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Re: Math: Absolute value (Modulus)
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27 May 2011, 07:15
someonear wrote: I am worried about values of x that are on the border of the ranges Say hypothetically for a particular set of equations we end with the identical 4 cases we have here. Now if say I have x=4 then the way I had come up with the ranges I will get a solution between 3 and 4 as x=4 exists in 3<= x <= 4. But if however I consider what was done in the OP then x=4 exists in the range x>=4 Granted either way I have a solution but will it be a biggie if I fail to show exactly in which range the solution, if it does,exists
Your aim is to get the solution. You create the ranges to help yourself solve the problem. It doesn't matter at all in which range you consider the border value to lie. Say, when x = 4, (4  x) = 0. You solve saying that in the range 3<= x<4, (4  x) is positive and in the range x>= 4, (4  x) is negative. At the border value i.e. x = 4, (4  x) = 0. There is no negative or positive at this point. Hence it doesn't matter where you include the '='. Put it wherever you like. I just like to go in a regular fashion like walker did above. Include the first point in the first range, the second one in the second range (but not the third one i.e. 3 <= x < 4) and so on.
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Re: Math: Absolute value (Modulus)
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21 Jun 2011, 00:01
Hi Walker,
Can you please explain this?
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:
How do we get 3 key points and 4 conditions?



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Re: Math: Absolute value (Modulus)
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21 Jun 2011, 03:28
There are 3 points where one of the modules is zero: 1)x+3=0 > x = 3 2)4x=0 > x = 4 3)8+x=0 > x = 8 Those 3 points divide the number line by 4 pieces: 1) inf, 8 2) 8,3 3) 3, 4 4) 4, +inf and for each condition we are solving the equation separately.
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Re: Math: Absolute value (Modulus)
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10 May 2012, 19:11
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)
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11 May 2012, 03:59
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)
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27 Oct 2012, 03:08
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)
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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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Re: Math: Absolute value (Modulus)
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28 Oct 2012, 08:45
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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Re: Math: Absolute value (Modulus)
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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)
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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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Re: Math: Absolute value (Modulus)
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14 May 2013, 15:42
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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Re: Math: Absolute value (Modulus)
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14 May 2013, 21:15
nikhil007 wrote: 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? First of all, if you do get a question with multiple mods and if you want to be prepared for it, using algebra will be far more time consuming than the approaches discussed in my blog. But nevertheless, you should understand it properly. When you have an equation with x in it, you solve by taking x to one side and everything else to the other. What happens when you have mods in it? Say x = 4, you still haven't got the value of x. You have the value of x only. So you need to remove the mod. Now there are rules to remove the mod. x= x (mod removed) when x is >= 0, x= x (mod removed) when x < 0 So x = 4 to remove the mod, I need to know whether x is positive or negative. If x >= 0, x = x so x = 4 = x We get that x is 4 If x < 0, x = x so x = 4 = x hence x = 4 So if we are looking for a positive value, then it is 4 and if we are looking for a negative value, it is 4. Similarly, when you have x+4 + x  3 = 10 (just an example), you need to remove the mods to solve for x. But to remove mods (which are around the entire factors x4 and x3 and not just around x), you need to know whether (x+ 4) and (x  3) (the thing inside the mod) are positive/negative. So you split it into ranges: x > 3 Put any value greater than 3 in (x+4), (x+4) will remain positive. Put any value greater than 3 in (x  3), (x  3) will remain positive. So when x > 3, we can remove the mods without any modification: (x + 4) + (x3) = 10 x = 9/2 Since 9/2 is greater than 3, this value of x is acceptable. 4 < x< 3 For these values of x, (x+4) will always be positive but (x3) will be negative. So x  3 = (x3) (x + 4)  (x3) = 10 You don't have any such value for x x < 4 For these values of x, (x+4) and (x3) will be negative. So x  3 = (x3) and x+4 = (x+4) (x + 4)  (x3) = 10 x = 11/2 Since 11/2 is less than 4, this value of x is also acceptable. I have discussed how to deal with such questions logically here: http://www.veritasprep.com/blog/2011/01 ... spartii/As for question with factors that are multiplied (discussed in the 3 links given above), We know how to deal with (xa)(xb)(xc) > 0 type of questions so we try to bring it that form. 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 \(2x(x\frac{5}{2})(x6) <0\) (x0) is nothing but x. I put as (x0) to make it consistent to the (xa)(xb).... form to help you remember that you have to take 0 as a transition point too.
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Re: Math: Absolute value (Modulus)
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15 May 2013, 06:30
Karishma, Kudos given for the post, thanks for explaining in detail, I agree that we would be better off plugging number on such a ques, but things get tricky when it comes to DS I basically covered this from MGmat guides and I can handle a simple Mod like x2>5 what I learnt is simply take 2 conditions, x2>5 and 2x>5 and solve for 2 set of x, however the book never taught me this 3 step method. so I have to dig it in here.
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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 12:35
(This question is from a GMAT club study book. It can be found here: mathabsolutevaluemodulus86462.html) x^24 = 1. What is x? Solution: There are 2 conditions: a) (x^24)\geq0 > x \leq 2 or x\geq2. x^24=1 > x^2 = 5. x e {\sqrt{5}, \sqrt{5}} and both solutions satisfy the condition. b) (x^24)<0 > 2 < x < 2. (x^24) = 1 > x^2 = 3. x e {\sqrt{3}, \sqrt{3}} and both solutions satisfy the condition. Why do we set these problems up as >= or <= 1? I would solve this problem as follows: x^24 = 1 x^24 = 1 ==> x^2 = 5 ==> x = \sqrt{5} OR (x^24) = 1 ==> x^2 +4 = 1 ==> x^2 = 3 ==> X^2 = 3 ==> x = \sqrt{3} Thanks!



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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 12:46
The first method is the correct one and will always give you the correct results. Consider however the following case \(x+5=4\), at glance this equation has no solution because \(x+5\) cannot be less than 0. But I wanna take it as example: With the first method you'll find if \(x>5\) \(x+5=4\), \(x=9\), out of the interval => it's not a solution if \(x<5\) \(x5=4\), \(x=1\) out of the interval => it's not a solution With the second method \(x+5=4\), \(x=9\) \((x+5)=4\), or \(x=1\) those seem valid... but the equation we know that has no solution. Main point: the first method works always, do not rely on the other one. The second one does not take into consideration the intervals, so it might not work Hope it's clear
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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 15:39
Hmmmm...I'm not sure I follow. This is how I solved the problem. x^24=1 x^2  4 =1 OR x^2 + 4 = 1 SO x^2=5 ==> x=+/ √5 OR x^2=3 ==> x^2=3 ==> x=+/ √3 So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head! Zarrolou wrote: The first method is the correct one and will always give you the correct results.
Consider however the following case
\(x+5=4\), at glance this equation has no solution because \(x+5\) cannot be less than 0. But I wanna take it as example:
With the first method you'll find if \(x>5\) \(x+5=4\), \(x=9\), out of the interval => it's not a solution
if \(x<5\) \(x5=4\), \(x=1\) out of the interval => it's not a solution
With the second method \(x+5=4\), \(x=9\) \((x+5)=4\), or \(x=1\) those seem valid... but the equation we know that has no solution.
Main point: the first method works always, do not rely on the other one. The second one does not take into consideration the intervals, so it might not work
Hope it's clear



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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 16:20
WholeLottaLove wrote: Hmmmm...I'm not sure I follow.
This is how I solved the problem.
x^24=1
x^2  4 =1 OR x^2 + 4 = 1
SO
x^2=5 ==> x=+/ √5 OR x^2=3 ==> x^2=3 ==> x=+/ √3
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
The original solution used \(\geq{}\) and \(\leq{}\) to define the intervals. It solved the cases: \(x^24=1\) so \(x=+\sqrt{5}\), and then it check weather those numbers are in the interval \(2<x<2\). Both are inside so both are valid solutions Then the other case \(x^2+4=1\) so \(x=+\sqrt{3}\), and then check the interval it is considering in this scenario \(x<2\) and \(x>2\), both are inside the intervals so both are valid solutions As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment. If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage  you find the solutions, but you do not check if they are possible or not. In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the \(x+5=4\) example) Hope that what I mean is clear
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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 16:57
So, in other words, if I were to plug in /+ √5 into x^2 = 5 it will yield me a result between 2<x<2? And where does 2<x<2 come from??? Zarrolou wrote: WholeLottaLove wrote: Hmmmm...I'm not sure I follow.
This is how I solved the problem.
x^24=1
x^2  4 =1 OR x^2 + 4 = 1
SO
x^2=5 ==> x=+/ √5 OR x^2=3 ==> x^2=3 ==> x=+/ √3
So, I believe that is the right answer but I'm still not sure about how the greater than/less than signs come into play. Sorry for the thick head!
The original solution used \(\geq{}\) and \(\leq{}\) to define the intervals. It solved the cases: \(x^24=1\) so \(x=+\sqrt{5}\), and then it check weather those numbers are in the interval \(2<x<2\). Both are inside so both are valid solutions Then the other case \(x^2+4=1\) so \(x=+\sqrt{3}\), and then check the interval it is considering in this scenario \(x<2\) and \(x>2\), both are inside the intervals so both are valid solutions As I said before the correct method to solve abs values always checks if the result obtained is inside the interval is considering at that moment. If you do not double check weather the solution you find is inside the interval, you are likely to commit errors in the problem. What I am trying to say is that your method is incomplete: it lacks the last passage  you find the solutions, but you do not check if they are possible or not. In this example all 4 solutions are possible, so the last step does nothing; but if one of the solution were not valid, you "incomplete" method would not be albe to detect it. ( as I showed you in the \(x+5=4\) example) Hope that what I mean is clear



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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 17:16
WholeLottaLove wrote: So, in other words, if I were to plug in /+ √5 into x^2 = 5 it will yield me a result between 2<x<2? And where does 2<x<2 come from???
2<x<2 is the interval in which the function is negative, bare with me: Take the function x^24=1 1) Define where it is positive and where is negative => \(x^24>0\) if x<2 and x>2 So if \(x<2\) or \(x>2\) is positive, if \(2<x<2\) is negative 2)Study each case on its own: \(x^24=1\) \(x=+\sqrt{5}\), are those results valid? Are they in the interval we are considering? Are they in the x<2 or x>2 interval? \(\sqrt{5}\) is less than 2, and \(+\sqrt{5}\) is more than 2. So they are valid solutions because they are in the intervals we are considering \(x^2+4=1\) \(x=+\sqrt{3}\), are those results valid? same as above Yes they are valid because they are numbers between \(2\) and \(2\)(the interval we are considering now, in which abs is negative => x^2+4)
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Re: Math: Absolute value (Modulus)
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04 Jun 2013, 17:37
So, I need to find the positive and negative values of x^24=1 (i.e. x^24=1 and x^2+4=1) and which x values make x^24 positive and x^2+4 negative? Thanks for putting up with my slowness in picking up these concepts! Zarrolou wrote: WholeLottaLove wrote: So, in other words, if I were to plug in /+ √5 into x^2 = 5 it will yield me a result between 2<x<2? And where does 2<x<2 come from???
2<x<2 is the interval in which the function is negative, bare with me: Take the function x^24=1 1) Define where it is positive and where is negative => \(x^24>0\) if x<2 and x>2 So if \(x<2\) or \(x>2\) is positive, if \(2<x<2\) is negative 2)Study each case on its own: \(x^24=1\) \(x=+\sqrt{5}\), are those results valid? Are they in the interval we are considering? Are they in the x<2 or x>2 interval? \(\sqrt{5}\) is less than 2, and \(+\sqrt{5}\) is more than 2. So they are valid solutions because they are in the intervals we are considering \(x^2+4=1\) \(x=+\sqrt{3}\), are those results valid? same as above Yes they are valid because they are numbers between \(2\) and \(2\)(the interval we are considering now, in which abs is negative => x^2+4)




Re: Math: Absolute value (Modulus)
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