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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 8348 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 the correct ordering
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14 Aug 2016, 08:36
if i take value 9/10 it satisfies 2nd ordering but if i take 2/3 it doesnt satisfies 2nd ordering. why?
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Re: If x is positive, which of the following could be the correct ordering
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16 Aug 2016, 01:27
sananoor wrote: if i take value 9/10 it satisfies 2nd ordering but if i take 2/3 it doesnt satisfies 2nd ordering. why? The question says "...could be the correct ordering of..." So if even 1 positive value of x satisfies the ordering, it is included. Since 9/10 satisfies the second ordering, it is included in the ordering. A transition point here is 1/sqrt(2). That is why values less than 1/sqrt(2) (such as 2/3) behave differently from values more than 1/sqrt(2) (such as 9/10). Check 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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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 the correct ordering
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21 Feb 2018, 13:16
butterfly wrote: If x is positive, which of the following could be the correct ordering of 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 only (E) I II and III If 0 < x < 1, for example, x = 1/2, then 1/x = 2, 2x = 1 and x^2 = 1/4. We see that x^2 < 2x < 1/x. Roman numeral I could be true. Also, if x = 3/4; then 1/x = 4/3, 2x = 3/2 and x^2 = 9/16. We see that x^2 < 1/x < 2x. Roman numeral II could also be true. For Roman numeral III, observe that x^2 > 2x is equivalent to x^2  2x > 0 and we can factor out the x to obtain x(x  2) > 0. Since x is positive, we can divide each side by x and we will get x  2 > 0; in other words, x > 2. On the other hand, if 1/x > 2x, then 1/x  2x > 0; or, equivalently, (1  2x^2)/x > 0. Since x is positive, we can multiply each side by x to obtain 1  2x^2 > 0, which is equivalent to 2x^2 < 1. Then, x^2 < 1/2 and we know that this is only possible when x < 1. Thus, the inequalities x^2 > 2x and 1/x > 2x cannot hold simultaneously for a positive x. Therefore, Roman numeral III is not possible. In conclusion, I and II could be true. Answer: D
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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)
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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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If x is positive, which of the following could be the correct ordering
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15 May 2019, 06:47
butterfly wrote: If x is positive, which of the following could be the correct ordering of 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 only (E) I II and III Method: Plug Using Transition Points x = + Identify Transition Points: 1/x = 2x => 1 = 2x^2 => x = 1/sqrt(2) 2x = x^2 => x^2  2x = 0 => x(x2) = 0 => x = 0 or 2 (0 is not possible since x = +) x^2 = 1/x => x^3 = 1 => x = 1 Therefore, transition points = 1/sqrt(2), 1, 2 Need to check numbers between those transition points Range: Value of x, 1/x, 2x, x^2, Order < 1/sqrt(2): 1/10, 10, 1/5, 1/100 x^2 < 2x < 1/x => I 1/sqrt(2) to 1: 0.99, 1/0.99, 1.98, ~1 x^2 < 1/x < 2x => II 1 to 2: 3/2, 2/3, 3, 9/4 1/x < x^2 < 2x (Not given) > 2: 5/2, 2/5, 5, 25/4 1/x < 2x < x^2 (Not given) Hence, only I & II could be true. ANSWER: D Method: Using Inequalities (along with wavy line) Split complex inequality into 2 simpler inequalities to see whether there is an overlap. If there is, that complex inequality could be true, otherwise not. x = + 1st Inequality (I): a) x^2 < 2x => x^2  2x < 0 => x(x2) < 0 => 0 < x < 2 b) 2x < 1/x => 2x^2 < 1 => 1/sqrt(2) < x < 1/sqrt(2) => 0 < x < 1/sqrt(2) [since x=+] Overlap: 0 < x < 1/sqrt(2) => I could be true 2nd Inequality (II): a) x^2 < 1/x => x^3 < 1 => x < 1 b) 1/x < 2x => 1 < 2x^2 => x < 1/sqrt(2) or x > 1/sqrt(2) => x > 1/sqrt(2) [since x=+] Overlap: 1/sqrt(2) < x < 1 => II could be true 3rd Inequality (III): a) 2x < x^2 => 0 < x^2  2x => 0 < x(x2) => x < 0 or x > 2 => x > 2 [since x=+] b) x^2 < 1/x => x^3 < 1 => x < 1 Overlap: None => III could not be true ANSWER: D This Q has been explained in depth here: https://gmatclub.com/forum/inequalities ... l#p1582961



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Re: If x is positive, which of the following could be correct ordering of
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18 Sep 2019, 08:58
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, if x = 1/2, then 1/x = 2, 2x = 1, and x^2 = 1/4; i.e. x^2 < 2x < 1/x. Thus, Roman numeral I is possible. We eliminate answer choice A. Since we eliminated the answer choice “none” and since 1/x is greater than x^2 in every Roman numeral; it must be true that x < 1. For Roman numeral II, in order for 1/x < 2x to hold, we must have 2x^2 > 1, which is equivalent to x^2 > 1/2. This implies that x > 1/√2 or x < 1/√2. The latter is not possible because x is positive. Since √2 is roughly 1.41, we see that if we let x = 1/1.4 = 5/7, then we have 1/x = 7/5, 2x = 10/7, and x^2 = 25/49. In this case, we have x^2 < 1/x < 2x and thus, Roman numeral II is also possible. Finally, for Roman numeral III to be true, we must have x^2 > 2x, or equivalently, x^2  2x > 0. Factoring the left hand side, we get x(x  2) > 0. In order for the product of x and (x  2) to be positive, either both of them must be positive or both of them must be negative. If both x and (x  2) are positive, then x > 2; but then x^2 > 1/x. If both of them are negative, then x < 0 but this contradicts the fact that x is positive. Therefore, Roman numeral III is not possible. Answer: D
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Re: If x is positive, which of the following could be correct ordering of
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20 Sep 2019, 10:59
Bunuel how to decide which ranges to be checked ? Bunuel wrote: If x is positive, which of the following could be the correct ordering of 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 only (E) I II and III
First note that we are asked "which of the following COULD be the correct ordering" not MUST be. Basically we should determine relationship between \(x\), \(\frac{1}{x}\) and \(x^2\) in three areas: \(0<1<2<\).
\(x>2\)
\(1<x<2\)
\(0<x<1\)
When \(x>2\) > \(x^2\) is the greatest and no option is offering this, so we know that x<2. If \(1<x<2\) > \(2x\) is greatest then comes \(x^2\) and no option is offering this.
So, we are left with \(0<x<1\): In this case \(x^2\) is least value, so we are left with:
I. \(x^2<2x<\frac{1}{x}\) > can \(2x<\frac{1}{x}\)? Can \(\frac{2x^21}{x}<0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
II. \(x^2<\frac{1}{x}<2x\) > can \(\frac{1}{x}<2x\)? The same here \(\frac{2x^21}{x}>0\), the expression \(2x^21\) can be negative or positive for \(0<x<1\). (You can check it either algebraically or by picking numbers)
Answer: D.
Second condition: \(x^2<\frac{1}{x}<2x\)
The question is which of the following COULD be the correct ordering not MUST be.
Put \(0.9\) > \(x^2=0.81\), \(\frac{1}{x}=1.11\), \(2x=1.8\) > \(0.81<1.11<1.8\). Hence this COULD be the correct ordering.
Hope it's clear.



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Re: integers and inequalities
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