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

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.

Answer: C.
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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.

Answer: C.

Hope it's clear.
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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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Hi Bunuel,
can you please explain:
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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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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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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\(x=|3x-2|\)

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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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..

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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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!

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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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.

Bunuel, I have a question on the part in red. Shouldn't it actually be:

We know that for \(|x|\):
When \(x<{0}\), then \(|x|=-x\); (I have changed the "less than or equal to" to only "less than")
When \(x\geq{0}\), then \(|x|=x\).

Because we should consider 2 cases -
a) greater than or equal to zero
AND
b) less than zero. [Not less than or equal to zero]

In the part B of your solution we are also considering the case where \(x=1\), right? If this is the case, how can \(|x-1|\) be \(-x +1\)? At \(x=1\), I am guessing \(|x-1|\) = \(x-1\).

No, it works with = sign as well: |0|=0=-0.

If x=1, then |x-1|=0 and -x+1=0 too.
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For Such questions the first and most important question is TO THINK, for what vale of x, will |x| + |x -1| = 1 be true

The ranges of values to be checked are
1) between 0 and 1
2) between 0 and -1
3) Greater than 1
4) Less than -1

After a few hit and trial you may comfortably arrive at the range 0 ≤ x ≤ 1 for which the above expression will be correct

Statement 1: x ≥ 0

x may be between 0 and 1 (Answer to the question 'YES') OR
x may be Greater than 1 (Answer to the question 'NO')
NOT SUFFICIENT

Statement 2: x ≤ 1

x may be between 0 and 1 (Answer to the question 'YES') OR
x may be Less than 0 (Answer to the question 'NO')
NOT SUFFICIENT

Combining the two Statements
x ≥ 0 and x ≤ 1
i.e. 0 ≤ x ≤ 1
SUFFICIENT

Answer: option C

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x ≥ 0
(2) x ≤ 1

If we modify the question, the range becomes 0≤x≤1, and the answer becomes (C), as the range of conditions 1 and 2 is 0≤x≤1

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x ≥ 0
(2) x ≤ 1

In general, when absolute values appear in addition, the answer is between. That is, just like the question above, when |x|+|x-1|=1?, the between is 0<=x<=1, which is yes and therefore sufficient. In 1) & 2), 0<=x<=1 -> yes, which is sufficient. Therefore, the answer is C.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.