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Math: Absolute value (Modulus)

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New post 15 May 2010, 12:27
thanks for the wonderfull effort
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New post 30 Aug 2010, 15:39
thank you, very useful
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New post 03 Sep 2010, 17:42
Thanks Walker for wonderful explanation. +1 from me.
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New post 15 Sep 2010, 02:58
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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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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New post 15 Sep 2010, 04:15
shrouded1 wrote:
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



>>>>Ooops..Thanks dude..U made that look so easy :)
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New post 15 Sep 2010, 04:37
thanks for bumping this post ; i had not noticed this and just read through the absolute notes by walker!....
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New post 23 Sep 2010, 01:58
How to print this document?
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New post 29 Sep 2010, 08:55
great resource. thanks!
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New post 08 Nov 2010, 09:34
Quote:
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


I tried to solve this question by 3 steps method to improve my shaky concepts. Please clarify

A) |X| < 3
Negative: x<0 => -x<0 => x>0
Positive: x>0

solving for new equation
|x|<3 => -x<3 or x>-3
|x|<3 => x<3

This means x lies between -3 and 3 or -3<x<3
Hence it doesn't satisfy our question 1<x<9


B) |X+5| < 4

Negative: (x+5)<0 => x>-5
positive: (x+5) > 0 => x>-5

solving for new equation
-x-5<4 => x>-9
x+5<4 => x< -1

This means x lies between -9 and -1 or -9<x<-1
hence it doesn't satisfy the equation 1<x<9


C) |x-1|<9

Negative: (x-1)<0 => x>1
Positive: (x-1)>0 => x>1

solving for new equation
-x+1<9 => x>-8
x-1<9 => x<8

it means x lies between -8 and 8 or -8<x<8
it doesn't satisfy the equation 1<x<9

D)|-5+x|<4

Negative: (-5+x)<0 => x>5
Positive: (-5+x) >0 => x>5

solving for new equation
-5+x<4 => 5-x<4 => x>1
-5+x<4 => x<9
This means x>1 but less than 9 or 1<x<9
Hence it proved the relation.

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

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New post 08 Nov 2010, 23:38
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I have ppointed out where you are going wrong :
(final solution you get is correct)

Try using the shortcu method I showed above, it uses the exact same concept as you did, but it is faster

shrive555 wrote:
Quote:
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


I tried to solve this question by 3 steps method to improve my shaky concepts. Please clarify

A) |X| < 3
Negative: x<0 => -x<0 => x>0
Positive: x>0

solving for new equation
|x|<3 => -x<3 or x>-3
|x|<3 => x<3

This means x lies between -3 and 3 or -3<x<3
Hence it doesn't satisfy our question 1<x<9


B) |X+5| < 4

Negative: (x+5)<0 => x>-5
positive: (x+5) > 0 => x>-5

solving for new equation
-x-5<4 => x>-9
x+5<4 => x< -1

This means x lies between -9 and -1 or -9<x<-1
hence it doesn't satisfy the equation 1<x<9


C) |x-1|<9

Negative: (x-1)<0 => x>1 This should be x<1
Positive: (x-1)>0 => x>1

solving for new equation
-x+1<9 => x>-8
x-1<9 => x<8 This should be x<10

it means x lies between -8 and 8 or -8<x<8 x between -8 & 10
it doesn't satisfy the equation 1<x<9

D)|-5+x|<4

Negative: (-5+x)<0 => x>5 This should be x<5
Positive: (-5+x) >0 => x>5

solving for new equation
-5+x<4 => 5-x<4 => x>1 This is the case x<5, your final solution is correct, x>1
-5+x<4 => x<9
This means x>1 but less than 9 or 1<x<9
Hence it proved the relation.

:shock: :roll:

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New post 09 Nov 2010, 10:48
Thanks shrouded !! you're right ! The method you mentioned is faster and i'm solving question on that method.
+1
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New post 29 Nov 2010, 21:04
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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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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New post 01 Dec 2010, 10:49
walker wrote:
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.


Thanks for the explanation Walker. So would this then be correct:

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. |x+3|is always - because x<-8 but l4-xl will always be positive because 4-x= a +, l8+xl will also be negative as well given x<-8

b) -8 \leq x < -3. -(x+3) - (4-x) = (8+x) --> x = -15. on this one lx+3l again can only be negative and l4-xl will be still always be negative but given the range l8+xl may only be postiive hence l8+xl has to be positive or non-negative (o)?

c) -3 \leq x < 4. (x+3) - (4-x) = (8+x) --> x = 9. so on this term first one is positive second term continues to be positive and 3rd term is positive.

d) x \geq 4. (x+3) + (4-x) = (8+x) --> x = -1. on this one first term obviously positive second term is negative therefore the sign changes from - to + so from -(4-x) to +(4-x) since this will always be negative for given x range? And a negative times a negative gets us to a positive?

I just want to make sure I am thinking about this correctly. Thanks walker!
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New post 01 Dec 2010, 11:19
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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New post 01 Dec 2010, 13:21
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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
Ques1.jpg [ 14.57 KiB | Viewed 14422 times ]


If |x - 5| < 4, now we are looking for points at a distance less than 4 away from 5.
Attachment:
Ques2.jpg
Ques2.jpg [ 4.35 KiB | Viewed 14410 times ]

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New post 01 Dec 2010, 18:35
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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New post 01 Dec 2010, 19:28
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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
Ques2.jpg [ 3.08 KiB | Viewed 14596 times ]

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New post 02 Dec 2010, 07:42
Excellent thanks for you patience and explanation Karishma. I can only thank you through kudos!
Re: Math: Absolute value (Modulus)   [#permalink] 02 Dec 2010, 07:42

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