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# Is |x| < 1 ?

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Is |x| < 1 ? [#permalink]

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31 Jul 2014, 09:55
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Is |x| < 1 ?

(1) x/|x| < x

(2) x/|x| < 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 31 Jul 2014, 10:12, edited 1 time in total.
Edited the question.

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Re: Is |x| < 1 ? [#permalink]

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31 Jul 2014, 10:32
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Is |x| < 1 ?

Is |x| < 1 --> is -1 < x < 1.

(1) x/|x| < x.

Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES.

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO.

Not sufficient.

(2) x/|x| < 1. This simply means that x is a negative number: if x is positive then x/|x| = 1, if x is negative x/|x| = -1 < 1. Not sufficient.

(1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient.

Answer: C.

Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source.

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Re: Is |x| < 1 ? [#permalink]

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01 Sep 2015, 08:52
Bunuel wrote:
Is |x| < 1 ?

Is |x| < 1 --> is -1 < x < 1.

(1) x/|x| < x.

Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES.

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO.

Not sufficient.

(2) x/|x| < 1. This simply means that x is a negative number: if x is positive then x/|x| = 1, if x is negative x/|x| = -1 < 1. Not sufficient.

(1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient.

Answer: C.

Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source.

Hi Bunuel
I fail to understand above solution.
We have to find whether -1< x<1
From 1 ) we know that x/|x| < x holds true for -1<x<0 and x>1 so we know that x is not in range 0<x<1
is this not sufficient to answer NO ....?
Again from 2) We know -infinity < x<0 then answer is NO
Sorry, could you please tell me what am I missing...how are you getting YES and NO for each?

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Re: Is |x| < 1 ? [#permalink]

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01 Sep 2015, 09:24
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anupamadw wrote:
Bunuel wrote:
Is |x| < 1 ?

Is |x| < 1 --> is -1 < x < 1.

(1) x/|x| < x.

Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES.

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO.

Not sufficient.

(2) x/|x| < 1. This simply means that x is a negative number: if x is positive then x/|x| = 1, if x is negative x/|x| = -1 < 1. Not sufficient.

(1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient.

Answer: C.

Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source.

Hi Bunuel
I fail to understand above solution.
We have to find whether -1< x<1
From 1 ) we know that x/|x| < x holds true for -1<x<0 and x>1 so we know that x is not in range 0<x<1
is this not sufficient to answer NO ....?
Again from 2) We know -infinity < x<0 then answer is NO
Sorry, could you please tell me what am I missing...how are you getting YES and NO for each?

Let me try to explain.

From 1,

x/|x| < x ---> 2 cases:

--- when x $$\geq$$0 --> |x| = x ---> x/x < x --> x > 1 . Thus a "no" for is |x| < 1

--- when x < 0 --> |x| = -x ---> x/-x < x --> x > -1 . For x = 0.5, then a "yes" for |x|<1 but if x = 5, then a "no" for is |x|<1. Thus NOT sufficient.

From 2,

x/|x| < 1 , again 2 cases,

--- when x $$\geq$$0 --> |x| = x ---> x/x < 1 --> 1<1 . Thus x $$\geq$$0 is not a possible scenario.

--- when x <0 --> |x| = -x ---> x/-x < 1 --> 1>-1 . This is true for ALL x and thus x <0 satisfies this. For x = -0.5, you get a "yes" for is |x|<1 but for x = -4, you get a "no" for is |x| <1 . Thus NOT sufficient.

Alternately, if you want you can also test cases to come up with ranges for x. But this might be a bit more time consuming.

For your analysis of statement 1, the text in red is not correct. Refer to the explanation mentioned above.

For statement 2, you are correct that x<0 but you need to check whether -1<x<0 is satisfied or not. You get 2 different answers when you take x = -0.5 or x = -4.

Hope this helps.
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Is |x| < 1 ? [#permalink]

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01 Sep 2015, 10:54
I have tried to use the line method to solve this problem.
1. x/!x!<xSince !x! is always positive, I can multiply it each side and get x<x.!x! and after substracting x-x.!x!<0
Now taking x as common factor, x(1-!x!)<0
x will have three factors:0,1,-1
Now refer to the attached figure to get the region where the given condition will be true.
Since the inequality is of "less than 0" type, we shold consider "negative portion"
we will get two possibilities which are -1<x<0 and x>1. Hence not suffecient.

2. x/!x!<1. Repeating the same logic above we get x-!x!<0. This will be true only if x<0. Still not suffeceint.

Combining both statements, we can eliminate the portion x>1, which we got in statement 1.
Leaving us only with -1<x<0, which tells us that !x! will always be a fraction less than 1. Hence suffeceint.
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Re: Is |x| < 1 ? [#permalink]

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08 Oct 2017, 20:06
Engr2012 wrote:
anupamadw wrote:
Bunuel wrote:
Is |x| < 1 ?

Is |x| < 1 --> is -1 < x < 1.

(1) x/|x| < x.

Two cases:
A. $$x<0$$ --> $$\frac{x}{-x}<x$$ --> $$-1<x$$. Since we consider the case when $$x<0$$, then we'd have $$-1<x<0$$. Answer YES.

B. $$x>0$$ --> $$\frac{x}{x}<x$$ --> $$1<x$$. Answer NO.

Not sufficient.

(2) x/|x| < 1. This simply means that x is a negative number: if x is positive then x/|x| = 1, if x is negative x/|x| = -1 < 1. Not sufficient.

(1)+(2) Since from (2) we have that x is negative, then we have case A from (1), which gives an YES answer to the question. Sufficient.

Answer: C.

Please tag the questions properly (this is NOT an algebra question) and copy the questions EXACTLY as they appear in the source.

Hi Bunuel
I fail to understand above solution.
We have to find whether -1< x<1
From 1 ) we know that x/|x| < x holds true for -1<x<0 and x>1 so we know that x is not in range 0<x<1
is this not sufficient to answer NO ....?
Again from 2) We know -infinity < x<0 then answer is NO
Sorry, could you please tell me what am I missing...how are you getting YES and NO for each?

Let me try to explain.

From 1,

x/|x| < x ---> 2 cases:

--- when x $$\geq$$0 --> |x| = x ---> x/x < x --> x > 1 . Thus a "no" for is |x| < 1

--- when x < 0 --> |x| = -x ---> x/-x < x --> x > -1 . For x = 0.5, then a "yes" for |x|<1 but if x = 5, then a "no" for is |x|<1. Thus NOT sufficient.

From 2,

x/|x| < 1 , again 2 cases,

--- when x $$\geq$$0 --> |x| = x ---> x/x < 1 --> 1<1 . Thus x $$\geq$$0 is not a possible scenario.

--- when x <0 --> |x| = -x ---> x/-x < 1 --> 1>-1 . This is true for ALL x and thus x <0 satisfies this. For x = -0.5, you get a "yes" for is |x|<1 but for x = -4, you get a "no" for is |x| <1 . Thus NOT sufficient.

Alternately, if you want you can also test cases to come up with ranges for x. But this might be a bit more time consuming.

For your analysis of statement 1, the text in red is not correct. Refer to the explanation mentioned above.

For statement 2, you are correct that x<0 but you need to check whether -1<x<0 is satisfied or not. You get 2 different answers when you take x = -0.5 or x = -4.

Hope this helps.

Statement 2

Aren't we told that x lies between 1 and -1 ?

then why are we considering x=-4

Pls help

Thanks in advance

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Re: Is |x| < 1 ?   [#permalink] 08 Oct 2017, 20:06
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