Tag: rational and irrational numbers

Quizzes (6) Questions (185)

Questions Related to rational and irrational numbers

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which one of the following set of surds is correct sequence of ascending order of their values?

  1. $\displaystyle \sqrt[4]{10},\sqrt[3]{6},\sqrt{3}$
  2. $\displaystyle \sqrt{3},\sqrt[4]{10},\sqrt[3]{6},$
  3. $\displaystyle \sqrt{3},\sqrt{10},\sqrt[3]{6},$
  4. $\displaystyle \sqrt[4]{10},\sqrt{3},\sqrt[3]{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sqrt[4]{10},\sqrt[3]{6},\sqrt{3}$
The order of the given irrational numbers are 2,3,4.
LCM of (2,3,4)=12
Now convert each irrational number as of order 12
$\sqrt[4]{10}=\sqrt[12]{10^3}=\sqrt[12]{1000}$
$\sqrt[3]{6}=\sqrt[12]{6^4}=\sqrt[12]{1296}$
$\sqrt{3}=\sqrt[12]{3^6}=\sqrt[12]{729}$
Hence, ascending order$\sqrt{3}<\sqrt[4]{10}<\sqrt[3]{6}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which is the greatest out of the following ?

  1. $\displaystyle \sqrt[3]{1.728}$
  2. $\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}$
  3. $\displaystyle \left ( \frac{1}{2} \right )^{-2}$
  4. $\displaystyle \frac{17}{8}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\Rightarrow  \sqrt[3]{1.728}=1.2$

$\Rightarrow \cfrac{\sqrt{3}-1}{\sqrt{3}+1}=\cfrac{(\sqrt{3}-1)^{2}}{(\sqrt{3}+1)(\sqrt{3}-1)}=\cfrac{3+1-2\sqrt{3}}{3-1}$
$ =\cfrac{4-2\sqrt{3}}{2}=2-\sqrt{3}$
$ =2-1.732=0.268$

$\Rightarrow \left ( \cfrac{1}{2} \right )^{-2}=2^{2}=4$

$\Rightarrow \cfrac{17}{8}=2.2125$

$ \therefore \left ( \cfrac{1}{2} \right )^{-2}$ is the greatest. 
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which is greater $\displaystyle (\sqrt{7}+\sqrt{10})$ or $\displaystyle (\sqrt{3}+\sqrt{19})$?

  1. $\displaystyle \sqrt{7}+\sqrt{10}$
  2. $\displaystyle \sqrt{3}+\sqrt{19}$

  3. Both are equal

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(\sqrt{7}+\sqrt{10})$
$2.6457+3.1622=5.8079$
$(\sqrt{3}+\sqrt{19})$
$1.732+4.358=6.090$
Hence $(\sqrt{3}+\sqrt{19})$is greater.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

$\displaystyle \sqrt[4]{3},\sqrt[6]{10},\sqrt[12]{25}$, when arranged in descending order will be 

  1. $\displaystyle \sqrt[4]{3},\sqrt[6]{10},\sqrt[12]{25}$
  2. $\displaystyle \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$
  3. $\displaystyle \sqrt[6]{10},\sqrt[12]{25},\sqrt[4]{3}$
  4. $\displaystyle \sqrt[4]{3},\sqrt[12]{25},\sqrt[6]{10}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

LCM of $4, 6$ and $12 = 12$.
$ \therefore $ Raising each of the given number to power $12$, we have 
$ (3^{1/4})^{12},(10^{1/6})^{12},(25^{1/12})^{12}$
$= 3^{3},10^{2},25$
$= 27, 100, 25$
Arranging in descending order, the numbers are $ 100, 27, 25$
$\Rightarrow \sqrt[6]{10},\sqrt[4]{3},\sqrt[12]{25}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The greatest amongst the the values $0.7 + \sqrt { 0.16 } ,  1.02 - \displaystyle\frac { 0.6 }{ 24 } ,   1.2 \times 0.83$ and $\sqrt { 1.44 } $ is

  1. $0.7+\sqrt { 0.16 } $
  2. $1.02-\displaystyle\frac { 0.6 }{ 24 } $
  3. $1.2\times 0.83$
  4. $\sqrt { 1.44 } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$0.7+ \sqrt { 0.16 } = 0.7 +0.4 = 1.1$ 

$1.02-\displaystyle  \frac { 0.6 }{ 24 } = 1.02 - 0.025 = 1.175$
$1.2 \times 0.83 = 0.996 $
$\sqrt { 1.44 } = 1.2$

$\therefore $ the greatest is $\sqrt { 1.44 } $.

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Which of the following is smallest?

  1. $\sqrt [4]{5}$
  2. $\sqrt [5]{4}$
  3. $\sqrt {4}$
  4. $\sqrt {3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let us rewrite the given set of magnitudes $\sqrt [ 4 ]{ 5 } ,\sqrt [ 5 ]{ 4 } ,\sqrt { 4 },\sqrt {3}$ as follows:


$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1 }{ 4 }  }\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 5 }  }\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1 }{ 2 }  }\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1 }{ 2 }  }$

 
We now take the LCM of the denominators of the powers to make the denominators same, then the above magnitudes will be:

$\sqrt [ 4 ]{ 5 } ={ \left( 5 \right)  }^{ \dfrac { 1\times 5 }{ 4\times 5 }  }={ \left( 5 \right)  }^{ \dfrac { 5 }{ 20 }  }={ \left( { 5 }^{ 5 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 3125 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 3125 } \\ \sqrt [ 5 ]{ 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 4 }{ 5\times 4 }  }={ \left( 4 \right)  }^{ \dfrac { 4 }{ 20 }  }={ \left( { 4 }^{ 4 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 256 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 256 } \\ \sqrt { 4 } ={ \left( 4 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 4 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 4 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 1048576 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 1048576 } \\ \sqrt { 3 } ={ \left( 3 \right)  }^{ \dfrac { 1\times 10 }{ 2\times 10 }  }={ \left( 3 \right)  }^{ \dfrac { 10 }{ 20 }  }={ \left( { 3 }^{ 10 } \right)  }^{ \dfrac { 1 }{ 20 }  }={ \left( 2187 \right)  }^{ \dfrac { 1 }{ 20 }  }=\sqrt [ 20 ]{ 2187 }$     

Now, the descending order is as shown below:

$\sqrt [ 20 ]{ 1048576 } >\sqrt [ 20 ]{ 3125 } >\sqrt [ 20 ]{ 2187 } >\sqrt [ 20 ]{ 256 } \\ \Rightarrow \sqrt { 4 } >\sqrt [ 4 ]{ 5 } >\sqrt { 3 } >\sqrt [ 5 ]{ 4 }$ 

Hence, the smallest magnitude is $\sqrt [ 5 ]{ 4 }$.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Arrange the following surds in ascending order of their magnitudes: $\sqrt{5},\sqrt [ 3 ]{ 11 } ,2\sqrt [ 6 ]{ 3 } $

  1. $\sqrt [ 3 ]{ 11 } > \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $
  2. $\sqrt [ 3 ]{ 11 } < \sqrt{5}< 2\sqrt [ 6 ]{ 3 } $
  3. $\sqrt [ 3 ]{ 11 } > \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $
  4. $\sqrt [ 3 ]{ 11 } < \sqrt{5}> 2\sqrt [ 6 ]{ 3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\sqrt{5} = 5^{1/2}$

$\sqrt[3]{11} = 11^{1/3}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{1/6}$

L.C.M of the denominators of the exponents is 12.

So,

$\sqrt{5} = 5^{\frac{1}{2}\times\frac{6}{6}} = \sqrt [12]{5^6}=\sqrt[12]{15625}$

$\sqrt[3]{11} = 11^{\frac{1}{3}\times\frac{4}{4}} = \sqrt [12]{11^4} =\sqrt[12]{14641}$

$2\sqrt[6]{3} = \sqrt[6]{12}= 12^{\frac{1}{6}\times\frac{2}{2}} = \sqrt[12]{12^2} =\sqrt[12]{144}$

Hence, the Ascending order is $2\sqrt[6]{3}, \sqrt[3]{11},\sqrt{5}$

Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

Write $\displaystyle \sqrt[4]{6},\sqrt{2},\sqrt[3]{4}$ in ascending order

  1. $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$
  2. $\displaystyle \sqrt[4]{6}$, $\sqrt{2}$ and $\displaystyle \sqrt[3]{4}$
  3. $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$ and $\sqrt[4]{6}$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\displaystyle 6^\cfrac 14,2^\cfrac 12,4^\cfrac 13$
The surds are of the order $4, 2$ and $3$ respectively The L.C.M. is $12$ So, we change each surd of the order $12$
The terms now are $\displaystyle \left ( 6^{3} \right )^\cfrac {1}{12},\left ( 2^{6} \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 4^{4} \right )^\cfrac{1}{12}$
$\displaystyle \Rightarrow \left ( 216 \right )^\cfrac{1}{12},\left ( 64 \right )^\cfrac{1}{12}$ and $\displaystyle \left ( 256 \right )^\cfrac{1}{12}$
$\displaystyle \therefore $ Ascending order is $\displaystyle \sqrt{2},\sqrt[4]{6}$ and $\displaystyle \sqrt[3]{4}$