GMAT Prep Absolute question

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GMAT Prep Absolute question

by Mo2men » Mon Sep 12, 2016 2:49 am
If x≠0, is |x| < 1 ?

(1) x^2 < 1
(2) |x| < 1/x

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by GMATGuruNY » Mon Sep 12, 2016 3:29 am
Mo2men wrote:If x≠0, is |x| < 1 ?

(1) x^2 < 1
(2) |x| < 1/x
Statement 1:
Since x is nonzero, only a positive or negative fraction between -1 and 1 will satisfy x²<1.
The absolute value of a positive or negative fraction must be less than 1.
SUFFICIENT.

Statement 2:
Only a positive fraction between 0 and 1 will satisfy |x| < 1/x.
The absolute value of a positive fraction must be less than 1.
SUFFICIENT.

The correct answer is D.
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by MartyMurray » Mon Sep 12, 2016 3:32 am
Question: If x ≠ 0, is |x| < 1?

Rephrase: Is -1 < x < 1

Statement 1: x² < 1

If x² < 1, then -1 < x < 1.

Sufficient.

Statement 2: |x| < 1/x

Try some numbers. There are six categories that could be tried.

x < -1, x = -1, -1 < x < 0, 0 < x < 1, x = 1, 1 < x

-2: |-2| < 1/-2 This is impossible as the left side is positive and the right side is negative. A positive number is not less than a negative number.

No negative numbers work. Try the positive categories

1/2: |1/2| < 1/(1/2) This is true.

1: |1| < 1/1 This is not true.

2: |2| < 1/2 This is not true, and illustrates the fact that for any x such that x > 1, Statement 2 will not be true.

The only one that works is 1/2.

|1/2| < 1

Sufficient.

The correct answer is D.

There is a trick to Statement 2.

One may think that one can multiply both sides by x and get x|x| < 1.

However that idea is not entirely correct.

In working with the inequality have to consider the possibility that x could be negative, in which case when we multiply both sides by x, the sign of the inequality would have to be reversed.

In considering the case in which x is negative, when we multiply both sides by x, we get x|x| > 1.

Since a negative number multiplied by an absolute value must generate a negative product, if x is negative, x|x| > 1 is not true.

So once again we have showed that x cannot be negative.

We also have to consider the case in which x is positive. If x is positive when we multiply we get x|x| < 1.

That statement is true when x is a fraction.

So 0 < x < 1, and once again we have showed that Statement 2 is sufficient.
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by Mo2men » Mon Sep 12, 2016 3:48 am
GMATGuruNY wrote:
Mo2men wrote:If x≠0, is |x| < 1 ?

(1) x^2 < 1
(2) |x| < 1/x
Statement 1:
Since x is nonzero, only a positive or negative fraction between -1 and 1 will satisfy x²<1.
The absolute value of a positive or negative fraction must be less than 1.
SUFFICIENT.

Statement 2:
Only a positive fraction between 0 and 1 will satisfy |x| < 1/x.
The absolute value of a positive fraction must be less than 1.
SUFFICIENT.

The correct answer is D.
Dear GMATGuru,

In statement 2,

if I want to use critical point. I struggle when x<0

is it -x<1/x or -x<1/-x?

Should replace every 'x' with '-x' or only 'x' inside modulus??

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by GMATGuruNY » Mon Sep 12, 2016 7:01 am
Mo2men wrote: Dear GMATGuru,

In statement 2,

if I want to use critical point. I struggle when x<0

is it -x<1/x or -x<1/-x?

Should replace every 'x' with '-x' or only 'x' inside modulus??
Since an absolute value cannot be less than a negative, |x| < 1/x is valid only if x is nonnegative.
Thus, we can simplify as follows:
x < 1/x
x² < 1
x < 1.
Since x≠0, 0 < x < 1.
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by Jeff@TargetTestPrep » Mon Sep 12, 2016 4:48 pm
Mo2men wrote:If x≠0, is |x| < 1 ?

(1) x^2 < 1
(2) |x| < 1/x
We are given that x is not equal to 0 and we need to determine whether the absolute value of x is less than 1.

Statement One Alone:

x^2 < 1

We can take the square root of both sides of the inequality x^2 < 1:

√(x^2) < √1.

Since √(x^2) = |x|, |x| < 1.

Statement one alone is sufficient. Eliminate answer choices B, C and E.

Statement Two Alone:

|x| < 1/x

Since |x| will always be positive, in order for 1/x to be greater than |x|, x must be positive. Furthermore, the only way |x| < 1/x holds true is if x is a proper fraction. Thus, x must be a positive proper fraction. Since a positive proper fraction is between zero and one, the absolute value of x is less than one. Statement two alone is also sufficient.

Answer: D

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by Matt@VeritasPrep » Thu Sep 15, 2016 7:35 pm
We could start by rephrasing the prompt:

Is -1 < x < 1?

S1:

x*x < 1

It's probably easiest to see if we plug in numbers. If x ≥ 1, this won't work. If -1 ≥ x it won't work either. So we must have -1 < x < 1, and this answers our question! Sufficient.

S2:

|x| < 1/x

If x is positive, this is the same as statement one! We multiply both sides by x, and get |x| * x < 1.

If x is negative, this is impossible, since the left side is positive and the right side is negative.

So we must have x positive, and each statement is sufficient.