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 for the students of class 8 to excel in their examination. These solution help students to frame a better understanding of the topic. To ease the fear of maths, we at BYJU’S provide NCERT Solution for Class 8 Maths Chapter 12 Exponents and Powers. Student can download the NCERT Solution for class 8 Maths Chapter 12 pdf or can view it online by following the link. 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 maths chapter 12.

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