Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 2512:30 AM EDT
-01:30 AM EDT
At one point, she believed GMAT wasn’t for her. After scoring 595, self-doubt crept in and she questioned her potential. But instead of quitting, she made the right strategic changes. The result? A remarkable comeback to 695. Check out how Saakshi did it.
Apr 2610:00 AM EDT
-11:00 AM EDT
The Target Test Prep course represents a quantum leap forward in GMAT preparation, a radical reinterpretation of the way that students should study. Try before you buy with a 5-day, full-access trial of the course for FREE!
Apr 2711:00 AM EDT
-12:00 PM EDT
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors
Apr 2808:00 AM PDT
-11:00 AM PDT
Whether you are just beginning your MBA journey or fine-tuning your path to a 700+ score, GMAT Day is designed to give you a competitive edge.
19 Nov 2025, 23:17
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
69% (01:40) correct
31%
(01:33)
wrong
based on 278
sessions
History
Date
Time
Result
Not Attempted Yet
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 \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).
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 III
I. If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\).
II. If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(\frac{1}{x}\).
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 III
20 Nov 2025, 02:17
Kudos
Bookmarks
Let’s analyze the statements:
I) If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\)
If \(x^2\) is greater than \(x\), then it follows that \(x>1\) or \(x<0\).
For \(x>1\):
Assume, that \(x\) is the positive integer \(5\). Then we can say that \(5^3=125\) and therefore greater than \(5^2=25\). (True)
For \(x<0\):
Assume, that x is the negative integer \(-5\). Then \((-5)^2=25\) and therefore greater than \(-5\). But \((-5)^3=-125\) and therefore smaller than \((-5)^2\). (False)
This statement is not always correct. (We can eliminate answer choice A and E)
II) If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(1/x\)
If \(x\) is greater than \(\frac{1}{x}\), it means \(x>1\) or \(-1<x<0\)
For \(x>1\):
Take a number, for example \(2\). \(2^2=4\) and therefore greater than \frac{1}{2}. This also works for any number \(x>2\). (True)
For \(-1<x<0\):
In general, a negative number which is squared becomes positive. Therefore, it \(x^2\) will always be larger than the negative fraction \(\frac{1}{x}\). (True)
This statement is always correct. (We can eliminate answer choice C)
III) If \(x^3\) is greater than \(x^2\), \(x^4\) is greater than \(x^3\)
Taking the first part of the statement, we can conclude that \(x>1\). Then, no matter which number we take larger than \(1\), \(x^4\) will be greater than \(x^3\). (e.g.: \(3^4=81\) and is greater than \(3^3=27\))
This statement is always correct.
The correct answer is D.
I) If \(x^2\) is greater than \(x\), \(x^3\) is greater than \(x^2\)
If \(x^2\) is greater than \(x\), then it follows that \(x>1\) or \(x<0\).
For \(x>1\):
Assume, that \(x\) is the positive integer \(5\). Then we can say that \(5^3=125\) and therefore greater than \(5^2=25\). (True)
For \(x<0\):
Assume, that x is the negative integer \(-5\). Then \((-5)^2=25\) and therefore greater than \(-5\). But \((-5)^3=-125\) and therefore smaller than \((-5)^2\). (False)
This statement is not always correct. (We can eliminate answer choice A and E)
II) If \(x\) is greater than \(\frac{1}{x}\), \(x^2\) is greater than \(1/x\)
If \(x\) is greater than \(\frac{1}{x}\), it means \(x>1\) or \(-1<x<0\)
For \(x>1\):
Take a number, for example \(2\). \(2^2=4\) and therefore greater than \frac{1}{2}. This also works for any number \(x>2\). (True)
For \(-1<x<0\):
In general, a negative number which is squared becomes positive. Therefore, it \(x^2\) will always be larger than the negative fraction \(\frac{1}{x}\). (True)
This statement is always correct. (We can eliminate answer choice C)
III) If \(x^3\) is greater than \(x^2\), \(x^4\) is greater than \(x^3\)
Taking the first part of the statement, we can conclude that \(x>1\). Then, no matter which number we take larger than \(1\), \(x^4\) will be greater than \(x^3\). (e.g.: \(3^4=81\) and is greater than \(3^3=27\))
This statement is always correct.
The correct answer is D.
21 Nov 2025, 01:33
Kudos
Bookmarks
ExpertsGlobal5
Let's analyze each statement individually.
Statement I: If \(x^2 > x\), then \(x^3 > x^2\)
First, determine the range of \(x\) for which \(x^2 > x\).
\(x^2 - x > 0\)
\(x(x - 1) > 0\)
This inequality holds true when \(x > 1\) OR \(x < 0\).
Now, let's test a value from the range \(x < 0\).
Let \(x = -2\).
Condition: \((-2)^2 > -2 \implies 4 > -2\) (True).
Result check: Is \((-2)^3 > (-2)^2\)?
\(-8 > 4\) (FALSE).
Since it fails for negative numbers, Statement I is not always correct.
Statement II: If \(x > \frac{1}{x}\), then \(x^2 > \frac{1}{x}\)
First, solve \(x > \frac{1}{x}\).
\(x - \frac{1}{x} > 0\)
\(\frac{x^2 - 1}{x} > 0\)
The critical points are -1, 0, and 1. Testing the intervals, this inequality is true when:
1. \(-1 < x < 0\)
2. \(x > 1\)
Now check if \(x^2 > \frac{1}{x}\) holds in these two regions:
Region 1 (\(x > 1\)):
Here, \(x\) is positive. \(x^2\) is positive and \(\frac{1}{x}\) is a positive fraction less than 1. A number greater than 1 squared is definitely greater than a proper fraction. (True).
Region 2 (\(-1 < x < 0\)):
Here, \(x\) is negative.
\(x^2\) is positive.
\(\frac{1}{x}\) is negative.
A positive number is always greater than a negative number. So, \(x^2 > \frac{1}{x}\) is always True.
Statement II is always correct.
Statement III: If \(x^3 > x^2\), then \(x^4 > x^3\)
First, solve \(x^3 > x^2\).
Since \(x^2\) is always positive (for \(x \neq 0\)), we can divide both sides by \(x^2\) without flipping the inequality sign.
\(\frac{x^3}{x^2} > \frac{x^2}{x^2}\)
\(x > 1\).
So the condition implies that \(x\) must be greater than 1.
Now check the result \(x^4 > x^3\) given that \(x > 1\).
Divide by \(x^3\) (which is positive since \(x > 1\)):
\(x > 1\).
This is consistent with the condition.
Statement III is always correct.
Conclusion: Statements II and III are correct.
Answer: D
