NCERT Solutions Class 7 Mathematics Chapter 13 Exponents and Power have been provided below and is also available in Pdf for free download. The NCERT solutions for Class 7 Mathematics have been prepared as per the latest syllabus, NCERT books and examination pattern suggested in Class 7 by CBSE, NCERT and KVS. Questions given in NCERT book for Class 7 Mathematics are an important part of exams for Class 7 Mathematics and if answered properly can help you to get higher marks. Refer to more Chapter-wise answers for NCERT Class 7 Mathematics and also download more latest study material for all subjects. Chapter 13 Exponents and Power is an important topic in Class 7, please refer to answers provided below to help you score better in exams
Chapter 13 Exponents and Power Class 7 Mathematics NCERT Solutions
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Chapter 13 Exponents and Power NCERT Solutions Class 7 Mathematics
Exercise 13.1
Q.1) Find the value of:
(i) 26 (ii) 93 (iii) 112 (iv) 54
Sol.1) (i) 26
2 × 2 × 2 × 2 × 2 × 2 = 64
(ii) 93
9 × 9 × 9 = 729
(iii) 112
11 × 11 = 121
(iv) 54
5 × 5 × 5 × 5 = 65
Q.2) Express the following in exponential form:
(i) 6 × 6 × 6 × 6 (ii) 𝑡 × 𝑡 (iii) b × b × b × b (iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a (vi) a × a × a × c × c × c × c × 𝑑
Sol.2) (i) 6 × 6 × 6 × 6
64
(ii) 𝑡 × 𝑡
𝑡2
(iii) b × b × b × b
b4
(iv) 5 × 5 × 7 × 7 × 7
52 × 73
(v) 2 × 2 × a × a
22 × a2
(vi) a × a × a × c × c × c × c × 𝑑
c3 × c4 × 𝑑
Q.3) Express each of the following numbers using exponential notation:
(i) 512 (ii) 343 (iii) 729 (iv) 3125
Sol.3) i) 512
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
(ii) 343
= 7 × 7 × 7 = 73
(iii) 729
= 3 × 3 × 3 × 3 × 3 × 3 = 36
(iv) 3125
= 5 × 5 × 5 × 5 × 5 = 55
Q.4) Identify the greater number, wherever possible, in each of the following:
(i) 43 and 34 (ii) 53 or 35 (iii) 28 or 82
(iv) 1002 or 2100 (v) 210 or 102
Sol.4) (i) 43 and 34
43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Since 64 < 81
Hence, 43 < 34
Thus, 34 is greater than 43.
(ii) 53 or 35
53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
Hence, 53 < 35
Since, 125 < 243
Thus, 35 is greater than 53.
(iii) 28 or 82
28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8 × 8 = 64
Hence, 28 > 82
Since, 256 > 64 Thus, 28is greater than 82.
(iv) 1002 or 2100
= 100 × 100 = 10,000
= 2 × 2 × 2 × 2 × 2 × … . .14 t𝑖m𝑒𝑠 × … × 2 = 16,384 × … . .× 2
Since, 10,000 < 16,384 × … . .× 2
Thus, 2100is greater than 1002.
(v) 210 or 102
= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
= 10 × 10 = 100
Since, 1,024 > 100
Thus, 210 > 102
Q.5) Express each of the following as product of powers of their prime factors:
(i) 648 (ii) 405 (iii) 540 (iv) 3,600
Sol.5) (i) 648
= 2 × 2 × 2 × 2 × 3 × 3 × 3
= 24 × 33
(ii) 405
= 3 × 3 × 3 × 3 × 5
= 34 × 5
(iii) 540
= 2 × 2 × 3 × 3 × 3 × 5
= 22 × 33 × 5
(iv) 3,600
= 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 24 × 32 × 52
Q.6) Simplify:
(i) 2 × 103 (ii) 72 × 22 (iii) 23 × 5 (iv) 3 × 44
(v) 0 × 102 (vi) 52 × 33 (vii) 24 × 32 (viii) 32 × 104
Sol.6) i) 2 × 103
= 2 × 10 × 10 × 10
= 2 × 1000 = 2000
(ii) 72 × 22
= 7 × 7 × 2 × 2 = 196
(iii) 23 × 5
= 2 × 2 × 2 × 5 = 40
(iv) 3 × 44
= 3 × 4 × 4 × 4 × 4 = 768
(v) 0 × 102
= 0 × 10 × 10 = 0
(vi) 52 × 33
= 5 × 5 × 3 × 3 × 3 = 675
(vii) 24 × 32
= 2 × 2 × 2 × 2 × 3 × 3 = 144
(viii) 32 × 104
= 3 × 3 × 10 × 10 × 10 × 10 = 90,000
Q.7) Simplify:
i) (−4)3
ii) (−3) × (−2)3
iii) (−3)2 × (−5)2
iv) (−2)3 × (−10)3
Sol.7) i) (−4)3
= (−4) × (−4) × (−4) = −64
ii) (−3) × (−2)3
= (−3) × (−2) × (−2) = 24
iii) (−3)2 × (5)2
= (−3) × (−3) × (−5) × (−5) = 225
iv) (−2)3 × (−10)3
= (−2) × (−2) × (−2) × (−10) × (−10) × (−10) = 8000
Q.8) Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108 (ii) 4 × 1014; 3 × 1017
Sol.8) (i) 2.7 × 1012; 1.5 × 108
because exponent on 10 is larger in case of first number
On comparing the exponents of base 10,
2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014; 3 × 1017
because exponent on 10 is smaller in case of first number.
On comparing the exponents of base 10
4 × 1014 < 3 × 1017
Exercise 13.2
Q.1) Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 × 38 (ii) 615 ÷ 610 (iii) a3 × a2 (iv) 7𝑥 × 72
(v) (52)3 × 53 (vi) 25 × 55 (vii) a4 × b4 (viii) (34)3
(ix) (220 ÷ 215) × 23 (x) 8t ÷ 82
Sol.1) (i) 32 × 34 × 38 = (3)2+4+8 (am × an = am+n)
= 314
(ii) 615 ÷ 610 = (6)15−10 (am ÷ an = am−n)
= 65
(iii) a3 × a2 = a(3+2) (am × an = am+n)
= a5
(iv) 7x × 72 = 7𝑥+2 (am × an = am+n)
(v) (52)3 ÷ 53
= 52×3 ÷ 53 (am)n = amn
= 56 ÷ 53 (am ÷ an = am−n)
= 56−3 (am ÷ an = am−n)
= 53
(vi) 25 × 55
= (2 × 5)5 [am × bm = (a × b) m ]
= 105
(vii) a4 × b4
= (ab)4 [am × bm = (a × b) m ]
(viii) (34)3 = 34×3 = 312 (am)n = amn
(ix) (220 ÷ 215) × 23
= (220−15) × 23 (am ÷ an = am−n)
= 25 × 23
= (25+3) = 28 (am × an = am+n)
(x) 8t ÷ 82 = 8t−2 (am ÷ an = am−n)
Q.2) Simplify and express each of the following in exponential form:
i) 23×34×4/3×32 ii) [(52)3 × 54] ÷ 57 iii) 254 ÷ 53 iv) 3×72×118/21×11
v) 37/34×33 vi) 20 + 30 + 40 vii) 20 × 30 × 40 viii) (30 + 20) × 50
ix) 28×α5/43×α3 x) ( α5/α3) × α8 xi) 45×α8b3/45×α5b2 xii) (23 × 2)2
Sol.2) i) 23×34×4/3×32
= 23×34×22/3×25 = 23+2×34/3×52 [αm × α𝑛 = αm+𝑛]
= 25×34/3×25 = 25−5 × 34−3 [αm ÷ α𝑛 = αm−𝑛]
= 20 × 33 = 1 × 33 = 33
ii) [(52)3 × 54] ÷ 57
= [56 × 54] ÷ 57 [(αm)𝑛 = αm×𝑛]
= [56+4] ÷ 57 [αm × α𝑛 = αm+𝑛]
= 510 ÷ 57
= 510−7 = 53 [αm ÷ α𝑛 = αm−𝑛]
iii) 254 ÷ 53
= (52)4 ÷ 53 = 58 ÷ 53 [(αm)𝑛 = αm×𝑛]
= 58−3 = 55 [αm ÷ α𝑛 = αm−𝑛]
iv) 3×72×118/21×11
= 3×72×118/3×7×113
= 31−1 × 72−1 × 118−3 [αm ÷ α𝑛 = αm−𝑛]
= 30 × 71 × 115 = 7 × 115
v) 37/34×33 = 37
34+3 = 37/37 [αm × α𝑛 = αm+𝑛]
= 37−7 = 30 = 1 [αm ÷ α𝑛 = αm−𝑛]
vi) 20 + 30 + 40 [α0 = 1]
= 1 + 1 + 1 = 3
vii) 20 × 30 × 40 [α0 = 1]
= 1 × 1 × 1 = 1
viii) (30 + 20) × 50 [α0 = 1]
= (1 + 1) × 1 = 2 × 1 = 2
ix) 28×α5/43×α3
= 28×α5/(22)3×α3 = 28×α5/26×α3 [(αm)𝑛 = αm×𝑛]
= 28−6 × α5−2 = 22 × α2 [αm ÷ α𝑛 = αm−𝑛]
= (2α)2 [αm × b𝑚 = (α × b)𝑚]
x) (α5/α3) × α8
= (α5−3) × α8 = α2 × α8 [αm ÷ α𝑛 = αm−𝑛]
= α2+8 = α10 [αm × α𝑛 = αm+𝑛]
xi) 45×α8b3/45×α5b2
= 45−5 × α8−5 × b3−2 = 40 × α3 × b [αm ÷ α𝑛 = αm−𝑛]
= 1 × α3 × b = α3b [α0 = 1]
xii) (23 × 2)2
= (23+1)2 = (24)2 [αm × α𝑛 = αm+𝑛]
= 24×2 = 28
Q.3) Say true or false and justify your answer:
i) 10 × 1011 = 10011 ii) 23 > 52 iii) 23 × 32 = 65 iv) 30 = (1000)0
Sol.3) i) 10 × 1011 = 10011
LHS 101+11 = 1012 and RHS (102)11 = 1022
False, because first number will have 12 zeroes while the second number will have 22 zeroes at
the end.
Since, 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆
∴ given statement is false.
ii) 23 > 52
LHS 23 = 8 and RHS 52 = 25
Since, 𝐿𝐻𝑆 ≠ 𝑅𝐻𝑆
False, because first number is 8 while second number is 25.
∴ given statement is false.
iii) 23 × 32 = 65
LHS 23 × 32 = 8 × 9 = 72 and RHS 65 = 7,776
Since, LHS is not greater than RHS
∴ given statement is false.
iv) 30 = (1000)0
LHS 30 = 1 and RHS (1000)0 = 1
Since, 𝐿𝐻𝑆 = 𝑅𝐻𝑆
True, because both are equal to 1.
∴ given statement is true.
Q.4) Express each of the following as a product of prime factors only in exponential form:
i) 108 × 192 ii) 270 iii) 729 × 64 iv) 768
Sol.4) i) 108 × 192
= (22 × 33) × (26 × 3)
= 22+6 × 33+1
= 28 × 34
= 25−5 × 35−5 × 55−5
= 20 × 30 × 3^ 0
= 1 × 1 × 1 = 1
Exercise 13.3
Q.1) Write the following numbers in the expanded forms :
279404, 3006194, 2806196, 120719, 20068
Sol.1) (i) 2,79,404
= 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4
= 2 × 100000 + 7 × 10000 + 9 × 1000 + 4 × 100 + 0 × 10 + 4 × 1
= 2 × 105 + 7 × 104 + 9 × 103 + 4 × 102 + 0 × 101 + 4 × 100
(ii) 30,06,194
= 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4
= 3 × 1000000 + 0 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 4 × 1
= 3 × 106 + 0 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 10 + 4 × 100
(iii) 28,06,196
= 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6
= 2 × 1000000 + 8 × 100000 + 0 × 10000 + 6 × 1000 + 1 × 100 + 9 × 10 + 6 × 1
= 2 × 106 + 8 × 105 + 0 × 104 + 6 × 103 + 1 × 102 + 9 × 10 + 6 × 100
(iv) 1,20,719
= 1,00,000 + 20,000 + 0 + 700 + 10 + 9
= 1 × 100000 + 2 × 10000 + 0 × 1000 + 7 × 100 + 1 × 10 + 9 × 1
= 1 × 105 + 2 × 104 + 0 × 103 + 7 × 102 + 1 × 101 + 9 × 100
(v) 20,068
= 20,000 + 00 + 00 + 60 + 8
= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1
= 20 × 104 + 0 × 103 + 0 × 102 + 6 × 101 + 8 × 100
Q.2) Find the number from each of the following expanded forms :
i) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
ii) 4 × 105 + 5 × 104 + 3 × 102 + 2 × 100
iii) 3 × 104 + 7 × 102 + 5 × 100
iv) 9 × 105 + 2 × 102 + 3 × 101
Sol.2) i) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
= 8 × 10000 + 6 × 1000 + 0 × 100 + 4 × 10 + 5 × 1
= 80000 + 6000 + 0 + 40 + 5
= 86,045
ii) 4 × 105 + 5 × 104 + 3 × 102 + 2 × 100
= 4 × 100000 + 0 × 10000 + 5 × 1000 + 3 × 100 + 0 × 10 + 2 × 1
= 400000 + 0 + 5000 + 3000 + 0 + 2
= 4,05,302
iii) 3 × 104 + 7 × 102 + 5 × 100
= 3 × 10000 + 0 × 1000 + 7 × 100 + 0 × 10 + 5 × 1
= 30000 + 0 + 700 + 0 + 5
= 30,705
iv) 9 × 105 + 2 × 102 + 3 × 101
= 9 × 100000 + 0 × 10000 + 0 × 1000 + 2 × 100 + 3 × 10 + 0 × 1
= 900000 + 0 + 0 + 200 + 30 + 0
= 9,00,230
Q.3) Express the following numbers in standard form:
(i) 5,00,00,000 (ii) 70,00,000 (iii) 3,18,65,00,000
(iv) 3,90,878 (v) 39087.8 (vi) 3908.78
Sol.3) (i) 5,00,00,000 = 5 × 1,00,00,000 = 5 × 107
(ii) 70,00,000 = 7 × 10,00,000 = 7 × 106
(iii) 3,18,65,00,000 = 31865 × 100000
= 3.1865 × 10000 × 100000 = 3.1865 × 109
(iv) 3,90,878 = 3.90878 × 100000 = 3.90878 × 105
(v) 39087.8 = 3.90878 × 10000 = 3.90878 × 104
(vi) 3908.78 = 3.90878 × 1000 = 3.90878 × 103
Q.4) Express the number appearing in the following statements in standard form:
(a) The distance between Earth and Moon is 384,000,000 m.
(b) Speed of light in vacuum is 300,000,000 m/s. (c) Diameter of Earth id 1,27,56,000 m.
(d) Diameter of the Sun is 1,400,000,000 m. (e) In a galaxy there are on an average
100,000,000,0000 stars.
(f) The universe is estimated to be about 12,000,000,000 years old.
(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated to be
300,000,000,000,000,000,000 m.
(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8
gm.
(i) The Earth has 1,353,000,000 cubic km of sea water.
(j) The population of India was about 1,027,000,000 in march, 2001.
Sol.4) (a) The distance between Earth and Moon
= 384,000,000 m = 384 × 1000000 m
= 3.84 × 100 × 1000000 = 3.84 × 108 m
(b) Speed of light in vacuum
= 300,000,000 m/s = 3 × 100000000 m/s
= 3 × 108 m/s
(c) Diameter of the Earth
= 1,27,56,000 m = 12756 × 1000 m
= 1.2756 × 10000 × 1000 m = 1.2756 × 107 m
(d) Diameter of the Sun
= 1,400,000,000 m = 14 × 100,000,000 m
= 1.4 × 10 × 100,000,000 m = 1.4 × 109 m
(e) Average of Stars
= 100,000,000,000 = 1 × 100,000,000,000
= 1 × 1011
(f) Years of Universe
= 12,000,000,000 years = 12 × 1000,000,000 years
= 1.2 × 10 × 1000,000,000 years = 1.2 × 1010 years
(g) Distance of the Sun from the center of the Milky Way Galaxy
= 300,000,000,000,000,000,000 m
= 3 × 100,000,000,000,000,000,000 m
= 3 × 1020 m
(h) Number of molecules in a drop of water weighing 1.8 𝑔m
= 60,230,000,000,000,000,000,000 = 6023 × 10,000,000,000,000,000,000
= 6.023 × 1000 × 10,000,000,000,000,000,000 = 6.023 × 1022
(i) The Earth has Sea water
= 1,353,000,000 km3 = 1,353 × 1000000 km3
= 1.353 × 1000 × 1000,000 km3 = 1.353 × 109 km3
(j) The population of India
= 1,027,000,000 = 1027 × 1000000
= 1.027 × 1000 × 1000000
= 1.027 × 109
| NCERT Solutions Class 7 Mathematics Chapter 1 Integers |
| NCERT Solutions Class 7 Mathematics Chapter 2 Fractions and Decimals |
| NCERT Solutions Class 7 Mathematics Chapter 3 Data Handling |
| NCERT Solutions Class 7 Mathematics Chapter 4 Simple Equations |
| NCERT Solutions Class 7 Mathematics Chapter 5 Lines and angles |
| NCERT Solutions Class 7 Mathematics Chapter 6 Triangle and its properties |
| NCERT Solutions Class 7 Mathematics Chapter 7 Congruence of Triangle |
| NCERT Solutions Class 7 Mathematics Chapter 8 Comparing Quantities |
| NCERT Solutions Class 7 Mathematics Chapter 9 Rational Numbers |
| NCERT Solutions Class 7 Mathematics Chapter 10 Practical Geometry |
| NCERT Solutions Class 7 Mathematics Chapter 11 Perimeter and Area |
| NCERT Solutions Class 7 Mathematics Chapter 12 Algebraic Expressions |
| NCERT Solutions Class 7 Mathematics Chapter 13 Exponents and Power |
| NCERT Solutions Class 7 Mathematics Chapter 14 Symmetry |
| NCERT Solutions Class 7 Mathematics Chapter 15 Visualizing Solid Shapes |
NCERT Solutions Class 7 Mathematics Chapter 13 Exponents and Power
The above provided NCERT Solutions Class 7 Mathematics Chapter 13 Exponents and Power is available on our website www.studiestoday.com for free download in Pdf. You can read the solutions to all questions given in your Class 7 Mathematics textbook online or you can easily download them in pdf. The answers to each question in Chapter 13 Exponents and Power of Mathematics Class 7 has been designed based on the latest syllabus released for the current year. We have also provided detailed explanations for all difficult topics in Chapter 13 Exponents and Power Class 7 chapter of Mathematics so that it can be easier for students to understand all answers. These solutions of Chapter 13 Exponents and Power NCERT Questions given in your textbook for Class 7 Mathematics have been designed to help students understand the difficult topics of Mathematics in an easy manner. These will also help to build a strong foundation in the Mathematics. There is a combination of theoretical and practical questions relating to all chapters in Mathematics to check the overall learning of the students of Class 7.
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