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.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
26 May 2017, 04:41
Kudos
Bookmarks
Is \(x < 0\)?
(1) \(x^{3} + 1 < 0\)
(2) \(x^{3} + x + 1 = 0\)
(1) \(x^{3} + 1 < 0\)
(2) \(x^{3} + x + 1 = 0\)
26 May 2017, 04:41
Kudos
Bookmarks
Official Solution:
For inequality questions, one of the most important thing is if the range of the question includes the range of the condition, that condition is sufficient. If you look at the original condition, there is 1 variable \((x)\) and in order to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In the case of con 1), you get \(x^{3} +1 < 0\), \((x + 1) (x^{2} - x + 1) < 0\), and \(x^{2} - x + 1 > 0\). Here, it is always true, and it becomes \(x + 1 < 0\), then \(x < - 1 < 0\), so the range of the question includes the range of the condition, hence it is sufficient.
In the case of con 2), \(x^{3} + x + 1 = 0\) is not factored. The most important thing here is "CMT 4(B: if you get A or B too easily, consider D)". Using CMT 4(B), you get yes, hence it is sufficient. This is because, from \(x^{3} + x = -1\), \(x(x^{2} + 1) = -1\), or \(x = \frac{-1}{x^{2} + 1}\), \(x^{2} + 1 > 0\) is always true, so \(x=\frac{-1}{x^{2} + 1} < 0\), hence yes. The answer is D.
Answer: D
For inequality questions, one of the most important thing is if the range of the question includes the range of the condition, that condition is sufficient. If you look at the original condition, there is 1 variable \((x)\) and in order to match the number of variables to the number of equations, there must be 1 equation. Therefore, D is most likely to be the answer.
In the case of con 1), you get \(x^{3} +1 < 0\), \((x + 1) (x^{2} - x + 1) < 0\), and \(x^{2} - x + 1 > 0\). Here, it is always true, and it becomes \(x + 1 < 0\), then \(x < - 1 < 0\), so the range of the question includes the range of the condition, hence it is sufficient.
In the case of con 2), \(x^{3} + x + 1 = 0\) is not factored. The most important thing here is "CMT 4(B: if you get A or B too easily, consider D)". Using CMT 4(B), you get yes, hence it is sufficient. This is because, from \(x^{3} + x = -1\), \(x(x^{2} + 1) = -1\), or \(x = \frac{-1}{x^{2} + 1}\), \(x^{2} + 1 > 0\) is always true, so \(x=\frac{-1}{x^{2} + 1} < 0\), hence yes. The answer is D.
Answer: D
Moderator:
