|
|
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.
22 Jun 2022, 02:30
Kudos
Bookmarks
C
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:
66% (01:21) correct
34%
(00:58)
wrong
based on 35
sessions
History
Date
Time
Result
Not Attempted Yet
Is it true that \(\sqrt[3]{a}<a\)?
(1) \(a < 0\)
(2) \(a > -1\)
Are You Up For the Challenge: 700 Level Questions
(1) \(a < 0\)
(2) \(a > -1\)
Are You Up For the Challenge: 700 Level Questions
22 Jun 2022, 03:07
Kudos
Bookmarks
Bunuel
Concept : Inequalities
For most of the test takers inequalities is a pain point, But the fact is that we need to solve them if we are targeting for Q48+.
One of the easy approach to solve such questions is to select the numbers intelligently
Given \(\sqrt[3]{a}<a\)
Statement 1 : \(a < 0\)
Let \(-1 < a < 0\), So a can be -1/2;
\(\sqrt[3]{-1/2}< -1/2\)
\(-\sqrt[3]{1/2}< -1/2\) => \(\sqrt[3]{1/2}>1/2\) This is true
Let \(a < -1\) , So a can be -2;
\(\sqrt[3]{-2}< -2\)
\(-\sqrt[3]{2}< -2\) => \(\sqrt[3]{2}>2\) This is false
Hence it is inconsistent, A,D are eliminated
Statement 2: \(a > -1\)
So a can be 1,0, or -1/2..
We already know that incase of -1/2... \(\sqrt[3]{a}<a\) This is true from the previous statement.
If a = 0, \(\sqrt[3]{a}<a\) This is false
Hence B can be eliminated.
Using both Statement 1 : \(a < 0\) and Statement 2: \(a > -1\)
So a has to be -1<a<0; In this case it will always be true as we saw incase of statement 1(-1/2) use case.
Hence C is the correct answer.
22 Jun 2022, 03:35
Kudos
Bookmarks
(1) a<0
If \(a = -1\), then \(\sqrt[3]{a} = 1\). a = ∛a
If \( a = -\frac{1}{27}\), then \( \sqrt[3]{a} = - \frac{1}{3}\). a > ∛a
If \(a = -27\), then \( \sqrt[3]{a} = -3\). a < ∛a
Three different possible solutions and therefore this statement is insufficient
\(\)
(2) a>(−1)
If \(a = 1\), then \(\sqrt[3]{a} = 1\). a = ∛a
If \(a = \frac{1}{27}\), then \(\sqrt[3]{a} = \frac{1}{3}\). a < ∛a
If \(a = 27\), then \(\sqrt[3]{a} = 3\) . a > ∛a
Three different possible solutions and therefore this statement is insufficient
(1)+(2)
Together we get: \(0 > a > −1 \), which means that a has to be a negative fraction. From the calculations under the first statement: When \( a = -\frac{1}{27}\), then \( \sqrt[3]{a} = -\frac{1}{3}\).
Therefore, \(a > \sqrt[3]{a}\)
sufficient
Answer C
If \(a = -1\), then \(\sqrt[3]{a} = 1\). a = ∛a
If \( a = -\frac{1}{27}\), then \( \sqrt[3]{a} = - \frac{1}{3}\). a > ∛a
If \(a = -27\), then \( \sqrt[3]{a} = -3\). a < ∛a
Three different possible solutions and therefore this statement is insufficient
\(\)
(2) a>(−1)
If \(a = 1\), then \(\sqrt[3]{a} = 1\). a = ∛a
If \(a = \frac{1}{27}\), then \(\sqrt[3]{a} = \frac{1}{3}\). a < ∛a
If \(a = 27\), then \(\sqrt[3]{a} = 3\) . a > ∛a
Three different possible solutions and therefore this statement is insufficient
(1)+(2)
Together we get: \(0 > a > −1 \), which means that a has to be a negative fraction. From the calculations under the first statement: When \( a = -\frac{1}{27}\), then \( \sqrt[3]{a} = -\frac{1}{3}\).
Therefore, \(a > \sqrt[3]{a}\)
sufficient
Answer C
