Mathematics Part i Solutions Solutions for Class 9 Math Chapter 2 - Real Numbers

Answer:

$\left(\mathrm{i}\right) \frac{13}{5}$

Since, $5=2^0\times 5^1$

$\Rightarrow$ The denominator is in the form of $2^m\times 5^n$, where m and n are non-negative integers.

So, the decimal form of $\frac{13}{5}$ will be terminating type.

$\left(\mathrm{ii}\right) \frac{2}{11}$

Since, $11=2^0\times 5^0\times {11}^1$

$\Rightarrow$ The denominator is not in the form of $2^m\times 5^n$, where m and n are non-negative integers.

So, the decimal form of $\frac{2}{11}$ will be non-terminating recurring type.

$\left(\mathrm{iii}\right) \frac{29}{16}$

Since, $16=2^4\times 5^0$

$\Rightarrow$ The denominator is in the form of $2^m\times 5^n$, where m and n are non-negative integers.

So, the decimal form of $\frac{29}{16}$ will be terminating type.

$\left(\mathrm{iv}\right) \frac{17}{125}$

Since, $125=2^0\times 5^3$

$\Rightarrow$ The denominator is in the form of $2^m\times 5^n$, where m and n are non-negative integers.

So, the decimal form of $\frac{17}{125}$ will be terminating type.

$\left(\mathrm{v}\right) \frac{11}{6}$

Since, $6=2^1\times 5^0\times 3^1$

$\Rightarrow$ The denominator is not in the form of $2^m\times 5^n$, where m and n are non-negative integers.

So, the decimal form of $\frac{11}{6}$ will be non-terminating recurring type.

Answer:

$\left(\mathrm{i}\right) \frac{127}{200}=\frac{127}{200}\times \frac{5}{5}=\frac{635}{1000}=0.635$

$\left(\mathrm{ii}\right) \frac{25}{99}=\frac{4}{4}\times \frac{25}{99}=\frac{1}{4}\times \frac{100}{99}=\frac{1}{4}\times 1.010101...=0.2525...=0.\overline{25}$​

$\left(\mathrm{iii}\right) \frac{23}{7}=3.2857142857...=3.\overline{285714}$​

$\left(\mathrm{iv}\right) \frac{4}{5}\times \frac{2}{2}=\frac{8}{10}=0.8$​

$\left(\mathrm{v}\right) \frac{17}{8}=\frac{17}{8}\times \frac{125}{125}=\frac{2125}{1000}=2.125$​

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Let} x=0.6\mathbf{°} ...\left(1\right) \\ x=0.666... \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 10, \mathrm{we} \mathrm{get} \\ 10x=6.666... ...\left(2\right) \\ \mathrm{Subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ 9x=6 \\ \therefore x=\frac{6}{9} \\ \mathrm{So}, 0.6\mathbf{°}=\frac{2}{3} \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{Let} x=0.\overline{37} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 100, \mathrm{we} \mathrm{get} \\ 100x=37.\overline{37} ...\left(2\right) \\ \mathrm{Subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ 99x=37 \\ \therefore x=\frac{37}{99} \\ \mathrm{So}, 0.\overline{37}=\frac{37}{99} \end{gathered}$$​

$$\begin{gathered} \left(\mathrm{iii}\right) \mathrm{Let} x=3.\overline{17} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 100, \mathrm{we} \mathrm{get} \\ 100x=317.\overline{17} ...\left(2\right) \\ \mathrm{Subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ 99x=314 \\ \therefore x=\frac{314}{99} \\ \mathrm{So}, 3.\overline{17}=\frac{314}{99} \end{gathered}$$​

$$\begin{gathered} \left(\mathrm{iv}\right) \mathrm{Let} x=15.\overline{89} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 100, \mathrm{we} \mathrm{get} \\ 100x=1589.\overline{89} ...\left(2\right) \\ \mathrm{Subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ 99x=1574 \\ \therefore x=\frac{1574}{99} \\ \mathrm{So}, 3.\overline{17}=\frac{1574}{99} \end{gathered}$$​

$$\begin{gathered} \left(\mathrm{v}\right) \mathrm{Let} x=2.\overline{514} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 1000, \mathrm{we} \mathrm{get} \\ 1000x=2514.\overline{514} ...\left(2\right) \\ \mathrm{Subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ 999x=2512 \\ \therefore x=\frac{2512}{999} \\ \mathrm{So}, 2.\overline{514}=\frac{2512}{999} \end{gathered}$$​

Answer:

Let us assume that 4$\sqrt{2}$  is a rational number.

$\Rightarrow$4$\sqrt{2}$ = $\frac{p}{q}$, where p and q are the integers and q $\ne$ 0.

$\Rightarrow$$\sqrt{2}$ = $\frac{p}{4q}$

Since, p, q and 4 are integers. So, $\frac{p}{4q}$ is a rational number.

$\Rightarrow$$\sqrt{2}$ is also a rational number.

but this contradicts the fact that $\sqrt{2}$ is an irrational number.

This contradiction has arisen due to the wrong assumption that 4$\sqrt{2}$ is a rational number.

Hence, 4$\sqrt{2}$ is an irrational number.

Answer:

Let us assume that 3 + $\sqrt{5}$  is a rational number.

$\Rightarrow$3 + $\sqrt{5}$ = $\frac{p}{q}$, where p and q are the integers and q $\ne$ 0.

$\Rightarrow$$\sqrt{5}$ = $\frac{p}{q}-3=\frac{p-3q}{q}$

Since, p, q and 3 are integers. So, $\frac{p-3q}{q}$ is a rational number.

$\Rightarrow$$\sqrt{5}$ is also a rational number.

but this contradicts the fact that $\sqrt{5}$ is an irrational number.

This contradiction has arisen due to the wrong assumption that 3 + $\sqrt{5}$ is a rational number.

Hence, 3 + $\sqrt{5}$​ is an irrational number.

Answer:

(i) Steps of construction for $\sqrt{5}$:

Step 1: Draw a number line. Mark O as the zero on the number line.

Step 2: At point A, draw AB $\perp$ OA such that AB = 1 unit.

Step 3: With point O as the centre and radius OB, draw an arc intersecting the number line at point P.

Thus, P is the point for $\sqrt{5}$ on the number line.



(ii) Steps of construction for $\sqrt{10}$:

Step 1: Draw a number line. Mark O as the zero on the number line.

Step 2: At point A, draw AB $\perp$ OA such that AB = 1 unit.

Step 3: With point O as the centre and radius OB, draw an arc intersecting the number line at point C.

Thus, C is the point for $\sqrt{10}$​ on the number line.

Answer:

Answer:

$\left(\mathrm{i}\right) \sqrt[3]{7}=7^{\frac{1}{3}}$

The order of the surd is 3.

$\left(\mathrm{ii}\right) 5 \sqrt{12}=5\times {12}^{\frac{1}{2}}$

The order of the surd is 2.

$\left(\mathrm{iii}\right) \sqrt[4]{10}={10}^{\frac{1}{4}}$

The order of the surd is 4.

$\left(\mathrm{iv}\right) \sqrt{39}={39}^{\frac{1}{2}}$

The order of the surd is 2.

$\left(\mathrm{v}\right) \sqrt[3]{18}={18}^{\frac{1}{3}}$

The order of the surd is 3.

Answer:

(i) Since, $\sqrt[3]{51}= {\left(3\times 17\right)}^{\frac{1}{3}}$

So, $\sqrt[3]{51}$ is a surd.

(ii) Since, $\sqrt[4]{16}=\sqrt[4]{2^4}=2$

So, $\sqrt[4]{16}$ is not a surd.

(iii) Since, $\sqrt[5]{81}=\sqrt[5]{3^4}={\left(3^4\right)}^{\frac{1}{5}}=3^{\frac{4}{5}}$

So, $\sqrt[5]{81}$ is a surd.

(iv) Since, $\sqrt{256}=\sqrt{{16}^2}=16$

So, $\sqrt{256}$ is not a surd.

(v) Since, $\sqrt[3]{64}=\sqrt[3]{4^3}=4$

So, $\sqrt[3]{64}$ is not a surd.

(vi) Since, $\sqrt{\frac{22}{7}}={\left(\frac{22}{7}\right)}^{\frac{1}{2}}$

So, $\sqrt{\frac{22}{7}}$ is a surd.

Answer:

(i) $\sqrt{52}, 5\sqrt{13}$

Since, $\sqrt{52}=\sqrt{4\times 13}=2\sqrt{13}$

So, $\sqrt{52}, 5\sqrt{13}$ is like surds.

(ii) $\sqrt{68}, 5\sqrt{3}$

Since, $\sqrt{68}=\sqrt{4\times 17}=2\sqrt{17}$

So, $\sqrt{68}, 5\sqrt{3}$ is unlike surds.

(iii) $4\sqrt{18}, 7 \sqrt{2}$

Since, $4\sqrt{18}=4\sqrt{9\times 2}=4\times 3\sqrt{2}=12\sqrt{2}$

So, $4\sqrt{18}, 7 \sqrt{2}$ is like surds.

(iv) $19\sqrt{12}, 6\sqrt{3}$

Since, $19\sqrt{12}=19\sqrt{4\times 3}=19\times 2\sqrt{3}=38\sqrt{3}$

So, $19\sqrt{12}, 6\sqrt{3}$ is like surds.

(v) $5\sqrt{22},7\sqrt{33}$

Since, $5\sqrt{22}=5\sqrt{2\times 11} \mathrm{and} 7\sqrt{33}=7\sqrt{3\times 11}$

So, $5\sqrt{22},7\sqrt{33}$ is unlike surds.

(vi) $5\sqrt{5} , \sqrt{75}$

Since, $\sqrt{75}=\sqrt{25\times 3}=5\sqrt{3}$

So, $5\sqrt{5} , \sqrt{75}$ is like surds.

Answer:

(i) $\sqrt{27}=\sqrt{9\times 3}=3\sqrt{3}$

(ii) $\sqrt{50}=\sqrt{25\times 2}=5\sqrt{2}$

(iii) $\sqrt{250}=\sqrt{25\times 10}=5\sqrt{10}$

(iv) $\sqrt{112}=\sqrt{16\times 7}=4\sqrt{7}$

(v) $\sqrt{168}=\sqrt{4\times 42}=2\sqrt{42}$ 

Answer:

(i) $7\sqrt{2}, 5\sqrt{3}$

Since, $7\sqrt{2}=\sqrt{49\times 2}=\sqrt{98} \mathrm{and} 5\sqrt{3}=\sqrt{25\times 3}=\sqrt{75}$

So, $7\sqrt{2} > 5\sqrt{3}$

(ii) $\sqrt{247},\sqrt{274}$

$\sqrt{247} < \sqrt{274}$

(iii) $2\sqrt{7},\sqrt{28}$

Since, $2\sqrt{7}=\sqrt{4\times 7}=\sqrt{28}$

So, $2\sqrt{7} = \sqrt{28}$

(iv) $5\sqrt{5}, 7\sqrt{2}$

Since, $5\sqrt{5}=\sqrt{25\times 5}=\sqrt{125} \mathrm{and} 7\sqrt{2}=\sqrt{49\times 2}=\sqrt{98}$

So, $5\sqrt{5} > 7\sqrt{2}$

(v) $4\sqrt{42}, 9\sqrt{2}$

Since, $4\sqrt{42}=\sqrt{16\times 42}=\sqrt{672} \mathrm{and} 9\sqrt{2}=\sqrt{81\times 2}=\sqrt{162}$

So, $4\sqrt{42} > 9\sqrt{2}$

(vi) $5\sqrt{3}, 9$

Since, $5\sqrt{3}=\sqrt{25\times 3}=\sqrt{75} \mathrm{and} 9=\sqrt{81}$

So, $5\sqrt{3} < 9$

(vii) $7, 2\sqrt{5}$

Since, $7=\sqrt{49} \mathrm{and} 2\sqrt{5}=\sqrt{4\times 5}=\sqrt{20}$

So, $7 > 2\sqrt{5}$

Answer:

(i) $5\sqrt{3}+8\sqrt{3}=13\sqrt{3}$

(ii) $9\sqrt{5 }-4\sqrt{5}+\sqrt{125}=5\sqrt{5}+\sqrt{25\times 5}=5\sqrt{5}+5\sqrt{5}=10\sqrt{5}$

(iii) $7\sqrt{48}-\sqrt{27}-\sqrt{3}=7\sqrt{16\times 3}-\sqrt{9\times 3}-\sqrt{3}=7\times 4\sqrt{3}-3\sqrt{3}-\sqrt{3}=28\sqrt{3}-3\sqrt{3}-\sqrt{3}=24\sqrt{3}$

(iv) $\sqrt{7} -\frac{3}{5}\sqrt{7} +2\sqrt{7}=3\sqrt{7}-\frac{3}{5}\sqrt{7}= \frac{15}{5}\sqrt{7}-\frac{3}{5}\sqrt{7}=\frac{12}{5}\sqrt{7}$

Answer:

(i) $3\sqrt{12} \times \sqrt{18}=3\sqrt{4\times 3}\times \sqrt{9\times 2}=6\sqrt{3}\times 3\sqrt{2}=18\sqrt{6}$

(ii) $3\sqrt{12} \times 7\sqrt{15}=3\sqrt{4\times 3}\times 7\sqrt{15}=6\sqrt{3}\times 7\sqrt{15}=42\sqrt{45}=42\sqrt{9\times 5}=126\sqrt{5}$  

(iii) $3\sqrt{8} \times \sqrt{5}=3\sqrt{4\times 2}\times \sqrt{5}=6\sqrt{2}\times \sqrt{5}=6\sqrt{10}$

(iv) $5\sqrt{8} \times 2\sqrt{8}=10\sqrt{64}=10\times 8=80$

Answer:

(i) $\sqrt{98} ÷ \sqrt{2}=\frac{\sqrt{98}}{\sqrt{2}}=\sqrt{\frac{98}{2}}=\sqrt{49}=7$

(ii) $\sqrt{125} ÷ \sqrt{50}=\frac{\sqrt{125}}{\sqrt{50}}=\sqrt{\frac{125}{50}}=\sqrt{\frac{5}{2}}$

(iii) $\sqrt{54} ÷ \sqrt{27}=\frac{\sqrt{54}}{\sqrt{27}}=\sqrt{\frac{54}{27}}=\sqrt{2}$

(iv) $\sqrt{310} ÷ \sqrt{5}=\frac{\sqrt{310}}{\sqrt{5}}=\sqrt{\frac{310}{5}}=\sqrt{62}$

Answer:

(i) $\frac{3}{\sqrt{5}}$

$$\begin{gathered} =\frac{3}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}} \\ =\frac{3\sqrt{5}}{{\left(\sqrt{5}\right)}^2} \\ =\frac{4\sqrt{5}}{5} \end{gathered}$$

(ii) $\frac{1}{\sqrt{14}}$

$$\begin{gathered} =\frac{1}{\sqrt{14}}\times \frac{\sqrt{14}}{\sqrt{14}} \\ =\frac{\sqrt{14}}{{\left(\sqrt{14}\right)}^2} \\ =\frac{\sqrt{14}}{14} \end{gathered}$$

(iii) $\frac{5}{\sqrt{7}}$

$$\begin{gathered} =\frac{5}{\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}} \\ =\frac{5\sqrt{7}}{{\left(\sqrt{7}\right)}^2} \\ =\frac{5\sqrt{7}}{7} \end{gathered}$$

(iv)  $\frac{6}{9\sqrt{3}}$

$$\begin{gathered} =\frac{2}{3\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =\frac{2\sqrt{3}}{3\times 3} \\ =\frac{2\sqrt{3}}{9} \end{gathered}$$

(v) $\frac{11}{\sqrt{3}}$

$$\begin{gathered} =\frac{11}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =\frac{11\sqrt{3}}{{\left(\sqrt{3}\right)}^2} \\ =\frac{11\sqrt{3}}{3} \end{gathered}$$

Answer:

(i)

$$\begin{gathered} \sqrt{3} \left(\sqrt{7} - \sqrt{3}\right) \\ =\sqrt{3}\times \sqrt{7}-\sqrt{3}\times \sqrt{3} \\ =\sqrt{21}-\sqrt{9} \\ =\sqrt{21}-3 \end{gathered}$$

(ii)

$$\begin{gathered} \left(\sqrt{5} - \sqrt{7}\right) \sqrt{2} \\ =\sqrt{5}\times \sqrt{2} - \sqrt{7}\times \sqrt{2} \\ =\sqrt{10} - \sqrt{14} \end{gathered}$$

(iii)

$$\begin{gathered} \left(3\sqrt{2} - \sqrt{3}\right) \left(4\sqrt{3} - \sqrt{2}\right) \\ =3\sqrt{2}\times 4\sqrt{3} - 3\sqrt{2}\times \sqrt{2} - \sqrt{3}\times 4\sqrt{3} + \sqrt{3}\times \sqrt{2} \\ =12\sqrt{6} - 3\sqrt{4} - 4\sqrt{9} + \sqrt{6} \\ =13\sqrt{6} - 3\times 2 - 4\times 3 \\ =13\sqrt{6} - 6 - 12 \\ =13\sqrt{6} - 18 \end{gathered}$$

Answer:

(i) $\frac{1}{\sqrt{7} +\sqrt{2}}$

$$\begin{gathered} =\frac{1}{\left(\sqrt{7}+\sqrt{2}\right)}\times \frac{\left(\sqrt{7}-\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)} \\ =\frac{\sqrt{7}-\sqrt{2}}{{\left(\sqrt{7}\right)}^2-{\left(\sqrt{2}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\sqrt{7}-\sqrt{2}}{7-2} \\ =\frac{\sqrt{7}-\sqrt{2}}{5} \end{gathered}$$

(ii) $\frac{3}{2\sqrt{5} - 3\sqrt{2}}$

$$\begin{gathered} =\frac{3}{\left(2\sqrt{5}-3\sqrt{2}\right)}\times \frac{\left(2\sqrt{5}+3\sqrt{2}\right)}{\left(2\sqrt{5}+3\sqrt{2}\right)} \\ =\frac{3\left(2\sqrt{5}+3\sqrt{2}\right)}{{\left(2\sqrt{5}\right)}^2-{\left(3\sqrt{2}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{3\left(2\sqrt{5}+3\sqrt{2}\right)}{20-18} \\ =\frac{3}{2}\left(2\sqrt{5}+3\sqrt{2}\right) \end{gathered}$$

(iii) $\frac{4}{7 + 4\sqrt{3}}$

$$\begin{gathered} =\frac{4}{\left(7+4\sqrt{3}\right)}\times \frac{\left(7-4\sqrt{3}\right)}{\left(7-4\sqrt{3}\right)} \\ =\frac{4\left(7-4\sqrt{3}\right)}{{\left(7\right)}^2-{\left(4\sqrt{3}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{4\left(7-4\sqrt{3}\right)}{49-48} \\ =28-16\sqrt{3} \end{gathered}$$

(iv)  $\frac{\sqrt{5} - \sqrt{3}}{\sqrt{5}+ \sqrt{3}}$

$$\begin{gathered} =\frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}+\sqrt{3}\right)}\times \frac{\left(\sqrt{5}-\sqrt{3}\right)}{\left(\sqrt{5}-\sqrt{3}\right)} \\ =\frac{{\left(\sqrt{5}-\sqrt{3}\right)}^2}{{\left(\sqrt{5}\right)}^2-{\left(\sqrt{3}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{{\left(\sqrt{5}\right)}^2+{\left(\sqrt{3}\right)}^2-2\left(\sqrt{5}\right)\left(\sqrt{3}\right)}{5-2} \\ =\frac{5+3-2\sqrt{15}}{3} \\ =\frac{8-2\sqrt{15}}{3} \end{gathered}$$

Answer:

(i) $\left|15 - 2\right|=\left|13\right|=13$

(ii) $\left| 4 - 9\right|=\left|-5\right|=5$

(iii) $\left|7\right| \times \left| -4\right|=7\times 4=28$

Answer:

(i) $\left|3x-5\right|=1$
$$\begin{gathered} \Rightarrow 3x-5=\pm 1 \\ \Rightarrow 3x-5=1, \mathrm{or}, 3x-5=-1 \\ \Rightarrow 3x=1+5, \mathrm{or}, 3x=-1+5 \\ \Rightarrow 3x=6, \mathrm{or}, 3x=4 \\ \therefore x=2, \mathrm{or}, x=\frac{4}{3} \end{gathered}$$

(ii) $\left| 7 - 2x\right| = 5$
$$\begin{gathered} \Rightarrow 7-2x=\pm 5 \\ \Rightarrow 7-2x=5, \mathrm{or}, 7-2x=-5 \\ \Rightarrow 2x=7-5, \mathrm{or}, 2x=7+5 \\ \Rightarrow 2x=2, \mathrm{or}, 2x=12 \\ \therefore x=1, \mathrm{or}, x=6 \end{gathered}$$

(iii) $\left|\frac{ 8 - x}{2}\right| = 5$
$$\begin{gathered} \Rightarrow \frac{8-x}{2}=\pm 5 \\ \Rightarrow \frac{8-x}{2}=5, \mathrm{or}, \frac{8-x}{2}=-5 \\ \Rightarrow 8-x=10, \mathrm{or}, 8-x=-10 \\ \Rightarrow x=8-10, \mathrm{or}, x=8+10 \\ \therefore x=-2, \mathrm{or}, x=18 \end{gathered}$$

(iv) $\left| 5 + \frac{x}{4}\right| = 5$
$$\begin{gathered} \Rightarrow 5+\frac{x}{4}=\pm 5 \\ \Rightarrow 5+\frac{x}{4}=5, \mathrm{or}, 5+\frac{x}{4}=-5 \\ \Rightarrow \frac{x}{4}=5-5, \mathrm{or}, \frac{x}{4}=-5-5 \\ \Rightarrow \frac{x}{4}=0, \mathrm{or}, \frac{x}{4}=-10 \\ \therefore x=0, \mathrm{or}, x=-40 \end{gathered}$$

Answer:

Answer:

(i ) 0.555

$$\begin{gathered} =\frac{555}{1000} \\ =\frac{111}{200} \end{gathered}$$

(ii) $29.\overline{568}$

$$\begin{gathered} \mathrm{Let} x=29.\overline{568} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 1000, \mathrm{we} \mathrm{get} \\ 1000x=29568.\overline{568} ...\left(2\right) \\ \mathrm{By} \mathrm{subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ \Rightarrow 999x=29539 \\ \Rightarrow x=\frac{29539}{999} \\ \therefore 29.\overline{568}=\frac{29539}{999} \end{gathered}$$

(iii) 9.315 315 ...

$$\begin{gathered} \mathrm{Since}, 9.315 315 ...=9.\overline{315} \\ \mathrm{Let} x=9.\overline{315} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 1000, \mathrm{we} \mathrm{get} \\ 1000x=9315.\overline{315} ...\left(2\right) \\ \mathrm{By} \mathrm{subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ \Rightarrow 999x=9306 \\ \Rightarrow x=\frac{9306}{999} \\ \therefore 9.\overline{315}=\frac{1034}{111} \end{gathered}$$

(iv) 357.417417...

$$\begin{gathered} \mathrm{Since}, 357.417 417 ...=357.\overline{417} \\ \mathrm{Let} x=357.\overline{417} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 1000, \mathrm{we} \mathrm{get} \\ 1000x=357417.\overline{417} ...\left(2\right) \\ \mathrm{By} \mathrm{subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ \Rightarrow 999x=357060 \\ \Rightarrow x=\frac{357060}{999} \\ \therefore 357.\overline{417}=\frac{119020}{333} \end{gathered}$$

(v) $30.\overline{219}$

$$\begin{gathered} \mathrm{Let} x=30.\overline{219} ...\left(1\right) \\ \mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 1000, \mathrm{we} \mathrm{get} \\ 1000x=30219.\overline{219} ...\left(2\right) \\ \mathrm{By} \mathrm{subtracting} \left(1\right) \mathrm{from} \left(2\right), \mathrm{we} \mathrm{get} \\ \Rightarrow 999x=30189 \\ \Rightarrow x=\frac{30189}{999} \\ \therefore 30.\overline{219}=\frac{10063}{333} \end{gathered}$$

Answer:

(i) $\frac{-5}{7}=-0.\overline{714285}$

(ii) $\frac{9}{11}=0.\overline{81}$

(iii) $\sqrt{5}=2.23606797...$

(iv) $\frac{121}{13}=9.\overline{307692}$

(v) $\frac{29}{8}=3.625$

Answer:

Let us assume that 5 + $\sqrt{7}$ is a rational number.

$$\begin{gathered} \Rightarrow 5+\sqrt{7}=\frac{p}{q}, \mathrm{where} p \mathrm{and} q \mathrm{are} \mathrm{two} \mathrm{integers} \mathrm{and} q\ne 0 \\ \Rightarrow \sqrt{7}=\frac{p}{q}-5=\frac{p-5q}{q} \end{gathered}$$

Since, $p, q \mathrm{and} 5 \mathrm{are} \mathrm{integers}, \mathrm{so} \frac{p-5q}{q} \mathrm{is} \mathrm{a} \mathrm{rational} \mathrm{number}$.

$\Rightarrow$ $\sqrt{7}$ is also a rational number.

But this contradicts the fact that $\sqrt{7}$ is an irrational number.

This contradiction has arisen due to our assumption that 5 + $\sqrt{7}$ is a rational number.

Hence, 5 + $\sqrt{7}$ is an irrational number.

Answer:

(i) $\frac{3}{4}\sqrt{8}=\frac{3}{4}\times \sqrt{4\times 2}=\frac{3}{4}\times 2\sqrt{2}=\frac{3}{2}\sqrt{2}$

(ii) $-\frac{5}{9}\sqrt{45}=-\frac{5}{9}\sqrt{9\times 5}=-\frac{5}{9}\times 3\sqrt{5}=-\frac{5}{3}\sqrt{5}$

Answer:

$$\begin{gathered} \left(\mathrm{i}\right) \mathrm{Since}, \sqrt{32}=\sqrt{16\times 2}=4\sqrt{2} \\ \mathrm{So}, \sqrt{2} \mathrm{is} \mathrm{the} \mathrm{simplest} \mathrm{form} \mathrm{of} \mathrm{rationalising} \mathrm{factor}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{ii}\right) \mathrm{Since}, \sqrt{50}=\sqrt{25\times 2}=5\sqrt{2} \\ \mathrm{So}, \sqrt{2} \mathrm{is} \mathrm{the} \mathrm{simplest} \mathrm{form} \mathrm{of} \mathrm{rationalising} \mathrm{factor}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iii}\right) \mathrm{Since}, \sqrt{27}=\sqrt{9\times 3}=3\sqrt{3} \\ \mathrm{So}, \sqrt{3} \mathrm{is} \mathrm{the} \mathrm{simplest} \mathrm{form} \mathrm{of} \mathrm{rationalising} \mathrm{factor}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{iv}\right) \mathrm{Since}, \frac{3}{5}\sqrt{10}=\frac{3}{5}\sqrt{2\times 5} \\ \mathrm{So}, \sqrt{10} \mathrm{is} \mathrm{the} \mathrm{simplest} \mathrm{form} \mathrm{of} \mathrm{rationalising} \mathrm{factor}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{v}\right) \mathrm{Since}, 3\sqrt{72}=3\sqrt{36\times 2}=3\times 6\sqrt{2}=18\sqrt{2} \\ \mathrm{So}, \sqrt{2} \mathrm{is} \mathrm{the} \mathrm{simplest} \mathrm{form} \mathrm{of} \mathrm{rationalising} \mathrm{factor}. \end{gathered}$$

$$\begin{gathered} \left(\mathrm{vi}\right) \mathrm{Since}, 4\sqrt{11}=4\sqrt{11\times 1} \\ \mathrm{So}, \sqrt{11} \mathrm{is} \mathrm{the} \mathrm{simplest} \mathrm{form} \mathrm{of} \mathrm{rationalising} \mathrm{factor}. \end{gathered}$$

Answer:

(i) $\frac{4}{7}\sqrt{147} + \frac{3}{8}\sqrt{192} - \frac{1}{5}\sqrt{75}$ 

$$\begin{gathered} = \frac{4}{7}\sqrt{49\times 3} + \frac{3}{8}\sqrt{64\times 3} - \frac{1}{5}\sqrt{25\times 3} \\ =\frac{4}{7}\times 7\sqrt{3} + \frac{3}{8}\times 8\sqrt{3} - \frac{1}{5}\times 5\sqrt{3} \\ =4\sqrt{3}+3\sqrt{3}-\sqrt{3} \\ =6\sqrt{3} \end{gathered}$$

(ii)  $5\sqrt{3}+2\sqrt{27}+\frac{1}{\sqrt{3}}$

$$\begin{gathered} =5\sqrt{3}+2\sqrt{9\times 3}+\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =5\sqrt{3}+6\sqrt{3}+\frac{\sqrt{3}}{3} \\ =11\sqrt{3}+\frac{\sqrt{3}}{3} \\ =\frac{33\sqrt{3}+\sqrt{3}}{3} \\ =\frac{34\sqrt{3}}{3} \end{gathered}$$

(iii) $\sqrt{216}-5\sqrt{6}+\sqrt{294}-\frac{3}{\sqrt{6}}$

$$\begin{gathered} =\sqrt{36\times 6}-5\sqrt{6}+\sqrt{49\times 6}-\frac{3}{\sqrt{6}}\times \frac{\sqrt{6}}{\sqrt{6}} \\ =6\sqrt{6}-5\sqrt{6}+7\sqrt{6}-\frac{3\sqrt{6}}{6} \\ =8\sqrt{6}-\frac{\sqrt{6}}{2} \\ =\frac{18\sqrt{6}-\sqrt{6}}{2} \\ =\frac{17\sqrt{6}}{2} \end{gathered}$$

(iv) $4\sqrt{12}-\sqrt{75}-7\sqrt{48}$

$$\begin{gathered} =4\sqrt{4\times 3}-\sqrt{25\times 3}-7\sqrt{16\times 3} \\ =8\sqrt{3}-5\sqrt{3}-28\sqrt{3} \\ =-25\sqrt{3} \end{gathered}$$

(v) $2\sqrt{48}-\sqrt{75}-\frac{1}{\sqrt{3}}$

$$\begin{gathered} =2\sqrt{16\times 3}-\sqrt{25\times 3}-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}} \\ =8\sqrt{3}-5\sqrt{3}-\frac{\sqrt{3}}{3} \\ =3\sqrt{3}-\frac{\sqrt{3}}{3} \\ =\frac{9\sqrt{3}-\sqrt{3}}{3} \\ =\frac{8\sqrt{3}}{3} \end{gathered}$$

Answer:

(i) $\frac{1}{\sqrt{5}}$

$$\begin{gathered} =\frac{1}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}} \\ =\frac{\sqrt{5}}{5} \end{gathered}$$

(ii) $\frac{2}{3\sqrt{7}}$

$$\begin{gathered} =\frac{2}{3\sqrt{7}}\times \frac{\sqrt{7}}{\sqrt{7}} \\ =\frac{2\sqrt{7}}{3\times 7} \\ =\frac{2\sqrt{7}}{21} \end{gathered}$$

(iii) $\frac{1}{\sqrt{3}-\sqrt{2}}$

$$\begin{gathered} =\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)}\times \frac{\left(\sqrt{3}+\sqrt{2}\right)}{\left(\sqrt{3}+\sqrt{2}\right)} \\ =\frac{\left(\sqrt{3}+\sqrt{2}\right)}{{\left(\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\left(\sqrt{3}+\sqrt{2}\right)}{3-2} \\ =\sqrt{3}+\sqrt{2} \end{gathered}$$

(iv) $\frac{1}{3\sqrt{5}+2\sqrt{2}}$

$$\begin{gathered} =\frac{1}{\left(3\sqrt{5}+2\sqrt{2}\right)}\times \frac{\left(3\sqrt{5}-2\sqrt{2}\right)}{\left(3\sqrt{5}-2\sqrt{2}\right)} \\ =\frac{\left(3\sqrt{5}-2\sqrt{2}\right)}{{\left(3\sqrt{5}\right)}^2-{\left(2\sqrt{2}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{\left(3\sqrt{5}-2\sqrt{2}\right)}{45-8} \\ =\frac{3\sqrt{5}-2\sqrt{2}}{37} \end{gathered}$$

(v) $\frac{12}{4\sqrt{3}-\sqrt{2}}$

$$\begin{gathered} =\frac{12}{\left(4\sqrt{3}-\sqrt{2}\right)}\times \frac{\left(4\sqrt{3}+\sqrt{2}\right)}{\left(4\sqrt{3}+\sqrt{2}\right)} \\ =\frac{12\left(4\sqrt{3}+\sqrt{2}\right)}{{\left(4\sqrt{3}\right)}^2-{\left(\sqrt{2}\right)}^2} \left[\left(a+b\right)\left(a-b\right)=a^2-b^2\right] \\ =\frac{12\left(4\sqrt{3}+\sqrt{2}\right)}{48-2} \\ =\frac{12\left(4\sqrt{3}+\sqrt{2}\right)}{46} \\ =\frac{6\left(4\sqrt{3}+\sqrt{2}\right)}{23} \end{gathered}$$