12 Days of Christmas GMAT Competition - Day 9: If x is positive, which : Problem Solving (PS) - Page 2

**Bambi63** · Dec 22, 2022

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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**GauthamManoj** · Dec 22, 2022

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

= 2
x = 0.5

= 0.25

= 0.125

<

< x <

In the second range, the higher the power, the higher the value
Let x = 1.5

= 0.6
x = 1.5

= 2.25

= 3.3

< x <

<

I & III only

**snowbee15** · Dec 22, 2022

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

**Bhartii** · Dec 22, 2022

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.

**LeopardLiu** · Dec 22, 2022

Answer is D. x can be either 0<x<1 or x>= 1. So only I and III are possible outcomes

**rxb266** · Dec 22, 2022

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 (12.55 KiB) Viewed 308 times
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**PranjalJ** · Dec 22, 2022

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 ?

I.

II.

III.

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

**limuengaoey** · Dec 22, 2022

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

**kavitaverma** · Dec 22, 2022

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

?

I.

II.

III.

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

**rn1112** · Dec 22, 2022

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

**rajtare** · Dec 22, 2022

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 ?

I.

II.

III.

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

- Holds true

let x= 0.1
x^3 = 0.001, x^2 = 0.01, 1/x = 10

- Holds true

Correct answer should be D.

**wadhwakaran** · Dec 22, 2022

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

**divyadna** · Dec 22, 2022

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: Option D is the answer IMO; IMG_20221223_094911__01.jpg (2.57 MiB) Viewed 235 times
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**Stanindaw** · Dec 22, 2022

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 ?

I.

II.

III.

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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**rvgmat12** · Dec 22, 2022

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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**rahulp11** · Dec 22, 2022

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

?

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

I.

- possible when x>1

II.

- Not possibile in either of the two cases

III.

- 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

**szcz** · Dec 23, 2022

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.

**kratos0906** · Dec 23, 2022

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.

**Lenel** · Dec 23, 2022

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

,

I.

II.

III.

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

If

then

for example Let

I holds

If

then

for example Let

III holds

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

Ans : D

**nikhil553** · Dec 23, 2022

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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