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Math: Absolute value (Modulus)

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Updated on: 15 May 2019, 20:55
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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 (9 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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Spoiler: :: Images
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Originally posted by walker on 06 Nov 2009, 19:49.
Last edited by Bunuel on 15 May 2019, 20:55, edited 29 times in total.
UPDATED.
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Re: Math: Absolute value (Modulus)  [#permalink]

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06 Nov 2009, 21:02
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Re: Math: Absolute value (Modulus)  [#permalink]

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13 Nov 2009, 23:07
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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Re: Math: Absolute value (Modulus)  [#permalink]

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14 Nov 2009, 04:53
4
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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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15 Sep 2010, 03:58
Can any one explain the below problem once again I didnt understand even thought it was explained in the above posts..

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

Stuck between C & D
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Re: Math: Absolute value (Modulus)  [#permalink]

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15 Sep 2010, 04:03
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ravitejapandiri wrote:
Can any one explain the below problem once again I didnt understand even thought it was explained in the above posts..

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

Stuck between C & D

An easy way to interpret these problems is |x-a|<b means that x is within b units of a or mathematically a-b < x < a+b

So these options mean :
A. 0-3<x<0+3
B. -5-4<x<-5+4
C. 1-9<x<1+9

D. 5-4<x<5+4
E. -3-5<x<-3+5
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Re: Math: Absolute value (Modulus)  [#permalink]

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29 Nov 2010, 22:04
Let’s consider following examples,

Example #1 I am not understanding this example and really struggling with modulus? Can someone please elaborate and explain in further detail? From this post I can't see how I would use this on every modulus problem?
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) $$x < -8$$. $$-(x+3) - (4-x) how did we get -(x+3) here?= -(8+x)$$ --> $$x = -1$$. 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)

I am totally lost with this post and also with other modulus problems I looked up in Gmat club thank you very much for your help in advance!!!!!
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Re: Math: Absolute value (Modulus)  [#permalink]

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30 Nov 2010, 07:09
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1
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)

if x < -8, (x + 3) is always negative. So, modulus is non-negative and we need to change a sign: |x+3| = - (x+3) for x<-8
For example, if x = -10,
|-10+3| = |-7| = 7
-(-10+3) = -(-7) = 7

In other words, |x| = x if x is positive and |x|=-x if x is negative.
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01 Dec 2010, 12:19
Sorry walker one more question on the below as I review this
Thinking of inequality with modulus as a segment at the number line.

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

Why would be looking for 4 on the right side? shouldn't this be 5 as the midpoint? And why do we want the left side to be 0 at x=5? Id like to understand this a little better

II. Converting inequalities with modulus into 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)

How did you convert lx+3l>3 into (-inf,-6) and (0,+inf)

Sorry these maybe simple questions but I just want to grasp the concept firmly
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Re: Math: Absolute value (Modulus)  [#permalink]

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01 Dec 2010, 14:21
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A couple of figures to see what modulus means.

It is quite convenient to remember the definition of modulus. |x| is the distance of x from 0 on the number line.
So if |x| = 1, we are looking for points which are at a distance 1 away from 0.
If |x| < 1, we are looking for points which are at a distance less than 1 away from 0.
If |x| > 1, we are looking for points which are at a distance more than 1 away from 0.
Attachment:

Ques1.jpg [ 14.57 KiB | Viewed 66056 times ]

If |x - 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Attachment:

Ques2.jpg [ 4.35 KiB | Viewed 66059 times ]

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Re: Math: Absolute value (Modulus)  [#permalink]

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01 Dec 2010, 19:35
Great thanks for the explanation on that Karishma - very very very helpful.

Can you also help me with how we converted lx+3l>3 into (-inf,-6) and (0,+inf)?

Is this because x+3>=0 (non-negative) and negative x+3<0 (negative). So we get to x+3=3 so x=0 which fits condition 1 above and this will go to infinity

and then

-(x+3)=3 so you get -x=6 or x=-6 which fits condition two above so x<-6 so therefore it goes to negative infinity?

I think I am close. appreciate the help.
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01 Dec 2010, 20:28
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gettinit wrote:
Great thanks for the explanation on that Karishma - very very very helpful.

Can you also help me with how we converted lx+3l>3 into (-inf,-6) and (0,+inf)?

Is this because x+3>=0 (non-negative) and negative x+3<0 (negative). So we get to x+3=3 so x=0 which fits condition 1 above and this will go to infinity

and then

-(x+3)=3 so you get -x=6 or x=-6 which fits condition two above so x<-6 so therefore it goes to negative infinity?

I think I am close. appreciate the help.

Yes, lx+3l>3 gives us two cases

Case 1: x + 3 >= 0 or x > = -3
Then (x + 3) > 3 or we can say x > 0

Case 2: x + 3 < 0 or x < -3
Then -(x + 3) > 3 or we can say x < -6

So either x > 0 which translates to (0, inf) or x < -6 which translates to (-inf, -6)

OR consider that lx+3l>3 means distance of x from -3 is more than 3.
If you go to 3 steps to right from -3, you reach 0. Anything after than is ok.
If you go 3 steps to left from -3, you reach -6. Anything to its left is ok.
Attachment:

Ques2.jpg [ 3.08 KiB | Viewed 66146 times ]

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15 Jan 2011, 23:52
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rahul321 wrote:
I've understood everything about Absolute Values in these posts except for Example #1 in walker's first post. I was hoping someone could help me better understand the logic behind the solution.

First, I think what i'm struggling with most is the fact that we have absolute values on both the left and right hand side of the equation. How do you deal with those?

Second, I noticed that the three key points mentioned here are -8, -3, 4. I get that these three values for $$x$$ equate the value inside each modulus to $$0$$. But why do we need the value of each modulus to be $$0$$?

Finally, I don't understand how we derived the 4 conditions. With the first one, for example, if $$x<-8$$, let's say $$x=-10$$, then $$|8+(-10)|=2$$ not $$0$$. I'm completely lost here.

I am assuming that you have understood that mod is nothing but distance on the number line. If so, then you can easily solve the question without equations.
I re arrange the eg to get |x + 3| = |x + 8| + |x - 4|
(Inside the mod, 4-x is same as x-4 since mod will always be non-negative)
What I get from this question is: "I want the point on the number line whose distance from -3 is equal to the sum of its distances from -8 and from 4."
i.e. the green distance is equal to red distance + blue distance.
Attachment:

Ques5.jpg [ 3.7 KiB | Viewed 53088 times ]

Is there any such point possible? If x is between -8 and 4, you see above that it is not possible. If x < -8 or x > 4, see the diagrams below and it will be obvious that there is no such point. In the first case, the blue distance itself is greater than the green distance and in the second case, the red distance itself is greater than the green distance. Since there is no such point on the number line, no such value of x exists.
Attachment:

Ques6.jpg [ 7.64 KiB | Viewed 53104 times ]

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16 Jan 2011, 00:08
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Of course it can be done using algebra as well. It doesn't matter how many mods there are. you always deal with them in the same way.
|x|= x when x is >= 0,
|x|= -x when x < 0

|x - 2|= (x - 2) when x - 2 >= 0 (or x >= 2),
|x - 2|= -(x - 2) when (x-2) < 0 (or x < 2)

Then you solve the equations using both conditions given above. That is the importance of the points.
So if you have:
|x - 2|= |x + 3|

You say, |x - 2|= (x - 2) when x >= 2.
|x - 2|= -(x - 2) when x < 2
|x + 3| = (x + 3) when x >= -3
|x + 3| = -(x + 3) when x < -3

Now, x can be either greater than 2, between -3 and 2 or less than -3.
So you solve for these 3 cases:
Case 1: x >= 2
(x - 2) = (x + 3)
-2 = 3
No solution

Case 2: -3 <= x < 2
-(x - 2) = (x + 3)
x = -1/2 which lies between -3 and 2
So this is a solution to the equation

Case 3: x <= -3
-(x - 2) = -(x + 3)
2 = -3
No solution

Similarly you can solve for as many terms as you want.
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Re: Math: Absolute value (Modulus)  [#permalink]

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09 Feb 2011, 04:41
i have a question related to Example 1:

|x+3| - |4-x| = |8+x|

-3 \leq x < 4
x \geq 4.
if we take /4-x/ than we have 4-x >=0 so x >=4
and 4-x<0 4<=0
i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o
i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4
i hope i made myself clear
thanks
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09 Feb 2011, 05:47
3
tinki wrote:
i have a question related to Example 1:

|x+3| - |4-x| = |8+x|

-3 \leq x < 4
x \geq 4.
if we take /4-x/ than we have 4-x >=0 so x >=4
and 4-x<0 4<=0
i see in the solution that you switched x-4 and wrote x-4>= 0 and x-4 <o
i understand that it is perfectly ok to do so. my question: when do we have to know when to switch ? because if not switched than we shall have x >=4 if switched x<=4
i hope i made myself clear
thanks

You can switch inside the mod when you like! When you feel switching makes it easier for you to handle.
We are used to getting terms in the form |x-4|, not |4-x|... so you switch.. it doesn't matter at all... let me show with a simpler example:

Question: |4-x| > 5

Case 1: When 4 - x >= 0 i.e. when x <= 4
4 - x > 5
or x < -1
So solution is x < -1

Case 2: When 4 - x <= 0 i.e. when x >= 4
-(4-x) > 5
x > 9
So solution is x > 9

Answer: x is either less than -1 or greater than 9.

Now switch:
Question: |x-4| > 5

Case 1: When x - 4 >= 0 i.e. when x >= 4
x - 4 > 5
or x > 9
Solution is x > 9

Case 2: When x - 4 <= 0 i.e. when x <= 4
-(x - 4) > 5
x < -1
So solution is x < -1

Answer: x is either less than -1 or greater than 9.

The same two cases in both the questions.. same answer in both...
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20 Feb 2011, 05:19
1
worldogvictor wrote:
Hi,
In the original post, in the section GMAC books, there are some prob numbers stated. Can you please explain how they relate to the post? How one will use those numbers is what I am asking.

The given problem numbers are official questions on the concept of Modulus.
You need to have those books to be able to access the given questions. The numbers give you the question numbers e.g. in OG12 in sample problem solving questions, question no 22 (on page 155) tests you on Mods.
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Re: Math: Absolute value (Modulus)  [#permalink]

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26 May 2011, 14:34
walker wrote:

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)

This is a great topic.However i have a little confusion on the range of numbers that are defined in the 4 scenarios.
As we are looking at x+3 >=0 we get x>=-3
Similarly when x+3 <0 we have x<-3

Based on these the ranges I have for the 4 cases
1. x<-8
2. -8<= x <-3
3. -3<= x <= 4
4. x> 4
the last two cases are whats different .
This obviously stems from solving 4-x >= 0 and 4-x<0 Could someone explain what I am doing wrong here? thanks in advance
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27 May 2011, 05:00
2
1
someonear wrote:
walker wrote:

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)

This is a great topic.However i have a little confusion on the range of numbers that are defined in the 4 scenarios.
As we are looking at x+3 >=0 we get x>=-3
Similarly when x+3 <0 we have x<-3

Based on these the ranges I have for the 4 cases
1. x<-8
2. -8<= x <-3
3. -3<= x <= 4
4. x> 4
the last two cases are whats different .
This obviously stems from solving 4-x >= 0 and 4-x<0 Could someone explain what I am doing wrong here? thanks in advance

You don't need to solve anything to get these four ranges.
You see that the points where the signs will vary are -8, 4 and -3.
To cover all the numbers on the number line, the ranges are:
x is less than -8, then between -8 and -3, then between -3 and 4 and then greater than 4. It doesn't matter where you put the '='. In each range the sign of the terms will be different. After assigning the proper signs, you will get a value for x and you have to check if the value lies in the range you were considering. If it does, it is a solution, else it is not.
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Re: Math: Absolute value (Modulus)  [#permalink]

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27 May 2011, 06:44
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VeritasPrepKarishma wrote:
someonear wrote:
walker wrote:

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)

This is a great topic.However i have a little confusion on the range of numbers that are defined in the 4 scenarios.
As we are looking at x+3 >=0 we get x>=-3
Similarly when x+3 <0 we have x<-3

Based on these the ranges I have for the 4 cases
1. x<-8
2. -8<= x <-3
3. -3<= x <= 4
4. x> 4
the last two cases are whats different .
This obviously stems from solving 4-x >= 0 and 4-x<0 Could someone explain what I am doing wrong here? thanks in advance

You don't need to solve anything to get these four ranges.
You see that the points where the signs will vary are -8, 4 and -3.
To cover all the numbers on the number line, the ranges are:
x is less than -8, then between -8 and -3, then between -3 and 4 and then greater than 4. It doesn't matter where you put the '='. In each range the sign of the terms will be different. After assigning the proper signs, you will get a value for x and you have to check if the value lies in the range you were considering. If it does, it is a solution, else it is not.

I am worried about values of x that are on the border of the ranges
Say hypothetically for a particular set of equations we end with the identical 4 cases we have here.
Now if say I have x=4 then the way I had come up with the ranges I will get a solution between -3 and 4 as x=4 exists in -3<= x <= 4. But if however I consider what was done in the OP then x=4 exists in the range x>=4
Granted either way I have a solution but will it be a biggie if I fail to show exactly in which range the solution, if it does,exists
Re: Math: Absolute value (Modulus)   [#permalink] 27 May 2011, 06:44

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