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Re: If x is positive, which of the following could be correct [#permalink]
28 Jul 2012, 19:24

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 ?

Re: If x is positive, which of the following could be correct [#permalink]
29 Jul 2012, 00:37

4

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 4084 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

Re: If x is positive, which of the following could be correct [#permalink]
29 Jul 2012, 21:58

1

This post received KUDOS

Expert's post

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. _________________

Re: If x is positive, which of the following could be correct [#permalink]
07 Aug 2012, 23:50

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

(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.

(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. _________________

Re: If x is positive, which of the following could be correct [#permalink]
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. _________________

Re: If x is positive, which of the following could be correct [#permalink]
02 Oct 2013, 20:21

1

This post received KUDOS

Expert's post

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 2244 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. _________________

Re: If x is positive, which of the following could be correct [#permalink]
14 Feb 2014, 07:49

Bunuel wrote:

imhimanshu wrote:

Hi Bunnel,

Could you please provide a reasoning to the below text... how did you find the range...Pls help

The reasoning is that in these ranges x (2x), 1/x and x^2 are ordered differently:

For \(x>2\) --> \(x^2\) has the largest value. Since no option offers this we know that \(x\) cannot be more that 2; For \(1<x<2\) --> \(2x\) has the largest value, then comes \(x^2\). Since no option offers this we know that \(x\) cannot be from this range either;

So, we are left with last range: \(0<x<1\). In this case \(x^2\) has the least value. Options, I and II offer this, so we should concentrate on them and test the values of x from 0 to 1.

Hope it's clear.

Hi Bunuel.

How do you know which numbers to pick? I mean for example 0.9 for the second statement. Any clues?

Re: If x is positive, which of the following could be correct [#permalink]
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!

Re: If x is positive, which of the following could be correct [#permalink]
24 Feb 2014, 01:12

1

This post received KUDOS

Expert's post

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!

Re: If x is positive, which of the following could be correct [#permalink]
30 Jul 2014, 19:40

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.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...