Is the number x positive?

Question

(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.

BrentGMATPrepNow · Accepted Answer

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.
 https://i.imgur.com/JqOEJ0c.png

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.
 https://i.imgur.com/LdssvzM.png 
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

gmatsheeba · Answer

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.

neelzwiz · Answer

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 x-1 = = 0.2 => x = 1.2. so x +ve

Take choice 2.

let x -1 = - 0.4 => x = 0.4, so X +ve.
let x-1 = -2.4 => x = -1.4, so X -ve.

Hence the ans is A.

Hope this helps.

mau5 · Answer

|x| denotes the distance between 0 and x on the number line. F.S 1 states that |0-(x-1)|<|0-x| --> |x-1|<|x| --> Square on both sides, and subtract --> 1*(2x-1)>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.

mohnish104 · Answer

Forgot to substitute the fractional value

((((

Blackbox · Answer

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.

Bunuel · Answer

For (2) consider the values of x from -1/2 (not inclusive) to 0 (inclusive) to get that (2) is NOT in fact sufficient.

Blackbox · Answer

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 :roll:

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?

WoundedTiger · Answer

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..

mvictor · Answer

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.

samridhi30 · Answer

What if x=0 in the first case?

sahilvijay · Answer

===========>>>>>>>>>>> 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

farsar · Answer

Thank you for this answer! Its by far the clearest. I loved this approach.

5JUwiu4 · Answer

GMATPrepNow

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.

ruchik · Answer

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.

hope it helped.

Please give kudos if you like the explanation.

BrentGMATPrepNow · Answer

We could have x = 0.8, and x-1 = -0.2
In this case, 0 is closer to x-1 than to x.

Cheers, 
Brent

bumpbot · Answer

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Is the number x positive?

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