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# Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1

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Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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11 Aug 2017, 02:48
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Is x negative?

(1) |1 - x| < 1
(2) |1 + x| > 1

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Joined: 02 Nov 2015
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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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11 Aug 2017, 03:33
Statement 1 : it says that x is a fraction and positive.
Thus sufficient.

Statement 2: x can be negative as well as positive. Not sufficient.

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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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11 Aug 2017, 08:54
|1-x| <1
taking critical point +1 for reference, if x>1. -(1-x)<1 i.e. x<2 and if x<1. (1-x)<1 i.e. x>0 which means statement 1 is sufficient. A, D are under consideration.
|1+x|>1
taking -1 as critical point, if x>-1 then (1+x)>1 yields x>0 and if x<-1 then -(1+x)>1 yields x<-2, hence x is both +ve and -ve. So D is ruled out

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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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12 Aug 2017, 03:39
if x>0 or x<2 then since x<2 it can be negative also?
first statement gives x>0 or x<2. How is A sufficient?
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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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12 Aug 2017, 03:45
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Bounce1987 wrote:
if x>0 or x<2 then since x<2 it can be negative also?
first statement gives x>0 or x<2. How is A sufficient?

x > 0 and x < 2 are simultaneous conditions, so you should consider them together as 0 < x < 2.

|1 - x| < 1

-1 < 1 - x < 1

Subtract 1 from all three parts: -2 < -x < 0.

Multiply by -1 and flip the signs as we multiply by negative number: 2 > x > 0.
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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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12 Aug 2017, 05:20
Top Contributor
Bunuel wrote:
Is x negative?

(1) |1 - x| < 1
(2) |1 + x| > 1

Is x negative?

(1) |1 - x| < 1 ;So -1< 1-x<1 ,or -2<-x<0 ,or 0<x<2 ,Sufficient
(2) |1 + x| > 1 ;So 1+x>1 or 1+x<-1 ,So x>0 ,or x<-2 ,Not Sufficient

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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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17 Aug 2017, 09:46
sahilvijay wrote:
|1-x| <1
taking critical point +1 for reference, if x>1. -(1-x)<1 i.e. x<2 and if x<1. (1-x)<1 i.e. x>0 which means statement 1 is sufficient. A, D are under consideration.
|1+x|>1
taking -1 as critical point, if x>-1 then (1+x)>1 yields x>0 and if x<-1 then -(1+x)>1 yields x<-2, hence x is both +ve and -ve. So D is ruled out

I got the same solution, but I had one question.

When trying to do the solution via the 3-step approach from here: https://gmatclub.com/forum/math-absolut ... 86462.html, I had an issue with statement #2. I'll show my work for both statements though.

Statement 1: |1-x|<1
a) Positive: 1-x<1 --> 0<x
b) Negative: (1-x)>-1 --> 2>x

0<x<2, thus sufficient

For Statement 2: |1+x|>1

a) Positive: if (1+x)>=0, so if x>1, then we can rewrite the equation as 1+x>1, x>0
b) Negative: if (1+x)<0, so if x<1, then I'm not sure how to write the equation because if x is -1<x<1, then the inequality (1+x)<-1 doesn't hold true. but if x<-1, then the inequality (1+x)<-1 does hold true.

I think I am confusing concepts, but my head is spinning trying to figure it out. Does the 3 step approach not work in this case? Am I even doing it correctly? Thanks.
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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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17 Aug 2017, 09:57
brandon7 wrote:
sahilvijay wrote:
|1-x| <1
taking critical point +1 for reference, if x>1. -(1-x)<1 i.e. x<2 and if x<1. (1-x)<1 i.e. x>0 which means statement 1 is sufficient. A, D are under consideration.
|1+x|>1
taking -1 as critical point, if x>-1 then (1+x)>1 yields x>0 and if x<-1 then -(1+x)>1 yields x<-2, hence x is both +ve and -ve. So D is ruled out

I got the same solution, but I had one question.

When trying to do the solution via the 3-step approach from here: https://gmatclub.com/forum/math-absolut ... 86462.html, I had an issue with statement #2. I'll show my work for both statements though.

Statement 1: |1-x|<1
a) Positive: 1-x<1 --> 0<x
b) Negative: (1-x)>-1 --> 2>x

0<x<2, thus sufficient

For Statement 2: |1+x|>1

a) Positive: if (1+x)>=0, so if x>1, then we can rewrite the equation as 1+x>1, x>0
b) Negative: if (1+x)<0, so if x<1, then I'm not sure how to write the equation because if x is -1<x<1, then the inequality (1+x)<-1 doesn't hold true. but if x<-1, then the inequality (1+x)<-1 does hold true.

I think I am confusing concepts, but my head is spinning trying to figure it out. Does the 3 step approach not work in this case? Am I even doing it correctly? Thanks.

Let me know why your head is spinning, See my solution above and let me know which part you did not understand, i will be happy to clear your doubt.
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Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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17 Aug 2017, 10:27
sahilvijay wrote:
brandon7 wrote:
sahilvijay wrote:
|1-x| <1
taking critical point +1 for reference, if x>1. -(1-x)<1 i.e. x<2 and if x<1. (1-x)<1 i.e. x>0 which means statement 1 is sufficient. A, D are under consideration.
|1+x|>1
taking -1 as critical point, if x>-1 then (1+x)>1 yields x>0 and if x<-1 then -(1+x)>1 yields x<-2, hence x is both +ve and -ve. So D is ruled out

I got the same solution, but I had one question.

When trying to do the solution via the 3-step approach from here: https://gmatclub.com/forum/math-absolut ... 86462.html, I had an issue with statement #2. I'll show my work for both statements though.

Statement 1: |1-x|<1
a) Positive: 1-x<1 --> 0<x
b) Negative: (1-x)>-1 --> 2>x

0<x<2, thus sufficient

For Statement 2: |1+x|>1

a) Positive: if (1+x)>=0, so if x>1, then we can rewrite the equation as 1+x>1, x>0
b) Negative: if (1+x)<0, so if x<1, then I'm not sure how to write the equation because if x is -1<x<1, then the inequality (1+x)<-1 doesn't hold true. but if x<-1, then the inequality (1+x)<-1 does hold true.

I think I am confusing concepts, but my head is spinning trying to figure it out. Does the 3 step approach not work in this case? Am I even doing it correctly? Thanks.

Let me know why your head is spinning, See my solution above and let me know which part you did not understand, i will be happy to clear your doubt.

Thanks. While I was have the following question about your solution below, can you also answer the questions in my post as well? It'd also be really helpful to understand what I am doing wrong in my post.

For statement 2: Why do you say if x>-1, then (1+x)>1? And then if x<-1 then you put the negative sign? How does the critical point of -1 factor into here? I don't see where that's apparent.
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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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17 Aug 2017, 11:25
If |1+x| > 1 for eg
Then
Plot a number line
Put the modulus value =0 to get critical points and plot on number line

Now see here critical point is -1

Now put any value more than -1 like 3,4 etc the modulus value will always be positive as it has to be
Leaving as it is.

But let us suppose x<-1
Then modulus if i let the sign as it is outside brackets
The. For eg for value of x = -4, -5
And value less than -1 will make value inside the bracket as negative

Like using -3
-3+1 = -2
Now we know modulus always open and makes value positive
As it is the distance from number line and can not be negative

Therefore it is must to put a negative outside the value when x<-1 And when i open the bracket.

I hope i have explained well.

Hit kudos if you like the explanation.

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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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21 Aug 2017, 11:42
Bunuel wrote:
Bounce1987 wrote:
if x>0 or x<2 then since x<2 it can be negative also?
first statement gives x>0 or x<2. How is A sufficient?

x > 0 and x < 2 are simultaneous conditions, so you should consider them together as 0 < x < 2.

|1 - x| < 1

-1 < 1 - x < 1

Subtract 1 from all three parts: -2 < -x < 0.

Multiply by -1 and flip the signs as we multiply by negative number: 2 > x > 0.

Hi Bunuel,

When I squared both the options *since both sides are positive), I am getting different result. Can you please share such solution when both options are squared and still we get option A as correct answer.
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Posts: 54496
Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1  [#permalink]

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21 Aug 2017, 13:04
msk0657 wrote:
Bunuel wrote:
Bounce1987 wrote:
if x>0 or x<2 then since x<2 it can be negative also?
first statement gives x>0 or x<2. How is A sufficient?

x > 0 and x < 2 are simultaneous conditions, so you should consider them together as 0 < x < 2.

|1 - x| < 1

-1 < 1 - x < 1

Subtract 1 from all three parts: -2 < -x < 0.

Multiply by -1 and flip the signs as we multiply by negative number: 2 > x > 0.

Hi Bunuel,

When I squared both the options *since both sides are positive), I am getting different result. Can you please share such solution when both options are squared and still we get option A as correct answer.

|1 - x| < 1

1 - 2x + x^2 < 1

x^2 - 2x < 0

x(x - 2) < 0

The roots are 0 and 2. "<" sign indicates that the solution lies between the roots: 0 < x < 2.
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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1   [#permalink] 21 Aug 2017, 13:04
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