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I hate absolute value...this makes it easier! Thank you!



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24 Nov 2011, 11:17
Nice Topic and nice explanation... +1 kudos...
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28 Feb 2012, 20:40
Great Post !! Thanks to contributors esp Karishma



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02 Mar 2012, 13:01
Great topic. Absolute value always confuses me.
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Let’s consider following examples,
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) \(x < 8\). \((x+3)  (4x) = (8+x)\) > \(x = 1\). 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)
Two Questions: Can this only be done when we all absolute values in the equation. Could we have done it for \(x+3  5 = 8+x\)? States the conditions are 3 and 8?
Second, How did you know whether to put a negative or make positive the terms in the conditions?
For example in (a) you made \((x+3) and (8+x)\) negative in (b) you made \((x+3)\) negative and everything positive.
What gives?
Thank you!
How do you know



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Re: Math: Absolute value (Modulus) [#permalink]
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11 May 2012, 03:59
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1) yes, you can use the same approach for x+35=8+x 2) let say we have condition x < 8. Then x + 8 =  (x+8). Why do we have "" here? Because (x+8) is always negative at x<8 and we need to add "" to get a positive value. Actually, it's the definition of the absolute value: x = x for x >=0 x = x for x<0
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Re: Math: Absolute value (Modulus) [#permalink]
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17 May 2012, 04:32
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walker wrote: 1) yes, you can use the same approach for x+35=8+x
Thanks a lot for the explanation, had the same question, as the guy above. Kudos to you. MGMAT doesnt explain absolute value as good, as u do. thanks again



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Re: Math: Absolute value (Modulus) [#permalink]
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27 Oct 2012, 03:08
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Here is a GMAT DS question from the topic Inequalities. Tests concepts of modulus. Question Is a > b? 1. 1/(a  b) > 1/(b  a) 2. a + b < 0 Correct Answer : Choice C. Both statements together are sufficient. Explanatory Answer We need to determine whether a is greater than b. The answer to this question will be a conclusive 'yes' if a > b. The answer will be a conclusive 'n' if a <= b Let us evaluate statement 1 1/(a  b) > 1/(b  a) We can rewrite the same inequality as 1/(a  b) > 1(a  b). If a number is greater than the negative of the number, the number has to be a positive number. So, we can conclude that a  b > 0 or a > b. If a > b, a may or may not be greater than b. For e.g, a = 4, b = 2. Then a > b. For positive a and b, when a > b, a > b. Let us look at a counter example. a = 2 and b = 10. a > b. But a < b. Hence, we cannot conclude from statement 1 whether a > b. Statement 1 is NOT sufficient. Let us evaluate statement 2 a + b < 0 Either both a and b are negative or one of a or b is negative. If only one of the two numbers is negative, then the magnitude of the negative number is greater than the magnitude of the negative number is greater than the magnitude of the positive number. For e.g., a = 3 and b = 4. a + b < 0, a < b Here is a counter example: a = 4 and b = 3. a + b < 0 and a > b. So, statement 2 is NOT sufficient. Let us combine the two statements. We know a > b from statement 1 and a + b < 0 from statement 2. If both a and b are negative, and we know that a > b, then a < b. Note in negative numbers, lesser the magnitude, greater the number. If one of 'a' or 'b' is negative, as a > b, a has to be positive and b has to be negative. The sum of a + b < 0. Therefore, the magnitude of a has to be lesser than the magnitude of b. So, we can conclude that a < b. Hence, by combining the two statements we can conclude that a is not greater than b. So, the two statements taken together are sufficient to answer the question. Choice C is the correct answer Here is an alternative explanation for the same. From statement 1 we know a  b > 0. From statement 2 we know a + b < 0. So, (a  b)(a + b) < 0 Or a^2  b^2 < 0 or a^2 < b^2 If a^2 < b^2, we can conclude that a < b.
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Re: Math: Absolute value (Modulus) [#permalink]
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27 Oct 2012, 14:17
[quote="walker"]
How to decide the signs of equality in the first example given :
i.e after determining check points when to use < or <= , how to decide on this and in which interval ? (referring to 2nd example explained above)
why we use x<8 in (a) why not x<=8 , how to decide upon this and similarly in other subsequent intervals
In second example:
How in a) x<=2 or x>=2 , when it should be x>= 2 , 2
How in b) 2<x<2 , when it should be x< 2 , 2
Pl. help me in understanding the above concepts.



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Re: Math: Absolute value (Modulus) [#permalink]
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27 Oct 2012, 15:26
1) We can use < or <= as it doesn't matter because in a key point the modulus turns 0 and you can open it at any key point with negative or positive sign. For example, x = 4 So, you can do: 1) x>=0; x = 4 2) x<0; x = 4 or 1) x>0; x = 4 2) x<=0; x = 4  2) x^24 >= 0  > x^2 >= 4 > x >= 2 or x <= 2
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Re: Math: Absolute value (Modulus) [#permalink]
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27 Oct 2012, 21:05
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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)
hi ,
could you kindly explain more about the shift change in above eqn...
since x<8 we have modified the sign of all the x from + to  but what about (4x). I am bit confused.



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Re: Math: Absolute value (Modulus) [#permalink]
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28 Oct 2012, 06:55
when x<8, (4x) is always positive and we don't need to modify the sign when we open it.
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Re: Math: Absolute value (Modulus) [#permalink]
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28 Oct 2012, 07:23
walker wrote: when x<8, (4x) is always positive and we don't need to modify the sign when we open it. sorry Walker, I still don't get it when we consider x<8 then we made all the x to be negative so multiplied each with negative terms.. is that (4x) is already having negative for x so we didn't have to multiply with negative here. Sorry i am really novice for this kind of problem.. please also let me know if there is any post related with this



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breakit wrote: walker wrote: when x<8, (4x) is always positive and we don't need to modify the sign when we open it. sorry Walker, I still don't get it when we consider x<8 then we made all the x to be negative so multiplied each with negative terms.. is that (4x) is already having negative for x so we didn't have to multiply with negative here. Sorry i am really novice for this kind of problem.. please also let me know if there is any post related with this You might want to check out these posts where I have discussed these concepts in detail: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/
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20 Nov 2012, 04:11
Great post. It would be nice if we have some sample questions also. (apart from few mentioned in original post from og)
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28 Nov 2012, 21:16
(Optional) The following illustration may help you understand how to open modulus at different conditions. Image
Can someone pls. help explain if there is a quicker way to understand how to open modulus for different conditions...figuring out looking at the illustration takes too much time for me...any help is appreciated..



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Re: Math: Absolute value (Modulus) [#permalink]
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22 Jan 2013, 03:18
Hello everyone!! I have some problem to understand this section, I don't understand the first example: I have understood how find the key number (8,3,4) but I haven't understood how you have set up the four conditions, in the a) point you set up x<8 and then you find x=1, how can you find x=1? do you substitute a number below 8 in the equalities or do you do other operation? can you clarify this point please? Thank you so much!



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Re: Math: Absolute value (Modulus) [#permalink]
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13 May 2013, 12:43
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.: \(x+3  4x = 8+x\). 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!!!!! like the gentlemen above, I continue to be puzzled by this post. I tried searching for answer in this topic post itself, but couldn't get a compilation to all my questions, can any expert please make me understand this, here are my queries. 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: I get this part, perfectly fine, basic goal to arrive at points is to make every value within modulus "0"so we have 3 key points and 4 solutions Ie, 3+3= 0 for 1st modulus sign so key point here is 3, 44=0 for second modulus sign, so key point here is 4 88 in third modulus, so key point here is 8 Therefore on a number line it will be 3 points something like this \((8)\)\((3)\)\((4)\) second step: Quote: A. 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)
I do understand in first Bracket \((x+3)\), since we are testing X against x < 8[/m], so we need to make \(X\) here. as per Walkers quote walker wrote: 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. but my Question is If we eventually want to see a negative X inside the bracket than why \( (4x)\)? as in this case X will turn positive after opening the bracket 2nd EQ \(8 \leq x < 3\) \((x+3)  (4x)\) = \((8+x)\) again in 2nd equation my doubt is why do we have the \((8+X)\) as non negative, I mean it should be same as \((8+x)\), like in 1st test case. as X is still negative. in this test case? Ofcourse this is fine if I can get answer to my 1st query, if we have to make X negative than this is not ok. in 3rd test case Quote: 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.) in this case X can be negative or positive, so why don't we put \((x+3)\) here? rather than \((X+3)\) ? Quote: 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) Again in the equation above we are testing against positive X test point, than why \(+(4X)\), I think it should be \((4X)\) to turn X into positive after opening the brackets.? All of the above questions may sound stupid, but I need to understand this, as inequalities is my weak topic.
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Re: Math: Absolute value (Modulus) [#permalink]
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14 May 2013, 09:51
nikhil007 wrote: 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.: \(x+3  4x = 8+x\). 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!!!!! like the gentlemen above, I continue to be puzzled by this post. I tried searching for answer in this topic post itself, but couldn't get a compilation to all my questions, can any expert please make me understand this, here are my queries. 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: I get this part, perfectly fine, basic goal to arrive at points is to make every value within modulus "0"so we have 3 key points and 4 solutions Ie, 3+3= 0 for 1st modulus sign so key point here is 3, 44=0 for second modulus sign, so key point here is 4 88 in third modulus, so key point here is 8 Therefore on a number line it will be 3 points something like this \((8)\)\((3)\)\((4)\) second step: Quote: A. 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)
I do understand in first Bracket \((x+3)\), since we are testing X against x < 8[/m], so we need to make \(X\) here. as per Walkers quote walker wrote: 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. but my Question is If we eventually want to see a negative X inside the bracket than why \( (4x)\)? as in this case X will turn positive after opening the bracket 2nd EQ \(8 \leq x < 3\) \((x+3)  (4x)\) = \((8+x)\) again in 2nd equation my doubt is why do we have the \((8+X)\) as non negative, I mean it should be same as \((8+x)\), like in 1st test case. as X is still negative. in this test case? Ofcourse this is fine if I can get answer to my 1st query, if we have to make X negative than this is not ok. in 3rd test case Quote: 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.) in this case X can be negative or positive, so why don't we put \((x+3)\) here? rather than \((X+3)\) ? Quote: 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) Again in the equation above we are testing against positive X test point, than why \(+(4X)\), I think it should be \((4X)\) to turn X into positive after opening the brackets.? All of the above questions may sound stupid, but I need to understand this, as inequalities is my weak topic. In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
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VeritasPrepKarishma wrote: In a post above, I have given the links to 3 posts which explain the process in detail (including the reasoning behind the process). Check those out to get answers to your questions.
Hi Karishma I went through your post on the blog, but to be frank found this post of your more helpfull VeritasPrepKarishma wrote: 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 (x2) < 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
Ok, Now after literally banging my head for 3 hrs and reading you blog articles back and forth, I get it that to make an EQ in (XK) format we manipulate it by taking tive sign out but I guess in this example its this concept that we need to understand x= x when x is >= 0, x= x when x < 0 ok, so based on this understanding I will take a fresh shot, please let me know what's wrong Quote: 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) In this test case, since we will always have x+3 negative we put a tive sign outside because modulus will turn it into non negative, so to do that we take multiply it by (1), is this understanding correct? and since we are ok with (4x), because we will again get 4x positive with a negative x, the tive sign outside the bracket will make sure its always tive when out of the Modulus. However to be frank, a little confusion here is, as you mentioned in the blog, why don't we try to convert it into (xk) format? in RHS we have (8x) because again we want 8x to turn out a negative number so we put (8x) to make it always negative, let me know if I got it correctly. Quote: 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.) again, I get it why LHS is that way, however I still don't get it why we don't have (8X) as we need to make sure that the result of this bracket is tive so 8x = (8x) Quote: 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.) I still dont get it, if we test x against both tive and positive scenario, why is that we just have 1 equation? in my view we should split it in 2 eq. to test against both negative and positive value. Quote: 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) (x+3) since we don't have a negative value of X this bracket will always be positive, we don't need a tive sign outside, is this the reason? (4x) again since a >4 will always make it positive we don't need a tive sign outside the bracket, is this the reason? (8+x) again same reason as above for this? I also had one doubt in your blog question. Complication No 3: on this post http://www.veritasprep.com/blog/2012/07 ... nsparti/(2x^3 + 17x^2 – 30x) > 0 This is how I understand it, \(x(2x^2 + 17x  30) > 0\) (just took out x common) ok x(2x – 5)(6 – x) > 0(factoring the quadratic) ok 2x(x – 5/2)(1)(x – 6) > 0 (take 2 common) > I think in this you took out 1 common to make the second bracket = (xk) format? 2(x–0)(x–5/2)(x–6) < 0 (multiply both sides by 1)> how did you arrive at 2(x0)? i think it should be just \(2x(x\frac{5}{2})(x6) <0\)
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