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

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:


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.

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: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/05 ... on-points/

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/if-x-is-posi ... l#p1748102

Answer is clear but how did you decide the ranges. Question only mentions that X is positive. But you use three ranges X>2, 2<X<1 and 0<X<1. When I was solving this question, I used only two ranges 0<X<1 and X>1. Would be great if you could explain why did we check for X>2.

Hi divakarns

The critical ranges on positive sides are
1) 0 < x < 1
2) x > 1

However, this question is really beautiful cause it prompts us to think of values within the range 0 < x < 1

For x^2 > x, x must be either negative or x > 1 so third one is definitely out cause 1/x in that range can NOT be greater than x^2

Tricky case is Part II

There we need to think for what values of x in range 0 < x < 1, \(\frac{1}{x}<2x\) and that's possible only when x is closer to 1 than 0 i.e. evalues close to 1 such as 0.9

I hope this Help!

Video solution from Quant Reasoning starts at 25:46
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

BrentGMATPrepNow thanks for your reply. It makes much more sense now.
One thing I'm confused is why is different numbers need to be picked for option I , II , III and not one number picking then check which option will satisfy it as in other GMAT questions?
Therefore how will I know when to pick one number or to pick different numbers for each option choice as in this case?
Once again thanks for your time in advaned and great helps always.

The main idea here is that the ordering of 1/x, 2x, and x^2 depends on the value of x.
For example, if x = 3, then 1/x < 2x < x^2
Conversely, if x = -1, then 2x < 1/x < x^2
And so on.

The question is asking us which of the three orderings are possible.

Does that help?

Hi IanStewart,

Here I am troubling you again.

This question has intrigued me time and time again in terms of what it actually tests. We know now that GMAC is not interested in testing our maths skills but rather reasoning skills. So, how does one go about solving this in a manner where he/she does not have to write down a bunch of inequalities, do number picking and square root approximations and then reach to the answer.

As there's no one better than you when it comes to maths reasoning, I wonder how you'd look at such questions and how you'd go about solving them.

I haven't looked at this question in ages, but I posted a solution back in 2009 on BTG: www.beatthegmat.com/question-from-gmat- ... 32641.html . Since you're asking something of a philosophical question, I'll post some philosophical thoughts before an actual solution:

• If you read the first post in that BTG thread, I think it becomes clear why this question is difficult for many test takers: a test taker who just plugs a generic set of numbers into each inequality won't usually see how item II can be true. Back then (and it's probably still the case) many prep books taught an inequality "strategy" that involved plugging -2, -1/2, 1/2 and 2 into any inequality. That's actually a good strategy for one very specific type of inequality, and it's a bad strategy for almost every other inequality question. This official question seems designed to foil that strategy, and to distinguish between test takers who know how to reason through an inequality, and test takers who have just memorized some number-picking technique without understanding when to use it and when to do something else.

• In GMAT algebra questions, we'll often need to reason algebraically. You say the GMAT tests "reasoning skills" and not "maths skills", and that's true depending what you mean by "maths skills". To me, reasoning is math, so those skills are almost the same thing. But many people think "math" is "calculation and algebra", and the GMAT isn't interested in testing if you're a human calculator, or if you can complete twenty algebraic steps in two minutes. So this is an algebra question, and we'll need to do some algebraic steps to answer it, but if we approach the problem in the best way none of those steps will be complicated.

• inequality questions tend only to be tricky to manipulate algebraically when negative numbers are involved. If you see an inequality like x^3 < 9x, then if x can be negative, we have to be pretty careful when we analyze it. But if we know x is positive, this kind of inequality is pretty straightforward: if x > 0, we can safely just divide by x on both sides (without worrying about whether to reverse the inequality). So it then becomes x^2 < 9, or x < 3, and 0 < x < 3 is the set of positive solutions to the inequality.

• to the actual question in this thread: in any similar question, I'd normally quickly imagine extreme values of x, to see if any items are obviously true. So I'd quickly imagine that x = 1000, say, which doesn't work in I, II or III, and I'd quickly imagine that x = 1/1000, which does work in I. So I can be true, but you can just as easily investigate I algebraically.

• for II, since x is positive, we can multiply by x on both sides of x^2 < 1/x, so x^3 < 1, and x < 1. Similarly, the inequality 1/x < 2x becomes 1 < 2x^2, and x > 1/√2. So we see this inequality will work when 1/√2 < x < 1

• for III, again since x is positive, we can divide 2x < x^2 by x on both sides, to learn x > 2. We can multiply x^2 < 1/x by x on both sides to learn x^3 < 1, or x < 1. Since x > 2 and x < 1 cannot both be true, this inequality works for no positive value of x, and the answer is I and II only.

Hello, PyjamaScientist. I happened to come across this question by chance, combing through the New Posts section in search of a question I had not attempted. Since I was sitting outside enjoying a sunny day, I did not want to go in and grab my noteboard and markers, so I used a little number sense to reason my way to the correct answer. It took me 1:48, as you can see in the attached screenshot.


Please note that I am not disagreeing with the excellent response you have received by IanStewart above. I would defer to Ian in any mathematical area the exam tests. I am more of a scientist at heart (perhaps like you, if your namesake is any indication), someone who enjoys testing an idea in search of an underlying truth or tendency. That said, the following is how I answered the question with confidence in under 2 minutes. If I am repeating something that someone else has written above, it is because I have not looked at the community dialogue yet (apart from your post and the response to it, of course).


Statement I: Any time I see positive and x^2 as the lowest value in an inequality, I think of a fraction between 0 and 1, because anything greater than 1 will grow larger by squaring. Confession (in light of the above post): I did choose 1/2 to test. I did so because I did not want to work with anything complex in my head, and it "looked" right.

\((\frac{1}{2})^2<2(\frac{1}{2})<\frac{1}{(0.5)}\)

\(\frac{1}{4}<1<2\) TRUE

I dumped answer choices (A) and (C).

Statement II: I wanted to focus on the first half of the compound inequality, specifically on how I could get these two unknowns to be as close as possible to each other while being mindful of the fact that 2x had to be greater than both. They could not get any closer to each other than 1:

\((1)^2=\frac{1}{(1)}\)

Now, since x does not have to be an integer, I reasoned that if it were something close to 1, such 0.999, the difference would be good enough for the inequality:

\((0.999)^2 < 1\) and

\(\frac{1}{(0.999)} > 1\)

I did not bother figuring out the exact values. A little number sense was good enough. Of course, 2x would be close to 2 using the same input value.

Since Statement II could be TRUE, I eliminated answer choice (B).

Statement III: I pursued the same line of logic as before, focusing on the first half of the compound inequality:

\(2x<x^2\)

I wanted to get these values to be as close as possible to each other before considering the rest. Of course, if x = 2, we would get an equation:

\(2(2)=(2)^2\)

Thus, I wanted to pursue something close to 2 that would make the inequality true, something like 2.001.

\(2(2.001)<(2.001)^2\)

Again, the exact values were unimportant. But when I looked at the rest of the compound inequality, I could see that my input value would not work:

\(\frac{1}{(2.001)}\)

That would be close to a half, nowhere near as great as 4. I did not bother going any further, since I could only complicate matters by revisiting the earlier inequality (e.g., testing fractions). I selected answer choice (D).

In retrospect, I suppose my tutoring experience kicked in, and I pursued the line of reasoning that Ian outlined in one of his bullet points: "quickly imagine extreme values of x, to see if any items are obviously true." Extreme values do not have to be gigantic or minuscule. They can simply be numbers that you know would produce similar but slightly different outcomes. For instance, if we are not allowed to test 1, what about 0.99999999... or 1.00000000... with a 1 at the end, numbers that would effectively behave like a 1? These are the sorts of extremes I often pursue, since they are easy to work with in my head and can reveal larger truths or tendencies in many algebra or inequality problems.

If you have read this far, thank you for paying attention to the response no one asked for. I am not a Quant Expert, even if I aspire to be one.

- Andrew

IanStewart,
Thank you for such an insightful post. It really is a pleasure to read your posts, similar to watching Hans Zimmer weave a musical score.
Though I have been doing good in Quant these days, my lack of number picking ability dwarfs my confidence. So, it is always a blessing to learn and observe how you, and now Andrew too, go about solving such questions.
Besides the wonderful post that you wrote, I also got to see a much younger and handsome you (not that you are not anymore). That profile picture on BTG from 2009 took me by surprise. Did you ever cosplay as Johnny Depp? Because the resemblance is quite startling.

AndrewN,
I wish you do not have to spend another beautiful sunny day working through compound inequalities. And you surely should start calling yourself a Quant expert now. The way you went about picking numbers was delightful and astute.
And please tag me with (@). Because there was a chance that I could've missed your brilliant post. And yes I am a scientist by profession (Geoscientist to be precise).

Hi KarishmaB, the graph make sense for the ordering easier to understand but how did you derive the number to plot on this graph please? Thanks for your time in advanced.

To draw any graph, put in some values of x and get corresponding values of y. Trying out x = 0, 1, -1, 10 and -10 and y = 0 if it hasn't been obtained from a previous value of x usually gives us a fair idea of what the graph looks like). For many well-known functions, we do "learn from experience" what the graph looks like.

Here we are drawing 3 graphs:
y = 2x (a straight line with slope 2 passing through 0, 0 )
y = x^2 (an upwards open parabola with minimum at 0, 0)
y = 1/x ( Passing through 1, 1 asymptotic to X and Y axis)

If you are unsure, try putting in some values of x and plot points with corresponding values of y for each of these. See what graph you get.