Ncert Mathematics Book Class 7 Chapter 13 Exercise Solution
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers| PDF Download
Here you will find Chapter 13 Exponents and Powers Class 7 Maths NCERT Solutions which are helpful resources that can help you not only cover the entire syllabus but also provide in depth analysis of the topics. You can download PDF of Chapter 13 Exponents and Powers NCERT Solutions from this page that can help you in scoring good marks in the exams. It will serve as beneficial tool that can be used to recall various questions any time.
Class 7 NCERT Solutions can be used to enrich knowledge and make lessons for learners more exciting. It is very challenging to score good marks in tests that is why we have prepared NCERT Solutions.
Exercise 13.1
1. Find the value of:
(i) 26
(ii) 93
(iii) 112
(iv) 54
Answer
(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 = 625
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
Answer
(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 = a3 × c4 x d
3. Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Answer
(i) 512
512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 29
| 2 | 512 |
| 2 | 256 |
| 2 | 128 |
| 2 | 64 |
| 2 | 32 |
| 2 | 16 |
| 2 | 8 |
| 2 | 4 |
| 2 | 2 |
| 1 |
(ii) 343
343 = 7 × 7 × 7 = 73
(iii) 729
729 = 3 × 3 × 3 × 3 × 3 × 3 = 36
| 3 | 729 |
| 3 | 243 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
(iv) 3125
3125 = 5 × 5 × 5 × 5 × 5 = 55
| 5 | 3125 |
| 5 | 625 |
| 5 | 125 |
| 5 | 25 |
| 5 | 5 |
| 1 |
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
Answer
(i) 43 = 4 × 4 × 4 = 64
34 = 3 × 3 × 3 × 3 = 81
Since 64 < 81
Thus, 34 is greater than 43.
(ii) 53 = 5 × 5 × 5 = 125
35 = 3 × 3 × 3 × 3 × 3 = 243
Since, 125 < 243
Thus, 34 is greater than 53.
(iii) 28= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
82 = 8×8 = 64
Since, 256 > 64
Thus, 28 is greater than 82.
(iv) 1002 = 100 × 100 = 10,000
2100 = 2 × 2 × 2 × 2 × 2 .......14 times ....× 2 = 16,384 ....× 2
Since, 10,000 < 16,384 ....×2
Thus, 2100 is greater than 1002.
(v) 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1,024
102 = 10×10 = 100
Since, 1,024 > 100
Thus, 210 >102
5. Express each of the following as product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3,600
Answer
(i) 648 = 23 × 34
| 2 | 648 |
| 2 | 324 |
| 2 | 162 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
(ii) 405 = 5 × 34
| 5 | 405 |
| 3 | 81 |
| 3 | 27 |
| 3 | 9 |
| 3 | 3 |
| 1 |
(iii) 540 = 22 × 33 × 5
| 2 | 540 |
| 2 | 270 |
| 3 | 135 |
| 3 | 45 |
| 3 | 15 |
| 5 | 5 |
| 1 |
(iv) 3,600 = 24 × 32 × 52
| 2 | 3600 |
| 2 | 1800 |
| 2 | 900 |
| 2 | 450 |
| 3 | 225 |
| 3 | 75 |
| 5 | 25 |
| 5 | 5 |
| 1 |
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
Answer
(i) 2 × 103 = 2 × 10 × 10 × 10 = 2,000
(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) 53 × 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
7. Simplify:
(i) (-4)3
(ii) (-3)×(-2)3
(iii) (-3)2 × (-5)2
(iv) (-2)3 × (-10)3
Answer
(i) (-4)3 = (-4)×(-4)×(-4) = -64
(ii) (-3)×(-2)3 =(-3)×(-2)×(-2)×(-2) = 24
(iii) (-3)2 × (-5)2 = (-3)×(-3)×(-5) × (-5) = 225
(iv) (-2)3×(-10)3 =(-2)×(-2)×(-2)×(-10)×(-10)×(-10)
8. Compare the following numbers:
(i) 2.7 × 1012; 1.5 × 108
(ii) 4 × 1014; 3 × 1017
Answer
(i) 2.7 × 1012 and 1.5 × 108
On comparing the exponents of base 10,
2.7 × 1012 > 1.5 × 108
(ii) 4 × 1014 and 3 × 1017
On comparing the exponents of base 10,
4 × 1014 < 3 × 1017
1. Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 × 34 x 38
(ii) 615 ÷ 610
(iii) a3 × a2
(iv) 7 × x 72
(v) (52)2 ÷ 53
(vi) 25 × 55
(vii) a4 × b4
(viii) (34)3
(ix) [220 ÷ 215) × 23
(x) 8t × 82
Answer
(i) 32 × 34 × 38 = 3(2+4+8) = 314 [∵ am× an = am+n]
(ii) 615 ÷ 610 = 615-10 = 65 [∵ am ÷ an = am-n]
(iii) a3 × a2 = a3+2 = a5 [∵ am × an = a m+n]
(iv) 7x × 72 = 7x+2 [∵ am × an = am+n]
(v) (52)2 ÷ 53 = 52×3 ÷ 53 = 56 ÷ 53 [∵ (am)n = am×n]
= 56-3 = 53 [∵ am ÷ an = am-n]
(vi) 25 × 25 = (2×5)5= 105 [∵ am×bm = (a×b)m]
(vii) a4 × b4 = (a×b)4 [∵ am×bm = (a×b)m]
(viii) (34)3 = 34×3 = 312 [∵ (am)n = am×n]
(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]
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 × a 5 / 43 × a3
(x) (a5 / a3) × a8
(xi) 45 × a8b3 / 45 × a5b2
(xii) (23 × 2)3
Answer
(i) 23 × 34 × 4 / 3× 32 = 23 × 34 × 22 / 3 × 25 = 23+2 × 34 / 3×25 [∵ am× an = am+n]
= 25 × 34 / 3 × 25 = 25-5 × 34-3 [∵ am ÷ an = am-n]
= 20 × 33 = 1× 33 = 33
(ii) [(52)3 × 54] ÷ 57 [∵ (am)n = am×n]
= [56+4] ÷ 57 = 510 ÷ 57 [∵ am× an = am+n]
= 510-7 = 53 [∵ am ÷ an = am-n]
(iii) 254 ÷ 53 = (52)4 ÷ 53 = 58 ÷ 53 [∵ (am)n = am×n]
= 58-3 = 55 [∵ am ÷ an = am-n]
(iv) 3 × 72 × 118 / 21 × 112 = 3 × 72 × 118 / 3 × 7 × 113 = 31-1 × 72-1 × 118-3 [∵ am ÷ an = am-n]
= 30 × 71 × 115 = 7 × 115
(v) 37 / 34 × 33 = 37 / × 34+3 = 37 / 37 [∵ am× an = am+n]
= 37-7 = 30 = 1 [∵ am× an = am+n]
(vi) 20 + 30 + 40 + 1+1+1 = 3 [∵ a0= 1]
(vii) 20 × 30 × 40 = 1×1×1 = 1 [∵ a0= 1]
(viii) (30 + 20) × 50 = (1+1)×1 = 2×1 = 2 [∵ a0= 1]
(ix) 28 × a5 / 43 × a3 = 28 × a5/ (22)3 × a3 = 28×a5 / 26 × a3 [∵ (am)n = am×n]
= 28-6 × a5-2 = 22 × a2 [∵ am ÷ an = am-n]
= (2a)2 [∵ am×bm = (a×b)m]
(x) (a5 / a3) × a8 = (a 5-3) × a8 = a2 × a8 [∵ am ÷ an = am-n]
= a2+8 = a10 [∵ am× an = am+n]
(xi) 45 × a8b3 / 45 × a5b2 = 45-5 × a8-5 × b3-2 = 40 × a3 × b [∵ am ÷ an = am-n]
= 1 × a3 × b = a3b [∵ a0= 1]
(xii) (23 × 2)2 = (23+1)2 = (24)2 [∵ am× an = am+n]
= 24×2 = 28
3. Say true or false and justify your answer:
(i) 10 x 1011 = 10011
(ii) 23 > 52
(iii) 23× 32 = 6s
(iv) 30 = (1000)0
Answer
(i) 10×1011 = 10011
L.H.S. 101+11 = 1012 and R.H.S. (102)11=1022
Since, L.H.S. ≠ R.H.S.
Therefore, it is false.
(ii) 23 > 52
L.H.S 23 = 8 and R.H.S. 52 = 25
Since, L.H,S. is not greater than R.H.S.
Therefore, it is false,
(iii) 23 ×32 = 65
L.H.S. 23 × 32 = 8 x 9 = 72 and R.H.S. 65 =7.776
Since, L.H.S. ≠ R.H.S.
Therefore, it is false.
(iv) 30 = (1000)0
L.H.S. 30 = 1 and R.H.S. (1000)0 = 1
Since, L.H.S. = R.H.S.
Therefore, it is true.
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
Answer
(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2 × 3)
= (22 × 33) × (26 × 3)
= 26+2 × 33+1 (am × an = am+n)
= 28 × 34
(ii) 270 = 2 × 3 × 3 × 3 × 5 = 2 × 33 × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2)
= 36 × 25
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 28 × 3
5. Simplify:
(i) (25)2 × 73 / 83 × 7
(ii) 25 × 52 × t8 / 103 × t4
(iii) 35 × 105 × 25 × 57 × 65
Answer
(i) (25)2 × 73 / 83 × 7 = 25×2 × 73 / (23)3 × 7
= 210 × 73 / 29 × 7
= 210-9 × 73-1
= 2 × 72
= 2 × 49
= 98
(ii) 25 × 52 ×t8 / 103 - t4 = 52 × 52 × t8 / (5×2)3 × t4
= 52+2 × t8-4/23×33
= 54 × t4 / 23 × 53
= 54-3 × t4 / 23
= 5t4 / 8
(iii) 35 × 105 × 25 / 57 × 65= 35 ×(2×5)5 × 52 / 57 × (2×3)5
= 35 × 25 × 55 × 52 / 57 × 25 × 35
= 35 × 25 × 55+2 / 57 × 25 × 35
= 35 × 25 × 57 / 57 × 25 × 35
= 25-5 × 35-5 × 55-5
= 20 × 30 × 50
= 1 × 1 × 1
= 1
1. Write the following numbers in the expanded form:
(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068
Answer
(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) 3006194
= 3000000 + 0 + 0 + 6000 + 100 + 90 + 4
= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 4
= (3 × 106) + (0 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (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+1x102+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 x 103 + 7 × 102 + 1 x 101 + 9 × 100
(v) 20,068= 20,000 + 00 + 00 + 60 + 8
= 2 × 10000 + 0 × 1000 + 0 × 100 + 6 × 10 + 8 × 1
= 2×104 + 0 × 103 + 0 × 102 + 6 x 101 + 8 × 100
2. Find the number from each of the following expanded forms:
(a) 8 × 104 + 6 × 103 + 0 × 102 + 4 × 101 + 5 × 100
(b) 4 × 105 + 5 × 103 + 3 × 102 + 2 × 100
(c) 3 × 103 + 7 × 102 + 5 × 100
(d) 9 × 105 + 2 × 102 + 3 × 101
Answer
(a) 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
(b) 4 × 105 + 5 × 103 + 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
(c) 3 × 104 + 7 × 102 + 5 × 100
= 3 × 10000 + 0 × 1000 + 7 × 100 + 0 × 10 + 5 × 1
= 30000 + 0 + 700 + 0 + 5
= 30,705
(d) 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
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
Answer
(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
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.
Answer
(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 = 1,27,50,000 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 centre 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.
Go Back To NCERT Solutions for Class 7 Maths
NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers
Chapter 13 NCERT Solutions will help you in improving the marks in the examinations and have edge over your classmates. An exponent or power is a mathematical representation that indicates the number of times that a number is multiplied by itself.
• If a number is multiplied by itself m times, then it can be written as: a x a x a x a x a...m times = am
• The factors of a product can be expressed as the powers of the prime factors of 100.
• The value of an exponential number with a negative base raised to the power of an even number is positive.
You can get exerciswise NCERT Solutions from the links given below that will help you in understanding the key concepts of the chapter properly and increase concentration among students. These NCERT Solutions are updated according to the latest NCERT Class 8 Maths textbook and syllabus.
- Exercise 13.1 Chapter 13 Class 7 Maths NCERT Solutions
- Exercise 13.2 Chapter 13 Class 7 Maths NCERT Solutions
- Exercise 13.3 Chapter 13 Class 7 Maths NCERT Solutions
Class 7 Maths NCERT Solutionsare prepared by subject matter experts of Studyrankers which will provide good experience and provide opportunities to learn new things. It will help you in identify, analyze, and then rectify the mistakes.
NCERT Solutions for Class 7 Maths Chapters:
| Chapter 1 Integers |
| Chapter 2 Fractions and Decimals |
| Chapter 3 Data Handling |
| Chapter 4 Simple Equations |
| Chapter 5 Lines and Angles |
| Chapter 6 The Triangle and its Properties |
| Chapter 7 Congruence of Triangles |
| Chapter 8 Comparing Quantities |
| Chapter 9 Rational Numbers |
| Chapter 10 Practical Geometry |
| Chapter 11 Perimeter and Area |
| Chapter 12 Algebraic Expressions |
| Chapter 14 Symmetry |
| Chapter 15 Visualising Solid Shapes |
FAQ on Chapter 1 3 Exponents and Powers
How many exercises are there in Chapter 13 Exponents and Powers Class 7 Maths NCERT Solutions?
There are total 3 exercises in the Chapter 13 Exponents and Powers which are very important for the purpose of examinations. One should prepare and revise the questions given before any tests and improve their understandings properly.
Write in the standard form: the distance between Earth and Moon is 384,000 km.
Using the standard form, write number 73984 in expanded form.
7 × 104 + 3 ×103 + 9 × 102 + 8 × 101 + 4 × 100.
Write in the standard form: speed of light in vacuum is 300,000,000 m/s.
Ncert Mathematics Book Class 7 Chapter 13 Exercise Solution
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