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NCERT Solutions for Class 7 Maths - Exponents and Powers- 1
Exercise 11.1
Q1: Find the value of:
(i) 26
(ii) 93
(iii) 112
(iv) 54
Ans:
(i) 26 = 2 x 2 x 2 x 2 x 2 x 2 = 64
(ii) 93 = 9 x 9 x 9 = 729
(iii) 112 = 11 x 11= 121
(iv) 54 = 5 x 5 x 5 x 5 = 625
Q2: Express the following in exponential form:
(i) 6 x 6 x 6 x 6
(ii) t x t
(iii) b x b x b x b
(iv) 5 x 5 x 7 x 7 x 7
(v) 2 x 2 x a x a
(vi) a x a x a x c x c x c x c x d
Ans:
(i) 6 x 6 x 6 x 6 = 64
(ii) t x t = t2
(iii) b x b x b x b = b4
(iv) 5 x 5 x 7 x 7 x 7 = 52 x 73
(v) 2 x 2 x a x a = 22 x a2
(vi) a x a x a x c x c x c x c x d = a3 x c4 x d
Q3: Express each of the following numbers using exponential notation:
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Ans:
(i) 512
512 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 29
(ii) 343
343 = 7 x 7 x 7 = 73 
(iii) 729
729 = 3 x 3 x 3 x 3 x 3 x 3 = 36 
(iv) 3125
3125 = 5 x 5 x 5 x 5 x 5 = 55 
Q4: 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
Ans:
(i) 43 = 4 x 4 x 4 = 64
34 = 3 x 3 x 3 x 3 = 81
Since 64 < 81
Thus, 34 is greater than 43.
(ii) 53 = 5 x 5 x 5 = 125
35 = 3 x 3 x 3 x 3 x 3 = 243
Since, 125 < 243
Thus, 35 is greater than 53.
(iii) 28= 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 256
82 = 8 x 8 = 64
Since, 256 > 64
Thus, 28 is greater than 82.
(iv) 1002 = 100 x 100 = 10,000
2100 = 2 x 2 x 2 x 2 x 2 x.......14 times x....x 2 = 16,384 x....x 2
Since, 10,000 < 16,384 x....x2
Thus, 2100 is greater than 1002.
(v) 210 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 = 1,024
102 = 10x10 = 100
Since, 1,024 > 100
Thus, 210 greater than 102
Q5: Express each of the following as product of powers of their prime factors:
(i) 648
(ii) 405
(iii) 540
(iv) 3,600
Ans:
(i) 648 = 23 x 34
(ii) 405 = 5 x 34 
(iii) 540 = 22 x 33 x 5 
(iv) 3,600 = 24 x 32 x 52 
Q6: Simplify:
(i) 2 x 103
(ii) 72 x 22
(iii) 23 x 5
(iv) 3 x 44
(v) 0 x 102
(vi) 52 x 33
(vii) 24 x 32
(viii) 32 x 104
Ans:
(i) 2 x 103 = 2 x 10 x 10 x 10 = 2,000
(ii) 72 x 22 = 7 x 7 x 2 x 2 = 196
(iii) 23 x 5 = 2 x 2 x 2 x 5 = 40
(iv) 3 x 44 = 3 x 4 x 4 x 4 x 4 = 768
(v) 0 x 102 = 0 x 10 x 10 = 0
(vi) 52 x 33 = 5 x 5 x 3 x 3 x 3 = 675
(vii) 24 x 32 = 2 x 2 x 2 x 2 x 3 x 3 = 144
(viii) 32 x 104 = 3 x 3 x 10 x 10 x 10 x 10 = 90,000
Q7: Simplify:
(i) (-4)3
(ii) (-3)x(-2)3
(iii) (-3)2 x (-5)2
(iv) (-2)3 x (-10)3
Ans:
(i) (-4)3 = (-4) x (-4) x (-4) = -64
(ii) (-3) x (-2)3 =(-3) x (-2) x (-2) x (-2) = 24
(iii) (-3)2 x (-5)2 = (-3) x (-3) x (-5) x (-5) = 225
(iv) (-2)3x (-10)3 =(-2) x (-2) x (-2) x (-10) x (-10) x (-10) = 8,000
Question 8: Compare the following numbers:
(i) 2.7 x 1012; 1.5 x 108
(ii) 4 x 1014; 3 x 1017
Answer 8:
(i) 2.7 x 1012 and 1.5 x 108
On comparing the exponents of base 10,
2.7 x 1012 > 1.5x108
(ii) 4 x 1014 and 3 x 1017
On comparing the exponents of base 10,
4 x 1014 < 3 x 1017
Exercise 11.2
Question 1: Using laws of exponents, simplify and write the answer in exponential form:
(i) 32 x 34 x 38
(ii) 615 ÷ 610
(iii) a3 x a2
(iv) 7x x 72
(v) (52)2 ÷ 53
(vi) 25 x 55
(vii) a4 x b4
(viii) (34)3
(ix) [220 ÷ 215) x 23
(x) 8t x 82
Answer 1:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
(viii) 
(ix) 
(x) 
Q2: Simplify and express each of the following in exponential form:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
(viii) 
(ix) 
(x) 
(xi) 
(xii) 
Ans:
(i) 
(ii) 
(iii) 
(iv) 
(v) 
(vi) 
(vii) 
(viii) 
(ix) 
(x) 
(xi) 
(xii) 
Q3: Say true or false and justify your answer:
(i) 10 x 1011 = 10011
(ii) 23 > 52
(iii) 23x 32 = 65
(iv) 30 = (1000)0
Ans:
(i) 10x1011 = 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 x32 = 65
L.H.S. 23 x 32 = 8 x 9 = 72 and R.H.S. 65 =7776
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.
Q4: Express each of the following as a product of prime factors only in exponential form:
(i) 108 x 192
(ii) 270
(iii) 729 x 64
(iv) 768
Ans:
(i) 108 × 192
= (2 × 2 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 3)
= (2² × 3³) × (2⁶ × 3)
= 2⁸ × 3⁴
(ii) 270 = 2 × 3 × 3 × 3 × 3 × 5 = 2 × 3³ × 5
(iii) 729 × 64 = (3 × 3 × 3 × 3 × 3 × 3) × (2 × 2 × 2 × 2 × 2 × 2)
= 3⁶ × 2⁶
(iv) 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 = 2⁸ × 3
Question 5: Simplify:
(i) 
(ii) 
(iii) 
Answer 5:
(i) 
(ii) 
(iii) 
Exercise 11.3
Q1: Write the following numbers in the expanded form:(i) 279404
(ii) 3006194
(iii) 2806196
(iv) 120719
(v) 20068
Ans:
(i) 2,79,404
= 2,00,000 + 70,000 + 9,000 + 400 + 00 + 4
= 2x100000 + 7x 10000+ 9x1000 + 4 x 100+ 0x 10 + 4x1
= 2x105+7x104 +9x103 + 4 x102+ 0x101+4x100
(ii) 30,06,194
= 30,00,000 + 0 + 0 + 6,000 + 100 + 90 + 4
= 3 x 1000000 + 0 x 100000 + 0 x 10000 + 6 x 1000 + 1x100 + 9x10 + 4x1
= 3x106 + 0x105 +0x104 +6 x103+1x102 +9x10+4x100
(iii) 28,06,196
= 20,00,000 + 8,00,000 + 0 + 6,000 + 100 + 90 + 6
= 2x1000000 + 8x 100000+ 0x 10000 +6 x 1000 + 1 x 100 + 9 x 10 + 6 x 1
= 2 x106+8x105+0x104 + 6x103+1x102+9x10+6x100
(iv) 1,20,719
= 1,00,000 + 20,000 + 0 + 700 + 10 + 9
= 1 x 100000 + 2 x 10000 + 0x 1000 + 7x100 + 1x 10+9x1
= 1 x 105 + 2 x 104 + 0 x 103 + 7 x 102 + 1 x 101 + 9 x 100
(v) 20,068
= 20,000 + 00 + 00 + 60 + 8
= 2 x 10000 + 0 x 1000 + 0 x 100 + 6 x 10 + 8 x 1
Q2: Find the number from each of the following expanded forms:
(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100
(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100
(c) 3 x 104 + 7 x 102 + 5 x 100
(d) 9 x 105 + 2 x 102 + 3 x 101
Ans:
(a) 8 x 104 + 6 x 103 + 0 x 102 + 4 x 101 + 5 x 100
= 8 x 10000 + 6 x 1000 + 0 x 100 + 4 x 10 + 5 x 1
= 80000 + 6000 + 0 + 40+5
= 86,045
(b) 4 x 105 + 5 x 103 + 3 x 102 + 2 x 100
= 4 x 100000 + 0 x 10000 + 5 x 1000 + 3 x 100 + 0 x 10 + 2 x 1
= 400000 + 0 + 5000 + 3000 + 0 + 2
= 4,05,302
(c) 3 x 104 + 7 x 102 + 5 x 100
= 3 x 10000 + 0 x 1000 + 7 x 100 + 0 x 10 + 5 x 1
= 30000 + 0 + 700 + 0 + 5
= 30,705
(d) 9x 105 + 2 x 102 + 3 x 101
= 9 x 100000 + 0 x 10000 + 0 x 1000 + 2 x 100 + 3 x 10 + 0 x 1
= 900000 + 0 + 0 + 200 + 30 + 0
= 9,00,230
Q3: 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 3:
(i) 5,00,00,000 = 5 x 1,00,00,000 = 5 x 107
(ii) 70,00,000 = 7 x 10,00,000 = 7 x 106
(iii) 3,18,65,00,000 = 31865 x 100000
= 3.1865 x 10000 x 100000 = 3.1865 x 109
(iv) 3,90,878 = 3.90878 x 100000 = 3.90878 x 105
(v) 39087.8 = 3.90878 x 10000 = 3.90878 x 104
(vi) 3908.78 = 3.90878 x 1000 = 3.90878 x 103
Q4: 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 is 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.
Ans :
(a) The distance between Earth and Moon = 384,000,000 m
= 384x 1000000 m
= 3.84 x 100 x 1000000
= 3.84x108 m
(b) Speed of light in vacuum = 300,000,000 m/s
= 3 x 100000000 m/s
= 3x108 m/s
(c) Diameter of the Earth = 1,27,56,000 m
= 12756 x 1000 m
= 1.2756 x 10000 x 1000 m
= 1.2756 x 107 m
(d) Diameter of the Sun = 1,400,000,000 m
= 14 x 100,000,000 m
= 1.4 x 10 x 100,000,000 m
= 1.4 x 109 m
(e) Average of number of stars in a galaxy = 100, 000, 000,000
= 1 x 100,000,000,000
= 1x1011
(f) Years of Universe = 12,000,000,000 years
= 12 x 1000,000,000 years
= 1.2 x 10 x 1000,000,000 years
= 1.2 x 1010 years
(g) Distance of the Sun from the centre of the Milky Way Galaxy
= 300,000,000,000,000,000,000 m
= 3 x 100,000,000,000,000,000,000 m
= 3 x 1020 m
(h) Number of molecules in a drop of water weighing 1.8 gm
= 60,230,000,000,000,000,000,000
= 6023 x 10,000,000,000,000,000,000
= 6.023 x 1000 x 10,000,000,000,000,000,000
= 6.023 x1022
(i) The Earth has Sea water = 1,353,000,000 km3
= 1,353 x 1000000 km3
= 1,353 x 1000 x 1000,000 km3
= 1.353x109 km3
(j) The population of India = 1,027,000,000
= 1027x1000000 = 1.027x1000x1000000
= 1.027x109
The document NCERT Solutions for Class 7 Maths - Exponents and Powers- 1 is a part of the Class 7 Course Mathematics (Maths) Class 7.
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FAQs on NCERT Solutions for Class 7 Maths - Exponents and Powers- 1
| 1. What are the basic rules of exponents? |  |
| 2. How do you simplify expressions with exponents? | |
Ans. To simplify expressions with exponents, apply the exponent rules systematically. Combine like terms by adding or subtracting the exponents as per the product and quotient rules, and ensure to carry out operations in the correct order. For example, to simplify \(2^3 \times 2^2\), use the product of powers rule: \(2^{3+2} = 2^5 = 32\).
| 3. What is the significance of zero and negative exponents? | |
Ans. Zero exponent states that any non-zero number raised to the power of zero equals one: \(a^0 = 1\). Negative exponents indicate the reciprocal of the base raised to the opposite positive exponent: \(a^{-n} = \frac{1}{a^n}\). These concepts help in simplifying expressions and solving equations involving exponents.
| 4. How can I solve problems involving fractional exponents? | |
Ans. Fractional exponents can be expressed as roots. For instance, \(a^{\frac{m}{n}}\) can be rewritten as \(\sqrt[n]{a^m}\). To solve problems, convert the fractional exponent into its radical form and then simplify as needed. For example, \(x^{\frac{1}{2}} = \sqrt{x}\).
| 5. Can you provide an example of an exponential growth problem? | |
Ans. Yes! An example of an exponential growth problem is: If a bacteria culture doubles every hour and starts with 100 bacteria, how many will there be after 5 hours? The formula is \(N = N_0 \times 2^t\), where \(N_0\) is the initial amount, and \(t\) is the time in hours. Thus, \(N = 100 \times 2^5 = 100 \times 32 = 3200\) bacteria after 5 hours.
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