Tag: exponents

Quizzes (9) Questions (51)

Questions Related to exponents

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

Compare the following pairs of surds. $\sqrt[8]{80}, \sqrt[4]{40}$    

  1. $\sqrt[8]{80} < \sqrt[4]{40}$
  2. $\sqrt[8]{80} \neq \sqrt[4]{40}$
  3. $\sqrt[8]{80} = \sqrt[4]{40}$
  4. $\sqrt[8]{80} > \sqrt[4]{40}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

   $\sqrt[8]{80}, \sqrt[4]{40}$
$={80}^{\frac{1}{8}}, {40}^{\frac{1}{4}}$
$={80}^{\frac{1}{8}}, {40}^{\frac{2}{8}}$
$={80}^{\frac{1}{8}}, {1600}^{\frac{1}{8}}$
Now,
   $80<1600$
$=>{80}^{\frac{1}{8}}<{1600}^{\frac{1}{8}}$
$=>\sqrt[8]{80}< \sqrt[4]{40}$

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

Which among the following numbers is the greatest?
$\displaystyle 0.07+\sqrt{0.16},\sqrt{1.44},1.2\times 0.83,1.02-\frac{0.6}{24}$

  1. $\displaystyle \sqrt {1.44}$
  2. $\displaystyle 0.07+\sqrt{0.16}$
  3. $1.2\times 0.83$
  4. $1.02-\dfrac{0.6}{24}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$\Rightarrow 0.07+\sqrt{0.16}=0.07+0.4=0.47$
$\Rightarrow \sqrt{1.44}=1.2$
$\Rightarrow 1.2 \times 0.83 = 0.996$
$\Rightarrow 1.02-\cfrac{0.6}{24}=1.02-\cfrac {6}{240}=1.02-0.025=0.995$
$ \therefore$ The greatest number is $1.2$ i.e. $\sqrt {1.44}$
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

State whether the following equality is true or false:

$\displaystyle \frac{2\sqrt{3}}{\sqrt{5}} = $$\displaystyle \frac{2\sqrt{15}}{\sqrt{5}}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rationalizing factor of  $\displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$ is $\sqrt{5}$


$\therefore \displaystyle \frac{2\sqrt{3}}{\sqrt{5}}$  $=\displaystyle \frac{2\sqrt{3}\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}$

$= \dfrac{2\sqrt{15}}{5}$

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

Determine the order relation between the following pairs of ratios.

$\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}}, \frac{2\sqrt{2}}{3\sqrt{3}}$

  1. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} > \frac{2\sqrt{2}}{3\sqrt{3}}$
  2. $\displaystyle \frac{3\sqrt{3}}{2\sqrt{2}} < \frac{2\sqrt{2}}{3\sqrt{3}}$

  3. Cannot be determined

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

$\dfrac{3\sqrt{3}}{2\sqrt{2}}=\dfrac{3\times 1.73214}{2\times 1.41429} = \dfrac{5.19642}{2.85828}
=1.82151$
$\dfrac{2\sqrt{2}}{3\sqrt{3}}=\dfrac{2\times 1.41429}{3\times 1.73214}=\dfrac{2.85828}{5.19642}=0.55004$
$\therefore \dfrac{3\sqrt{3}}{2\sqrt{2}} >\dfrac{2\sqrt{2}}{3\sqrt{3}}$

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

Compare the following pairs of surds $\sqrt[8]{12}, \sqrt[4]{6}$

  1. $\sqrt[8]{2} < \sqrt[4]{6}$
  2. $\sqrt[8]{8} < \sqrt[4]{6}$
  3. $\sqrt[8]{12} < \sqrt[4]{6}$
  4. $\sqrt[8]{12} < \sqrt[4]{4}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To compare, convert to a common root: sqrt[8]{12} is 12^(1/8) and sqrt[4]{6} is 6^(1/4) = 6^(2/8) = 36^(1/8). Since 12 < 36, sqrt[8]{12} < sqrt[4]{6}.

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

Compare the following pair of surds:

$\sqrt[3]{6}, \sqrt[4]{8}$

  1. $\sqrt[3]{6} > \sqrt[4]{8}$
  2. $\sqrt[3]{6} > \sqrt[4]{4}$
  3. $\sqrt[3]{6} > \sqrt[3]{8}$
  4. $\sqrt[3]{4} > \sqrt[4]{8}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Convert to a common root (12th root): sqrt[3]{6} = 6^(4/12) = 1296^(1/12). sqrt[4]{8} = 8^(3/12) = 512^(1/12). Since 1296 > 512, the first is larger.

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

Arrange the following in ascending order of magnitude: 

$\displaystyle \sqrt[3]{4}, \sqrt[4]{5}, \sqrt{3}$ 

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

Convert to 12th roots: sqrt[3]{4} = 4^(4/12) = 256^(1/12). sqrt[4]{5} = 5^(3/12) = 125^(1/12). sqrt{3} = 3^(6/12) = 729^(1/12). Ordering 125 < 256 < 729 gives the correct sequence.

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

What is the least value of $a$ in $ \displaystyle\frac{\sqrt 2+\sqrt 3}{\sqrt{2+3}} < a$?

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 2+3 }  } =\dfrac { \sqrt { 2 } +\sqrt { 3 }  }{ \sqrt { 5 }  } =\dfrac { (\sqrt { 2 } +\sqrt { 3 } )\times \sqrt { 5 }  }{ 5 } =\dfrac { 7.02 }{ 5 } \\ =1.40$
$\Rightarrow 1.40<a$
So, least integer value of $a$ is $2$.
Hence, option B is correct.
Multiple choice comparison of irrational numbers surds and law of surds rational and irrational numbers exponents maths

The greatest among $\displaystyle \sqrt[6]{3}$, $\displaystyle \sqrt{2}$, $\displaystyle \sqrt[3]{4}$, $\displaystyle \sqrt[4]{5}$ is--

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

$\displaystyle \therefore $ $\displaystyle \sqrt[6]{3}$ = $\displaystyle \left ( 3 \right )^{\dfrac{1}{6}}$ = $\displaystyle \left ( 3^{2} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 9 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt{2}$ = $\displaystyle \left ( 2 \right )^{\dfrac{1}{2}}$ = $\displaystyle \left ( 2^{6} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 64 \right )^{\dfrac{1}{12}}$
$\displaystyle \sqrt[3]{4}$ = $\displaystyle \left ( 4 \right )^{\dfrac{1}{3}}$ = $\displaystyle \left ( 4^{4} \right )^{\dfrac{1}{12}}$ = $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$

$\sqrt[4]{5}=(5)^{\dfrac{1}{4}}=(5^{3})^{\dfrac{1}{12}}=(125)^{\dfrac{1}{12}}$
$\displaystyle \therefore $ The greatest number is $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$ = $\displaystyle \sqrt[3]{4}$