Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 : GMAT Data Sufficiency (DS)
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# Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0

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Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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15 Aug 2009, 13:58
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Is |x| < 1?

(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0
[Reveal] Spoiler: OA

Last edited by Bunuel on 02 Mar 2012, 20:00, edited 1 time in total.
Edited the question and added the OA
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Re: Is |x| < 1? [#permalink]

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15 Aug 2009, 21:37
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C

1)
if x>1, solving the equality we get x=3.
if -1<x<1, we get x=1/3.
if x<-1, we get x=3.

Not suff
2)
we get x is not equal to 3.

Combining, x can only be 1/3
OA?

Last edited by Economist on 15 Aug 2009, 22:01, edited 1 time in total.
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Re: Is |x| < 1? [#permalink]

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15 Aug 2009, 21:51
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is -1<x<1 ?

1) |x + 1| = 2|x - 1|

two possibilities:
a) x+1=2(x-1) or x=3 (consider x+1 and x-1 to be of same sign)
b) x+1=2(1-x) or x=1/3 (consider x+1 and x-1 to be of different signs)
not sufficient, x may or may not be between -1 and 1

2) |x - 3| ≠ 0
=> x≠3
not sufficient, x may or may not be between -1 and 1

together, x≠3 so x is 1/3 which is between -1 and 1
sufficient

hence, C
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Re: Is |x| < 1? [#permalink]

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23 Oct 2009, 19:18
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tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it helps.
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Re: Is |x| < 1? [#permalink]

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23 Oct 2009, 19:23
I got C too substituting values and evaluating each of those.

thanks Bunuel and Economist for elaborations
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09 Nov 2009, 21:26
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Expert's post
|x|<1 means x e (-1,1)

1)|x+1|=2|x-1|
There are 2 key points (x=-1 and x=1) where one of the expressions under modules passes 0 and change its sign.
x<-1: -(x+1)=-2(x-1) --> -x-1=-2x+2 ---> x = 3. But x=3 does not satisfy x<-1 condition.
-1<=x<1: x+1 = -2(x-1) ---> x+1= -2x+2 --> x=1/3. It satisfies the condition.
x>=1: x+1=2x-2 ---> x=3. It satisfies the condition.
Insufficient

2) |x - 3| ≠ 0
x≠3
Insufficient

1&2) Only x=1/3 satisfies both statements. Sufficient.

If you find it useful/useless, let me know
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09 Nov 2009, 21:32
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Answer is C - you beat me to it walker

For statement 1, you need to actually break it down to 2 separate statements to account for both scenarios:

Option #1 ->
x + 1 = 2(x-1)
x + 1 = 2x -2
x = 2x -3
3 + x = 2x
3 = x {or also Option #2}

Option #2 -> x + 1 = -2(x-1)
x + 1 = -2x +2
x = -2x + 1
3x = 1
x = 1/3

Statement (1) is insufficient because |x| could be 3 or 1/3. So the answer is sometimes yes, sometimes no which means insufficient.

Statement (2) is insufficient because we are told that |x -3| ≠ 0. This means that as long as x ≠ 3, then we're ok. There are far too many options so we certainly have a sometimes yes, and sometimes no.

But together, from Statement (1) we have either 3 or 1/3 and from statement (2) we have everything BUT 3. So the only overlap we have for possible values to make both statements true is 1/3. So the answer is Yes, |x| IS less than 1 and we have enough info to answer the question.

Is |x|< 1?
(1) |x + 1| = 2|x - 1|
(2) |x - 3| ≠ 0

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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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16 Jan 2012, 07:48
+1 C

However, I used using other methods: a) Squaring both sides of the equation because |x| = $$(x^2)$$^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem?
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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16 Jan 2012, 13:12
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metallicafan wrote:
+1 C

However, I used using other methods: a) Squaring both sides of the equation because |x| = $$(x^2)$$^(1/2), b) Combining different scenarios between these two possibilities: (x+1)>0, (x+1)<0, (x-1)>0, (x-1)<0. Therefore, I get four possible scenarios.

Bunuel, could you explain the logic behing your method to solve this problem?

It's about expanding the absolute value in different ranges. Discussed in Walker's post on absolute values: math-absolute-value-modulus-86462.html

If you're more comfortable with "squaring" method (note that it's not always applicable) then you can apply it to the first statement too (and not only to the stem).

Is $$|x| < 1$$?

Is $$|x| < 1$$ --> is $$-1<x<1$$?

(1) $$|x + 1| = 2|x - 1|$$ --> square both sides: $$(x+1)^2=4(x-1)^2$$ --> $$x^2+2x+1=4x^2-8x+4$$ --> $$3x^2-10x+3=0$$ --> $$x=\frac{1}{3}$$ or $$x=3$$ --> first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$ --> just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it's clear.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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24 Apr 2012, 03:04
Hi Bunuel ,
I also used the squaring method....
In what all situations is the squaring method not applicable ???
Can you please explain why ?
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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24 Apr 2012, 04:43
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shikhar wrote:
Hi Bunuel ,
I also used the squaring method....
In what all situations is the squaring method not applicable ???
Can you please explain why ?

Sure, for example it's not always applicable for inequalities.

A. We can raise both parts of an inequality to an even power if we know that both parts of an inequality are non-negative (the same for taking an even root of both sides of an inequality).
For example:
$$2<4$$ --> we can square both sides and write: $$2^2<4^2$$;
$$0\leq{x}<{y}$$ --> we can square both sides and write: $$x^2<y^2$$;

But if either of side is negative then raising to even power doesn't always work.
For example: $$1>-2$$ if we square we'll get $$1>4$$ which is not right. So if given that $$x>y$$ then we can not square both sides and write $$x^2>y^2$$ if we are not certain that both $$x$$ and $$y$$ are non-negative.

B. We can always raise both parts of an inequality to an odd power (the same for taking an odd root of both sides of an inequality).
For example:
$$-2<-1$$ --> we can raise both sides to third power and write: $$-2^3=-8<-1=-1^3$$ or $$-5<1$$ --> $$-5^2=-125<1=1^3$$;
$$x<y$$ --> we can raise both sides to third power and write: $$x^3<y^3$$.

So for our question we can not square x/|x|< x as we don't know the sign of either of side.

Hope it helps.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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28 Apr 2012, 09:42
Nice explaination Bunuel, Thanks! 1+
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Re: Is |x| < 1? [#permalink]

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21 May 2012, 21:35
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it helps.

How is the sign applied for different range?can someone help please
X<-1
-x-1=2(-x+1)

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

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

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21 May 2012, 22:18
sdpkind wrote:
Bunuel wrote:
tejal777 wrote:
I am having trouble understanding how |x| < 1 translates to -1<x<1.Am missing something:

We have two case,
a. x>0 eg: x=5, |x|=5 so here |x|>0.Therefore,x < 1
b. when x<0,x=-5,|x| is still 5 how does this become x>-1??

Is $$|x| < 1$$?

Is $$|x| < 1$$, means is $$x$$ in the range (-1,1) or is $$-1<x<1$$ true?

(1) $$|x + 1| = 2|x - 1|$$
Two key points: $$x=-1$$ and $$x=1$$ (key points are the values of x when absolute values equal to zero), thus three ranges to check:
---------{-1}--------{1}---------

A. $$x<-1$$ (blue range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$, not OK, as this value is not in the range we are checking ($$x<-1$$);
B. $$-1\leq{x}\leq{1}$$ (green range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$ --> $$x=\frac{1}{3}$$. OK, as this value is in the range we are checking ($$-1\leq{x}\leq{1}$$);
C. $$x>1$$ (red range) --> $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(x-1)$$ --> $$x=3$$. OK, as this value is in the range we are checking ($$x>1$$).

So we got TWO values of $$x$$ (two solutions): $$\frac{1}{3}$$ and $$3$$, first is in the range (-1,1) but second is out of the range. Not sufficient.

(2) $$|x - 3|\neq{0}$$
Just says that $$x\neq{3}$$. But we don't know whether $$x$$ is in the range (-1,1) or not.

(1)+(2) $$x=\frac{1}{3}$$ or $$x=3$$ AND $$x\neq{3}$$ --> means $$x$$ can have only value $$\frac{1}{3}$$, which is in the range (-1,1). Sufficient.

Hope it helps.

How is the sign applied for different range?can someone help please
X<-1
-x-1=2(-x+1)

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

X>1
x+1=2(x-1)

Absolute value properties:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|={-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|={some \ expression}$$. For example: $$|5|=5$$;

Applying this to $$|x + 1| = 2|x - 1|$$.

If $$x<-1$$ then $$x+1<0$$ and $$x-1<0$$ so $$|x + 1| =-(x+1)$$ and $$2|x - 1|=-2(x-1)$$ which means that $$|x + 1| = 2|x - 1|$$ becomes: $$-x-1=2(-x+1)$$ --> $$x=3$$.

Similarly for other ranges.

For more check Absolute Value chapter for Math Book: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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22 May 2012, 19:28
Hi Bunuel,

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)"
Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here?
How are you deciding the sign and coming here: x+1=2(-x+1)
I couldn't understand this

I have seen this post: http://gmatclub.com/forum/math-absolute-value-modulus-86462.html as you referred above but ctill couldn't understand it.

I would really appreciate if you can throw a little light on this.

Thanks,
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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22 May 2012, 22:05
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hey kartik,

i also had the same issue initially...then i went thru the link Bunuel has given and it toook me a little time to get it....

so seee.....
the Question states

|x+1| = 2|x-1|
if we take the number line

--------------{-1}----------------{1}---------------

|x+1| if u see will be negative till -1 and after that it will be positive
|x-1| will be negative till 1 and will be positive after that.....

Now we have 3 areas where we need to check

x<-1
-1<=x<=1
x>1

If u check the number line till {-1} both the expression's value will be negative......so when we solve x<-1 we take both of them as Negative

but when we solve -1<=x<=1 the expression |x+1| is positive as its sign changed on x=-1 but the other one |x-1| is still Negative as its sign would change on {1}.
therefore when we solve of this range we take |x+1|= x+1 and 2|x-1| = -2(x-1) = 2(-x+1)

in the third one both are Positive os both the expressions will be positive....

I hope this help you...as it also took me a lot of time to get this concept in my head.........
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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22 May 2012, 22:30
kartik222 wrote:
Hi Bunuel,

A. x<-1 (blue range) --> |x + 1| = 2|x - 1| becomes: -x-1=2(-x+1) --> x=3, not OK, as this value is not in the range we are checking (x<-1);
B. -1\leq{x}\leq{1} (green range) --> |x + 1| = 2|x - 1| becomes: x+1=2(-x+1) --> x=\frac{1}{3}. OK, as this value is in the range we are checking (-1\leq{x}\leq{1});
C. x>1 (red range) --> |x + 1| = 2|x - 1| becomes: x+1=2(x-1) --> x=3. OK, as this value is in the range we are checking (x>1).

If understand correctly for (A) you are putting '-' in front of both Left/Right Hand Side Equation. Hence coming to "-x-1=2(-x+1)"
Similarly in (C) I guess you are keeping the Left/Right Hand Side Equation both as '+'.

What about (B)?? value of x is between -1 and 1. How did you decide signs' here?
How are you deciding the sign and coming here: x+1=2(-x+1)
I couldn't understand this

I have seen this post: http://gmatclub.com/forum/math-absolute-value-modulus-86462.html as you referred above but ctill couldn't understand it.

I would really appreciate if you can throw a little light on this.

Thanks,

This issue is discussed here: is-x-1-1-x-1-2-x-1-2-x-82478.html#p1088086

Absolute value properties:
When $$x\leq{0}$$ then $$|x|=-x$$, or more generally when $$some \ expression\leq{0}$$ then $$|some \ expression|\leq{-(some \ expression)}$$. For example: $$|-5|=5=-(-5)$$;

When $$x\geq{0}$$ then $$|x|=x$$, or more generally when $$some \ expression\geq{0}$$ then $$|some \ expression|\leq{some \ expression}$$. For example: $$|5|=5$$;

Applying this to $$|x + 1| = 2|x - 1|$$.

If $$-1\leq{x}\leq{1}$$ then $$x+1>0$$ and $$x-1<0$$ so $$|x + 1| =x+1$$ and $$2|x - 1|=-2(x-1)$$ which means that $$|x + 1| = 2|x - 1|$$ becomes: $$x+1=2(-x+1)$$.

Hope it's clear.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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30 Aug 2012, 19:12
good explanation sir thanks
a lot learnt new concept in this
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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31 Aug 2012, 01:36
I'm looking for absolute number problems. I've looked around the forum but i can't find any problem sets.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0 [#permalink]

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31 Aug 2012, 01:42
qweert wrote:
I'm looking for absolute number problems. I've looked around the forum but i can't find any problem sets.

Check our Question Banks: viewforumtags.php

PS questions on Absolute Values: search.php?search_id=tag&tag_id=58
DS questions on Absolute Values: search.php?search_id=tag&tag_id=37

Tough inequality and absolute value questions: inequality-and-absolute-value-questions-from-my-collection-86939.html

Theory on absolute values: math-absolute-value-modulus-86462.html

Hope it helps.
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Re: Is |x| < 1? (1) |x + 1| = 2|x - 1| (2) |x - 3| 0   [#permalink] 31 Aug 2012, 01:42

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