If x is positive, which of the following could

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If x is positive, which of the following could be the 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 only
(E) I II and III

OA is D

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by theCodeToGMAT » Thu Oct 03, 2013 6:31 am
Statement 1:
x^2 < 2x --> x < 2
2x < 1/x --> x < 0.7

Statement 2:
x^2 < 1/x ---> x < 1
1/x < 2x ---> x > 0.7

Statement 3:
2x < x^2 ----> x > 2
x^2 < 1/x ----> x < 1

Only, Statement 3 doesn't give the overlapping set..

So, Answer [spoiler]{D}[/spoiler]
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by Brent@GMATPrepNow » Thu Oct 03, 2013 6:56 am
rakeshd347 wrote:If x is positive, which of the following could be the correct ordering of 1/x, 2x and x²?
I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

(A) None
(B) I only
(C) III only
(D) I and II only
(E) I II and III
Let's start by plugging in some positive values of x and see what we get.

x = 1/2
1/x = 2
2x = 1
x² = 1/4
So, we get x² < 2x < 1/x
This matches statement I.

x = 3/4
1/x = 4/3
2x = 3/2
x² = 9/16
So, we get x² < 1/x < 2x
This matches statement II

x = 3
1/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 < x
Take x² < 1/x and multiply both sides by x to get x^3 < 1, which means x < 1
Hmmm, 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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by GMATGuruNY » Thu Oct 03, 2013 9:20 am
If x is positive, which of the following could be the correct ordering of 1/x, 2x, and x²?

I. x² < 2x < 1/x
II. x² < 1/x < 2x
III. 2x < x² < 1/x

a. None
b. I
c. III
d. I and II
e. I, II, and III
Determine the CRITICAL POINTS by setting the expressions equal to each other:

1/x = 2x
2x² = 1
x² = 1/2
x = √(1/2) = 1/√2 ≈ 1/1.4 ≈ 10/14 ≈ 5/7.

1/x = x²
x^3 = 1
x = 1.

2x = x²
x=2
(We can divide by x because x>0.)

The critical points are x=5/7, x=1, x=2.
These critical points indicate where two of the expressions are EQUAL.
Thus, to the left and right of each critical point, the value of one expression must be GREATER than the value of another.

To determine which of I, II and II could be true, plug in values to the left and right of each critical point.
Start with the range that many test-takers will fail to consider: 5/7 < x < 1.

5/7 < x < 1:
If x = 3/4, then:
1/x = 4/3.
x² = 9/16.
2x = 3/2.
Since x² < 1/x < 2x, we know that II could be true.
Eliminate A, B and C.

In statement III, 2x<x², which implies that 2<x.
But if x>2, then 1/x cannot be the greatest of the three values.
Thus, III is not possible.
Eliminate E.

The correct answer is D.
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by [email protected] » Thu Oct 03, 2013 10:56 am
Hi rakeshd347,

Each of the attached explanations can be used to find the correct answer, so rather than provide another explanation, I thought that I'd offer some insight into these types of questions.

Most Roman Numeral questions are based on Number Properties (the little rules that define how numbers "work", interact, etc.); if you know these rules, then you can move through Number Property questions rather quickly.

Here, we're told that X is POSITIVE, but that doesn't necessarily mean that X is an integer. We have to be THOROUGH to get the correct answer.

X could be ANYTHING POSITIVE, including fractions, 1, other integers, etc. so we have to do enough work to prove which Roman Numerals are possible and which is not.

The number properties that I see:
1) When X = 1, X^2 and 1/X are equal; since we're dealing with inequalities, using 1 is NOT an option
2) When X = a positive fraction, X^2 makes the number SMALLER; 1/X makes the number BIGGER
3) When X = an Integer > 1, X^2 makes the number BIGGER, 1/X makes the number SMALLER
4) When X = anything, 2X makes the number TWICE as BIG

With these rules and TESTing some values, you can easily get the correct answer.

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