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Is the number x positive? [#permalink]
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18 Sep 2013, 06:15
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Is the number x positive? (1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1.
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Re: Is the number x positive? [#permalink]
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18 Sep 2013, 06:24
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rrsnathan wrote: Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1. Ans should be A. 1. x=2, x1 =1 Yes x=1, x1 =2 Yes 2. x=3,x+1 =2 No x=1/3, x+1=2/3 Yes Hence A is sufficient. Please press KUDOS if my post HELPED !!



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Re: Is the number x positive? [#permalink]
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18 Sep 2013, 06:35
Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1.
Take choice 1,
let x 1 = 0.6 => x = 0.6. so X +ve. let x1 = = 0.2 => x = 1.2. so x +ve
Take choice 2.
let x 1 =  0.4 => x = 0.4, so X +ve. let x1 = 2.4 => x = 1.4, so X ve.
Hence the ans is A.
Hope this helps.



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Re: Is the number x positive? [#permalink]
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18 Sep 2013, 06:47
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rrsnathan wrote: Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1. x denotes the distance between 0 and x on the number line. F.S 1 states that 0(x1)<0x > x1<x > Square on both sides, and subtract >\(1*(2x1)>0 > x>\frac{1}{2}\). Sufficient. F.S 2 states that x<x+1 > Just as above, this gives > 1*(2x+1)>0 > \(x>\frac{1}{2}\). Insufficient. A.
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Re: Is the number x positive? [#permalink]
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09 Oct 2013, 02:45
Forgot to substitute the fractional value ((((



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Re: Is the number x positive? [#permalink]
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03 Nov 2014, 02:04
Bumping up for any other explanation. This is how I approached it 
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1
(1) assume x= 2; => x1 = 3. (Given answer choice is not true/satisfied, so x can't be negative) Consider x= 22; => x1 =21 (Given answer choice is true/satisfied)
So, it is true that x is positive.
(2) same assumption, x=32 => x+1 = 31(Given answer choice is not true/satisfied, so x can't be negative) Consider x=1 => x+1 = 2 (Given answer choice is true/satisfied)
So, it is true that x is positive.



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Re: Is the number x positive? [#permalink]
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03 Nov 2014, 02:16
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Blackbox wrote: Bumping up for any other explanation. This is how I approached it 
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1
(1) assume x= 2; => x1 = 3. (Given answer choice is not true/satisfied, so x can't be negative) Consider x= 22; => x1 =21 (Given answer choice is true/satisfied)
So, it is true that x is positive.
(2) same assumption, x=32 => x+1 = 31(Given answer choice is not true/satisfied, so x can't be negative) Consider x=1 => x+1 = 2 (Given answer choice is true/satisfied)
So, it is true that x is positive. For (2) consider the values of x from 1/2 (not inclusive) to 0 (inclusive) to get that (2) is NOT in fact sufficient.
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Re: Is the number x positive? [#permalink]
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05 Nov 2014, 00:50
Bunuel wrote: For (2) consider the values of x from 1/2 (not inclusive) to 0 (inclusive) to get that (2) is NOT in fact sufficient. I am not sure what you meant by 1/2 (not inclusive) and 0 (inclusive) and I'd appreciate if you can shed some light on it. But yeah, I messed up not considering fractions. So ... For option 2  if x were 1/2 then x+1 would be 1/2. Given statement of 0 closer to x than x+1 is busted with two answers (0 is as much closer to 1/2 as it is to 1/2). Hence option 2 has multiple answers. It goes to show again how much I suck at picking magic numbers (or numbers at all). I have to keep reminding myself to test for fractions too. Ugh. Is there any other general approach to this problem besides picking numbers?



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Re: Is the number x positive? [#permalink]
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05 Nov 2014, 01:04
Blackbox wrote: Bunuel wrote: For (2) consider the values of x from 1/2 (not inclusive) to 0 (inclusive) to get that (2) is NOT in fact sufficient. I am not sure what you meant by 1/2 (not inclusive) and 0 (inclusive) and I'd appreciate if you can shed some light on it. But yeah, I messed up not considering fractions. So ... For option 2  if x were 1/2 then x+1 would be 1/2. Given statement of 0 closer to x than x+1 is busted with two answers (0 is as much closer to 1/2 as it is to 1/2). Hence option 2 has multiple answers. It goes to show again how much I suck at picking magic numbers (or numbers at all). I have to keep reminding myself to test for fractions too. Ugh. Is there any other general approach to this problem besides picking numbers? Hello Blackbox... What Bunuel intends to says consider value of x between \(1/2<x\leq{0}\).. For option 2  if x were 1/2 then x+1 would be 1/2. Given statement of 0 closer to x than x+1 is busted with two answers (0 is as much closer to 1/2 as it is to 1/2). Hence option 2 has multiple answers. That is why Bunuel says consider a number greater than 1/2.. Consider x=1/3...so x=1/3 and x+1 =2/3 : x is closer to 0 then x+1.. Look at mau5 method above...it is easier to understand and follow..
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Re: Is the number x positive? [#permalink]
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02 Feb 2016, 19:18
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rrsnathan wrote: Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1. 1. A is positive. otherwise, if x is negative, then negative minus negative makes an even bigger negative, and the new number would be even farther from 0. 2. it can be x=1 and x+1 =2, and 0 is closer to x, or it can be: x=0.1 and x+1=0.9. x is closer to x than to x+1. since we have 2 outcomes, B is not sufficient. A.



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Re: Is the number x positive? [#permalink]
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09 Feb 2017, 12:07
gmatsheeba wrote: rrsnathan wrote: Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1. Ans should be A. 1. x=2, x1 =1 Yes x=1, x1 =2 Yes 2. x=3,x+1 =2 No x=1/3, x+1=2/3 Yes Hence A is sufficient. Please press KUDOS if my post HELPED !! What if x=0 in the first case?



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Is the number x positive? [#permalink]
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Updated on: 16 Apr 2018, 12:30
rrsnathan wrote: Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1. Another approach is the sketch the cases on a number line. Target question: Is x positive? Statement 1: On the number line, 0 is closer to x – 1 than to x. First, recognize that x1 will always be to the left of x. Second, recognize that there are 3 possible ways to place x1 and x with relation to zero. If zero is closer to x1 than to x, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios. If case #1 is true, we can see that x must be positiveIf case #3 is true, we can see that x must be positiveSince both possible cases yield the same answer to the target question, we can answer the target question with certainty. So, statement 1 is SUFFICIENT Statement 2: On the number line, 0 is closer to x than to x + 1.Recognize that x+1 will always be to the right of x. Also recognize that there are 3 possible ways to place x and x+1 with relation to zero. If zero is closer to x than to x+1, then we can rule out case #2, leaving us with cases #1 and #3 as possible scenarios. If case #1 is true, we can see that x is negativeIf case #3 is true, we can see that x is positiveSince the two possible cases yield different answers to the target question, we cannot answer the target question with certainty. So, statement 2 is NOT SUFFICIENT Answer = Cheers, Brent
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Is the number x positive? [#permalink]
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22 Feb 2018, 11:00
samridhi30 wrote: gmatsheeba wrote: rrsnathan wrote: Is the number x positive?
(1) On the number line, 0 is closer to x – 1 than to x. (2) On the number line, 0 is closer to x than to x + 1. Ans should be A. 1. x=2, x1 =1 Yes x=1, x1 =2 Yes 2. x=3,x+1 =2 No x=1/3, x+1=2/3 Yes Hence A is sufficient. Please press KUDOS if my post HELPED !! What if x=0 in the first case? ===========>>>>>>>>>>> 1) x1<x square both side 12x<0 x>0.5 => x>0.5 => A or D 2) x<x+1 square both 0<2x+1 x> 0.5 but as we know x can never be negative , but it can be zero 0 => x>=0 so x=0 and x>0 are solution therfore insufficient A is winner
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