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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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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. Sent from my Lenovo TAB S850LC using GMAT Club Forum mobile app



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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
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1x <1 taking critical point +1 for reference, if x>1. (1x)<1 i.e. x<2 and if x<1. (1x)<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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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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Re: Is x negative? (1) 1  x < 1 (2) 1 + x > 1 [#permalink]
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12 Aug 2017, 05:20
Bunuel wrote: Is x negative?
(1) 1  x < 1 (2) 1 + x > 1 Is x negative? (1) 1  x < 1 ;So 1< 1x<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 SufficientCorrect Answer A
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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: 1x <1 taking critical point +1 for reference, if x>1. (1x)<1 i.e. x<2 and if x<1. (1x)<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 3step approach from here: https://gmatclub.com/forum/mathabsolut ... 86462.html, I had an issue with statement #2. I'll show my work for both statements though. Statement 1: 1x<1a) Positive: 1x<1 > 0<x b) Negative: (1x)>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
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brandon7 wrote: sahilvijay wrote: 1x <1 taking critical point +1 for reference, if x>1. (1x)<1 i.e. x<2 and if x<1. (1x)<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 3step approach from here: https://gmatclub.com/forum/mathabsolut ... 86462.html, I had an issue with statement #2. I'll show my work for both statements though. Statement 1: 1x<1a) Positive: 1x<1 > 0<x b) Negative: (1x)>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: 1x <1 taking critical point +1 for reference, if x>1. (1x)<1 i.e. x<2 and if x<1. (1x)<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 3step approach from here: https://gmatclub.com/forum/mathabsolut ... 86462.html, I had an issue with statement #2. I'll show my work for both statements though. Statement 1: 1x<1a) Positive: 1x<1 > 0<x b) Negative: (1x)>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
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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. Hit kudos if you like the explanation. Sent from my iPhone using GMAT Club Forum
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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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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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