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22 Jun 2022, 02:30
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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
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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
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Bunuel
Is it true that \(\sqrt[3]{a}<a\)?
(1) \(a < 0\)
(2) \(a > -1\)
Are You Up For the Challenge: 700 Level Questions[/b][/url]
(1) \(a < 0\)
(2) \(a > -1\)
Are You Up For the Challenge: 700 Level Questions[/b][/url]
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
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(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
