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15 Dec 2022, 07:00
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D
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Difficulty:
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Question Stats:
75% (01:56) correct
25%
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based on 405
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12 Days of Christmas GMAT Competition with Lots of Fun
Which of the following is (are) always correct ?
I. If x^2 is greater than x, x^3 is greater than x^2.
II. If x is greater than 1/x, x^2 is greater than 1/x.
III. If x^3 is greater than x^2, x^4 is greater than x^3.
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and II
Which of the following is (are) always correct ?
I. If x^2 is greater than x, x^3 is greater than x^2.
II. If x is greater than 1/x, x^2 is greater than 1/x.
III. If x^3 is greater than x^2, x^4 is greater than x^3.
A. I only
B. II only
C. III only
D. II and III only
E. I, II, and II
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limuengaoey
Joined: 09 Jun 2021
Last visit: 07 Feb 2025
Posts: 38
Given Kudos: 239
Location: Thailand
Schools: Judge '26 (A$) LBS '26 (A)
GMAT 1: 680 Q49 V34
GPA: 3.59
15 Dec 2022, 07:26
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I) If x^2 is greater than x, x^3 is greater than x^2.
x^2 >x
x^2-x>0
x(x-1)>0
hence x<0 or x>1 therefore if x = -5
x^3 = -125 and x^2 = 25 therefore -125 is not greater than 25 then I if false
II. If x is greater than 1/x, x^2 is greater than 1/x.
x>1/x
x-1/x>0
(x^2-1)/x > 0 x cannot be 0
hence (x+1)(x-1)>0
x<-1 or x>1
but x<-1 cannot make x>1/x therefore x can be only in range x>1 suppose x= 1.5
x^2 = 2.25 and 1/x = 2/3 therefore it is true
III. If x^3 is greater than x^2, x^4 is greater than x^3.
x^3>x^2
x^3-x^2>0
x^2(x-1)>0
x>1 therefore x^4>x^3
only II and III are correct hence D is the answer
x^2 >x
x^2-x>0
x(x-1)>0
hence x<0 or x>1 therefore if x = -5
x^3 = -125 and x^2 = 25 therefore -125 is not greater than 25 then I if false
II. If x is greater than 1/x, x^2 is greater than 1/x.
x>1/x
x-1/x>0
(x^2-1)/x > 0 x cannot be 0
hence (x+1)(x-1)>0
x<-1 or x>1
but x<-1 cannot make x>1/x therefore x can be only in range x>1 suppose x= 1.5
x^2 = 2.25 and 1/x = 2/3 therefore it is true
III. If x^3 is greater than x^2, x^4 is greater than x^3.
x^3>x^2
x^3-x^2>0
x^2(x-1)>0
x>1 therefore x^4>x^3
only II and III are correct hence D is the answer
General Discussion
15 Dec 2022, 07:36
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This is how I approached:
I)
We can have two cases here.
If x=-2, x^2>x, but x^3<x^2, so I out. Elim A and E.
III) x^3>x^2. X must be positive integer greater than 1, because if x is negative then x^2>x^3. Moreover, X cannot be a pos fraction because again x^2>x^3.
Therefore, if x is pos integer>1, then x^4>x^3.
Elim B.
II) x>1/x
For this to be true, x is either negative fraction, or x is a positive integer>1. X cannot be negative integer, otherwise 1/x>x, and likewise X cannot be a positive fraction because the reciprocal of a fraction is greater than fraction.
If x is negative fraction, clearly x^2>1/x, as that is positive.
If x is positive integer>1, clearly x^2>1/x, as one is integer and the other a fraction.
Elim C.
II and III is the choice (D)
I)
We can have two cases here.
If x=-2, x^2>x, but x^3<x^2, so I out. Elim A and E.
III) x^3>x^2. X must be positive integer greater than 1, because if x is negative then x^2>x^3. Moreover, X cannot be a pos fraction because again x^2>x^3.
Therefore, if x is pos integer>1, then x^4>x^3.
Elim B.
II) x>1/x
For this to be true, x is either negative fraction, or x is a positive integer>1. X cannot be negative integer, otherwise 1/x>x, and likewise X cannot be a positive fraction because the reciprocal of a fraction is greater than fraction.
If x is negative fraction, clearly x^2>1/x, as that is positive.
If x is positive integer>1, clearly x^2>1/x, as one is integer and the other a fraction.
Elim C.
II and III is the choice (D)
