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Re: Math: Absolute value (Modulus) [#permalink]
 15 May 2010, 13:27
thanks for the wonderfull effort
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Re: Math: Absolute value (Modulus) [#permalink]
30 Aug 2010, 16:39
thank you, very useful
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Re: Math: Absolute value (Modulus) [#permalink]
03 Sep 2010, 18:42
Thanks Walker for wonderful explanation. +1 from me.
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Re: Math: Absolute value (Modulus) [#permalink]
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]
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+9D. 5-4<x<5+4E. -3-5<x<-3+5
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Re: Math: Absolute value (Modulus) [#permalink]
15 Sep 2010, 05:15
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+9D. 5-4<x<5+4E. -3-5<x<-3+5>>>>Ooops..Thanks dude..U made that look so easy 
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Re: Math: Absolute value (Modulus) [#permalink]
15 Sep 2010, 05:37
thanks for bumping this post ; i had not noticed this and just read through the absolute notes by walker!....
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Re: Math: Absolute value (Modulus) [#permalink]
23 Sep 2010, 02:58
How to print this document?
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Re: Math: Absolute value (Modulus) [#permalink]
29 Sep 2010, 09:55
great resource. thanks!
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Re: Math: Absolute value (Modulus) [#permalink]
08 Nov 2010, 10:34
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.  
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Re: Math: Absolute value (Modulus) [#permalink]
09 Nov 2010, 00:38
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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<1Positive: (x-1)>0 => x>1 solving for new equation -x+1<9 => x>-8 x-1<9 => x<8 This should be x<10it means x lies between -8 and 8 or -8<x<8 x between -8 & 10it doesn't satisfy the equation 1<x<9 D)|-5+x|<4 Negative: (-5+x)<0 => x>5 This should be x<5Positive: (-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. _________________
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Re: Math: Absolute value (Modulus) [#permalink]
09 Nov 2010, 11:48
Thanks shrouded !! you're right ! The method you mentioned is faster and i'm solving question on that method. +1
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Re: Math: Absolute value (Modulus) [#permalink]
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]
30 Nov 2010, 07:09
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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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Re: Math: Absolute value (Modulus) [#permalink]
01 Dec 2010, 11:49
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<-8b) -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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Re: Math: Absolute value (Modulus) [#permalink]
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]
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 9977 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 9967 times ]
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Re: Math: Absolute value (Modulus) [#permalink]
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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Re: Math: Absolute value (Modulus) [#permalink]
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 10154 times ]
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Re: Math: Absolute value (Modulus) [#permalink]
02 Dec 2010, 08:42
Excellent thanks for you patience and explanation Karishma. I can only thank you through kudos!
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Re: Math: Absolute value (Modulus)
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02 Dec 2010, 08:42
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