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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink]

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18 Aug 2013, 06:15

btg9788 wrote:

Shouldn't it be mandatory that X is an integer? Or that is assumed implicitly?

No you should not assume it to be a integer, if it is not mentioned. The problem here is the range is such that even if you pick a non integer value, you will get the answer as 1.

After combining we have 0<=x<=1

For example if x=0.33

|0.3|+|1-0.3|=.3+.7=1 . Hope it is clear
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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink]

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29 Sep 2013, 13:20

Bunuel wrote:

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.

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 \(|x-1|\) = \(x-1\).

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.

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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Re: Is |x| + |x -1| = 1? (1) x 0 (2) x 1 [#permalink]

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04 Jun 2014, 05:40

wow! That is a lot of discussion. My solution will baffle you all. Basically he is asking if the sum of the distance between 0 and x & x and 1 equal to one.

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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I would choose the graphical method for this problem.

Statements 1 and 2 are clearly insufficient on their own. However taken together we see that x lies between 1 and 0. |x| represents distance from Zero and |x-1| represents distance from 1.

now |x|+|x-1| represents total distance between 1 and 0.

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.

Exactly the way I did it, just confused by the huge discussion around this if that is sufficient enough....?!

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?

hii Bunuel i am not able to understand how do you decide the check points for such question .

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.

Bunuel,

please help me.

highlighted part: if \(x<=0--->-x-x+1=1-----> x=0\). then it becomes valid, right ?. then doesn't it change our Answer?

or when we have to consider "equal sign". \(x<=0,0<x<1 ( or ) x<0, 0<=x<1\)

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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.
_________________

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.

Hi Bunuel, I have a question in selecting ranges. Why are you selecting the range as X<0, 0≤X≤1, X>1 and not X<0, 0≤X<1 ,X≥1? If I choose the second way, solution X = 1 becomes valid in the range X≥1.

Hi Bunuel, I have a question in selecting ranges. Why are you selecting the range as X<0, 0≤X≤1, X>1 and not X<0, 0≤X<1 ,X≥1? If I choose the second way, solution X = 1 becomes valid in the range X≥1.

Let me try to answer your question.

The reason why we are taking \(x<0\) , \(0 \leq x \leq 1\) and \(x>1\) as the ranges to consider is because |x| = |x-0| and as you are given |x-1| and |x|, these expressions change their "nature" at points 0 and 1. So you need to understand what happens to |x| and |x-1| around 0 and 1.

You can take the ranges as x<0, 0≤x<1 ,x≥1, in which case you will get the following 3 cases:

1. x<0 , |x|+|x-1|=1 --> -x-x+1=1 ---> -2x=0 --> x= 0 (not possible as we have assumed that x<0). So x<0 is not a valid option.

2. 0≤x<1, |x|+|x-1|=1 --> x-x+1=1 ---> 1=1 --> this range satisfies the values and hence should be considered.

3. x≥1, |x|+|x-1|=1 --> x+x-1=1 ---> 2x=2 ---> x= 1 --> this range satisfies the values and hence should be considered. As you are getting a particular value here, you need to check for x>1 in particular as you might not get consistent value.

For checking, consider x=5, |5|+|5-1| = 5+4 = 9 \(\neq\) 1. Thus x>1 is NOT a valid range. Hence the range in question becomes \(0 \leq x \leq 1\)

Alternate solution, |x| and |x-1| can be interpreted as distance of 'x' from 0 and 1 respectively. So in effect, the question is asking you what is the range of variable 'x' for which we get sum of its distances from 0 and 1 equal to 1 unit .

When you draw the number line, you see that this sum of distances of x for 0 and 1 will be = 1 ONLY when \(0 \leq x \leq 1\) .

Both statements when combined give this information and hence C is the correct answer.

Hope this helps.

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Hi Bunuel, I have a question in selecting ranges. Why are you selecting the range as X<0, 0≤X≤1, X>1 and not X<0, 0≤X<1 ,X≥1? If I choose the second way, solution X = 1 becomes valid in the range X≥1.

Let me try to answer your question.

The reason why we are taking \(x<0\) , \(0 \leq x \leq 1\) and \(x>1\) as the ranges to consider is because |x| = |x-0| and as you are given |x-1| and |x|, these expressions change their "nature" at points 0 and 1. So you need to understand what happens to |x| and |x-1| around 0 and 1.

You can take the ranges as x<0, 0≤x<1 ,x≥1, in which case you will get the following 3 cases:

1. x<0 , |x|+|x-1|=1 --> -x-x+1=1 ---> -2x=0 --> x= 0 (not possible as we have assumed that x<0). So x<0 is not a valid option.

2. 0≤x<1, |x|+|x-1|=1 --> x-x+1=1 ---> 1=1 --> this range satisfies the values and hence should be considered.

3. x≥1, |x|+|x-1|=1 --> x+x-1=1 ---> 2x=2 ---> x= 1 --> this range satisfies the values and hence should be considered. As you are getting a particular value here, you need to check for x>1 in particular as you might not get consistent value.

For checking, consider x=5, |5|+|5-1| = 5+4 = 9 \(\neq\) 1. Thus x>1 is NOT a valid range. Hence the range in question becomes \(0 \leq x \leq 1\)

Alternate solution, |x| and |x-1| can be interpreted as distance of 'x' from 0 and 1 respectively. So in effect, the question is asking you what is the range of variable 'x' for which we get sum of its distances from 0 and 1 equal to 1 unit .

When you draw the number line, you see that this sum of distances of x for 0 and 1 will be = 1 ONLY when \(0 \leq x \leq 1\) .

Both statements when combined give this information and hence C is the correct answer.

Hope this helps.

Thanks for the awesome response! I have another question. Is there a quick way to decide between X<0, 0≤X≤1, X>1 and X<0, 0≤X<1 ,X≥1 at the start of the question?

Thanks for the awesome response! I have another question. Is there a quick way to decide between X<0, 0≤X≤1, X>1 and X<0, 0≤X<1 ,X≥1 at the start of the question?

It is dangerous to start generalizing quant concepts without actually looking a the question. The 'convention' is to take x<0, 0≤x≤1, x>1 as this will not make us commit the mistake of bringing x>1 into the picture as you had proposed.

Generally, for absolute value questions, |x-a| should immediately make you realize that you need to look at values less than a, =a and > a.

Thanks for the awesome response! I have another question. Is there a quick way to decide between X<0, 0≤X≤1, X>1 and X<0, 0≤X<1 ,X≥1 at the start of the question?

It is dangerous to start generalizing quant concepts without actually looking a the question. The 'convention' is to take x<0, 0≤x≤1, x>1 as this will not make us commit the mistake of bringing x>1 into the picture as you had proposed.

Generally, for absolute value questions, |x-a| should immediately make you realize that you need to look at values less than a, =a and > a.

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.
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