(i) \( (300000)^{2} \times(20000)^{4}\)
\(=\left(3.0 \times 10^{5}\right)^{2} \times\left(2.0 \times 10^{4}\right)^{4} \)
\(=3^{2} \times 10^{10} \times 2^{4} \times 10^{16} \)
\(=9 \times 16 \times 10^{10+16} \)
\(=144 \times 10^{26} \)
\(=1.44 \times 10^{28} \)
(ii) (0.000001)¹¹ ÷ (0.005)³
\( =\left(1.0 \times 10^{-6}\right)^{11} \div\left(5.0 \times 10^{-3}\right)^{3} \)
\( =\frac{1.0 \times 10^{-66}}{125.0 \times 10^{9}} \)
\( =\frac{1000.0 \times 10^{-69}}{125.0 \times 10^{-9}} \)
\( =8.0 \times 10^{-69} \times 10^{9} \)
\( =8.0 \times 10^{-60} \)
(iii) \(\left\{(0.00003)^{6} \times(0.00005)^{4}\right\} \div\left\{(0.009)^{3} \times(0.05)^{2}\right\}\)
=\(\frac{(3.0\times10^{-5})^6\times(5.0\times10^{-5})^4}{(9.0\times10^{-3})^3\times(5.0\times10^{-2})^2}\)
=\(\frac{3^6\times10^{-30}\times5^4\times10^{20}}{9^3\times10^{-9}\times5^{2}\times10^{-4}}=\frac{3^6\times5^4\times10^{-30-20}}{(3^{2})^3\times10^{-9-4}\times5^{2}}=\frac{ ̶3̶^6̶\times5^4\times10^{-50}}{ ̶3̶^6̶\times5^2\times10^{-13}}\)
\( =5^{4-2} \times 10^{-50+13} \)
\(=5^{2} \times 10^{-37} \)
\(=25 \times 10^{-37} \)
\(=2.5 \times 10^{1} \times 10^{-37}\)
\(=2.5 \times 10^{-36} \)

(i) Yes, the sum of two surds is a rational number.
Example \( 3 \sqrt{4}+2 \sqrt{4} =5 \sqrt{4} =\sqrt{5^{2} \times 4}=\sqrt{25 \times 4} =\sqrt{100}=10 \) rational number.
(ii) ) Yes, the difference of two surds is a rational number
Example  \(2 \sqrt{9}-\sqrt{9}=\sqrt{9}=3\), a rational number
(iii) ) Yes, the product of two surds is a rational number.
Example \(\sqrt{2} \times \sqrt{2}=\sqrt{4}=2\), a rational number
(iv) ) Yes, the quotient of two surds is a rational number.
 \(\sqrt{75} \div \sqrt{3}=\sqrt{\frac{75}{3}}=\sqrt{25}=5 \), a rational number

(i) \( =\sqrt{8 \times 5}-\sqrt{4 \times 5} =\sqrt{4 \times 2 \times 5}-\sqrt{4 \times 5} \)
\( =2 \sqrt{2 \times 5}-2 \sqrt{5} \)
\( =2 \sqrt{2} \times \sqrt{5}-2 \sqrt{5} \)
\( =2 \times 2.236(1.414-1) \)
\( =4.472 \times 0.414 =1.852 \)
(ii) \(\sqrt{300}+\sqrt{90}-\sqrt{8}=\sqrt{3\times100}+\sqrt{9\times10}+\sqrt{4\times2}\)
\(=10\sqrt{3}+3\sqrt{10}+2\sqrt{2}\)
= 10\(\times\)1.732 + 3\(\times\)3.162 + 2\(\times\)1.414
= 17.32 + 9.486 + 2.828 = 23.978

\(\frac { 23 }{ 10 } \)is 2.3 and\(\frac { 12 }{ 5 } \) is 2.4
You need an irrational number greater than 2.3 but less than 2.4
One such irrational number is
2.301001000100001000001000000100000001……..
observe that it is non-terminating and non-recurring.

Thus, by division method \(\sqrt { 3 } \) = 1.7320508…
It is found that the square root of every positive non perfect square number is an irrational number  \(\sqrt { 2 } ,\sqrt { 3 } ,\sqrt { 5 } ,\sqrt { 6 } ,\sqrt { 7 } \)..... are all irrational numbers.

Let x = 0.00010001 ............. \(\rightarrow\) (1)
10000x = 1.00010001 .......... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 10000x - x = 1.00010001 ......... (-)
                                   0.00010001 .........
                                 ____________
                     9999x = 1
      \(x=\frac{1}{9999}\)      

Let x = \(3.1\overline { 7 } \) = 3.1777 .......  (1)
Here period of decimal is 2, multiply equation (1) by 100
10x = 31.777 .........  (2)
(2)-(1)

\( =\frac{143}{45} \)

Let \(2.\overline { 327 } \) = 2.327327327 ..... \(\rightarrow\) (1)
Here period of decimal is 3, multiply equation (1) by 1000
1000 x = 2327.327327 ..... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 1000x - x = 999x = 2325
 \(x=\frac{2325}{999}\) (or) \(775\over 333\)

Let x = 0.242424..... \(\rightarrow\) (1)
100 x = 24.2424 ...... \(\rightarrow\) (2)
(2) - (1) \(\Rightarrow\) 100x - x = 24.2424 
99x = 24
x = \(24\over 99\) (or) \(8\over 33\)
x = \(8\over 33\)