Tag: surds and law of surds

Questions Related to surds and law of surds

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

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. $p\leq q$

Show answer
Correct Option: B
Explanation:

(B) $\frac{p}{q}=\frac{\sqrt{32}-\sqrt{24}}{\sqrt{50}-\sqrt{48}}\times \frac{\sqrt{50}+\sqrt{48}}{\sqrt{50}+\sqrt{48}}$

$=\frac{(4\sqrt{2}-2\sqrt{6})(5\sqrt{2}+4\sqrt{3})}{2}$

$=(2\sqrt{2}-\sqrt{6})(5\sqrt{2}+4\sqrt{3})> 1$

$\therefore p> q$

If $x=\sqrt{2}+1,    y=\sqrt{17}-\sqrt{2}$, then:

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

  1. $x< y$

  2. $x > y$

  3. $x=y$

  4. $x\geq y$

Show answer
Correct Option: A
Explanation:

(A) Given, $x=\sqrt{2}+1$ and $y=\sqrt{17}-\sqrt{2}$

$\dfrac{x}{y}=\dfrac{\sqrt{2}+1}{\sqrt{17}-\sqrt{2}}\times \dfrac{\sqrt{17}+\sqrt{2}}{\sqrt{17}+\sqrt{2}}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{17-2}$

$=\dfrac{(\sqrt{2}+1)(\sqrt{17}+\sqrt{2})}{15}< 1$

$\therefore x< y$

Arrange the following in ascending order of magnitude: $\displaystyle \sqrt[4]{90}, \sqrt[3]{10}, \sqrt{6}$

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

  1. $\displaystyle \sqrt{3} < \sqrt[4]{10} < \sqrt[3]{6}$

  2. $\displaystyle \sqrt{3} > \sqrt[4]{10} > \sqrt[3]{6}$

  3. $\displaystyle \sqrt{3} > \sqrt[4]{10} < \sqrt[3]{6}$

  4. $\displaystyle \sqrt{3} < \sqrt[4]{10} > \sqrt[3]{6}$

Show answer
Correct Option: A

$if\,A\, = \sqrt 7  - \sqrt 6 \,and\,B = \,\sqrt 6  - \sqrt {5,} \,then\,$

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

  1. $A > B$

  2. $A = B$

  3. $A < B\,$

  4. $A \geqslant B$

Show answer
Correct Option: C
Explanation:
Given numbers are $A=\sqrt{7}-\sqrt{6}$ and $B=\sqrt{6}- \sqrt{5}$
Let $x =\sqrt{5}$ and $y= \sqrt{7}$ then by
$A.M$ and $G.M$
$\dfrac{x+y}{2} \le \sqrt{\dfrac{x^{2}+y^{2}}{2}}$
$\Rightarrow \dfrac{\sqrt{5}+ \sqrt{7}}{2} \le \sqrt{6}$
$\Rightarrow \sqrt{5}+ \sqrt{7} \le 2 \sqrt{6}$
$\Rightarrow \sqrt{7}- \sqrt{6} \le \sqrt{6} - \sqrt{5}$

Which of the following numbers is the least ?
$\displaystyle (0.5)^{2},\sqrt{0.49},\sqrt[3]{0.008},0.23$

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

  1. $\displaystyle (0.5)^{2}$

  2. $\displaystyle \sqrt{0.49}$

  3. $\displaystyle \sqrt[3]{0.008}$

  4. 0.23

Show answer
Correct Option: C
Explanation:

$ (0.5)^{2}=0.25$
$\sqrt{0.49}=0.7;$
$ \sqrt[3]{0.008}=\sqrt[3]{.2^3}=0.2$
$0.23$
Arranging in ascending order the numbers are $0.2< 0.23< 0.25< 0.7$
$ \therefore \sqrt[3]{0.008}=0.2$ is the least

The greatest number among $\displaystyle \sqrt[3]{2},\sqrt{3},\sqrt[3]{5}$ and $1.5$ is 

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

  1. $\displaystyle \sqrt[3]{2}$

  2. $\displaystyle \sqrt{3}$

  3. $\displaystyle \sqrt[3]{5}$

  4. $1.5$

Show answer
Correct Option: B
Explanation:

LCM of $3, 2 = 6$
Given numbers are $ \sqrt[3]{2},\sqrt{3},\sqrt[3]{5}, 1.5$ i.e,
$ 2^{1/3},3^{1/2},5^{1/3},1.5$
$ \therefore $ Raising each number to power $6$, we get
$ (2^{1/3})^{6},(3^{1/2})^{6},(5^{1/3})^{6}, (1.5)^{6}$

$= 2^{2},3^{3},5^{2}, \left(\cfrac{3}{2}\right)^{6}$
$=4,27,25,\cfrac{729}{64}$
Of all these numbers, $27$ is the greatest.
$ \Rightarrow \sqrt{3}$ is the greatest. 

The smallest of $\displaystyle \sqrt{8}+\sqrt{5},\sqrt{7}+\sqrt{6},\sqrt{10}+\sqrt{3}$ and $\displaystyle \sqrt{11}+\sqrt{2}$ is 

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

  1. $\displaystyle \sqrt{8}+\sqrt{5}$

  2. $\displaystyle \sqrt{7}+\sqrt{6}$

  3. $\displaystyle \sqrt{10}+\sqrt{3}$

  4. $\displaystyle \sqrt{11}+\sqrt{2}$

Show answer
Correct Option: D
Explanation:

$\displaystyle \sqrt{8}+\sqrt{5}=2.83+2.24=5.07$
$\displaystyle \sqrt{7}+\sqrt{6}=2.65+2.45=5.09$
$\displaystyle \sqrt{10}+\sqrt{13}=3.16+3.61=6.77$
$\displaystyle \sqrt{11}+\sqrt{12}=3.32+1.41=4.73$
$\displaystyle \therefore $ Smallest is $\displaystyle \sqrt{11}+\sqrt{2}$

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

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

  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. $\displaystyle \sqrt[4]{10},\sqrt{3},\sqrt[3]{6}$

Show answer
Correct Option: B
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$\sqrt{3}<\sqrt[4]{10}<\sqrt[3]{6}$

Which is the greatest out of the following ?

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

  1. $\displaystyle \sqrt[3]{1.728}$

  2. $\displaystyle \frac{\sqrt{3}-1}{\sqrt{3}+1}$

  3. $\displaystyle \left ( \frac{1}{2} \right )^{-2}$

  4. $\displaystyle \frac{17}{8}$

Show answer
Correct Option: C
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. 

$4\sqrt{18}$ $=$ $12\sqrt{2}$
State true or false

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

  1. True

  2. False

Show answer
Correct Option: A
Explanation:

Consider $4\sqrt { 18 }$ and factorize it as follows:

 
$4\sqrt { 18 } =4\sqrt { 2\times 3\times 3 } =4\sqrt { 2\times 3^{ 2 } } =4\times 3\sqrt { 2 } =12\sqrt { 2 }$
 
Hence, $4\sqrt { 18 } =12\sqrt { 2 }$