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|x - 1|

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Source: — Data Sufficiency |
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by pops » Sun Mar 28, 2010 6:34 am
neoreaves wrote:Is |x - 1| less than 1 ?

1). (x - 1)^2 less than and equal to 1
2). (x^2) - 1 greater than 0
1. alone:
since, (x-1)^2 <= 1
=> |x-1| <= 1
(always remember whenever you take root on both side of inequality, its absolute value that comes out i.e.
(a-b)^2 <= c^2
=> |a-b| <= |c|)
hence, yes |x-1| is less than 1!

2. alone:
since, x^2 -1 >=0
=> x^2>=1
=> either x >= 1 or x <= -1
hence, not sure whether |x-1| would be less than 1
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by dxgamez » Sun Mar 28, 2010 4:35 pm
hi pops,

why is the statement 2, the inequality sign is >=?

I thought st 2 stated (x^2) - 1 is greater than 0. Thus

(x^2) - 1 > 0

I thought x must always be more than 1 or less than -1 cos of the square. As such |x-1| is always more than 1. Sufficient, right?
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by pops » Sun Mar 28, 2010 10:06 pm
thanks for pointing my mistake.. the inequality should not have =sign.. its pure inequality

so, now
2 alone:
since, x^2 -1 >0
=> x^2>1
=> either x > 1 or x < -1

still, not sure whether |x-1| would be less than 1
take x=1.5, then |x-1| = 0.5 < 1
take x=2.5, then |x-1|=1.5 > 1

hence insufficient
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by srinivasarajui » Mon Mar 29, 2010 3:59 am
So we can conclude the answer is A?
Srinu
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by papumba2011 » Mon Mar 29, 2010 12:15 pm
pops wrote:
neoreaves wrote:Is |x - 1| less than 1 ?

1). (x - 1)^2 less than and equal to 1
2). (x^2) - 1 greater than 0
1. alone:
since, (x-1)^2 <= 1
=> |x-1| <= 1
(always remember whenever you take root on both side of inequality, its absolute value that comes out i.e.
(a-b)^2 <= c^2
=> |a-b| <= |c|)
hence, yes |x-1| is less than 1!

2. alone:
since, x^2 -1 >=0
=> x^2>=1
=> either x >= 1 or x <= -1
hence, not sure whether |x-1| would be less than 1
I think we missed the '=' sign in the first statement.
If (x-1)^2 <=1
this implies |x -1|<= 1 i.e. -1<=|x-1|<=1
Since |x-1| can be equal to 1 we cannot say surely that |x-1| is less than 1. Insufficient.

From stmt 2

(x^2) - 1 > 0

Hence x must always be more than 1 or less than -1 cos of the square. As such |x-1| is always more than 1. Sufficient.

Hence answer is B.
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by neoreaves » Mon Mar 29, 2010 10:47 pm
papumba2011 wrote:
pops wrote:
neoreaves wrote:Is |x - 1| less than 1 ?

1). (x - 1)^2 less than and equal to 1
2). (x^2) - 1 greater than 0
1. alone:
since, (x-1)^2 <= 1
=> |x-1| <= 1
(always remember whenever you take root on both side of inequality, its absolute value that comes out i.e.
(a-b)^2 <= c^2
=> |a-b| <= |c|)
hence, yes |x-1| is less than 1!

2. alone:
since, x^2 -1 >=0
=> x^2>=1
=> either x >= 1 or x <= -1
hence, not sure whether |x-1| would be less than 1
I think we missed the '=' sign in the first statement.
If (x-1)^2 <=1
this implies |x -1|<= 1 i.e. -1<=|x-1|<=1
Since |x-1| can be equal to 1 we cannot say surely that |x-1| is less than 1. Insufficient.

From stmt 2

(x^2) - 1 > 0

Hence x must always be more than 1 or less than -1 cos of the square. As such |x-1| is always more than 1. Sufficient.

Hence answer is B.

I dont think it is B ...

(x^2) - 1 = (x+1)(x-1) > 0 --> take negative numbers x = -5 --> (-5 + 1) (-5 - 1) = (-4)(-6) = 24 > 0

we can also try x = 5 --< (5+1)(5-1) = (6)(4) = 24 Thus insufficient


Now that we know A and B are insufficient lets try C

(x - 1)^2 < = 1 ... thus (x-1) is either 1 or a fraction

and (x^2) - 1 > 0 ..Thus |x| > 1

Lets try x < 0 |x| > 1 --> lets say x = -1.1 then x - 1 = -1.1 -1 = -2.2 thus x -1 < 1 but this doesnt satisfy 1) where x- 1 has to be fraction or 1. Thus not possible

Lets try x > 0 |x| > 1 --> x = 1.1 --> x - 1 = 0.1 thus x - 1 is a fraction

Thus IMO C
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by eaakbari » Mon Mar 29, 2010 11:26 pm
I think Neoreves is right but Is that the OA?
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by akhpad » Tue Mar 30, 2010 3:06 am
|x - 1| < 1 ? Please remember that here equal to is not mentioned.
-1 < x-1 < 1
0 < x < 2

Statement 1:
(x - 1)^2 <= 1
-1 <= (x - 1) <= 1
0 <= x <= 2
|x - 1| <= 1
Here, we have two options < and =.
Insufficient

Statement 2:
(x^2) - 1 > 0
x^2 > 1
-1 > x > 1
So part of the region comes under 0 < x < 2 and part of the region comes outside
Insufficient

Statement 1 and 2
0 <= x <= 2 and -1 > x > 1
result into 1 < x <= 2
here, 1 < x < 2 is satisfying but x = 2 not
Insufficient

Ans E
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by dxgamez » Tue Mar 30, 2010 7:01 pm
i agree with akhp77.

when x=2, it satisfies both statement 1 and 2. but not the question stem.

however when x=1.5, it satisfies all st1, 2, and question stem.

thus insufficient.
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by Shawshank » Wed Mar 31, 2010 5:16 am
Guys.. I question from my side

Is |x - 1| less than 1 ?

1). (x - 1)^2 less than and equal to 1
2). (x^2) - 1 greater than 0

I am still not clear why the 2 condition is insufficient

x2 - 1 > 0

Then x can be greater than or less than 1 BUT,,, it is not equal to 1

SO !x-1! has to be greater than 1 right,,

please correct me..

IMO -- B
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by akhpad » Wed Mar 31, 2010 5:26 am
Ok, look here

Statement 2:
(x^2) - 1 > 0
x^2 > 1
-1 > x > 1
-2 > x-1 > 0
|x-1| > 0
|x-1| > 2

It gives me 2 solution

|x-1| > 2 is fine
|x-1| > 0 can not conclude whether |x-1| < 1 or not.
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by Shawshank » Wed Mar 31, 2010 7:55 am
akhp77 wrote:Ok, look here

Statement 2:
(x^2) - 1 > 0
x^2 > 1
-1 > x > 1
-2 > x-1 > 0
|x-1| > 0
|x-1| > 2

It gives me 2 solution

|x-1| > 2 is fine
|x-1| > 0 can not conclude whether |x-1| < 1 or not.
Thanks Man
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