# NCERT Solutions Class 8 Maths Chapter 12

## NCERT Solutions for Class 8 Mathematics Chapter 12 – Exponents and Powers

Class 8, 9, and 10 Mathematics is crucial for students as it builds the foundation for Class 11 and 12. Mathematics will be a central subject for many higher education courses. It is used in our day-to-day lives and also increases our problem-solving skills and has proven beneficial in our lives.

Chapter 12 of Class 8 Mathematics – Exponents and Powers helps one learn how to write large numbers in the simplest form. Students will learn to convert the long numbers into their powers and learn about various laws of exponents in this chapter, which will be used in calculations.

Extramarks’ website helps students score well in their board examinations and many other entrance examinations. Many students struggle due to a lack of clarification of the topics and concepts. As a result, they score low in their examinations. Extramarks’’ NCERT Solutions for Class 8 Mathematics Chapter 12 are made by Mathematics subject experts with many years of teaching experience. As a result, it helps students in improving their grades.

Lakhs of students have relied on the Extramarks’ platform for studying and preparing for their examinations. Extramarks provides comprehensive study materials, including NCERT Solutions, chapter notes, revision notes, important questions and answers, etc. So once students register on the Extramarks’ website, they don’t have to worry about gathering different study material from various sources. Extramarks’ is a one-stop solution for their studies.

## Key Topics Covered In NCERT Solutions for Class 8 Mathematics Chapter 12

In the chapter Exponents and Power, we will cover the topics like converting a larger number into Exponential form. You will also learn about the Power of Negative Exponents, the uses of exponents to express small numbers in standard form and the Law of Exponents. You will also be provided with some examples to understand more about powers. You can refer to NCERT Solutions for Class 8 Mathematics Chapter 12, available on the Extramarks’ website.

Extramarks’ website is a reliable and trusted source to look upon. One can regularly check the website to remain updated about the most recent CBSE syllabus and any curriculum-related news.

You will have a thorough understanding of exponents and powers, how they are used in calculations, and how accurate they are once you have finished this chapter. As a result, you’ll develop excellent mathematical skills and have quick, accurate understanding.

### Introduction

In this chapter, We’ll study exponent. Exponents will be used to reduce greater numbers to smaller values. You can reduce a greater number to a smaller one in this way.

For example, there is a large number 1500000000000000 that we can convert into Exponents and write as 15 x 10^15.

To get further practise, please refer to our NCERT Solutions for Class 8 Mathematics Chapter 12. Students can register on the Extramarks’ website and get full access to our study materials..

### Power with Negative Exponents

This part of the chapter covers the nature of power and expressed in two ways.

• Positive Exponents,

for example, a^m, where a=Base, m=Power

• Negative Exponents,

for example, a^(-m), where a=Base, m=Power

• It’s also called the Multiplicative Inverse of a^m.

Since many Class 8 students mistake the power of exponents, solving exponential equations can be challenging. Exponents have been clearly explained with several examples in our NCERT Solutions for Class 8 Mathematics Chapter 12 by our academic experts. To completely comprehend the topic of exponents, it is advised that students consult our study materials.

### Law of Exponents

In the above section, we covered the positive exponents and negative exponents. In this section you will learn about the process and law applied to positive and negative exponents. The theory is applied to non-zero integers.

• Positive exponents

In this section, there are two exponents whose base is the same for both. Then we multiply both the exponents, and both the powers will be added.

for example, a^b x a^y = a^(b+y)

• Negative Exponents

In this section, there are two exponents whose base is the same. We divide both the exponents. The power will be subtracted from each other. For example, a^b / a^y = a^(b-y).

For furtherunderstanding about the law of exponents and to clear their doubts, students can visit the Extramarks’ website and refer to our NCERT Solutions for Class 8 Mathematics Chapter 12.

Use of Exponents to Express Small Numbers in Standard Form

We use some examples of exponents to make a large number more convenient and reliable.

Observe the following facts.

• The distance between the earth and the sun is 1,49,600,000,000 m.
• The light speed is 300,000,000 m/sec.
• The thickness of the class VIII Mathematics book is 20 mm.
• The height of Mount Everest is 8848 m.
• The Sun’s average radius is 695000 km.

Comparing very large and very small numbers.

In this part, we can compare the distance from the earth to the sun and from the earth to the moon.

Our NCERT Solutions for Class 8 Mathematics Material 12 provide several solved examples and learning techniques to help students study the chapter more easily. Students can register on the Extramarks’ website and get full access to our NCERT Solutions.

NCERT Solutions for Class 8 Mathematics Chapter 12 Exercises & Solutions.

Mathematics is fun, but it also requires a lot of practice. Extramarks’ understands the importance of regularly solving questions to score good marks in Mathematics exams. So we have included hundreds of questions in our study materials.

In this chapter, students have learned about Exponents and Powers. Students can refer to our NCERT Solutions for Class 8 Mathematics Chapter 12 to access all study materials related to the Exponents and Powers chapter. In our study material, we give a lot of questions to practise, covering all the theoretical and conceptual-based questions. Our solutions also cover all the past years’ CBSE board exam questions. Students will start to clear their doubts and get clarity on all topics by completing these exercises.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions for Class 8 Mathematics Chapter 12:

• Chapter 12: Exercise 12.1 Question and answers
• Chapter 12: Exercise 12.2 Question and answers
• Exponents and Powers Class 8 – Exercise and Solutions

Extramarks’ website, the online platform trusted by millions of students and teachers, is the right source for all types of solutions and exercises. You can visit the website anytime to access any study material you want.

Along with Class 8 Mathematics solutions, you can explore NCERT Solutions on our Extramarks’ website for all primary and secondary classes.

• NCERT Solutions Class 1
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• NCERT Solutions Class 12

## NCERT Exemplar Class 12 Mathematics

NCERT Exemplar books contain many practice questions and cover all the topics in the NCERT textbook. NCERT Exemplars are available for the subjects of Mathematics and Science from Class 6 to 12. It makes answering questions easier for students. Some of them are directly asked in entrance exams like JEE, NEET, WBJEE, etc.

These books play an important role in building the foundation and boosting the students’ confidence. They develop a more thinking-oriented perspective that facilitates solving the maximum numerical problems with improved accuracy. As a result, the NCERT Exemplar has been beneficial to students at every stage of their preparation.

NCERT Solution also provides lots of questions to practise for your examinations. You can score well by referring to both the resources. To avail our study material, you can visit the Extramarks’ website and register.

### Key Features of NCERT Solutions Class 8 Mathematics Chapter 12

NCERT Solutions Class 8 Mathematics Chapter 12 is available on our Extramarks’ website. You can include it in your preparation to stand out in your class.

Below are a few of the reasons why students should choose our NCERT Solutions:

• Our study material is prepared by the experts who have specialised knowledge in Mathematics.
• Our study material language is simple and easy for a Class 8 student to understand.
• Our study material covers study notes for the full chapter topics, sub-topics and concepts. It also includes revision notes and questions that will help students in exam preparation.
• In our study material, we cover the past years’ questions that students will require to score well in the CBSE Board examinations.

$\begin{array}{l}\text{Q.1 Evaluate.}\\ {\text{(i) 3}}^{\text{-2}}{\text{(ii) (-4)}}^{\text{-2}}\text{(iii)}{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{-5}}\end{array}$

Ans-

$\begin{array}{l}\left(\mathrm{i}\right){3}^{–2}=\frac{1}{{3}^{2}}=\frac{1}{9}\text{\hspace{0.17em}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ \\ \left(\mathrm{ii}\right){\text{(-4)}}^{\text{-2}}=\frac{1}{{\left(-4\right)}^{2}}=\frac{1}{16}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ \\ \left(\mathrm{iii}\right){\left(\frac{1}{2}\right)}^{–5}=\frac{1}{{2}^{-5}}={2}^{5}=2×2×2×2×2=32\end{array}$ $\begin{array}{l}\text{}\end{array}$

$\begin{array}{l}\text{Q.2}\end{array}$

Simplify and express the result in power notation

with positive exponent.(i) (- 4)5 ÷ (- 4)8 (ii) (123)2 (iii) (– 3)4×(53)4 (iv) (3– 7÷ 3 – 10) × 3 – 5 (v) 2 – 3 × (- 7)– 3

Ans-

$\begin{array}{l}{\text{(i) (- 4)}}^{\text{5}}{\text{÷ (- 4)}}^{\text{8}}\\ {\text{=(- 4)}}^{\text{5-8}}\text{\hspace{0.17em}}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=(- 4)}}^{\text{-3}}\\ \text{=}\frac{\text{1}}{{\left(\text{-4}\right)}^{\text{3}}}\text{\hspace{0.17em}}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \\ \text{(ii)}{\left(\frac{\text{1}}{{\text{2}}^{\text{3}}}\right)}^{\text{2}}\\ \text{=}\frac{\text{1}}{{\left({\text{2}}^{\text{3}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{1}}{{\text{2}}^{\text{6}}}\text{}\left[\text{By using}{\left({\text{a}}^{\text{m}}\right)}^{\text{n}}{\text{=a}}^{\text{mn}}\right]\end{array}$ $\begin{array}{l}\text{(iii)}{\left(-3\right)}^{4}×{\left(\frac{5}{3}\right)}^{4}\\ ={\left(-1\right)}^{4}×{3}^{4}×\frac{{5}^{4}}{{3}^{4}}\\ ={\left(-1\right)}^{4}×{5}^{4}\\ ={5}^{4}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(-1\right)}^{4}=1\right]\\ \end{array}$ $\begin{array}{l}{\text{(iv) (3}}^{\text{– 7}}{\text{÷ 3}}^{\text{– 10}}{\text{) × 3}}^{\text{– 5}}\\ {\text{=3}}^{\left(\text{– 7}\right)\text{–}\left(\text{-10}\right)}{\text{× 3}}^{\text{– 5}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=3}}^{\text{3}}{\text{× 3}}^{\text{– 5}}\\ {\text{=3}}^{\left(\text{3}\right)\text{+}\left(\text{-5}\right)}\text{}\left[{\text{By using a}}^{\text{m}}{\text{×a}}^{\text{n}}{\text{=a}}^{\text{m+n}}\right]\\ {\text{=3}}^{\text{-2}}\\ \text{=}\frac{\text{1}}{{\text{3}}^{\text{2}}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \\ {\text{(v) 2}}^{\text{– 3}}{\text{× (- 7)}}^{\text{-3}}\\ \text{=}\frac{\text{1}}{{\text{2}}^{\text{3}}}\text{×}\frac{\text{1}}{{\left(\text{-7}\right)}^{\text{3}}}\\ \text{=}\frac{\text{1}}{{\left[\text{2×}\left(\text{-7}\right)\right]}^{\text{3}}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{×b}}^{\text{m}}{\text{=ab}}^{\text{m}}\right]\\ \text{=}\frac{\text{1}}{{\left(\text{-14}\right)}^{\text{3}}}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{Q.3}\end{array}$

Find the value of:(i) (30 + 41) × 22(ii)(21 × 41) ÷ 22iii(12)2+(13)2+(14)2(iv) (3-1 + 4-1 + 5-1)0 v {(23)2}2

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\text{ }\left(3\mathrm{°}+{4}^{-1}\right)×{2}^{2}\\ =\left(1+\frac{1}{4}\right)×{2}^{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{0}\text{=1,}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{5}{4}×4\\ =5\\ \\ \left(\mathrm{ii}\right)\text{ }\left({2}^{-1}×{4}^{-1}\right)÷{2}^{-2}\\ =\left({2}^{-1}×{\left({2}^{2}\right)}^{-1}\right)÷{2}^{-2}\text{}\\ ={2}^{-1}×\left({2}^{-2}\right)÷{2}^{-2}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left({\mathrm{a}}^{\mathrm{m}}\right)}^{\text{n}}{\text{=a}}^{\text{mn}}\right]\\ ={2}^{-1+\left(-2\right)}÷{2}^{-2}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}×{\mathrm{a}}^{\text{n}}{\text{=a}}^{\text{m+n}}\right]\end{array}$ $\begin{array}{l}{\text{=2}}^{\text{-3}}{\text{÷2}}^{\text{-2}}\\ {\text{=2}}^{\text{-3-(-2)}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=2}}^{\text{-3+2}}\\ {\text{=2}}^{\text{-1}}\\ \text{=}\frac{\text{1}}{\text{2}}\\ \\ \text{(iii) }{\left(\frac{\text{1}}{\text{2}}\right)}^{\text{-2}}\text{+}{\left(\frac{\text{1}}{\text{3}}\right)}^{\text{-2}}\text{+}{\left(\frac{\text{1}}{\text{4}}\right)}^{\text{-2}}\\ \text{=}{\left(\frac{\text{2}}{\text{1}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{3}}{\text{1}}\right)}^{\text{2}}\text{+}{\left(\frac{\text{4}}{\text{1}}\right)}^{\text{2}}\text{ }\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ {\text{=2}}^{\text{2}}{\text{+3}}^{\text{2}}{\text{+4}}^{\text{2}}\\ \text{=4+9+16}\\ \text{=29}\\ \\ {\text{(iv) (3}}^{\text{-1}}{\text{+ 4}}^{\text{-1}}{\text{+ 5}}^{\text{-1}}{\text{)}}^{\text{0}}\text{}\\ \text{=}{\left(\frac{\text{1}}{\text{3}}\text{+}\frac{\text{1}}{\text{4}}\text{+}\frac{\text{1}}{\text{5}}\right)}^{\text{0}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \text{=1}\left[{\text{By using a}}^{\text{0}}\text{=1}\right]\end{array}$ $\begin{array}{l}\left(\mathrm{v}\right){\left\{{\left(\frac{-2}{3}\right)}^{-2}\right\}}^{2}\\ ={\left\{{\left(\frac{3}{-2}\right)}^{2}\right\}}^{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ ={\left\{\frac{{3}^{2}}{{\left(-2\right)}^{2}}\right\}}^{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}=\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{b}}^{\mathrm{m}}}\right]\\ ={\left(\frac{9}{4}\right)}^{2}\\ =\frac{81}{16}\end{array}$

$\begin{array}{l}\text{Q.4}\end{array}$

Simplify and express the result in power notation with positive exponent.

(i) (- 4)5 ÷ (- 4)8 (ii) (123)2 (iii) (– 3)4×(53)4 (iv) (3– 7÷ 3 – 10) × 3 – 5 (v) 2 – 3 × (- 7)– 3

Ans-

$\begin{array}{l}{\text{(i) (- 4)}}^{\text{5}}{\text{÷ (- 4)}}^{\text{8}}\\ {\text{=(- 4)}}^{\text{5-8}}\text{\hspace{0.17em}}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=(- 4)}}^{\text{-3}}\\ \text{=}\frac{\text{1}}{{\left(\text{-4}\right)}^{\text{3}}}\text{\hspace{0.17em}}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \\ \text{(ii)}{\left(\frac{\text{1}}{{\text{2}}^{\text{3}}}\right)}^{\text{2}}\\ \text{=}\frac{\text{1}}{{\left({\text{2}}^{\text{3}}\right)}^{\text{2}}}\\ \text{=}\frac{\text{1}}{{\text{2}}^{\text{6}}}\text{}\left[\text{By using}{\left({\text{a}}^{\text{m}}\right)}^{\text{n}}{\text{=a}}^{\text{mn}}\right]\end{array}$ $\begin{array}{l}\text{(iii)}{\left(-3\right)}^{4}×{\left(\frac{5}{3}\right)}^{4}\\ ={\left(-1\right)}^{4}×{3}^{4}×\frac{{5}^{4}}{{3}^{4}}\\ ={\left(-1\right)}^{4}×{5}^{4}\\ ={5}^{4}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(-1\right)}^{4}=1\right]\\ \end{array}$ $\begin{array}{l}{\text{(iv) (3}}^{\text{– 7}}{\text{÷ 3}}^{\text{– 10}}{\text{) × 3}}^{\text{– 5}}\\ {\text{=3}}^{\left(\text{– 7}\right)\text{–}\left(\text{-10}\right)}{\text{× 3}}^{\text{– 5}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{÷a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ {\text{=3}}^{\text{3}}{\text{× 3}}^{\text{– 5}}\\ {\text{=3}}^{\left(\text{3}\right)\text{+}\left(\text{-5}\right)}\text{}\left[{\text{By using a}}^{\text{m}}{\text{×a}}^{\text{n}}{\text{=a}}^{\text{m+n}}\right]\\ {\text{=3}}^{\text{-2}}\\ \text{=}\frac{\text{1}}{{\text{3}}^{\text{2}}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \\ {\text{(v) 2}}^{\text{– 3}}{\text{× (- 7)}}^{\text{-3}}\\ \text{=}\frac{\text{1}}{{\text{2}}^{\text{3}}}\text{×}\frac{\text{1}}{{\left(\text{-7}\right)}^{\text{3}}}\\ \text{=}\frac{\text{1}}{{\left[\text{2×}\left(\text{-7}\right)\right]}^{\text{3}}}\text{}\left[{\text{By using a}}^{\text{m}}{\text{×b}}^{\text{m}}{\text{=ab}}^{\text{m}}\right]\\ \text{=}\frac{\text{1}}{{\left(\text{-14}\right)}^{\text{3}}}\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{Q.5}\end{array}$

Evaluate

i81×5324

ii(51×21)×61

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\frac{{8}^{-1}×{5}^{3}}{{2}^{-4}}\\ =\frac{{2}^{4}×{5}^{3}}{{8}^{1}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{{2}^{4}×{5}^{3}}{{2}^{3}}\end{array}$ $\begin{array}{l}={2}^{4-3}×{5}^{3}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\text{}\\ =2×125\\ =250\text{}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{ }\left({5}^{–1}×{2}^{–1}\right)×{6}^{–1}\\ =\left(\frac{1}{5}×\frac{1}{2}\right)×\frac{1}{6}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{1}{10}×\frac{1}{6}\\ =\frac{1}{60}\end{array}$ 

$\begin{array}{l}\text{Q.6}\end{array}$

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

Ans-

$\begin{array}{l}{\text{5}}^{\text{m}}{\text{÷ 5}}^{\text{-3}}{\text{= 5}}^{\text{5}}\\ ⇒{\text{5}}^{\text{m-(-3)}}{\text{= 5}}^{\text{5}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\text{n}}{\text{=a}}^{\text{m-n}}\right]\\ ⇒{\text{5}}^{\text{m+3}}{\text{= 5}}^{\text{5}}\\ ⇒\mathrm{m}+3=5\text{}\left(\text{Since the bases are same,\hspace{0.17em}\hspace{0.17em}their}\\ \text{powers must be equal.)}\\ ⇒\mathrm{m}=5-3\\ ⇒\mathrm{m}=2\end{array}$

$\begin{array}{l}\text{Q.7}\end{array}$

Evaluate

$\left(\mathrm{i}\right){\left\{{\left(\frac{1}{3}\right)}^{–1}–{\left(\frac{1}{4}\right)}^{–1}\right\}}^{–1}\left(\mathrm{ii}\right){\left(\frac{5}{8}\right)}^{–7}×{\left(\frac{8}{5}\right)}^{–4}$

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\text{}{\left\{{\left(\frac{\text{1}}{\text{3}}\right)}^{\text{-1}}\text{–}{\left(\frac{\text{1}}{\text{4}}\right)}^{\text{-1}}\right\}}^{\text{-1}}\\ \text{=}{\left\{\left(\frac{\text{3}}{\text{1}}\right)\text{–}\left(\frac{\text{4}}{\text{1}}\right)\right\}}^{\text{-1}}\text{}\left[{\text{By using a}}^{\text{-m}}\text{=}\frac{\text{1}}{{\text{a}}^{\text{m}}}\right]\\ \text{=}{\left\{\text{3-4}\right\}}^{\text{-1}}\\ {\text{= (-1)}}^{\text{-1}}\\ \text{=}\frac{\text{1}}{\left(\text{-1}\right)}\\ \text{= -1}\end{array}$

$\begin{array}{l}\left(\mathrm{ii}\right)\text{}{\left(\frac{5}{8}\right)}^{–7}×{\left(\frac{8}{5}\right)}^{–4}\\ =\frac{{5}^{-7}}{{8}^{-7}}×\frac{{8}^{-4}}{{5}^{-4}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\left(\frac{\mathrm{a}}{\mathrm{b}}\right)}^{\mathrm{m}}=\frac{{\mathrm{a}}^{\mathrm{m}}}{{\mathrm{b}}^{\mathrm{m}}}\right]\end{array}$

$\begin{array}{l}=\frac{{8}^{7}}{{5}^{7}}×\frac{{5}^{4}}{{8}^{4}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{-\mathrm{m}}=\frac{1}{{\mathrm{a}}^{\mathrm{m}}}\right]\\ =\frac{{8}^{7-4}}{{5}^{7-4}}\left[\mathrm{By}\text{}\mathrm{using}\text{}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\\ =\frac{{8}^{3}}{{5}^{3}}\\ =\frac{512}{125}\end{array}$

$\begin{array}{l}\text{Q.8}\end{array}$

Simplify

$\text{(i)}\frac{{\text{25×t}}^{\text{-4}}}{{\text{5}}^{\text{-3}}{\text{×10×t}}^{\text{-8}}}\text{(t ≠ 0) (ii)}\frac{{\text{3}}^{\text{-5}}{\text{×10}}^{\text{-5}}\text{×125}}{{\text{5}}^{\text{-7}}{\text{×6}}^{\text{-5}}}\text{}$

Ans-

$\begin{array}{l}\left(\mathrm{i}\right)\frac{25×{\mathrm{t}}^{-4}}{{5}^{–3}×10×{\mathrm{t}}^{–8}}\left(\mathrm{t}\ne 0\right)\\ =\frac{{5}^{2}×{\mathrm{t}}^{-4}}{{5}^{–3}×5×2×{\mathrm{t}}^{–8}}\\ =\frac{{5}^{2}×{\mathrm{t}}^{-4}}{{5}^{–3+1}×2×{\mathrm{t}}^{–8}}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{\mathrm{m}}×{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]\\ =\frac{{5}^{2}×{\mathrm{t}}^{-4}}{{5}^{–2}×2×{\mathrm{t}}^{–8}}\end{array}$ $\begin{array}{l}=\frac{{5}^{2+2}×{\mathrm{t}}^{-4+8}\text{}}{2}\text{}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\\ =\frac{{5}^{4}×{\mathrm{t}}^{4}\text{}}{2}\\ =\frac{625{\mathrm{t}}^{4}}{2}\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)\frac{{3}^{–5}×{10}^{–5}×125}{{5}^{–7}×{6}^{–5}}\\ =\frac{{3}^{–5}×{\left(2×5\right)}^{–5}×{5}^{3}}{{5}^{–7}×{\left(2×3\right)}^{–5}}\\ ={3}^{–5+5}×{2}^{-5+5}×{5}^{-5+3+7}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{\mathrm{m}}÷{\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}-\mathrm{n}}\right]\\ ={3}^{0}×{2}^{0}×{5}^{5}\left[\mathrm{By}\text{}\mathrm{using}\text{:}{\mathrm{a}}^{0}={\mathrm{a}}^{1}\right]\\ ={5}^{5}\end{array}$

$\begin{array}{l}\text{Q.9}\end{array}$

Express the following numbers in standard form.

(i) 0.0000000000085 (ii) 0.00000000000942

(iii) 6020000000000000 (iv) 0.00000000837

(v) 31860000000

Ans-

The standard form of a number is

$\begin{array}{l}\mathrm{N}×{10}^{\mathrm{power}},\text{where N is number greater than or}\\ \text{equal to 1 but less than 10.}\end{array}$ $\begin{array}{l}\left(\text{i}\right)\text{The standard form of}0.00000000000\text{85}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}.\text{5}×\text{1}{0}^{-\text{12}}\\ \left(\text{ii}\right)\text{The standard form of}0.00000000000\text{942}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{9}.\text{42}×\text{1}{0}^{-\text{12}}\\ \left(\text{iii}\right)\text{The standard form of 6}0\text{2}0000000000000=\text{6}.0\text{2}×\text{1}{0}^{\text{15}}\\ \left(\text{iv}\right)\text{The standard form of}0.00000000\text{837}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{8}.\text{37}×\text{1}{0}^{-\text{9}}\\ \left(\text{v}\right)\text{The standard form of 3186}0000000\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\text{3}.\text{186}×\text{1}{0}^{\text{1}0}\end{array}$

$\begin{array}{l}\text{Q.10}\end{array}$

Express the following numbers in 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

Ans-

$\begin{array}{l}\left(\text{i}\right)\text{3}.0\text{2}×\text{1}{0}^{-\text{6}}=\frac{3.02}{{10}^{6}}=\frac{302×{10}^{-2}}{1000000}=\text{}0.00000302\\ \left(\mathrm{ii}\right)\text{}4.5\text{}×\text{}{10}^{4}=\text{}4.5×10000=45000\\ \left(\mathrm{iii}\right)\text{}3\text{}×\text{}{10}^{-8}=\text{}\frac{3}{{10}^{8}}=0.00000003\end{array}$ $\begin{array}{l}\left(\mathrm{iv}\right)\text{}1.0001\text{}×\text{}{10}^{9}=\text{}1000100000\\ \left(\mathrm{v}\right)\text{}5.8\text{}×\text{}{10}^{12}=\text{}5800000000000\\ \left(\mathrm{vi}\right)3.61492\text{}×\text{}{10}^{6}=\text{}3614920\end{array}$

$\begin{array}{l}\text{Q.11}\end{array}$

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 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

Ans-

The standard form the numbers are as follows:

$\begin{array}{l}\left(\text{i}\right)\frac{1}{1000000}=\frac{1}{{10}^{6}}=\text{}1\text{}×\text{}{10}^{-6}\\ \left(\mathrm{ii}\right)\text{}0.000,000,000,000,000,000,16=\frac{1.6\text{}×\text{}10}{{10}^{20}}=1.6\text{}×\text{}{10}^{-19}\\ \left(\mathrm{iii}\right)\text{}0.0000005\text{}=\frac{5}{10000000}=\frac{5}{{10}^{7}}=\text{}5\text{}×\text{}{10}^{-7}\\ \left(\mathrm{iv}\right)\text{}0.00001275\text{}=\frac{1.275×{10}^{3}}{{10}^{8}}=\text{}1.275\text{}×\text{}{10}^{-5}\\ \left(\mathrm{v}\right)\text{}0.07\text{}=\frac{7}{100}=\text{}7\text{}×\text{}{10}^{-2}\end{array}$

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

$\begin{array}{l}\text{Thickness of each book}=\text{2}0\text{mm}\\ \text{Hence},\text{thickness of 5 books}=\left(\text{5}×\text{2}0\right)\text{mm}=\text{1}00\text{mm}\\ \text{Thickness of each paper sheet}=0.0\text{16 mm}\\ \text{Hence},\text{thickness of 5 paper sheets}=\left(\text{5}×\text{}0.0\text{16}\right)\text{mm}=0.0\text{8}0\text{mm}\\ \text{Total thickness of the stack}=\text{Thickness of 5 books}+\text{Thickness of}\\ \text{5 paper sheets}\\ =\text{}\left(\text{1}00+0.0\text{8}0\right)\text{mm}=\text{1}00.0\text{8 mm}=\text{1}.000\text{8}×\text{1}{0}^{\text{2}}\text{mm}\end{array}$

1. Why should I use NCERT Solutions for Class 8 Mathematics Chapter 12?

NCERT Solutions for Class 8 Mathematics Chapter 12 covers all the topics and concepts in the chapter and the past year’s questions.

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3. What is an Exponent?

The exponent indicates how many number of times it’s multiplied by its own number.

For example, 2^3, Exponent = 3 & base= 2