Math: Absolute value (Modulus) : GMAT Quantitative Section
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# Math: Absolute value (Modulus)

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06 Nov 2009, 18:49
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ABSOLUTE VALUE
(Modulus)

This post is a part of [GMAT MATH BOOK]

created by: walker
edited by: bb, Bunuel

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Definition

The absolute value (or modulus) $$|x|$$ of a real number x is x's numerical value without regard to its sign.

For example, $$|3| = 3$$; $$|-12| = 12$$; $$|-1.3|=1.3$$

Graph:

Important properties:

$$|x|\geq0$$

$$|0|=0$$

$$|-x|=|x|$$

$$|x|+|y|\geq|x+y|$$

$$|x|\geq0$$

How to approach equations with moduli

It's not easy to manipulate with moduli in equations. There are two basic approaches that will help you out. Both of them are based on two ways of representing modulus as an algebraic expression.

1) $$|x| = \sqrt{x^2}$$. This approach might be helpful if an equation has × and /.

2) |x| equals x if x>=0 or -x if x<0. It looks a bit complicated but it's very powerful in dealing with moduli and the most popular approach too (see below).

3-steps approach:

General approach to solving equalities and inequalities with absolute value:

1. Open modulus and set conditions.
To solve/open a modulus, you need to consider 2 situations to find all roots:
• Positive (or rather non-negative)
• Negative

For example, $$|x-1|=4$$
a) Positive: if $$(x-1)\geq0$$, we can rewrite the equation as: $$x-1=4$$
b) Negative: if $$(x-1)<0$$, we can rewrite the equation as: $$-(x-1)=4$$
We can also think about conditions like graphics. $$x=1$$ is a key point in which the expression under modulus equals zero. All points right are the first condition $$(x>1)$$ and all points left are second condition $$(x<1)$$.

2. Solve new equations:
a) $$x-1=4$$ --> x=5
b) $$-x+1=4$$ --> x=-3

3. Check conditions for each solution:
a) $$x=5$$ has to satisfy initial condition $$x-1>=0$$. $$5-1=4>0$$. It satisfies. Otherwise, we would have to reject x=5.
b) $$x=-3$$ has to satisfy initial condition $$x-1<0$$. $$-3-1=-4<0$$. It satisfies. Otherwise, we would have to reject x=-3.

3-steps approach for complex problems

Let’s consider following examples,

Example #1
Q.: $$|x+3| - |4-x| = |8+x|$$. How many solutions does the equation have?
Solution: There are 3 key points here: -8, -3, 4. So we have 4 conditions:

a) $$x < -8$$. $$-(x+3) - (4-x) = -(8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not less than -8)

b) $$-8 \leq x < -3$$. $$-(x+3) - (4-x) = (8+x)$$ --> $$x = -15$$. We reject the solution because our condition is not satisfied (-15 is not within (-8,-3) interval.)

c) $$-3 \leq x < 4$$. $$(x+3) - (4-x) = (8+x)$$ --> $$x = 9$$. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

d) $$x \geq 4$$. $$(x+3) + (4-x) = (8+x)$$ --> $$x = -1$$. We reject the solution because our condition is not satisfied (-1 is not more than 4)

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Example #2
Q.: $$|x^2-4| = 1$$. What is x?
Solution: There are 2 conditions:

a) $$(x^2-4)\geq0$$ --> $$x \leq -2$$ or $$x\geq2$$. $$x^2-4=1$$ --> $$x^2 = 5$$. x e {$$-\sqrt{5}$$, $$\sqrt{5}$$} and both solutions satisfy the condition.

b) $$(x^2-4)<0$$ --> $$-2 < x < 2$$. $$-(x^2-4) = 1$$ --> $$x^2 = 3$$. x e {$$-\sqrt{3}$$, $$\sqrt{3}$$} and both solutions satisfy the condition.

(Optional) The following illustration may help you understand how to open modulus at different conditions.

Answer: $$-\sqrt{5}$$, $$-\sqrt{3}$$, $$\sqrt{3}$$, $$\sqrt{5}$$

Tip & Tricks

The 3-steps method works in almost all cases. At the same time, often there are shortcuts and tricks that allow you to solve absolute value problems in 10-20 sec.

I. Thinking of inequality with modulus as a segment at the number line.

For example,
Problem: 1<x<9. What inequality represents this condition?

A. |x|<3
B. |x+5|<4
C. |x-1|<9
D. |-5+x|<4
E. |3+x|<5
Solution: 10sec. Traditional 3-steps method is too time-consume technique. First of all we find length (9-1)=8 and center (1+8/2=5) of the segment represented by 1<x<9. Now, let’s look at our options. Only B and D has 8/2=4 on the right side and D had left site 0 at x=5. Therefore, answer is D.

II. Converting inequalities with modulus into a range expression.
In many cases, especially in DS problems, it helps avoid silly mistakes.

For example,
|x|<5 is equal to x e (-5,5).
|x+3|>3 is equal to x e (-inf,-6)&(0,+inf)

III. Thinking about absolute values as the distance between points at the number line.

For example,
Problem: A<X<Y<B. Is |A-X| <|X-B|?
1) |Y-A|<|B-Y|
Solution:

We can think of absolute values here as the distance between points. Statement 1 means than the distance between Y and A is less than that between Y and B. Because X is between A and Y, |X-A| < |Y-A| and at the same time the distance between X and B will be larger than that between Y and B (|B-Y|<|B-X|). Therefore, statement 1 is sufficient.

Pitfalls

The most typical pitfall is ignoring the third step in opening modulus - always check whether your solution satisfies conditions.

Official GMAC Books:

The Official Guide, 12th Edition: PS #22; PS #50; PS #130; DS #1; DS #153;
The Official Guide, Quantitative 2th Edition: PS #152; PS #156; DS #96; DS #120;
The Official Guide, 11th Edition: DT #9; PS #20; PS #130; DS #3; DS #105; DS #128;

Generated from [GMAT ToolKit]

Resources

Absolute value DS problems: [search]
Absolute value PS problems: [search]

Fig's post with absolute value problems: [Absolute Value Problems]

A lot of questions as well as a separate topic of PrepGame on absolute value is included in GMAT ToolKit 2

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Last edited by walker on 02 Jun 2013, 14:01, edited 27 times in total.
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Re: Math: Absolute value (Modulus) [#permalink]

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06 Nov 2009, 20:02
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Great post!! Kudos!!

If some problems links with categorization (like 700 level, 600-700 level) are posted, it will be great. I was in search of range approach for solving modules problems, I got some stuff here. Thanks again!
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06 Nov 2009, 20:06
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Great post walker!!! + 1 to you.

PS. I have just made 2 corrections above.
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06 Nov 2009, 20:26
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hgp2k wrote:
PS. I have just made 2 corrections above.

+1
Thanks
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06 Nov 2009, 20:34
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Hussain15 wrote:
If some problems links with categorization (like 700 level, 600-700 level) are posted, it will be great. I was in search of range approach for solving modules problems, I got some stuff here. Thanks again!

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07 Nov 2009, 16:06
kudos!
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Re: Math: Absolute value (Modulus) [#permalink]

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13 Nov 2009, 22:07
Hi Walker, Thanks for posting this.

You've written a property that:
|X + Y| >= |X| + |Y|
Is the same true for negative?
|X - Y| <= |X| - |Y|
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14 Nov 2009, 03:53
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yangsta8 wrote:
Hi Walker, Thanks for posting this.

You've written a property that:
|X + Y| >= |X| + |Y|
Is the same true for negative?
|X - Y| <= |X| - |Y|

|X + Y| <= |X| + |Y|

|X - Y| >=|X| - |Y|
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Re: Math: Absolute value (Modulus) [#permalink]

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13 Dec 2009, 04:40
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Not that it matters in any way, but can you correct the "witch" under 1b in the initial post. Other than that: awesome post!
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Re: Math: Absolute value (Modulus) [#permalink]

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15 Dec 2009, 22:45
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Hi Walker!
I found a mistake in your expl.
In 3-step approach for more than one module: (d) is not -9, it's -1
Correct me if I'm wrong.
Thank you.
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Re: Math: Absolute value (Modulus) [#permalink]

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15 Dec 2009, 23:08
Igor010 wrote:
Hi Walker!
I found a mistake in your expl.
In 3-step approach for more than one module: (d) is not -9, it's -1
Correct me if I'm wrong.
Thank you.

Thanks! You are right
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21 Dec 2009, 20:49
Thank you! Thank you! Thank you! This is the clearest explanation of absolute value that I've come across thus far.
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23 Dec 2009, 19:42
This is exactly what I was looking for past few days. Great Post.
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Re: Math: Absolute value (Modulus) [#permalink]

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27 Dec 2009, 03:26
Sorry, but please clarify why is it not...
a) x < -8. -(x+3) + (4-x) = -(8+x)

The -ve sign is given in the question stem, once we take the solution -(4-x) with the -ve sign in the equation, then two -ves should be positive, shouldn't it?
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27 Dec 2009, 06:55
wonderful post walker
kudos..
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Re: Math: Absolute value (Modulus) [#permalink]

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27 Dec 2009, 07:05
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arjunrampal wrote:
Sorry, but please clarify why is it not...
a) x < -8. -(x+3) + (4-x) = -(8+x)

The -ve sign is given in the question stem, once we take the solution -(4-x) with the -ve sign in the equation, then two -ves should be positive, shouldn't it?

hai arjunrampal..

when x< -8, (x+3) and (8+x) are both negative, while (4-x) is positive. (if you want to confirm this, u can plug in values for x and try)

when x<0 |x| = -x ( i.e. |x| = - (-ve x), ultimately modulus x is positive. i hope this point is clear)
when x>0 |x| = x

so for x< -8, |(4-x)| remains positive while modulus of the other two expressions become negative.

hope it is clear...
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27 Dec 2009, 09:22
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Example #2
Q.: $$|x^2-4| = 1$$. What is x?
Solution: There are 2 conditions:

a) $$(x^2-4)\geq0$$ --> $$x \leq -2$$ or $$x\geq2$$. $$x^2-4=1$$ --> $$x^2 = 5$$. x e {$$-\sqrt{5}$$, $$\sqrt{5}$$} and both solutions satisfy the condition.

b) $$(x^2-4)<0$$ --> $$-2 < x < 2$$. $$-(x^2-4) = 1$$ --> $$x^2 = 3$$. x e {$$-\sqrt{3}$$, $$\sqrt{3}$$} and both solutions satisfy the condition.

Answer: $$-\sqrt{5}$$, $$-\sqrt{3}$$, $$\sqrt{3}$$, $$\sqrt{5}$$
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Re: Math: Absolute value (Modulus) [#permalink]

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27 Dec 2009, 10:32
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arjunrampal wrote:
Sorry, but please clarify why is it not...
a) x < -8. -(x+3) + (4-x) = -(8+x)

The -ve sign is given in the question stem, once we take the solution -(4-x) with the -ve sign in the equation, then two -ves should be positive, shouldn't it?

Thanks for the question. I've added illustration to the example as well as one more example and hope they help
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27 Dec 2009, 11:16
Many thanks for the illustration. This is now made clear. Kudos!
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26 Apr 2010, 04:56
Thank you walker. This post helped me understand the concepts of absolute value
+1
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Re: Math: Absolute value (Modulus)   [#permalink] 26 Apr 2010, 04:56

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