NCERT Solutions For Class 8 Maths Chapter 12

Ncert Solutions For Class 8 Maths Chapter 12 PDF Free Download

NCERT Solutions for class 8 Maths chapter 12 Exponents and Powers is available here to practice for the final exam. This is a crucial chapter for the students of class 8 to learn, as the concepts explained here are widely used in upcoming higher classes. Therefore, these NCERT solutions help students to frame a better understanding of the topic and make them through the concepts. These learning materials are designed by subject experts as per the CBSE syllabus (2018-2019) prescribed by the board.

To ease the complexity of fundamentals of Exponents and Powers, we at BYJU’S provide NCERT Solutions for Class 8 Maths 12th Chapter. We have provided comprehensive study materials, which students can easily download and practice all the exercise-wise NCERT exemplar problems with solutions to score good marks in the final exams.

Class 8 Maths NCERT Solutions for Exponents and Powers

Exponents are applied in various algebra questions, therefore, it is essential to understand the rules of exponents. There are 2 easy rules of exponents, one can master it.

  • First, any amount raised to the power of “1” equals itself.
  • Secondly, 1 raised to any power is one.

The operations performed in exponents or the basic concepts of this topic are;

  • Multiplication of 2 powers having an 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 it’s reciprocal raised to the opposite positive power.

Many times, there are numbers which are very large and difficult to be represented or 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. Learn more of these concepts with BYJU’S and share your doubts with us, so that we can reply to your queries with the help of our responsive team.

NCERT Solutions Class 8 Maths Chapter 12 Exercises

The students can download and understand all the exercise-wise NCERT problems on Exponents and Powers from the links provided below:

The students can either download the PDF or can access online to practice class 8th NCERT solutions for Maths subject. Solving sample papers and previous year question papers are also a good practice as they give you an idea for types of questions asked in the main exam and marks contained by individual chapters.


Q.1. (i)\( 4^{-3}\)

(ii)\( (-5)^{-5}\)

(iii)\( (\frac{1}{2})^{-4}\)


(i)\( 4^{-2}\)

\( (4)^{-2}=\frac{1}{4^{2}}=\frac{1}{16}\)


(ii)\( (-5)^{-2}\)

\( (-5)^{-2}=\frac{1}{(-5)^{2}}=\frac{1}{25}\)


(iii)\( (\frac{1}{2})^{-4}\)

\( (\frac{1}{2})^{-2}=(\frac{2}{1})^{2}=(2)^{2}=4\)

Formula: \( a^{-m}=\frac{1}{a^{m}}\)


Q.2. (i)\( (-5)^{4}\div (-5)^{7}\)

(ii)\( (\frac{1}{4^{3}})^{2}\)

(iii)\( (-4)^{3}\times (\frac{5}{4})^{3}\)

(iv)\( (2^{-7}\div 2^{-10})\times 2^{-5}\)

(v)\( 3^{-3}\times (-7)^{-3}\)


(i)\( (-5)^{4}\div (-5)^{7}\)

\( (-5)^{4}\div (-5)^{7}=(-5)^{4-7}\\ \\ =(-5)^{-3}\\ \\ =\frac{1}{(-5)^{3}}\)


(ii)\( (\frac{1}{4^{3}})^{2}\)

\( (\frac{1}{4^{3}})^{2}=\frac{1^{2}}{(4^{3})^{2}}\\ \\ =\frac{1}{4^{3\times 2}}\\ \\ =\frac{1}{4^{6}}\)


(iii)\( (-4)^{3}\times (\frac{5}{4})^{3}\)

\( (-4)^{2}\times (\frac{5}{4})^{2}=(-4)^{2}\times \frac{5^{2}}{4^{2}}\\ \\ =[(-1)^{2}\times 4^{2}]\times \frac{5^{2}}{4^{2}}\\ \\ =4^{2-2}\times 5^{2}=4^{0}\times 5^{2}=5^{2}\)


(iv)\( (2^{-7}\div 2^{-10})\times 2^{-5}\)

\((2^{-7}\div 2^{-10})\times 2^{-5}=2^{-7-(-10)}\times 2^{-5}\\ \\ =2^{-7+10}\times 2^{-5}\\ \\ =2^{3}\times 2^{-5}=2^{3-5}\\ \\ =2^{-2}=\frac{1}{2^{2}}\)


(v)\( 3^{-3}\times (-7)^{-3}\)

\(3^{-3}\times (-7)^{-3}=\frac{1}{3^{3}}\times \frac{1}{(-7)^{3}}\\ \\ =\frac{1}{[3\times (-7)]^{3}}=\frac{1}{(-21)^{3}}\)



1.\((a^{m})^{n}=a^{m\times n}\)


3.\((ab)^{m}=a^{m} b^{m}\)

4.\(a^{m}\div a^{n}=a^{m-n}\)

5.\(a^{m}\times a^{n}=a^{m+n}\)


Q.3. Find the value of:

(i)\((3^{0}+4^{-1})\times 2^{2}\)

(ii)\((2^{-1}\times 4^{-1})\div 2^{-2}\)





(i)\((3^{0}+4^{-1})\times 2^{2}\) \((1+\frac{1}{4})\times 2^{2}\\ \\ =(\frac{4+1}{4})\times 2^{2}=\frac{5}{4}\times 2^{2}\\ \\ =5\times 2^{2-2}\\ \\ =5\times 2^{0}=5\times 1=5\)


(ii)\((2^{-1}\times 4^{-1})\div 2^{-2}\) \((2^{-1}\times 4^{-1})\div 2^{-2}=(\frac{1}{2^{1}}\times \frac{1}{4^{1}})\div 2^{-2}\\ \\ =(\frac{1}{2}\times \frac{1}{2^{2}})\div 2^{-2}=\frac{1}{2^{3}}\div 2^{-2}\\ \\ =2^{-3}\div 2^{-2}=2^{-3-(-2)}=2^{-3+2}=2^{-1}\\ \\ =\frac{1}{2}\)


(iii)\((\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}\) \((\frac{1}{2})^{-2}+(\frac{1}{3})^{-2}+(\frac{1}{4})^{-2}=(2^{-1})^{2}+(3^{-1})^{2}+(4^{-1})^{2}\\ \\ =2^{-1\times (-2)}+3^{-1\times (-2)}+4^{-1\times (-2)}\\ \\ =2^{2}+3^{2}+4^{2}=4+9+16=29\)


(iv)\((3^{-1}+4^{-1}+5^{-1})^{0}\) \((3^{-1}+4^{-1}+5^{-1})^{0}=(\frac{1}{3}+\frac{1}{4}+\frac{1}{5})^{0}\\ \\ =(\frac{20+15+12}{60})^{0}=(\frac{47}{60})^{0}= 1\)


(v)\([(\frac{-2}{3})^{-2}]^{2}\) \([(\frac{-2}{3})^{-2}]^{2}=(\frac{-2}{3})^{-2\times 2}\\ \\ =(\frac{-2}{3})^{-4}=(\frac{-3}{2})^{4}=\frac{81}{16}\)


Q.4. Evaluate:

(i)\(\frac{8^{-1}\times 5^{3}}{2^{-4}}\)

(ii)\((5^{-1}\times 2^{-1})\times 6^{-1}\)


(i)\(\frac{8^{-1}\times 5^{3}}{2^{-4}}\) \(\frac{8^{-1}\times 5^{3}}{2^{-4}}=\frac{(2^{3})^{-1}\times 5^{3}}{2^{-4}}=\frac{2^{-3}\times 5^{3}}{2^{-4}}\\ \\ =2^{-3(-4)}\times 5^{3}=2^{-3+4}\times 5^{3}\\ \\ =2\times 125=250\)


(ii)\((5^{-1}\times 2^{-1})\times 6^{-1}\) \((5^{-1}\times 2^{-1})\times 6^{-1}=(\frac{1}{5}\times \frac{1}{2})\times \frac{1}{6}\\ \\ =\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}\)


Q.5. Find the value of m for which\(5^{m}\div 5^{-3}=5^{5}\).


\(5^{m}\div 5^{-3}=5^{5}\\ \\ \Rightarrow 5^{m-(-3)}=5^{5}\\ \\ \Rightarrow 5^{m+3}=5^{5}\) \(5^{m}\div 5^{-3}=5^{5}\\ \\ \Rightarrow 5^{m-(-3)}=5^{5}\\ \\ \Rightarrow 5^{m+3}=5^{5}\\ \\ \Rightarrow m+3=5\\ \\ \Rightarrow m=5-3\\ \\ \Rightarrow m=2\)


Q.6. Evaluate:

(i) \([(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}\)

(ii)\((\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}\)


(i) \([(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}\) \([(\frac{1}{3})^{-1}-(\frac{1}{4})^{-1}]^{-1}=[(\frac{3}{1})^{1}-(\frac{4}{1})^{1}]\, (\ because a^{-m}=\frac{1}{m})\\ \\ =[3-4]=-1\)


(ii)\((\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}\) \((\frac{5}{8})^{-7} \times (\frac{8}{5})^{-4}=\frac{5^{-7}}{8^{-7}}\times \frac{8^{-4}}{5^{-4}}\\ \\ =5^{-7(-4)}\times 8^{-4(-7)}\\ \\ =5^{-7+4}\times 8^{-4+7}=5^{-3}\times 8^{3}=\frac{8^{3}}{5^{3}}\\ \\ =\frac{512}{125}\)


Q.7. Simplify:

(i)\(\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\, \, (t\neq 0)\)


(ii)\(\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}\)


(i)\(\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}\, \, (t\neq 0)\) \(\frac{25\times t^{-4}}{5^{-3}\times 10\times t^{-8}}=\frac{5^{2}\times t^{-4}}{5^{-3}\times 5\times2\times t^{-8}}\\ \\ =\frac{5^{2-(-3)-1}\times t^{-4(-8)}}{2}\\ \\ =\frac{5^{2+3-1}\times t^{-4+8}}{2}=\frac{5^{4}\times t^{4}}{2}\\ \\ =\frac{625}{2}t^{4}\)


(ii)\(\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}\) \(\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}=\frac{3^{-5}\times (2\times 5)^{-5}\times 5^{3}}{5^{-7}\times (2\times 3)^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5}\times 5^{3}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =\frac{3^{-5}\times 2^{-5}\times 5^{-5+3}}{5^{-7}\times 2^{-5}\times 3^{-5}}=\frac{3^{-5}\times 2^{-5}\times 5^{-2}}{5^{-7}\times 2^{-5}\times 3^{-5}}\\ \\ =3^{-5-(-5)}\times 2^{-5-(-5)}\times 5^{-2-(-7)}\\ \\ =3^{-5+5}\times 2^{-5+5}\times 5^{-2+7}=3^{0}\times 2^{0}\times 5^{5}=1\times 1\times 3125\\ \\ =3125\)



Q.1. Converts the numbers into standard form:









\(0.0000000000085=0.0000000000085\times \frac{10^{12}}{10^{12}}=8.5\times 10^{-12}\)



\(0.00000000000942 =0.00000000000942 \times \frac{10^{12}}{10^{12}}=9.42\times 10^{-12}\)



\(6020000000000000 =6020000000000000 \times \frac{10^{15}}{10^{15}}=6.02\times 10^{15}\)



\(0.00000000837 =0.00000000837 \times \frac{10^{9}}{10^{9}}=8.37\times 10^{-9}\)



\(31860000000 =31860000000 \times \frac{10^{10}}{10^{10}}=3.186\times 10^{10}\)


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

(i)\(3.02\times 10^{-6}\)

(ii)\(4.5\times 10^{4}\)

(iii)\(3\times 10^{-8}\)

(iv)\(1.0001\times 10^{9}\)

(v) \(5.8\times 10^{12}\)

(vi)\(3.61492\times 10^{6}\)



(i)\(3.02\times 10^{-6}\)

\(3.02\times 10^{-6}=\frac{3.02}{10^{6}}=0.00000302\)


(ii)\(4.5\times 10^{4}\)

\(4.5\times 10^{4}=4.5\times 10000=45000\)


(iii) \(3\times 10^{-8}\)

\(3\times 10^{-8}=\frac{3}{10^{8}}=0.00000003\)


(iv) \(1.0001 \times 10^{9}\)

\(1.0001 \times 10^{9}=1000100000\)


(v) \(5.8\times 10^{12}\)

\(5.8\times 10^{12}=5.8\times 1000000000000=5800000000000\)


(vi) \(3.61492\times 10^{6}\)

\(3.61492\times 10^{6}=3.61492 \times 1000000 =3614920\)


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

(i) 1 micron is equal to \(\frac{1}{1000000}\) 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.



(i)1 micron =\(\frac{1}{1000000}=\frac{1}{10^{6}}=1\times 10^{-6}\: m\)


(ii) Charge of an electron is 0.00000000000000000028 coulombs.

\(= 0.00000000000000000028\times \frac{10^{19}}{10^{19}}=2.8\times 10^{-19}\: coulomb\)


(iii)Size of bacteria = 0.0000005 =\(\frac{5}{10000000}=\frac{5}{10^{7}}=5\times 10^{-7\: m}\)


(iv)Size of a plant cell is 0.00001275 m =\(0.00001275\times \frac{10^{5}}{10^{5}}=1.275\times 10^{-5}\: m\)


(v) Thickness of a thick paper = 0.07 mm =\(\frac{7}{100}\: mm=\frac{7}{10^{2}}=7\times 10^{-2}\: mm\)


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?


Thickness of one book = 20 mm

Thickness of 6 books =\(20\times 6=120\: mm\)

Thickness of one paper = 0.016 mm

Thickness of 5 papers =\(0.016\times 5=0.08\: mm\)

Total thickness of a stack =120+0.08

=120.08 mm=\(120.08\times \frac{10^{2}}{10^{2}}=1.2008\times 10^{2}\: mm\)

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