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Re: If x is positive, which of the following could be correct ordering of
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02 Oct 2013, 08:40
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.



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Re: If x is positive, which of the following could be correct ordering of
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02 Oct 2013, 20:21
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 7214 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.
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Re: If x is positive, which of the following could be correct ordering of
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22 Feb 2014, 07:48
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!



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Re: If x is positive, which of the following could be correct ordering of
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24 Feb 2014, 01:12
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/
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Re: If x is positive, which of the following could be correct ordering of
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29 Dec 2015, 15:01
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.
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Re: If x is positive, which of the following could be correct ordering of
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03 Jan 2016, 22:06
dep91 wrote: 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. x(x2)<0 gives us 0 < x < 2 Check this post for details on this: http://www.veritasprep.com/blog/2012/06 ... efactors/
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Re: If x is positive, which of the following could be correct ordering of
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26 Aug 2016, 01:35
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 I) Let's break this ordering into two parts: X^2 < 2X > X < 2 2X < 1/X ; 2 X^2 < 1 ; X^2 < 1/2 ; X < \(\sqrt{2}\)/2, which roughly is 0.7. X < 0.7 > Hence, if we plug X = 1/2, we will satisfy the ordering > 1/4 < 1 < 2 II) Let's break this ordering into two parts: X^2 < 1/X ; X^3 < 1 ; X < 1 1/X < 2X ; 2 X^2 > 1; X^2 > 1/2 ; X > \(\sqrt{2}\)/2, which roughly is 0.7. 0.7 < X < 1 > Hence, if we plug X = 9/10, we will satisfy the ordering > 81/100 < 10/9 < 18/10 III) Let's break this ordering into two parts: 2X < X^2 ; X > 2 X^2 < 1/X ; X^3 < 1 ; X < 1 Both inequalities contradict each other. If we satisfy one inequality, we cannot satisfy the other. So: If X = 3 > We satisfy 2X < X^2, but not X^2 < 1/X. If X = 1/2 > We satisfy X^2 < 1/X, but not 2X < X^2. Hence, we cannot satisfy this ordering. Answer: D



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Re: If x is positive, which of the following could be correct ordering of
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12 Oct 2016, 09:06
Bunuel wrote: ykaiim wrote: IMO B. for 0<x<1, only statement I holds. Brunuel, if u put x=1/2: II. II. x^2<1/x<2x >>>>> will not hold true. x^2 = 1/4, 1/x=2 and 2x=1 then this expression will not hold. 1/4<2<1 [Incorrect]
If x=1/9 then: x^2=1/81, 1/x=9 and 2x=2/9 1/81<9<2/9 [Incorrect]
Let's check the III option for above values: III. 2x<x^2<1/x For x=1/2: 1<1/4<2 [Incorrect] For x=1/9: 2/9<1/81<9 [Incorrect]
So, B should be the correct answer. Please check. OA IS D.Algebraic approach is given in my solution. Here is number picking: I. \(x^2<2x<\frac{1}{x}\) > \(x=\frac{1}{2}\) > \(x^2=\frac{1}{4}\), \(2x=1\), \(\frac{1}{x}=2\) > \(\frac{1}{4}<1<2\). Hence this COULD be the correct ordering. II. \(x^2<\frac{1}{x}<2x\) > \(x=0.9\) > \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) > \(0.81<1.11<1.8\). Hence this COULD be the correct ordering. III. \(2x<x^2<\frac{1}{x}\) > \(x^2\) to be more than \(2x\), \(x\) must be more than 2 (for positive \(xes\)). But if \(x>2\), then \(\frac{1}{x}\) is the least value from these three and can not be more than \(2x\) and \(x^2\). So III can not be true. Thus I and II could be correct ordering and III can not. Answer: D. how did u choose the numbers for plugging in. i guess thats the trick in inequalities



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Re: If x is positive, which of the following could be correct ordering of
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14 Oct 2016, 01:11
vsvikas wrote: how did u choose the numbers for plugging in. i guess thats the trick in inequalities Use transition points. Discussed here: https://www.veritasprep.com/blog/2013/0 ... onpoints/
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Re: If x is positive, which of the following could be correct ordering of
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25 Jul 2017, 09:55
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 We need to equate these expressions first. Of course, we can only equate two of them at a time. So we have three equations to solve, 1/x = 2x, 1/x = x^2, and 2x = x^2. 1) 1/x = 2x 2x^2 = 1 x^2 = 1/2 x = √(1/2) = (√2)/2 ≈ 1.4/2 = 0.7 2) 1/x = x^2 x^3 = 1 x = ∛1 = 1 3) 2x = x^2 Dividing both sides by x (since we know x > 0), we have: 2 = x From the three equations above, we see that x = (√2)/2, 1, and 2. These numbers are critical since they make two of the three expressions equal to one another. Thus, we need to consider all the values that are not exactly these numbers in order to determine the order of these expressions. That is, we need to consider the following intervals: i) 0 < x < (√2)/2 ii) (√2)/2 < x < 1 iii) 1 < x < 2 iv) x > 2 However, for each of these intervals, we can just pick a representative number (for example, in 1 < x < 2, we can pick 1.5) to determine the order of these expressions. i) 0 < x < (√2)/2 Since (√2)/2 ≈ 0.7, we can let x = ½. Then 1/x = 2, 2x = 1, and x^2 = ¼. Thus, we have x^2 < 2x < 1/x, and hence Roman numeral I could be true. ii) (√2)/2 < x < 1 We can let x = ¾. Then 1/x = 4/3, 2x = 3/2, and x^2 = 9/16. Thus, we have x^2 < 1/x < 2x, and hence Roman numeral II could be true. iii) 1 < x < 2 We can let x = 3/2. Then 1/x = ⅔, 2x = 3, and x^2 = 9/4. Thus, we have 1/x < x^2 < 2x. (However, this is not one of the given Roman numerals. iv) x > 2 We can let x = 3. Then 1/x = 1/3, 2x = 6, and x^2 = 9. Thus, we have 1/x < 2x < x^2. However, this is not one of the given Roman numerals. We see that only Roman numerals I and II could be true. Answer: D
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Re: If x is positive, which of the following could be correct ordering of
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17 Jan 2018, 15:29
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 start by PLUGGING IN some positive values of x and see what we get. x = 1/21/x = 2 2x = 1 x² = 1/4 So, we get x² < 2x < 1/x This matches statement I. x = 3/41/x = 4/3 2x = 3/2 x² = 9/16 So, we get x² < 1/x < 2x This matches statement II x = 31/x = 1/3 2x = 6 x² = 9 So, we get 1/x < 2x < x² NO MATCHES At this point, the correct answer is either D or E. If you're pressed for time, you might have to guess. Alternatively, you can use some algebra to examine statement III (2x < x² < 1/x) Notice that there are 2 inequalities here ( 2x < x² and x² < 1/x) Take 2x < x² and divide both sides by x to get 2 < xTake x² < 1/x and multiply both sides by x to get x^3 < 1, which means x < 1Hmmm, so x is greater than 2 AND less than 1. This is IMPOSSIBLE, so statement III cannot be true. Answer = D Cheers, Brent
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Re: If x is positive, which of the following could be correct ordering of
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15 Feb 2018, 14:46
Bunuel, do you know of any similar questions that you could share? Preferably from the Official Guide?



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Re: If x is positive, which of the following could be correct ordering of
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If x is positive, which of the following could be correct ordering of
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14 Oct 2018, 04:42
Can't any one or two numbers(that are same) be used in all the three statements? The problem is plugging number in statement II is not satisfying. Pls explain by putting in same numbers in all statements. (I got that Statment 3 is not satisfying) Posted from my mobile device
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Re: If x is positive, which of the following could be correct ordering of
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15 Oct 2018, 02:32
topper97 wrote: Can't any one or two numbers(that are same) be used in all the three statements? The problem is plugging number in statement II is not satisfying. Pls explain by putting in same numbers in all statements. (I got that Statment 3 is not satisfying)
Posted from my mobile device You cannot plug in same numbers in each statement and expect to get the answer. The requirement of each statement is different. Hence, we need to find transition points to ensure that we get the answer. This is explained in the link given here: https://gmatclub.com/forum/ifxisposi ... l#p1748102
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Re: If x is positive, which of the following could be correct ordering of
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15 Nov 2018, 20:46
Hi Bunuel, bbhow can I find more practice questions which are of this type?



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Re: If x is positive, which of the following could be correct ordering of
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Re: If x is positive, which of the following could be correct ordering of
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28 Jan 2019, 00:03
Bunuel wrote: can \(2x<\frac{1}{x}\)? Can \(\frac{2x^21}{x}<0\) Hello Bunuel, we have: 2x < 1/x From this, can't we directly say: 2x^2 < 1 (by multiplying both sides of the inequality by x)? Why do we have to say: (2x^21)/x < 0 ?



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Re: If x is positive, which of the following could be correct ordering of
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