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

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17 00:00

Difficulty:   95% (hard)

Question Stats: 36% (02:41) correct 64% (02:23) wrong based on 180 sessions

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If $$x ≠ 0$$, is $$|x|<1$$?

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

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

I have a hard time to answer this question. Could someone please help to explain?

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

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1
2
ziyuenlau wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

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

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

I have a hard time to answer this question. Could someone please help to explain?

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
-1< x < 0. => Unique answer. Answer C.

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

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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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GMAT 1: 620 Q46 V29 Re: If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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

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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]

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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.

Greetings Bunuel

in analyzing statement 2,

you mentioned : "If x < 0, we'll have x/(-x) < x --> -1 < x"

my question: why the inequality is not flipped?
isn't it right to flip the from '<' to '> ?

I think I am confused in this point, please elaborate. _________________
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Re: If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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Mahmoudfawzy83 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.

Greetings Bunuel

in analyzing statement 2,

you mentioned : "If x < 0, we'll have x/(-x) < x --> -1 < x"

my question: why the inequality is not flipped?
isn't it right to flip the from '<' to '> ?

I think I am confused in this point, please elaborate. We are not dividing the entire inequality by x, we are simply reducing x/(-x) by x: x/(-x) = -(x/x) = -1.
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Re: If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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Bunuel

Please let me know where I am going wrong in my approach with statement 1.

x^2 / |x| > x

we know that the LHS will always be positive.

So, x^2 / |x| > 0 (its not >= 0 since x cannot be 0)

x^2 / |x| > 0

x^2 > 0

|x| > 0

so, x>0 and x<0. Hence, x can be any non zero number.

With this approach, i get answer choice E.

I also understand the approaches mentioned by you and others in the thread. But, I am not able to understand where I am going wrong. Please help.
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Re: If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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GMATaspirant641 wrote:
Bunuel

Please let me know where I am going wrong in my approach with statement 1.

x^2 / |x| > x

we know that the LHS will always be positive.

So, x^2 / |x| > 0 (its not >= 0 since x cannot be 0)

x^2 / |x| > 0

x^2 > 0

|x| > 0

so, x>0 and x<0. Hence, x can be any non zero number.

With this approach, i get answer choice E.

I also understand the approaches mentioned by you and others in the thread. But, I am not able to understand where I am going wrong. Please help.

Yes, $$\frac{x^2}{|x|} > 0$$ is true of all x's except 0. But we don't have $$\frac{x^2}{|x|} > 0$$, we have $$\frac{x^2}{|x|} > x$$ and as shown above it won't hold true if x is positive. For example, if x = 2, then $$\frac{x^2}{|x|} =2$$, so $$\frac{x^2}{|x|} = x$$.
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If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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hazelnut wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

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

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

I have a hard time to answer this question. Could someone please help to explain?

Hi chetan2u Bunuel

By adding two equations, we get -

$$\frac{x}{{|x|}}(x-1) > 0$$

Here, x can be -1,2.

Can you please advise where I am going wrong.
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Re: If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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hazelnut wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

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

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

I have a hard time to answer this question. Could someone please help to explain?

Why didn't i use the values used in A, to negate B , Nevertheless inline is my approach

|x| < 1, will mean that -1<x<1

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

when you use -0.5, the value will answer the statement and will answer the question as Yes

but when you use -2, the value will answer the statement and will answer the question as No

4/2 > -2

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

when you use -0.5 , the value will answer the statement and will answer the question as Yes

-0.5 /0.5 < -0.5

But when you use 2, the value will answer the statement and will answer the question as No

Now when we combine both the statements for both the statements to be true together

When you use x =-0.5, equality holds good

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

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rahul16singh28 wrote:
hazelnut wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

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

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

I have a hard time to answer this question. Could someone please help to explain?

Hi chetan2u Bunuel

By adding two equations, we get -

$$\frac{x}{{|x|}}(x-1) > 0$$

Here, x can be -1,2.

Can you please advise where I am going wrong.

We cannot add the inequalities the way you did.

You can only apply subtraction when their signs are in the opposite directions:

If $$a>b$$ and $$c<d$$ (signs in opposite direction: $$>$$ and $$<$$) --> $$a-c>b-d$$ (take the sign of the inequality you subtract from).
Example: $$3<4$$ and $$5>1$$ --> $$3-5<4-1$$.

You can only add inequalities when their signs are in the same direction:

If $$a>b$$ and $$c>d$$ (signs in same direction: $$>$$ and $$>$$) --> $$a+c>b+d$$.
Example: $$3<4$$ and $$2<5$$ --> $$3+2<4+5$$.

For more check Manipulating Inequalities.
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Re: If x≠0, is |x|<1? (1) x^2/|x| > x  [#permalink]

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hazelnut wrote:
If $$x ≠ 0$$, is $$|x|<1$$?

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

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

Here's my reasoning:

Statement 1:$$x^2 > x{|x|}$$

X can be both a negative integer and a negative decimal. Thus we get a YES and a NO for the stem above.

Statement 2: this tells us $$x < x{|x|}$$
X can be a positive integer or a negative fraction. Once again a YES and a NO for the stem above.

Combining the two the overlapping set is a negative fraction. Hence we can conclude $$|x|<1$$ Re: If x≠0, is |x|<1? (1) x^2/|x| > x   [#permalink] 08 Feb 2019, 21:17
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