# ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 4 Exponents and Powers Ex 4.3

## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 4 Exponents and Powers Ex 4.3

Question 1.
Write the following numbers in the standard form (or scientific notation):
(i) 530.7
(ii) 3908.78
(iii) 39087.8
(iv) 2.35
(v) 3,43,000
(vi) 70,00,000
(vii) 3,18,65,00,000
(viii) 893,000,000
(ix) 70,040,000,000
Solution:
(i) 530.7 = 5.307 × 102
(ii) 3908.78 = 3.90878 × 103
(iii) 39087.8 = 3.90878 × 104
(iv) 2.35 = 2.35 × 100 {a0 = 1}
(v) 3,43,000 = 3.43000 × 105
(vi) 70,00,000 = 7.0 × 106
(vii) 3,18,65,00,000 = 3.186500000 × 109
(viii) 893,000,000 = 8.93 × 108
(ix) 70,040,000,000 = 7.004 × 1010

Question 2.
Write the following numbers in usual decimal notation:
(i) 4.7 × 103
(ii) 1.205 × 105
(iii) 1.234 × 106
(iv) 4.87 × 107
(v) 6.05 × 108
(vi) 9.083 × 1011
Solution:
(i) 4.7 × 103 = 4700
(ii) 1.205 × 105 = 120500
(iii) 1.234 × 106 = 1234000
(iv) 4.87 × 107 = 48700000
(v) 6.05 × 108 = 605000000
(vi) 9.083 × 1011 = 908300000000

Question 3.
Express the numbers appearing in the following statements in scientific notation (or standard form):
(i) The distance between the earth and the moon is 384,000,000 m.
(ii) The diameter of the sun is 1,400,000,000m.
(iii) The universe is estimated to be about 12,0000,000 years old.
(iv) In a galaxy, there are on an average 100,000,000,000 stars.
(v) The distance of the sun from the centre of milky way galaxy is estimated to be 300,000,000,000,000,000 km.
(vi) The astronomical distance of unit is a light year (a light year is a distance travelled by the light in one year). A light year is about 9,467,500,000,000 kilometers.
Solution:
(i) The distance between the earth and the moon is 384,000,000 m = 3.84 × 108 m
(ii) The diameter of the sun is 1,400,000,000 m = 1.4 × 108 m
(iii) The universe is estimated to be about 12,000,000,000 years old = 1.2 × 1010
(iv) In a galaxy there are on an average 100000000000 stars = 1.0 × 1011
(v) The distance of the sun from the centre of milky way galaxy is estimated to be 300000000000000000 km = 3.0 × 1017
(vi) The astronomical distance of unit is a light year (a light year is a distance travelled by the light in one year). A light year is about
9467500000000 kilometers = 9.4675 × 1012 km

Question 4.
Compare the following numbers:
(i) 4.3 × 1014; 3.01 × 1017
(ii) 1.439 × 1012; 1.4335 × 1012
Solution:
(i) 4.3 × 1014; 3.01 × 1017
3.01 × 1017 > 4.3 × 1014 {17 > 14}
(ii) 1.439 × 1012; 1.4335 × 1012
1.439 > 1.4335
1.439 × 1012 > 1.4335 × 1012

Question 5.
Write the following numbers in the expanded Exponential form:
(i) 279404
(ii) 3006194
(iii) 28061906
Solution:
(i) 279404 = 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102 + 4 × 100
(ii) 3006194 = 3 × 106 + 6 × 103 + 1 × 102 + 9 × 101 + 4 × 100
(iii) 28061906 = 2 × 107 + 8 × 106 + 6 × 105 + 1 × 103 + 9 × 102 + 6 × 100

Question 6.
Find the number from each of the expanded form:
(i) 3 × 104 + 7 × 102 + 5 × 100
(ii) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
(iii) 8 × 107 + 3 × 104 + 7 × 103 + 5 × 102 + 8 × 101
Solution:
(i) 3 × 104 + 7 × 102 + 5 × 100 = 30705
(ii) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
= 400000 + 5000 + 300 + 2 × 1
= 405302
(iii) 8 × 107 + 3 × 104 + 7 × 103 + 5 × 102 + 8 × 101
= 80000000 + 30000 + 7000 + 500 + 80
= 80037580