Is |x| + |x -1| = 1? : GMAT Data Sufficiency (DS)
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# Is |x| + |x -1| = 1?

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23 Oct 2009, 08:55
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Is |x| + |x -1| = 1?

(1) x ≥ 0
(2) x ≤ 1
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Re: Absolute values DS questions... [#permalink]

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23 Oct 2009, 20:26
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I will go with B

As the absolute value can not be negative , so for x <= 1
x can be in the range of 0 to 1 ( inclusive)

putting values :-

|0| + |0 -1| = 1
|0.65| + |0.65 -1| = 0.65 + 0.35 = 1

I intially thought C , but realised my mistake
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Re: Absolute values DS questions... [#permalink]

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23 Oct 2009, 21:31
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This one is very tricky!

Is |x| + |x -1| = 1?
(1) x ≥ 0
(2) x ≤ 1

Q is $$|x| + |x -1| = 1$$. Let's check when this equation holds true. We should consider three ranges (as there are two check points $$x=0$$ and $$x=1$$):

A. $$x<0$$ --> $$-x-x+1=1$$ --> $$x=0$$, but this solution is not valid as we are checking the range $$x<0$$;

B. $$0\leq{x}\leq{1}$$ -->$$x-x+1=1$$ --> $$1=1$$, which is true. That means that for ANY value from the range $$0\leq{x}\leq{1}$$, equation $$|x| + |x -1| = 1$$ holds true.

C. $$x>1$$ --> $$x+x-1=1$$ --> $$x=1$$, but this solution is not valid as we are checking the range $$x>1$$.

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

Statements:
(1) $$x\geq{0}$$. Not sufficient, as $$x$$ must be also $$\leq{1}$$;
(2) $$x\leq{1}$$. Not sufficient, as $$x$$ must be also $$\geq{0}$$;

(1)+(2) $$0\leq{x}\leq{1}$$, exactly the range we needed. Sufficient.

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Re: Absolute values DS questions... [#permalink]

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23 Oct 2009, 22:35
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Hi Bunuel,
0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true.

we cannot derive anything in this interval, does it mean that all values in this interval satisfy the equation ? This is something new for me...do you have any links for this? I thought, since we cannot derive anything, this interval is also out of scope.

Though, I got the answer by some quick number substitutions.
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Re: Absolute values DS questions... [#permalink]

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23 Oct 2009, 22:56
Economist wrote:
Hi Bunuel,
0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true.

we cannot derive anything in this interval, does it mean that all values in this interval satisfy the equation ? This is something new for me...do you have any links for this? I thought, since we cannot derive anything, this interval is also out of scope.

Though, I got the answer by some quick number substitutions.

Well knew that this part needs more explanation.

When $$x$$ is in the range $$0\leq{x}\leq{1}$$, equation $$|x|+|x-1|=1$$ will become: $$x-x+1=1$$ --> $$1=1$$. Which is true, indeed $$1=1$$. But what does that mean? This means that when $$x$$ is in this range, equation takes the form of $$x-x+1=1$$ and value of $$x$$ does not affects the equation as it cancels out. OR in other words any $$x$$ from this range makes equation to hold true.

You can try some number picking from this range to see that.

Hope it's clear. Please tell me if it needs more clarification.

BTW what answer did you get?
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Re: Absolute values DS questions... [#permalink]

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25 Oct 2009, 14:08
bunuel /economist,

Can you please clear my concept here:-

Is |x| + |x -1| = 1? i thought that an absolute can not be negative , so x can not be less than 0

like in case of |x| = 3x – 2 , absolute value can not be negative

why did you consider x <0 ? can you please explain or point me to some document , may be this is a basic question , but damn i am confused
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25 Oct 2009, 15:25
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ichha148 wrote:
bunuel /economist,

Can you please clear my concept here:-

Is |x| + |x -1| = 1? i thought that an absolute can not be negative , so x can not be less than 0

like in case of |x| = 3x – 2 , absolute value can not be negative

why did you consider x <0 ? can you please explain or point me to some document , may be this is a basic question , but damn i am confused

When we have equation of a type: $$|x|=3x-2$$, we should consider two cases:

1. $$x<0$$ --> $$|x|=-x$$ --> $$-x=3x-2$$ --> $$x=\frac{1}{2}$$, not a vaild solution for this range as we are considering $$x<0$$. Which means that when $$x<0$$ equation $$|x|=3x-2$$ has no real roots.
2. $$x\geq{0}$$ --> $$|x|=x$$ --> $$x=3x-2$$ --> $$x=1$$, good as $$x>0$$.
Final answer: equation $$|x|=3x-2$$ has one root $$x=1$$.

BUT the problem above can be solved in another way:
You should notice following: $$3x-2$$ is equal to "something" and that "something", as it is an absolute value, cannot be negative. So, we conclude that $$3x-2$$ cannot be negative. $$3x-2\geq{0}$$ --> $$x\geq{\frac{2}{3}}$$. After that when we definitely know that $$x\geq{\frac{2}{3}}>0$$, we can check only ONE range for $$x$$, which is $$x\geq{\frac{2}{3}}>0$$ and write $$|x|=x$$. So, equation will become $$x=3x-2$$ --> $$x=1$$.

But if it were: $$x=|3x-2|$$, we can definitely say that $$x\geq{0}$$, but we cannot write $$x=3x-2$$. We should still check two ranges:
1. $$0\leq{x}\leq{\frac{2}{3}}$$ --> $$x=-3x+2$$ --> $$x=\frac{1}{2}$$;
2. $$\frac{2}{3}<x$$ -->$$x=3x-2$$ --> $$x=1$$.
So, this equation has two roots 1/2 and 1.

Third case: $$|x|=|3x-2|$$. In this case we should consider 3 ranges:
1. $$x<0$$ --> $$-x=-3x+2$$ --> $$x=1$$, not good as $$x<0$$;
2. $$0\leq{x}\leq{\frac{2}{3}}$$ --> $$x=-3x+2$$ --> $$x=\frac{1}{2}$$;
3. $$\frac{2}{3}<x$$ --> $$x=3x-2$$ --> $$x=1$$.
Again two roots 1/2 and 1.

Back to our original question:
Is $$|x|+|x-1|=1$$?
(1) x ≥ 0
(2) x ≤ 1

Absolute value, $$|x|$$, cannot be negative, BUT $$x$$ in it can be. $$|x|\geq{0}$$ but $$x$$ can be $$-3<0$$ --> $$|-3|=3>0$$.

Basically here we have the sum of two absolute values, or the sum of two non negative values totaling 1. But again $$x$$ in it can take negative values and still these two can give us 1 as their sum.

Q: is $$|x|+|x-1|=1$$?
This equation can be true in some ranges of $$x$$ and false in another, or can be true/false in all of the ranges. Thus we should check ALL ranges for $$x$$ to answer the question. How to do this? There are two crucial points when absolute values $$|x|$$ and $$|x -1|$$ flip signs, two check points 0 ($$x=0$$) and 1 $$(x-1=0$$, $$x=1$$). Thus three ranges must be checked:

1. $$x<0$$ --> $$-x-x+1=1$$ --> $$x=0$$. Not good as $$x<0$$;

2. $$0\leq{x}\leq{1}$$ --> $$x-x+1=1$$ --> $$1=1$$. Which means that for ANY value from the range $$0\leq{x}\leq{1}$$, equation $$|x|+|x-1|=1$$ holds true.

3. $$x>1$$ --> $$x+x-1=1$$ --> $$x=1$$. Not good as $$x>1$$.

Note that we aren't considering the statements (1) and (2) yet. We are just checking in which range the equation $$|x|+|x-1|=1$$ holds true. And from 1. 2. and 3. we get that it's true ONLY in the range $$0\leq{x}\leq{1}$$ and not true in all other ranges. But at this stage we still don't know in which range $$x$$ is.

Moving to the statements:
(1) x ≥ 0. Not sufficient, as x must be also <=1
(2) x ≤ 1. Not sufficient, as x must be also >=0.

(1)+(2) $$0\leq{x}\leq{1}$$, exactly the range we needed. Sufficient.

Hope it's clear.
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25 Oct 2009, 17:48
bunuel - thanks for explaning this in such a detail ,i really appreciate this ,+1 to you

1 more doubt from your explaination and hope you don't mind in explaining

But if it were:x= |3x – 2|, we can definitely say that x>=0, but we can not write x=3x-2. We should check two ranges:
0<=x<=2/3 --> x=-3x+2 --> x=1/2
2/3<x --> x=3x-2 --> x=1.
So, this equation has two roots 1/2 and 1.

How did you reach at these ranges - the way i think is |3x-2| can be either +ve or -ve

3x-2 >0 ---> x >2 /3
3x-2 <0 ---> x < 2/3 --> how did you changed this range to 0<= x <= 2/3
is this because absolute value can not be negative , but why the equal sign
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Re: Absolute values DS questions... [#permalink]

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25 Oct 2009, 18:20
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ichha148 wrote:
bunuel - thanks for explaning this in such a detail ,i really appreciate this ,+1 to you

1 more doubt from your explaination and hope you don't mind in explaining

But if it were:x= |3x – 2|, we can definitely say that x>=0, but we can not write x=3x-2. We should check two ranges:
0<=x<=2/3 --> x=-3x+2 --> x=1/2
2/3<x --> x=3x-2 --> x=1.
So, this equation has two roots 1/2 and 1.

How did you reach at these ranges - the way i think is |3x-2| can be either +ve or -ve

3x-2 >0 ---> x >2 /3
3x-2 <0 ---> x < 2/3 --> how did you changed this range to 0<= x <= 2/3
is this because absolute value can not be negative , but why the equal sign

$$x=|3x-2|$$

Left Hand Side is $$x$$;
RHS is $$|3x-2|$$;

RHS is absolute value, which means that it's never negative. If $$RHS\geq{0}$$ (more than or equal to zero), hence LHS, $$x$$, must be also $$\geq{0}$$. $$x\geq{0}$$.

Actually we should check $$x=|3x-2|$$ for two ranges.
1. $$3x-2\leq{0}$$ --> $$x\leq{\frac{2}{3}}$$ (Note I'm saying that $$3x-2\leq{0}$$, not $$|3x-2|\leq{0}$$, the something that is in || can be negative, but the absolute value of $$3x-2$$ which is $$|3x-2|$$ can not.)
2. $$3x-2>0$$ --> $$x>\frac{2}{3}$$ (I put = in first, because $$|3x-2|$$ can be zero thus zero must be included in either of ranges we check)

So TWO ranges $$x\leq{\frac{2}{3}}$$ and $$x>\frac{2}{3}$$.

BUT since we already determined that $$x\geq{0}$$, we could narrow the first range $$x\leq{\frac{2}{3}}$$ to $$0\leq{x}\leq{\frac{2}{3}}$$. In fact if you write just $$x\leq{\frac{2}{3}}$$ it won't affect anything in this case, I just wanted to demonstrate what you can notice at first glance when looking at the equations of these type. Meaning, when we have $$x=|3x-2|$$, it should be clear that as RHS is absolute value, thus LHS, $$x$$ must be more or equal to zero, $$x\geq{0}$$.

The way you've written is almost right:
$$3x-2>0$$ ---> $$x>\frac{2}{3}$$;
$$3x-2<0$$ ---> $$x<\frac{2}{3}$$;

You should just add $$=$$ sign in one of them, as to include possibility that $$3x-2$$ can be equal to 0.
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Re: Absolute values DS questions... [#permalink]

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25 Oct 2009, 19:08
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I got C by substituting some numbers:

Is |x| + |x -1| = 1? lets call this the equation.

(1) x ≥ 0
(2) x ≤ 1

S1) X>=0 when x= 1, equation holds true, when x =0 equation holds true,but when x = 3 equation doesn't hold true - insuff

S2) x<=1 when x=0 equation holds true, x=-1 equation doesn't hold true - insuff

Combining S1 and S2, we have 0<=x<=1

and using 0, 1 for x we know equation holds true, try 0.5, 0.3 and 0.7 for equation , still holds true - hence C is correct.

I like Bunuel's explanation though - as almost always..
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Re: Absolute values DS questions... [#permalink]

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26 Oct 2009, 18:22
Bunuel wrote:
ichha148 wrote:
bunuel - thanks for explaning this in such a detail ,i really appreciate this ,+1 to you

1 more doubt from your explaination and hope you don't mind in explaining

But if it were:x= |3x – 2|, we can definitely say that x>=0, but we can not write x=3x-2. We should check two ranges:
0<=x<=2/3 --> x=-3x+2 --> x=1/2
2/3<x --> x=3x-2 --> x=1.
So, this equation has two roots 1/2 and 1.

How did you reach at these ranges - the way i think is |3x-2| can be either +ve or -ve

3x-2 >0 ---> x >2 /3
3x-2 <0 ---> x < 2/3 --> how did you changed this range to 0<= x <= 2/3
is this because absolute value can not be negative , but why the equal sign

x= |3x – 2|

Left Hand Side=x
RHS=|3x – 2|

RHS is absolute value, which means that it's never negative. If RHS>=0 (more or equal), hence LHS x, must be also >=0. x>=0.

Actually we should check x= |3x – 2| for two ranges.
1. 3x-2<=0, x<=2/3 (Note I'm saying that 3x-2<=0, not |3x – 2|<=0, the something that is in || can be negative, but the absolute value of 3x-2 which is |3x – 2| can not.)
2. 3x-2>0, x>2/3 (I put = in first, because |3x – 2| can be zero thus zero must be included in either of ranges we check)

So TWO ranges x<=2/3 and x>2/3.

BUT since we already determined that x>=0, we could narrow the first range x<=2/3 to 0<=x<=2/3. In fact if you write just x<=2/3 it won't affect anything in this case, I just wanted to demonstrate what you can notice at first glance when looking at the equations of these type. Meaning, when we have x= |3x – 2|, it should be clear that as RHS is absolute value, thus LHS x must be more or equal to zero, x>=0.

The way you've written is almost right:
3x-2 >0 ---> x >2 /3
3x-2 <0 ---> x < 2/3

You should just add = sign in one of them, as to include possibility that 3x-2 can be equal to 0.

Thanks again Bunuel , you are awesome
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Re: Absolute values DS questions... [#permalink]

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26 Oct 2009, 20:15
could it get any better. Thank a lot for this explaination
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18 Sep 2010, 02:27
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Bunuel,

I am confused here..
"B. 0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true."

I am confused that how first x is +ive and second one -ve...after we take condition 0<=x<=1?
Pls, explain.

Thanks!
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18 Sep 2010, 02:44
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samark wrote:
Bunuel,

I am confused here..
"B. 0<=x<=1 --> x-x+1=1 --> 1=1. Which means that for ANY value from the range 0<=x<=1, equation |x| + |x -1| = 1 holds true."

I am confused that how first x is +ive and second one -ve...after we take condition 0<=x<=1?
Pls, explain.

Thanks!

We know that for $$|x|$$:
When $$x\leq{0}$$, then $$|x|=-x$$;
When $$x\geq{0}$$, then $$|x|=x$$.

We have $$|x| + |x -1| = 1$$.

Now for the range: $$0\leq{x}\leq{1}$$ --> $$|x|=x$$ (as $$x$$ in given range is positive) and $$|x-1|=-(x-1)=-x+1$$ (as expression $$x-1$$ in the given range is negative, to check this try some $$x$$ from this range, let $$x=-0.5$$ then $$x-1=0.5-1=-0.5=negative$$). So $$|x| + |x -1| = 1$$ in this range becomes: $$x-x+1=1$$ --> $$1=1$$, which is true. That means that for ANY value from the range $$0\leq{x}\leq{1}$$, equation $$|x| + |x -1| = 1$$ holds true.

Hope it's clear.
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18 Sep 2010, 02:56
Oh..so it comes from: $$|x-1|=-(x-1)=-x+1$$ (as expression $$x-1$$ in the given range is negative).

Thanks a ton..I shud've realized it before.. and all the best for ur G.mat!
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Re: Absolute values DS questions... [#permalink]

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18 Sep 2010, 06:02
cano wrote:
How do you know that the number is an integer?

Are you referring to the original question? If yes then given equation $$|x| + |x -1| = 1$$ holds true ONLY in the range $$0\leq{x}\leq{1}$$, so it's not necessary for $$x$$ to be an integer, it can be for example: 1/2 or 3/4, basically any fraction in the range (0,1).
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Re: Absolute values DS questions... [#permalink]

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18 Sep 2010, 06:09
cano wrote:
How do you know that the number is an integer?

Cano,

We are not assuming any integer values. Whether, value of x is integer or fractions(decimals)...the explanation above proves the answer. What we are dealing herewith are ranges after knowing boundary values..(as you can see in Bunuel explanation), viz:
A. x<0 B. 0<=x<=1 C. x>1

We find that only B. 0<=x<=1 holds true..after plugging in original statement. Until now, we haven't seen statements (1) and (2) yet. We just checked out of which of three range the equation "|x| + |x -1| = 1 holds true". So, our mission now becomes to verify whether x falls between range 0 to 1 or out of this range (As it is a Yes or No qus)

Statement 1: told us x ≥ 0. But, what if it is greater than 2. e.g: x=8/6
Statement 2: told us x ≤ 1. But, what if it is less than 0..e.g x=-3.7

Combining both 1 +2 tells that 0<=x<=1. So, x value can be either 0, 0.6 or 1. (integer or decimal)
This proves that |x| + |x -1| = 1. Choice C.

Cheers!
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Re: Absolute values DS questions... [#permalink]

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18 Sep 2010, 16:07
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I got C. I plugged in numbers for each statement.
If x=0, then true. If x=1, then true. If x=2, then not true. S1 not sufficient
If x=1, then true. If x=-2, then not true. S2 not sufficient
if x is between 0 and 1 inclusive that means we plug in fractions (plus we already know that it's true for 0 and 1). No matter what fraction x represents 1-x will always give the value needed to add to x to make it = 1. Thus C is sufficient.
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink]

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18 Nov 2012, 00:00
Hi buneul,

in the second range why do we take the less than equal to sign

0<=x<=1?

When i was going through certain other problems , we do not consider <= or >= in all cases.

Please let me know when we have to consider the <= range
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink]

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17 Aug 2013, 15:13
Shouldn't it be mandatory that X is an integer? Or that is assumed implicitly?
Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1   [#permalink] 17 Aug 2013, 15:13

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