Tag: maths

Questions Related to maths

Determine the order relation between the following pairs of ratios.

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

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

  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

Show answer
Correct Option: A
Explanation:

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

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

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

  1. $\sqrt[8]{2} < \sqrt[4]{6}$

  2. $\sqrt[8]{8} < \sqrt[4]{6}$

  3. $\sqrt[8]{12} < \sqrt[4]{6}$

  4. $\sqrt[8]{12} < \sqrt[4]{4}$

Show answer
Correct Option: C

Compare the following pair of surds:

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

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

  1. $\sqrt[3]{6} > \sqrt[4]{8}$

  2. $\sqrt[3]{6} > \sqrt[4]{4}$

  3. $\sqrt[3]{6} > \sqrt[3]{8}$

  4. $\sqrt[3]{4} > \sqrt[4]{8}$

Show answer
Correct Option: A

Arrange the following in ascending order of magnitude: 

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

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

  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. $\displaystyle \sqrt[4]{5} < \sqrt[3]{4} > \sqrt{3}$

Show answer
Correct Option: A

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

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

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: B
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.

Which of the following is the greatest?

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

  1. $\sqrt 2$

  2. $\sqrt 3$

  3. $\sqrt 4$

  4. $\sqrt 5$

Show answer
Correct Option: D
Explanation:

$\sqrt 5$ > $\sqrt 4$ > $\sqrt 3$ > $\sqrt 2$


So, option D is correct.

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

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

  1. $\displaystyle \sqrt[6]{3}$

  2. $\displaystyle \sqrt{2}$

  3. $\displaystyle \sqrt[3]{4}$

  4. $\displaystyle \sqrt[4]{5}$

Show answer
Correct Option: C
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 $\displaystyle \left ( 256 \right )^{\dfrac{1}{12}}$ = $\displaystyle \sqrt[3]{4}$

Which of the following is smallest?

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

  1. $ \displaystyle \sqrt[4]{5} $

  2. $ \displaystyle \sqrt[5]{4} $

  3. $ \displaystyle \sqrt{4} $

  4. $ \displaystyle \sqrt{3} $

Show answer
Correct Option: B
Explanation:

Since, fifth root of four is smallest.

Option $B$ is correct.

Which one of the following is the smallest surd?

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

  1. $\sqrt 4$

  2. $\sqrt {27}$

  3. $\sqrt 9$

  4. $\sqrt 5$

Show answer
Correct Option: A
Explanation:
As we know, $4 < 5 < 9 < 27$
So, $\sqrt 4$ < $\sqrt 5$ < $\sqrt 9$ < $\sqrt {27}$
Hence, the answer is $\sqrt{4}$.

If $p=\sqrt{32}-\sqrt{24}$ and $q=\sqrt{50}-\sqrt{48}$, then:

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

  1. $p< q$

  2. $p> q$

  3. $p=q$

  4. $p\le q$

Show answer
Correct Option: B
Explanation:

$p= \sqrt{32} - \sqrt{24}$

$p= 5.656-4.898 = 0.758$

$q = \sqrt{50} + \sqrt{48}$

$q= 7.071- 6.928 = 0.089$

$\therefore p> q$