought they contradict each other

Source: another GMAT site
Statement 1: x� = |x|.
Case 1: x� = x.
x� - x = 0.
x(x³ - 1) = 0.
x = 0 or x=1.

Case 2: x� = -x.
x� + x = 0.
x(x³ + 1) = 0.
x = 0 or x= -1.
Since it's possible that x=-1, x=0, or x=1, INSUFFICIENT.

Statement 2: x² > x.
x² - x > 0.
x(x-1) > 0.

The critical points are x=0 and x=1.
These are the only values where x² = x.
When x is any other value, x² > x or x² < x.
Thus, there are 3 ranges to consider: x<0, 0<x<1, and x>1.

To determine the range of x, test one value to the left and right of each critical point.
When x = -1, x² > x. Thus, x<0 is part of the range.
When x = 1/2, x² < x. Thus, 0<x<1 is not part of the range.
When x = 2, x² > x. Thus, x>1 is part of the range.
Thus, two ranges satisfy statement 2:
x<0 or x>1.
INSUFFICIENT.

Statements 1 and 2 combined:
Of the three values that satisfy statement 1, only x=-1 is included in the ranges that satisfy statement 2 (x<0 or x>1).
Thus, x=-1.
SUFFICIENT.

The correct answer is C.
Oh, ya, I missed X > 1 for statement 2 and luckily still got the correct solution. Haste makes waste! Updated my work. Answer unchanged.