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Re: Inequalities trick
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22 Apr 2014, 23:42
VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 0, draw the points 2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x  7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x  7) and (x  4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be 2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x  a)(x  b)/(x  c)(x  d) < 0 (x + 2)(x  1)/(x 4)(x  7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. Hi Karishma, I have a query. I have following question x^3  4x^5 < 0 I can define this as (1+2x).x^3(12x). now I have roots 1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of 1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks



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Re: Inequalities trick
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23 Apr 2014, 03:43
PathFinder007 wrote: I have a query. I have following question
x^3  4x^5 < 0
I can define this as (1+2x).x^3(12x). now I have roots 1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of 1/2 and 0 I will get value as 0.
So How I will define them in graph and what range I will consider for this inequality.
Thanks
The factors must be of the form (x  a)(x  b) .... etc x^3  4x^5 < 0 x^3 * (1  4x^2) < 0 x^3 * (1  2x) * (1 + 2x) < 0 x^3 * (2x  1) * (2x + 1) > 0 (Note the sign flip because 12x was changed to 2x  1) x^3 * 2(x  1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and 1/2. ____________  1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or 1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/The links will give you the theory behind this method in detail.
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Re: Inequalities trick
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03 Jul 2014, 23:36
VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 0, draw the points 2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x  7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x  7) and (x  4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be 2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x  a)(x  b)/(x  c)(x  d) < 0 (x + 2)(x  1)/(x 4)(x  7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. can we do modulus questions with the help of this method? Also please suggest when we decide to take the + ve curve values or ve values.



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06 Jul 2014, 19:58
GmatDestroyer2013 wrote: VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 0, draw the points 2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x  7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x  7) and (x  4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be 2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x  a)(x  b)/(x  c)(x  d) < 0 (x + 2)(x  1)/(x 4)(x  7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. can we do modulus questions with the help of this method? Also please suggest when we decide to take the + ve curve values or ve values. This method is only for inequalities with factors such that they are of the form (xa). You should check out the links of the 3 posts I have given in my post above. They discuss the theory in detail and tell you why the regions are positive or negative. Also, they tell you how to handle complications such as factors of the form (a  x) or (ax + b) etc. When you have all factors of the form (xa) (or when you convert them to this form), the rightmost region is always positive. To solve absolute value questions you have a few different methods that you can use. http://www.veritasprep.com/blog/2014/06 ... thegmat/http://www.veritasprep.com/blog/2011/01 ... edoredid/http://www.veritasprep.com/blog/2011/01 ... htomods/http://www.veritasprep.com/blog/2011/01 ... spartii/
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Re: Inequalities trick
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22 Sep 2014, 21:31
VeritasPrepKarishma wrote: Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain. If (xa)(xb)(xc)(xd) < 0, we can draw the points a, b, c and d on the number line. e.g. Given (x+2)(x1)(x7)(x4) < 0, draw the points 2, 1, 7 and 4 on the number line as shown. Attachment: doc.jpg This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive. When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x  7) will be negative since x < 7 in this region. All other factors are still positive. When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x  7) and (x  4) are negative, but negative x negative is positive... and so on till you reach the leftmost section. Since we are looking for values of x where the expression is < 0, here the solution will be 2 < x < 1 or 4< x < 7 It should be obvious that it will also work in cases where factors are divided. e.g. (x  a)(x  b)/(x  c)(x  d) < 0 (x + 2)(x  1)/(x 4)(x  7) < 0 will have exactly the same solution as above. Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion. Hi Karishma Thanks for your valuable tips. Can you explain the logic when the expression has a denominator or when factors are divided. I did read the solution given below, but can you please explain the logic? Or if you can share a link in which you have explained expressions with denominators, I would be grateful Thanks



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Re: Inequalities trick
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22 Sep 2014, 21:48
alphonsa wrote: Hi Karishma Thanks for your valuable tips. Can you explain the logic when the expression has a denominator or when factors are divided. I did read the solution given below, but can you please explain the logic? Or if you can share a link in which you have explained expressions with denominators, I would be grateful Thanks Here: http://www.veritasprep.com/blog/2012/07 ... spartii/Last question here handles factors in the denominator.
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Re: Inequalities trick
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16 Oct 2014, 11:10
gurpreetsingh wrote: I learnt this trick while I was in school and yesterday while solving one question I recalled. Its good if you guys use it 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "" curves and the ans is : (a < x < b) , (c < x < d) and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
If you can not figure out how and why, just remember it. Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively. Hi Could you please explain this question to me with this logic? The explanation at ( rangeforvariablexinagiveninequality109468.html ) starts with  +  + logic instead of +  +  as advised above....



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Re: Inequalities trick
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17 Oct 2014, 09:21
mayankpant wrote: gurpreetsingh wrote: I learnt this trick while I was in school and yesterday while solving one question I recalled. Its good if you guys use it 12 times to get used to it.
Suppose you have the inequality
f(x) = (xa)(xb)(xc)(xd) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "" curves and the ans is : (a < x < b) , (c < x < d) and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have  +  + If f(x) has four factors then the graph will have +  +  +
If you can not figure out how and why, just remember it. Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively. Hi Could you please explain this question to me with this logic? The explanation at ( rangeforvariablexinagiveninequality109468.html ) starts with  +  + logic instead of +  +  as advised above.... There are two methods to go about it. Either decide whether the rightmost region is positive or negative depending on your factors or ALWAYS take rightmost region to be positive and adjust your factors accordingly i.e. bring all factors to the form (xa). The second method is discussed in detail in the links given here: inequalitiestrick9148260.html#p1358533
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Re: Inequalities trick
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05 Nov 2014, 21:40
mrinal2100 wrote: if = sign is included with < then <= will be there in solution like for (x+2)(x1)(x7)(x4) <=0 the solution will be 2 <= x <= 1 or 4<= x <= 7
in case when factors are divided then the numerator will contain = sign
like for (x + 2)(x  1)/(x 4)(x  7) < =0 the solution will be 2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite
correct me if i am wrong Can you please explain how the solution will be infinite ? when we take = into consideration.



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10 Nov 2014, 22:11
vpranavanshu91 wrote: mrinal2100 wrote: if = sign is included with < then <= will be there in solution like for (x+2)(x1)(x7)(x4) <=0 the solution will be 2 <= x <= 1 or 4<= x <= 7
in case when factors are divided then the numerator will contain = sign
like for (x + 2)(x  1)/(x 4)(x  7) < =0 the solution will be 2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite
correct me if i am wrong Can you please explain how the solution will be infinite ? when we take = into consideration. Given inequality: \(\frac{(x + 2)(x  1)}{(x 4)(x  7)} <= 0\) What happens if we put x = 4? We get \(\frac{(x + 2)(x  1)}{(4 4)(x  7)} <= 0\) \(\frac{(x + 2)(x  1)}{0} <= 0\) Division by 0 is not defined. So x cannot be 4 and neither can it be 7.
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Re: Inequalities trick
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11 Nov 2014, 20:02
VeritasPrepKarishma  Could you please help me find solution for this equation using the graph method? \(x^2  4 > 3x\). I know Bunuel and others have replied with different solutions but could you show me how this can be solved using graphs?



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11 Nov 2014, 23:45
Blackbox wrote: VeritasPrepKarishma  Could you please help me find solution for this equation using the graph method? \(x^2  4 > 3x\). I know Bunuel and others have replied with different solutions but could you show me how this can be solved using graphs? Draw the diagram. x^2 is a quadratic with minimum at (0, 0). To get x^2  4, move the curve down 4 units on the y axis. Take the mod by flipping whatever is below x axis to its reflection above x axis. Attachment:
Ques3.jpg [ 15.92 KiB  Viewed 7170 times ]
Before the first intersection of the line 3x with the hump of x^2  4, x^2  4 is greater. After this intersection, the line 3x is greater. Note that 3x will intersect the curve again and after that, x^2  4 will again be greater than 3x. So all we need to do is find the intersections. To get first point of intersection: (x^2  4) = 3x x^2 + 3x 4 = 0 x = 1, 4 To get second point of intersection: x^2  4 = 3x x^2  3x  4 = 0 x = 4, 1 Ignore negative x values since our intersections are in first quadrant only. Between x = 1 and x = 4, 3x is greater than x^2  4. But when x < 1 or x > 4, x^2  4 > 3x.
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Re: Inequalities trick
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12 Nov 2014, 22:43
vpranavanshu91 wrote: For statement I (1) (12x)(1+x)<0 Once you get this into the req form 2 (x1/2) (x+1) > 0 sol +++++ 1  1/2 +++++ As Karishma pointed out ponce in the req form, the rightmost will always be positive and the alternating will happen from there. So sol for this x> 1/2 and x<1 Integers greater than 1/2 and less than 1 , thus x may be >= 1. Unsure Thus Insufficient. For Statement II (2) (1x)(1+2x)<0 Once you get this into the req form 2(x+1/2) (x1) > 0 sol +++++ 1/2  1 +++++ So sol for this x>1 and x< 1/2 Integers greater than 1 and less than 1/2 , thus x may be >= 1. Unsure Thus Insufficient. Combining Both the statements x<1 and x>1 Thus Integers for this range will give x > 1 Thus Sufficient. Hope this helps. I am not very confident of my solution though. Its my first solution gmatclubResponding to a pm: Yes, on the whole, the solution is fine. A couple of things: Question: Is x > 1? Rewrite: Is x>1 or x<1? x is an integer. "So sol for this x> 1/2 and x<1 " should be "So sol for this x> 1/2 OR x<1 " x cannot be both greater than 1/2 AND less than 1. It will be only one of those two. x could be 1 here so you are right that this statement alone is not sufficient. Similar analysis for statement 2 too. Both together, we get that x > 1 or x < 1 so sufficient together.
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Re: Inequalities trick
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05 Jan 2015, 05:18
I had read this post a while ago, and not heeded the advice (i thought it was beyond my understanding). But I came across quite a few questions where this could be used (links that I saw in the discreet charm of DS post), and realized knowing this could be blessing. But to make sure i understood right, here's a paraphrase: In cases of inequality regarding factors of a function, in terms of zero and depending on the sign of inequality, a few shortcuts (of which, the logic has been explained by veritasprepkarishma) can be employed. The roots of the inequality, once arranged in the ascending order, maybe on a number line even, can be assigned + and  signs by starting from + in the right most portion of the line(which is openended), alternating between +and  till the last portion (i.e. the left most one) Now, here is where the sign '<'/'>' matters: if the sign is < then I am to consider the ranges where the  sign lies (set by the roots marking such a range) and if the sign is > then i am to consider ranges that contain + sign. Is this correct? This applies only for inequalities of zero, right? Also, in the "now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution.....", what should I combine? I don't think we are talking about combining + and  here, are we? Am I to combine the ranges and place the variable(in this case 'x') in this range? Thanks in advance



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Re: Inequalities trick
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05 Jan 2015, 19:37
deeuk wrote: I had read this post a while ago, and not heeded the advice (i thought it was beyond my understanding). But I came across quite a few questions where this could be used (links that I saw in the discreet charm of DS post), and realized knowing this could be blessing. But to make sure i understood right, here's a paraphrase: In cases of inequality regarding factors of a function, in terms of zero and depending on the sign of inequality, a few shortcuts (of which, the logic has been explained by veritasprepkarishma) can be employed. The roots of the inequality, once arranged in the ascending order, maybe on a number line even, can be assigned + and  signs by starting from + in the right most portion of the line(which is openended), alternating between +and  till the last portion (i.e. the left most one) Now, here is where the sign '<'/'>' matters: if the sign is < then I am to consider the ranges where the  sign lies (set by the roots marking such a range) and if the sign is > then i am to consider ranges that contain + sign. Is this correct? Yes, but remember, the factors of the inequality should be of the form (axb), (cxd) etc. If they are not in this form, convert them to this form using the techniques explained in the links given here: inequalitiestrick9148260.html#p1265455Quote: This applies only for inequalities of zero, right?
Yes, but you will be able to bring most GMAT inequalities in the 'zero on the right hand side' form. Quote: Also, in the "now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution.....", what should I combine? I don't think we are talking about combining + and  here, are we? Am I to combine the ranges and place the variable(in this case 'x') in this range? Thanks in advance Are you talking about a particular question here? Kindly give the link.
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Updated on: 10 Jan 2015, 20:41
Just came across this useful discussion. VeritasPrepKarishma has given a very lucid explanation of how this “wavy line” method works. I have noticed that there is still a little scope to take this discussion further. So here are my two cents on it. I would like to highlight an important special case in the application of the Wavy Line Method When there are multiple instances of the same root:Try to solve the following inequality using the Wavy Line Method: \((x1)^2(x2)(x3)(x4)^3 < 0\) To know how you did, compare your wavy line with the correct one below. Did you notice how this inequality differs from all the examples above? Notice that two of the four terms had an integral power greater than 1. How to draw the wavy line for such expressions?Let me directly show you how the wavy line would look and then later on the rule behind drawing it. Attachment: File comment: Observe how the wave bounces back at x = 1.
bounce.png [ 10.4 KiB  Viewed 7241 times ]
Notice that the curve bounced down at the point x = 1. (At every other root, including x = 4 whose power was 3, it was simply passing through them.) Can you figure out why the wavy line looks like this for this particular inequality? (Hint: The wavy line for the inequality \((x1)^{38}(x2)^{57}(x3)^{15}(x4)^{27} < 0\) Is also the same as above) Come on! Give it a try. If you got it right, you’ll see that there are essentially only two rules while drawing a wavy line. (Remember, we’ll refer the region above the number line as positive region and the region below the number line as negative region.) How to draw the wavy line?1. How to start: Start from the top right most portion. Be ready to alternate (or not alternate) the region of the wave based on how many times a point is root to the given expression. 2. How to alternate: In the given expression, if the power of a term is odd, then the wave simply passes through the corresponding point (root) into the other region (to –ve region if the wave is currently in the positive region and to the +ve region if the wave is currently in the negative region). However, if the power of a term is even, then the wave bounces back into the same region. Now look back at the above expression and analyze your wavy line. Were you (intuitively) using the above mentioned rules while drawing your wavy line? SolutionOnce you get your wavy line right, solving an inequality becomes very easy. For instance, for the above inequality, since we need to look for the space where the above expression would be less than zero, look for the areas in the wavy line where the curve is below the number line. So the correct solution set would simply be {3 < x < 4} U {{x < 2} – {1}} In words, it is the Union of two regions – region1 between x = 3 and x = 4 and region2 which is x < 2, excluding the point x = 1. Food for ThoughtNow, try to answer the following questions: 1. Why did we exclude the point x = 1 from the solution set of the last example? (Easy Question) 2. Why do the above mentioned rules (especially rule #2) work? What is/are the principle(s) working behind the curtains? Foot Note: Although the post is meant to deal with inequality expressions containing multiple roots, the above rules to draw the wavy line are generic and are applicable in all cases.  Krishna
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08 Jan 2015, 04:44
VeritasPrepKarishma wrote: deeuk wrote: I had read this post a while ago, and not heeded the advice (i thought it was beyond my understanding). But I came across quite a few questions where this could be used (links that I saw in the discreet charm of DS post), and realized knowing this could be blessing. But to make sure i understood right, here's a paraphrase: In cases of inequality regarding factors of a function, in terms of zero and depending on the sign of inequality, a few shortcuts (of which, the logic has been explained by veritasprepkarishma) can be employed. The roots of the inequality, once arranged in the ascending order, maybe on a number line even, can be assigned + and  signs by starting from + in the right most portion of the line(which is openended), alternating between +and  till the last portion (i.e. the left most one) Now, here is where the sign '<'/'>' matters: if the sign is < then I am to consider the ranges where the  sign lies (set by the roots marking such a range) and if the sign is > then i am to consider ranges that contain + sign. Is this correct? Yes, but remember, the factors of the inequality should be of the form (axb), (cxd) etc. If they are not in this form, convert them to this form using the techniques explained in the links given here: inequalitiestrick9148260.html#p1265455Quote: This applies only for inequalities of zero, right?
Yes, but you will be able to bring most GMAT inequalities in the 'zero on the right hand side' form. Quote: Also, in the "now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution.....", what should I combine? I don't think we are talking about combining + and  here, are we? Am I to combine the ranges and place the variable(in this case 'x') in this range? Thanks in advance Are you talking about a particular question here? Kindly give the link. When (axb) is NOT the form, convert it to such a form, and the root will be b/a. Got it. And no. I was referring to this post itself. I mean, the first post of this thread.



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Re: Inequalities trick
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08 Jan 2015, 20:01
deeuk wrote: When (axb) is NOT the form, convert it to such a form, and the root will be b/a. Got it. And no. I was referring to this post itself. I mean, the first post of this thread.
Yes, you will take the constant a out and will be left with (x  b/a) as a factor. Also, if a factor is (b  x), multiply both sides by 1 to get (x  b). The inequality sign will flip in this case. These and more complications are discussed in the links mentioned in my post above.
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Re: Inequalities trick
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15 Jan 2015, 22:54
deeuk wrote: VeritasPrepKarishma wrote: deeuk wrote: Also, in the "now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution.....", what should I combine? I don't think we are talking about combining + and  here, are we? Am I to combine the ranges and place the variable(in this case 'x') in this range? Thanks in advance
Are you talking about a particular question here? Kindly give the link. And no. I was referring to this post itself. I mean, the first post of this thread. I was referring to the first post of this thread. inequalitiestrick91482.html



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Re: Inequalities trick
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16 Jan 2015, 00:30
deeuk wrote: Also, in the "now if f(x) < 0 consider curve having "" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution.....", what should I combine? I don't think we are talking about combining + and  here, are we? Am I to combine the ranges and place the variable(in this case 'x') in this range? I was referring to the first post of this thread. inequalitiestrick91482.htmlYou combine all the + ranges in case of f(x) > 0 to get the possible values of x. In case of f(x) < 0, you combine all  ranges together to get the possible value of x. If you combine all + and  ranges together, you will get the entire number line. That wouldn't make sense, now would it?
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Re: Inequalities trick &nbs
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