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

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New post 11 Aug 2017, 02:33
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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New post 11 Aug 2017, 07:54
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|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

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

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New post 12 Aug 2017, 02:39
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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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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New post 12 Aug 2017, 04:20
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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

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

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New post 17 Aug 2017, 08:46
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

Answer is A.


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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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

Answer is A.


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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New post 17 Aug 2017, 09:27
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

Answer is A.


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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New post 17 Aug 2017, 10:25
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Reply for your query
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
So we are not adding any additional sign
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.

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

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New post 21 Aug 2017, 10:42
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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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1 [#permalink]

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New post 21 Aug 2017, 12:04
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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Re: Is x negative? (1) |1 - x| < 1 (2) |1 + x| > 1   [#permalink] 21 Aug 2017, 12:04
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