Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents) Exercise 5B

Selina Solutions Concise Maths Class 7 Chapter 5 Exponents (Including Laws of Exponents) Exercise 5B explains the laws of exponents and more about indices. Product Law, Quotient Law and Power law are the main laws, which are discussed in brief under this exercise. The solutions PDF helps students analyse the areas of weaknesses and work on them for further progress. The solutions PDF can be referred by the students while solving textbook problems to clear doubts instantly. Selina Solutions Concise Maths Class 7 Chapter 5 Exponents (Including Laws of Exponents) Exercise 5B PDF links are given below for free download.

Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents) Exercise 5B Download PDF

 

selina sol concise maths class 7 ch5 ex 5b 01
selina sol concise maths class 7 ch5 ex 5b 02
selina sol concise maths class 7 ch5 ex 5b 03
selina sol concise maths class 7 ch5 ex 5b 04
selina sol concise maths class 7 ch5 ex 5b 05
selina sol concise maths class 7 ch5 ex 5b 06
selina sol concise maths class 7 ch5 ex 5b 07
selina sol concise maths class 7 ch5 ex 5b 08
selina sol concise maths class 7 ch5 ex 5b 09
selina sol concise maths class 7 ch5 ex 5b 10
selina sol concise maths class 7 ch5 ex 5b 11
selina sol concise maths class 7 ch5 ex 5b 12
selina sol concise maths class 7 ch5 ex 5b 13
selina sol concise maths class 7 ch5 ex 5b 14
selina sol concise maths class 7 ch5 ex 5b 15

 

Access other exercises of Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents)

Exercise 5A Solutions

Access Selina Solutions Concise Maths Class 7 Chapter 5: Exponents (Including Laws of Exponents) Exercise 5B

1. Fill in the blanks:

(i) In 52 = 25, base = ………. and index = …………

(ii) If index = 3x and base = 2y, the number = ………

Solution:

(i) In 52 = 25, base = 5 and index = 2.

(ii) If index = 3x and base = 2y, the number = 2y3x.

2. Evaluate:

(i) 28 ÷ 23
(ii) 2 28
(iii) (26)0
(iv) (3o)6
(v) 83 x 8-5 x 84
(vi) 5x 53 ÷ 55
(vii) 54 ÷ 53 x 55
(viii) 44 ÷ 43 x 40
(ix) (35 x 47 x 58)0

Solution:

(i) 28 ÷ 23

It can be written as

= 28/ 23

On further calculation

= 2 8-3

= 25

(ii) 2 28

It can be written as

= 23/ 28

On further calculation

= 2 3-8

So we get

= 2 -5

= 1/25

(iii) (26)0

It can be written as

= 2 6 × 0

On further calculation

= 20

So we get

= 1

(iv) (3o) 6

It can be written as

= 3 0 × 6

On further calculation

= 30

So we get

= 1

(v) 83 x 8-5 x 84

It can be written as

= 8 3 + 4 – 5

On further calculation

= 8 7 – 5

So we get

= 82

(vi) 5x 53 ÷ 55

It can be written as

= (5x 53)/ 55

On further calculation

= 5 4 + 3 – 5

So we get

= 5 7 – 5

= 52

(vii) 54 ÷ 53 x 55

It can be written as

= 54/ 53 x 55

On further calculation

= 5 4 – 3 + 5

So we get

= 56

(viii) 44 ÷ 43 x 40

It can be written as

= 44/ (43 x 40)

On further calculation

= 44/ (43 x 1)

So we get

= 44/43

= 44-3

= 41

= 4

(ix) (35 x 47 x 58)0

It can be written as

= 35 × 0 × 47 × 0 × 58 × 0

On further calculation

= 30 × 40 × 50

So we get

= 1 × 1 × 1

= 1

3. Simplify, giving answers with positive index:

(i) 2b6. b3. 5b4

(ii) x2y3. 6x5y. 9x3y4

(iii) (-a)5 (a2)

(iv) (-y)2 (-y)3

(v) (-3)2 (3)3

(vi) (-4x) (-5x2)

(vii) (5a2b) (2ab2) (a3b)

(viii) x2a + 7. x2a – 8

(ix) 3y. 32. 3-4

(x) 24a. 23a. 2-a

(xi) 4x2y2 ÷ 9x3y3

(xii) (102)3 (x8)12

(xiii) (a10)10 (16)10

(xiv) (n2)2 (-n2)3

(xv) – (3ab)2 (-5a2bc4)2

(xvi) (-2)2 × (0)3 × (3)3

(xvii) (2a3)4 (4a2)2

(xviii) (4x2y3)3 ÷ (3x2y3)3

Selina Solutions Concise Maths Class 7 Chapter 5 Image 14

Solution:

(i) 2b6. b3. 5b4

It can be written as

= 2 × 5 × b6 + 3 + 4

On further calculation

= 10 b13

(ii) x2y3. 6x5y. 9x3y4

It can be written as

= 6 × 9 × x2 + 5 + 3 × y3 + 1 + 4

On further calculation

= 54 x10 y8

(iii) (-a)5 (a2)

It can be written as

= (-1 × a)5 × a2

On further calculation

= (-1)5 × a5 + 2

So we get

= – 1 × a7

= – a7

(iv) (-y)2 (-y)3

It can be written as

= (-1 × y)2. (-1 × y)3

On further calculation

= (-1)2. y2. (-1)3 × y3

So we get

= 12 + 3. y2 + 3

= 15 y5

= y5

(v) (-3)2 (3)3

It can be written as

= (-1 × 3)2. (3)3

On further calculation

= (-1)2 × 32. 33

So we get

= (-1)2. 32 + 3

= 1. 35

= 35

(vi) (-4x) (-5x2)

It can be written as

= (-1 × 4 × x). (-1 × 5 × x2)1

On further calculation

= (- 1 × 4 × x). (-1 × 5 × x2)

So we get

= – 1 × – 1 × 4 × 5 × x1 + 2

Here

= – 11 + 1. 41. 51 x3

= 20 x3

(vii) (5a2b) (2ab2) (a3b)

It can be written as

= 5 × 2 × a2 + 1 + 3 × b1 + 2 + 1

On further calculation

= 10 a6b4

(viii) x2a + 7. x2a – 8

It can be written as

= x2a + 7 + 2a – 8

On further calculation

= x4a – 1

(ix) 3y. 32. 3-4

It can be written as

= 3y. 32/34

On further calculation

= 3y. (3 × 3)/ (3 × 3 × 3 × 3)

So we get

= 3y × 1/32

= 3y – 2

(x) 24a. 23a. 2-a

It can be written as

= 2 4a + 3a – a

On further calculation

= 2 7a – a

So we get

= 26a

(xi) 4x2y2 ÷ 9x3y3

It can be written as

= 4x2y2/ 9x3y3

On further calculation

= 4x2 – 3 y2 – 3/ 9

So we get

= 4x-1y-1/ 9

= 4/9xy

(xii) (102)3 (x8)12

It can be written as

= 102 × 3. x8 × 12

On further calculation

= 106 x96

(xiii) (a10)10 (16)10

It can be written as

= a10 × 10. 16 × 10

On further calculation

= a100. 160

So we get

= a100

(xiv) (n2)2 (-n2)3

It can be written as

= n2 × 2. (-n)2 × 3

On further calculation

= n4. (-n)6

So we get

= – n4 – 16 n6

= – n4 + 6

= – n 10

(xv) – (3ab)2 (-5a2bc4)2

It can be written as

= – (32a2b2) × (-1)2 × 52a2 × 2b2c4 × 2

On further calculation

= – (32a2b2) (52a4b2c8)

So we get

= – 32. 52. a2 + 4 b2 + 2 c8

= – 225a6b4c8

(xvi) (-2)2 × (0)3 × (3)3

It can be written as

= 4 × 0 × 27

On further calculation

= 0

(xvii) (2a3)4 (4a2)2

It can be written as

= (2a3)4 (22a2)2

On further calculation

= 24 a3 × 4. 22 × 2. a2 × 2

So we get

= 24 a12. 24 a4

Here

= 24 + 4. a12 + 4

= 28 a16

We get

= 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × a16

= 256 a16

(xviii) (4x2y3)3 ÷ (3x2y3)3

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 15

Selina Solutions Concise Maths Class 7 Chapter 5 Image 16

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 17

Selina Solutions Concise Maths Class 7 Chapter 5 Image 18

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 19

Selina Solutions Concise Maths Class 7 Chapter 5 Image 20

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 21

Selina Solutions Concise Maths Class 7 Chapter 5 Image 22

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 23

Selina Solutions Concise Maths Class 7 Chapter 5 Image 24

4. Simplify and express the answer in the positive exponent form:

Selina Solutions Concise Maths Class 7 Chapter 5 Image 25

Solution:

Selina Solutions Concise Maths Class 7 Chapter 5 Image 26

= – (3)3 – 1 26 – 4

= – (3)2 22

= – 3222

Selina Solutions Concise Maths Class 7 Chapter 5 Image 27

Selina Solutions Concise Maths Class 7 Chapter 5 Image 28

Selina Solutions Concise Maths Class 7 Chapter 5 Image 29

(iv) – 128/2187

So we get

Selina Solutions Concise Maths Class 7 Chapter 5 Image 30

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 31

Selina Solutions Concise Maths Class 7 Chapter 5 Image 32

(vi) (a3b-5)-2

It can be written as

= a 3 x -2 b-5 x -2

So we get

= a-6 b10

= b10/ a6

5. Evaluate:

(i) 6-2 ÷ (4-2 × 3-2)

Selina Solutions Concise Maths Class 7 Chapter 5 Image 33

(iii) 53 × 32 + (17)0 × 73

(iv) 25 × 150 + (-3)3 – (2/7)-2

(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3

(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)

Solution:

(i) 6-2 ÷ (4-2 × 3-2)

It can be written as

= (1/6)2 ÷ (1/4)2 × (1/3)2

On further calculation

= 1/36 ÷ 1/16 × 1/9

So we get

= 1/36 ÷ 1/144

= 1/36 × 144/1

= 4

Selina Solutions Concise Maths Class 7 Chapter 5 Image 34

Selina Solutions Concise Maths Class 7 Chapter 5 Image 35

It can be written as

Selina Solutions Concise Maths Class 7 Chapter 5 Image 36

Selina Solutions Concise Maths Class 7 Chapter 5 Image 37

(iii) 53 × 32 + (17)0 × 73

It can be written as

= 5 × 5 × 5 × 3 × 3 + (17)0 × 7 × 7 × 7

On further calculation

= 125 × 9 + 1 × 343

So we get

= 1125 + 343

= 1468

(iv) 25 × 150 + (-3)3 – (2/7)-2

It can be written as

= 2 × 2 × 2 × 2 × 2 × 1 + (-3) × (-3) × (-3) – (7/2) × (7/2)

By further calculation

= 32 × 1 – 27 – 49/4

Here we get LCM = 4

= (32 × 4)/ (1 × 4) – (27 × 4)/ (1 × 4) – 49/ (4 × 1)

So we get

= (128 – 108 – 49)/ 4

By subtraction

= -29/ 4

= – 7 1/4

(v) (22)0 + 2-4 ÷ 2-6 + (1/2)-3

It can be written as

= (4)0 + (1/2)4 ÷ (1/2)6 + (2/1)3

By further calculation

= 1 + (1/2 × 1/2 × 1/2 × 1/2) ÷ (1/2 × 1/2 × 1/2 × 1/2 × 1/2 × 1/2) + (2/1 × 2/1 × 2/1)

So we get

= 1 + (1/2 × 1/2 × 1/2 × 1/2 × 2 × 2 × 2 × 2 × 2 × 2) + 8

On further simplification

= 1 + 4 + 8

= 13

(vi) 5n × 25n-1 ÷ (5n-1 × 25n-1)

It can be written as

= 5n × 25n-1 × 1/ (5n-1 × 25n-1)

By further calculation

= 5n × 1/ 5n-1

So we get

= 5n – n + 1

= 51

6. If m = – 2 and n = 2; find the value of:

(i) m2 + n2 – 2mn

(ii) mn + nm

(iii) 6m-3 + 4n2

(iv) 2n3 – 3m

Solution:

(i) m2 + n2 – 2mn

It is given that

m = – 2 and n = 2

Substituting the values we get

= (-2)2 + 22 – 2 (-2) (2)

By further calculation

= 4 + 4 – (-8)

So we get

= 8 + 8

= 16

= 24

Selina Solutions Concise Maths Class 7 Chapter 5 Image 38

(ii) mn + nm

It is given that m = – 2 and n = 2

Substituting the values we get

= (-2)2 + (2)-2

We can write it as

= 4 + 1/2 × 1/2

We get the LCM = 4

= (4 × 4)/ (1 × 4) + 1/4

So we get

= (16 + 1)/ 4

= 17/ 4

= 4 ¼

(iii) 6m-3 + 4n2

It is given that

m = – 2 and n = 2

Substituting the values

= 6 (-2)-3 + 4 (2)2

It can be written as

= 6 × 1/ -2 × 1/ -2 × 1/ -2 + 4 × 2 × 2

So we get

= – 3/4 + 16

Here the LCM = 4

= (-3 + 16 × 4)/ 4

By calculation

= (-3 + 64)/ 4

= – 61/4

= 15 ¼

(iv) 2n3 – 3m

It is given that

m = – 2 and n = 2

By substituting the values

= 2 (2)3 – 3 (-2)

It can be written as

= 2 × (2 × 2 × 2) – 3 × (-2)

By further calculation

= 16 – 3 × (-2)

So we get

= 16 + 6

= 22

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class