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.
23 Sep 2023, 05:15
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:57) correct
38%
(02:09)
wrong
based on 45
sessions
History
Date
Time
Result
Not Attempted Yet
23 Sep 2023, 07:53
Kudos
Bookmarks
Bunuel
\(x^4-x^3+x^2-x > 0\)
\(x^3(x-1)+x(x-1) > 0\)
\((x-1)(x^3+x) > 0\)
\(x(x-1)(x^2+1) > 0\)
\(x^2 + 1\) is always positive, hence for the expression to hold true, \((x)(x-1) > 0\)
Case 1: \(x > 0\)
If \(x > 0, x - 1 > 0.\)
Therefore \(x > 1\)
Case 2: \(x < 0\)
If \(x < 0, x - 1 < 0.\)
Therefore \(x < 1\).
Combining the information above, either \(x > 1\) or \(x < 0\)
Statement 1
x > 1
We know from the analysis above that if x > 1, the expression holds true. Hence, we can conclude that the information provided in Statement 1 is sufficient to answer the question.
We can eliminate B, C, and E.
Statement 2
\(|x^3|>|x|\)
This statement tells us that on a number-line \(x\) lies in the highlighted region.
------------- -1 ------------- 0 ------------- 1 -------------
In both regions, x is either less than -1 (satisfies the condition \(x < 0\)) or x is greater than 1 (satisfies the condition \(x > 1\)). Hence, this statement is also sufficient to answer the question.
Option D
23 Sep 2023, 07:59
Kudos
Bookmarks
Is \(x^4-x^3+x^2-x>0\)?
\(x(x^3 - x^2 + x - 1)>0\)
So either both x and \(x^3 - x^2 + x - 1\) are positive or both are negative for it to be true.
Only problem is when
0 < x < 1
as it leaves \(x^3 - x^2 + x - 1\) negative giving us a NO.
(1) \(x>1\)
x = 2
16 - 8 + 4 - 2 > 0
10 > 0
YES
x = 3/2
81/16 - 27/8 + 9/4 - 3/2 > 0
~5 - ~3.5 + 2.25 - 1.5 > 0
~ 2.25 > 0
SUFFICIENT.
(2) \(|x^3|>|x|\)
Taking clue from St. 1
x = 2 gives YES
x = - 2 gives YES too
Here x cannot be a fraction e.g. x = 1/2 or -1/2
Hence
SUFFICIENT.
Answer D.
\(x(x^3 - x^2 + x - 1)>0\)
So either both x and \(x^3 - x^2 + x - 1\) are positive or both are negative for it to be true.
Only problem is when
0 < x < 1
as it leaves \(x^3 - x^2 + x - 1\) negative giving us a NO.
(1) \(x>1\)
x = 2
16 - 8 + 4 - 2 > 0
10 > 0
YES
x = 3/2
81/16 - 27/8 + 9/4 - 3/2 > 0
~5 - ~3.5 + 2.25 - 1.5 > 0
~ 2.25 > 0
SUFFICIENT.
(2) \(|x^3|>|x|\)
Taking clue from St. 1
x = 2 gives YES
x = - 2 gives YES too
Here x cannot be a fraction e.g. x = 1/2 or -1/2
Hence
SUFFICIENT.
Answer D.
