12 Days of Christmas GMAT Competition - Day 9: If x is positive, which

Question

12 Days of Christmas GMAT Competition with Lots of Fun

If x is positive, which of the following may be the correct ordering of  \frac{1}{x}, \ x, \ x^2, \ x^3  ?
 
I.  \frac{1}{x} < x < x^2 < x^3

II.  \frac{1}{x} < x^3 < x^2 < x

III.  x^3 < x^2 < x < \frac{1}{x}

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

gaurav1511 · Answer

IMO, option D is the answer Given: x is positive, x>0 Case 1: If 01, then 1/x

TBT · Answer

Given x is positive: *Makes it a little bit easy as it removes the 0 to -1 range

* Try no from range 0 to 1 , and other big positive no. Eg. x=0.1 , x=2, x=10 We will get 1 and 3 correct. As x is positive 2 is not possible. Ans: D

Kinshook · Answer

Asked: If x is positive, which of the following may be the correct ordering of  \frac{1}{x}, \ x, \ x^2, \ x^3  ?
 
I.  \frac{1}{x} < x < x^2 < x^3 
If x=2
1/x = 1/2 < x=2 < x^2=4 < x^3 = 8
MAY BE TRUE

II.  \frac{1}{x} < x^3 < x^2 < x 
x^3 < x^2 < x is possible only when 0<x<1
But 1/x > x in that interval
CAN NOT BE TRUE

III.  x^3 < x^2 < x < \frac{1}{x} 
If x = 1/2
x^3 = 1/8 < x^2 = 1/4 < x=1/2 < 1/x = 2
MAY BE TRUE

A. I only
B. II only
C. I and II only
D. I and III only
E. I, II, and III

IMO D

kguduri · Answer

The questions asks which of the following may be the correct ordering.So just one instance that satisfies the conditions would be enough. I. \frac{1}{x}x From values x>1, x^{3}> x^{2} but \frac{1}{x}

Lizaza · Answer

For any positive number in the given situation, there's a split regarding its position to 1:

When \(x<1\), with the growth of powers the number will decrease, so \(x^{2,3,4...} < x\) and \(\frac{1}{x}\) will always be higher than 1, so also higher than X
When \(x>1\), with the growth of powers the number will increase, so \(x^{2,3,4...} > x\) and \(\frac{1}{x}\) will always be lower than 1, so also lower than X

Now let's look at the answer options.
For Option I, any integer above 1 is rationally working - it's easily possible.
For Option III, any decimal below 1 is rationally working - it's easily possible as well.

However, there's no situation that makes Option II work. When powers are lower than the original number, it's a fraction. When you divide 1 by a fraction, you will always go above 1.
It's a paradox, therefore, there's no such number for X.

The answer is D) I and III only.

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