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 practise Maths regularly. Students of Class 7 are suggested to solve NCERT Solutions 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 Chapter 13 – Exponents and Powers of NCERT Solutions for Class 7 Maths. These exercises cover 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
ncert sol class 7 math ch 13 ex 1
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
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
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 210 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

= 34 × 5

 

(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,

72 ≠ 7776

LHS ≠ RHS

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.

 


 

Frequently Asked Questions on NCERT Solutions for Class 7 Maths Chapter 13

Where to download NCERT Solutions for Class 7 Maths Chapter 13?

NCERT Solutions for Class 7 Maths Chapter 13 can be downloaded at BYJU’S website. It can be avail in free PDF. To download go to BYJU’S NCERT Solutions website/select the Class 7/opt the subject as Maths/click on to the desired chapter that is Chapter 13 Some Applications of Trigonometry.

What are the applications of learning NCERT Solutions for Class 7 Maths Chapter 13?

Applications of learning NCERT Solutions for Class 7 Maths Chapter 13 are finding the
1. How are Expressions Formed
2. Terms of An Expression
3. Like and Unlike Terms
4. Monomials, Binomials, Trinomials and Polynomials
5. Addition and Subtraction of Algebraic Expressions
6. Finding The Value of An Expression
7. Using Algebraic Expressions – Formulas and Rules.
By learning these concepts students will be able to answer all the questions based on algebraic expressions as well as it may help in writing class tests and board exams.

Is NCERT Solutions for Class 7 Maths Chapter 13 enough for board exams?

Yes, you have to learn all the questions thoroughly and practice them for more time. Once you are done with solving all the questions then you can refer to other reference books and questions provided NCERT Exemplar textbooks.

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