Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 18 Jun 2012
Posts: 39

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
28 Jul 2012, 20:24
The question asked "what could be the correct ordering" means it asked for the possibilities. What is question asked "what must be the correct ordering" ? In that case would we be required to choose an option which is true for all scenarios ?



Director
Joined: 22 Mar 2011
Posts: 612
WE: Science (Education)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
29 Jul 2012, 01:37
5
This post received KUDOS
Please, refer to the attached drawing, in which the three graphs \(y=1/x,\) \(y=2x,\) and \(y=x^2\) are depicted for \(x>0\). The exact values for A, B, and C can be worked out, but they are not important to establish the order of the three algebraic expressions. So, the correct orderings are: If \(x\) between 0 and A: \(x^2<2x<1/x\) If \(x\) between A and B: \(x^2<1/x<2x\) If \(x\) between B and C: \(1/x<x^2<2x\) If \(x\) greater than C: \(1/x<2x<x^2\) We can see that only the first two of the above options are listed as answers (I and II). Answer: D.
Attachments
3Graphs.jpg [ 15.15 KiB  Viewed 6520 times ]
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
29 Jul 2012, 22:58
smartmanav wrote: The question asked "what could be the correct ordering" means it asked for the possibilities. What is question asked "what must be the correct ordering" ? In that case would we be required to choose an option which is true for all scenarios ? The ordering will be different for different values of x so the question cannot ask for a single correct ordering.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 23 Oct 2010
Posts: 383
Location: Azerbaijan
Concentration: Finance

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
08 Aug 2012, 00:50
given that x>0 1 ) x^2 < 2X<1/X lets just check x^2 < 2X => 0 <x<2 ; 2X<1/X => 0<x<1/\sqrt{2} these 2 inequities do not conradict each other. so, 1) is ok 2) x^2 <1/X< 2X check them  x^2 <1/X => 0<x<1 ; 1/X< 2X => x> 1/\sqrt{2} these 2 inequities do not conradict each other. so, 2) is ok 3) 2X< x^2 <1/X check them  2X< x^2 => x>2 ; x^2 <1/X => x<1 these 2 inequities conradict each other. so, 3) is not ok p.s. dont know why such symbols as sqroot , fraction ets dont work.
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
I am still on all gmat forums. msg me if you want to ask me smth



Moderator
Joined: 01 Sep 2010
Posts: 3218

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
13 Dec 2012, 07:28



VP
Joined: 09 Jun 2010
Posts: 1418

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
14 Dec 2012, 00:30
positiveness or negativeness is important to inequalit because x is positive we can multiple both sides of inequality with x and keep the same mark.for example x^2<2x<1/x is the same as x^3<2x^2<1 (if x is negative we have to change the mark of the inequality. this question is not relevant to that cases) now solve 2 inequality independently . this can be done quick. this questions can be done in less than 3 minutes. Other method takes longer time and in fact is not good. pls, comment.
_________________
visit my facebook to help me. on facebook, my name is: thang thang thang



Intern
Joined: 28 Aug 2012
Posts: 20
Location: United States
Concentration: General Management, Leadership
GPA: 3.37
WE: Information Technology (Consulting)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
11 May 2013, 02:05
Hey Karishma, I feel below highlighted part is not correct. Please check. If I am wrong, please explain. thanks. VeritasPrepKarishma wrote: Vavali wrote: If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?
(I) \(x^2 < 2x < \frac{1}{x}\) (II) \(x^2 < \frac{1}{x} < 2x\) (III) \(2x < x^2 < \frac{1}{x}\)
(a) none (b) I only (c) III only (d) I and II (e) I, II and III Let's look at this question logically. There will be some key takeaways here so don't focus on the question and the (long) solution. Focus on the logic. First of all, we are just dealing with positives so life is simpler. To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\). When \(x < 2, x^2 < 2x\) When \(x > 2, x^2 > 2x\) Similarly \(1/x = x^2\) when \(x = 1\) When \(x < 1, 1/x > x^2\). When \(x > 1, 1/x > x^2\) Are you sure this is correct? I think.. we can use x=4 here, then 1/4>16 .. which is not correct. Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\) When \(x < 1/\sqrt{2}, 1/x > 2x\) When \(x > 1/\sqrt{2}, 1/x < 2x\) So now you know that: If \(x < 1/\sqrt{2}\), \(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\) So \(x^2 < 2x < 1/x\) is possible. If \(1/\sqrt{2} < x < 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. If \(x > 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. (Same as above) For no positive values of x is the third relation possible.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
11 May 2013, 03:12
yogirb8801 wrote: Hey Karishma, I feel below highlighted part is not correct. Please check. If I am wrong, please explain. thanks. VeritasPrepKarishma wrote: Vavali wrote: If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?
(I) \(x^2 < 2x < \frac{1}{x}\) (II) \(x^2 < \frac{1}{x} < 2x\) (III) \(2x < x^2 < \frac{1}{x}\)
(a) none (b) I only (c) III only (d) I and II (e) I, II and III Let's look at this question logically. There will be some key takeaways here so don't focus on the question and the (long) solution. Focus on the logic. First of all, we are just dealing with positives so life is simpler. To compare two terms e.g. \(x^2\) and \(2x\), we should focus on the points where they are equal. \(x^2 = 2x\) holds when \(x = 2\). When \(x < 2, x^2 < 2x\) When \(x > 2, x^2 > 2x\) Similarly \(1/x = x^2\) when \(x = 1\) When \(x < 1, 1/x > x^2\). When \(x > 1, 1/x > x^2\) Are you sure this is correct? I think.. we can use x=4 here, then 1/4>16 .. which is not correct. Going on, \(1/x = 2x\) when \(x = 1/\sqrt{2}\) When \(x < 1/\sqrt{2}, 1/x > 2x\) When \(x > 1/\sqrt{2}, 1/x < 2x\) So now you know that: If \(x < 1/\sqrt{2}\), \(1/x > 2x, 1/x > x^2\) and \(x^2 < 2x\) So \(x^2 < 2x < 1/x\) is possible. If \(1/\sqrt{2} < x < 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. If \(x > 1\) \(1/x < 2x, 1/x > x^2\) So \(x^2 < 1/x < 2x\) is possible. (Same as above) For no positive values of x is the third relation possible. That is a typo. If you notice, for every case, the relation is opposite on the opposite sides of the equality value. So the relation that holds in x < 1 will be opposite to the relation that holds when x > 1. That did mess up the entire explanation. Good you pointed it out. I have edited the original post.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Math Expert
Joined: 02 Sep 2009
Posts: 39745

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
05 Jul 2013, 02:26



Intern
Joined: 15 Jul 2013
Posts: 1

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
15 Sep 2013, 10:47
1
This post was BOOKMARKED
lylya4 wrote: Vavali wrote: If x is positive, which of the following could be correct ordering of 1/x, 2x, and x^2?
(I) X^2 < 2x < 1/x (II) x^2 < 1/x < 2x (III) 2x < x^2 < 1/x
(a) none (b) I only (c) III only (d) I and II (e) I, II and III could be correct orderingSo if we can find any example that satisfy the inequation, that statement will be correct (I) x = 0.1 => 0.01 < 0.2 < 10 (II) x= 1/2 => 1/4 < 1/2 < 1 (III) 2x < x^2 <=> x ( 2 x) < 0, x > 0 then x > 2with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1 So (III) can't happen The answer is D if x = 1/2, 1/x = 2 .. wrong example used



Senior Manager
Joined: 07 Sep 2010
Posts: 327

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
02 Oct 2013, 09:40
Hello Bunuel, Can you show us graphical approach to this question. I was able to draw the graph for all three equations and intersection points, however, I was not able to negate the third ordering. Would you please help me out. Thanks imhimanshu P.S  How can I post graphs here.
_________________
+1 Kudos me, Help me unlocking GMAT Club Tests



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
02 Oct 2013, 21:21
imhimanshu wrote: Hello Bunuel, Can you show us graphical approach to this question. I was able to draw the graph for all three equations and intersection points, however, I was not able to negate the third ordering. Would you please help me out.
Thanks imhimanshu P.S  How can I post graphs here. Here is the graph: Attachment:
Ques3.jpg [ 11.4 KiB  Viewed 4695 times ]
III. 2x < x^2 < 1/x For 2x to be less than x^2, the graph of 2x should lie below the graph of x^2. This happens when the graph of 2x is the red line. For x^2 to be less than 1/x at the same time, the graph of x^2 should lie below the graph of 1/x in the region of the red line. But in the region of the red line, the graph of x^2 is never below the graph of 1/x. It will never be because graph of 1/x is going down toward y = 0 while graph of x^2 is going up toward y = infinity. Hence this inequality will not hold for any region.
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Intern
Joined: 09 Dec 2013
Posts: 31

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
22 Feb 2014, 08:48
I picked numbers: 1/2, 1, 3/2, 2, 3
However, it didn occur to me that I must look something like 0.9. Request experts to help me understand the logic behind picking such numbers. Have my actual GMAT in 10 days, any help would be immensely valuable!
Thanks!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7449
Location: Pune, India

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
24 Feb 2014, 02:12
1
This post received KUDOS
Expert's post
1
This post was BOOKMARKED
abdb wrote: I picked numbers: 1/2, 1, 3/2, 2, 3
However, it didn occur to me that I must look something like 0.9. Request experts to help me understand the logic behind picking such numbers. Have my actual GMAT in 10 days, any help would be immensely valuable!
Thanks! I have answered your query using this very question here: http://www.veritasprep.com/blog/2013/05 ... onpoints/
_________________
Karishma Veritas Prep  GMAT Instructor My Blog
Get started with Veritas Prep GMAT On Demand for $199
Veritas Prep Reviews



Senior Manager
Joined: 10 Mar 2013
Posts: 277
GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
30 Jul 2014, 20:40
What a horrific problem?! I think a fast and simple way is to graph all functions and compare all vertically. Doing this, it can be easy to see that III is impossible.



Manager
Joined: 17 Apr 2013
Posts: 66
Location: United States
Concentration: Other, Finance
GPA: 2.76
WE: Analyst (Real Estate)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
13 Aug 2014, 11:39
lylya4 wrote: Vavali wrote: If x is positive, which of the following could be correct ordering of 1/x, 2x, and x^2?
(I) X^2 < 2x < 1/x (II) x^2 < 1/x < 2x (III) 2x < x^2 < 1/x
(a) none (b) I only (c) III only (d) I and II (e) I, II and III could be correct orderingSo if we can find any example that satisfy the inequation, that statement will be correct (I) x = 0.1 => 0.01 < 0.2 < 10 (II) x= 1/2 => 1/4 < 1/2 < 1(III) 2x < x^2 <=> x ( 2 x) < 0, x > 0 then x > 2with x > 2 ==> x^2 < 1/x <=> x^3 < 1 <=> x < 1 So (III) can't happen The answer is D double check your calculation for x= 1/2 the equation won't hold true 1/0.5 = 2
_________________
Please +1 KUDO if my post helps. Thank you.



BSchool Forum Moderator
Joined: 28 Nov 2014
Posts: 912
Concentration: Strategy
GPA: 3.71

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
11 Aug 2015, 00:41
This is how I approached it Try breaking the eqn and solve Eq 1) x^2 < 2x < 1/x x^2 < 2x and 2x < 1/x or, x < 2 and x^2 < 1/2 (As X is positive, we can multiply) x < ( 1/root 2 ) satisfies both eqn. 2) x^2 < 1/x < 2x x^2 < 1/x and 1/x < 2x or, x^3 < 1 and x2 > 1/2 1 > x > ( 1/root 2 ) satisfies both eqn (Eg: Root 3 / 2 ) 3) 2x < x^2 < 1/x 2x < x^2 and x^2 < 1/x or, 2 < x and x^3 < 1 Not possible So, 4) I and 2 only Thanks



Director
Joined: 10 Mar 2013
Posts: 597
Location: Germany
Concentration: Finance, Entrepreneurship
GPA: 3.88
WE: Information Technology (Consulting)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
29 Dec 2015, 16:01
Vavali wrote: If x is positive, which of the following could be correct ordering of \(\frac{1}{x}\), \(2x\), and \(x^2\)?
(I) \(x^2 < 2x < \frac{1}{x}\) (II) \(x^2 < \frac{1}{x} < 2x\) (III) \(2x < x^2 < \frac{1}{x}\)
(a) none (b) I only (c) III only (d) I and II (e) I, II and III first, let's get rid of at least one x in the given expressions, as x is positive just multiply the expressions by x and we'll get (I) \(x^3 < 2x^2 < 1\)as \(2x^2\) < 1 we must pick a value which is < 1, let's pick 1/2 and it works. COULD BE (II) \(x^3 < 1< 2x^2\)Here we can see that \(x^3\)<1 so we must pick a value <1 BUT which will make \(2x^2\) >1 if possible. Let's pick 0.9 and it works also here. COULD BE (III) \(2x^2 < x^3 < 1\)We must pick a value < 1 BUT as we've already seen, if we pick a fraction < 1 we cannot make \(2x^2 < x^3\), in the above cases it \(2x^2 was > x^3\) each time we picked a fraction < 1 Hope it helps.
_________________
When you’re up, your friends know who you are. When you’re down, you know who your friends are.
Share some Kudos, if my posts help you. Thank you !
800Score ONLY QUANT CAT1 51, CAT2 50, CAT3 50 GMAT PREP 670 MGMAT CAT 630 KAPLAN CAT 660



Intern
Joined: 04 Sep 2015
Posts: 15
WE: Asset Management (Investment Banking)

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
29 Dec 2015, 22:13
Alternatively, we can just divide each term by x (since x > 0) and arrive at: 1/x^2 , 2 and x Much easier to pick number this way :D



Intern
Joined: 04 Oct 2015
Posts: 2

Re: If x is positive, which of the following could be correct ordering of [#permalink]
Show Tags
03 Jan 2016, 03:02
Hi Experts, In solving x^2<2x=>x^22x<0=>x(x2)<0=>x<0 or x<2
why did we ignore x<0 is it bcz we are told in question that x is positive number...it is so then why can't it be ignored in statement 3...btw I understood the number picking methodology, just little bit curious...Thanks in advance.




Re: If x is positive, which of the following could be correct ordering of
[#permalink]
03 Jan 2016, 03:02



Go to page
Previous
1 2 3
Next
[ 48 posts ]




