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(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.
First, recognize that x-1 will always be to the left of x.
Second, recognize that there are 3 possible ways to place x-1 and x with relation to zero.
Target question:Is x positive?
Statement 1: On the number line, 0 is closer to x – 1 than to x. If zero is closer to x-1 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 positive If case #3 is true, we can see that x must be positive Since 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 negative If case #3 is true, we can see that x is positive Since 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
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; => x-1 = -3. (Given answer choice is not true/satisfied, so x can't be negative) Consider x= 22; => x-1 =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)
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; => x-1 = -3. (Given answer choice is not true/satisfied, so x can't be negative) Consider x= 22; => x-1 =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.
_________________
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?
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..
_________________
“If you can't fly then run, if you can't run then walk, if you can't walk then crawl, but whatever you do you have to keep moving forward.”
(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.
(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, x-1 =1 Yes x=-1, x-1 =-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) |x-1|<|x| square both side 1-2|x|<0 |x|>0.5 => x>0.5 => A or D
2) |x|<|x+1| square both 0<2|x|+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
_________________
Give Kudos for correct answer and/or if you like the solution.
(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.
First, recognize that x-1 will always be to the left of x.
Second, recognize that there are 3 possible ways to place x-1 and x with relation to zero.
Target question:Is x positive?
Statement 1: On the number line, 0 is closer to x – 1 than to x. If zero is closer to x-1 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 positive If case #3 is true, we can see that x must be positive Since 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 negative If case #3 is true, we can see that x is positive Since 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: A
Cheers, Brent
Thank you for this answer! Its by far the clearest. I loved this approach.
Loved the explanation, but I think I'm missing something.
In statement 1 it states " On the number line, 0 is closer to x – 1 than to x."
You explain x could be on the right of zero and x-1 to the left (0 in the middle). If this is true x must be less than 1 to jump to the negative side of the number line. In this case isn't x and x-1 equidistant from 0? So how can 0 be closer to x-1 than to x?
Loved the explanation, but I think I'm missing something.
In statement 1 it states " On the number line, 0 is closer to x – 1 than to x."
You explain x could be on the right of zero and x-1 to the left (0 in the middle). If this is true x must be less than 1 to jump to the negative side of the number line. In this case isn't x and x-1 equidistant from 0? So how can 0 be closer to x-1 than to x?
Thanks in advance.
Hi, The way i looked at option 1 was 0 is closer to x-1 than x. Now if x is negative, x-1 is going to be more negative and hence going to be farther away from 0. So if we attack the problem in reverse away it might be easier. Can we find a negative number x, such that x-1 is closer to 0 than x? the answer is no hence the option is sufficient.
Loved the explanation, but I think I'm missing something.
In statement 1 it states " On the number line, 0 is closer to x – 1 than to x."
You explain x could be on the right of zero and x-1 to the left (0 in the middle). If this is true x must be less than 1 to jump to the negative side of the number line. In this case isn't x and x-1 equidistant from 0? So how can 0 be closer to x-1 than to x?
Thanks in advance.
We could have x = 0.8, and x-1 = -0.2 In this case, 0 is closer to x-1 than to x.
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; => x-1 = -3. (Given answer choice is not true/satisfied, so x can't be negative) Consider x= 22; => x-1 =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.
Hi Banuel,
If we put x=1/2 in statement 1 then?
gmatclubot
Re: Is the number x positive?
[#permalink]
19 Apr 2019, 03:55
close
46% (01:44) correct 
((((
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?
