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

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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 09:42
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Case 1
We need to consider X as positive and greater than 1
Then option I is correct.
Case 2 - X is positive and lies between 0 and 1
Then option III is correct.

Hence D. I and III only

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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 09:43
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It is given that X>0 (positive)
Two ranges for X>0 is 0<x<1 & 1<x<∞

In the first range, the higher the power, the lower the value
let x = 0.5
\(\frac{1}{x}\) = 2
x = 0.5
\(x^2\) = 0.25
\(x^3\) = 0.125
\(x^3\) < \(x^2\) < x < \(\frac{1}{x}\)

In the second range, the higher the power, the higher the value
Let x = 1.5
\(\frac{1}{x}\) = 0.6
x = 1.5
\(x^2\) = 2.25
\(x^3\) = 3.3
\(\frac{1}{x}\) < x < \(x^2\) < \(x^3\)

I & III only
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 10:27
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X = + is a fixed situation

we can have x = fraction or x= integer

Situation 1: x = fraction

let x = 1/2
1/x = 2
x^2 = 1/4
x^3 = 1/8

Answer = x^3 < x^2< x < 1/x which is Option III

Situation 2: x = integer

let x = 2
x^2 = 4
x^3 = 8
1/x = 1/2

Answer = 1/x < x < x^2 < x^3 which is Option I

Option 2 is just not possible under the given situations so it is eliminated

Therefore answer is D I and III only
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 12:29
C option. There can be 2 type of X i.e integer or fraction.
For integer x, condition1 is right
for fraction x , condition 3 is fitting.
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 12:31
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Answer is D. x can be either 0<x<1 or x>= 1. So only I and III are possible outcomes
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 12:35
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If x is positive, which of the following may be the correct ordering of 1/x, x, x^2, x^3?

When x is positive, there are 2 ranges that we need to consider as shown in the image below:



Case 1:
Range of x (0,1)

Let x = 0.5

1/x = 2
x^2 = 0.25
x^3 = 0.125

Therefore, x^3< x^2 < x < 1/x is possible

Case 2:
Range of x (1,∞)

Let x = 2

1/x = 0.5
x^2 = 4
x^3 = 8

Therefore, 1/x < x < x^2 < x^3 is possible

Answer: Option D
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PS range.jpg
PS range.jpg [ 12.55 KiB | Viewed 670 times ]

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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 12:55
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Bunuel wrote:
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


 


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x is positive so there are 2 possible cases:
1. 0 < x < 1 => let's take an example x = 0.1 1/x = 10 x^2 = 0.01 and x^3 = 0.001 => Statement III is true
2. x > 1 => let's take an example x = 2 x^3 = 8 x^2 = 4 1/x = 0.5 => Statement I is true

Answer should be D
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 13:07
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if x=2

1/x = 1/2, x=2,x^2=4,x^3=8
hence 1/x<x<x^2<x^3 --> I can be true

if x=1/2
1/x = 2, x=1/2,x^2 = 1/4, x^3=1/8
hence x^3<x^2<x^1/x ---> III can be true

Therefore, I and III can be true which is D
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 16:32
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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
------------------------------------------------
IMO D

x is positive.

I. Possible when x>1. For eg - 2

II. x is positive. x^3 < x^2 < x only when 0<x<1. But in this case 1/x has tp be greater than 1. So not possible

III. possible when 0<x<1. Eg -> x=1/2
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 17:36
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If x > 1. then option (III) will be true

example; 5^3 > 5^2 > 5 > 1/5

If 0<x<1. then option (I) will be true

example; 1/0.9 > 0.9 > 0.9^2 > 0.9^3

So, (D) will be correct choice
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 19:09
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Bunuel wrote:
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


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

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let x =2
x^3 =8 , x^2=4 , 1/x = 0.5

0.5<2<4<8
\(\frac{1}{x} < x < x^2 < x^3\) - Holds true

let x= 0.1
x^3 = 0.001, x^2 = 0.01, 1/x = 10
\(x^3 < x^2 < x < \frac{1}{x}\) - Holds true

Correct answer should be D.
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 19:47
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for 0<x<1 (let's take x=1/5)
1/x=5
x=0.2
x^2= 0.04
x^3= 0.008

therefore, statement III may be true

for x>1
1/x=0.5
x=2
x^2=4
x^3=8

therefore, statement I may be true
Option D
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 21:21
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File comment: Option D is the answer IMO
IMG_20221223_094911__01.jpg
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 21:44
Bunuel wrote:
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


 


This question was provided by Experts'Global
for the 12 Days of Christmas Competition

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x is positive
when x lies between 0 and 1
x^3 is smallest and 1/x is greatest

when x lies between 1 and infinity
x^3 is greatest and 1/x is smallest

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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 22:40
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Option D is correct

I is true if x>1

x3 > x2 > x > 1/x

III is true if 0<x<1

x3 < x2 < x < 1/x

I and III may be true but II can't be

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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   22 Dec 2022, 23:25
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If x is positive, which of the following may be the correct ordering of \(\frac{1}{x}, \ x, \ x^2, \ x^3\) ?

x is positive we have to consider 2 cases. Case I: x>1 ; Case II: 0<x<1

I. \(\frac{1}{x} < x < x^2 < x^3\) - possible when x>1

II. \(\frac{1}{x} < x^3 < x^2 < x\) - Not possibile in either of the two cases

III. \(x^3 < x^2 < x < \frac{1}{x}\) - possible when 0<x<1

A. I only
B. II only
C. I and II only
D. I and III only - Correct
E. I, II, and III
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   23 Dec 2022, 00:52
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Since x>0, only 2 ranges need to be checked.
Range 1: 1>x>0
Here: 1/x>x>x^2>x^3
Range 2: x>1
Here: x^3>x^2>x>1/x
Therefore Option D.
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   23 Dec 2022, 02:39
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Since x is a positive number, there are two possible cases.

First, 0<x<1 .
In this case, x^3<x^2<x<1/x

Second, x>1.
In this case, 1/x<x<x^2<x^3

Hence, IMO D- I and III.
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   23 Dec 2022, 03:16
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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

If \(x>1\) then \(\frac{1}{x}<x<x^{2}<x^{3}\)
for example Let \(x=2\)
\(x^{3}=8, x^{2}=4, x=2, \frac{1}{x}=0.5\)
I holds

If \(0<x<1\) then \(x^{3}<x^{2}<x<\frac{1}{x}\)
for example Let \(x=0.2\)
\(x^{3}=0.008, x^{2}=0.04, x=0.2, \frac{1}{x}=5\)
III holds

II won't hold for any value of x>0

Ans : D
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink] 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which   23 Dec 2022, 03:28
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If x is positive, which of the following may be the correct ordering of 1/x, x, x^2, x^3 ?

I. 1/x<x<x^2<x^3 --> possible for all values above 1

II. 1/x<x^3<x^2<x --> not possible for positive values ( possible when -1<X<0 )

III. x^3<x^2<x<1/x --> possible for +ve value below 1

ans D
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Re: 12 Days of Christmas GMAT Competition - Day 9: If x is positive, which [#permalink]
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