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#### A massless string connects two pulley of masses ' $2 \mathrm{~kg}$' and '$1 \mathrm{~kg}$' respectively as shown in the figure.The heavier pulley is fixed and free to rotate about its central axis while the other is free to rotate as well as translate. Find the acceleration of the lower pulley if the system was released from the rest. [Given, $g=10 \mathrm{~m} / \mathrm{s}^2$]Option: 1 $\frac{4}{3} \mathrm{~gm} / \mathrm{s}^2$Option: 2 $\frac{3}{2} \mathrm{~gm} / \mathrm{s}^2$Option: 3 $\frac{3}{4} \mathrm{~gm} / \mathrm{s}^2$Option: 4 $\frac{2}{3} \mathrm{~gm} / \mathrm{s}^2$

Not understanding sir

#### Two particles of mass 5kg and 10kg respectively are attached to the two ends of a rigid rod of length 1m with negligible mass. The centre of mass of the system from the 5kg particle is nearly at a distance of: Option: 1  33 cm   Option: 2 50 cm Option: 3 67 cm Option: 4 80 cm

$\text { The position of } COM \text { from } m _{1} \text { mass is } \frac{ m _{2} r }{\left( m _{1}+ m _{2}\right)}$

Here, m1 = 5  kg

r = 1 m

m2 = 10 kg

By putting this value we will get 0.666667 meter

or we can write = 67 cm

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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 $= \sqrt{1+8+27}$
$=\sqrt{36}= 6$
Hence the answer is 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
Hence the answer is 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
Hence the answer is 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 $\sqrt[3]{\frac{1}{3}}$

(vi) Answer. – 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
$\because \frac{\left ( a \right )^{m}}{\left ( a \right )^{n}}= \left ( a \right )^{m-n}$
$= 2^{\frac{3+4+5}{3}}= 2^{\frac{12}{3}}$
$= 2^{4}= 16$

Hence the answer is 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
Hence the answer is 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 :
$= \sqrt{2}\cdot \sqrt{3}$
$= 1\cdot 414\times 1\cdot 732= 2\cdot 449$
Hence the answer is 2.449

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
Hence the answer is 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
Hence the answer is 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
Hence the answer is 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 : $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}= a+\frac{7}{11}\sqrt{5b}$

(i) Answer.   a = 11
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
$= \left ( \frac{9+18+7}{3} \right )\sqrt{3}$
$= \frac{34}{3}\sqrt{3}$
$= 19\cdot 63$

Hence the answer is 19.63

(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
Hence the answer is 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 ……

(iv)Answer. Rational number: 0
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

(vii) Answer.  Rational number: 3
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

(ix) Answer. Rational number: 1
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……..