Author 
Message 
TAGS:

Hide Tags

CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Inequalities trick [#permalink]
Show Tags
16 Mar 2010, 10:11
86
This post received KUDOS
259
This post was BOOKMARKED
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.
Attachments
1.jpg [ 6.73 KiB  Viewed 51501 times ]
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
22 Oct 2010, 06:33
67
This post received KUDOS
Expert's post
89
This post was BOOKMARKED
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 [ 7.9 KiB  Viewed 50569 times ]
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.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Inequalities trick [#permalink]
Show Tags
11 Mar 2011, 06:49
20
This post received KUDOS
11
This post was BOOKMARKED
vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote:
How you have decided on the first sign of the graph?Why it is ve if it has three factors and +ve when four factors?
Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Yaxis for four values and starts from ve Yaxis for three values. What if the equation you mentioned is (x+2)(x1)(x7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x1)(x7) < 0 Here the roots are; 2,1,7 Arrange them in ascending order; 2,1,7; These are three points where the wave will alternate. The ranges are; x<2 2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(10001)(10007) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7> ve between 2 and 1> +ve Before 2 > ve Since the inequality has the less than sign; consider only the ve side of the graph; 1<x<7 or x<2 is the complete range of x that satisfies the inequality.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Manager
Joined: 29 Sep 2008
Posts: 143

Re: Inequalities trick [#permalink]
Show Tags
22 Oct 2010, 11:45
12
This post received KUDOS
16
This post was BOOKMARKED
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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
11 Mar 2011, 19:57
vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Yaxis for four values and starts from ve Yaxis for three values. What if the equation you mentioned is (x+2)(x1)(x7)<0,will the last two ranges be excluded or the graph will also change?
Ok, look at this expression inequality: (x+2)(x1)(x7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x  1) will be positive and (x7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when 2 , x < 1 and negative when x < 2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < 2. Now let me add another factor: (x+8)(x+2)(x1)(x7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and  signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax  b), not (b  ax)... e.g. (x+2)(x1) (7  x)<0 Convert this to: (x+2)(x1) (x7)>0 (Multiply both sides by '1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Status: On...
Joined: 16 Jan 2011
Posts: 188

Re: Inequalities trick [#permalink]
Show Tags
10 Aug 2011, 17:01
7
This post received KUDOS
16
This post was BOOKMARKED
WoW  This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help. 1) CORE CONCEPT@gurpreetsingh  Suppose you have the inequality f(x) = (xa)(xb)(xc)(xd) < 0
Arrange the NUMBERS in ascending order from left to right. a<b<c<d 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. 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. Note: Make sure that the factors are of the form (ax  b), not (b  ax)...example  (x+2)(x1)( 7  x)<0 Convert this to: (x+2)(x1)(x7)>0 (Multiply both sides by '1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. 2) Variation  ODD/EVEN POWER@ulm/Karishma  if we have even powers like (xa)^2(xb) we don't need to change a sign when jump over "a". This will be same as (xb)We can ignore squares BUT SHOULD consider ODD powersexample  2.a (xa)^3(xb)<0 is the same as (xa)(xb) <0 2.b (x  a)(x  b)/(x  c)(x  d) < 0 ==> (x  a)(x  b)(xc)^1(xd)^1 <0 is the same as (x  a)(x  b)(x  c)(x  d) < 0 3) Variation <= in FRACTION@mrinal2100  if = sign is included with < then <= will be there in solutionlike for (x+2)(x1)(x7)(x4) <=0 the solution will be 2 <= x <= 1 or 4<= x <= 7 BUT if it is a fraction the denominator in the solution will not have = SIGNexample  3.a (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 4) Variation  ROOTS @Karishma  As for roots, you have to keep in mind that given \(\sqrt{x}\), x cannot be negative. \(\sqrt{x}\) < 10 implies 0 < \(\sqrt{x}\) < 10 Squaring, 0 < x < 100 Root questions are specific. You have to be careful. If you have a particular question in mind, send it. Refer  inequalitiesandroots118619.html#p959939 Some more useful tips for ROOTS....I am too lazy to consolidate<5> THESIS  @gmat1220  Once algebra teacher told me  signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane. I will save this future references.... Please add anything that you feel will help. Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post
_________________
Labor cost for typing this post >= Labor cost for pushing the Kudos Button http://gmatclub.com/forum/kudoswhataretheyandwhywehavethem94812.html



eGMAT Representative
Joined: 04 Jan 2015
Posts: 726

Inequalities trick [#permalink]
Show Tags
06 Jan 2015, 03:35
7
This post received KUDOS
Expert's post
17
This post was BOOKMARKED
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 5612 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
_________________
 '4 out of Top 5' Instructors on gmatclub  70 point improvement guarantee  www.egmat.com



ISB Thread Master
Joined: 20 Jul 2015
Posts: 99
Location: India
Concentration: Marketing, General Management
GMAT 1: 720 Q49 V40 GMAT 2: 720 Q50 V38 GMAT 3: 760 Q50 V42
GPA: 3.8
WE: Engineering (NonProfit and Government)

Re: Inequalities trick [#permalink]
Show Tags
07 Nov 2015, 23:34
4
This post received KUDOS
Manali888 wrote: if (k – 5)(k – 1)(k – 6) < 0 or (k – 5)(k – 1)(k – 6) > 0 what will be range for both? I have tried method suggested by gurpreetsingh, but got confused when to take ">" and "<". can anyone explain? It will be more helpful if anyone can upload an image of the solution. Thanks. See the attached solution. Hope its clear!
Attachments
IMG_20151108_115517448[1].jpg [ 1.56 MiB  Viewed 1179 times ]



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
22 Oct 2010, 12:46
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
mrinal2100: Kudos to you for excellent thinking!
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
09 Sep 2013, 23:35
3
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
karannanda 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 Gurpreet, Thanks for the wonderful method. I am trying to understand it so that i can apply it in tests. Can you help me in applying this method to the below expression to find range of x. x^3 – 4x^5 < 0? I am getting the roots as 1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help. Before you apply the method, ensure that the factors are of the form (x  a)(x  b) etc \(x^3  4x^5 < 0\) \(x^3 ( 1  4x^2) < 0\) \(x^3(1  2x) (1 + 2x) < 0\) \(4x^3(x  1/2)(x + 1/2) > 0\) (Notice the flipped sign. We multiplied both sides by 1 to convert 1/2  x to x  1/2) Now the transition points are 0, 1/2 and 1/2 so put + in the rightmost region. The solution will be x > 1/2 or 1/2 < x< 0. Check out these posts discussing such complications: http://www.veritasprep.com/blog/2012/06 ... efactors/http://www.veritasprep.com/blog/2012/07 ... nsparti/http://www.veritasprep.com/blog/2012/07 ... spartii/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
23 Jul 2012, 03:13
Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\(\frac {(x+2)(x1)}{(x4)(x7)}\) were \(4\leq x \leq 7\), than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Current Student
Joined: 04 May 2015
Posts: 75
Concentration: Strategy, Operations
WE: Operations (Military & Defense)

Re: Inequalities trick [#permalink]
Show Tags
27 Aug 2015, 17:33
2
This post received KUDOS
vihavivi wrote: VeritasPrepKarishma wrote: vihavivi wrote: Can somebody please explain for me why this one doesnt work that way?
Which of the following is a value of x for which x^11  x^9 > 0 ?
A. 2 B. 1 C. 1/2 D. 1/2 E. 1
x11−x9>0 > x9(x2−1)>0 > (x+1)x9(x−1)>0 > roots are 1, 0 and 1 > −1<x<0 or x>1. Only C fits.
Answer: C. But it does. Everything you have done is correct (ignoring the formatting issues). I knew the answer makes sense, but if I apply the inequalities trick into this, with 3 roots 1, 0, and 1 and for the f(x) >0, shouldn't it be x <1 and 0 <x<1? Thanks Hi Vihavivi, I think you have your inequalities curve flipped upside down. When you have the roots plotted on your line, it should always open up positive on the right and everything alternates from there. There was a previous post on the first page by fluke where he suggest plugging in large numbers to intuitively highlight what's going on. In your case being \(x^9(x+1)(x1)\) just choose say x=1000... In this case all three components will be positiive...\((1000^x)(10001)(1000+1)\) will surely yield a positive answer. Use this as a reminder that in these cases the inequalities line will open up positive to the right. Have a look at my freehand below. If you find this useful maybe throw me my first Kudos???
Attachments
File comment: Show's this equations "inequalities trick" line
IMG_20150828_3395.jpg [ 50.19 KiB  Viewed 1386 times ]
_________________
If you found my post useful, please consider throwing me a Kudos... Every bit helps



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
24 Sep 2015, 23:07
mrslee wrote: Dear All, Is x a negative number? (1) 9x > 10x (2) is positive. GMAT OG claims A is sufficient. But, what if x is a fraction number? Such as 2/90. In this case 2/90 also satisfies ST 1 .. What is the exact solution for this? 2/90 does not satisfy statement 1. No positive value satisfies statement 1. 9*2/90 > 10*2/90 18/90 > 20/90  this is false
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Inequalities trick [#permalink]
Show Tags
19 Mar 2010, 12:59
1
This post received KUDOS
ttks10 wrote: Can u plz explainn the backgoround of this & then the explanation.
Thanks i m sorry i dont have any background for it, you just reread it again and try to implement whenever you get such question and I will help you out in any issue. sidhu4u wrote: I have applied this trick and it seemed to be quite useful. Nice to hear this....good luck.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Math Forum Moderator
Joined: 20 Dec 2010
Posts: 2010

Re: Inequalities trick [#permalink]
Show Tags
26 May 2011, 20:55
1
This post received KUDOS
chethanjs 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 tell me why the solution gets infinite for 4<=x<=7 ? Thanks. (x 4)(x  7) is in denominator. Making x=4 or 7 would make the denominator 0 and the entire function undefined. Thus, the range of x can't be either 4 or 7. 4<=x<=7 would be wrong. 4<x<7 is correct because now we removed "=" sign.
_________________
~fluke
GMAT Club Premium Membership  big benefits and savings



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
08 Aug 2011, 11:59
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
Asher wrote: gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (xa)^2(xb) we don't need to change a sign when jump over "a". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a This way of solving inequalities actually makes it soo much easier. Thanks gurpreetsingh and karishma However, i am confused about how to solve inequalities such as: (xa)^2(xb) and also ones with root value. could someone please explain. When you have (xa)^2(xb) < 0, the squared term is ignored because it is always positive and hence doesn't affect the sign of the entire left side. For the left hand side to be negative i.e. < 0, (x  b) should be negative i.e. x  b < 0 or x < b. Similarly for (xa)^2(xb) > 0, x > b As for roots, you have to keep in mind that given \(\sqrt{x}\), x cannot be negative. \(\sqrt{x}\) < 10 implies 0 < \(\sqrt{x}\) < 10 Squaring, 0 < x < 100 Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
11 Aug 2011, 22:57
1
This post received KUDOS
Expert's post
2
This post was BOOKMARKED
sushantarora wrote: hey , can u please tel me the solution for this ques
a car dealership sells only sports cars and luxury cars and has atleast some of each type of car in stock at all times.if exactly 1/7 of sports car and 1/2 of luxury cars have sunroofs and there are exactly 42 cars on the lot.what is the smallest number of cars that could have roofs?
ans 11 Please put questions in new posts. Put it in the same post only if it is totally related or a variation of the question that we are discussing. Now for the solution: There are 42 cars on the lot. 1/7 of sports cars and 1/2 of luxury cars have sunroofs. This means that 1/7 of number of sports cars and 1/2 of number of luxury cars should be integers (You cannot have 1.5 cars with sunroofs, right?) We want to minimize the sunroofs. Since 1/2 of luxury cars have sunroofs and only 1/7 of sports cars have them, it will be good to have fewer luxury cars and more sports cars. Best would be to have all sports cars. But, the question says there are some of each kind at any time. So let's say there are 2 luxury cars (since 1/2 of them should be an integer value). But 1/7 of 40 (the rest of the cars are sports cars) is not an integer number. Let's instead look for the multiple of 7 that is less than 42. The multiple of 7 that is less than 42 is 35. So we could have 35 sports cars. But then, 1/2 of 7 (since 42  35 = 7 are luxury cars) is not an integer. The next smaller multiple of 7 is 28. This works. 1/2 of 14 (since 42  28 = 14 are luxury cars) is 7. So we can have 14 luxury cars and 28 sports cars. That is the maximum number of sports cars that we can have. 1/7 of 28 sports cars = 4 cars have sunroofs 1/2 of 14 luxury cars = 7 cars have sunroofs So at least 11 cars will have sunroofs.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Manager
Joined: 16 Feb 2012
Posts: 230
Concentration: Finance, Economics

Re: Inequalities trick [#permalink]
Show Tags
22 Jul 2012, 03:03
1
This post received KUDOS
VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\(\frac {(x+2)(x1)}{(x4)(x7)}\) were \(4\leq x \leq 7\), than the lower part would be equal to zero,thus making it impossible to calculate the whole equation.
_________________
Kudos if you like the post!
Failing to plan is planning to fail.



CEO
Status: Nothing comes easy: neither do I want.
Joined: 12 Oct 2009
Posts: 2783
Location: Malaysia
Concentration: Technology, Entrepreneurship
GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35

Re: Inequalities trick [#permalink]
Show Tags
18 Oct 2012, 05:17
1
This post received KUDOS
1
This post was BOOKMARKED
GMATBaumgartner wrote: gurpreetsingh wrote: ulm wrote: in addition: if we have smth like (xa)^2(xb) we don't need to change a sign when jump over "a". yes even powers wont contribute to the inequality sign. But be wary of the root value of x=aHi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ? Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. If the powers are even then the inequality won't be affected. eg if u have to find the range of values of x satisfying (xa)^2 *(xb)(xc) >0 just use (xb)*(xc) >0 because xa raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.
_________________
Fight for your dreams :For all those who fear from Verbal lets give it a fight
Money Saved is the Money Earned
Jo Bole So Nihaal , Sat Shri Akaal
Support GMAT Club by putting a GMAT Club badge on your blog/Facebook
GMAT Club Premium Membership  big benefits and savings
Gmat test review : http://gmatclub.com/forum/670to710alongjourneywithoutdestinationstillhappy141642.html



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7450
Location: Pune, India

Re: Inequalities trick [#permalink]
Show Tags
18 Oct 2012, 10:27
GMATBaumgartner wrote: Hi Gurpreet, Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above. In addition, you can check out this post: http://www.veritasprep.com/blog/2012/07 ... spartii/I have discussed how to handle powers in it.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews




Re: Inequalities trick
[#permalink]
18 Oct 2012, 10:27



Go to page
1 2 3 4 5 6 7 8
Next
[ 146 posts ]




