utkalnayak wrote:Is |x| < 1 ?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| > 0
Alternate approach:
Question stem, rephrased:
Is x between -1 and 1?
|a-b| = the distance between a and b.
|a+b| = |a - (-b)| = the distance between a and -b.
Statement 1: |x + 1| = 2|x - 1|
|x - (-1)| = 2|x-1|.
In words:
The distance between x and -1 is twice the distance between x and 1.
Case 1: x is BETWEEN -1 AND 1
-1.........0...x......1
Here, the distance between x and -1 is twice the distance between x and 1.
Case 2: x is to the RIGHT OF 1
-1.........0.........1..................x
Here, the distance between x and -1 is twice the distance between x and 1
Since x is between -1 and 1 in Case 1 but to the right of 1 in Case 2, INSUFFICIENT.
Statement 2: |x - 3| > 0
In words:
The distance between x and 3 is greater than 0.
Implication:
x can be ANY OTHER VALUE THAN 3, with the result that the distance between x and 3 will be greater than 0.
Since it's possible that x is between -1 and 1 or NOT between -1 and 1, INSUFFICIENT.
Statements combined:
Always consider how the two statements might AFFECT EACH OTHER.
Statement 2 indicates x≠3.
In the number line for Case 2, x is to the right of 1 and might be equal to 3.
My suspicion:
The value that yields Case 2 is x=3.
Plugging x=3 into |x + 1| = 2|x - 1|, we get:
|3 + 1| = 2|3 - 1|
4 = 4.
Implication: The value that yields Case 2 is x=3.
Since statement 2 indicates that x≠3, Case 2 is not possible.
Thus, only Case 1 is possible, implying that x is between -1 and 1.
SUFFICIENT.
The correct answer is
C.
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