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Tamilnadu Samacheer Kalvi 9th 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 × 1011
(ii) 2000.57 = 2.00057 × 1013
(iii) 0.0000006000 = 6.0 × 10-7
(iv) 0.0009000002 = 9.000002 × 10-4

Question 2.
Write the following numbers in decimal form:
(i) 3.459 × 106
(ii) 5.678 × 104
(iii) 1.00005 × 10-5
(iv) 2.530009 × 10-7
Solution:
(i) 3.459 × 106 = 3459000
(ii) 5.678 × 104 = 56780
(iii) 1.00005 × 10-5 = 0.0000100005
(iv) 2.530009 × 10-7 = 0.0000002530009

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.8

Question 3.
Represent the following numbers in scientific notation:
(i) (300000)2 × (20000)4
(ii) (0.000001)11 ÷ (0. 005)3
(iii) {(0.00003)6 × (0.00005)4 } ÷ {(0. 009)3 × (0.05)2 }
Solution:
(i) (300000)2 x (20000)4 = (3.0 x 105)2 x (2.0 x 104)4
= 32 × 1010 × 24 × 1016
= 9 × 16 × 1010+16
= 144 × 1026 = 1.44 × 1028

(ii) (0.000001)11 ÷ (0. 005)3
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 1
= 1 x× 10-66+9 × 53
= 125 × 10-57
= 1.25 × 102 × 10-57
= 1.25 × 10(-57+2)
= 1.25 × 10-55

(iii) {(0.00003)6 x (0.00005)4} ÷ {(0. 009)3x (0.05)2}
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 2
= 54-2 × 10-50+13
= 52 × 10-37 = 25 × 10-37 = 2.5 × 101 × 10-37 = 2.5 × 10-36

Samacheer Kalvi 9th Maths Solutions Chapter 2 Real Numbers Ex 2.8

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 × 109
(ii) One light year means the distance 9460528400000000 km = 9.4605284 × 1015 km
(iii) Mass of an electron is 0.000 000 000 000 000 000 000 000 000 00091093822 kg. = 9.1093822 × 10-31 kg

Question 5.
Simplify:
(i) (2.75 × 107) + (1.23 × 108)
(ii) (1.598 × 1017) – (4.58 × 1015)
(iii) (1.02 × 1010) × (1.20 × 10-3)
(iv) (8.41 × 104) ÷ (4.3 × 105)
Solution:
(i) 2.75 x 107 = 27500000
1.23 x 108 – 123000000
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 3

(ii) (1.598 × 1017) – (4.58 × 1015)
1.598 × 1017 = 159800000000000000
4.58 × 1015= 4580000000000000
(1.598 × 1017) – (4.58 × 1015)
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 4

(iii) (1.02 × 1010) x (1.20 × 10-3) = 1.02 × 1010 × 1.20 × 10-3
= 1.02 × 1.20 × 1010-3 = 1.224 × 107

(iv) (8.41 x 104) ÷ (4.3 x 105)
Samacheer Kalvi 9th Maths Chapter 2 Real Numbers Ex 2.8 5
= 1.95581395 × 101 × 102-4
= 1.95581395 × 101 × 10-2
= 1.95581395 × 10-1
= 0. 195581395

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