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#### (i) Simplify :- (13 + 23 + 33)1/2 (ii) Simplify :- $\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}$ (iii) Simplify :- $\left ( \frac{1}{27} \right )^{-\frac{2}{3}}$ (iv) Simplify :- $\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}$ (v) Simplify :- $\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}$ (vi) Simplify :-$64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )$ (vii) Simplify :- $\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$

Solution.     (13 + 23 + 33)1/2
We know that
13 = 1.1.1 = 1
23 = 2.2.2 = 8
33 = 3.3.3 = 27
Putting these values we get
(13 + 23 + 33)1/2 $= \sqrt{1+8+27}$
$=\sqrt{36}= 6$

(ii) Answer.  $\frac{2025}{64}$

Solution.$\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}$
We know that

8 = 2.2.2 = 23
32 = 2.2.2.2.2 = 25
$\left ( \frac{3}{5} \right )^{4}\left ( \frac{8}{5} \right )^{-12}\left ( \frac{32}{5} \right )^{6}= \left ( \frac{3}{5} \right )^{4}\left ( \frac{2^{3}}{5} \right )^{-12}\left ( \frac{2^{5}}{5} \right )^{6}$
$= \frac{3^{4}\left ( 2^{3} \right )^{-12}\left ( 2^{5} \right )^{6}}{5^{4}5^{-12}5^{6}}$                $\because \left ( \frac{a}{b} \right )^{m}= \frac{a^{m}}{b^{m}}$
$= \frac{3^{4}2^{-36}2^{30}}{5^{4}5^{-12}5^{6}}$                            $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= \frac{3^{4}\times 2^{-36+30}}{5^{4-12+6}}$                        $\because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}$
$= \frac{3^{4}\times 2^{-6}}{5^{-2}}$
$= \frac{3^{4}\times 5^{2}}{2^{6}}$                                   $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \frac{81\times 25}{64}$
$= \frac{2025}{64}$
Hence the answer is $\frac{2025}{64}$

Solution. Given $\left ( \frac{1}{27} \right )^{-\frac{2}{3}}$
We know that
27 = 3.3.3 = 33
$\left ( \frac{1}{27} \right )^{-\frac{2}{3}}= \left ( \frac{1}{3^{3}} \right )^{-\frac{2}{3}}$
$= \left ( 3^{3} \right )^{\frac{2}{3}}$            $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \left ( 3 \right )^{3\times \frac{2}{3}}$         $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$

= 32 = 9

Solution. Given $\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}$
We know that
$625= \left ( 25 \right )\left ( 25 \right )= 5\cdot 5\cdot 5\cdot 5= 5^{4}$
$\left [ \left \{ \left ( 625 \right )^{-\frac{1}{2}} \right \}^{-\frac{1}{4}} \right ]^{2}= \left [ \left \{ \left ( \left ( 5 \right )^{4} \right )^{-\frac{1}{2}} \right \} ^{-\frac{1}{4}}\right ]^{2}$
$= 5^{4\times \frac{-1}{2}\times \frac{-1}{4}\times 2}$        $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$

= 51 = 5

(v) Answer.      $\sqrt[3]{\frac{1}{3}}$
Solution. We have $\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}$
Now we know that
9 = 3.3 = 32
27 = 3.3.3 = 33
$\frac{9^{\frac{1}{3}}\times 27^{\frac{1}{2}}}{3^{\frac{1}{6}}\times 9^{-\frac{2}{3}}}= \frac{\left ( 3^{2} \right )^{\frac{1}{3}}\times \left ( 3^{3} \right )^{\frac{1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3^{2} \right )^{\tfrac{-2}{3}}}$
$= \frac{\left ( 3\right )^{2\times \frac{1}{3}}\times \left ( 3 \right )^{3\times \frac{-1}{2}}}{\left ( 3 \right )^{\frac{1}{6}}\times \left ( 3 \right )^{\tfrac{-2}{3}}}$        $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= \frac{\left ( 3 \right )^{\frac{2}{3}-\frac{3}{2}}}{\left ( 3 \right )^{\frac{1}{6}-\frac{2}{3}}}$              $\because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}$
$= \frac{3^{\frac{4-9}{6}}}{3^{\frac{1-4}{6}}}$
$=\frac{3^{-\frac{5}{6}}}{3^{-\frac{3}{6}}}$
$= 3^{\frac{-5}{6}-\left ( \frac{-3}{6} \right )}$           $\because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}$
$= 3^{-\frac{2}{6}}$
$= \left ( \frac{1}{3} \right )^{\frac{1}{3}}$                 $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \sqrt[3]{\frac{1}{3}}$

Hence the answer is $\sqrt[3]{\frac{1}{3}}$

Solution. We have ,$64^{-\frac{1}{3}}\left ( 64^{\frac{1}{3}}-64^{\frac{2}{3}} \right )$
We know that 64 =4.4.4=43
$= \left ( 4^{3} \right )^{\frac{-1}{3}}\left \{ \left ( \left ( 4^{3} \right ) ^{\frac{1}{3}}-\left ( 4^{3} \right ) ^{\frac{2}{3}}\right )\right \}$
$= \left ( 4 \right )^{3\times \frac{-1}{3}}\left \{ \left ( \left ( 4 \right )^{3\times \frac{1}{3}}-\left ( 4 \right )^{3\times \frac{2}{3}} \right ) \right \}$      $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= 4^{-1}\left ( 4-4^{2} \right )$
$= \frac{1}{4}\left ( 4-16 \right )$                                                        $\because \left ( a \right )^{-m}= \left ( \frac{1}{a} \right )^{m}$
$= \frac{1}{4}\left ( -12 \right )$

= – 3
Hence the answer is – 3

Solution.
Given ,$\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}$
We know that
8 = 2.2.2 = 23
16 = 2.2.2.2 = 24
32 = 2.2.2.2.2 = 25
$\frac{8^{\frac{1}{3}}\times 16^{\frac{1}{3}}}{32^{-\frac{1}{3}}}= \frac{\left ( 2^{3} \right )^{\frac{1}{3}}\times\left ( 2^{4} \right )^{\frac{1}{3}}}{\left ( 2^{5} \right )^{-\frac{1}{3}}}$
$= \frac{2^{3\times\frac{1}{3}}\times2^{4\times\frac{1}{3}}}{2^{5\times\frac{-1}{3}}}$              $\because \left ( \left ( a \right )^{m} \right )^{n}= \left ( a \right )^{mn}$
$= 2^{1+\frac{4}{3}+\frac{5}{3}}$                             $\because \left ( a \right )^{m}\left ( a \right )^{n}= \left ( a \right )^{m+n}$  and
$\because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}$
$= 2^{\frac{3+4+5}{3}}= 2^{\frac{12}{3}}$
$= 2^{4}= 16$

#### (i) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{4}{\sqrt{3}}$ (ii) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{6}{\sqrt{6}}$ (iii) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{\sqrt{10}-\sqrt{5}}{\sqrt{2}}$ (iv) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{2}{2+\sqrt{2}}$ (v) Rationalize the denominator in each of the following and hence evaluate by taking $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$ and $\sqrt{5}= 2\cdot 236$, upto three places of decimal : $\frac{1}{\sqrt{3}+\sqrt{2}}$

Solution. Given: $\frac{4}{\sqrt{3}}$
Rationalising,
$\frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}= \frac{4\sqrt{3}}{3}$
(Given that $\sqrt{3}= 1.732$)
$= \frac{4\times 1\cdot 732}{3}$
= 2.3093

Solution.  Given: $\frac{6}{\sqrt{6}}$
Rationalising,
$\frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}= \frac{6\sqrt{6}}{6}$
$= \frac{6\times \sqrt{2}\sqrt{3}}{6}$
Putting the given values,
$\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$
We get :
$= \sqrt{2}\cdot \sqrt{3}$
$= 1\cdot 414\times 1\cdot 732= 2\cdot 449$

Solution.   Given that $\frac{\sqrt{10}-\sqrt{5}}{2}$
This can be written as

$\frac{\sqrt{2}\times \sqrt{5}-\sqrt{5}}{2}$
Now putting the given values,

$\sqrt{2}= 1\cdot 414,\sqrt{5}= 2\cdot 236$
We get :
$\Rightarrow \frac{1\cdot 414\times 2\cdot 236-2\cdot 236}{2}$
= 0.462852

Solution.  Given: $\frac{\sqrt{2}}{2+\sqrt{2}}$
Rationalising,
$\frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}$
Using   (a – b) (a + b) = a2 – b2
$= \frac{\sqrt{2}\left ( 2-\sqrt{2} \right )}{2^{2}-\sqrt{2}^{2}}$
$= \frac{2\sqrt{2}-2}{4-2}$
$= \frac{2\left ( \sqrt{2}-1 \right )}{2}$
$= \sqrt{2}-1$

Putting the given value of $\sqrt{2}= 1\cdot 414$
We get
= 1.414 – 1
= 0.414

Solution. Given that $\frac{1}{\sqrt{3}+\sqrt{2}}$
Rationalising,
$\frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
Using   (a – b) (a + b) = a2 – b2
$= \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}^{2}-\sqrt{2}^{2}}$
$= \frac{\sqrt{3}-\sqrt{2}}{3-2}$
$= \sqrt{3}-\sqrt{2}$

Putting the given values, $\sqrt{2}= 1\cdot 414,\sqrt{3}= 1\cdot 732$
We get,
= 1.732 – 1.414
= 0.318

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#### (i) Find the values of a in each of the following : $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}$ (ii) Find the values of a in the following : $\frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}$(iii) Find the values of b in the following : $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$ (iv) Find the values of a and b in the following : $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}$

Solution.   We have, $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}= a-6\sqrt{3}$
LHS = $\frac{5+2\sqrt{3}}{7+4\sqrt{3}}$
Rationalising the denominator, we get:
$= \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}$
$= \frac{\left ( 5+2\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}{\left ( 7+4\sqrt{3} \right )\left ( 7-4\sqrt{3} \right )}$
{Using (a – b) (a + b) = a2 – b2}

$= \frac{35+14\sqrt{3}-20\sqrt{3}-24}{7^{2}-\left ( 4\sqrt{3} \right )^{2}}$
$= \frac{11-6\sqrt{3}}{49-48}$
$= 11-6\sqrt{3}$
Now RHS $= a-6\sqrt{3}$
$\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}$
$\Rightarrow 11-6\sqrt{3}= a-6\sqrt{3}$
$\Rightarrow a= 11$
Hence a = 11 is the required answer

(ii)Answer.   $a= \frac{9}{11}$
Solution.  Given that, $\frac{3-\sqrt{5}}{3+2\sqrt{5}}= a\sqrt{5}-\frac{19}{11}$

LHS = $\frac{3-\sqrt{5}}{3+2\sqrt{5}}$
Rationalising the denominator, we get:
LHS $= \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}}$
{Using (a – b) (a + b) = a2 – b2}
$= \frac{9-3\sqrt{5}-6\sqrt{5}+10}{3^{2}-\left ( 2\sqrt{5} \right )^{2}}$
$= \frac{19-9\sqrt{5}}{9-20}$
Now RHS $= a\sqrt{5}-\frac{19}{11}$
$\Rightarrow \frac{9\sqrt{5}}{11}-\frac{19}{11}= a\sqrt{5}-\frac{19}{11}$
Comparing both , we get
$\Rightarrow a= \frac{9}{11}$
Hence $a= \frac{9}{11}$is the correct answer

(iii) Answer:$b = -\frac{5 }{6}$

Solution:

Given:

$\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$

LHS = $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}$

Rationalize

= $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} \times \frac{3 \sqrt{2}+2 \sqrt{3}}{3 \sqrt{2}+2 \sqrt{3}}$

=$\frac{\sqrt{2}(3 \sqrt{2}+2 \sqrt{3})+\sqrt{3}(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2})^{2}-(2 \sqrt{3})^{2}}$

= $\frac{6+2 \sqrt{6}+3 \sqrt{6}+6}{18-12}$

= $2+\frac{5 \sqrt{6}}{6}$

$2+\frac{5 \sqrt{6}}{6}=2-b \sqrt{6}$

$b = -\frac{5 }{6}$

(iv) Answer. a = 0, b = 1
Solution.         Given,
$\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}$
LHS $= \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}$
$=\frac{\left ( 7+\sqrt{5} \right )\times\left ( 7+\sqrt{5} \right )-\left ( 7-\sqrt{5} \right ) \times \left ( 7-\sqrt{5} \right )}{\left (7 -\sqrt{5} \right )\left (7 +\sqrt{5} \right )}$
Using (a – b) (a + b) = a2 – b2
(a + b)2 = a2 + b2 + 2ab
(a - b)2 = a2 + b2 - 2ab
$= \frac{\left ( 7^{2} +\sqrt{5}^{2}+2\cdot 7\cdot \sqrt{5}\right )-\left ( 7^{2} +\sqrt{5}^{2}-2\cdot 7\cdot \sqrt{5} \right )}{7^{2}-\sqrt{5}^{2}}$
$= \frac{\left ( 49+5+14\sqrt{5} \right )-\left ( 49+5-14\sqrt{5} \right )}{49-5}$
$= \frac{54+14\sqrt{5}-54+14\sqrt{5}}{44}$
$= \frac{28\sqrt{5}}{44}$
RHS $= a+\frac{7}{11}\sqrt{5b}$
Now LHS = RHS
$\Rightarrow \frac{28\sqrt{5}}{44}= a+\frac{7}{11}\sqrt{5b}$
$\Rightarrow 0+\left ( \frac{4}{4} \right )\frac{7}{11}\sqrt{5}= a+\frac{7}{11}\sqrt{5b}$

$\Rightarrow$ a = 0, b = 1
Hence the answer is a = 0, b = 1

#### (i) Rationalise the denominator of the following : $\frac{2}{3\sqrt{3}}$ (ii)Rationalise the denominator of the following : $\frac{\sqrt{40}}{\sqrt{3}}$ (iii) Rationalise the denominator of the following : $\frac{3+\sqrt{2}}{4\sqrt{2}}$ (iv)Rationalise the denominator of the following :$\frac{16}{\sqrt{41}-5}$ (v) Rationalise the denominator of the following : $\frac{2+\sqrt{3}}{2-\sqrt{3}}$ (vi) Rationalise the denominator of the following  : $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$ (vii) Rationalise the denominator of the following : $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$(viii) Rationalise the denominator of the following : $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$(ix) Rationalise the denominator of the following : $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

(i) Answer.  $\frac{2\sqrt{3}}{9}$
Solution.         We have, $\frac{2}{3\sqrt{3}}$
Rationalising the denominator, we get:
$\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2\sqrt{3}}{3\sqrt{3}\sqrt{3}}$
$= \frac{2\sqrt{3}}{9}$
Hence the answer is $\frac{2\sqrt{3}}{9}$

(ii) Answer.  $\frac{2\sqrt{30}}{3}$
Solution. We have ,$\frac{\sqrt{40}}{\sqrt{3}}$
We know that, 40 = (2) (2) (10)
$\frac{\sqrt{40}}{\sqrt{3}}= \frac{\sqrt{2\cdot 2\cdot 10}}{\sqrt{3}}= \frac{2\sqrt{10}}{\sqrt{3}}$

Rationalising the denominator, we get:
$= \frac{2\sqrt{10}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
$= \frac{2\sqrt{10}\sqrt{3}}{\sqrt{3}\sqrt{3}}$
$= \frac{2\sqrt{30}}{3}$

Hence the answer is: $\frac{2\sqrt{30}}{3}$

(iii) Answer.  $\frac{3\sqrt{2}+2}{8}$
Solution.  We have $\frac{3+\sqrt{2}}{4\sqrt{2}}$
Rationalising the denominator, we get:
$\frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$
$= \frac{\left ( 3+\sqrt{2} \right )\sqrt{2}}{4\sqrt{2}\sqrt{2}}$
$= \frac{3\sqrt{2}+2}{8}$
Hence the answer is $\frac{3\sqrt{2}+2}{8}$

(iv) Answer. $\sqrt{41}+5$
Solution. We have $\frac{16}{\sqrt{41}-5}$
Rationalising the denominator, we get:
$\frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41}-5 \right )\sqrt{41}+5}$
Using  the identity (a – b) (a + b) = a2 – b2
We get:
$= \frac{16\left ( \sqrt{41}+5 \right )}{\left ( \sqrt{41} \right )^{2}-\left ( 5 \right )^{2}}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{41-25}$
$= \frac{16\left ( \sqrt{41}+5 \right )}{16}$
$= \sqrt{41}+5$
Hence the answer is $\sqrt{41}+5$

(v) Answer.    $7+4\sqrt{3}$
Solution. We have, $\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Rationalising the denominator, we get:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2-\sqrt{3}}$
$= \frac{\left ( 2+\sqrt{3} \right )^{2}}{\left ( 2-\sqrt{3} \right )\left ( 2+\sqrt{3} \right )}$

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
$= \frac{2^{2}+\left ( \sqrt{3} \right )^{2}+2\cdot 2\cdot \sqrt{3}}{2^{2}-\left ( \sqrt{3} \right )^{2}}$
$= \frac{4+3+4\sqrt{3}}{4-3}$
$= 7+4\sqrt{3}$
Hence the answer is $7+4\sqrt{3}$

(vi)Answer.  $3\sqrt{2}-2\sqrt{3}$
Solution.   We have, $\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
Rationalising the denominator, we get:
$= \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$
$= \frac{\sqrt{6}\left ( \sqrt{3}-\sqrt{2} \right )}{\left ( \sqrt{2}+\sqrt{3} \right )\left ( \sqrt{3}-\sqrt{2} \right )}$
Using  the identity (a – b) (a + b) = a2 – b2
We get:
$= \frac{\sqrt{18}-\sqrt{12}}{\left ( \sqrt{3} \right )^{2}\left ( \sqrt{2} \right )^{2}}$
$= \frac{\sqrt{3\cdot 3\cdot 2}-\sqrt{2\cdot 2\cdot 3}}{3-2}$
$= \frac{3\sqrt{2}-2\sqrt{3}}{1}$
$= 3\sqrt{2}-2\sqrt{3}$
Hence the answer $3\sqrt{2}-2\sqrt{3}$

(vii) Answer. $5+2\sqrt{6}$
Solution. We have, $\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

Rationalising the denominator, we get:
$\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}$
$= \frac{\left ( \sqrt{3} +\sqrt{2}\right )^{2}}{\left ( \sqrt{3} -\sqrt{2} \right )\left ( \sqrt{3} +\sqrt{2} \right )}$

Using   (a – b) (a + b) = a2 – b2
and      (a + b)2 = a2 + b2 + 2ab
$= \frac{\left ( \sqrt{3} \right )^{2}+\left ( \sqrt{2} \right )^{2}+2\sqrt{3}\sqrt{2}}{\left ( \sqrt{3} \right )^{2}-\left ( \sqrt{2} \right )^{2}}$
$= \frac{3+2+2\sqrt{6}}{3-2}$
$= 5+2\sqrt{6}$
Hence the answer is $5+2\sqrt{6}$

(viii)

Answer: $9+2 \sqrt{15}$

Solution:

We have $\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$

Rationalize

$=\frac{3 \sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}} \times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{3 \sqrt{5}(\sqrt{5}+\sqrt{3})+\sqrt{3}(\sqrt{5}+\sqrt{3})}{(\sqrt{5})^{2}-(\sqrt{3})^{2}}$

$=\frac{15+3 \sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4 \sqrt{15}}{2}$

$=9+2 \sqrt{15}$

(ix) Answer: $\frac{9+4 \sqrt{6}}{15}$

Solution:

We have $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$

$=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{16 \times 3}+\sqrt{9 \times 2}}=\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$

Rationalize

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

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

$=\frac{48-12 \sqrt{6}+20 \sqrt{6}-30}{30}$

$=\frac{18+8 \sqrt{6}}{30}$

$=\frac{9+4 \sqrt{6}}{15}$

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#### (i) Simplify the following : $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$ (ii) Simplify the following : $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$ (iii) Simplify the following : $4\sqrt{12}\times 7\sqrt{6}$ (iv) Simplify the following : $4\sqrt{28}\div 3\sqrt{7}\div 3\sqrt{7}$ (v) Simplify the following : $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$ (vi) Simplify the following : $\left ( \sqrt{3}-\sqrt{2} \right )^{2}$ (vii) Simplify the following : $\sqrt[4]{81}- 8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{255}$ (viii) Simplify the following : $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$ (ix) Simplify the following : $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$

(i) Answer. $\sqrt{5}$
Solution.    $\sqrt{45}-3\sqrt{20}+4\sqrt{5}$
We know that,
45 = $3\times 3\times 5$
20 = $2\times 2\times 5$
So we get
$\sqrt{3\times3\times5 }-3\sqrt{2\times2\times5}+4\sqrt{5}$
$= 3\sqrt{5}-3\left ( 2\sqrt{5} \right )+4\sqrt{5}$
$= 3\sqrt{5}-6\sqrt{5}+4\sqrt{5}$
$= 7\sqrt{5}-6\sqrt{5}$
$= \sqrt{5}$
Hence the answer is $\sqrt{5}$

(ii)  Answer. $\frac{7\sqrt{6}}{12}$
Solution. We have, $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$
We know that,

$24= 6\times 4= 3\times 2\times 2\times 2$
$54= 9\times 6= 3\times 3\times 3\times 2$
So we get
$\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}= \frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}$
$= \frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}$
Taking LCM (3,4) = 12

$= \frac{3\sqrt{6}+4\sqrt{6}}{12}$
$= \frac{7\sqrt{6}}{12}$

(iii) Answer.  $\sqrt[28]{2^{18} \times 3^{11}}$
Solution.   We have
$\sqrt[4]{12}\times \sqrt[7]{6}$
We know that
12 = $2\times 2\times 3$
6 = $2\times 3$
So we get,
=$\sqrt[4]{2\times 2\times 3}\times \sqrt[7]{2\times 3}$
$=2^{1 / 4} \cdot 2^{1 / 4} \cdot 3^{1 / 4} \cdot 2^{1 / 7} \cdot 3^{1 / 7}$
$=2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}} \times 3^{\frac{1}{4}+\frac{1}{7}}$

$=2^{9 / 14} \times 3^{11 / 28}$

=$\sqrt[28]{2^{18} \times 3^{11}}$
Hence the number is $\sqrt[28]{2^{18} \times 3^{11}}$.

(iv)Answer.    $\frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$
Solution.   We have, $4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}$
We know that
28 = $4\times 7$
So we can write,
$4\sqrt{28}\div 3\sqrt{7}\div \sqrt[3]{7}= \left [ \frac{4\sqrt{28}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$
= $\left [ \frac{4\sqrt{4\times 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$

= $\left [ \frac{4\times 2\sqrt{ 7}}{3\sqrt{7}} \right ]\div 7^{\frac{1}{3}}$

$= \frac{8}{3}\div 7^{\frac{1}{3}}$
$= \frac{8}{\left ( 3\times 7^{\frac{1}{3}} \right )}$
$= \frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$
Hence the answer is $\frac{8}{\left ( 3\times \sqrt[3]{7} \right )}$

(v) Answer.  $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$
We know that
27 = $3\times 3\times 3$
So, $3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}$ = $3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}$

$= 3\sqrt{3}+2\left ( 3\sqrt{3} \right )+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}$
(Rationalising the denominator)

$= 3\sqrt{3}+6\left ( \sqrt{3} \right )+\frac{7\sqrt{3}}{3}$

$= \left ( 3+6+\frac{7}{3} \right )\sqrt{3}$         (Taking $\sqrt{3}$ common)
Now LCM (1,1,3) = 3
$= \left ( \frac{9+18+7}{3} \right )\sqrt{3}$
$= \frac{34}{3}\sqrt{3}$
$= 19\cdot 63$

(vi) Answer. $5-2\sqrt{6}$
Solution. Given, $\left ( \sqrt{3}-\sqrt{2} \right )^{2}$
We know that (a + b)2 = a2 – 2ab + b2
Comparing the given equation with the identity, we get:

$\left ( \sqrt{3}-\sqrt{2} \right )^{2}= \left ( \sqrt{3} \right )^{2}-2\left ( \sqrt{3} \right )\left ( \sqrt{2} \right )+\left ( \sqrt{2} \right )^2$
= 3 + 2 – $2\sqrt{3\times 2}$
$= 5-2\sqrt{6}$
Hence the answer is $5-2\sqrt{6}$

Solution. We have, $\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
We know that
81 = $3\times 3\times3\times3$
216 = $6\times 6\times6$
32 = $2\times 2\times2\times2\times2$
225 = $15\times 15$
So,$\sqrt[4]{81}-8 \sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225}$
$= \sqrt[4]{3\times3\times3\times3 }-8\sqrt[3]{6\times6\times6}+15\sqrt[5]{2\times2\times2\times2\times2}+\sqrt{15\times15}$
= 3 – 8 × 6 + 15 × 2 + 15
= 3 – 48 + 30 + 15
= – 45 + 45
= 0

(viii) Answer.   $\frac{5}{2\sqrt{2}}$

Solution.   We have, $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$
We know that, 8 =$2\times 2\times 2$
So,

$\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$ $= \frac{3}{\sqrt{2\times 2\times 2}}+\frac{1}{\sqrt{2}}$

$= \frac{3}{2\sqrt{2}}+\frac{1}{\sqrt{2}}$

$= \frac{3}{2\sqrt{2}}+\frac{2}{2\sqrt{2}}$
$= \frac{5}{2\sqrt{2}}$
Hence the answer is $\frac{5}{2\sqrt{2}}$

(ix) Answer.     $\frac{\sqrt{3}}{2}$
Solution.     We have, $\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}$
LCM (3,6) = 6

$\frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}= \frac{4\sqrt{3}}{6}-\frac{\sqrt{3}}{6}$

$= \frac{4\sqrt{3}-\sqrt{3}}{6}$

$= \frac{3\sqrt{3}}{6}$
$= \frac{\sqrt{3}}{2}$
Hence the answer is $\frac{\sqrt{3}}{2}$.

#### (i) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$. 0.2 (ii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.0.888……. (iii)  Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.$5.\bar{2}$    (iv) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.$0\cdot \overline{001}$ (v) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$.0.2555…… (vii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$..00323232….. (viii) Express the following in the form $\frac{p}{q}$, where p and q are integers and $q\neq 0$..404040……..

(i) Answer.     $\frac{1}{5}$
Solution.   We know that
0.2 can be written as $\frac{2}{10}$
Now,
$\frac{2}{10}= \frac{1}{5}$
Hence the answer is $\frac{1}{5}$

(ii) Answer.    $\frac{8}{9}$
Solution.    Let x = 0.888…..        .…(i)
Multiply RHS and LHS by 10
10 x = 8.88…….         …(ii)
Subtracting equation (i) from (ii)
We get
10x – x = 8.8 – 0.8
$\Rightarrow 9x= 8$
$\Rightarrow x= \frac{8}{9}$
Hence answer is  $\frac{8}{9}$

(iii) Answer.  $\frac{47}{9}$
Solution.  Let x = $5\cdot \bar{2}$          …eq. (1)
Multiply by 10 on both sides
10x = $52\cdot \bar{2}$                  …eq (2)

Subtracting equation (1) from (2)
We get
10x – x = $52\cdot \bar{2}$ – $5\cdot \bar{2}$
$\Rightarrow$ 9x = 47
$\Rightarrow$ x = $\frac{47}{9}$
Hence the answer is $\frac{47}{9}$

(iv) Answer.   $\frac{1}{999}$
Solution.    Let x = $0\cdot \overline{001}$              …. Eq. (1)
Multiply by 1000 on both sides
1000 x = $1\cdot \overline{001}$            …eq.(2)
Subtracting eq. (1) from (2)
We get
1000 x – x = $1\cdot \overline{001}$ – $1\cdot \overline{001}$

$\Rightarrow$ 999x = 1
$\Rightarrow$ x = $\frac{1}{999}$
Hence the answer is $\frac{1}{999}$

(v) Answer.   $\frac{23}{90}$

Solution. Let x = 0.2555 …..    …eq.(1)
Multiply by 10 on both sides
10x = 2.555…                         …eq.(2)
Multiply by 100 on both sides
100x = 25.55…                       …eq. (3)
Subtracting eq. (2) from (3),
We get
100x – 10x = 25.555… – 2.555…

$\Rightarrow$ 90x = 23
$\Rightarrow$ x = $\frac{23}{90}$
Hence the answer is $\frac{23}{90}$

(vii) Answer.    $\frac{8}{2475}$
Solution.   Let    x = 0.00323232…..      …eq.(1)
Multiply with 100 on both sides.
We get
100x = 0.3232…                …eq(2)
Multiply again with 100 on both sides,
10000x = 32.3232…                           …eq(3)
Equation (3) – (2)
we get,
10000 x – 100x = 32.3232… – 0.3232…
$\Rightarrow$  9900x = 32
$\Rightarrow$ x = $\frac{32}{9900}$
x = $\frac{8}{2475}$
Hence the answer is  $\frac{8}{2475}$

(viii) Answer.     $\frac{40}{99}$
Solution. Let x = 0.404040…….           …(1)
Multiplying by 100 on both sides
we get
100x = 40.40…                       …(2)
Subtracting equation (1) from (2)
we get
100x – x = 40.40 – 0.40
$\Rightarrow$99x = 40
$\Rightarrow$  x = $\frac{40}{99}$
Hence the answer is $\frac{40}{99}$

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#### (i) Represent geometrically the following numbers on the number line : $\sqrt{4\cdot 5}$ (ii) Represent geometrically the following numbers on the number line : $\sqrt{5\cdot 6}$ (iii) Presentation of $\sqrt{8\cdot 1}$ on number line : (iv) Presentation of $\sqrt{2\cdot 3}$ on number line:

(i) Solution. AB = 4.5 units, BC = 1 unit

OC = OD = $\frac{5\cdot 5}{2}$ = 2.75 units
OD2 = OB2 + BD2
$\left ( \frac{4\cdot 5}{2} \right )^{2}= \left ( \frac{4\cdot 5}{2} -1\right )^{2}+\left ( BD \right )^{2}$
$\Rightarrow BD^{2}= \left ( \frac{4\cdot 5+1}{2} \right )^{2}- \left ( \frac{4\cdot 5-1}{2} \right )^{2}$
$\Rightarrow BD^{2}= 4. 5$
$\Rightarrow BD= \sqrt{4. 5}$

So the length of BD will be the required one so mark an arc of length BD on number line, this will result in the required length.

(ii) Solution.  Presentation of $\sqrt{5. 6}$ on number line.
Mark the distance 5.6 units from a fixed point A on a given line to obtain a point B such that AB = 5.6 units. From B mark a distance of 1 unit and mark a new point C. Find the mid point of AC and mark that point as O. Draw a semicircle with center O and radius OC. Draw a line
perpendicular to AC passing through B and intersecting the semicircle at O. Then BD = $\sqrt{5\cdot 6}$

(iii) Solution
Mark the distance 8.1 units from a fixed point A on a given line to obtain a point B such that AB = 8.1 units. From B mark a distance of 1 unit and mark the new point AB. Find the mid point of AC and mark a point as O. Draw a semi circle with point O and radius OC. Draw a line perpendicular to AC passing through B and intersecting
the semicircle at D. Then BD -
$\sqrt{8. 1}$

(iv) Solution
Mark the distance 2.3 unit from a fixed point A on a given line. To obtain a point B such that AB = 2.3 units. From B mark a distance of 1 unit and mark a new point as C. Find the mid point of AC and mark the point asO. Draw a line perpendicular to AC passing through B and intersecting the semicircle at D. Then BD = $\sqrt{2. 3}$

#### (i)Insert a rational number and an irrational number between the following. 2 and 3 (ii)Insert a rational number and an irrational number between the following.0.1 and 0.1 (iii)Insert a rational number and an irrational number between the following.$\frac{1}{3}\, and\, \frac{1}{2}$ (iv)Insert a rational number and an irrational number between the following.$\frac{-2}{5}\, and\, \frac{1}{2}$ (v)Insert a rational number and an irrational number between the following.0.15 and 0.16 (vi)Insert a rational number and an irrational number between the following.$\sqrt{2}$ and $\sqrt{3}$ (vii)Insert a rational number and an irrational number between the following.2.357 and 3.121 (viii)Insert a rational number and an irrational number between the following..0001 and .001 (ix)Insert a rational number and an irrational number between the following.3.623623 and 0.484848 (x) Insert a rational number and an irrational number between the following.6.375289 and 6.375738

(i) Answer.  Rational number: $\frac{5}{2}$

Irrational number: 2.040040004 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.

Between 2 and 3
Rational number: 2.5 = $\frac{25}{10}$$= \frac{5}{2}$

and irrational number : 2.040040004

(ii) Answer. Rational number: $\frac{19}{1000}$

Irrational number 0.0105000500005 ……..
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0 and 0.1:
0.1 can be written as 0.10
Rational number: 0.019 = $\frac{19}{1000}$
and irrational number 0.0105000500005

(iii)Answer. Rational number $\frac{21}{60}$
Irrational number : 0.414114111 ……
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions. They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
LCM of 3 and 2 is 6.
We can write $\frac{1}{3}$ as $\frac{1\times 20}{3\times 20}= \frac{20}{60}$

and $\frac{1}{2}$ as $\frac{1\times 30}{3\times 30}= \frac{30}{60}$
Also, $\frac{1}{3}$ = 0.333333….
And  $\frac{1}{2}= 0\cdot 5$
So, rational number between $\frac{1}{3}$  and $\frac{1}{2}$ is $\frac{21}{60}$
and irrational number : 0.414114111 ……

Irrational number: 0.151551555 …….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.

$\frac{-2}{5}= -0\cdot 4$ and $\frac{1}{2}= -0\cdot 5$
Rational number between -0.4 and 0.5 is 0
And irrational number: 0.151551555 …….

(v) Answer. Rational number: $\frac{151}{1000}$
Irrational number: 0.151551555 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0.15 and 0.16
Rational number : 0.151 = $\frac{151}{1000}$
and irrational number 0.151551555

(vi) Answer.  Rational number: $\frac{3}{2}$
Irrational number: 1.585585558 ………
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions. They are non-terminating non-recurring in nature.
Between $\sqrt{2}\, and\, \sqrt{3}$

$\sqrt{2}= 1\cdot 414213562373$
$\sqrt{3}= 1\cdot 732050807568$
Rational number: 1.5 = $\frac{3}{2}$
and irrational number: 1.585585558

Irrational number: 3.101101110………
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 2.357 and 3.121
Rational number: 3
Irrational number: 3.101101110………
(viii) Answer.  Rational number: $\frac{2}{10000}$
Irrational number: 0.000113133133 ……….
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 0.0001 and 0.001
Rational number: 0.0002 = $\frac{2}{10000}$
Irrational number: 0.000113133133

Irrational number: 1.909009000 ……
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 3.623623 and 0.484848
A rational number between 3.623623 and 0.484848 is 1.
An irrational number between 3.623623 and 0.484848 is 1.909009000 ……

(x) Answer. A rational number is $\frac{63753}{10000}$
An irrational number is 6.375414114111……..
Solution.
Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. Also, both p and q should be rational when the fraction is expressed in the simplest form.
Irrational numbers are real numbers which cannot be represented as simple fractions.
They are non-terminating non-recurring in nature. It means they keep on continuing after decimal point and do not have a particular pattern/sequence after the decimal point.
Between 6.375289 and 6.375738:
A rational number is 6.3753 = $\frac{63753}{10000}$
An irrational number is 6.375414114111……..

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#### Represent the following numbers on the number line. 7, 7.2, $-\frac{3}{2},-\frac{12}{5}$

Solution.
Firstly we draw a number line whose mid-point is O. Mark positive numbers on right hand side of O and negative numbers on left hand side of O.

(i) Number 7 is a positive number. So we mark a number 7 on the right
hand side of O, which is at a 7 units distance form zero.

(ii) Number 7.2 is a positive number. So, we mark a number 7.2 on the right hand side of O, which is 7.2 units distance from zero.
(iii) Number $-\frac{3}{2}$ or – 1.5 is a negative number so, we mark a number 1.5 on the left hand side of zero, which is at 1.5 units distance from zero.

(iv) Number $-\frac{12}{5}$ or –2.4 is a negative number. So, we mark a number 2.4 on the left hand side of zero, which is at 2.4 units distance from zero.

#### If a = 2 + $\sqrt{3}$ then find the value of $a-\frac{1}{a}$

Answer.          $2\sqrt{3}$
Solution.   Given that a = $2\sqrt{3}$

$\therefore$ We have $\frac{1}{a}= \frac{1}{2+\sqrt{3}}$
Rationalising,
$\Rightarrow \frac{1}{a}= \frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$
Using   (a – b) (a + b) = a2 – b2

$\Rightarrow \frac{1}{a}= \frac{2-\sqrt{3}}{2^{2}-\sqrt{3}^{2}}= \frac{2-\sqrt{3}}{4-3}$
$\Rightarrow \frac{1}{a}=2-\sqrt{3}$
Now, $a- \frac{1}{a}=2+\sqrt{3}-\left ( 2-\sqrt{3} \right )$

$\Rightarrow a- \frac{1}{a}=2\sqrt{3}$

Hence the answer is $2\sqrt{3}$