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Is |x| < 1 ? GMATPrep

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Source: — Data Sufficiency |
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by massi2884 » Mon May 07, 2012 6:23 am
Hi all,
regarding statement 1, I know that |A| = |B| --> A = B and A = -B
But I'm wondering why we don't check if the results are in the intervals given by the absolute values, which are x > -1 and x > 1.

Thanks.
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by neelgandham » Mon May 07, 2012 7:16 am
Is |x| < 1 ?
1) |x + 1| = 2|x - 1|

x + 1 = 2(x - 1) or x + 1 = -2(x - 1)
x + 1 = 2x - 2 or x + 1 = 2 - 2x
x = 3 or 3x = 1
x = 3 or x = 1/3
So |x| can be > 1(when x = 3) or < 1 (when x =1/3). It is insufficient to answer the question with Statement I
2) |x - 3| ≠ 0
Implies x !=3
It is insufficient to answer the question with statement II
Statement I + Statement II
x = 3 or x = 1/3 - Statement I
x != 3 - Statement II. Combining both the statements we get: x =1/3 and |x|<1. It is sufficient to answer the question. IMO C
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by massi2884 » Mon May 07, 2012 10:31 am
Thanks. Can you please explain why here we don't check whether the values are in the ranges given by the absolute values (x > -1 and x > 1), as we do for Inequalities?

Thank you
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by darontan » Tue May 08, 2012 6:52 pm
hi massi2884, the absolute value |x|<1 means -1<x<1
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by Anurag@Gurome » Tue May 08, 2012 8:19 pm
massi2884 wrote:Is |x| < 1 ?

1) |x + 1| = 2|x - 1|
2) |x - 3| ≠ 0

OA C
Question is: Is |x| < 1 OR is -1 < x < 1 true?

(1) |x + 1| = 2|x - 1|
Square both sides, (x + 1)² = 4(x - 1)² or x² + 2x + 1 = 4x² - 8x + 4 or 3x² - 10x + 3 = 0 or x = 1/3 or x = 3.
So, two valid values of x are 1/3 and 3, from which 1/3 lies in the range -1 < x < 1 but 3 does not lies in this range; NOT sufficient.

(2) |x - 3| ≠ 0 or x ≠ 3 but we have no clue whether x is in the range (-1, 1) or not; NOT sufficient.

Combining (1) and (2), x = 1/3, or x = 3 and x ≠ 3 implies x can have only one value, that is, 1/3, which is in the range (-1,1); SUFFICIENT.

The correct answer is C.
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