## Tamilnadu State Board Class 9 Maths Solutions Chapter 2 Real Numbers Ex 2.8

Question 1.

Represent the following numbers in the scientific notation:

(i) 569430000000

(ii) 2000.57

(iii) 0.0000006000

(iv) 0.0009000002

Solution:

(i) 569430000000 = 5.6943 x 10^{11}

(ii) 2000.57 = 2.00057 x 10^{13}

(iii) 0.0000006000 = 6.0 x 10^{-7}

(iv) 0.0009000002 = 9.000002 x 10^{-4}

Question 2.

Write the following numbers in decimal form:

(i) 3.459 x 10^{6}

(ii) 5.678 x 10^{4}

(iii) 1.00005 x 10^{-5}

(iv) 2.530009 x 10^{-7}

Solution:

(i) 3.459 x 10^{6} = 3459000

(ii) 5.678 x 10^{4} = 56780

(iii) 1.00005 x 10^{-5} = 0.0000100005

(iv) 2.530009 x 10^{-7} = 0.0000002530009

Question 3.

Represent the following numbers in scientific notation:

(i) (300000)^{2} x (20000)^{4}

(ii) (0.000001)^{11} ÷ (0. 005)^{3}

(iii) {(0.00003)^{6} x (0.00005)^{4} } ÷ {(0. 009)^{3} x (0.05)^{2} }

Solution:

(i) (300000)^{2} x (20000)^{4} = (3.0 x 10^{5})^{2} x (2.0 x 10^{4})^{4}

= 3^{2} x 10^{10} x 2^{4} x 10^{16}

= 9 x 16 x 10^{10} + 16

= 144 x 10^{26} = 1.44 x 10^{28}

(ii) (0.000001)^{11} ÷ (0. 005)^{3}

= 1 x 10^{-66+9} x 5^{3} = 125 x 10^{-57}

= 1.25 x 10^{2} x 10^{-57}

= 1.25 x 10^{(-57+2)}

= 1.25 x 10^{-55}

(iii) {(0.00003)^{6} x (0.00005)^{4}} ÷ {(0. 009)^{3}x (0.05)^{2}}

= 54^{4-2} x 10^{-5+13}

= 5^{2} x 10^{-37} = 25 x 10^{-37} = 2.5 x 10^{1} x 10^{-37} = 2.5 x 10^{-36}

Question 4.

Represent the following information in scientific notation:

(i) The world population is nearly 7000,000,000.

(ii) One light year means the distance 9460528400000000 km.

(iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg.

Solution:

(i) The world population is nearly 7000,000,000 = 7.0 x 10^{9}

(ii) One light year means the distance 9460528400000000 km = 9.4605284 x 10^{15} km

(iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg. = 9.1093822 x 10^{-31} kg

Question 5.

Simplify:

(i) (2.75 x 10^{7}) + (1.23 x 10^{8})

(ii) (1.598 x 10^{17}) – (4.58 x 10^{15})

(iii) (1.02 x 10^{10}) x (1.20 x 10^{-3})

(iv) (8.41 x 10^{4}) ÷ (4.3 x 10^{5})

Solution:

(i) 2.75 x 10^{7} = 27500000

1.23 x 108 – 123000000

(2.75 x 10^{7}) + (1.23 x 10^{8}) = 27500000 + 123000000

= 150500000

= 1.505 x 10^{8}

(ii) (1.598 x 10^{17}) – (4.58 x 10^{15})

1.598 x 10^{17} = 159800000000000000

4.58 x 10^{15}= 4580000000000000

(1.598 x 10^{17}) – (4.58 x 10^{15})

= 159800000000000000

(-)4580000000000000

= 155220000000000000

= 1.5522 x 10^{17}

(iii) (1.02 x 10^{10}) x (1.20 x 10^{-3}) = 1.02 x 10^{10} x 1.20 x 10^{-3}

= 1.02 x 1.20 x 10^{10-3} = 1.224 x 10^{7}

(iv) (8.41 x 10^{4}) ÷ (4.3 x 10^{5})

= 1.95581395 x 101 x 10^{2-4}

= 1.95581395 x 10^{1} x 10^{-2}

= 1.95581395 x 10^{-1}

= 0. 195581395