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# If x≠0, is |x|<1? (1) x^2/|x| > x

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Joined: 14 Nov 2016
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If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink]

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18 Feb 2017, 04:24
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Question Stats:

39% (01:34) correct 61% (01:41) wrong based on 93 sessions

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If $$x ≠ 0$$, is $$|x|<1$$?

(1) $$\frac{x^2}{{|x|}} > x$$

(2) $$\frac{x}{{|x|}} < x$$

[Reveal] Spoiler: OA

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Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink]

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18 Feb 2017, 06:50
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ziyuenlau wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

(1) $$\frac{x^2}{{|x|}} > x$$

(2) $$\frac{x}{{|x|}} < x$$

Hi,

is $$|x|<1$$? => is $$-1 < x < 1$$?

St 1:
Case1: x>0 => |x| = x.
$$\frac{x^2}{{|x|}} = \frac{x^{2}}{x} = x > x$$ => No solution.

Case2: x<0 => |x| = -x
$$\frac{x^2}{{|x|}} = \frac{x^{2}}{-x} = -x > x$$ => All negative values will satisfy this equation.

Hence, not sufficient.

St 2:
Case 1: x>0 => |x| = x

$$\frac{x}{{|x|}} = \frac{x}{x} = 1 < x$$ => solution x>1.

case 2: x<0 => |x| = -x

$$\frac{x}{{|x|}} = \frac{x}{-x} = -1 < x$$ => solution -1 < x < 0.

From this statement, we have two sets of solutions. Hence, not sufficient.

By St 1 and st 2, we have

Hope this helps.
Math Expert
Joined: 02 Sep 2009
Posts: 43893
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink]

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18 Feb 2017, 07:06
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If $$x ≠ 0$$, is $$|x|<1$$?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) $$\frac{x^2}{{|x|}} > x$$ --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

(2) $$\frac{x}{{|x|}} < x$$:

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that $$\frac{x}{{|x|}} < x$$ is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

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Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink]

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28 Feb 2017, 00:41
Bunuel wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) $$\frac{x^2}{{|x|}} > x$$ --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

(2) $$\frac{x}{{|x|}} < x$$:

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that $$\frac{x}{{|x|}} < x$$ is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Hello Bunuel, could you please elaborate more on the (1) condition, I can't grasp how did you get that x is -ve.
Thx
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Joined: 02 Sep 2009
Posts: 43893
Re: If x≠0, is |x|<1? (1) x^2/|x| > x [#permalink]

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28 Feb 2017, 04:52
Alexey1989x wrote:
Bunuel wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

The question asks whether -1 < x < 1 (x ≠ 0).

(1) $$\frac{x^2}{{|x|}} > x$$ --> reduce the LHS by |x|: |x| > x. This implies that x is negative. Not sufficient.

(2) $$\frac{x}{{|x|}} < x$$:

If x < 0, we'll have x/(-x) < x --> -1 < x. Since we consider x < 0, then we'll have -1 < x < 0.
If x > 0, we'll have x/x < x --> 1 < x.

So, we have that $$\frac{x}{{|x|}} < x$$ is true for -1 < x < 0 and x > 1. Not sufficient.

(1)+(2) Intersection of the ranges from (1) and (2) is -1 < x < 0. Thus the answer to the question is YES. Sufficient.

Hello Bunuel, could you please elaborate more on the (1) condition, I can't grasp how did you get that x is -ve.
Thx

We have |x| > x.

Now, if x were non-negative, then |x| would be equal to x, for example, if x=2, then |x| = |2| = 2 = x. Only for negative x we can have |x| > x. For example, consider x = -2: |x| = |-2| = 2 > x = 2.
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Re: If x≠0, is |x|<1? (1) x^2/|x| > x   [#permalink] 28 Feb 2017, 04:52
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