NCERT Solutions For Class 8 Maths Chapter 12

NCERT Solutions Class 8 Maths Exponents and Powers

NCERT Solutions for class 8 Maths chapter 12 Exponents and Powers is crucial chapter for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. Exponents are applied in various algebra questions, therefore, it is essential to understand the rules of exponents. There are 2 easy "rules of 1" to master. First, any amount raised to the power of "1" equals itself. Secondly, 1 raised to any power is one. Multiplication of 2 powers having identical base can be simplified by adding them. For raising a power to a power we simply need to multiply the exponents. Two powers can be divided into the same base by subtracting the exponents. Any non-zero number raised to the power of 0 = 1. Any non-zero number raised to a negative power is equal to its reciprocal raised to the opposite positive power.


Several times large numbers are tough to understand, read, operate, therefore, superscripts are applied to transform massive numbers into the modest form with the assistant of powers and exponents. Exponents describe the number of times the base is multiplied. whereas, when it comes to power, it comprises of 2 parts, i.e. the exponent and a base number. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 12 Exponents and Powers. We have provided a comprehensive study material for  NCERT Solution for Class 8 Maths Chapter 12 Exponents and Powers. The students can download and understand all the exercise-wise NCERT problems on Exponents and Powers from the links provided below.

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NCERT Solutions Class 8 Maths Chapter 12 Exercises

Exercise-12.1

Q.1. (i)43

(ii)(5)5

(iii)(12)4

Solution:

(i)42

(4)2=142=116

 

(ii)(5)2

(5)2=1(5)2=125

 

(iii)(12)4

(12)2=(21)2=(2)2=4

Formula: am=1am

 

Q.2. (i)(5)4÷(5)7

(ii)(143)2

(iii)(4)3×(54)3

(iv)(27÷210)×25

(v)33×(7)3

Solution:

(i)(5)4÷(5)7

(5)4÷(5)7=(5)47=(5)3=1(5)3

 

(ii)(143)2

(143)2=12(43)2=143×2=146

 

(iii)(4)3×(54)3

(4)2×(54)2=(4)2×5242=[(1)2×42]×5242=422×52=40×52=52

 

(iv)(27÷210)×25

(27÷210)×25=27(10)×25=27+10×25=23×25=235=22=122

 

(v)33×(7)3

33×(7)3=133×1(7)3=1[3×(7)]3=1(21)3

 

Formulae:

1.(am)n=am×n

2.(ab)m=ambm

3.(ab)m=ambm

4.am÷an=amn

5.am×an=am+n

 

Q.3. Find the value of:

(i)(30+41)×22

(ii)(21×41)÷22

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

(iv)(31+41+51)0

(v)[(23)2]2

Solution:

(i)(30+41)×22

(1+14)×22=(4+14)×22=54×22=5×222=5×20=5×1=5

 

(ii)(21×41)÷22

(21×41)÷22=(121×141)÷22=(12×122)÷22=123÷22=23÷22=23(2)=23+2=21=12

 

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

(12)2+(13)2+(14)2=(21)2+(31)2+(41)2=21×(2)+31×(2)+41×(2)=22+32+42=4+9+16=29

 

(iv)(31+41+51)0

(31+41+51)0=(13+14+15)0=(20+15+1260)0=(4760)0=1

 

(v)[(23)2]2

[(23)2]2=(23)2×2=(23)4=(32)4=8116

 

Q.4. Evaluate:

(i)81×5324

(ii)(51×21)×61

Solution:

(i)81×5324

81×5324=(23)1×5324=23×5324=23(4)×53=23+4×53=2×125=250

 

(ii)(51×21)×61

(51×21)×61=(15×12)×16=110×16=160

 

Q.5. Find the value of m for which5m÷53=55.

Solution:

5m÷53=555m(3)=555m+3=55 5m÷53=555m(3)=555m+3=55m+3=5m=53m=2

 

Q.6. Evaluate:

(i) [(13)1(14)1]1

(ii)(58)7×(85)4

Solution:

(i) [(13)1(14)1]1

[(13)1(14)1]1=[(31)1(41)1]( becauseam=1m)=[34]=1

 

(ii)(58)7×(85)4

(58)7×(85)4=5787×8454=57(4)×84(7)=57+4×84+7=53×83=8353=512125

 

Q.7. Simplify:

(i)25×t453×10×t8(t0)

 

(ii)35×105×12557×65

Solution:

(i)25×t453×10×t8(t0)

25×t453×10×t8=52×t453×5×2×t8=52(3)1×t4(8)2=52+31×t4+82=54×t42=6252t4

 

(ii)35×105×12557×65

35×105×12557×65=35×(2×5)5×5357×(2×3)5=35×25×55×5357×25×35=35×25×55+357×25×35=35×25×5257×25×35=35(5)×25(5)×52(7)=35+5×25+5×52+7=30×20×55=1×1×3125=3125

 

Exercise-12.2

Q.1. Converts the numbers into standard form:

(i)0.0000000000085

(ii)0.00000000000942

(iii)6020000000000000

(iv)0.00000000837

(v)31860000000

 

Solution:

(i)0.0000000000085

0.0000000000085=0.0000000000085×10121012=8.5×1012

 

(ii)0.00000000000942

0.00000000000942=0.00000000000942×10121012=9.42×1012

 

(iii)6020000000000000

6020000000000000=6020000000000000×10151015=6.02×1015

 

(iv)0.00000000837

0.00000000837=0.00000000837×109109=8.37×109

 

(v)31860000000

31860000000=31860000000×10101010=3.186×1010

 

Q.2. Convert the following numbers in usual form:

(i)3.02×106

(ii)4.5×104

(iii)3×108

(iv)1.0001×109

(v) 5.8×1012

(vi)3.61492×106

 

Solution:

(i)3.02×106

3.02×106=3.02106=0.00000302

 

(ii)4.5×104

4.5×104=4.5×10000=45000

 

(iii) 3×108

3×108=3108=0.00000003

 

(iv) 1.0001×109

1.0001×109=1000100000

 

(v) 5.8×1012

5.8×1012=5.8×1000000000000=5800000000000

 

(vi) 3.61492×106

3.61492×106=3.61492×1000000=3614920

 

Q.3. Conert the given numbers in the following statements in standard form:

(i) 1 micron is equal to 11000000 m.

(ii)Charge of an electron is 0.000,000,000,000,000,000,28 coulomb.

(iii)Size of bacteria is 0.0000005 m.

(iv)Size of a plant cell is 0.00001275 m.

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

 

Solution:

(i)1 micron =11000000=1106=1×106m

 

(ii) Charge of an electron is 0.00000000000000000028 coulombs.

=0.00000000000000000028×10191019=2.8×1019coulomb

 

(iii)Size of bacteria = 0.0000005 =510000000=5107=5×107m

 

(iv)Size of a plant cell is 0.00001275 m =0.00001275×105105=1.275×105m

 

(v) Thickness of a thick paper = 0.07 mm =7100mm=7102=7×102mm

 

Q.4. In a stack there are 6 books each of thickness 20 mm and 5 paper sheets each of thickness 0.016 mm. What is the total thickness of the stack?

Solution:

Thickness of one book = 20 mm

Thickness of 6 books =20×6=120mm

Thickness of one paper = 0.016 mm

Thickness of 5 papers =0.016×5=0.08mm

Total thickness of a stack =120+0.08

=120.08 mm=120.08×102102=1.2008×102mm

The students can also download the NCERT Solution for class 8 Maths Chapter 12 pdf or can view it online by following the link mentioned above. These solution are provided in detailed manner, where one can find step-by-step solution to all the questions of NCERT Solutions for class 8 mathschapter 12.

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