NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

 

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

 

Introduction:

Numbers in exponential form obey certain laws, which are :

For any non-zero integers a and b and whole numbers m and n,

a) am× an= a m+ n

b) am ÷ an =am-n

c) (am)= a m n

d) am × bm = (ab)m

e) am ÷ bm = \(\displaystyle {{(\frac{a}{b})}^{m}}\)

f) a0 = 1

g) (-1)even number = 1

h) (-1)odd  number = -1

NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.1

NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

NCERT Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.3

 

Class 7 Maths Exercise 11.2

(Page-181)

Q1. Using laws of exponents, simplify and write the answer in exponential form:

i) 32× 34× 38

ii)615 ÷ 610

iii)a3 × a2

iv) 7x × 72

v) (52)3 ÷ 53

vi)  25 × 55

vii)  a4 × b4

viii) ( 34)3

ix) (220 ÷ 215) × 23

x)  8t × 82

Solution: 

i)  3(2+4+8)

= 314

ii)  6(15-10)

= 65

iii) a(3+2)

= a5

iv) 7(x+2)

v) 5(2×3) ÷ 53

= 56 ÷ 53

= 5(6-3)

= 53

vi) ( 2×5)5

= 105

vii) (a × b)4

viii)  3(4×3)

= 312

ix)  2(20-15) × 23

= 25× 23

= 2(5+3)

= 28

x) 8(t-2)

 

Q2. Simplify and Express each of the following in exponential form:

i) \(\displaystyle \frac{{{{2}^{3}}\times {{3}^{4}}\times 4}}{{3\times 32}}\)

ii) ((52)×5457

iii) 254  ÷ 53

iv) \(\displaystyle \frac{{3\times {{7}^{2}}\times {{{11}}^{8}}}}{{21\times {{{11}}^{3}}}}\)

v) \(\displaystyle \frac{{{{3}^{7}}}}{{{{3}^{4}}\times {{3}^{3}}}}\)

vi)  20 + 30+ 40

vii)  20 × 30 ×40

viii) (30 + 2050

ix) \(\displaystyle \frac{{{{2}^{8}}\times {{a}^{5}}}}{{{{4}^{3}}\times {{a}^{3}}}}\)

x) \(\displaystyle (\frac{{{{a}^{5}}}}{{{{a}^{3}}}})\) × a8

xi) \(\displaystyle \frac{{{{4}^{5}}\times {{a}^{8}}{{b}^{3}}}}{{{{4}^{5}}\times {{a}^{5}}{{b}^{2}}}}\)

xii) (23× 2)2

Solution: 

i) \(\displaystyle \frac{{{{2}^{3}}\times {{3}^{4}}\times {{2}^{2}}}}{{3\times {{2}^{5}}}}\)

= \(\displaystyle \frac{{{{2}^{{(3+2)}}}\times {{3}^{4}}}}{{3\times {{2}^{5}}}}\)

= \(\displaystyle \frac{{{{2}^{5}}\times {{3}^{{(4-1)}}}}}{{{{2}^{5}}}}\)

= \(\displaystyle \frac{{{{2}^{5}}\times {{3}^{{(3)}}}}}{{{{2}^{5}}}}\)

= 2(5-5) × 33

= 20× 33

= 1× 33

= 33

ii) 5(2×3)×5457

= (56×5457

= 5(6+4) ÷ 57

= 510 ÷ 57

= 5(10-7)

= 53

iii) (52)4 ÷53

= 5(2×4) ÷53

= 58 ÷53

= 5(8-3)

= 55

iv) \(\displaystyle \frac{{3\times {{7}^{2}}\times {{{11}}^{8}}}}{{3\times 7\times {{{11}}^{3}}}}\)

= \(\displaystyle \frac{{3\times {{7}^{{(2-1)}}}\times {{{11}}^{8}}}}{{3\times {{{11}}^{3}}}}\)

= \(\displaystyle \frac{{3\times {{7}^{{(1)}}}\times {{{11}}^{{(8-3)}}}}}{3}\)

= 7× 115

v) \(\displaystyle \frac{{{{3}^{7}}}}{{{{3}^{{4+3}}}}}\)

= \(\displaystyle \frac{{{{3}^{7}}}}{{{{3}^{7}}}}\)

= 30

= 1

vi) 20 + 30+ 40

= 1+1+1

= 3

vii) 20 × 30 ×40

= 1×1×1

= 1

viii) (30 + 2050

=  (1 + 1)× 1

= (2)× 1

= 2

ix) \(\displaystyle \frac{{{{2}^{8}}\times {{a}^{5}}}}{{{{{({{2}^{2}})}}^{3}}\times {{a}^{3}}}}\)

= \(\displaystyle \frac{{{{2}^{8}}\times {{a}^{5}}}}{{{{2}^{6}}\times {{a}^{3}}}}\)

= \(\displaystyle \frac{{{{2}^{8}}\times {{a}^{{(5-3)}}}}}{{{{2}^{6}}}}\)

= \(\displaystyle \frac{{{{2}^{8}}\times {{a}^{2}}}}{{{{2}^{6}}}}\)

= 2(8-6) × a2

= 22 × a2

= (2a)2

x) \(\displaystyle ({{a}^{{5-3}}})\times a8\)

= a2× a8

= a2+8

= a10

xi) \(\displaystyle \frac{{{{4}^{{5-5}}}\times {{a}^{8}}{{b}^{3}}}}{{{{a}^{5}}{{b}^{2}}}}\)

= \(\displaystyle \frac{{{{4}^{0}}\times {{a}^{8}}{{b}^{3}}}}{{{{a}^{5}}{{b}^{2}}}}\)

= \(\displaystyle \frac{{1\times {{a}^{8}}{{b}^{3}}}}{{{{a}^{5}}{{b}^{2}}}}\)

= (a8-5)(b3-2)

= a3b1

= a3b

xii) (23× 2)2

= (23×222

= 26×22

= 26+2

= 28

 

Q3. Say true or false and justify your answer:

i) 10 × 1011  = 10011

ii) 23 > 52

iii) 23 × 32 = 65

iv) 30 = (1000)0

Solution: 

i) False

101 × 1011  = 10011

1012 = (102)11

1012  ≠ 1022

ii) False

23 > 52

2×2×2> 5×5

8<25 iii) False

23 × 32 = 65

8×9 = 7776

72 ≠ 7776

iv) true

30 = (1000)0

1 = 1

Q4. 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

Solution:

i) 108 × 192

 NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

108 × 192 = 2×2×3×3×3×2×2×2×2×2×2×3

= 28 × 34

ii) 

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

270 = 2×3×3×3×5

= 2 × 33×5

iii) 

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

729 × 64= 3×3×3×3×3×3×2×2×2×2×2×2

= 26 × 36

= (2×3)6

iv)

NCERT Solutions Class 7 Maths Chapter 11 Exponents and Powers Exercise 11.2

768 = 2×2×2×2×2×2×2×2×3

=28 × 3

 

Q5. Simplify:

i) \(\displaystyle \frac{{{{{({{2}^{5}})}}^{2}}\times {{7}^{3}}}}{{{{8}^{3}}\times \,7}}\)

ii) \(\displaystyle \frac{{25\times {{5}^{2}}\times {{t}^{8}}}}{{{{{10}}^{3}}\times {{t}^{4}}}}\)

iii) \(\displaystyle \frac{{{{3}^{5}}\times {{{10}}^{5}}\times 25}}{{{{5}^{7}}\times {{6}^{5}}}}\)

Solution:

i) \(\displaystyle \frac{{{{2}^{{10}}}\times {{7}^{3}}}}{{{{{({{2}^{3}})}}^{3}}\times 7}}\)

= \(\displaystyle \frac{{{{2}^{{10}}}\times {{7}^{3}}}}{{{{2}^{9}}\times 7}}\)

= 2(10-9) ×7(3-1)

= 2×72

= 2×49

=98

ii) \(\displaystyle \frac{{{{5}^{2}}\times {{5}^{2}}\times {{t}^{8}}}}{{{{{(2\times 5)}}^{3}}\times {{t}^{4}}}}\)

= \(\displaystyle \frac{{{{5}^{2}}\times {{5}^{2}}\times {{t}^{{8-4}}}}}{{{{{(2\times 5)}}^{3}}}}\)

= \(\displaystyle \frac{{{{5}^{{(2+2)}}}\times {{t}^{4}}}}{{{{{(2\times 5)}}^{3}}}}\)

= \(\displaystyle \frac{{{{5}^{{(4)}}}\times {{t}^{4}}}}{{{{2}^{3}}\times {{5}^{3}}}}\)

= \(\displaystyle \frac{{{{5}^{{(4-3)}}}\times {{t}^{4}}}}{{{{2}^{3}}}}\)

= \(\displaystyle \frac{{5{{t}^{4}}}}{8}\)

iii) \(\displaystyle \frac{{{{3}^{5}}\times {{{10}}^{5}}\times 25}}{{{{5}^{7}}\times {{6}^{5}}}}\)

= \(\displaystyle \frac{{{{3}^{5}}\times {{{(2\times 5)}}^{5}}\times {{5}^{2}}}}{{{{5}^{7}}\times {{{(2\times 3)}}^{5}}}}\)

= \(\displaystyle \frac{{{{3}^{5}}\times {{2}^{5}}\times {{5}^{5}}\times {{5}^{2}}}}{{{{5}^{7}}\times {{2}^{5}}\times {{3}^{5}}}}\)

= \(\displaystyle \frac{{{{3}^{{5-5}}}\times {{2}^{{5-5}}}\times {{5}^{{5+2}}}}}{{{{5}^{7}}}}\)

= \(\displaystyle \frac{{{{3}^{0}}\times {{2}^{0}}\times {{5}^{7}}}}{{{{5}^{7}}}}\)

= 1×1×5(7-7)

= 1×70

= 1×1

=1

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