NCERT Solutions Class 8 Maths Chapter 10 Exponents and Powers

Chapter 10 of Class 8 Maths, Exponents and Powers, provides a comprehensive understanding of key concepts such as an introduction to exponents, handling powers with negative exponents, and the important laws of exponents. To gain mastery over these topics, students can refer to the 'Exponents and Powers Class 8 NCERT Solutions,' which offers detailed step-by-step explanations. These solutions help solidify understanding and make it easier to tackle the exercises in NCERT Solutions for Class 8 Maths Exponents and Powers. Enhance your learning of Maths Exponents and Powers Class 8 with these well-structured solutions that boost confidence in solving problems.

1.0Download NCERT Solutions Class 8 Maths Chapter 10 Exponents and Powers: Free PDF

This article provides comprehensive solutions for Class 8 Maths Chapter 10, Exponents and Powers, fully aligned with NCERT guidelines. By practising these solutions, students can strengthen their mathematical foundation and enhance their problem-solving skills, ultimately leading to improved exam performance. For more detailed learning, students can download the NCERT Class 8 Maths Chapter 10 PDF Solutions, expertly prepared by ALLEN's subject specialists, using the link below.

2.0NCERT Solutions Class 8 Maths Chapter 10: Important Topics

Before diving into the Solutions for NCERT Class 8 Maths Chapter Exponents and Powers, let’s first take a quick overview of the key topics and subtopics covered in this chapter of the NCERT Class 8 Maths book.

Topics Covered in Class 8 Maths Chapter 10

Introduction: Exponents are a shorthand way to represent repeated multiplication of a number.

Powers with Negative Exponents: Negative exponents indicate the reciprocal of the base raised to the positive exponent.

Laws of Exponents: The laws of exponents provide rules for simplifying expressions involving powers.

Use of Exponents to Express Small Numbers in Standard Form: Standard form allows us to express very small or large numbers compactly using powers of ten.

Class 8 Maths Chapter 10 Exponents and Powers: Exercise Solutions

There are two exercises in chapter 10 (Exponents and Powers) of class 8 Maths. Students can find the split below:

Explore Exponents and Powers and learn how to solve various problems only on NCERT Solutions For Class 8 Maths.

3.0Benefits of Studying this Chapter

Studying Chapter 10, "Exponents and Powers," in Class 8 has several benefits for students as it provides foundational knowledge that is useful in higher mathematics and real-world applications. Here are the key benefits:

Simplification of Large Numbers: This chapter teaches students how to express and work with very large or very small numbers in a simplified form using exponents, making complex calculations easier to handle.

Foundation for Algebra: Understanding exponents and powers is crucial for future topics in algebra, such as polynomials, equations, and algebraic expressions. It builds the base for manipulating algebraic terms effectively.

Practical Application: The concept of exponents is widely used in scientific notations, measurements, and calculations involving large distances (like in astronomy) or small quantities (like in chemistry or physics).

Enhances Problem-Solving Skills: By learning the laws of exponents (product law, quotient law, power law, etc.), students develop critical problem-solving skills that are essential for competitive exams and higher-level mathematics.

Preparation for Higher Classes: The chapter prepares students for more advanced mathematical concepts such as logarithms, exponential functions, and powers of complex numbers in higher classes.

Improves Mental Calculation: The practice of using powers and exponents enhances students' ability to perform mental calculations quickly and efficiently.

Logical Thinking and Analysis: It helps in developing logical thinking as students analyze how exponents behave under different operations like multiplication, division, and raising powers.

4.0NCERT Questions with Solutions for Class 8 Maths Chapter 10 - Detailed Solutions

Exercise : 10.1

• ${\left(\frac{1}{2}\right)}^{-5}$ Sol. (i) $3^{-2}=\frac{1}{3^2}=\frac{1}{3\times 3}=\frac{1}{9}$ (ii) $(-4{)}^{-2}=\frac{1}{(-4{)}^2}=\frac{1}{(-1{)}^2\times (4{)}^2}=\frac{1}{1\times 16}=\frac{1}{16}$ (iii) ${\left(\frac{1}{2}\right)}^{-5}=2^5=2\times 2\times 2\times 2\times 2=32$
• Simplify and express the result in power notation with positive exponent. (i) $(-4{)}^5÷(-4{)}^8$ (ii) ${\left(\frac{1}{2^3}\right)}^2$ (iv) $\left(3^{-7}÷3^{-10}\right)\times 3^{-5}=\frac{3^{-7}}{3^{-10}}\times 3^{-5}$ $=3^{-7+10}\times 3^{-5}=3^3\times 3^{-5}$ $=3^{3+(-5)}=3^{3-5}=3^{-2}=\frac{1}{3^2}$ (v) $2^{-3}\times (-7{)}^{-3}$ $=-\frac{1}{2^3}\times \frac{1}{7^3}=-\frac{1}{(2\times 7{)}^3}=-\frac{1}{14^3}$
• Find the value of (i) $\left(3^0+4^{-1}\right)\times 2^2$ (ii) $\left(2^{-1}\times 4^{-1}\right)÷2^{-2}$ (iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}$ (iv) ${\left(3^{-1}+4^{-1}+5^{-1}\right)}^0$ (v) ${\left\{{\left(\frac{-2}{3}\right)}^{-2}\right\}}^2$ Sol. (i) $\left(3^0+4^{-1}\right)\times 2^2=\left(1+\frac{1}{4}\right)\times 4$ $=\frac{4+1}{4}\times 4=\frac{5}{4}\times 4=5$ (ii) $\left(2^{-1}\times 4^{-1}\right)÷2^{-2}=\left(\frac{1}{2}\times \frac{1}{4}\right)÷\frac{1}{2^2}=\frac{1}{8}÷\frac{1}{4}$ $=\frac{1}{8}\times \frac{4}{1}=\frac{1}{2}$ (iii) ${\left(\frac{1}{2}\right)}^{-2}+{\left(\frac{1}{3}\right)}^{-2}+{\left(\frac{1}{4}\right)}^{-2}=2^2+3^2+4^2$ $=4+9+16=29$ (iv) ${\left(3^{-1}+4^{-1}+5^{-1}\right)}^0$ $=1\left[\because a^0=1\right]$ (v) ${\left\{{\left(-\frac{2}{3}\right)}^{-2}\right\}}^2={\left\{{\left(-\frac{3}{2}\right)}^2\right\}}^2$ $={\left(-\frac{3}{2}\right)}^4=\frac{3^4}{2^4}=\frac{81}{16}$
• Evaluate: (i) $\frac{8^{-1}\times 5^3}{2^{-4}}$ (ii) $\left(5^{-1}\times 2^{-1}\right)\times 6^{-1}$ Sol. (i) $\frac{8^{-1}\times 5^3}{2^{-4}}=\frac{1}{8}\times 5^3\times 2^4=\frac{1}{8}\times 125\times 16$ $=125\times 2=250$ (ii) $\left(5^{-1}\times 2^{-1}\right)\times 6^{-1}$ $=\left(\frac{1}{5}\times \frac{1}{2}\right)\times \frac{1}{6}=\frac{1}{10}\times \frac{1}{6}=\frac{1}{60}$
• Find the value of $m$ for which $5^m÷5^{-3}=5^5$ Sol. We have, $5^m÷5^{-3}=5^5$ $\Rightarrow 5^{\mathrm{m}}÷\frac{1}{5^3}=5^5\Rightarrow 5^{\mathrm{m}}\times 5^3=5^5$ $\Rightarrow 5^{m+3}=5^5$ $\Rightarrow \mathrm{m}+3=5[\because$ if ${\mathrm{a}}^{\mathrm{m}}={\mathrm{a}}^{\mathrm{n}}$, then $\mathrm{m}=\mathrm{n}]$ $\Rightarrow m=5-3=2$
• Evaluate (i) ${\left\{{\left(\frac{1}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}$ (ii) ${\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-4}$ Sol. (i) ${\left\{{\left(\frac{1}{3}\right)}^{-1}-{\left(\frac{1}{4}\right)}^{-1}\right\}}^{-1}={\left(3^1-4^1\right)}^{-1}=(-1{)}^{-1}$ $=\frac{1}{(-1{)}^1}=\frac{1}{-1}=-1$ (ii) ${\left(\frac{5}{8}\right)}^{-7}\times {\left(\frac{8}{5}\right)}^{-4}$ $$\begin{gathered} ={\left(\frac{8}{5}\right)}^7\times {\left(\frac{8}{5}\right)}^{-4} \\ ={\left(\frac{8}{5}\right)}^{7+(-4)}={\left(\frac{8}{5}\right)}^{7-4}={\left(\frac{8}{5}\right)}^3=\frac{512}{125} \end{gathered}$$
• Simplify (i) $\frac{25\times {\mathrm{t}}^{-4}}{5^{-3}\times 10\times {\mathrm{t}}^{-8}}(\mathrm{t}\ne 0)$ (ii) $\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}$ Sol. (i) $\frac{25\times {\mathrm{t}}^{-4}}{5^{-3}\times 10\times {\mathrm{t}}^{-8}}=\frac{25\times 5^3}{10}\times {\mathrm{t}}^{-4+8}$ $=\frac{5\times 5^3}{2}\times t^4=\frac{625}{2}{t}^4$ (ii) $\frac{3^{-5}\times 10^{-5}\times 125}{5^{-7}\times 6^{-5}}=\frac{3^{-5}\times 10^{-5}\times 5^3}{5^{-7}\times 6^{-5}}$ $=\frac{3^{-5}\times \left(2^{-5}\times 5^{-5}\right)\times 5^3}{5^{-7}\times \left(2^{-5}\times 3^{-5}\right)}$ $=5^{-5}\times 5^3\times 5^7$ $=5^{-5+3+7}=5^5=3125$

Exercise : 10.2

• Express the following numbers in standard form (i) 0.0000000000085 (ii) 0.00000000000942 (iii) 6020000000000000 (iv) 0.00000000837 (v) 31860000000 Sol. (i) $0.0000000000085=\frac{85}{10000000000000}$ $$\begin{gathered} =\frac{85}{10^{13}}=\frac{8.5\times 10}{10^{13}}=8.5\times 10\times 10^{-13} \\ =8.5\times 10^{-12} \end{gathered}$$ (ii) 0.00000000000942 $=\frac{942}{100000000000000}$ $=\frac{942}{10^{14}}=\frac{9.42\times 100}{10^{14}}=9.42\times 10^2\times 10^{-14}$ $=9.42\times 10^{-12}$ (iii) 6020000000000000 $=6.02\times 10^{15}$ (iv) $0.00000000837=\frac{837}{100000000000}$ $=\frac{837}{10^{11}}=\frac{8.37}{10^9}=8.37\times 10^{-9}$ (v) 31860000000 $=3.186\times 10^{10}$
• Express 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) $3.61492\times 10^6$ Sol. (i) $3.02\times 10^{-6}=\frac{302}{100}\times \frac{1}{10^6}=\frac{302}{100000000}$ $=0.00000302$ (ii) $4.5\times 10^4=\frac{45}{10}\times 10^4=45\times 10^3=45000$ (iii) $3\times 10^{-8}=\frac{3}{10^8}=\frac{3}{100000000}$ $=0.00000003$ (iv) $1.0001\times 10^9$ $\frac{10001}{10000}\times 10^9=\frac{10001}{10^4}\times 10^9$ $=10001\times 10^{9-4}$ $=10001\times 10^5$ $=1000100000$
• (i) 1 micron (ii) Charge of an electron is 0.00000000000000000016 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 . Sol. (i) 1 micron $=\frac{1}{1000000}\mathrm{m}=1\times 10^{-6}\mathrm{m}$ (ii) Charge of an electron is $=0.00000000000000000016$ $=\frac{16}{100000000000000000000}$ $=\frac{16}{10^{20}}=\frac{1.6\times 10}{10^{20}}=\frac{1.6}{10^{19}}$ $=1.6\times 10^{-19}$ coulomb (iii) Size of a bacteria $=0.0000005\mathrm{m}$ $=\frac{5}{10000000}\mathrm{m}=\frac{5}{10^7}\mathrm{m}=5\times 10^{-7}\mathrm{m}$ (iv) Size of plant cell $=0.00001275\mathrm{m}$. $=\frac{1275}{100000000}=\frac{1.275}{10^8}\times 10^3$ $=1.275\times 10^{3-8}$ $=1.275\times 10^{-5}\mathrm{m}$ (v) Thickness of a thick paper $=0.07\mathrm{mm}$. $=\frac{7}{100}=\frac{7}{10^2}=7\times 10^{-2}\mathrm{mm}$
• 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? Sol. Thickness of one book $=20\mathrm{mm}$ $\therefore$ Thickness of 5 books $=5\times 20\mathrm{mm}=100\mathrm{mm}$ Thickness of 1 paper sheet $=0.016\mathrm{mm}$ Thickness of 5 paper sheets $=5\times 0.016\mathrm{mm}$ $=0.08\mathrm{mm}$ Total thickness $=100\mathrm{mm}+0.08\mathrm{mm}$ $=100.08\mathrm{mm}$ $=1.0008\times 10^2\mathrm{mm}$