NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers are given here in detail. Maths is a subject which requires understanding and reasoning skills accompanied by logic. Along with this, it also requires students to practice Maths regularly. Students of Class 7 are suggested to solve NCERT Solutions for Class 7 Maths Chapter 13 to strengthen their fundamentals and be able to solve questions that are usually asked in the examination. A total of 3 exercises are present in the chapter Chapter 13 – Exponents and Powers of NCERT for Class 7 Maths. These exercises covers various concepts mentioned in the chapter, which are given below.

  • Exponents
  • Laws of Exponents
  • Multiplying Powers With The Same Base
  • Dividing Powers With The Same Base
  • Taking Power of a Power
  • Multiplying Powers With The Same Base
  • Dividing Powers With The Same Exponents
  • Miscellaneous Examples Using The Laws of Exponents
  • Decimal Number System
  • Expressing Large Numbers in The Standard Form

Download the PDF of NCERT Solutions For Class 7 Maths Chapter 13 Exponents and Powers

 

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ncert sol class 7 math ch 13 ex 1
ncert sol class 7 math ch 13 ex 1
ncert sol class 7 math ch 13 ex 1
ncert sol class 7 math ch 13 ex 1
ncert sol class 7 math ch 13 ex 1
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ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
ncert sol class 7 math ch 13 ex 2
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ncert sol class 7 math ch 13 ex 3
ncert sol class 7 math ch 13 ex 3
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Access Exercises of NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

Exercise 13.1 Solutions

Exercise 13.2 Solutions

Exercise 13.3 Solutions

Access answers to Maths NCERT Solutions For Class 7 Chapter 13 – Exponents and Powers

Exercise 13.1 Page: 252

1. Find the value of:

(i) 26

Solution:-

The above value can be written as,

= 2 × 2 × 2 × 2 × 2 × 2

= 64

(ii) 93

Solution:-

The above value can be written as,

= 9 × 9 × 9

= 729

(iii) 112

Solution:-

The above value can be written as,

= 11 × 11

= 121

(iv) 54

Solution:-

The above value can be written as,

= 5 × 5 × 5 × 5

= 625

2. Express the following in exponential form:

(i) 6 × 6 × 6 × 6

Solution:-

The given question can be expressed in the exponential form as 64.

(ii) t × t

Solution:-

The given question can be expressed in the exponential form as t2.

(iii) b × b × b × b

Solution:-

The given question can be expressed in the exponential form as b4.

(iv) 5 × 5× 7 × 7 × 7

Solution:-

The given question can be expressed in the exponential form as 52 × 73.

 

(v) 2 × 2 × a × a

Solution:-

The given question can be expressed in the exponential form as 22 × a2.

(vi) a × a × a × c × c × c × c × d

Solution:-

The given question can be expressed in the exponential form as a3 × c4 × d.

3. Express each of the following numbers using exponential notation:

(i) 512

Solution:-

The factors of 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

So it can be expressed in the exponential form as 29.

(ii) 343

Solution:-

The factors of 343 = 7 × 7 × 7

So it can be expressed in the exponential form as 73.

(iii) 729

Solution:-

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

So it can be expressed in the exponential form as 36.

(iv) 3125

Solution:-

The factors of 3125 = 5 × 5 × 5 × 5 × 5

So it can be expressed in the exponential form as 55.

4. Identify the greater number, wherever possible, in each of the following?

(i) 43 or 34

Solution:-

The expansion of 43 = 4 × 4 × 4 = 64

The expansion of 34 = 3 × 3 × 3 × 3 = 81

Clearly,

64 < 81

So, 43 < 34

Hence 34 is the greater number.

(ii) 53 or 35

Solution:-

The expansion of 53 = 5 × 5 × 5 = 125

The expansion of 35 = 3 × 3 × 3 × 3 × 3= 243

Clearly,

125 < 243

So, 53 < 35

Hence 35 is the greater number.

(iii) 28 or 82

Solution:-

The expansion of 28 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256

The expansion of 82 = 8 × 8= 64

Clearly,

256 > 64

So, 28 > 82

Hence 28 is the greater number.

(iv) 1002 or 2100

Solution:-

The expansion of 1002 = 100 × 100 = 10000

The expansion of 2100

210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

Then,

2100 = 1024 × 1024 ×1024 × 1024 ×1024 × 1024 × 1024 × 1024 × 1024 × 1024 =

Clearly,

1002 < 2100

Hence 2100 is the greater number.

(v) 210 or 102

Solution:-

The expansion of 210 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024

The expansion of 102 = 10 × 10= 100

Clearly,

1024 > 100

So, 210 > 102

Hence 28 is the greater number.

5. Express each of the following as product of powers of their prime factors:

(i) 648

Solution:-

Factors of 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3

= 23 × 34

(ii) 405

Solution:-

Factors of 405 = 3 × 3 × 3 × 3 × 5

= 35 × 3

 

(iii) 540

Solution:-

Factors of 540 = 2 × 2 × 3 × 3 × 3 × 5

= 22 × 33 × 5

(iv) 3,600

Solution:-

Factors of 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5

= 24 × 32 × 52

6. Simplify:

(i) 2 × 103

Solution:-

The above question can be written as,

= 2 × 10 × 10 × 10

= 2 × 1000

= 2000

(ii) 72 × 22

Solution:-

The above question can be written as,

= 7 × 7 × 2 × 2

= 49 × 4

= 196

(iii) 23 × 5

Solution:-

The above question can be written as,

= 2 × 2 × 2 × 5

= 8 × 5

= 40

(iv) 3 × 44

Solution:-

The above question can be written as,

= 3 × 4 × 4 × 4 × 4

= 3 × 256

= 768

(v) 0 × 102

Solution:-

The above question can be written as,

= 0 × 10 × 10

= 0 × 100

= 0

(vi) 52 × 33

Solution:-

The above question can be written as,

= 5 × 5 × 3 × 3 × 3

= 25 × 27

= 675

(vii) 24 × 32

Solution:-

The above question can be written as,

= 2 × 2 × 2 × 2 × 3 × 3

= 16 × 9

= 144

(viii) 32 × 104

Solution:-

The above question can be written as,

= 3 × 3 × 10 × 10 × 10 × 10

= 9 × 10000

= 90000

7. Simplify:

(i) (– 4)3

Solution:-

The expansion of -43

= – 4 × – 4 × – 4

= – 64

(ii) (–3) × (–2)3

Solution:-

The expansion of (-3) × (-2)3

= – 3 × – 2 × – 2 × – 2

= – 3 × – 8

= 24

(iii) (–3)2 × (–5)2

Solution:-

The expansion of (-3)2 × (-5)2

= – 3 × – 3 × – 5 × – 5

= 9 × 25

= 225

(iv) (–2)3 × (–10)3

Solution:-

The expansion of (-2)3 × (-10)3

= – 2 × – 2 × – 2 × – 10 × – 10 × – 10

= – 8 × – 1000

= 8000

8. Compare the following numbers:

(i) 2.7 × 1012 ; 1.5 × 108

Solution:-

By observing the question

Comparing the exponents of base 10,

Clearly,

2.7 × 1012 > 1.5 × 108

(ii) 4 × 1014 ; 3 × 1017

Solution:-

By observing the question

Comparing the exponents of base 10,

Clearly,

4 × 1014 < 3 × 1017


Exercise 13.2 Page: 260

1. Using laws of exponents, simplify and write the answer in exponential form:

(i) 32 × 34 × 38

Solution:-

By the rule of multiplying the powers with same base = am × an = am + n

Then,

= (3)2 + 4 + 8

= 314

(ii) 615 ÷ 610

Solution:-

By the rule of dividing the powers with same base = am ÷ an = am – n

Then,

= (6)15 – 10

= 65

(iii) a3 × a2

Solution:-

By the rule of multiplying the powers with same base = am × an = am + n

Then,

= (a)3 + 2

= a5

(iv) 7x × 72

Solution:-

By the rule of multiplying the powers with same base = am × an = am + n

Then,

= (7)x + 2

 

(v) (52)3 ÷ 53

Solution:-

By the rule of taking power of as power = (am)n = amn

(52)3 can be written as = (5)2 × 3

= 56

Now, 56 ÷ 53

By the rule of dividing the powers with same base = am ÷ an = am – n

Then,

= (5)6 – 3

= 53

(vi) 25 × 55

Solution:-

By the rule of multiplying the powers with same exponents = am × bm = abm

Then,

= (2 × 5)5

= 105

(vii) a4 × b4

Solution:-

By the rule of multiplying the powers with same exponents = am × bm = abm

Then,

= (a × b)4

= ab4

(viii) (34)3

Solution:-

By the rule of taking power of as power = (am)n = amn

(34)3 can be written as = (3)4 × 3

= 312

(ix) (220 ÷ 215) × 23

Solution:-

By the rule of dividing the powers with same base = am ÷ an = am – n

(220 ÷ 215) can be simplified as,

= (2)20 – 15

= 25

Then,

By the rule of multiplying the powers with same base = am × an = am + n

25 × 23 can be simplified as,

= (2)5 + 3

= 28

(x) 8t ÷ 82

Solution:-

By the rule of dividing the powers with same base = am ÷ an = am – n

Then,

= (8)t – 2

2. Simplify and express each of the following in exponential form:

(i) (23 × 34 × 4)/ (3 × 32)

Solution:-

Factors of 32 = 2 × 2 × 2 × 2 × 2

= 25

Factors of 4 = 2 × 2

= 22

Then,

= (23 × 34 × 22)/ (3 × 25)

= (23 + 2 × 34) / (3 × 25) … [∵am × an = am + n]

= (25 × 34) / (3 × 25)

= 25 – 5 × 34 – 1 … [∵am ÷ an = am – n]

= 20 × 33

= 1 × 33

= 33

(ii) ((52)3 × 54) ÷ 57

Solution:-

(52)3 can be written as = (5)2 × 3 … [∵(am)n = amn]

= 56

Then,

= (56 × 54) ÷ 57

= (56 + 4) ÷ 57 … [∵am × an = am + n]

= 510 ÷ 57

= 510 – 7 … [∵am ÷ an = am – n]

= 53

(iii) 254 ÷ 53

Solution:-

(25)4 can be written as = (5 × 5)4

= (52)4

(52)4 can be written as = (5)2 × 4 … [∵(am)n = amn]

= 58

Then,

= 58 ÷ 53

= 58 – 3 … [∵am ÷ an = am – n]

= 55

(iv) (3 × 72 × 118)/ (21 × 113)

Solution:-

Factors of 21 = 7 × 3

Then,

= (3 × 72 × 118)/ (7 × 3 × 113)

= 31-1 × 72-1 × 118 – 3

= 30 × 7 × 115

= 1 × 7 × 115

= 7 × 115

(v) 37/ (34 × 33)

Solution:-

= 37/ (34+3) … [∵am × an = am + n]

= 37/ 37

= 37 – 7 … [∵am ÷ an = am – n]

= 30

= 1

(vi) 20 + 30 + 40

Solution:-

= 1 + 1 + 1

= 3

(vii) 20 × 30 × 40

Solution:-

= 1 × 1 × 1

= 1

(viii) (30 + 20) × 50

Solution:-

= (1 + 1) × 1

= (2) × 1

= 2

(ix) (28 × a5)/ (43 × a3)

Solution:-

(4)3 can be written as = (2 × 2)3

= (22)3

(52)4 can be written as = (2)2 × 3 … [∵(am)n = amn]

= 26

Then,

= (28 × a5)/ (26 × a3)

= 28 – 6 × a5 – 3 … [∵am ÷ an = am – n]

= 22 × a2

= 2a2 … [∵(am)n = amn]

(x) (a5/a3) × a8

Solution:-

= (a5 – 3) × a8 … [∵am ÷ an = am – n]

= a2 × a8

= a2 + 8 … [∵am × an = am + n]

= a10

(xi) (45 × a8b3)/ (45 × a5b2)

Solution:-

= 45 – 5 × (a8 – 5 × b3 – 2) … [∵am ÷ an = am – n]

= 40 × (a3b)

= 1 × a3b

= a3b

(xii) (23 × 2)2

Solution:-

= (23 + 1)2 … [∵am × an = am + n]

= (24)2

(24)2 can be written as = (2)4 × 2 … [∵(am)n = amn]

= 28

3. Say true or false and justify your answer:

(i) 10 × 1011 = 10011

Solution:-

Let us consider Left Hand Side (LHS) = 10 × 1011

= 101 + 11 … [∵am × an = am + n]

= 1012

Now, consider Right Hand Side (RHS) = 10011

= (10 × 10)11

= (101 + 1)11

= (102)11

= (10)2 × 11 … [∵(am)n = amn]

= 1022

By comparing LHS and RHS,

LHS ≠ RHS

Hence, the given statement is false.

(ii) 23 > 52

Solution:-

Let us consider LHS = 23

Expansion of 23 = 2 × 2 × 2

= 8

Now, consider RHS = 52

Expansion of 52 = 5 × 5

= 25

By comparing LHS and RHS,

LHS < RHS

23 < 52

Hence, the given statement is false.

(iii) 23 × 32 = 65

Solution:-

Let us consider LHS = 23 × 32

Expansion of 23 × 32= 2 × 2 × 2 × 3 × 3

= 72

Now, consider RHS = 65

Expansion of 65 = 6 × 6 × 6 × 6 × 6

= 7776

By comparing LHS and RHS,

LHS < RHS

23 < 52

Hence, the given statement is false.

(iv) 30 = (1000)0

Solution:-

Let us consider LHS = 30

= 1

Now, consider RHS = 10000

= 1

By comparing LHS and RHS,

LHS = RHS

30 = 10000

Hence, the given statement is true.

4. Express each of the following as a product of prime factors only in exponential form:

(i) 108 × 192

Solution:-

The factors of 108 = 2 × 2 × 3 × 3 × 3

= 22 × 33

The factors of 192 = 2 × 2 × 2 × 2 × 2 × 2 × 3

= 26 × 3

Then,

= (22 × 33) × (26 × 3)

= 22 + 6 × 33 + 1 … [∵am × an = am + n]

= 28 × 34

(ii) 270

Solution:-

The factors of 270 = 2 × 3 × 3 × 3 × 5

= 2 × 33 × 5

(iii) 729 × 64

The factors of 729 = 3 × 3 × 3 × 3 × 3 × 3

= 36

The factors of 64 = 2 × 2 × 2 × 2 × 2 × 2

= 26

Then,

= (36 × 26)

= 36 × 26

(iv) 768

Solution:-

The factors of 768 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3

= 28 × 3

5. Simplify:

(i) ((25)2 × 73)/ (83 × 7)

Solution:-

83 can be written as = (2 × 2 × 2)3

= (23)3

We have,

= ((25)2 × 73)/ ((23)3 × 7)

= (25 × 2 × 73)/ ((23 × 3 × 7) … [∵(am)n = amn]

= (210 × 73)/ (29 × 7)

= (210 – 9 × 73 – 1) … [∵am ÷ an = am – n]

= 2 × 72

= 2 × 7 × 7

= 98

(ii) (25 × 52 × t8)/ (103 × t4)

Solution:-

25 can be written as = 5 × 5

= 52

103 can be written as = 103

= (5 × 2)3

= 53 × 23

We have,

= (52 × 52 × t8)/ (53 × 23 × t4)

= (52 + 2 × t8)/ (53 × 23 × t4) … [∵am × an = am + n]

= (54 × t8)/ (53 × 23 × t4)

= (54 – 3 × t8 – 4)/ 23 … [∵am ÷ an = am – n]

= (5 × t4)/ (2 × 2 × 2)

= (5t4)/ 8

(iii) (35 × 105 × 25)/ (57 × 65)

Solution:-

105 can be written as = (5 × 2)5

= 55 × 25

25 can be written as = 5 × 5

= 52

65 can be written as = (2 × 3)5

= 25 × 35

Then we have,

= (35 × 55 × 25 × 52)/ (57 × 25 × 35)

= (35 × 55 + 2 × 25)/ (57 × 25 × 35) … [∵am × an = am + n]

= (35 × 57 × 25)/ (57 × 25 × 35)

= (35 – 5 × 57 – 7 × 25 – 5)

= (30 × 50 × 20) … [∵am ÷ an = am – n]

= 1 × 1 × 1

= 1


Exercise 13.3 Page: 263

1. Write the following numbers in the expanded forms:

279404

Solution:-

The expanded form of the number 279404 is,

= (2 × 100000) + (7 × 10000) + (9 × 1000) + (4 × 100) + (0 × 10) + (4 × 1)

Now we can express it using powers of 10 in the exponent form,

= (2 × 105) + (7 × 104) + (9 × 103) + (4 × 102) + (0 × 101) + (4 × 100)

3006194

Solution:-

The expanded form of the number 3006194 is,

= (3 × 1000000) + (0 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 4

Now we can express it using powers of 10 in the exponent form,

= (3 × 106) + (0 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (4 × 100)

2806196

Solution:-

The expanded form of the number 2806196 is,

= (2 × 1000000) + (8 × 100000) + (0 × 10000) + (6 × 1000) + (1 × 100) + (9 × 10) + 6

Now we can express it using powers of 10 in the exponent form,

= (2 × 106) + (8 × 105) + (0 × 104) + (6 × 103) + (1 × 102) + (9 × 101) + (6 × 100)

120719

Solution:-

The expanded form of the number 120719 is,

= (1 × 100000) + (2 × 10000) + (0 × 1000) + (7 × 100) + (1 × 10) + (9 × 1)

Now we can express it using powers of 10 in the exponent form,

= (1 × 105) + (2 × 104) + (0 × 103) + (7 × 102) + (1 × 101) + (9 × 100)

20068

Solution:-

The expanded form of the number 20068 is,

= (2 × 10000) + (0 × 1000) + (0 × 100) + (6 × 10) + (8 × 1)

Now we can express it using powers of 10 in the exponent form,

= (2 × 104) + (0 × 103) + (0 × 102) + (6 × 101) + (8 × 100)

2. Find the number from each of the following expanded forms:

(a) (8 ×10)4 + (6 × 10)3 + (0 × 10)2 + (4 × 10)1 + (5 × 10)0

Solution:-

The expanded form is,

= (8 × 10000) + (6 × 1000) + (0 × 100) + (4 × 10) + (5 × 1)

= 80000 + 6000 + 0 + 40 + 5

= 86045

(b) (4 ×10)5 + (5 × 10)3 + (3 × 10)2 + (2 × 10)0

Solution:-

The expanded form is,

= (4 × 100000) + (0 × 10000) + (5 × 1000) + (3 × 100) + (0 × 10) + (2 × 1)

= 400000 + 0 + 5000 + 300 + 0 + 2

= 405302

(c) (3 ×10)4 + (7 × 10)2 + (5 × 10)0

Solution:-

The expanded form is,

= (3 × 10000) + (0 × 1000) + (7 × 100) + (0 × 10) + (5 × 1)

= 30000 + 0 + 700 + 0 + 5

= 30705

(d) (9 ×10)5 + (2 × 10)2 + (3 × 10)1

Solution:-

The expanded form is,

= (9 × 100000) + (0 × 10000) + (0 × 1000) + (2 × 100) + (3 × 10) + (0 × 1)

= 900000 + 0 + 0 + 200 + 30 + 0

= 900230

3. Express the following numbers in standard form:

(i) 5,00,00,000

Solution:-

The standard form of the given number is 5 × 107

(ii) 70,00,000

Solution:-

The standard form of the given number is 7 × 106

(iii) 3,18,65,00,000

Solution:-

The standard form of the given number is 3.1865 × 109

(iv) 3,90,878

Solution:-

The standard form of the given number is 3.90878 × 105

(v) 39087.8

Solution:-

The standard form of the given number is 3.90878 × 104

(vi) 3908.78

Solution:-

The standard form of the given number is 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.

Solution:-

The standard form of the number appearing in the given statement is 3.84 × 108m.

(b) Speed of light in vacuum is 300,000,000 m/s.

Solution:-

The standard form of the number appearing in the given statement is 3 × 108m/s.

(c) Diameter of the Earth is 1,27,56,000 m.

Solution:-

The standard form of the number appearing in the given statement is 1.2756 × 107m.

(d) Diameter of the Sun is 1,400,000,000 m.

Solution:-

The standard form of the number appearing in the given statement is 1.4 × 109m.

(e) In a galaxy there are on an average 100,000,000,000 stars.

Solution:-

The standard form of the number appearing in the given statement is 1 × 1011 stars.

(f) The universe is estimated to be about 12,000,000,000 years old.

Solution:-

The standard form of the number appearing in the given statement is 1.2 × 1010 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.

Solution:-

The standard form of the number appearing in the given statement is 3 × 1020m.

(h) 60,230,000,000,000,000,000,000 molecules are contained in a drop of water weighing 1.8 gm.

Solution:-

The standard form of the number appearing in the given statement is 6.023 × 1022 molecules.

(i) The earth has 1,353,000,000 cubic km of sea water.

Solution:-

The standard form of the number appearing in the given statement is 1.353 × 109 cubic km.

(j) The population of India was about 1,027,000,000 in March, 2001.

Solution:-

The standard form of the number appearing in the given statement is 1.027 × 109.

 


 

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