Is the number x positive? : Data Sufficiency (DS)

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18 Sep 2013, 06:35

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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.

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.

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18 Sep 2013, 06:47

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rrsnathan

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-(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.

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09 Oct 2013, 02:45

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Forgot to substitute the fractional value

((((

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03 Nov 2014, 02:04

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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

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03 Nov 2014, 02:16

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Blackbox

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.

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05 Nov 2014, 00:50

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Bunuel

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 :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?

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05 Nov 2014, 01:04

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Blackbox

Bunuel

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 :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?

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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02 Feb 2016, 19:18

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rrsnathan

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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09 Feb 2017, 12:07

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gmatsheeba

rrsnathan

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, 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?

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22 Feb 2018, 11:00

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samridhi30

gmatsheeba

rrsnathan

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, 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

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GMATPrepNow

rrsnathan

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.

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.

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15 Apr 2019, 16:33

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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.

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16 Apr 2019, 02:43

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5JUwiu4

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.

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

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16 Apr 2019, 07:49

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5JUwiu4

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.

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

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