NCERT Solutions For Class 7 Maths Chapter 13

NCERT Solutions Class 7 Maths Exponents and Powers

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers includes solutions for all the questions provided in NCERT Class 7 Maths textbook. By practicing various questions from this book help students to develop a thorough understanding of the concepts as it provides more solved examples and all questions are explained in detail to help students in learning the concept in an easiest possible manner. NCERT Solutions for Class 7 Maths Chapter 13 are available in PDF format. Students can either view online or download these pdf files from our website BYJU’S. NCERT Solutions for Class 7 Maths chapter 13 pdf provides a strong foundation for every topic related to Exponents and Powers.

NCERT Solutions For Class 7 Maths Chapter 13 Exercises

Exercise-13.1

 

Q.1.Find the value of:

(i) 26

(ii) 93

(iii) 112

(iv) 54

Solution:

(i)26=2×2×2×2×2×2=64

(ii)93=9×9×9=729

(iii)112=11×11=121

(iv) 54=5×5×5×5=625

 

Q.2. Express the following in exponential form:

(i) 6×6×6×6

Solution: 6×6×6×6=64

(ii) t×t

Solution: t×t=t2

(iii) b×b×b×b

Solution: b×b×b×b=b4

(iv) 5×5×7×7×7

Solution: 5×5×7×7×7=52×73

(v) 2×2×a×a

Solution: 2×2×a×a=22×a2

(vi) a×a×a×c×c×c×c×d

Solution: a×a×a×c×c×c×c×d=a3×c4×d

 

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

(i) 512

(ii) 343

(iii) 729

(iv) 3125

Solution:

(i) 512

512=2×2×2×2×2×2×2×2×2=29

(ii)343

343=7×7×7=73

(iii)729

729=3×3×3×3×3×3=36

(iv)3125

3125=5×5×5×5×5=55

 

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

(i) 43and34

(ii) 53or35

(iii) 28or82

(iv) 1002and2100

Solution:

(i)43=4×4×4=6434=3×3×3×3=81

Since 64<81

So, 34 is greater than 43.

 

(ii)53=5×5×5=12535=3×3×3×3×3=243

Since 125<243

So, 35 is greater than 53.

 

(iii)28=2×2×2×2×2×2×2×2=25682=8×8=64

Since 64<256

So, 28 is greater than 82.

 

(iv) 1002=100×100=10,0002100=2×2×2×2×2×..14times×.×2=16,384×..×2

So, 2100 is greater than 1002.

 

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

(i) 648

(ii) 405

(iii) 540

Solution:

(i)648

648=23×34

(ii)405

405=5×34

(iii)540

540=22×33×5

 

Q.6. Simplify:

(i) 2×103

(ii) 72×22

(iii)23×5

(iv)3×44

(v)0×102

(vi)52×33

(vii)24×32

(viii)32×104

Solution:

(i)2×103=2×10×10×10=2000

(ii)72×22=7×7×2×2=196

(iii)23×5=2×2×2×5=40

(iv)3×44=3×4×4×4×4=768

(v)0×102=0×10×10=0

(vi)53×33=5×5×5×3×3×3=675

(vii)24×32=2×2×2×2×3×3=144

(viii)32×104=3×3×10×10×10×10=90,000

 

Q.7.Simplify:

(i)(4)3

(ii)(3)×(2)3

(iii) (3)2×(5)2

(iv)(2)3×(10)3

Solution:

(i)(4)3=(4)×(4)×(4)=64

(ii)(3)×(2)3=(3)×(2)×(2)×(2)=24

(iii)(3)2×(5)2=(3)×(3)×(5)×(5)=225

(iv)(2)3×(10)3=(2)×(2)×(2)×(10)×(10)×(10)=8000

 

Q.8.Compare the following numbers:

(i)2.7×1012;1.5×108

(ii)4×1014;3×1017

Solution:

(i)2.7×1012;1.5×108

On comparing the exponents of base 10,

2.7×1012>1.5×108

(ii)4×1014;3×1017

On comparing the exponents of base 10,

4×1014<3×1017

 

Exercise-13.2

Q.1.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)2÷53

(vi)25×55

(vii)a4×b4

(viii)(34)3

(ix)(220÷215)×23

(x)8t÷82

Solution:

(i)32×34×38=3(2+4+8)=314

(ii)615÷610=61510=65

(iii)a3×a2=a3+2=a5

(iv)7x×72=7x+2

(v)(52)2÷53=52×3÷53=56÷53=563=53

(vi)25×55=(2×5)5=105

(vii)a4×b4=(a×b)4

(viii)(34)3=34×3=312

(ix)(220÷215)×23=(22015)×23=25×23=25+3=28

(x) 8t÷82=8t2

Formulae:

(i)am×an=am+n

(ii)am÷an=amn

(iii)(am)n=am×n

(iv)am×bm=(a×b)m

 

Q.2.(i)23×34×43×32

(ii)[(52)3×54]÷57

(iii)254÷53

(iv) 3×72×11821×113

(v)3734×33

(vi)20+30+40

(vii)20×30×40

(viii)30+20×50

(ix)28×a543×a3

(x)(a5a3)×a8

(xi)45×a8b345×a5b2

(xii)(23×2)2

Solution:

(i)23×34×43×32=23×34×223×25=23+2×343×25=25×343×25=255×343=20×33=1×33=33

 

(ii)[(52)3×54]÷57=[56×54]÷57=[56+4]÷57=510÷57=5107=53

 

(iii)254÷53=(52)4÷53=58÷53=583=55

 

(iv) 3×72×11821×113=3×72×1183×7×113=311×721×1183=30×71×115=7×115

 

(v)3734×33=3734+3=3737=377=30=1

 

(vi)20+30+40=1+1+1=3

 

(vii)20×30×40=1×1×1=1

 

(viii)30+20×50=(1+1)×1=2×1=2

 

(ix)28×a543×a3=28×a5(22)3×a3=28×a526×a3=286×a52=22×a2=(2a)2

 

(x)(a5a3)×a8=(a53)×a8=a2×a8=a2+8=a10

 

(xi)45×a8b345×a5b2=455×a85×b32=40×a3×b=1×a3×b=a3b

 

(xii)(23×2)2=(23+1)2=(24)2=24×2=28

 

Q.3.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)10×1011=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)23×32=65

L.H.S. 23×32=8×9=72

And

R.H.S. 65=7,776

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.

 

Q.4. 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

108×192=(22×33)×(26×3)=22+6×33+1=28×34

 

(ii)270

270=2×35×5

 

(iii)729×64

729×64=36×26\

 

(iv)768

768=28×3

 

Q.5. Simplify:

(i)(25)2×7383×7

(ii)25×52×t8103×t4

(iii)35×105×2557×65

Solution:

(i)(25)2×7383×7

(25)2×7383×7=(25×2)×73(23)3×7=210×7329×7=2109×731=2×72=2×49=98

 

(ii)25×52×t8103×t4

25×52×t8103×t4=52×52×t8(5×2)3×t4=52+2×t8423×33=54×t423×53=543×t423=5t48

 

(iii)35×105×2557×65

35×105×2557×65=35×(2×5)5×5257×(2×3)5=35×25×55×5257×25×35=35×25×55+257×25×35=35×25×5757×25×35=255×355×555=20×30×50=1×1×1=1

 

Exercise-13.3

Q.1. Write the following numbers in the expanded form:

(i)279404

(ii)3006194

(iii)2806196

(iv)120719

(v)20068

Solution:

(i)279404=2,00,000+70,000+9,000+400+00+4

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

=2×105+7×104+9×103+4×102+0×101+4×100

 

(ii)3006194=30,00,000+0+0+6,000+100+90+4

=3×1000000+0×100000+0×10000+6×1000+1×100+9×10+4×1=3×106+0×105+0×104+6×103+1×102+9×101+4×100

 

(iii)2806196=20,00,000+8,00,000+0+6,000+100+90+6

=2×1000000+8×100000+0×10000+6×1000+1×100+9×10+6×1=2×106+8×105+0×104+6×103+1×102+9×101+6×100

 

(iv)120719=1,00,000+20,000+0+700+10+9

==1×100000+2×10000+0×1000+7×100+1×10+9×1=1×105+2×104+0×103+7×102+1×101+9×100

 

(v)20068=20,000+00+00+60+8

==2×10000+0×1000+0×100+6×10+8×1=2×104+0×103+0×102+6×101+8×100

 

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

(a)8×104+6×103+0×102+4×101+5×100

(b)4×105+5×103+3×102+2×100

(c) 3×104+7×102+5×100

(d)9×105+2×102+3×101

Solution:

(a)8×104+6×103+0×102+4×101+5×100

8×10000+6×1000+0×100+4×10+5×1=80000+6000+0+40+5=86,045

 

(b)4×105+5×103+3×102+2×100

4×100000+5×1000+3×100+2×1=400000+5000+300+2=4,05,302

 

(c) 3×104+7×102+5×100

3×10000+7×100+5×1=30000+700+5=30,705

 

(d)9×105+2×102+3×101

9×100000+2×100+3×10=900000+200+30=9,00,230

 

Q.3. 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

Solution:

(i)5,00,00,000 =5×1,00,00,000=5×107

(ii)70,00,000=7×10,00,000=7×106

(iii)3,18,65,00,000= 31865×100000=3.1865×10000×100000=3.1865×109

(iv)3,90,878 =3.90878×100000=3.90878×105

(v)39087.8 =3.90878×10000=3.90878×104

(vi)3908.78=3.90878×1000=3.90878×103

 

Q.4. 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 id 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.

Solution:

(i)The distance between Earth and Moon ==384,000,000m=384×1000000m=3.84×100×1000000=3.84×108m

(ii)Speed of light in vacuum=300,000,000m/s=3×100000000m/s=3×108m/s

(iii)Diameter of the Earth=1,27,56,000m=12756×1000m=1.2756×10000×1000m=1.2756×107m

(iv)Diameter of the Sun=1,400,000,000m=14×100,000,000m=1.4×10×100,000,000m=1.4×109m

(v)Average of Stars=100,000,000,000=1×100,000,000,000=1×1011

(vi)Years of Universe=12,000,000,000years=12×1000,000,000years=1.2×10×1000,000,000years=1.2×1010years

(vii)Distance of the Sun from the centre of the Milky Way Galaxy=300,000,000,000,000,000,000m=3×100,000,000,000,000,000,000m=3×1020m

(viii)Number of molecules in a drop of water weighing 1.8 gm=60,230,000,000,000,000,000,000=6023×10,000,000,000,000,000,000=6.023×1000×10,000,000,000,000,000,000=6.023×1022

(ix)The Earth has Sea water=1,353,000,000km33=1,353×1000000km3=1.353×1000×1000,000km3=1.353×109km3

(x) The population of India=1,027,000,000=1027×1000000=1.027×1000×1000000=1.027×109