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Re: Math: Absolute value (Modulus) [#permalink]
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15 May 2010, 13:27
thanks for the wonderfull effort



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30 Aug 2010, 16:39
thank you, very useful



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03 Sep 2010, 18:42
Thanks Walker for wonderful explanation. +1 from me.
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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. x1<9 D. 5+x<4 E. 3+x<5 Stuck between C & D



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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. x1<9 D. 5+x<4 E. 3+x<5 Stuck between C & D An easy way to interpret these problems is xa<b means that x is within b units of a or mathematically ab < x < a+b So these options mean : A. 03<x<0+3 B. 54<x<5+4 C. 19<x<1+9D. 54<x<5+4E. 35<x<3+5
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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. x1<9 D. 5+x<4 E. 3+x<5 Stuck between C & D An easy way to interpret these problems is xa<b means that x is within b units of a or mathematically ab < x < a+b So these options mean : A. 03<x<0+3 B. 54<x<5+4 C. 19<x<1+9D. 54<x<5+4E. 35<x<3+5>>>>Ooops..Thanks dude..U made that look so easy



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Re: Math: Absolute value (Modulus) [#permalink]
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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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23 Sep 2010, 02:58
How to print this document?



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29 Sep 2010, 09:55
great resource. thanks!



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Re: Math: Absolute value (Modulus) [#permalink]
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08 Nov 2010, 10:34
Quote: 1<x<9. What inequality represents this condition?
A. x<3 B. x+5<4 C. x1<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 x5<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) x1<9 Negative: (x1)<0 => x>1 Positive: (x1)>0 => x>1 solving for new equation x+1<9 => x>8 x1<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 => 5x<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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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. x1<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 x5<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) x1<9 Negative: (x1)<0 => x>1 This should be x<1Positive: (x1)>0 => x>1 solving for new equation x+1<9 => x>8 x1<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 => 5x<4 => x>1 This is the case x<5, your final solution is correct, x>15+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]
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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]
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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.: \(x+3  4x = 8+x\). How many solutions does the equation have? Solution: There are 3 key points here: 8, 3, 4. So we have 4 conditions:
a) \(x < 8\). \((x+3)  (4x) how did we get (x+3) here?= (8+x)\) > \(x = 1\). We reject the solution because our condition is not satisfied (1 is not less than 8)
b) \(8 \leq x < 3\). \((x+3)  (4x) = (8+x)\) > \(x = 15\). We reject the solution because our condition is not satisfied (15 is not within (8,3) interval.)
c) \(3 \leq x < 4\). \((x+3)  (4x) = (8+x)\) > \(x = 9\). We reject the solution because our condition is not satisfied (15 is not within (3,4) interval.)
d) \(x \geq 4\). \((x+3) + (4x) = (8+x)\) > \(x = 1\). 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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a) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). 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 nonnegative 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]
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01 Dec 2010, 11:49
walker wrote: a) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). 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 nonnegative 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  4x = 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)  (4x) = (8+x) > x = 1. x+3is always  because x<8 but l4xl will always be positive because 4x= a +, l8+xl will also be negative as well given x<8b) 8 \leq x < 3. (x+3)  (4x) = (8+x) > x = 15. on this one lx+3l again can only be negative and l4xl will be still always be negative but given the range l8+xl may only be postiive hence l8+xl has to be positive or nonnegative (o)? c) 3 \leq x < 4. (x+3)  (4x) = (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) + (4x) = (8+x) > x = 1. on this one first term obviously positive second term is negative therefore the sign changes from  to + so from (4x) to +(4x) 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]
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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. x1<9 D. 5+x<4 E. 3+x<5 Solution: 10sec. Traditional 3steps method is too timeconsume technique. First of all we find length (91)=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 15043 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 15031 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 (nonnegative) 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]
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01 Dec 2010, 20:28
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 (nonnegative) 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 15215 times ]
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02 Dec 2010, 08:42
Excellent thanks for you patience and explanation Karishma. I can only thank you through kudos!




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