NCERT Solutions for Class 9 Maths Chapter 1 - Number Systems

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Answer:

Whole numbers are the numbers from 0, 1, 2, 3...∞ . 
Natural numbers start from 1, 2, 3, 4, 5, 6... ∞. 
Integers contains all numbers from $-$∞ to $+$∞. 
Rational numbers are all number which can be written in the form of $\frac{p}{q}$ where p and q are integers.
Real numbers are all number which can or cannot be written in the form of $\frac{p}{q}$.
So, we can conclude that all rational numbers are real numbers .

Hence, the correct answer is option (C). 
 

Answer:

Between 2 rational numbers, there are infinitely many rational numbers.

Hence, the correct answer is option (C).

Answer:

Decimal representation of a rational number cannot be non-terminating non-repeating. 
 

Answer:

When it comes to a product of two irrational numbers, there can be two possible cases which are mentioned as follows:-
1. $\sqrt{a}\times \sqrt{a}=a$ , where $a$ is a rational number
    for example: $\sqrt{2}\times \sqrt{2}$ $=2$ 
2. $\sqrt{a }\times \sqrt{b}=\sqrt{ab},$ where $a$ and $b$ are rational numbers
     for example: $\sqrt{3}\times \sqrt{2}=\sqrt{6}$ and $\sqrt{6}$ is an irrational number 
     
Hence, the correct answer is option (D).

Answer:

Since, $\sqrt{2}$ is an irrational number, therefore its decimal expansion will be non-terminating non-recurring.
Hence, the correct answer is option (D)

Answer:

$\sqrt{\frac{4}{9}}=\frac{2}{3}$; which is a rational number.
$\sqrt{\frac{12}{3}}=\sqrt{4}=2$; which is a rational number.
$\sqrt{81}=9$ which is a rational number.
$\sqrt{7}$ is a non-terminating, non-repeating decimal expansion.

So, $\sqrt{7}$ is an irrational number.

Hence, the correct answer is option (C).

Answer:

0.4014001400014... is an irrational number because its decimal expansion is non-terminating and non-repeating.
Hence, the correct answer is option (D).

Answer:

Value of $\sqrt{2}$ = 1.414...
Value of $\sqrt{3}$ = 1.732...
$\frac{\sqrt{2}+\sqrt{3}}{2}$ is an irrational number.
$\frac{\sqrt{2}·\sqrt{3}}{2}=\frac{\sqrt{6}}{2}$ is an irrational number.
1.5 is a rational number and lies between 1.414... and 1.732..

Hence, the correct answer is option (C).

Answer:

Let x = 1.999...  (1)
Then, 10x = 19.999...  (2)
Subtracting (1) from (2), we get
10x – x = (19.999...) – (1.999...)
⇒ 9x = 18
⇒ $x=\frac{18}{9}=2$

Hence, the correct answer is option (C).

Answer:

$$\begin{gathered} 2\sqrt{3}+\sqrt{3}=\sqrt{3}(2+1) \\ =3\sqrt{3} \end{gathered}$$
Hence, the correct answer is option C.

Answer:

$$\begin{gathered} \sqrt{10}\times \sqrt{15}=\sqrt{(2\times 5)}\times \sqrt{(5\times 3)} \\ =\sqrt{2}\times \sqrt{5}\times \sqrt{5}\times \sqrt{3} \\ =5\sqrt{6} \end{gathered}$$

Hence, the correct answer is option B.

Answer:

$\frac{1}{\sqrt{7}-2}$ 
$$\begin{gathered} =\frac{1}{\sqrt{7}-2}\times \frac{\sqrt{7}+2}{\sqrt{7}+2} \\ =\frac{\sqrt{7}+2}{{\left(\sqrt{7}\right)}^2-2^2} \\ =\frac{\sqrt{7}+2}{7-4} \\ =\frac{\sqrt{7}+2}{3} \end{gathered}$$

Hence, the correct answer is option A.

Answer:

$\frac{1}{\sqrt{9}-\sqrt{8}}=\frac{1}{3-2\sqrt{2}}$
By Rationalizing the denominator, we get
$\begin{array}{rcl}\frac{1}{3-2\sqrt{2}}\times \left(\frac{3+2\sqrt{2}}{3+2\sqrt{2}}\right)& =& \frac{3+2\sqrt{2}}{3^2-{\left(2\sqrt{2}\right)}^2} \left[\because a^2-b^2=(a-b)(a+b)\right]\\ & =& \frac{3+2\sqrt{2}}{9-8}\\ & =& 3+2\sqrt{2}\end{array}$

Hence, the correct answer is option D.

Answer:

$$\begin{gathered} \frac{7}{3\sqrt{3}-2\sqrt{2}} \\ =\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \left(\frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}\right) \\ =\frac{7\left(3\sqrt{3}+2\sqrt{2}\right)}{{\left(3\sqrt{3}\right)}^2-{\left(2\sqrt{2}\right)}^2} \left[\because \left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ =\frac{7\left(3\sqrt{3}+2\sqrt{2}\right)}{27-8} \\ =\frac{7\left(3\sqrt{3}+2\sqrt{2}\right)}{19} \end{gathered}$$
Thus, we get 19 as the denominator, after rationalization of denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}}$.

Hence, the correct answer is option B.

Answer:

$$\begin{gathered} \frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}} \\ =\frac{\sqrt{16\times 2}+\sqrt{16\times 3}}{\sqrt{4\times 2}+\sqrt{3\times 4}} \\ =\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}} \\ =\frac{4\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)} \\ =2 \end{gathered}$$

Hence, the correct answer is option B.

Answer:

$\sqrt{\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)}$
$$\begin{gathered} =\sqrt{\frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}+1\right)}\times \frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)}} \\ =\sqrt{\frac{{\left(\sqrt{2}-1\right)}^2}{{\left(\sqrt{2}\right)}^2-1^2}} \\ =\sqrt{\frac{{\left(\sqrt{2}-1\right)}^2}{1}} \\ =\sqrt{2}-1 \end{gathered}$$
= 1.4142... – 1
= 0.4142 ...

Hence, the correct answer is option C.

Answer:

$\begin{array}{rcl}\sqrt[4]{\sqrt[3]{2^2}}& =& {\left[\sqrt[3]{2^2}\right]}^{\frac{1}{4}} \left[\because \sqrt[n]{a}=a^{\frac{1}{n}}\right]\\ & =& {\left[{\left[2^2\right]}^{\frac{1}{3}}\right]}^{\frac{1}{4}}\\ & =& 2^{\frac{2}{3}\times \frac{1}{4} \left[\because {\left(a^m\right)}^n= a^{mn}\right]}\\ & =& 2^{\frac{1}{6}}\end{array}$

Hence, the correct answer is option C.

Answer:

LCM of 3, 4 and 12 = 12
and $\sqrt[12]{32}=\sqrt[12]{2^5}$
Thus, the product of
$\begin{array}{rcl}\sqrt[3]{2}·\sqrt[4]{2}·\sqrt[12]{32}& =& \sqrt[12]{2^4·2^3·2^5}\\ & =& \sqrt[12]{2^{12}}\\ & =& 2^{12\times \frac{1}{12}} \left[\because \sqrt[m]{a^n}=a^{\frac{n}{m}}\right]\\ & =& 2\end{array}$

Hence, the correct answer is option B.

Answer:

$\begin{array}{rcl}\sqrt[4]{{\left(81\right)}^{-2}}& =& \sqrt[4]{\frac{1}{{\left(81\right)}^2}}\\ & =& \sqrt[4]{\frac{1}{9^4}}\\ & =& \frac{1}{9^{{\displaystyle \frac{4}{4}}}} \left[\because \sqrt[n]{a}=a^{\frac{1}{n}}\right]\\ & =& \frac{1}{9}\end{array}$

Hence, the correct answer is option A.

Answer:

(256)^0.16 × (256)^0.09
$$\begin{gathered} \mathit{=}\mathit{ }{\left(256\right)}^{0.16+0.09} \\ ={\left(256\right)}^{0.25} \\ = {\left(256\right)}^{\frac{25}{100}} \\ ={\left(256\right)}^{\frac{1}{4}} \\ \mathrm{Since}, 256={\left(2\right)}^8 \end{gathered}$$

$$\begin{gathered} \mathrm{Using} {\mathit{\left(}{a}^m\mathit{\right)}}^n\mathit{=}\mathit{ }{\mathit{\left(}a\mathit{\right)}}^{m\mathit{\times }n} \\ {=\left(2^8\right)}^{\frac{1}{4}} \\ = {\left(2\right)}^{8\times \frac{1}{4}} \\ ={\left(2\right)}^2 \\ =2\times 2 \\ = 4 \\ \mathrm{Hence}, \mathrm{the} \mathrm{correct} \mathrm{answer} \mathrm{is} \mathrm{option} \mathrm{A}. \end{gathered}$$

Answer:

$\begin{array}{rcl}{x}^{\frac{12}{7}}-x^{\frac{5}{7}}& =& x^{\frac{5}{7}+1}-x^{\frac{5}{7}}\\ & =& x^{\frac{5}{7}}x-x^{\frac{5}{7}} \left[\because a^{mn}=a^m-a^n\right]\\ & =& x^{\frac{5}{7}}(x-1)\ne x\\ \sqrt[12]{{\left(x^4\right)}^{\frac{1}{3}}}& =& {\left({\left(x^4\right)}^{\frac{1}{3}}\right)}^{\frac{1}{12}}\\ & =& x^{4\times \frac{1}{3}\times \frac{1}{12}}\\ & =& x^{\frac{1}{9}}\ne x\end{array}$

$\begin{array}{rcl}{\left(\sqrt{x^3}\right)}^{\frac{2}{3}}& =& {\left(x^3\right)}^{\frac{1}{2}\times \frac{2}{3}}\\ & =& {\left(x^3\right)}^{\frac{1}{3}}\\ & =& x^{3\times \frac{1}{3}}\\ & =& x^1\\ & =& x\end{array}$

$\begin{array}{rcl}{x}^{\frac{12}{7}}\times x^{\frac{7}{12}}& =& x^{\left(\frac{12}{7}+\frac{7}{12}\right)}\\ & =& x^{\frac{193}{84}}\\ & \ne & x\end{array}$

Hence, the correct answer is option C.

Answer:

Since it is given that x and y are a rational and irrational number.
Let, $x=4 \mathrm{and} y=\sqrt{2}$,

Answer:

No, xy can only be irrational if x ≠ a.
Let x be a non zero rational y be an irrational. Then for xy to be irrational. Assume that xy be a rational number. Now, quotient of two non-zero rational number is a rational number.
So, $\frac{xy}{x}$is a rational number y is a rational number.
But this contradicts the fact that y is an irrational number.
Thus, our assumption is wrong.
Hence, xy is an irrational number.
Exception occurs when x = 0, then xy = 0, i.e. rational number.

Answer:

(i) False, $\sqrt{2}$ is a an irrational number and 3 is a rational number. When we divide irrational number by non-zero rational number it will always give irrational number.
(ii) False, because between 2 consecutive integers there doesn't exists any other integer.
(iii) False, because between any 2 rational numbers there exist infinitely many rational numbers.
(iv) True, because there are infinitely many numbers which cannot be written in the form $\frac{p}{q}$, q ≠ 0, p, q both are integers and these numbers are called irrational numbers.
(v) False, say an irrational number be $\sqrt{3}$ and $\sqrt[4]{3}$
(a) ${\left(\sqrt{3}\right)}^2=3$, which is a rational number
(b) ${\left(\sqrt[4]{3}\right)}^2=\sqrt{3}$, which is not a rational number
Hence, square of an irrational number is not always a rational number
(vi) False, $\frac{\sqrt{12}}{\sqrt{3}}=\frac{\sqrt{4\times 3}}{\sqrt{3}}=\sqrt{4}=2$, which is rational number.
(vii) False, $\frac{\sqrt{15}}{\sqrt{3}}=\sqrt{\frac{15}{3}}=\sqrt{5}$, which is an irrational number.

Answer:

(i) $\sqrt{196}=\sqrt{{14}^2}=14$
Hence, it is a rational number.
(ii) $3\sqrt{18}=3\sqrt{3^2\times 2}=3\times 3\sqrt{2}=9\sqrt{2}$
Hence, it is an irrational number.
(iii) $\sqrt{\frac{9}{27}}=\sqrt{\frac{3\times 3}{3\times 3\times 3}}=\sqrt{\frac{1}{3}}=\frac{1}{\sqrt{3}}$
Hence, it is an irrational number.
(iv) $\frac{\sqrt{28}}{\sqrt{343}}=\sqrt{\frac{2\times 2\times 7}{7\times 7\times 7}}=\sqrt{\frac{2^2}{7^2}}=\frac{2}{7}$
Hence, it is a rational number.
(v) $-\sqrt{0.4}=-\sqrt{\frac{4}{10}}=\frac{-2}{\sqrt{10}}$
Hence, it is an irrational number.
(vi) $\frac{\sqrt{12}}{\sqrt{75}}=\frac{\sqrt{4\times 3}}{\sqrt{25\times 3}}=\sqrt{\frac{4}{25}}=\frac{2}{5}$
Hence, it is a rational number.
(vii) 0.5918, it is a number with terminating decimal. Thus, it can be represented in $\frac{p}{q}$ form
$0.5918=\frac{5918}{10000}$
Hence, it is a rational number.
(viii) $(1+\sqrt{5})-(4+\sqrt{5})=1-4+\sqrt{5}-\sqrt{5}$ = −3 which is a rational number.
(ix) 10.124124... is a number with non-terminating recurring decimal expansion.
Hence, it is a rational number.
(x) 1.010010001..., is a number with non-terminating non-recurring decimal expansion.
Hence, it is an irrational number.

Answer:

(i) x² = 5
Square root both sides, we get
$x=\pm \sqrt{5}$
Hence, the variable represents irrational number.
(ii) y² = 9
Square root both sides, we get
$y=\pm \sqrt{9}=\pm 3$
Hence, rational number
(iii) z² = 0.04
$\Rightarrow z^2=\frac{4}{100}$
Square root both sides, we get
$$\begin{gathered} z=\pm \sqrt{\frac{4}{100}}=\pm \frac{2}{10} \\ z=\pm \frac{1}{5} \end{gathered}$$
Hence, rational number

(iv) $u^2=\frac{17}{4}$
Square root both sides, we get
$u=\pm \sqrt{\frac{17}{4}}=\frac{\sqrt{17}}{2}$
Hence, irrational number.

Answer:

(i) Let x = –1 and y = –2
$\begin{array}{rcl}\mathrm{a} \mathrm{rational} \mathrm{number} \mathrm{between} x \& y & =& \frac{x+y}{2}\\ & =& \frac{-1+(-2)}{2}\\ & =& \frac{-3}{2}\end{array}$
$\begin{array}{rcl}\mathrm{rational} \mathrm{number} \mathrm{between} -1 \& \frac{-3}{2}& =& \frac{-1+\left({\displaystyle \frac{-3}{2}}\right)}{2}\\ & =& \frac{-2-3}{4}=\frac{-5}{4}\end{array}$

$\begin{array}{rcl}\mathrm{rational} \mathrm{number} \mathrm{between} \frac{-3}{2} \& -2& =& \frac{{\displaystyle \frac{-3}{2}}+(-2)}{2}\\ & =& \frac{-7}{4}\end{array}$

$\mathrm{Thus}, \mathrm{required} \mathrm{solution} \mathrm{is}\frac{-3}{2}, \frac{-5}{4}, \frac{-7}{4}.$

(ii) x = 0.1 and y = 0.11
$x=\frac{1}{10} \mathrm{and} y=\frac{11}{100}$
$\begin{array}{rcl}\mathrm{rational} \mathrm{number} \mathrm{between} \mathrm{them} & =& \frac{x+y}{2}\\ & =& \frac{{\displaystyle \frac{1}{10}}+{\displaystyle \frac{11}{100}}}{2}\\ & =& \frac{10+11}{200}\\ & =& \frac{21}{200}\end{array}$
$\begin{array}{rcl}\mathrm{rational} \mathrm{number} \mathrm{between} \frac{1}{10}\mathrm{and}\frac{21}{200}& =& \frac{{\displaystyle \frac{1}{10}}+{\displaystyle \frac{21}{200}}}{2}\\ & =& \frac{200+210}{4000}=\frac{41}{400}\end{array}$

$\begin{array}{rcl}\mathrm{rational} \mathrm{number} \mathrm{between} \frac{21}{200} \mathrm{and} \frac{11}{100}& =& \frac{{\displaystyle \frac{21}{200}}+{\displaystyle \frac{11}{100}}}{2}\\ & =& \frac{21+22}{400}\\ & =& \frac{43}{400}\end{array}$
$\mathrm{Hence}, \mathrm{required} \mathrm{solution} \mathrm{is}\frac{21}{200}, \frac{41}{400}, \frac{43}{400}$.

(iii) $x=\frac{5}{7} \mathrm{and} y=\frac{6}{7}$
Multiplying the numerator and denominator by 4.
$\frac{5\times 4}{7\times 4}=\frac{20}{28}$

$\frac{6\times 4}{7\times 4}=\frac{24}{28}$
Thus, the rational numbers between $\frac{5}{7} \mathrm{and} \frac{6}{7}$ are $\frac{21}{28},\frac{22}{28},\frac{23}{28}$.

Hence, the required numbers are $\frac{3}{4},\frac{11}{14},\frac{23}{28}$.

(iv) $x=\frac{1}{5} \mathrm{and} y=\frac{1}{4}$
Multiply x by 20 in numerator & denominator
$x=\frac{1}{5}\times \frac{20}{20}=\frac{20}{100}$

Multiply y by 25 in numerator & denominator

$y=\frac{1}{4}\times \frac{25}{25}=\frac{25}{100}$
Thus, 3 rational numbers between $\frac{20}{100} \mathrm{and} \frac{25}{100} \mathrm{are} \frac{21}{100}, \frac{22}{100}, \frac{23}{100}.$

Hence, the required numbers are $\frac{21}{100}, \frac{11}{50}, \frac{23}{100}$.

Answer:

(i) A rational number between 2 and 3 is 2.5. Irrational number is 2.050050005...
(ii) A rational number between 0 and 0.1 is 0.05. Irrational number is 0.056005600056...
(iii) A rational number between $\frac{1}{3}$ and $\frac{1}{2}$ is $\frac{5}{12}$ Irrational number is 0.4014001400014...
(iv) A rational number between $\frac{-2}{5}$ and $\frac{1}{2}$ is 0. Irrational number is 0.130130013... 
(v) A rational number between 0.15 and 0.16 is 0.155. Irrational number is 0.150150015...
(vi) A rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5. Irrational number is 1.00150015...
(vii) A rational number between 2.357 and 3.121 is 3. Irrational number is 2.707007...
(viii) A rational number between 0.0001 and 0.001 is 0.00014. Irrational number is 0.00010130013...
(ix) A rational number between 3.623673 and 0.484848 is 2. Irrational number is 1.90900...
(x) A rational number between 6.375289 and 6.375738 is 6.3755. Irrational number is 6.385321002100021...

Answer:

Answer:

(i) 5 = 2² +1²
In ΔABC
AB = 2 units, BC = 1 units
∠ABC = 90º

By Pythagoras theorem,
$\begin{array}{rcl}\mathrm{AC}& =& \sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2}\\ & =& \sqrt{2^2+1^2}\\ \mathrm{AC}& =& \sqrt{5}\end{array}$
Taking AC as radius we represent $\sqrt{5}$ on the number line with point P.
(ii) 10 = 3² + 1²
In ΔABC
AB = 3 units, BC = 1 unit
∠ABC = 90º

By Pythagoras theorem,
$\begin{array}{rcl}\mathrm{AC}& =& \sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2}\\ & =& \sqrt{3^2+1^2}\\ \mathrm{AC}& =& \sqrt{10}\end{array}$
Taking AC as radius we represent $\sqrt{10}$ on the number line with point P. 
(iii) 17 = 4² + 1
In ΔABC
AB = 4 units, BC = 1 unit
∠ABC = 90º

By Pythagoras theorem,
$\begin{array}{rcl}\mathrm{AC}& =& \sqrt{{\mathrm{AB}}^2+{\mathrm{BC}}^2}\\ & =& \sqrt{4^2+1^2}\\ \mathrm{AC}& =& \sqrt{17}\end{array}$
Taking AC as radius, we represent $\sqrt{17}$ on the number line with point P.

Answer:

Step 1: Draw line segment AB of length equal to the number inside the root and extend it to C such that BC = 1.
Step 2: Draw a semi circle with centre O. O being mid point of AC and radius is OA. 
Step 3: Draw a perpendicular line from B to cut the semi circle at D.
Step 4: Draw an arc with centre B and radius BD meeting AC produced at E.

(i)  

Point E represent $\sqrt{4.5}$ on number line.
(ii) 
Point E represent $\sqrt{5.6}$ on number line.
(iii) 
Point E represent $\sqrt{8.1}$ on number line.
(iv) 
Point E represent $\sqrt{2.3}$ on number line.

Answer:

(i) $0.2=\frac{2}{10}=\frac{1}{5}$

(ii) x = 0.888...      (1)
multiply by 10 on both sides
10x = 8.888...        (2)
Subtract (1) from (2)
9x = 8.888... – 0.888...
9x = 8
$\therefore x=\frac{8}{9}$

(iii) x = 5.222...     (1)
multiply by 10 on both sides
10x = 52.222...     (2)
Subtract (1) from (2)
9x = 52.222... – 5.222...
9x = 47
$x=\frac{47}{9}$

(iv) x = 0.001001...     (1)
multiply both sides by 1000
1000x = 1.001001...    (2)
Subtract (1) from (2)
$999x=1\Rightarrow x=\frac{1}{999}$

(v) x = 0.25555...    (1) 
multiply both sides by 10
10x = 2.555...         (2)
Again multiply both sides by 10
100x = 25.555...     (3)
Subtract (2) from (3)
100x – 10x = (25.555...) – (2.555...)
90x = 23
$\therefore x=\frac{23}{90}$

(vi) x = 0.1343434...     (1)
10x = 1.343434...            (2)
Again multiply both sides by 1000.
1000x = 134.3434...     (3)
Subtract (2) from (3)
1000x – 10x = (134.3434...) – (1.3434...)
990x = 133
$\therefore x=\frac{133}{990}$

(vii) x = 0.003232...     (1)
Multiply both sides by 100.
100x = 0.323232...       (2)
Again multiply both sides by 10000.
10000x = 32.3232..      (3)
Subtract (2) from (3)
10000x – 100x = (32.3232...) – (0.3232...)
9900x = 32
$\therefore x=\frac{32}{9900}=\frac{8}{2475}$

(viii) x = 0.404040...    (1)
Multiply both sides by 100.
100x = 40.404040...     (2)
Subtract (1) from (2)
100x – x = (40.4040...) – (0.4040...)
99x = 40
$\therefore x=\frac{40}{99}$

Answer:

Let x = 0.142857142857...

x = $0.\overline{142857}$ ......(1)

Multiplying equation (1) by 1000000

1000000 x = $142857.\overline{142857}$  .... (2)

Subtracting equation (1)  from equation (2)

10,00,000x $-$ x =   $142857.\overline{142857 }- 0.\overline{142857}$

999999x = 142857

x = $\frac{142857}{999999}$

x = $\frac{1}{7}$

Hence proved.
 

Answer:

 $\begin{array}{rcl}\left(\mathrm{i}\right) \sqrt{45}-3\sqrt{20}+4\sqrt{5}& =& \sqrt{3\times 3\times 5}-3\sqrt{2\times 2\times 5} +4\sqrt{5}\\ & =& 3\sqrt{5}-6\sqrt{5}+4\sqrt{5}\\ & =& 7\sqrt{5}-6\sqrt{5}\\ & =& \sqrt{5}\end{array}$

$\begin{array}{rcl}\left(\mathrm{ii}\right) \frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}& =& \frac{\sqrt{2\times 2\times 2\times 3}}{8}+\frac{\sqrt{3\times 3\times 3\times 2}}{9}\\ & =& \frac{2\sqrt{6}}{8}+\frac{3\sqrt{6}}{9}\\ & =& \frac{\sqrt{6}}{4}+\frac{\sqrt{6}}{3}\\ & =& \frac{3\sqrt{6}+4\sqrt{6}}{12}\\ & =& \frac{7\sqrt{6}}{12}\end{array}$

$\begin{array}{rcl}\left(\mathrm{iii}\right) \sqrt[4]{12}\times \sqrt[7]{6}& =& {\left(12\right)}^{\frac{1}{4}}\times {\left(6\right)}^{\frac{1}{7}} \left[\sqrt[n]{a}=a^{\frac{1}{n}}\right]\\ & =& {(2\times 2\times 3)}^{\frac{1}{4}}\times {(2\times 3)}^{\frac{1}{7}}\\ & =& 2^{\frac{1}{4}}. 2^{\frac{1}{4}} . 3^{\frac{1}{4}} . 2^{\frac{1}{7}}. 3^{\frac{1}{7}}\\ & =& 2^{\frac{1}{4}+\frac{1}{4}+\frac{1}{7}}. 3^{\frac{1}{4}+\frac{1}{7}} \left(x^m·x^n=x^{m+n}\right)\\ & =& 2^{\frac{9}{14}}· 3^{\frac{11}{28}}\\ & =& \sqrt[14]{2^9} \sqrt[28]{3^{11}} \\ & =& \sqrt[28]{2^{18}} \sqrt[28]{3^{11}}\\ & =& \sqrt[28]{2^{18}\times 3^{11}}\end{array}$

$$\begin{gathered} \left(\mathrm{iv}\right) 4\sqrt{28}÷3\sqrt{7}÷\sqrt[3]{7} \\ =4\sqrt{4\times 7}÷3\sqrt{7}÷\sqrt[3]{7} \\ =\left(\frac{8\sqrt{7}}{3\sqrt{7}}\right)÷7^{\frac{1}{3}} \\ =\left(\frac{8}{3}\right)÷\sqrt[3]{7} \\ =\left(\frac{8}{3\sqrt[3]{7}}\right) \end{gathered}$$

$\begin{array}{rcl}\left(\mathrm{v}\right) 3\sqrt{3}+2\sqrt{27}+\frac{7}{\sqrt{3}}& =& 3\sqrt{3}+2\sqrt{3\times 3\times 3}+\frac{7}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ & =& 3\sqrt{3}+6\sqrt{3}+\frac{7\sqrt{3}}{3}\\ & =& 9\sqrt{3}+\frac{7\sqrt{3}}{3}\\ & =& \frac{27\sqrt{3}+7\sqrt{3}}{3}\\ & =& \frac{34\sqrt{3}}{3}\end{array}$

$\begin{array}{rcl}\left(\mathrm{vi}\right) {\left(\sqrt{3}-\sqrt{2}\right)}^2& =& {\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2-2\sqrt{3}\times \sqrt{2} \left[{(a-b)}^2=a^2+b^2-2ab\right]\\ & =& 3+2-2\sqrt{6}\\ & =& 5-2\sqrt{6}\end{array}$

$$\begin{gathered} \left(\mathrm{vii}\right) \sqrt[4]{81}-8\sqrt[3]{216}+15\sqrt[5]{32}+\sqrt{225} \\ ={\left(81\right)}^{\frac{1}{4}}-8\times {\left(216\right)}^{\frac{1}{3}}+15\times {\left(32\right)}^{\frac{1}{5}}+\sqrt{{\left(15\right)}^2} \\ ={\left(3^4\right)}^{\frac{1}{4}}-8\times {\left(6^3\right)}^{\frac{1}{3}}+15\times {\left(2^5\right)}^{\frac{1}{5}}+15 \\ =3-8\times 6+15\times 2+15 \\ =3-48+30+15 \\ =48-48 \\ =0 \end{gathered}$$
$\begin{array}{rcl}\left(\mathrm{viii}\right) \frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}& =& \frac{3}{\sqrt{2\times 2\times 2}}+\frac{1}{\sqrt{2}}=\frac{3}{2\sqrt{2}}+\frac{1}{\sqrt{2}}\\ & =& \frac{3+2}{2\sqrt{2}}=\frac{5}{2\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}} [\mathrm{Rationalizing} \mathrm{denominators}]\\ & =& \frac{5\sqrt{2}}{2\times 2}=\frac{5\sqrt{2}}{4}\end{array}$

$\begin{array}{rcl}\left(\mathrm{ix}\right) \frac{2\sqrt{3}}{3}-\frac{\sqrt{3}}{6}& =& \frac{4\sqrt{3}-\sqrt{3}}{6}\\ & =& \frac{3\sqrt{3}}{6}\\ & =& \frac{\sqrt{3}}{2}\end{array}$
 

Answer:

(i) For rationalization multiply numerator and denominator by $\sqrt{3}.$
$\begin{array}{rcl}& \Rightarrow & \frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{2\sqrt{3}}{3\times 3}\\ & =& \frac{2\sqrt{3}}{9}\end{array}$
(ii) Multiply numerator and denominator by $\sqrt{3}.$
$\begin{array}{rcl}& \Rightarrow & \frac{\sqrt{40}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{40\times 3}}{{\left(\sqrt{3}\right)}^2}\\ & =& \frac{\sqrt{120}}{3}=\frac{2}{3}\sqrt{30}\end{array}$
(iii) Multiply numerator by $\sqrt{2}.$
$\begin{array}{rcl}& \Rightarrow & \frac{3+\sqrt{2}}{4\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3\sqrt{2}+2}{4\times 2}\\ & =& \frac{3\sqrt{2}+2}{8}\end{array}$
(iv) Multiply numerator and denominator by $\sqrt{41}+5$
$\begin{array}{rcl}& \Rightarrow & \frac{16}{\sqrt{41}-5}\times \frac{\sqrt{41}+5}{\sqrt{41}+5}=\frac{16\left(\sqrt{41}+5\right)}{{\left(\sqrt{41}\right)}^2-5^2}\\ & =& \frac{16\left(\sqrt{41}+5\right)}{41-25}=\frac{16\left(\sqrt{41}+5\right)}{16}\\ & =& \sqrt{41}+5\end{array}$
(v) Multiply numerator denominator by $2+\sqrt{3}$
$$\begin{gathered} \Rightarrow \frac{2+\sqrt{3}}{2-\sqrt{3}}\times \frac{2+\sqrt{3}}{2+\sqrt{3}}=\frac{{\left(2+\sqrt{3}\right)}^2}{2^2-{\left(\sqrt{3}\right)}^2} [\mathrm{Using}, a^2-b^2=(a-b) (a+b\left)\right] \\ =\frac{2^2+{\left(\sqrt{3}\right)}^2+2\times 2\sqrt{3}}{4-3}=7+4\sqrt{3} [\mathrm{Using} {(a+b)}^2=a^2+b^2+2ab] \end{gathered}$$

(vi) Multiply numerator and denominator by $\sqrt{2}-\sqrt{3}$
$$\begin{gathered} \Rightarrow \frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}\times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\frac{\sqrt{6}(\sqrt{2}-\sqrt{3})}{{\left(\sqrt{2}\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =\frac{\sqrt{6}\left(\sqrt{2}-\sqrt{3}\right)}{-1}=\sqrt{18}-\sqrt{12} \\ =3\sqrt{2}-2\sqrt{3} \end{gathered}$$

(vii) Multiply numerator and denominator by $\sqrt{3}+\sqrt{2}$
$$\begin{gathered} \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}}=\frac{{\left(\sqrt{3}+\sqrt{2}\right)}^2}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} [\mathrm{Using} a^2-b^2=(a-b) (a+b\left)\right] \\ =\frac{{\left(\sqrt{3}\right)}^2+{\left(\sqrt{2}\right)}^2+2\times \sqrt{3}\times \sqrt{2}}{3-2} [\mathrm{Using} {(a+b)}^2=a^2+b^2+2ab] \\ =\frac{3+2+2\sqrt{6}}{1}=5+2\sqrt{6} \end{gathered}$$

(viii) Multiply numerator and denominator by $\sqrt{5}+\sqrt{3}$
$$\begin{gathered} \Rightarrow \frac{3\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\times \frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\frac{\left(3\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2} \\ =\frac{15+3\sqrt{15}+\sqrt{15}+3}{5-3}=\frac{18+4\sqrt{15}}{2} \\ =9+2\sqrt{15} \end{gathered}$$

(ix) Multiply numerator and denominator by $4\sqrt{3}-3\sqrt{2}$
$$\begin{gathered} \mathrm{As}, \frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{48}+\sqrt{18}}=\frac{4\sqrt{3}+5\sqrt{2}}{\sqrt{16\times 3}+\sqrt{9\times 2}}=\frac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}} \\ \Rightarrow \frac{4\sqrt{3}+5\sqrt{2}}{4\sqrt{3}+3\sqrt{2}}\times \frac{4\sqrt{3}-3\sqrt{2}}{4\sqrt{3}-3\sqrt{2}}=\frac{4\sqrt{3}\left(4\sqrt{3}-3\sqrt{2}\right)+5\sqrt{2}\left(4\sqrt{3}-3\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(3\sqrt{2}\right)}^2} \\ =\frac{48-12\sqrt{6}+20\sqrt{6}-30}{30}=\frac{18+8\sqrt{6}}{30} \\ =\frac{9+4\sqrt{6}}{15} \end{gathered}$$

 

Answer:

(i) Given,
$\frac{5+2\sqrt{3}}{7+4\sqrt{3}}=a-6\sqrt{3}$
Rationalizing the denominator on LHS, by multiplying numerator & denominator by $(7-4\sqrt{3})$ we have,
$$\begin{gathered} \frac{5+2\sqrt{3}}{7+4\sqrt{3}}\times \frac{7-4\sqrt{3}}{7-4\sqrt{3}}=\frac{5(7-4\sqrt{3})+2\sqrt{3}(7-4\sqrt{3})}{7^2-{\left(4\sqrt{3}\right)}^2} \\ =\frac{35-20\sqrt{3}+14\sqrt{3}-24}{49-48} \\ =\frac{35-24-20\sqrt{3}+14\sqrt{3}}{1}=11-6\sqrt{3} \end{gathered}$$
Now as LHS = RHS
$$\begin{gathered} 11-6\sqrt{3}=a-6\sqrt{6} \\ \Rightarrow a=11 \end{gathered}$$
(ii) Given, $\frac{3-\sqrt{5}}{3+2\sqrt{5}}=a\sqrt{5}-\frac{19}{11}$
Rationalizing the denominator on LHS by multiplying numerator and denominator by $\left(3-2\sqrt{5}\right)$
$$\begin{gathered} \frac{3-\sqrt{5}}{3+2\sqrt{5}}\times \frac{3-2\sqrt{5}}{3-2\sqrt{5}} \\ =\frac{3(3-2\sqrt{5})-\sqrt{5}(3-2\sqrt{5})}{3^2-{\left(2\sqrt{5}\right)}^2} [\mathrm{Using} (a+b\left)\right(a-b) = a^2-b^2)] \\ =\frac{9-6\sqrt{5}-3\sqrt{5}+10}{9-4\times 5} \\ =\frac{19-9\sqrt{5}}{9-20} \\ =\frac{9\sqrt{5}-19}{11} \\ =\frac{9\sqrt{5}}{11}-\frac{19}{11} \\ \mathrm{Now}, \mathrm{LHS}=\mathrm{RHS} \\ \frac{9\sqrt{5}}{11}-\frac{19}{11}=a\sqrt{5}-\frac{19}{11} \\ \Rightarrow a=\frac{9}{11} \end{gathered}$$
(iii) Given, $\frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}=2-b\sqrt{6}$
Rationalizing the denominator on LHS by multiplying the numerator and denominator by $(3\sqrt{2}+2\sqrt{3})$
$$\begin{gathered} \frac{\sqrt{2}+\sqrt{3}}{3\sqrt{2}-2\sqrt{3}}\times \frac{3\sqrt{2}+2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}=\frac{\sqrt{2}\left(3\sqrt{2}+2\sqrt{3}\right)+\sqrt{3}\left(3\sqrt{2}+2\sqrt{3}\right)}{{\left(3\sqrt{2}\right)}^2-{\left(2\sqrt{3}\right)}^2} \\ =\frac{6+2\sqrt{6}+3\sqrt{6}+6}{18-12}=\frac{12+5\sqrt{6}}{6} \\ =2+\frac{5}{6}\sqrt{6} \\ \mathrm{Now}, \mathrm{LHS}=\mathrm{RHS} \\ 2+\frac{5}{6}\sqrt{6}=2-b\sqrt{6} \\ \Rightarrow b=\frac{-5}{6} \end{gathered}$$

(iv) Given,
$$\begin{gathered} \frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11}\sqrt{5}b \\ \Rightarrow \frac{{(7+\sqrt{5})}^2-{(7-\sqrt{5})}^2}{7^2-{\left(\sqrt{5}\right)}^2}=\frac{\left(49+5+14\sqrt{5}\right)-\left(49+5-14\sqrt{5}\right)}{49-5} \\ =\frac{49+5+14\sqrt{5}-49-5+14\sqrt{5}}{44}=\frac{28\sqrt{5}}{44} \\ =\frac{14\sqrt{5}}{22}=\frac{7\sqrt{5}}{11} \\ \mathrm{Now}, \mathrm{LHS}=\mathrm{RHS} \\ \frac{7\sqrt{5}}{11}=a+\frac{7\sqrt{5}}{11}b \\ \Rightarrow a=0 \mathrm{and} b=1 \end{gathered}$$
 

Answer:

$\begin{array}{rcl}\mathrm{Given}, a& =& 2+\sqrt{3}\\ \frac{1}{a}& =& \frac{1}{2+\sqrt{3}}\end{array}$
Rationalizing the denominator by $2-\sqrt{3}$
$$\begin{gathered} \frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{2-\sqrt{3}}{2^2-{\left(\sqrt{3}\right)}^2}=\frac{2-\sqrt{3}}{4-3} \\ \frac{1}{a}=2-\sqrt{3} \end{gathered}$$
$\begin{array}{rcl}\therefore a-\frac{1}{a}& =& \left(2+\sqrt{3}\right)-\left(2-\sqrt{3}\right)\\ & =& 2+\sqrt{3}-2+\sqrt{3}\\ & =& 2\sqrt{3}\end{array}$

Answer:

(i) Rationalizing the denominator by multiplying the numerator and denominator by $\sqrt{3}$
$$\begin{gathered} \Rightarrow \frac{4}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{4\sqrt{3}}{{\left(\sqrt{3}\right)}^2}=\frac{4\sqrt{3}}{3} \\ \mathrm{Now} \mathrm{given} \sqrt{3}=1.732 \\ \Rightarrow \frac{4\times 1.732}{3}=\frac{6.928}{3}=2.309 \end{gathered}$$
(ii) Multiplying the numerator and denominator by $\sqrt{6}.$
$\begin{array}{rcl}& \Rightarrow & \frac{6}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}}=\frac{6\sqrt{6}}{{\left(\sqrt{6}\right)}^2}\\ & =& \frac{6\sqrt{6}}{6}\\ & =& \sqrt{6}=\sqrt{3}\times \sqrt{2}\end{array}$
$$\begin{gathered} \mathrm{Given}, \sqrt{3}=1.732 \mathrm{and} \sqrt{2}=1.414 \\ \Rightarrow \sqrt{6}=1.732\times 1.414=2.449 \end{gathered}$$
 $$\begin{gathered} \left(\mathrm{iii}\right) \frac{\sqrt{10}-\sqrt{5}}{2}=\frac{\sqrt{5}\left(\sqrt{2}-1\right)}{2} [\because \sqrt{10}=\sqrt{2}\times \sqrt{5}] \\ \mathrm{Given}, \sqrt{5} = 2.236 \mathrm{and} \sqrt{2}=1.414 \\ =\frac{2.236(1.414-1)}{2}=\frac{2.236\times 0.414}{2} \\ =0.463 \end{gathered}$$
(iv) $\frac{2}{2+\sqrt{2}} ;$ multiply numerator and denominator by $2-\sqrt{2}$
$$\begin{gathered} \Rightarrow \frac{\sqrt{2}}{2+\sqrt{2}}\times \frac{2-\sqrt{2}}{2-\sqrt{2}}=\frac{\sqrt{2}\left(2-\sqrt{2}\right)}{2^2-{\left(\sqrt{2}\right)}^2} \\ =\frac{2\sqrt{2}-2}{4-2}=\frac{2\sqrt{2}-2}{2} \\ =\sqrt{2}-1 \\ \mathrm{Given}, \sqrt{2}=1.414 \\ \Rightarrow 1.414-1=0.414 \end{gathered}$$

(v) Multiply numerator and denominator by $\sqrt{3}-\sqrt{2}.$
$$\begin{gathered} \frac{1}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{\sqrt{3}-\sqrt{2}}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2}=\frac{\sqrt{3}-\sqrt{2}}{3-2} \\ =\sqrt{3}-\sqrt{2} \\ \mathrm{Given}, \sqrt{3}=1.732 \mathrm{and} \sqrt{2}=1.414 \\ \Rightarrow 1.732-1.414=0.318 \end{gathered}$$