absolute value computation

[utkalnayak]

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0

OA: C

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Hi utkalnayak,

This DS question is worded in a very specific way and has a number of interesting Number Properties that you can take advantage of.

We're asked if |X| < 1. This is a YES/NO question.

The wording of this question is interesting in that the types of numbers that would give you a YES answer are "limited" - only positive/negative FRACTIONS or the number 0 would give you a YES; all other possibilities would give you a NO. This gets me thinking that I will have to consider possibilities that are fractions.

Fact 1: |X+1| = 2|X-1|

Since this is an equation that involves absolute values, chances are that there are 2 solutions.

IF.....
X = 3, then the answer to the question is NO.

Remember what I noted about fractions earlier....
IF....
X = 1/3, then the answer to the question is YES.
Fact 1 is INSUFFICIENT

Fact 2: |X-3| > 0

This inequality excludes ONLY 1 value of X: 3
X CANNOT be 3....

IF....
X = 4, then the answer to the question is NO.

IF....
X = 1/3, then the answer to the question is YES.
Fact 2 is INSUFFICIENT

Combined, we know....
X is either 3 or 1/3
X CANNOT be 3
This leaves us with just one answer (1/3) and the answer to the question is ALWAYS YES.
Combined, SUFFICIENT.

Final Answer: C

GMAT assassins aren't born, they're made,
Rich

[GMATGuruNY]

Is |x| < 1 ?

(1) |x + 1| = 2|x - 1|

(2) |x - 3| > 0

Statement 1: |x + 1| = 2|x - 1|
Case 1: No signs changed
x+1 = 2(x-1)
x+1 = 2x - 2
3 = x.

Case 2: Signs changed in ONE of the absolute values
-(x+1) = 2(x-1)
-x-1 = 2x - 2
-3x = -1
x = 1/3.

Since |x|>1 in Case 1 but |x|<1 in Case 2, INSUFFICIENT.

Statement 2: |x - 3| > 0
Case 3: No signs changed
x-3 > 0
x > 3

Case 4: Signs changed in the absolute value
-(x-3) > 0
-x + 3 > 0
-x > -3
x < 3.

The resulting inequalities -- x<3 or x>3 -- imply that x≠3.
If x=0, then |x|<1.
If x=10, then |x|>1.
INSUFFICIENT.

Statements combined:
Statement 1 requires that x=3 or x=1/3.
Statement 2 requires that x≠3.
Thus, x=1/3, with the result that |x| < 1.
SUFFICIENT.

The correct answer is C.

[GMATGuruNY]

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.

[Matt@VeritasPrep]

Another algebraic trick is remembering that |x| = √x², and here it seems to make the question much easier.

|x| < 1 is equivalent to √x² < 1, or x² < 1, or -1 < x < 1. So the REAL question is "Is 1 > x > -1 ?"

S1 gives us

√(x+1)² = 2√(x-1)²
(x + 1)² = 4(x - 1)²
x² + 2x + 1 = 4(x² - 2x + 1)
x² + 2x + 1 = 4x² - 8x + 4
0 = 3x² - 10x + 3
0 = (3x - 1)(x - 3)

So x = 1/3 or x = 3. If x = 1/3, then 1 > x > -1, but if x = 3, x > 1, so this is INSUFFICIENT.

S2 gives us

√(x-3)² > 0
(x - 3)² > 0
x > 3 or 3 > x; in other words, x is NOT 3. This is INSUFFICIENT too.

Together, we know that x = 1/3, so we're set!