CBSE NCERT Solutions For Class 8 Maths Chapter 12 PDF - Embibe
  • Written By Khushbu
  • Last Modified 29-06-2022
  • Written By Khushbu
  • Last Modified 29-06-2022

NCERT Solutions Class 8 Maths Chapter 12

In this post, we have provided NCERT Solutions for Class 8 Maths Chapter 12. The 12th Chapter of CBSE Class 8 Mathematics, ‘Exponents and Powers,’ covers some of the most important concepts in mathematics. As a result, it is important that students have fast access to answers for the exercise problems presented in this chapter throughout their study. The solutions provided here were created by Embibe experts to assist ambitious students seeking for top grades in their exams. Students can use these NCERT Class 8 Mathematics Exercise Solutions to double-check their answers and evaluate their own performance during practise. Furthermore, students can acquire a variety of shortcuts and approaches for solving problems from the solutions.

NCERT Solutions For Class 8 Maths Chapter 12: Exponents And Powers

We have provided the CBSE Maths Class 8 NCERT Solutions for Chapter 12 below in downloadable format. Click on the name of the exercise to download PDF files containing the solutions to the questions in that exercise.

ExerciseTopics
12.1Powers With Negative Exponents
12.2Laws of Exponents
12.3Use of Exponents to Express Small Numbers in Standard Form

Also, Check:

CBSE Class 8 English SyllabusNCERT Class 8 English Book
CBSE Class 8 Maths SyllabusNCERT Class 8 Maths Book
CBSE Class 8 Science SyllabusNCERT Class 8 Science Book
CBSE Class 8 Social Science SyllabusNCERT Class 8 Social Science Book
CBSE Class 8 Hindi SyllabusNCERT Class 8 Hindi Book

NCERT Solutions For Class 8 Maths Chapter 12: Solved Exercises And In-Text Questions

Students can check Class 8 NCERT Solutions for Maths for each exercise and in-text questions on this page. Also, we have provided NCERT Class 8 Maths Chapter 12 PDF download link below these solutions:

DOWNLOAD CBSE CLASS 8 MATHS CHAPTER 12 SOLUTIONS PDF NOW

Download CBSE Class 8 Solutions for Maths for other chapters from the table below:

Maths NCERT Chapter 12 Exponents and Powers: Solutions

Exercise 12.1 Page No: 197

1. Evaluate:
(i) 3-2 (ii) (-4)-2 (iii) (1/2)-5  
Solution:

Exercise 12.1 

1. Evaluate the Following:

(i) 3−2

Ans: We have to evaluate 3−2

We will apply the identity of indices a−n=1/an , we get

3−2=1/32

⇒3−2 =1/(3×3 )

∴3−2=1/9

(ii) (−4)−2

Ans: We have to evaluate (−4)−2

We will apply the identity of indices a−n=1/an, we get

(−4)−2 = 1/(−4)2

⇒(−4)−2 = 1/(−4×−4)

∴(−4)−2 = 1/16

(iii) (1/2)−5

Ans: We have to evaluate (1/2)−5.

We will apply the identity of indices (1/a)m=1m/am, we get

(1/2)-5=1-5/2-5

⇒(1/2)-5=1/2-5

We can apply the identity 1/a-n=an, we get

⇒(1/2)-5=25

⇒(1/2)-5=2×2×2×2×2⇒(1/2)-5=2×2×2×2×2

∴(1/2)-5=32∴(1/2)-5=32

2. Simplify and Express the Result in Power Notation With Positive Exponent.

(i) (−4)5÷(−4)8

Ans: We have to simplify the expression (−4)5÷(−4)8.

According to the Quotient of Power rule of exponents when the bases are same in division we can subtract the powers. We get

am÷an = am-n

Now, applying the above identity to the given expression, we get

(−4)5÷(−4)8=45-8

⇒\({(-4)}^5\div{(-4)}^8=4^{-3}\)

We have to express the result with a positive exponent, we can write 

a−n=1/an

∴(−4)5÷(−4)8=1/43

(ii) (1/23)2 

Ans: We have to simplify the expression (1/23)2.

We know that we can apply the identity of indices (1/a)m=1m/am, we get

(1/23)2=12/((2)3)2

Now, by applying the identity power of power (am)n=amn(am)n=amn, we get

⇒ (1/23)2=12/(2)3×2

∴(1/23)2=12/26

(iii) (−3)4×(5/3)4

Ans: We have to simplify the expression (−3)4×(5/3)4.

We can apply the identity of indices (1/a)m=1m/am, we get

(−3)4×(5/3)4=(−3)4×(54/34)

⇒(−3)4×(5/3)4=(−1)4×34×(54/34)

⇒(−3)4×(5/3)4=(−1)4×54

⇒(−3)4×(5/3)4=(−1)4×54

⇒(−3)4×(5/3)4=1×54

∴(−3)4×(5/3)4=54

(iv) (3−7÷3−10)×3−5

Ans: We have to simplify the expression (3−7÷3−10)×3−5

According to the Quotient of Power rule of exponents when the bases are same in division we can subtract the powers. We get

am÷an=am−n

Now, applying the above identity to the given expression, we get

(3−7÷3−10)×3−5 = 3−7−(−10)×3−5

⇒(3−7÷3−10)×3−5=3−7+10×3−5

⇒(3−7÷3−10)×3−5=33×3−5

Now, according to the product of power rule of exponents

am×an=am+n

Now, applying the above identity , we get

⇒(3−7÷3−10)×3−5=33+(−5)

⇒(3−7÷3−10)×3−5=3−2

We have to express the result with positive exponent, we can write 

a−n=1/an∴(3−7÷3−10)×3−5=1/32

(v) 2−3×(−7)−3

Ans: We have to simplify the expression 2−3×(−7)−3.

We know that an×bn=(ab)n

Now, applying the above identity to the given expression, we get

⇒2−3×(−7)−3=(2×(−7))−3

⇒2−3×(−7)−3=(−14)−3

We have to express the result with positive exponent, we can write 

a−n=1/an

∴2−3×(−7)−3=1/(−14)3

3. Find the Value of Following:

(i) (30+4−1)×22

Ans: We have to find the value of (30+4−1)×22.

We will apply the identity (1/a)m=1m/am, we get

(30+4−1)×22=(30+1/41)×22

Now, we know that a0=1, we get

⇒(30+4−1)×22=(1+1/4)×22

⇒(30+4−1)×22=(4+1/4)×2×2

⇒(30+4−1)×22=(5/4)×4

∴(30+4−1)×22=5

(ii) (2−1×4−1)÷2−2

Ans: We have to find the value of (2−1×4−1)÷2−2.

The given expression can be written as 

(2−1×4−1)÷2−2=(2−1×(22)−1)÷2−2

We will apply the identity (am)n=amn, we get

⇒(2−1×4−1)÷2−2=(2−1×2−2)÷2−2

Now, we know that am×an=am+n, we get

⇒(2−1×4−1)÷2−2=(2−1+(−2))÷2−2

⇒(2−1×4−1)÷2−2=2−3÷2−2

We will apply the identity am÷an=am−nam÷an=am−n, we get

⇒(2−1×4−1)÷2−2=2−3−(−2)

⇒(2−1×4−1)÷2−2=2−3+2

⇒(2−1×4−1)÷2−2=2−1

∴(2−1×4−1)÷2−2=1/2

(iii) (1/2)−2+(1/3)−2+(1/4)−2

Ans: We have to find the value of (1/2)−2+(1/3)−2+(1/4)−2.

We will apply the identity (1/a)m=1m/am, we get

(1/2)−2+(1/3)−2+(1/4)−2=(1−2/2−2)+(1−2/3−2)+(1−2/4−2)

⇒(1/2)−2+(1/3)−2+(1/4)−2=(1/2−2)+(1/3−2)+(1/4−2

We can apply the identity 1/a−n=an, we get

⇒(1/2)−2+(1/3)−2+(1/4)−2=22+32+42

⇒(1/2)−2+(1/3)−2+(1/4)−2=2×2+3×3+4×4

⇒(1/2)−2+(1/3)−2+(1/4)−2=4+9+16

∴(1/2)−2+(1/3)−2+(1/4)−2=29

(iv) (3−1+4−1+5−1)0(3−1+4−1+5−1)0

Ans: We have to find the value of (3−1+4−1+5−1)0(3−1+4−1+5−1)0.

We know that a0=1a0=1, then we get the value of the given expression 

∴(3−1+4−1+5−1)0=1∴(3−1+4−1+5−1)0=1

(v) {(−23)−2}2{(−23)−2}2 

Ans: We have to find the value of {(−23)−2}2{(−23)−2}2.

We can apply the identity (am)n=amn(am)n=amn, we get

{(−23)−2}2=(−23)−2×2{(−23)−2}2=(−23)−2×2

⇒{(−23)−2}2=(−23)−4⇒{(−23)−2}2=(−23)−4 

Now, applying the identity (1a)m=1mam(1a)m=1mam, we get

⇒{(−23)−2}2=(−2−43−4)⇒{(−23)−2}2=(−2−43−4)

Now, we know that 1a−n=an1a−n=an, we get

⇒{(−23)−2}2=(34−24)⇒{(−23)−2}2=(34−24)

⇒{(−23)−2}2=(3×3×3×3−2×−2×−2×−2)⇒{(−23)−2}2=(3×3×3×3−2×−2×−2×−2)

∴{(−23)−2}2=8116∴{(−23)−2}2=8116

4. Evaluate the Following:

(i) 8−1×532−48−1×532−4 

Ans: We have to evaluate the given expression 8−1×532−48−1×532−4.

We can write the given expression as (23)−1×532−4(23)−1×532−4.

⇒8−1×532−4=2−3×532−4⇒8−1×532−4=2−3×532−4

Now, we know that am÷an=am−nam÷an=am−n, we get

⇒8−1×532−4=2−3−(−4)×53⇒8−1×532−4=2−3−(−4)×53

⇒8−1×532−4=2−3+4×53⇒8−1×532−4=2−3+4×53

⇒8−1×532−4=21×53⇒8−1×532−4=21×53

⇒8−1×532−4=2×5×5×5⇒8−1×532−4=2×5×5×5

∴8−1×532−4=250∴8−1×532−4=250

(ii) (5−1×2−1)×6−1(5−1×2−1)×6−1

Ans: We have to evaluate the given expression (5−1×2−1)×6−1(5−1×2−1)×6−1.

We know that a−n=1ana−n=1an, we get

(5−1×2−1)×6−1=(15×12)×16(5−1×2−1)×6−1=(15×12)×16

⇒(5−1×2−1)×6−1=110×16⇒(5−1×2−1)×6−1=110×16

∴(5−1×2−1)×6−1=160∴(5−1×2−1)×6−1=160

5. Find the value of mm for which 5m÷5−3=555m÷5−3=55 .

Ans: The given expression is 5m÷5−3=555m÷5−3=55.

Now, according to the Quotient of Power rule of exponents when the bases are same in division we can subtract the powers. We get

am÷an=am−nam÷an=am−n 

Applying the above identity to the given expression, we get

5m−(−3)=555m−(−3)=55

⇒5m+3=55⇒5m+3=55 

Since the bases are same on both sides, therefore the exponents must be equal to each other, we get

⇒m+3=5⇒m+3=5 

⇒m=5−3⇒m=5−3 

∴m=2∴m=2 

Therefore, we get the value of m=2m=2.

6. Evaluate the Following:

(i) {(13)−1−(14)−1}−1{(13)−1−(14)−1}−1 

Ans: Given expression is {(13)−1−(14)−1}−1{(13)−1−(14)−1}−1.

We know that a−n=1ana−n=1an, applying to the given expression we get{(13)−1−(14)−1}−1={(31)1−(41)1}−1{(13)−1−(14)−1}−1={(31)1−(41)1}−1⇒{(13)−1−(14)−1}−1={3−4}−1⇒{(13)−1−(14)−1}−1={3−4}−1⇒{(13)−1−(14)−1}−1=(−1)−1⇒{(13)−1−(14)−1}−1=(−1)−1

Again applying the identity a−n=1ana−n=1an, we get⇒{(13)−1−(14)−1}−1=1−1⇒{(13)−1−(14)−1}−1=1−1∴{(13)−1−(14)−1}−1=−1∴{(13)−1−(14)−1}−1=−1

(ii) (58)−7×(85)−4(58)−7×(85)−4 

Ans: Given expression is (58)−7×(85)−4(58)−7×(85)−4.

Now, applying the identity (1a)m=1mam(1a)m=1mam, we get

(58)−7×(85)−4=(5−78−7)×(8−45−4)(58)−7×(85)−4=(5−78−7)×(8−45−4)

Now, we know that a−n=1ana−n=1an, we get⇒(58)−7×(85)−4=8757×5484⇒(58)−7×(85)−4=8757×5484

Now, applying the identity am÷an=am−nam÷an=am−n, we get⇒(58)−7×(85)−4=87−457−4⇒(58)−7×(85)−4=87−457−4⇒(58)−7×(85)−4=8353⇒(58)−7×(85)−4=8353⇒(58)−7×(85)−4=8×8×85×5×5⇒(58)−7×(85)−4=8×8×85×5×5∴(58)−7×(85)−4=512125∴(58)−7×(85)−4=512125

7. Simplify the Given Expressions:

(i) 25×t−45−3×10×t−825×t−45−3×10×t−8 

Ans: Given expression 25×t−45−3×10×t−825×t−45−3×10×t−8 can be written as 25×t−45−3×10×t−8=52×t−45−3×5×2×t−825×t−45−3×10×t−8=52×t−45−3×5×2×t−8

⇒25×t−45−3×10×t−8=52×t−45−3+1×2×t−8⇒25×t−45−3×10×t−8=52×t−45−3+1×2×t−8 

Now, applying the identity am÷an=am−nam÷an=am−n, we get

⇒25×t−45−3×10×t−8=52−(−2)×t−4−(−8)2⇒25×t−45−3×10×t−8=52−(−2)×t−4−(−8)2

⇒25×t−45−3×10×t−8=54×t42⇒25×t−45−3×10×t−8=54×t42

⇒25×t−45−3×10×t−8=5×5×5×5×t42⇒25×t−45−3×10×t−8=5×5×5×5×t42

∴25×t−45−3×10×t−8=625t42∴25×t−45−3×10×t−8=625t42

(ii) 3−5×10−5×1255−7×6−53−5×10−5×1255−7×6−5

Ans: Given expression 3−5×10−5×1255−7×6−53−5×10−5×1255−7×6−5 can be written as 3−5×10−5×1255−7×6−5=3−5×(2×5)−5×535−7×(2×3)−53−5×10−5×1255−7×6−5=3−5×(2×5)−5×535−7×(2×3)−5

⇒3−5×10−5×1255−7×6−5=3−5×2−5×5−5×535−7×2−5×3−5⇒3−5×10−5×1255−7×6−5=3−5×2−5×5−5×535−7×2−5×3−5 

Now, applying the identity am÷an=am−nam÷an=am−n, we get

⇒3−5×10−5×1255−7×6−5=3−5−(−5)×2−5−(−5)×5−5+3−(−7)⇒3−5×10−5×1255−7×6−5=3−5−(−5)×2−5−(−5)×5−5+3−(−7)

⇒3−5×10−5×1255−7×6−5=30×20×55⇒3−5×10−5×1255−7×6−5=30×20×55

We know that a0=1a0=1, we get

⇒3−5×10−5×1255−7×6−5=1×1×52⇒3−5×10−5×1255−7×6−5=1×1×52

∴3−5×10−5×1255−7×6−5=55

Practice Latest 8th CBSE Book Questions Aligned to the New Syllabus With Quick Hints & Easy Solutions.

Exercise 12.2 Page No: 200

1. Express the following numbers in standard form.

(i) 0.0000000000085
(ii) 0.00000000000942
(iii) 6020000000000000
(iv) 0.00000000837
(v) 31860000000

Solution:
(i) 0.0000000000085 = 0.0000000000085×(1012/1012) = 8.5 ×10-12
(ii) 0.00000000000942 = 0.00000000000942×(1012/1012) = 9.42×10-12
(iii) 6020000000000000 = 6020000000000000×(1015/1015) = 6.02×1015
(iv) 0.00000000837 = 0.00000000837×(109/109) = 8.37×10-9
(v) 31860000000 = 31860000000×(1010/1010) = 3.186×1010

Practice 8th CBSE Exam Questions

2. Express the following numbers in the 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

Solution:

(i) 3.02×10-6 = 3.02/106 = 0 .00000302
(ii) 4.5×104 = 4.5×10000 = 45000
(iii) 3×10-8 = 3/108 = 0.00000003
(iv) 1.0001×109 = 1000100000
(v) 5.8×1012 = 5.8×1000000000000 = 5800000000000
(vi) 3.61492×10= 3.61492×1000000 = 3614920
3. 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?
 
Solution:
Thickness of one book = 20 mm
Thickness of 5 books = 20×5 = 100 mm
Thickness of one paper = 0.016 mm
Thickness of 5 papers = 0.016×5 = 0.08 mm
Total thickness of a stack = 100+0.08 = 100.08 mm
= 100.08×102/102 mm
= 1.0008 x 102 mm
4. Express the number appearing in the following statements in standard form.

(i) 1 micron is equal to 1/1000000 m.
(ii) Charge of an electron is 0.000, 000, 000, 000, 000, 000, 16 coulomb.
(iii) Size of 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

 
Solution :
(i) 1 micron = 1/1000000
= 1/106
= 1×10-6
(ii) Charge of an electron is 0.00000000000000000016 coulombs.
= 0.00000000000000000016×1019/1019
= 1.6×10-19 coulomb
(iii) Size of bacteria = 0.0000005
=  5/10000000 = 5/107 = 5×10-7 m
(iv) Size of a plant cell is 0.00001275 m
= 0.00001275×105/105
= 1.275×10-5m
(v) Thickness of a thick paper = 0.07 mm
0.07 mm = 7/100 mm = 7/102 = 7×10-2 mm

NCERT Solutions For Class 8 Maths Chapter 12: Chapter Summary

When we have to repeatedly multiply a number by itself, we raise it to power. This is known as the Exponent. The power in the exponent represents the number of times that we want to carry out the multiplication operation. Exponents have their own set of rules when it comes to carrying out arithmetic operations. In this chapter, we will learn about powers, exponents, and their peripheral rules. Students will get to learn about new concepts and this chapter will further strengthen their base of the concepts of exponents and powers.

How To Prepare For CBSE Class 8 Maths Exponents And Powers?

Students are advised that they develop a good understanding of the concepts discussed in this chapter. Make sure you utilize these solutions provided here in the best possible way. Solving the Exponents and Powers practice questions will help you know where you are standing and will also help you to cover the areas you are weak at. If you are not able to answer a question or get stuck somewhere, refer to the solutions provided here. Also, go back and learn the concepts related to it. You can also take CBSE Class 8 Maths Chapter 12 mock test on Embibe. It is available for free and will be of great help to you in your preparation.

CBSE Class 8 Maths Chapter 12 Important Questions

Important Questions From Exponents And Powers
  1. Simplify the following expression and express the result in positive power notation: (−4)5 ÷ (−4)8
  2. 3 multiplied for 15 times is written as?
  3. Find the value of (40 + 4-1) × 22
  4. Solve 3-4 and (½)-2
  5. Express 4-3 as a power with base 2.
  6. Calculate the missing value of “x” in the following expression: (11/9)3 × (9/11)6 = (11/9)2x-1
  7. 5 books and 5 paper sheets are placed in a stack. Find the total thickness of the stack if each book has a thickness of 20 mm and each sheet has a thickness of 0.016 mm.
  8. Simplify [25 x t-4]/[5-3 x 10 x t-8]
  9. If a new-born bear weighs 4kg, calculate how many kilograms a five-year-old bear weigh if its weight increases by the power of 2 in 5 years?
  10. Write 3.61492 x 106 in the usual form.
  11. Express 0.00000000837 in standard form.
  12. Evaluate: [{1/2}-1+{1/3}-1]-1
  13. Express 31860000000 in standard form.
  14. Evaluate: (-4)-3
  15. A dish holds 100 bacteria. It is known that the bacteria double in number every hour. How many bacteria will be present after each number of hours?
  16. Express the following large numbers in its scientific notation.
    (a) 650200000000
    (b) 301000000
  17. Express the following number as a product of powers of prime factors.
    (a) 1225
    (b) 3600
Attempt 8th CBSE Mock Tests With Deep-improvement Analysis, Designed to Match the Actual Exam.

Here we have provided some of the frequently asked questions related to this chapter:

Q. Can I clear the board exams using NCERT Solutions for Class 8 Maths Chapter 12?
A. Yes, NCERT Solutions for Class 8 Maths Chapter 12 is an important chapter of Class 8 Maths. NCERT Solutions for this chapter will help you in your preparation by acting as a reference for self-evaluation and by making you learn several Mathematics tricks and shortcuts for quick and easy calculations. This will help you to clear the Maths paper in board exams.
Q. How do NCERT Solutions for Class 8 Maths Chapter 12 helps students to learn exponents and powers?
A. NCERT Solutions for Class 8 Maths Chapter 12 has been curated by our expert faculty in an interactive manner that makes it easy for students to understand the concepts. Students can refer to these PDF files containing solutions to improve their speed in solving problems accurately.
Q. How many questions do the NCERT Solutions for Class 8 Maths Chapter 12 contain?
A. NCERT Solutions for Class 8 Maths 12th Chapter is given for 2 exercises. In the first exercise, 7 questions are answered and 4 questions are solved in the second exercise. So a total of 11 short answer questions are present in NCERT Solutions for Class 8 Maths Chapter 12.
Q. Why NCERT Solutions for Class 8 Maths Chapter 12 Exponents and Powers are important?
A. NCERT Solutions give a clear and comprehensive view of the concepts of Class 8 Maths Chapter 12 Exponents and Powers. It deals with representing very large numbers or very small numbers in a standard form.
Q. What are the Important Topics that are covered in NCERT Solutions Class 8 Maths Chapter 12?
A. NCERT Solutions Class 8 Maths Chapter 12 covers several important topics such as the concept of exponents, definition of powers that comprise negative exponents, different laws of exponents. All in all, it makes you familiar with the laws of exponents from it.

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