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Is |x| + |x-1| = 1?

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Is |x| + |x-1| = 1? [#permalink] New post 16 Mar 2010, 02:43
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A
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Question Stats:

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Is |x| + |x-1| = 1?

(1) x \geq 0
(2) x \leq 1
[Reveal] Spoiler: OA

Last edited by Bunuel on 08 Mar 2013, 02:25, edited 1 time in total.
Edited the question and added the OA.
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Re: Modulus [#permalink] New post 16 Mar 2010, 02:50
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abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1

Any explanations for the given OA..

[Reveal] Spoiler:
C


stmnt1. let x= 0 then we have l 0 l + l0-1l= 1. suff
x= 10> 0 then we have l 10 l + l10-1l not equal to 1. hence insuff

stmnt2. let x = 0< 1 then we have l 0 l + l0-1l= 1. suff
let x= -5 < 1 then we l -5 l + l-5 -1l not equal to 1. hence insuff

taking together we have 0 \leq x \leq 1
if x= 0 we have l 0 l + l0-1l= 1
if x = 1 we have l 1 l + l1-1l= 1
if x = 1/2 we have l 1/2 l + l1/2-1l= 1

hence C
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Re: Modulus [#permalink] New post 16 Mar 2010, 03:06
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abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1

Any explanations for the given OA..

[Reveal] Spoiler:
C


Algebraic approach:

Let's check in which ranges of x expression |x|+|x-1|=1 holds true:

Two check points for x: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. x<0 --> -x-x+1=1 --> x=0, not good as we are checking for x<0;
B. 0\leq{x}\leq{1} --> x-x+1=1 --> 1=1, which is true. This means that in this range given equation holds true for all x-es;
C. x>1 --> x+x-1=1 --> x=1, not good as we are checking for x>1.

So we got that the equation |x|+|x-1|=1 holds true only in the range 0\leq{x}\leq{1}.

(1) x \geq 0. Not sufficient.
(2) x \leq 1. Not sufficient.

(1)+(2) gives us the range 0\leq{x}\leq{1}, which is exactly the range for which given equation holds true. Sufficient.

Answer: C.

Hope it helps.
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Re: Modulus [#permalink] New post 16 Mar 2010, 05:31
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1

Any explanations for the given OA..

[Reveal] Spoiler:
C


|x|+|x-1|, there can be 4 cases

stmt1: x >= 0 => x is positive, if we apply this to eq 1, we cannot say that 2x-1 = 1 cos x can be >= 0 but less than 1 so insuff
stmt2: x<=1 => if we apply this to eq 1, we cannot say that 2x-1 = 1 cos x<=1 can lead to negative numbers also.

both stmt taken together 0<= x <= 1
x is a fraction between 0 to 1 so always x + 1-x = 1
in our case |x| will be x and |x - 1| is actually value of 1-x
so it will always be 1 this is why C both the stmts are reqd.
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Re: Modulus [#permalink] New post 16 Mar 2010, 07:21
Thanks so much all of you! the workings definately helped to understand the OA better..

Thanks Bunnel for the algebraic approach..:-)
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Re: Modulus [#permalink] New post 16 Mar 2010, 07:39
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1
Any explanations for the given OA.
[Reveal] Spoiler:
C

Either statement alone is not sufficient to answer this question.
Together x can be 0 or 1, in both cases statement is true hence sufficient.
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Re: Modulus [#permalink] New post 18 Mar 2010, 09:10
Bunuel wrote:
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1

Any explanations for the given OA..

[Reveal] Spoiler:
C


Algebraic approach:

Let's check in which ranges of x expression |x|+|x-1|=1 holds true:

Two check points for x: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. x<0 --> -x-x+1=1 --> x=0, not good as we are checking for x<0;
B. 0\leq{x}\leq{1} --> x-x+1=1 --> 1=1, which is true. This means that in this range given equation holds true for all x-es;
C. x>1 --> x+x-1=1 --> x=1, not good as we are checking for x>1.

So we got that the equation |x|+|x-1|=1 holds true only in the range 0\leq{x}\leq{1}.

(1) x \geq 0. Not sufficient.
(2) x \leq 1. Not sufficient.

(1)+(2) gives us the range 0\leq{x}\leq{1}, which is exactly the range for which given equation holds true. Sufficient.

Answer: C.

Hope it helps.




Can you please clarify as to how is x-x+1=1 derived inB when it is provided that 0\leq{x}\leq{1} that means X is +ve.


Thanks !!!
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Re: Modulus [#permalink] New post 18 Mar 2010, 09:21
Expert's post
mustdoit wrote:
Bunuel wrote:
abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1

Any explanations for the given OA..

[Reveal] Spoiler:
C


Algebraic approach:

Let's check in which ranges of x expression |x|+|x-1|=1 holds true:

Two check points for x: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. x<0 --> -x-x+1=1 --> x=0, not good as we are checking for x<0;
B. 0\leq{x}\leq{1} --> x-x+1=1 --> 1=1, which is true. This means that in this range given equation holds true for all x-es;
C. x>1 --> x+x-1=1 --> x=1, not good as we are checking for x>1.

So we got that the equation |x|+|x-1|=1 holds true only in the range 0\leq{x}\leq{1}.

(1) x \geq 0. Not sufficient.
(2) x \leq 1. Not sufficient.

(1)+(2) gives us the range 0\leq{x}\leq{1}, which is exactly the range for which given equation holds true. Sufficient.

Answer: C.

Hope it helps.




Can you please clarify as to how is x-x+1=1 derived inB when it is provided that 0\leq{x}\leq{1} that means X is +ve.


Thanks !!!


When x is in the range 0\leq{x}\leq{1} then |x|=x and |x-1|=-(x-1)=-x+1, thus |x|+|x-1|=1 in this range becomes x-x+1=1.

Hope it's clear.
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Re: Modulus [#permalink] New post 07 Mar 2013, 20:30
I have got the answer as E. Can someone please clarify, what is wrong in my approach ?

Statement 1: X>=0

I assume X as 0
|0|+|0-1|= |0|+|1|= 1 ------------> Equation satisfied

I assume X as 3/4
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 1 is not sufficient

Statement 2: X<=1

I assume X as 1
|1|+|1-1|= |1|+|0|= 1 --------------> Equation satisfied

I assume X as 3/4 again
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 2 is not sufficient

Taking together both the equation, I assume X as 0, 3/4 and 1

When X=0, the answer is 1--------> Equation satisfied
When X=3/4, the answer is 2---------> Equation unsatisfied
When X=1, the answer is 1----------> Equation satisfied

Hence answer choice E is correct.

Clarifications needed, please.
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Re: Modulus [#permalink] New post 07 Mar 2013, 20:44
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abhi758 wrote:
Is l x l + lx-1l= 1?
(1) x \geq 0
(2) x \leq 1

Any explanations for the given OA..

[Reveal] Spoiler:
C


You can also consider using the number line method discussed here: http://www.veritasprep.com/blog/2011/01 ... s-part-ii/

l x l + lx-1l= 1 implies that sum of distance of x from 0 and from 1 should be 1. We can easily see that the distance between 0 and 1 is 1. So, x can lie on 0, on 1 or anywhere in between 0 and 1. It cannot go beyond 1 or before 0 because then the sum of the distance will become larger than 1.
Both statements together tell you that 0 <= x <= 1 and hence, the equation will hold for all such values of x.
Answer (C)
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Re: Modulus [#permalink] New post 08 Mar 2013, 02:31
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Backbencher wrote:
I have got the answer as E. Can someone please clarify, what is wrong in my approach ?

Statement 1: X>=0

I assume X as 0
|0|+|0-1|= |0|+|1|= 1 ------------> Equation satisfied

I assume X as 3/4
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 1 is not sufficient

Statement 2: X<=1

I assume X as 1
|1|+|1-1|= |1|+|0|= 1 --------------> Equation satisfied

I assume X as 3/4 again
|3/4|+|(3/4)-1|= |3/4|+|1/4|= 2 ----------> Equation unsatisfied

Statement 2 is not sufficient

Taking together both the equation, I assume X as 0, 3/4 and 1

When X=0, the answer is 1--------> Equation satisfied
When X=3/4, the answer is 2---------> Equation unsatisfied
When X=1, the answer is 1----------> Equation satisfied

Hence answer choice E is correct.

Clarifications needed, please.


The correct answer is C, not E.

If x = 3/4, then Is |x| + |x-1| = |3/4| + |3/4 - 1| = 3/4 + 1/4 = 1.

Hope it helps.
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Re: Is |x| + |x-1| = 1? [#permalink] New post 08 Mar 2013, 03:41
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abhi758 wrote:
Is |x| + |x-1| = 1?

(1) x \geq 0
(2) x \leq 1


So |x| + |x-1| can only equal 1 when

|x| = x and |x-1| = 1-x . Thus, x>=0 AND x-1=<0. Which gives, 0<=x<=1.

C.
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Re: Is |x| + |x-1| = 1? [#permalink] New post 17 Jun 2013, 07:21
Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2
x+ -(x-1)=1
x+ -x+1=1
1=1

OR

x=2
x+x-1=1
2x=2
x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

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Re: Is |x| + |x-1| = 1? [#permalink] New post 17 Jun 2013, 07:41
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Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2
x+ -(x-1)=1
x+ -x+1=1
1=1

OR

x=2
x+x-1=1
2x=2
x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!


No.
In the first case for an x=\frac{1}{2} (0\leq{x}\leq{1}) you get that the equation is true 1=1 it's like saying ALWAYS TRUE.
But in the second case for an x>1 you get x=1, but since x>1 that is not true
Statement 1 is not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink] New post 17 Jun 2013, 08:09
Ok, this is how I look at it:

x>1
|x| + |x-1| = 1
x + (x-1) = 1
2x = 2
x = 1
x = 1 which ISN'T greater than 1. Insufficient

0≤x<1
x = 1/2
|x| + |x-1| = 1
x + -(x-1) =1
x+ -x +1 = 1
0=0
I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!


Zarrolou wrote:
WholeLottaLove wrote:
Is |x| + |x-1| = 1?

For #1 (x≥0) I see two possible cases:

x=1/2
x+ -(x-1)=1
x+ -x+1=1
1=1

OR

x=2
x+x-1=1
2x=2
x=1

Wouldn't that imply that for both cases of x (where we have positive and negative values inside the | | signs) the equation is true?

Thanks!


No.
In the first case for an x=\frac{1}{2} (0\leq{x}\leq{1}) you get that the equation is true 1=1 it's like saying ALWAYS TRUE.
But in the second case for an x>1 you get x=1, but since x>1 that is not true
Statement 1 is not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink] New post 17 Jun 2013, 08:14
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WholeLottaLove wrote:
Ok, this is how I look at it:

x>1
|x| + |x-1| = 1
x + (x-1) = 1
2x = 2
x = 1
x = 1 which ISN'T greater than 1. Insufficient

0≤x<1
x = 1/2
|x| + |x-1| = 1
x + -(x-1) =1
x+ -x +1 = 1
0=0
I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!


0=0, and 1=1,... means that that equation is always true. You should read the results as follow:
x>1
x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1)
0≤x<1
0=0. So in this case the equation in TRUE( because 0=0 always)

So from statement 1 you get two different results (F/T)=> not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink] New post 17 Jun 2013, 08:33
Okay, that's what I thought. Thank you.

Zarrolou wrote:
WholeLottaLove wrote:
Ok, this is how I look at it:

x>1
|x| + |x-1| = 1
x + (x-1) = 1
2x = 2
x = 1
x = 1 which ISN'T greater than 1. Insufficient

0≤x<1
x = 1/2
|x| + |x-1| = 1
x + -(x-1) =1
x+ -x +1 = 1
0=0
I'm not sure how to read this. Does 0=0 mean that the statement is sufficient or does it mean that x=0 and because x≥0 the statement is sufficient?

Thanks!


0=0, and 1=1,... means that that equation is always true. You should read the results as follow:
x>1
x = 1 which ISN'T greater than 1. So in this case the equation if FALSE (as 1 does not equal a value greater than 1)
0≤x<1
0=0. So in this case the equation in TRUE( because 0=0 always)

So from statement 1 you get two different results (F/T)=> not sufficient
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Re: Is |x| + |x-1| = 1? [#permalink] New post 30 Jan 2014, 05:02
I went through the following link x-3-4-x-8-x-how-many-solutions-does-the-equation-148996.html and understand the following below.

Quote:
So the '=' sign can be put with x > 0 or with x < 0. We usually put it with 'x > 0' for consistency.

When we are considering ranges, say,
x < -8 ------ x is less than -8
-8 <= x < -3 ------- x is greater than or equal to -8 but less than -3
-3 <= x < 4 ------- x is greater than or equal to -3 but less than 4
x >=4 -------- x is greater than or equal to 4

We need to include the transition points (-8, -3, 4) somewhere so we include them with greater than sign.

Mind you, we could have taken the ranges as
x <= -8
-8 < x <= -3
-3 < x <= 4
x > 4


The above explanation is easily understandable but it creates a small doubt if i compare it with this question
Is |x| + |x-1| = 1?
(1) x\geq{0}
(2) x\leq{1}

Quote:
Two check points for x: 0 and 1 (x=0 and x-1=0, x=1), thus three ranges to check:
A. x<0 --> -x-x+1=1 --> x=0, not good as we are checking for x<0;
B. 0\leq{x}\leq{1} --> x-x+1=1 --> 1=1, which is true. This means that in this range given equation holds true for all x-es;
C. x>1 --> x+x-1=1 --> x=1, not good as we are checking for x>1.

So we got that the equation |x|+|x-1|=1 holds true only in the range 0\leq{x}\leq{1}


Here, we have to include the transition points (0 and 1) but only once.
Are these okay, for B. 0\leq{x}<1 --> 1=1 not in the range.
for C. x\geq{1} --> x = 1. It is in the range.

Or

A. x\leq{0} --> x= 0, It is in the range.
B. 0<{x}\leq{1} --> 1=1 It is also in the range.

Are my ranges correct ? if yes, then i still don't feel close to the answer.

Please give your views on it.

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Re: Is |x| + |x-1| = 1? [#permalink] New post 15 Feb 2014, 21:18
My answer is C
its by plugging in method

1. From 1st equation , if x = 1 , then |x| + |x-1| = 1 , then its yes
if x = 5 , 9 # 1 , so No
so Equation 1 is No
2. From 2nd equation , of x = -1 , then its No
x = 1 its yes
so equation is No

3. using both , x = 0 then its Definite yes
So C.. :)

Is my approach is Right :?:
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Re: Is |x| + |x-1| = 1? [#permalink] New post 16 Feb 2014, 19:40
Expert's post
kanusha wrote:
My answer is C
its by plugging in method

1. From 1st equation , if x = 1 , then |x| + |x-1| = 1 , then its yes
if x = 5 , 9 # 1 , so No
so Equation 1 is No
2. From 2nd equation , of x = -1 , then its No
x = 1 its yes
so equation is No

3. using both , x = 0 then its Definite yes
So C.. :)

Is my approach is Right :?:


Using both, you get 0 <= x <= 1
x can lie anywhere from 0 to 1. Just trying the extreme values is not a good idea. You must try some values from the middle too i.e. x = 1/2, 1/4 etc.

Though establishing something by plugging in values is always tricky. You can try to negate a universal statement by looking for one value for which the statement doesn't hold but since you cannot try every value for which a statement must hold, its treacherous to depend on number plugging in this case. You should try to use logic.
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Re: Is |x| + |x-1| = 1?   [#permalink] 16 Feb 2014, 19:40
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