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10 Aug 2021, 04:45
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (02:08) correct
42%
(02:28)
wrong
based on 269
sessions
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Time
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{\(\sqrt[3]{4}\); \(\sqrt{3}\); \(\sqrt[6]{25}\); \(\sqrt[{12}]{289}\)}
What are the greatest and the least number, respectively from the list above ?
A. \(\sqrt[{12}]{289}\) and \(\sqrt[3]{4}\)
B. \(\sqrt{3}\) and \(\sqrt[3]{4}\)
C. \(\sqrt[6]{25}\) and \(\sqrt{3}\)
D. \(\sqrt[3]{4}\) and \(\sqrt[6]{25}\)
E. \(\sqrt[6]{25}\) and \(\sqrt[{12}]{289}\)
What are the greatest and the least number, respectively from the list above ?
A. \(\sqrt[{12}]{289}\) and \(\sqrt[3]{4}\)
B. \(\sqrt{3}\) and \(\sqrt[3]{4}\)
C. \(\sqrt[6]{25}\) and \(\sqrt{3}\)
D. \(\sqrt[3]{4}\) and \(\sqrt[6]{25}\)
E. \(\sqrt[6]{25}\) and \(\sqrt[{12}]{289}\)
Updated on: 25 Jun 2024, 19:37
Originally posted by BrushMyQuant on 10 Aug 2021, 09:35.
Last edited by BrushMyQuant on 25 Jun 2024, 19:37, edited 1 time in total.
Last edited by BrushMyQuant on 25 Jun 2024, 19:37, edited 1 time in total.
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Let's look at the powers of all the roots. Powers are
\(\frac{1}{3}\), \(\frac{1}{2}\), \(\frac{1}{6}\) and \(\frac{1}{12}\) for {\(\sqrt[3]{4}\); \(\sqrt{3}\); \(\sqrt[6]{25}\); \(\sqrt[{12}]{289}\)} respectively.
Since, all are roots of a number so all values are positive.
Theory: When comparing positive numbers it is same as comparing powers of those positive numbers
We need to take LCM of all the roots i.e. we need to take LCM of 2,3,6 and 12. We are doing this as we can raise all the fractions to LCM to convert them to integer powers (so that it is easier to calculate and compare them!)
LCM (2,3,6,12) = 12 [ [url=https://www.youtube.com/watch?v=7jPV6JD0908]Watch this video to learn how to find LCM using multiple methods ]
So, if we have to compare {\(\sqrt[3]{4}\); \(\sqrt{3}\); \(\sqrt[6]{25}\); \(\sqrt[{12}]{289}\)} then it is same as comparing \(12^th\) (LCM) power of these numbers
{\((\sqrt[3]{4})^{12}\); \((\sqrt{3})^{12}\); \((\sqrt[6]{25})^{12}\); \((\sqrt[{12}]{289})^{12}\)}
=> \(4^{12/3}\); \(3^{12/2}\); \(25^{12/6}\); \(289^{12/12}\);
=> \(4^4\); \(3^6\) ; \(25^2\); 289
=> 256; 729; 625; 289
So, greatest number is 729 -> \(\sqrt{3}\) and least is 256 -> \(\sqrt[3]{4}\)
So, Answer will be B
Hope it helps!
Watch the following video to MASTER Roots
\(\frac{1}{3}\), \(\frac{1}{2}\), \(\frac{1}{6}\) and \(\frac{1}{12}\) for {\(\sqrt[3]{4}\); \(\sqrt{3}\); \(\sqrt[6]{25}\); \(\sqrt[{12}]{289}\)} respectively.
Since, all are roots of a number so all values are positive.
Theory: When comparing positive numbers it is same as comparing powers of those positive numbers
We need to take LCM of all the roots i.e. we need to take LCM of 2,3,6 and 12. We are doing this as we can raise all the fractions to LCM to convert them to integer powers (so that it is easier to calculate and compare them!)
LCM (2,3,6,12) = 12 [ [url=https://www.youtube.com/watch?v=7jPV6JD0908]Watch this video to learn how to find LCM using multiple methods ]
So, if we have to compare {\(\sqrt[3]{4}\); \(\sqrt{3}\); \(\sqrt[6]{25}\); \(\sqrt[{12}]{289}\)} then it is same as comparing \(12^th\) (LCM) power of these numbers
{\((\sqrt[3]{4})^{12}\); \((\sqrt{3})^{12}\); \((\sqrt[6]{25})^{12}\); \((\sqrt[{12}]{289})^{12}\)}
=> \(4^{12/3}\); \(3^{12/2}\); \(25^{12/6}\); \(289^{12/12}\);
=> \(4^4\); \(3^6\) ; \(25^2\); 289
=> 256; 729; 625; 289
So, greatest number is 729 -> \(\sqrt{3}\) and least is 256 -> \(\sqrt[3]{4}\)
So, Answer will be B
Hope it helps!
Watch the following video to MASTER Roots
General Discussion
11 Aug 2021, 01:33
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\(\sqrt[3]{4}, \sqrt{3}, \sqrt[6]{25}, \sqrt[12]{289}\)
Make square root same for all by taking LCM (3, 2, 6, 12) = 12
\(\sqrt[3*4]{4^4}, \sqrt[2*6]{3^6}, \sqrt[6*2]{25^2}, \sqrt[12]{289}\)
\(\sqrt[12]{256}, \sqrt[12]{729}, \sqrt[12]{625}, \sqrt[12]{289}\)
\(Greatest: \sqrt[12]{729} = \sqrt{3}\)
\(Least: \sqrt[12]{256} = \sqrt[3]{4}\)
Answer B
Make square root same for all by taking LCM (3, 2, 6, 12) = 12
\(\sqrt[3*4]{4^4}, \sqrt[2*6]{3^6}, \sqrt[6*2]{25^2}, \sqrt[12]{289}\)
\(\sqrt[12]{256}, \sqrt[12]{729}, \sqrt[12]{625}, \sqrt[12]{289}\)
\(Greatest: \sqrt[12]{729} = \sqrt{3}\)
\(Least: \sqrt[12]{256} = \sqrt[3]{4}\)
Answer B
