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Math: Absolute value (Modulus)

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

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New post 23 Aug 2018, 01:37
How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.
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New post 23 Aug 2018, 02:12
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onyx12102 wrote:
How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.

______________
I'd say 1 or 2.
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 23 Aug 2018, 04:02
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onyx12102 wrote:
How frequent or how many absolute value questions can we approximately expect on the GMAT? I understand there may not be a sure way to know or guess but just looking for an estimate based on past and present experiences.


Not more than a couple. The problem is that often GMAT likes to combine topics. So, you might have a geometry question which will make you use some basic absolute value concept such as area bounded by the graph of |x| + |y| = 4. So a basic understanding of all topics is a good idea even if you plan to ignore some topics.
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 01 Sep 2018, 14:16
Quote:
2. Solve new equations:
a) x−1=4x−1=4 --> x=5
b) −x+1=4−x+1=4 --> x=-3

3. Check conditions for each solution:
a) x=5x=5 has to satisfy initial condition x−1>=0x−1>=0. 5−1=4>05−1=4>0. It satisfies. Otherwise, we would have to reject x=5.
b) x=−3x=−3 has to satisfy initial condition x−1<0x−1<0. −3−1=−4<0−3−1=−4<0. It satisfies. Otherwise, we would have to reject x=-3.


So...just to clarify...the answer can either be 5 or -3?

Just need the clarification on what this is saying
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New post 03 Sep 2018, 20:55
In the following example of 3-steps approach for complex problems


Example #1
Q.: |x+3|−|4−x|=|8+x||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<−8x<−8. −(x+3)−(4−x)=−(8+x)−(x+3)−(4−x)=−(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not less than -8) Can any one please explain me how x<-8 because whenever I am doing it I am getting x>=-8 for both |8+x| and -|8+x|

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

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

d) x≥4x≥4. (x+3)+(4−x)=(8+x)(x+3)+(4−x)=(8+x) --> x=−1x=−1. We reject the solution because our condition is not satisfied (-1 is not more than 4)
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 22 Sep 2018, 02:50
IN Example #1
Q.: |x+3|−|4−x|=|8+x||x+3|−|4−x|=|8+x|. How many solutions does the equation have?
C condition has typo error
c) −3≤x<4−3≤x<4. (x+3)−(4−x)=(8+x)(x+3)−(4−x)=(8+x) --> x=9x=9. We reject the solution because our condition is not satisfied (-15 is not within (-3,4) interval.)

its not -15 but 9
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 29 Sep 2018, 05:26
chetan2u Hi, sorry I keep bugging you but your explanations are easiest to understand :P
Can you explain to me the "trick" mentioned in the post? I don't get it :/ What are we doing after finding the mid-point?
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Math: Absolute value (Modulus)  [#permalink]

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New post 29 Sep 2018, 09:02
hibobotamuss wrote:
chetan2u Hi, sorry I keep bugging you but your explanations are easiest to understand :P
Can you explain to me the "trick" mentioned in the post? I don't get it :/ What are we doing after finding the mid-point?


Please mention the portion you are talking about..
No problems. Pl keep asking questions and I would reply to each whenever I get time :)
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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

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New post 29 Sep 2018, 09:24
"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."

This is the "trick" mentioned in the post, don't know what it means chetan2u
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 20 Nov 2018, 12:54
gettinit wrote:
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!!!!!



I didn't understand how the sign for each expression was determined.
I'd appreciate any help.
Thanks!!
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Re: Math: Absolute value (Modulus)  [#permalink]

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New post 26 Nov 2018, 15:07
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Re: Math: Absolute value (Modulus) &nbs [#permalink] 26 Nov 2018, 15:07

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