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NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Download PDF

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Download PDF

Edited By Ravindra Pindel | Updated on Feb 07, 2024 06:16 PM IST

NCERT Solutions for Exponents And Powers Class 7 - If someone asks you about the speed of bus or car, you may say its 40 km/hr but if you have asked about the speed of light in a vacuum, you will say its approximately 300,000,000 m/s. This is a very big number which is very difficult to write, interpret, and handle. To make any number to represent in an easy way there is a system called exponents. In this article, you will get CBSE NCERT solutions for Class 7 Maths chapter 13 Exponents and Powers.

This Story also Contains
  1. NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Important Formulae
  2. NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Important Points
  3. NCERT Solutions for Maths Chapter 13 Exponents and Powers Class 7
  4. NCERT for Maths Chapter 13 Exponents and Powers class 7 Topic 13.2 Exponents
  5. NCERT Solutions for Maths Chapter 13 Exponents and Powers Class 7Exercise 13.1
  6. NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.1
  7. NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.2
  8. NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.3
  9. NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Topic 13.3.4
  10. NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Topic 13.3.5
  11. NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2
  12. NCERT Solutions for Chapter 13 Maths Class 7 Exponents and Powers Topic 13.6
  13. NCERT Solutions for Chapter 13 Maths Class 7 Exponents and Powers Exercise 13.3
  14. Exponents and Powers Class 7 Maths Chapter 13- Topics
  15. Some Important Properties and Formulas from NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

Exponents can be used to represent very large numbers and small numbers as the power of a base number. In the above example, you can write 300,000,000 m/s as 3×108 . Important topics like laws of exponents, multiplying and dividing the power with the same base, multiplying and dividing the powers with the same exponents and expressing large numbers in the standard form are covered in NCERT. There are many questions from the above topics in the solutions of NCERT for Class 7 Maths chapter 13 Exponents and Powers that are explained in a step-by-step manner. It will be very easy for you to understand the concepts. In this chapter, there are 17 questions in the 3 exercises of the NCERT textbook. You will get detailed explanations of all these questions in the NCERT Solutions for Class 7 . Check NCERT Solutions from Classes 6 to 12 by clicking on the above link. Here you will get solutions to three exercises of this chapter.

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Important Formulae

am × an = am+n

am ÷ an = am-n

( am )n = amn

am × bm = (ab)m

am ÷ bn = (a÷b)m

a0 = 1 ( a, non-zero integer)

(-)even number = 1

(-)odd number = -1

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers - Important Points

Exponents: Exponents represent repeated multiplication of a number (base) by itself.

x4 where x is the base and 4 is the exponent. It is read as x raised to the power of 4 or fourth power of x.

Understanding Powers of 10:

  • Powers of 10 are commonly used in scientific notation and decimal places.
  • 10n represents a 1 followed by 'n' zeros.

Free download NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers PDF for CBSE Exam.

NCERT Solutions for Maths Chapter 13 Exponents and Powers Class 7

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NCERT Solutions for Class 7 Maths Chapter 13 Questions and Exercise

NCERT for Maths Chapter 13 Exponents and Powers class 7 Topic 13.2 Exponents

Question:(i) Express:

729 as a power of 3

Answer:

729 as a power of 3 is given as

729=3×3×3×3×3×3=36

Question:(ii) Express:

128 as a power of 2

Answer:

128 as a power of 2 can be given as

128=2×2×2×2×2×2×2=27

Question:(iii) Express:

343 as a power of 7

Answer:

343 as a power of 7 can be givena as

343=7×7×7=73

NCERT Solutions for Maths Chapter 13 Exponents and Powers Class 7Exercise 13.1

Question: 1(i) Find the value of:

(i)26

Answer:

the value of 26 is given by

2×2×2×2×2×2=64

Question: 1(ii) Find the value of:

(ii)93

Answer:

the value of (ii)93 is given by

9×9×9=729

Question:1(iii) Find the value of:

(iii)112

Answer:

the value of (iii)112 is given by

11×11=121

Question: 1(iv) Find the value of:

(iv)54

Answer:

the value of (iv)54 is given by

54=5×5×5×5=625

Question: 2 Express the following in exponential form:

(i)6×6×6×6 (ii)t×t (iii)b×b×b×b (iv)5×5×7×7×7

(v)2×2×a×a (vi)a×a×a×c×c×c×c×d

Answer:

(i)6×6×6×6 can be given as

64 .

(ii)t×t can be given as t2 .

(iii)b×b×b×b can be given as b4 .

(iv)5×5×7×7×7 can be given as

52×73 .

(v)2×2×a×a can be given as

22×a2 .

(vi)a×a×a×c×c×c×c×d can be given as

a3×c4×d .

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

(i) 512 (ii) 343 (iii) 729 (iv) 3125

Answer:

(i) 512

2
512
2
256
2
128
2
64
2
32
2
16
2
8
2
4
2
2

1

512=2×2×2×2×2×2×2×2×2=29

(ii) 343

7
343
7
49
7
7

1

343=7×7×7=73

(iii)729

3
729
3
243
3
81
3
27
3
9
3
3

1

729=3×3×3×3×3×3=36

(iv)3125

5
3125
5
625
5
125
5
25
5
5

1

3125=5×5×5×5×5=55

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

(i)43or34 (ii)53or35 (iii)28or82 (iv)1002or2100 (v)210or102

Answer:

(i)43or34

43=4×4×4=64

34=3×3×3×3=81

since 81>64

34 is greater than 43

(ii)53or35

53=5×5×5=125

35=3×3×3×3×3=243

since 243>125

35 is greater than 53

(iii)28or82

28=2×2×2×2×2×2×2×2=256

82=8×8=64

since 256>64

28 is greater than 82

(iv)1002or2100

1002=100×100=10000

2^{100}=2\times 2\times 2\times 2\times 2...........till\, 100\, times\, 2

since 2^{100}> 100^{2}

2^{100} is greater than 1002

(v)210or102

Double exponent: use braces to clarify

102=10×10=100

since 1024>100

Double exponent: use braces to clarify is greater than 102

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

(i) 648 (ii) 405 (iii) 540 (iv) 3,600

Answer:

(i) 648

2
648
2
324
2
162
2
81
3
27
3
9
3
3

1

648=23×34

(ii) 405

5
405
3
81
3
27
3
9
3
3

1

405=5×34

(iii)540

2
540
2
270
3
135
3
45
3
15
5
5

1

540=22×33×5

(iv) 3600

2
3600
2
1800
2
900
2
450
3
225
3
75
5
25
5
5

1

3600=24×32×52


Question: 6(i) Simplify:

(i)2×103

Answer:

(i)2×103

can be simplified as

2×10×10×10=2000

Question: 6(ii) Simplify:

(ii)72×22

Answer:

(ii)72×22

can be simplified as

7×7×2×2=196

Question: 6(iii) Simplify:

(iii)23×5

Answer:

(iii)23×5

can be simplified as

2×2×2×5=40

Question: 6(iv) Simplify:

(iv)3×44

Answer:

(iv)3×44

can be simplified as

3×4×4×4×4=768

Question: 6(v) Simplify:

(v)0×102

Answer:

(v)0×102

can be simplified as

0×10×10=0

Question: 6(vi) Simplify:

(vi)52×33

Answer:

(vi)52×33

can be simplified as

5×5×3×3×3=675

Question: 6(vii) Simplify:

(vii)24×32

Answer:

(vii)24×32

can be simplified as

2×2×2×2×3×3=144

Question: 6(v iii) Simplify:

(viii)32×104

Answer:

(viii)32×104

can be simplified as

3×3×10×10×10×10=90000

Question: 7(i) Simplify:

(i)(4)3

Answer:

(i)(4)3

can be simplified as

4×4×4=64

Question: 7(ii) Simplify:

(ii)(3)×(2)3

Answer:

(ii)(3)×(2)3

can be simplified as

3×2×2×2=24

Question: 7(iii) Simplify:

(iii)(3)2×(5)2

Answer:

(iii)(3)2×(5)2

can be simplified as

(3)×(3)×(5)×(5)=225

Question: 7(iv) Simplify:

(iv)(2)3×(10)3

Answer:

(iv)(2)3×(10)3

can be simplified as

(2)×(2)×(2)×10×10×10=8000

Question: 8 Compare the following numbers:

(i)2.7×1012;1.5×108 (ii)4×1014;3×1017

Answer:

(i)2.7×1012;1.5×108

on comparing exponents of base 10.

2.7×1012>1.5×108

(ii)4×1014;3×1017

on comparing exponents of base 10.

4×1014<3×1017

NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.1

Question:(i) Simplify and write in exponential form:

(i)25×23

Answer:

(i)25×23

can be simplified as

25+3=28

Question:(ii) Simplify and write in exponential form:

(ii)p3×p2

Answer:

(ii)p3×p2

can be simplified as

p(3+2)=p5

Question:(iii) Simplify and write in exponential form:

(iii)43×42

Answer:

(iii)43×42

can be simplified as

4(3+2)=45

Question:(iv) Simplify and write in exponential form:

(iv)a3×a2×a7

Answer:

(iv)a3×a2×a7

can be simplified as

Double exponent: use braces to clarify

Question:(v) Simplify and write in exponential form:

(v)53×57×512

Answer:

(v)53×57×512

can be simplified as

Double exponent: use braces to clarify

Question:(vi) Simplify and write in exponential form:

(vi)(4)100×(4)20

Answer:

(vi)(4)100×(4)20

can be simplified as

(4)(100+20)=(4)120

NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.2

Question: 1 Simplify and write in exponential form: (eg., 116÷112=114 )

(i)29÷23 (ii)108÷104 (iii)911÷97 (iv)2015÷2013 (v)713÷710

Answer:

(i)29÷23

can be simplified as 293=26

(ii)108÷104

can be simplified as 10(84)=104

(iii)911÷97

can be simplified as 9(117)=94

(iv)2015÷2013

can be simplified as 20(1513)=202

(v)713÷710

can be simplified as 7(1310)=73

NCERT Solutions for Class 7th Math Chapter 13 Exponents and Powers Topic 13.3.3

Question: Simplify and write the answer in exponential form:

(i)(62)4 (ii)(22)100 (iii)(750)2 (iv)(53)7

Answer:

(i)(62)4

can be simplified as 6(2×4)=68

(ii)(22)100

can be simplified as 2(2×100)=2200

(iii)(750)2

can be simplified as 7(50×2)=7100

(iv)(53)7

can be simplified as 5(3×7)=521

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Topic 13.3.4

Question: Put into another form using am×bm=(ab)m :

(i)43×23 (ii)25×b5 (iii)a2×t2 (iv)56×(2)6 (v)(2)4×(3)4

Answer:

(i)43×23

can be simplified as (4×2)3=83

(ii)25×b5

can be simplified as (2×b)5=(2b)5

(iii)a2×t2

can be simplified as (a×t)2=at2

(iv)56×(2)6

can be simplified as (5×(2))6=106

(v)(2)4×(3)4

can be simplified as ((2)×(3))4=64

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Topic 13.3.5

Question: Put into another form using am÷bm=(ab)m :

(i)45÷35 (ii)25÷b5 (iii)(2)3÷b3 (iv)p4÷q4 (v)56÷(2)6

Answer:

(i)45÷35

can be simplified as

(43)5

(ii)25÷b5

can be simplified as

(2b)5

(iii)(2)3÷b3

can be simplified as

(2b)3

(iv)p4÷q4

can be simplified as

(pq)4

(v)56÷(2)6

can be simplified as

(52)6

NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers Exercise 13.2

Question: 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)3÷53 (vi)25×55 (vii)a4×b4 (viii)(34)3

(ix)(220÷215)×23 (x)8t÷82

Answer:

(i)32×34×38

can be simplified as 3(2+4+8)=314

(ii)615÷610

can be simplified as 6(1510)=65

(iii)a3×a2

can be simplified as a(3+2)=a5

(iv)7x×72

can be simplified as 7(x+2)=7(x+2)

(v)(52)3÷53

can be simplified as 5(2×3)÷5(3)=56÷53=563=53

(vi)25×55

can be simplified as

(2×5)5=105

(vii)a4×b4

can be simplified as (ab)4

(viii)(34)3

can be simplified as 34×3=312

(ix)(220÷215)×23

can be simplified as 2(2015)×23=25×23=2(5+3)=28

(x)8t÷82

can be simplified as 8(t2)

Question:2(i) Simplify and express each of the following in exponential form:

(i)23×34×43×32

Answer:

(i)23×34×43×32

can be simplified as

=23×34×223×25

=2(3+2)×343×25

=25×343×25

=2(55)×3(41)

=33

Question: 2(ii) Simplify and express each of the following in exponential form:

(ii)((52)3×54)÷57

Answer:

(ii)((52)3×54)÷57

can be simplified as

=[5(2×3)×54]÷57

=[56×54]÷57

=[5(6+4)]÷57

=[510]÷57

=5107

=53

Question: 2(iii) Simplify and express each of the following in exponential form:

(iii)254÷53

Answer:

(iii)254÷53

can be simplified as

=(52)4÷53

=5(2×4)÷53

=58÷53

=5(83)

=55

Question: 2(iv) Simplify and express each of the following in exponential form:

(iv)3×72×11821×113

Answer:

(iv)3×72×11821×113

can be simplified as

=3×72×1183×7×113

=3(11)×7(21)×11(83)

=30×71×115

=71×115

Question: 2(v) Simplify and express each of the following in exponential form:

(v)3734×33

Answer:

(v)3734×33

can be simplified as

=3734+3

=3737

=3(77)

=30

=1

Question: 2(vi) Simplify and express each of the following in exponential form:

(vi)20+30+40

Answer:

(vi)20+30+40

can be simplified as

=1+1+1

=3

Question: 2(vii) Simplify and express each of the following in exponential form:

(vii)20×30×40

Answer:

(vii)20×30×40

can be simplified as

=1×1×1=1

Question: 2(viii) Simplify and express each of the following in exponential form:

(viii)(30+20)×50

Answer:

(viii)(30+20)×50

can be simplified as

=(1+1)×1

=2×1

=2

Question: 2(ix) Simplify and express each of the following in exponential form:

(ix)28×a543×a3

Answer:

(ix)28×a543×a3

can be simplified as

=28×a5(22)3×a3

=28×a526×a3

=2(86)×a(53)

=2(2)×a(2)

=(2a)2

Question: 2(x) Simplify and express each of the following in exponential form:

(x)(a5a3)×a8

Answer:

(x)(a5a3)×a8

can be simplified as

=a(53)×a8

=a(2)×a8

=a(2+8)

=a(10)

Question: 2(xi) Simplify and express each of the following in exponential form:

(xi)45×a8b345×a5b2

Answer:

(xi)45×a8b345×a5b2

can be simplified as

=4(55)×a(85)×b(32)

=40×a3×b1

=a3×b

Question:2(xii) Simplify and express each of the following in exponential form:

(xii)(23×2)2

Answer:

(xii)(23×2)2

can be simplified as

=(2^{3+1})^{2}

=(2(4))2

=2(4×2)

=28

Question: 3 Say true or false and justify your answer:

(i)10×1011=10011 (ii)23>52 (iii)23×32=65 (iv)30=(1000)0

Answer:

(i)10×1011=10011

can be simplified as

LHS:10(1+11)

=10(12)

Since , LHSRHS

Thus, it is false

(ii)23>52

can be simplified as

LHS=23=8

RHS=52=25

Since, LHSRHS

Thus, it is false

(iii)23×32=65

can be simplified as

LHS:23×32=8×9=72

RHS:65=7776

Since , LHSRHS

Thus, it is false

(iv)30=(1000)0

can be simplified as

LHS:30=1

RHS:10000=1

Since, LHS = RHS

Thus, it is true.

Question: 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

Answer:

(i)108×192

2
108
2
54
3
27
3
9
3
3

1


2
192
2
96
2
48
2
24
2
12
2
6
3
3

1


108×192=(22×33)×(26×3)

=2(2+6)×3(3+1)

=28×34


(ii)270

2
270
3
135
3
45
3
15
5
5

1

270=2×33×5


(iii)729×64

3
729
3
243
3
81
3
27
3
9
3
3

1


2
64
2
32
2
16
2
8
2
4
2
2

1

729×64=36×26


(iv)768

2
768
2
384
2
192
2
96
2
48
2
24
2
12
2
6
3
3

1


768=28×3

Question: 5(i) Simplify:

(i)(25)2×7383×7

Answer:

(i)(25)2×7383×7

can be simplified as

=2(5×2)×73(23)3×7

=210×73(2(3×3))×7

=210×73(29)×7

=2(109)×7(31)

=2×72

=2×49=98

Question: 5(ii) Simplify:

(ii)25×52×t8103×t4

Answer:

(ii)25×52×t8103×t4

can be simplified as

=52×52×t8(2×5)3×t4

=5(2+2)×t823×53×t4

=54×t823×53×t4

=543×t(84)23

=5×t423

=5t48

Question: 5(iii) Simplify:

(iii)35×105×2557×65

Answer:

(iii)35×105×2557×65

can be simplified as

=35×(2×5)5×5257×(2×3)5

=35×25×55×5257×25×35

=3(55)×2(55)×5(5+27)

=30×20×50

=1×1×1=1

NCERT Solutions for Chapter 13 Maths Class 7 Exponents and Powers Topic 13.6

Question: Expand by expressing powers of 10 in the exponential form:

(i) 172 (ii) 5,643 (iii) 56,439 (iv) 1,76,428

Answer:

(i) 172

172=100+70+2

=100+(7×10)+2

=1×102+(7×101)+2×100

(ii) 5,643

5643=5000+600+40+3

=5×1000+6×100+4×10+3

=5×103+6×102+4×101+3×100

(iii) 56,439

56439=50000+6000+400+30+9

=5×10000+6×1000+4×100+3×10+9

=5×104+6×103+4×102+3×101+9×100

(iv) 1,76,428

176428=100000+70000+6000+400+20+8

=1×100000+7×10000+6×1000+4×100+2×10+8

=1×105+7×104+6×103+4×102+2×101+8×100

NCERT Solutions for Chapter 13 Maths Class 7 Exponents and Powers Exercise 13.3

Question: 1 Write the following numbers in the expanded forms:

279404, 3006194, 2806196, 120719, 20068

Answer:

(i) 279404

279404=200000+70000+9000+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

3006194=3000000+0+0+6000+100+90+4

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

(iii) 2806196

2806196=2000000+800000+0+6000+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

120719=100000+20000+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

20068=20000+0+0+60+8

=2×10000+0×1000+0×100+6×10+8×1

=2×104+0×103+0×102+6×101+8×100

Question: 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

Answer:

(a)8×104+6×103+0×102+4×101+5×100

=8×10000+6×1000+0×100+4×10+5×1

=80000+6000+000+40+5

=86045

(b)4×105+5×103+3×102+2×100

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

=400000+00000+5000+300+00+2

=405302

(c)3×104+7×102+5×100

=3×10000+0×1000+7×100+0×10+5×1

=30000+0000+700+00+5

=30705

(d)9×105+2×102+3×101

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

=900000+00000+0000+200+30+0

=900230

Thus, the above problems are simplified in simpler forms.

Question: 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

Answer:

(i) 5,00,00,000

50000000=5×10000000=5×107

(ii) 70,00,000

7000000=7×1000000=7×106

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

3186500000=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

Question: 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 the Earth is 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 100,000,000,000 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.

Answer:

(a) The distance between Earth and Moon = 384,000,000 m

=384×1000000

=3.84×100×1000000

=3.84×108m

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

=3×100000000

=3×108

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

=12756×1000

=1.2756×10000×1000

=1.2756×107m

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

=14×100000000

=14×108m

=1.4×109m

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

=1×100000000000

=1×1011

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

=1.2×10000000000

=1.2×1010years


(g) The distance of the Sun from the centre of the Milky Way Galaxy is estimated = 300,000,000,000,000,000,000 m.

=3×100000000000000000000000000000

=3×1019

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

60,230,000,000,000,000,000,000

=6023×10000,000,000,000,000,000

=6023×1019

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

=1.353×1000000000

=1.353×109km3

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

=1.027×1000000000

=1.027×109

Exponents and Powers Class 7 Maths Chapter 13- Topics

  • 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 Exponents
  • Dividing Powers With The Same Exponents
  • Miscellaneous Examples Using The Laws Of Exponents
  • Decimal Number System
  • Expressing Large Numbers In The Standard Form

NCERT Solutions for Class 7 Maths Chapter Wise

Chapter No.

Chapter Name

Chapter 1

Integers

Chapter 2

Fractions and Decimals

Chapter 3

Data Handling

Chapter 4

Simple Equations

Chapter 5

Lines and Angles

Chapter 6

The Triangle and its Properties

Chapter 7

Congruence of Triangles

Chapter 8

Chapter 9

Rational Numbers

Chapter 10

Practical Geometry

Chapter 11

Perimeter and Area

Chapter 12

Algebraic Expressions

Chapter 13

Exponents and Powers

Chapter 14

Symmetry

Chapter 15

NCERT Solutions for Class 7 Subject Wise

Some Important Properties and Formulas from NCERT Solutions for Class 7 Maths Chapter 13 Exponents and Powers

For any non-zero integers, a and b and whole numbers m and n it obey certain properties given below

  •  am×an=am+n
  • aman=amn
  •  (am)n=amn
  • am×bm=(ab)m
  • am÷bm=(ab)m
  • a0=1
  • (-1)^{\text {even number }}=1
  • (1)addnumber=1

Tip- If you understood the fundamental of this chapter, you don't need to remember above these findings. You can simply derive these properties by the strong fundamentals of this chapter. You should try to solve all the NCERT questions including the practice questions given at the end of every topic. You can take help from the CBSE NCERT solution for class 7 maths chapter 13 exponents and powers if you are facing difficulties while solving the problems.

Happy Reading!!!

Also Check NCERT Books and NCERT Syllabus here:

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