NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers

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Updated at 2021-02-11


NCERT solutions for class 8 Maths Chapter 12 Exponents And Powers Exercise 12.1

1.Evaluate:
(i) 3?2

(ii) (?4)?2

(iii) (12)?5



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i) We have:
3?2=132=19[?a?n=1an]

(ii)We have:
(?4)?2=1(?4)2=116[?a?n=1an]

(iii)We have:
(12)?5=25=32[?(ab)?n=(ba)n]

2.Simplify and express the result in power notation with a positive exponent. (i)(?4)5÷(?4)8
(ii)(123)2

(iii)(?3)4×(53)4

(iv)(3?7÷3?10)×3?5

(v)2?3×(?7)?3



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:(?4)5÷(?4)8
=(?4)5?8=(?4)?3=1(?4)3[?am÷an=am?n]

=(?14)3

(ii)We have:(123)2

(123)2=(1)2(23)2=126=(12)6

(iii)We have:(?3)4×(53)4

=(?3)4×(53)4=(?3)4×(5)4(3)4

=(3)4×(5)4(3)4=(5)4

(iv)We have:(3?7÷3?10)×3?5

=(3?7÷3?10)×3?5=3?7?(?10)×3?5

=3?7+10×3?5=33×3?5=3(3?5)

=3?2=132=(13)2

(v)We have:2?3×(?7)?3

=2?3×(?7)?3=[2×(?7)]?3=(?14)?3

=?(14)?3=?1143=(?114)3

3.Find the value of
(i)30+4?1×22

(ii)2?1×4?1÷2?2

(iii)(12)?2+(13)?2+(14)?2

(iv)(3?1+4?1+5?1)0

(v)[(?23)?2]2



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:30+4?1×22

=30+4?1×22=(1+14)×4

=(4+14)×4=54×4=5

(ii)We have:2?1×4?1÷2?2

=2?1×4?1÷2?2=(12×14)÷122

=18÷14=18×41

=12

(iii)We have:(12)?2+(13)?2+(14)?2

=(12)?2+(13)?2+(14)?2=22+32+42

=4+9+16=29

(iv)We have:(3?1+4?1+5?1)0

=(3?1+4?1+5?1)0=1[?a0=1]

(v)We have:[(?23)?2]2

=[(?23)?2]2=[(?32)2]2

=(94)2=8116

4.Evaluate
(i) 8?1×532?4

(ii) (5?1×2?1)×6?1



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:8?1×532?4

=8?1×532?4=18×53×24

=18×125×16=125×2=250

(ii)We have:(5?1×2?1)×6?1

=(15×12)×16

=110×16=160

5.Find the value of m for which 5m÷5?3=55.



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

We have given: 5m÷5?3=55.

?5m?(?3)=55 [?am÷an=am?n]

?5m+3=55

Now, Comparing the powers of equal bases, we have m + 3 = 5

?m=53=2, i.e., m = 2

6.Evaluate:
(i)[(13)?1?(14)?1]?1

(ii)(58)?7×(85)?4



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:[(13)?1?(14)?1]?1

=(31?41)?1=(?1)?1=1(?1)1=1?1=?1

(ii)We have: (58)?7×(85)?4
=(85)7×(58)4=8757×5484

=87?457?4=8353=512125

7.Simplify :
(i)25×t?45?3×10×t?8(t?0)

(ii)3?5×10?5×1255?7×6?5



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:25×t?45?3×10×t?8(t?0)

=25×5310×t(?4+8)=5×532×t4

=6252t4
(ii)We have:3?5×10?5×1255?7×6?5

=57×65×12535×105

=57×(3×2)5×5335×(2×5)5=57×35×25×5335×25×55=57+3?5=55

NCERT solutions for class 8 Maths Chapter 12 Exponents And Powers Exercise 12.2

1.Express the following numbers in standard form:
(i) 0.0000000000085

(ii) 0.00000000000942

(iii) 6020000000000000

(iv) 0.00000000837

(v) 31860000000



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:0.0000000000085
=8510000000000000=8.5×101013

=8.5×10(1?13)=8.5×10?12

(ii)We have:0.00000000000942
=942100000000000000=9.42×1021014

=9.42×10(2?14)=9.42×10?12

(iii)We have:6020000000000000=6.02×1000000000000000

=6.02×1015

(iv)We have:0.00000000837

=837100000000000=8.37×1001011

=8.37×10(2?11)=8.37×10?9

(v)We have: 31860000000

=3.186×10000000000

=3.186×1010

Hence, the required standard from = 3.186×1010

2.Express the following numbers in usual form.
(i) 3.02×10?6

(ii) 4.5×104

(iii) 3×10?8

(iv) 1.0001×109

(v) 5.8×1012

(vi) 3.61492×106



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have:3.02×10?6

=302100×1106=302100000000

=302×10?8=0.00000302

Thus, we have 3.02×10?6=0.00000302

(ii)We have:4.5×104

=4510×104=45×103

=45000

Thus, we have 4.5×104=45000

(iii)We have:3×10?8

=3108=3100000000

=0.00000003

Thus, we have 3×10?8=0.00000003

(iv)We have: 1.0001×109

=1000110000×109=10001104×109

=10001×105=1000100000

Thus, we have1.0001×109=1000100000

(v)We have: 5.8×1012

=5810×1012=58×1011

=5800000000000

Thus, we have 5.8×1012=5800000000000

(vi)We have: 3.61492×106

=361492100000×106=361492105×106

=361492×10=3614920

Thus, we have 3.61492×106=3614920

3.Express the number appearing in the following statements in standard form.
(i) 1 micron is equal to 11000000 m.

(ii) Charge of an electron is 0.000,000,000,000,000,000,16 coulomb

(iii) Size of a bacteria is 0.0000005 m

(iv) Size of a plant cell is 0.00001275 m

(v) Thickness of a thick paper is 0.07 mm.



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

(i)We have 1 micron=11000000

=1106m=10?6

(ii)We know charge of an electron=0.000,000,000,000,000,000,16

161,000,000,000,000,000,000,00=1.6×101,000,000,000,000,000,000,00

=1.6×101020=1.6×101?20=1.6×10?19

(iii)We have size of bacteria=0.0000005m

=510000000m=0.5×10107m

=0.5×101?7m=0.5×10?6m

=5×10?7m

(iv)We have size of plant cell=0.00001275m

=1275100000000m=1.275×103108m

=1.275×103?8m=1.275×10?5m

=1.275×10?5m

(v)We have thickness of paper=0.07mm

=7100mm=0.7×10102mm

=0.7×101?2mm=0.7×10?1mm

=7×10?2mm

4.In a stack there are 5 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?



NCERT Solutions for Class 8 Maths Chapter 12 Exponents And Powers


Answer :

We have the thickness of books = 5×20=100mm

Also, the thickness of 5 paper sheets = 5×0.016mm=0.080mm.

Total thickness of the stack = 100mm+0.080mm=100.080mm=1.0008×102mm



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