Is |X| < 1 ?

[kamalakarthi]

Hi, This is from MCAT Question.

Can you please help me on solving this question.

Steps Followed :: I tried to rephrase the question , Essentially the question is -1<X<1 ?

St 1:: If X is +ve, X < 1
if X is _ve, -x/X < -x
so 1/x <-1 (Dividing -x on both the sides) Is this correct ?
-1< x or X > -1 NS

St 2:: X is +ve, It cannot happen |X| >x
but if x is negative , x < -x ..I got stuck here. and I thought This is also not sufficient

After this, I thought C is also not sufficient so I chose E . Can you please help., What is my mistake in Statement 2.

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Hi kamalakarthi,

This question can be solved by TESTing VALUES and Number Properties.

We're told that X cannot equal 0. We're asked if |X| is less than 1. This is a YES/NO question.

1) X/|X| < X

From a Number Property standpoint, there are specific values of X that will 'fit' this inequality:
-Any positive number greater than 1.
-Any negative fraction (re: -1 < X < 0)

For example, IF....
X = 2.... 2/|2| = 1 < 2 and the answer to the question is NO.
X = -1/2.... (-1/2)/|-1/2| = -1 < -1/2 and the answer to the question is YES.
Fact 1 is INSUFFICIENT

2) |X| > X

From a Number Property standpoint, there are specific values of X that will 'fit' this inequality:
-Any negative value

For example, IF....
X = -1/2.... |-1/2| > -1/2 and the answer to the question is YES.
X = -2... |-2| > -2 and the answer to the question is NO.
Fact 2 is INSUFFICIENT

Combined, there is only one specific type of value that will fit BOTH Facts: negative fractions (-1 < X < 0), so the answer to the question will be ALWAYS YES.
Combined, SUFFICIENT

Final Answer: C

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

[Matt@VeritasPrep]

Here's one way:

S1:

Since |x| must be > 0, we can safely multiply both sides by it:

x < x * |x|

Now subtract x from both sides:

0 < x * |x| - x

Now factor out x:

0 < x * (|x| - 1)

This gives us two possibilities:

(i) x AND |x| - 1 are both positive
OR
(ii) x and |x| - 1 are both negative

(i) reduces to |x| - 1 > 0, or |x| > 1, or x > 1
(ii) reduces to |x| - 1 < 0, or |x| < 1, or -1 < x < 0

So IF x is negative, then |x| < 1. This helps, but isn't sufficient on its own. (It does reduce the question to "Is x < 0?", though!)

S2:

|x| > x

This obviously won't be true if x ≥ 0 (since, if x ≥ 0, |x| = x). So x < 0. This is great ... but not on its own!

Together, though, we're set. S1 reduces the prompt to "Is x < 0?" and S2 answers that question in the affirmative.

[Matt@VeritasPrep]

Another way, if the first is too technical:

S1:

x > x / |x|

|x| * x > x

Now take two cases. If x > 0, then we divide both sides to get |x| > 1.

If x < 0, then we divide both sides to get |x| < 1.

So |x| will be < 1 if and only if x < 0.

S2 I'd treat the same way I did in my post above, and much like that one, we'll arrive at the same conclusion: S1 simplifies the question and S2 answers that simplified question, leaving us with an answer of C.

[kamalakarthi]

Thank you for the help.

[Matt@VeritasPrep]

Thank you for the help.

My pleasure!